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Page 1: Solucionario de wade
Page 2: Solucionario de wade

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OR

LUT ONS MANUAL Jan William Simek California Polytechnic State University

C CHEMISTRY SIXTH EDITION

L. G. Wade, Jr.

".�JL.-':'

Prentice Hall

Upper Saddle River, NJ 07458

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Assistant Editor: Carole Snyder Project Manager: Kristen Kaiser Executive Editor: Nicole Folchetti

Executive Managing Editor: Kathleen Schiaparelli Assistant Managing Editor: Becca Richter Production Editor: Kathryn O'Neill

Supplement Cover Manager: Paul Gourhan

Supplement Cover Designer: Joanne Alexandris

Manufacturing Buyer: Ilene Kahn Cover Image Credit: Joseph Galluccio (2004)

PEARSON Prentice

Hall

© 2006 Pearson Education, Inc.

Pearson Prentice Hall

Pearson Education, Inc.

Upper Saddle River, NJ 07458

All rights reserved. No part of this book may be reproduced in any form or by any means, without permission in writing from the publisher.

Pearson Prentice Hall™ is a trademark of Pearson Education, Inc.

The author and publisher of this book have used their best efforts in preparing this book. These efforts include the development, research, and testing of the theories and programs to determine their effectiveness. The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book. The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, the furnishing, performance, or use of these programs.

This work is protected by United States copyright laws and is provided solely for teaching courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. The work and materials from it should never be made available except by instructors using the accompanying text in their classes. All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials.

Printed in the United States of America

10 9 8 7 6 5

ISBN 0-13-147882-6

Pearson Education Ltd., London Pearson Education Australia Pty. Ltd., Sydney Pearson Education Singapore, Pte. Ltd. Pearson Education North Asia Ltd., Hong Kong Pearson Education Canada, Inc., Toronto Pearson Educaci6n de Mexico, S.A. de c.y. Pearson Education-Japan, Tokyo Pearson Education Malaysia, Pte. Ltd.

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TABLE OF CONTENTS

Preface ........................................................................................................................................ v

Symbols and Abbreviations ...................................................................................................... vii

Chapter 1

Chapter 2

Chapter 3

Chapter 4

Chapter 5

Chapter 6

Chapter 7

Chapter 8

Chapter 9

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 14

Chapter 15

Chapter 16

Chapter 17

Chapter 18

Chapter 19

Chapter 20

Chapter 21

Chapter 22

Chapter 23

Chapter 24

Chapter 25

Chapter 26

Appendix I: Appendix 2:

Introduction and Review ..... . ....... . ... . ................ . ....... ................. . .............. ............ 1 Structure and Properties of Organic Molecules .................................................. 25

Structure and Stereochemistry of Alkanes .................................... ...................... 45

The Study of Chemical Reactions . . . .................... . . ............................................. 59

Stereochemistry .... . .............. . ... . . . ................................ . ........ . ............................. 79

Alkyl Halides: Nucleophilic Substitution and Elimination ................. . ....... ....... 99

Structure and Synthesis of Alkenes ... . ..... . ........ . . .............................................. 135

Reactions of Alkenes . . . . ......... . . . ....... . . .. . ........................................................... 159

Alkynes ...... . . ........................ . .............................................. . ................. .......... 191

Structure and Synthesis of Alcohols .............. ....................... . .... . ...................... 211

Reactions of Alcohols .................. . ................................ ............ ....................... 229

Infrared Spectroscopy and Mass Spectrometry ........ . ............ ................ . ...... ..... 259

Nuclear Magnetic Resonance Spectroscopy ....... . . ................. . . ......... . .... . . . . . ...... 271

Ethers, Epoxides, and Sulfides ................................. . .......... ............................. 299

Conjugated Systems, Orbital Symmetry, and Ultraviolet Spectroscopy ............ 317

Aromatic Compounds ...... ................. . ........................... . . . ................................ 341

Reactions of Aromatic Compounds ............................................ ......... . ............ 365

Ketones and Aldehydes .............. .... . ................. ............................. .................. 401

Amines .................. . .... . . . ........... ....................................................................... 439

Carboxylic Acids ............................................................................................. 469

Carboxylic Acid Derivatives ...................................................... ...................... 499

Condensations and Alpha Substitutions of Carbonyl Compounds ... . ... . . ........... 537

Carbohydrates and Nucleic Acids ................................ . ........ ........ ................... 585

Amino Acids, Peptides, and Proteins ................................................... ............ 617

Lipids ... . ................................................... . ...................................................... 645

Synthetic Polymers . ... . ..... . . ............... . ........... .......... . ....... ................................. 659

Summary of IUPAC Nomenclature ... . ........ .......... . ................. .......... ................ 675

Summary of Acidity and Basicity ... . . . ........................... .................. . ... . ............ 689

iii

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PREFACE Hints for Passing Organic Chemistry

Do you want to pass your course in organic chemistry? Here is my best advice, based on over thirty years of observing students learning organic chemistry:

Hint #1: Do the problems. It seems straightforward, but humans, including students, try to take the easy way out until they discover there is no short-cut. Unless you have a measured IQ above 200 and comfortably cruise in the top 1 % of your class, do the problems. Usually your teacher (professor or teaching assistant) will recommend certain ones; try to do all those recommended. If you do half of them, you will be half-prepared at test time. (Do you want your surgeon coming to your appendectomy having practiced only ha l/the procedure?) And when you do the problems, keep this Solutions Manual CLOSED. A void looking at my answer before you write your answer-your trying and struggling with the problem is the most valuable part of the problem. Discovery is a major part of learning. Remember that the primary goal of doing these problems is not just getting the right answer, but understanding the material well enough to get right answers to the questions you haven't seen yet.

Hint #2 : Keep up. Getting behind in your work in a course that moves as quickly as this one is the Kiss of Death. For most students, organic chemistry is the most rigorous intellectual challenge they have faced so far in their studies. Some are taken by surprise at the diligence it requires. Don't think that you can study all of the material the couple of days before the exam -well, you can, but you won't pass. Study organic chemistry like a foreign language: try to do some every day so that the freshly­trained neurons stay sharp.

Hint #3: Get help when you need it. Use your teacher's office hours when you have difficulty. Many schools have tutoring centers (in which organic chemistry is a popular offering). Here's a secret: absolutely the best way to cement this material in your brain is to get together with a few of your fellow students and make up problems for each other, then correct and discuss them. When you write the problems, you will gain great insight into what this is all about.

Purpose of this Solutions Manual

So w hat is the point of this Solutions Manual? First, I can't do your studying for you. Second, since I am not leaning over your shoulder as you write your answers, I can't give you direct feedback on what you write and think-the print medium is limited in its usefulness. What I can do for you is: I) provide correct answers; the publishers, Professor Wade, Professor Kantorowski (my reviewer), and I have gone to great lengths to assure that what I have written is correct, for we all understand how it can shake a student's confidence to discover that the answer book flubbed up; 2) provide a considerable degree of rigor; beyond the fundamental requirement of correctness, I have tried to flesh out these answers, being complete but succinct; 3) provide insight into how to solve a problem and into where the sticky intellectual points are. Insight is the toughest to accomplish, but over the years, I have come to understand where students have trouble, so I have tried to anticipate your questions and to add enough detail so that the concept, as well as the answer, is clear.

It is difficult for students to understand or acknowledge that their teachers are human (some are more human than others). Since I am human (despite what my students might report), I can and do make mistakes. If there are mistakes in this book, they are my sole responsibility, and I am sorry. If you find one, PLEASE let me know so that it can be corrected in future printings. Nip it in the bud.

What's New in this edition?

Better answers! Part of my goal in this edition has been to add more explanatory material to clarify how to arrive at the answer. The possibility of more than one answer to a problem has been noted. The IUPAC Nomenclature appendix has been expanded to include bicyclics, heteroatom replacements, and the Cahn-Ingold-Prelog system of stereochemical designation.

Better graphics! The print medium is very limited in its ability to convey three-dimensional structural i�formation,

.a problem that has plagued organic chemists for over a century. I have added

some graphiCS created 1I1 the software, Chem3D®, to try to show atoms in space where that information is a key part of the solution. In drawing NMR spectra, representational line drawings have replaced rudimentary attempts at drawing peaks from previous editions.

Better jokes? Too much to hope for.

v

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Some Web Stuff

Prentice-Hall maintains a w eb site dedicated to the Wade text: try www.prenhall.com/wade. Two essential web sites providing spectra are listed on the bottom of p. 270.

Acknowledgments

No project of this scope is ever done alone. These are team efforts, and there are several people who have assisted and facilitated in one fashion or another who deserve my thanks.

Professor L. G. Wade, Jr., your textbook author, is a remarkable person. He has gone to extraordinary lengths to make the textbook as clear, organized, informative and insightful as possible. He has solicited and followed my suggestions on his text, and his comments on my solutions have been perceptive and valuable. We agreed early on that our primary goal is to help the students learn a fascinating and challenging subject, and all of our efforts have been directed toward that goal. I have appreciated our collaboration.

My new colleague, Dr. Eric Kantorowski, has reviewed the entire manuscript for accuracy and style. His diligence, attention to detail and chemical wisdom have made this a better manual. Eric stands on the shoulders of previous reviewers who scoured earlier editions for errors: Jessica Gilman, Dr. Kristen Meisenheimer, and Dr. Dan Mattern. Mr. Richard King has offered numerous suggestions on how to clarify murky explanations. I am grateful to them all.

The people at Prentice-Hall have made this project possible. Good books would not exist without their dedication, professionalism, and experience. Among the many people who contributed are: Lee Englander, who connected me with this project; Nicole Fo1chetti, Advanced Chemistry Editor; and Kristen Kaiser and Carole Snyder, Project Managers.

The entire manuscript was produced using ChemDraw®, the remarkable software for drawing chemical structures developed by CambridgeSoft Corp., Cambridge, MA . We, the users of sophisticated software like ChemDraw, are the beneficiaries of the intelligence and creativity of the people in the computer industry. We are fortunate that they are so smart.

Finally, I appreciate my friends who supported me throughout this project, most notably my good friend of almost forty years, Judy Lang. The students are too numerous to list, but it is for them that all this happens.

Jan William Simek Department of Chemistry and Biochemistry Cal Poly State University San Luis Obispo, CA 93407 Email: [email protected]

DEDICATION

To my inspirational chemistry teachers:

Joe Plaskas, who made the batter;

Kurt Kaufman, who baked the cake;

Carl Djerassi, who put on the icing;

and to my parents:

Ervin J. and Imilda B. Simek,

who had the original concept.

vi

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SYMBOLS AND ABBREVIA TIONS

Below is a list of symbols and abbreviations used in this Solutions Manual, consistent with those used in the textbook by Wade . (Do not expect all of these to make sense to you now. You wil l learn them throughout your study of organic chemistry.) BONDS

-

111111111111

ARROWS -

.. ..

a single bond a double bond a triple bond a bond in three dimensions, coming out of the paper toward the reader a bond in three dimensions, going behind the paper away from the reader a stretched bond, in the process of forming or breaking

in a reaction, shows direction from reactants to products signifies equilibrium (not to be confused with resonance) signifies resonance (not to be confused with equilibrium) shows direction of electron movement: the arrowhead with one barb shows movement of one electron; the arrowhead with two barbs shows movement of a pair of electrons shows polarity of a bond or molecule, the arrowhead signifying the more negative end of the dipole

SUBSTITUENT GROUPS Me a methyl group, CH3 Et an ethyl group, CH2CH3 Pr a propyl group, a three carbon group (two possible arrangements) Bu a butyl group, a four carbon group (four possible arrangements) R the general abbreviation for an alkyl group (or any substituent group not under scrutiny) Ph a phenyl group, the name of a benzene ring as a substituent, represented:

< » or

Ar the general name for an aromatic group

continued on next page

0-

vii

Page 9: Solucionario de wade

Symbols and Abbreviations, continued SUBSTITUENT GROUPS, continued

o II

Ac an acetyl group: CH3 - C -Cy a cyclohexyl group: 0-

o

Ts tosyl, or p-toluenesulfonyl group: CH3 � S-"==1- II

o o I I

Boc a t-butoxycarbonyl group (amino acid and peptide chemistry): (CH3)3C -0 - C -

Z, or Cbz

a carbobenzoxy (benzyloxycarbonyl) group (amino acid and peptide chemistry): 0

REAGENTS AND SOLVENTS

DCC dicyclohexylcarbodiimide o- N= C= N-Q DMSO dimethylsulfoxide

o II S

H3C ....... 'CH3

ether diethyl ether, CH3CH20CH2CH3

Cl 0 MCPBA meta-chloroperoxyhenzo;c ac;d < > g - 0 - OH

MVK

NBS

o methyl vinyl ketone g

H3C ....... �

N -bromosucci nimide O�O N I Br

continued on next page

viii

< )- CH2-0 - g-

Page 10: Solucionario de wade

Symbols and Abbreviations, continued

REAGENTS AND SOLVENTS, continued PCC pyridinium chlorochromate, Cr03· HCI • N� ')

CH3 H H H CH3 I I I I I

Sia2BH disiamylborane H - C - C - B - C - C - H I I I I CH3 CH3 CH3 CH3

THF tetrahydrofuran o SPECTROSCOPY IR NMR MS UV ppm Hz MHz TMS s, d, t, q nm mJz 8 A v

OTHER a, ax

e, eq

HOMO

LUMO

NR

0, m,p

infrared spectroscopy nuclear magnetic resonance spectroscopy mass spectrometry ultraviolet spectroscopy parts per million, a unit used in NMR hertz, cycles per second, a unit of frequency megahertz, millions of cycles per second tetramethylsilane, (CH3)4Si, the reference compound in NMR singlet, doublet, triplet, quartet, referring to the number of peaks an NMR absorption gives nanometers, 10-9 meters (usually used as a unit of wavelength) mass-to-charge ratio, in mass spectrometry in NMR, chemical shift value, measured in ppm wavelength frequency

axial (in chair forms of cyclohexane) equatorial (in chair forms of cyclohexane) highest occupied molecular orbital lowest unoccupied molecular orbital no reaction artha, meta, para (positions on an aromatic ring) when written over an arrow: "heat"; when written before a letter: "change in" partial positive charge, partial negative charge energy from electromagnetic radiation (light) specific rotation at the D line of sodium (589 nm)

ix

Page 11: Solucionario de wade

1 - 1 Na

Mg

AI

Si

CHAPTER I-INTRODUCTION AND REVIEW

1 s22s22p63s 1

1 s22s22p63s2

Is22s22p63s23px 1

1 s22s22p63s23px 13py I

P

S

CI

Ar

1 s22s22p63s23px 13py 13pz 1

1 s22s22p63s23px 23py 13pz 1

1 s22s22p63s23px 23py 23pz 1

1 s22s22p63s23px 23py 23pz 2

1-2 In this book, lines between atom symbols represent covalent bonds between those atoms . Nonbonding electrons are indicated with dots. H H H

•• + I I I

(a) H - N- H I

(b) H - O - H (c) H - O - H (d) H - C - C - C - H

H

H H I I

(e) H - C - C - N - H I I I H H H

H I

H :0: H I I I

(h) H - C - C - C - H

1-3

I I I H H H

H H I • • I

(f) H -C - O - C - H I • • I H H

I I I I H H H H

H H I I

(g) H-C - C - F: I I H H

(i) H - B - H I

(j ) : F- B - F: - - I

H : F: . .

The compounds in (i) and (j) are unusual in that boron does not have an octet of electrons-normal for boron because it has only three valence electrons .

(a) :N N: (b) H - C N: (c) H - O - N= O (d) O == C =O

(e) H -C=N- H I

H H

:0 : I I

(f) H - C - O - H (g) H -C == C - CI : I I ••

H H

(h) H -N=N- H

H I

(i) I

H-C == C - C - H (j ) H - C==C==C-H (k) H - C C - C - H

1-4

I I I H H H

(a) G}J -NQ)

8 (e) H - C=N - H I H

(b)

(f)

H - C NeD

C)J(D

I I H H

(c) 88(.:) H - O - N=O

0) 0) • II ·8

O - H H - C -8 8 (g) H - C == C - CleD

I I 8 H H

There are no unshared electron pairs in parts (i), (j) , and (k).

1

(d)

I H

0) 0) o == c=o 0) 0)

88 (h) H - N=N-H

Page 12: Solucionario de wade

1-5 The symbols "8+" and "8-" indicate bond polarity by showing partial charge. (In the arrow symbolism, the arrow should point to the partial negative charge.) � � � � � �

(a) C -C1 (b) C - O (c) C - N

1-6

(a)

8+ 8- 8- 8+ (g) N - O (h) N - S

Non-zero formal charges are shown beside the atoms. H H 1 1 + . . -

H - C - O-H (b) H - N - H : C1: 1 1+ 1 H H H

8- 8+ (i) N - B

H 1

H H 1+ 1

(c) H - C - N - C - H 1 1 1 H H H

. . -

:Cl:

In (b) and (c), the chlorine is present as chloride ion. There is no covalent bond between chlorine and other atoms in the formula.

H - • • I + H

1-(d) Na+ :O - C - H (e) H - C - H (f) H - C-H

1 (g) Na+ H - B - H

(h) Na+

I H

H 1 -H - B - C - N: I H

H H H ,1/ C H

- • • 1 / ·O- C - C - H • • • I '

C H /1' H H H

1 H

H H , /

H

(i) H - C - O- C - H / 1 + ' H • • _ • • H

(I)

:F-B-F: 1 : F:

. .

H - C = O- H 1 + H

U)

I H

H 1 + H - O-N-H I H

As shown in (d), (g), (h), and (k), alkali metals like sodium and potassium form only ionic bonds, never covalent bonds.

1-7 Resonance forms in which all atoms have full octets are the most significant contributors . In resonance forms, ALL ATOMS KEEP THEIR POSITIONS-ONLY ELECTRONS ARE SHOWN IN DIFFERENT POSITIONS.

:0 : "

(a) :O- C - O: .. ..

:0 : " . . -

(b) :O - N - O: .. ..

+

. . -

: 0: - . . 1 :O - C=O

. . -

:0 : - . . 1 : O - N=O +

..

...

2

. . -

: 0: 1 • • -

O= C - O:

: 0: I • • -

O=N - O: +

Page 13: Solucionario de wade

1-7 continued . . -

(c) :O-N=O .. .. O=N-O:

+ +

(d) H-C=C-C-H .. .. H-C-C=C-H I I I I I I H H H H H H

(e) H - C = C - C - H .. .. H - C - C=C - H I I I I I I H H H H H H

(f) Sulfur can have up to 12 electrons around it because it has d orbitals accessible .

(g)

: 0: I I

:O - S- O: II

: 0:

H I

H - C - H I H :0 : I I

H - C - C+ I \

H :0 : I

H - C-H I H

..

. . -

: 0: I

.. .. :O - S=O ..

..

I I : 0:

H I

:0: - • • I I :O- S=O

I :0:

H I

H - C - H I +0: 1/

H- C - C .. I \ H :0 :

I H - C - H

I H

..

. . -:0 : I

O==S-O: .. ..

I I :0 :

. . -:0: I .. .. O==S==O I

:0 :

H I

H - C -H I H :0 : I I

.. H - C - C I \\ H +0:

I H - C - H

I H

/

:0 : I I • • -O=S- O: I

:0 : . .

1 -8 Major resonance contributors would have the lowest energy. The most important factors are: maximize ful l octets; maximize bonds; put negative charge on electronegative atoms; minimize charge separation.

• • - + . .- + +

(a) H - C - N=O .. .. H - C - N-O: .. .. H - C=N - O: (negative charge on I I I I I I I electronegative H :0 : H :0 : H :0 : atoms) mInor minor major

3

Page 14: Solucionario de wade

1 -8 continued H H H

(b)

(c)

(d)

(e)

\ + C=C - N=O

/ 1 1 H H :0:

• .. \ + C= C - N - O: .. .­

/ 1 11 H H :0 :

\ + + C - C = N - O: / 1 1 H H :0:

major major minor These two forms have equivalent energy and are major because they have ful l octets, more bonds, and less charge separation than the minor contributor.

+ H - C = O - H .. • H - C - O- H

1 + 1 H H

major mmor (octets , more bonds)

+ + H - C=N=N: .. .- H - C - N N:

1 1 H H

major (negative charge minor on electronegative atom)

H - C - C N: .. • H - C=C=N: 1 1 H H minor major (negative charge

on electronegative atom) + +

(f) H - N - C - C=C - N - H " • H-N=C - C=C - N - H

(g)

1 1 1 1 1 1 1 1 1 1 H H H H H H H H H H

mmor major

t these two forms are major contributors because al l atoms have ful l octets

+ • • + H - N - C = C - C- N - H .. • H - N - C= C - C = N - H

1 1 1 1 1 1 1 1 1 1 H H H H H H H H H H

mmor major . . -

:0 : :0 : : 0: :0 : :0 : :0 : I I • • I I 1 I I I I 1

H - C - C - C - H----- H - C = C - C - H ----- H - C - C = C - H 1- 1 1 H H H

mITIor major major these two have equivalent energy and are major because the negative charge is on the more electronegative oxygen atom

4

Page 15: Solucionario de wade

1 -8 continued

(h) :0 :

I I H - C - N - H

I major H

(no charge separation)

. . -:0 :

I + ...... II----I .. � H - C = N

- H I

minor H

1 -9 Your Lewis structures may appear different from these. As long as the atoms are connected in the same order and by the same type of bond, they are equivalent structures. For now, the exact placement of the atoms on the page is not significant. A Lewis structure is "complete" with unshared electron pairs shown.

(a)

(c)

H H H H H H H H H I I I I I I I I I

H - C - C - C - C - C - C - H (b) H - C - C - C - Cl: I I I I I I I I I H H H H C H H C H / 1 ' H H H / 1 ' H H H

H H :0 : H H H :0: I I I I / I I II

H-C - C - C - C=C (d) H - C - C - C-H I I I \ I I H H H H H H

Always be alert for the implied double or triple bond. Remember that C has to have four bonds, nitrogen has three bonds, oxygen has two bonds, and hydrogen has one bond. The only exceptions to these valence rules are structures with formal charges.

H :0: I I I

(e) H - C - C - C N: H H H , 1 /

H C :0: , I I I I

H (f) H - C - C - C - O - H

H H :0: H H I I I I I I

/ I H C

(g) H-C - C - C - C - C - H / 1 ' H H H

1 - 1 0 (a)

I I I I H H H H

Complete Lewis structures show all atoms, bonds, and unshared electron pairs. H H (b) H H H (c) H H I H

, / / , I 1/ H C C - H H - C C - H , / , / , / H - C C - H H H C, I I H I 1/ C-H H - C� • • �C - H _ I ' / ' N ' , -C - C H H

I H H I \ H H

...... C C ...... H / , • • ,....... , H 0 H

5

H H I I C - C II \\

C, . . /C / N ' H H I H

Page 16: Solucionario de wade

1 - 1 0 continued

(g) H :0: H I I I I

H /C:--.., /,C-C - H 'c "c I

I I I H / C, 'iC,

H C H I H

·0· ( ) H H · · e H ,, / I I (f) H H

" /C, /C, H - C C H

I I I H-C, ........ C,

/ C H H / "-H H

(h) H :0: H H H

H ,, / • • " /C .... ,...-0 H - C ' C/·· I I

H - C, /.C, / C/' H H

I H

I I I I I I H-C - C-C-C-C-H

I I I I H H H H

1 - 1 1 There is often more than one correct way to write condensed structural formulas . You must often make inferences about what a condensed formula means according to valence rules, especial ly in structures with C=O as shown in parts (a) and (d) . (a) CH3COCH2CH2CH3 (the 0 has a double bond to the

carbon preceding it)

(d) CH3CH2CH(CH3)CH2CHO (the 0 has a double bond to the

c arbon preceding it)

(b) (CH3hCHCH2CH20H or CH3CH(CH3)CH2CH20H

(c) (CH3hCHCH2CH(OH)CH3 or CH3CH(CH3)CH2CH(OH)CH3

1 - 1 2 If the percent values do not sum to 1 00%, the remainder must be oxygen. Assume 1 00 g of sample; percents then translate directly to grams of each element. There are usually many possible structures for a molecular formula. Yours may be different from the examples shown here. (a) 40 .0 g C = 3.33 moles C 3 .33 moles = 1 C 1 2 .0 g/mole

6.67 g H 6.60 moles H 3 .33 moles = 1 .98 == 2 H 1 .0 1 g/mole =

53.33 gO = 3.33 moles ° 3 .33 moles = 1 ° 16 .0 g/mole empirical formula = � c::::::> empirical weight = 30

molecular weight = 90, three times the empirical weight c:::::>

three times the empirical formula = molecular formula = I C3H603 I 6

some possible structures: H ° H I II I

HO - C-C-C - OH I I H H

OH

HoAOH MANY other structures possible.

Page 17: Solucionario de wade

1 - 1 2 continued (b) 32.0 g C

1 2.0 g/mole 6.67 g H

l.01 g/mole

= 2 .67 moles C 1. 34 moles = 1.99 ::= 2 C

6.60 moles H -:- 1.34 moles = 4.93 5 H

IS.7 g N = 1 . 34 moles N -:- 1 . 34 moles = 1 N 1 4 .0 g/mole

42.6 g a _ . 1 6 .0 g/mole - 2 .66 moles a -;- 1 . 34 moles = 1 .99 ::= 2 a

empirical formula = I C2H sN02 I c:::::::> empirical weight = 75

molecular weight = 75 , same as the empirical weight c:::::::>

empirical formula = molecular formula = C2HsN02 I (c) 37 .2 g C

= 3 . 1 0 moles C -:- 1 . 55 moles = 2 C 1 2.0 g/mole

(d)

7 .75 g H = 7 .67 moles H -:- 1 . 5 5 moles = 4.95 ::= 5 H 1 .0 1 g/mole

55 .0 g CI _ . _

35.45 g/mole - 1 . 55 moles Cl -;- 1 . 5 5 moles - 1 CI

empirical formula = I C2H sCl I c:::::::> empirical weight = 64.46

molecular weight = 64, same as the empirical weight c:::::::>

empirical formula = molecular formula = I C2HsCI I 3S.4 g C

1 2 .0 g/mole = 3 .20 moles C 1 .60 moles 2 C

1 .�·��/�:le = 4 .75 moles H -:- 1 .60 moles = 2 .97 3 H

3��4� ��ole = 1 .60 moles CI -:- 1 .60 moles = 1 Cl

empirical formula = I C2H 3Cl I c:::::::> empirical weight = 62.45

molecular weight = 1 25 , twice the empirical weight c:::::::> twice the empirical formula = molecular formula = ""-C-4-H-6-C -12---o.,

7

some possible structures: H H I I

H- C- C-N02 I I H H

H a H \ I I I N- C- O - C- H

/ I H H

MANY other structures possible.

There is only one structure possible with this molecular formula:

H H I I

H - C - C - Cl I I H H

some possible structures: H H I I

H - C- C==C- C- H I I I I H Cl Cl H

ClUC]

Cl� MANY other Cl structures possi ble.

Page 18: Solucionario de wade

1 - 1 3 (a) 5 .00 g HBr x 1 mole HEr

80.9 g HBr = 0.06 1 8 moles HBr

(b)

1 - 1 4

0.06 1 8 moles HEr 0.06 1 8 moles H30 + ( 1 00% dissociated)

0.0618 moles H30 + 1 00 mL x 1000 mL

1 L = 0.6 1 8 moles H30 +

1 L solution

pH = - 10gIO [H30+] = - log lo (0.6 1 8) = �

1 . 50 g NaOH x 1 mole NaOH 40.0 g NaOH

= 0.0375 moles NaOH

0 .0375 moles NaOH 0.0375 moles -OH ( 1 00% dissociated)

0.0375 moles -OH 50. mL

1 x 1 0-14 [ H 0+] = 3 [-OH]

x

=

1 000 mL l L

1 x 1 0-14 0.75

= 0.75 moles -OH 1 L solution

= 1 . 3 3 x 1 0-14

pH = -loglO [H30+] = -log lo (1 . 3 3 x 1 0-14 )

= 0.75 M

(the number of decimal places in a pH value is the number of significant figures)

(a) By definition, an acid is any species that can donate a proton. Ammonia has a proton bonded to nitrogen, so ammonia can be an acid (although a very weak one). A base is a proton acceptor, that is, it must have a pair of electrons to share with a proton; in theory, any atom with an unshared electron pair can be a base. The nitrogen in ammonia has an unshared electron pair so ammonia is basic. In water, ammonia is too weak an acid to give up its proton; instead, it acts as a base and pulls a proton from water to a small extent. (b) water as an acid: H2O + NH3 -- -OH + NH4+

water as a base: H2O + HCl H3O+ + Cl-(c) methanol as an acid: CH30H + NH3 CH3O- + NH + 4

methanol as a base: CH30H + H2SO4 CH3OH2+ + HS04-

8

Page 19: Solucionario de wade

1 - 1 5 (a) HCOOH + -CN .- HCOO- + HCN FAVORS --

(b)

(c)

(d)

(e)

(f)

1 - 1 6

stronger stronger weaker weaker PRODUCTS acid base base acid pKa 3.76 pKa 9.22

CH3COO- + CH30H -- CH3COOH + CH3O- FAVORS ..

weaker weaker stronger stronger REACTANTS

base acid acid base pKa 1 5 . 5 pKa 4 .74

.- CH30- Na+ NH3 FAVORS CH30H + NaNH2 -- + stronger stronger weaker weaker PRODUCTS

acid base base acid pKa 1 5. 5 pKa 33

.- FAVORS Na+ -OCH3 + HCN -- HOCH3 + NaCN stronger stronger weaker weaker PRODUCTS

base acid acid base pKa 9.22 pKa 1 5 . 5

.- H3O+ CI-HCl + H2O -- + FAVORS stronger stronger weaker weaker PRODUCTS acid base acid base The first reaction in Table 1 -5 shows the Keq for this reaction is 1 60, favoring products.

H30+ .-+ CH3O- -- H2O + CH30H FAVORS stronger stronger weaker weaker PRODUCTS

acid base base acid pKa-1 .7 pKa 1 5. 5 The seventh reaction in Table 1 -5 shows the Keq for the reverse of this reaction is 3 .2 x 1 0-16. Therefore, Keq for this reaction as written must be the inverse, or 3. 1 x 1 015, strongly favoring products.

:0: I I

CH - C - O - H 3 • •

:0: I I

CH - C - O-H 3 +\ H

Protonation of the double-bonded oxygen gives three resonance forms (as shown in Solved Problem 1 -5 (c)); protonation of the single-bonded oxygen gives only one . In general , the more resonance forms a species has, the more stable it is, so the proton would bond to the oxygen that gives a more stable species, that is, the double-bonded oxygen.

9

Page 20: Solucionario de wade

1 - l 7 In Solved Problem 1 -4, the structures of ethanol and methylamine are shown to be similar to methanol and ammonia, respectively. We must infer that their acid-base properties are also similar . (a) This problem can be viewed in two ways. 1 ) Quantitatively, the pKa values determine the order of acidity. 2) Qualitatively, the stabilities of the conjugate bases determine the order of acidity (see Solved Problem 1 -4 for structures): the conjugate base of acetic acid, acetate ion, is resonance-stabilized, so acetic acid is the most acidic; the conjugate base of ethanol has a negative charge on a very electronegative oxygen atom; the conjugate base of methylamine has a negative charge on a mi ldly electronegative nitrogen atom and is therefore the least stabilized, so methy lamine is the least acidic.

acetic acid > ethanol > methylamine pKa 4.74 pKa"" 1 5 . 5 pKa ""

33 strongest acid weakest acid

(b) Ethoxide ion is the conjugate base of ethanol, so it must be a stronger base than ethanol; Solved Problem 1 -4 indicates ethoxide is analogous to hydroxide in base strength. Methylamine has pKb 3.36. The basicity of methylamine is between the basicity of ethoxide ion and ethanol.

ethoxide ion > methylamine > ethanol strongest base weakest base

1 - 1 8 Curved arrows show electron movement, as described in text section 1 - 1 4.

stronger acid stronger base equilibrium favors PRODUCTS

'0'

__ .. CH3CH2-�: conjugate base weaker base

H

. .

+ CH3-�-H H

conjugate acid weaker acid

'11' .� � • • (b) CH3CH2 - C - 0 - H + CH3 - N - CH3 • • I

--

1 + CH3 - � - CH3

(c)

stronger acid H stronger base

equilibrium favors PRODUCTS

• O' .. � 0. '11' • • CH3 - O - H + H - O - S - O - H • • I I stronger base : 0·:

equilibrium favors PRODUCTS

stronger acid

--

:O - S- O - H ........ l----I .. �

i-·· :�: .. • • I I • •

: 0:

H conjugate acid weaker acid

H 1 + .. CH - O-H 3

• •

conjugate acid weaker acid

:0 : I I

O==S- O-H I

: 0: . .

+

..

conjugate base, weaker base 10

conjugate base weaker base

:0:- r . . , .. O==S- O - H • • II • •

:0 :

Page 21: Solucionario de wade

1 - 1 8 continued (d) - .. � 0·

Na+ :O - H + H-S-H . . stronger base stronger acid

equilibrium favors PRODUCTS

--- .. H - O - H + Na+ . .

:S-H conjugate acid weaker acid

conjugate base weaker base

(e) H � I + /'

___

"

• • -CH3 - �� H + CH3-O: H stronger base

--- CH3 - � - H +

H conjugate base weaker base

CH - O-H 3 • • conjugate acid weaker acid stronger acid

equilibrium favors PRODUCTS

stronger acid stronger base conjugate acid weaker acid

II • • -i :0:

. . : 0: � I • • . ......... CH -C = O

equilibrium favors PRODUCTS

+ CH3-C - 0: • • - t conjugate base 3 • • weaker base

(g) :0:

~ :0: :0: :0:- }

I I .� - . . I I I I • • I CH - C - O - H + :O-S-CH ....... f---l ... � 0==S-CH3 ....... f---... _ 0==S-CH3 3 . . • • I I

3 • • I • • II weaker acid : 0 : : 0 : : 0 : weaker base

equilibrium favors REACTANTS

: 0: :0: I I I

CH3 -C -�: ...... f---l ... � CH3 -C = � conjugate base stronger base

:0: I I

H - O-S-CH • • II 3

: 0: conjugate acid stronger acid

1 - 1 9 Solutions for (a) and (b) are presented in the Sol ved Problem in the text. Here, the newly formed bonds are shown in bold. H

• • I _

(c) H - B - H + CH -O - CH " H - B - H 1'-

3 /" 3 1+

(d)

acid H �base CH -O - CH 3 • • 3

co: I I • • CH -C - H + :O - H 3 � • • acid base

. . :0: I

CH3-r - H :O - H

1 1

Page 22: Solucionario de wade

1 - 1 9 continued (e) Bronsted-Lowry--c-proton transfer

• • _ } � :�: i -� :�: H :0:

H - C - C - H + :O - H �==� H - C - C - H .... -... H - b == b - H + . ) /.. .. k� base

(f)

1 -20

acid

.. � () CH3 - � - H + CH3 - S=!:

H acid base

. . -+ :Cl :

.. H - O- H

Learning organic chemistry is similar to learning a foreign language: new vocabulary, new grammar (the reactions), some new concepts, and even a new alphabet (the symbolism of chemistry) . This type of definition question is intended to help you review the vocabulary and concepts in each chapter. Al l of the definitions and examples are presented in the Glossary and in the chapter, so this So lutions Manual wil l not repeat them. Use these questions to evaluate your comprehension and to guide your review of the important concepts in the chapter.

1 -2 1 (a) CARBON! (b) oxygen (c) phosphorus (d) chlorine

1 -22

valence e- --- 1 2 3 4 5 6 7 8 H He (2e-) Li Be B C N 0 F Ne

p S C l Br I

1 -23 (a) ionic only (b) covalent (H-O-) and ionic (Na+ -OH) (c) covalent (H-C and C-Li), but the C-Li bond is strongly polarized

(d) covalent only (e) covalent (H-C and C-O-) and ionic (Na+ -OCH3)

(f) covalent (H-C and C=O and C-O- ) and ionic (HC02 - Na+) (g) covalent only

12

Page 23: Solucionario de wade

1 -24

(a) : Cl C l: .. , .. / .. P

I :Cl:

: C l Cl: .. " / ..

• • p ............ : C l /' I Cl:

:Cl:

: Cl Cl: . . , .. / .. (b) N

I :Cl:

: Cl Cl: . . , / ..

• • N ........ . . : c!'" I Cl: . . : CI: . •

CANNOT EXIST NCls violates the octet rule ; nitrogen can have no more than eight electrons (or four atoms) around it. Phosphorus, a third-row element, can have more than eight electrons because phosphorus can used orbitals in bonding, so PCl s is a stable, isolable compound. 1 -25 Your Lewis structures may look different from these. As long as the atoms are connected in the same order and by the same type of bond, they are equivalent structures. For now, the exact placement of the atoms on the page is not significant.

(a)

(d)

(g)

1 -26

H H H ,1/ H C H , 1+ / . . -

H - N - N-H (b) H - N=N - H (c) H - C - N - C - H / I " :CI : I I H H

H H :0: I I I

H C H /1" H H H

H :0: H I I I I I

H - C - C N: (e) H-C - C - H (f) H - C-S-C - H I H

: 0:

I H

H I

I I H H

H :0: H I I I I I I

H - O-S-O-H (h) H - C - N=C = O (i) H - C-O-S-O-C - H I I

: 0:

H / H :N H I I I I

H - C - C - C - H I I H H

H H :0:

I H

(k)

I I I I . •

H H H ,,1/ H C

" I H - C-C - N=O

/ I H C /1" H H H

I I I I H :0: H

H :0: H :0: I I I I I I (a) H - C-C = C-C-C = C-C - O-H (b) :N C-C - C - C - C-H

I I I I I I H H H H H H

13

I I H H

Page 24: Solucionario de wade

1-26 continued

(c)

. .

H - O: H :0: I I I I

H-C = C-C-C-C-O-H I I I I ••

H H H H

1-27 In each set below, the second structure is a more correct line formula. Since chemists are human (surprise!), they will take shortcuts where possible; the first structure in each pair uses a common abbreviation, either COOH or CHO. Make sure you understand that COOH does not stand for C-O-O-H. Likewise for CHO.

(a) � /'... � /' � � ..... COOH

OR

(c) �COOH OH

OR �OH

OH 0

1 -28 H H H H

OH

o

(b) N C�CHO o

OR N - CTIY H

o 0

CHO OR

H H H

H

(a) 1 I 1 1

H - C - C - C - C - H I I 1 I H H H H

1 1 I and H-C - C - C-H

I 1 1 H C H

/ 1 '

these are the only two possibilities, but your structures may appear different-making models will help you visualize these structures

(b) H H H I I 1

H - C - C - N: I I I H H H

H H H

H H H I I I

and H - C - N - C - H I I H H

14

these are the only two possibilities, but your structures may appear different-making models will help you visualize these structures

Page 25: Solucionario de wade

1 -28 continued (c) There are several other possibilities as wel l . Your answer may be correct even if it does not appear here. Check with others in your study group.

H H H H H H I I I I I I

:O - C - C-C - O: :O - C-C - O - C - H I I I I I I I I I H H H H H H H H H

H :0: I I I

(d) H - C-C - H I H

1 -29

H-C = C-O - H I I H H

H H H I I I

:0: / \ H - C - C - H

I I H H

H H H I I I

(a) only three possible structures

O - C-C-C - H H - C - C - O - C - H I I I I H H H H HOCH2CH2CH3

(b) There are several other possibilities as wel l .

H H 0 H 0 H

I I I H H H CH3CH20CH3

H

H H H I I I

:O-C-C - C - H I I I I H H :0: H

I H

These are the only three structures with this molecular formula.

H H H I I I

H - C-C - C - H I I I H 0 H

I H

CH3CH(OH)CH3

H I I II

H - C - C - C - H I II I

H - C - C - C - H I

H - C=C-C - O - H I

H - C= C - C - H I I H H CH3CH2CHO

H I

H - C = C - O-C - H I I I H H H

H2C = CHOCH3

I I H H CH3COCH3

I I I H H H

H2C = CHCH20H

I I I H 0 H

I H

H2C = C(OH)CH3

1-30 General rule: molecular formulas of stable hydrocarbons must have an even number of hydrogens. The formula CH2 does not have enough atoms to bond with the four orbitals of carbon.

one carbon: H I

H - C - H I H CH4

two carbons: H - C C -H

H I

three carbons: H -C C - C - H I

H - C = C - H I I

C2H4 H H

H I

H - C = C-C - H I I I

C3H4 H C3H6 H H H 15

H H I I

H - C-C - H I I

C2H6 H H

H H H I I I

H-C - C - C - H I I I

C3HS H H H

Page 26: Solucionario de wade

H H (b) H ..... � _ � _H

H I \ H ..... C C .... / ,0 0/ " H N H

I H

H H H :0: I I I "

� H - N - C-C-C-C-O - H I I I I H H H H

(g) H H \ I C-C :0: H, II \\ "

(e)

H - C-C C-S==O

1 -32 (a) CsHsN

(f) C9H1SO (b) C4H9N

(g) C7Hs03S

/ \ / I H C = C :0: I \ I H H H

(c) C4H9NO (h) C6H603

H H \ I

/C, ,....H :N C,

H II H I H \ /C, I H

H - C C - H H H \ I I I,.... H

H - C-C C H I 1 " 0 0 / /

(h)

O - C - C-H H H \ "

:0: ·0· II H

0 , , 0

C_HH

I \ H H

H , /C, I /C,oo C C O-H \\ I C-C-H / I H H

1 -33 (a) 1 00% - 62 .0% C - 1 0.4% H = 27.6% oxygen

(c) some possible structures-MANY other structures possible:

62.0 g C _ . _ _

1 2.0 g/mole - 5. l 7 moles C -;- l . 73 moles - 2 .99 - 3 C

1 0.4 g H = 1 0 . 3 moles H -;- l . 73 moles = 5 .95 == 6 H l . 0 1 g/mole

1��06 g7�o le = l . 73 moles ° -;- l . 73 moles = 1 °

(b) empirical formula = I C3H60 I c:::::> empirical weight = 5 8

molecular weight = 1 17, about double the empirical weight q double the empirical formula = molecular formula =

C6H1202 I 16

H H H " H , "C, ' H-C C-O-H

I I H-C C-O-H " 'c" , H '" H H H H H H H H °

I I I I I II

H-C-C-C-C-C-C-O-H I I I I I

H H H H H H H H H H 0

I 1 I I I II

H-O-C-C-C-C-C - C-H I I I I 1

H H H H H

H H H H 0 H I I I I II I

H-C-C-C-C-C-O-C-H I I I I I

H H H H H

Page 27: Solucionario de wade

1-34 Non-zero formal charges are shown by the atoms. + • • - +

(a) H - C = N = N: ...... I----I .. � H - C - N - N: 1 1 H H

H + 1 +

H H I 1

(b) H H H , 1 /

H C , + 1 • • H - C - N - O: / I H C

(c) H - C = C-C - H (d) H - C-N=O (e) H - C-O - C - H I +1 1

/1' H H H 1 1 1 H H H

1 1 H :0:

. . -H C H

/ 1 ' H H H

1-3 5 The symbols "8+" and "8-" indicate bond polarity by showing partial charge. Electronegativity differences greater than or equal to 0 . 5 are considered large.

8+ 8- 8- 8+ 8- 8+ 8+ 8- 8+ 8-(a) C - CI (b) C - H (c) C-Li (d) C - N (e) C - O

large small large small large

8- 8+ 8- 8+ 8- 8+ 8- 8+ 8+ 8-(f) C-B (g) C - Mg (h) N - H (i) O - H (j ) C - Br

large large large large small

1-36 Resonance forms must have atoms in identical positions. If any atom moves position, it is a different structure. (a) different compounds-a hydrogen atom has changed position (b) resonance forms-only the position of electrons is different (c) resonance forms-only the position of electrons is different (d) resonance forms-only the position of electrons is different (e) different compounds-a hydrogen atom has changed position (f) resonance forms-only the position of electrons is different (g) resonance forms-only the position of electrons is different (h) different compounds-a hydrogen atom has changed position (i) resonance forms-only the position of electrons is different (j ) resonance forms-only the position of electrons is different 1-37 (a) H :0:

I I I H - C-C - C - H

(b) :0:

I 1 H H

.. ..

. . H :0: i I

H - C - C=C - H

:0:

I 1 H H

:0: I I

H - C - C = C-C - H I I I

I I I H H H

....... I----J .. � H - C-C - C = C - H ....... f----J .. � H - C = C-C = C - H I 1 I 1 1 I H H H H H H

17

Page 28: Solucionario de wade

1 -37 continued (c)

(d)

+ < > CH2 4 • O CH, - +( ) CH,

0- 0/ � /; CH2 .. . � CH2

0+ + + ... 0----0 + � �

.. (e) < > �: 4 • 0 0 ---- -:0= 0 .

. . . - /-

(f)

< > �: -o� + + ( >-H 4 • CN-H - CN+-H

+ (g) 0

0 .. •

+ 0 ---- 0 .. � o 0+ ..

. . (h) Q _ Q.�_ Q (i )

:0 : :0 : :0 : .. -+ CH3 - C = C - C = C - C - CH3 -----+ CH3 - C = C - C - C = C-CH

I I I I I 3

I I I I I H H H H H + / H H H H H

CH3 - C - C = C - C = C -CH3 I I I I I H H H H H

(j) no resonance fonns-the charge must be on an atom next to a double or triple bond, or next to a non­bonded pair of electrons, in order for resonance to delocalize the charge

18

Page 29: Solucionario de wade

1-3 8 . . . . .. (a) O==S-O: .. • :O-S==O .. • O==S==O

+ +

(b) 0=0-0: .. • :0-0=0 + +

(c) The last resonance form of S02 has no equivalent form in 03 . Sulfur, a third row element, can have more than eight electrons around it because of d orbitals, whereas oxygen, a second row element, must adhere strictly to the octet rule.

1-39 (a)

H NH

#3 � ..

#1 NH #2 � • • II • • /

CH3 - N - C - NH2 I H I �H+to#2 ,H+to#3 �

NH I I + + I I I

CH3 - N - C - NH2 +

NH2 I I CH3-N - C - NH3 I H

I H CH3 - � - C - NH2

no other resonance forms H t no other resonance forms

NH2 NH2 NH2 + I I I +

CH3 - N = C - NH2 ....... t---'l... CH3 - N - C - NH2 ....... t---'l.... CH3 - N - C = NH2 I I + I H H H

(b) Protonation at nitrogen #3 gives four resonance forms that delocalize the positive charge over all three nitrogens and a carbon-a very stable condition. Nitrogen #3 will be protonated preferentially, which we interpret as being more basic.

1 -40 (a) CH3 - C - C N: ...... t----I .... CH3-C == C = N:

I I • •

H H minor major (negative charge

on electronegative atom) . . -

:0 : :0 : :0 : I + I I I (b) CH3 - C==C - C - CH3 ...... t----I ... CH -C - C==C - CH ...... t----I ... CH3 - C - C==C - CH3

I I 3

+ I I 3

I I H H H H H H

minor mmor

19

major-full octets, no charge separation

Page 30: Solucionario de wade

1 -40 continued (c) :0 : :0 :

. . -:0: :0:

. . -:0: :0:

II -. . I I CH3 - C - ? - C - CH3

I I I I I I ........ 1-----1 ... CH3 - C == C - C - CH3 ....... 1-----1... CH3 - C - C == C - CH3

I I H

minor H H

\ major major } �----------- ------�--_/ Y negative charge on electronegative atoms-equal energy

(d) + CH3 - C - C == C - N ==0 -

I I I I • • • • - + +

CH3 - C == C - C - N =0 - CH3 - C == C - C == N - 0 : I I I I • • I I I I • •

H H H :0:

minor H H H :0 : H H H :0 :

minor major-negative charge on electrone gati ve atoms

NOTE: The two structures below are resonance forms, varying from the first two structures in part (d) by the different positions of the double bonds in the N02. Usually, chemists omit drawing the second form of the N02 group although we all understand that its presence is implied. It is good idea to draw all the resonance forms until they become second nature. The importance o/ understanding resonance forms cannot be overemphasized.

+ • • - + CH3 - C - C == C - N -0: ....... f---.� CH3 - C == C - C - N -0:

I I I I I • • I I I I I • • H H H :0: H H H :0:

+

NH2 NH2 I I I

NH2 I + (e) CH3CH2 - C - NH2 ...... I---J.,.. CH3CH2 - C - NH2 ....... I---J.� + CH3CH2 - C == NH2

mmor \.. major-full octets major-ful l octet:;

1 -4 1 (a) +

y equal energy

+ + CH3 - C - CH3

I CH3 - C - 0 - CH3 ...... I---l.,� CH3 - C == 0 - CH3

I • • I • • H H H

no resonance stabilization more stable-resonance stabilized

�) + + CH2 == C - C - CH3 ....... f---l.,� CH2 - C == C - CH3

I I I I H H H H

more stable-resonance stabilized

20

H I +

CH2 == C - C - CH2 I I H H

no resonance stabilization

Page 31: Solucionario de wade

1 -4 1 continued (c) H - C - CH3

\ H

H - C - C - N: \ H

... .. H - C = C = N: \ H

no resonance stabi l i zation more stable-resonance stabi l i zed

(e)

more stable-resonance stabilized

CH3 - N1 - CH3

CH3 - C - CH3 +

........ I----I .. �

+

CH3 - � - CH3 CH3 - C - CH3

+ Cc ./ CH2 \ ' H C

C...:/ ' H \ H

no resonance stabil ization H \

CH3 - C - CH3 I

CH3 - C - CH3 +

more stable-resonance stabilized no resonance stabi l ization

1 -42 These pKa values from the text, Table 1 -5, and Appendix 5 provide the answers. The lower the pKa ' the stronger the acid.

least acidic most acidic NH3 3 3

< < <

1 -43 Conjugate bases of the weakest acids will be the strongest bases. The pKa values of the conjugate acids are listed here. (The relative order of the first two was determined from the pKa values of sulfuric acid and protonated acedic acid in Appendix 5 of the textbook.)

least basic

from �6 . 1 from �5 from 4.74 from 1 5. 5 from 1 5 .7 1 -44 (a) pKa = � log l O Ka = � log l O (5 . 2 x 1 0 - 5 ) = 4.3 for phenylacetic acid

for propionic acid, pKa 4. 87 : Ka = 1 0 - 4. 87 = 1 .35 x 1 0 - 5

(b) phenylacetic acid is 3.9 times stronger than propionic acid 5 .2 x 1 0 - 5

= 3 .9 1 . 3 5 x l 0 - S

most basic

from 33

(c) < > CH2COO- + CH3CH2COOH ... weaker acid

Equi l ibrium favors the weaker acid and base.

-- < > CH2COOH + CH3CH2COO­stronger acid

In this reaction, reactants are favored.

2 1

Page 32: Solucionario de wade

1 -45 The newly formed bond is shown in bold. . . -

(a) . . - (' . . CH3 - O: + CH3-CI: .. CH3 - �- CH3 . . � . . + :CI:

(b)

(c)

nucleophile Lewis base

electrophile Lewis acid

CH - O - CH + H - O - H 3 +1 ) 3 • •

H3c�ucleoPhile I hO I Lewis base e ectrop I e

Lewis acid

c� : � H - C-H + :N - H

'----/ 1 H

electrophile Lewis acid

nucleophile Lewis base

nucleophile Lewis base

electrophile Lewis acid

·0· � :0: ' , , ' � "

(e) CH -C - CH + H-O-S-OH 3 3 �. " . . nucleophile Lewis base :0:

electrophile Lewis acid

nucleophile electrophile Lewis base Lewis acid This may also be written in two steps: the C-Cl bond breaks.

: R� - . .

CH3 - ? - CH3 + H +� - H CH3

. . -:0 :

1 H - C - H

1+ H - N - H 1 H

H + 1 . . -

CH3 -N - CH2CH3 1

+ :CI: H

+ :O - H "

:0:

---t .. � CH3-C-CH3 + "

:O-S-OH . . " . .

CI 1 -

+ :Cl - Al - Cl 1 Cl

:0:

association of the CI with AI, and a second step where

. . -: 0:

I (g) CH3 - C - CH2 + : O - H

'--I � J • • ---t .. � CH3 - C == CH2 + H - O - H

H� nucleophile electrophile Lewis base Lewis acid

22

Page 33: Solucionario de wade

1 -45 continu� (h) F-B - F CH2 = CH2

I F

electrophile Lewis acid

nucleophile Lewis base

F - I +

F - f- CH2 - CH2 F

+ � + (i) BF3 - CH2 - CH2 + CH2 = CH2 .. BF3 - CH2 - CH2 - CH2 - CH2

1 -46

electrophile Lewis acid

nucleophile Lewis base

(a) H2S04 + CH3COO- HS04- + CH3COOH +

CH3COO- + (CH3hN - H (b) CH3COOH + (CH3hN: o (e) < ) g- O- H + �OH

o < ) g - o- + H20

+ (d) (CH3hN - H + -OH

o I I

(e) HO - C - OH + 2 -0H

(f) H20 + NH3

(g) HCOOH + CH30-

1 -47

(CH3hN: + H20

o I I

-O-C - O- + 2 H20

HO- + +NH4

HCOO- + CH30H

(a) CH3CH2 - O - H + CH3-Li -----l.� CH3CH2 - 0- Li+ + CH4 (b) The conjugate acid of CH3Li is CH4 Table 1 -5 gives the pKa of CH4 as > 40, one of the weakest acids known . The conjugate base of one of the weakest acids known must be one of the strongest bases known.

23

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1 -48 From the amounts of CO2 and H20 generated, the mi lligrams of C and H in the original sample can be determined, thus giving by difference the amount of oxygen in the 5.00 mg sample. From these values, the empirical formula and empirical weight can be calculated.

(a) how much carbon in 1 4 .54 mg CO2

1 mmole C 1 2.0 1 mg C 1 mmole CO2 14. 54 mg CO2 x x X = 3 .968 mg C 44 .0 1 mg CO2 1 mmole CO2 1 mmole C

how much h)::drogen in 3.97 mg H2Q 1 mmole H2O 2 mmoles H 1 .008 mg H

x X = 3.97 mg H20 x 1 8. 0 1 6 mg H2O 1 mmole H2O 1 mmole H 0.444 mg H

how much ox)::gen in 5 .00 mg estradiol 5 . 00 mg estradiol - 3 .968 mg C - 0.444 mg H = 0.59 mg °

calculate empirical formula 3.968 mg C 1 2.0 1 mg/mole

0.444 mg H 1 .008 mg/mole

0. 59 mg ° 16.00 mg/mole

= 0. 3304 mmoles C

= 0.440 mmoles H

= 0.037 mmoles °

empirical formula = �

0.037 mmoles = 8.93 == 9 C

0.037 mmoles = 1 1 . 9 "" 1 2 H

0.037 mmoles = 1 °

empirical weight = 1 36

(b) molecular weight = 272, exactly twice the empirical weight twice the empirical formula = molecular formula =

24

Page 35: Solucionario de wade

CHAPTER 2-STRUCTURE AND PROPERTIES OF ORGANIC MOLECULES

2- 1 The fundamental principle of organic chemistry is that a molecule's chemical and physical properties depend on the molecule's structure : the structure-function or structure-reactivity correlation. It is essential that you understand the three-dimensional nature of organic molecules, and there is no better device to assist you than a molecular model set . You are strongl y encouraged to use models regularly when reading the text and working the problems.

(a) requires use of models H H (b) , f The wedge bonds represent bonds coming out of the plane of the paper

H C H toward you. "C /" """C /" ./ The dashed bonds represent bonds going behind the plane of the paper. J \ J\./ H H H H

2-2 (a) The hybridization of oxygen is sp3 since it has two sigma bonds and two pairs of non bonding electrons. The reason that the bond angle of 1 04.5 ° is less than the perfect tetrahedral angle of 1 09. 5 ° is that the lone pairs in the two sp3 orbitals are repelling each other more strongly than the electron pairs in the sigma bonds, thereby compressing the bond angle.

repulsion (0 OO" " H

. H ) compression

(b) The electrostatic potential map for water shows that the hydrogens have low electron potential (blue), and the area of the unshared electron pairs in sp3 orbitals has high electron potential (red).

high electron (0' potential (red)

OO,�) low electron

potential (blue)

2-3 Each double-bonded atom is sp2 hybridized with bond angles about 1 20 ° ; geometry around sp2 atoms is trigonal planar . In (a), all four carbons and the two hydrogens on the sp2 carbons are al l in one plane . Each carbon on the end is sp3 hybridized with tetrahedral geometry and bond angles about 1 09°. In (b), the two carbons, the nitrogen, and the two hydrogens on the sp2 carbon are all in one plane . The CH3 carbon is sp3 hybridized with tetrahedral geometry and bond angles about 1 09° .

(b)

2-4 The hybridization of the nitrogen and the triple-bonded carbon are sp, giving linear geometry (C-C-N are linear) and a bond angle around the triple-bonded carbon of 1 80° . The CH3 carbon is sp3 hybridized, tetrahedral, with bond angles about 1 09°.

25

Page 36: Solucionario de wade

2-5 (a) linear, bond angle 1 80° • • • •

o == C == o • • • •

+ + + Sp2 Sp Sp2

(b) all atoms are sp3 ; tetrahedral geometry and bond angles of 1 09° around each atom not a bond-shows not a bqnd-;-shows lone Rair coming \ • ',. � � lon� pair gomg H H

I • • I H - C - O - C - H

I • • I H H

out 01 paper "'--- / behmd paper H , ....... 0, ..... H

C C l ", A '" H H H H

(c) all atoms are sp3 ; tetrahedral geometry and bond angles of 1 09° around each atom H H

" . . / H

H �' ..... H C H - C - N - C - H

/ I " H C H I ........-- not a bond-shows

H N '/ . lone pair going behind paper / 1 ' H H H '-... C /' , " " .

J '=:., C - H H

"H J \

H H (d) trigonal planar around the carbon, bond angles 1 20° ; tetrahedral around the single-bonded oxygen, bond angle 1 09° :0:

,.....-- ·0 · 3 I . . r sp

I I • •

sp2 H - C - O- H � . .

(e) carbon and nitrogen both sp, l inear, bond angle 1 80° H - C N :

I I /C " • • / H

H ° • •

all three atoms in a line

al l atoms in one plane

(0 trigonal planar around the sp2 carbons, bond angles 1 20° ; around the sp3 carbon, tetrahedral geometry and 1 09° angles sp2

H I

H - C - C == C - H I I I H H H

(g) trigonal planar, bond angle about 1 20°

. . . . . .-0 == 0 - 0 · • • + • • -

H\

1 \ /H

C == C / \

H/C " H \,' H sp3 H

(the other resonance form of ozone shows that BOTH end oxygens must be sp2_ see Solved Problem 2-8)

26

Page 37: Solucionario de wade

2-6 Carbon-2 is sp hybridized. If the p orbitals making the pi bond between C- l and C-2 are in the plane of the paper (putting the hydrogens in front of and behind the paper) , then the other p orbital on C-2 must be perpendicular to the plane of the paper, making the pi bond between C-2 and C-3 perpendicular to the paper. Thi s necessarily places the hydrogens on C-3 in the plane of the paper. (Models wil l surely help.)

model of perpendicular n bonds

H H " '" 1 2 3 / ' C == C == C

H ' t "'H sp

2-7 For clarity, electrons in sigma bonds are not shown . (a) carbon and oxygen are both sp2 hybridized

empty orbital

(b) oxygen and both carbons are sp2 hybridized H 1 200 H

\\ . . \

. . C == O ...... I---l .. � C - O: . 1 . . II . .

H -·C - H - C 1 2�\H \H

27

One pair of electrons on oxygen is always in an sp2 orbital . The other pair of electrons is shown in a p orbital in the first resonance form, and in a pi bond in the second resonance form.

Page 38: Solucionario de wade

2-7 continued (c) the nitrogen and the carbon bonded to it are sp hybridized ; the left carbon is sp2

H 1 800 H

1 200 (\c � N: II � �c-- c N: / . . /

H H

oil

(d) the boron and the oxygens bonded to it are sp2 hybridized

H - O: H H - O: H H - O H H - O: H \ I \ I + \\ I \ I B - O : II .. B = O : II � B - O : .. � B - O : I • • I + I · •

+ II • •

H - O : H - O : H - O : H - O . . . . . . . .

POe SP'Q( empty p H - O (j"" " """'" 0 P

PO�o'' B -006 -

H -OO� sp'

sp2

PG/.:\ Sp2

H -oo�n " " " " " ,:,'

:"

(.:) p

'" _ V H

H _���/�OO� +

sp2 • •

POe sP2

H {J'Q""""'"""" ",\JD ', _ H

. . B - O""'" P VtQD�

H -OO� sp'

sp' ! -

28

Page 39: Solucionario de wade

2-8 Very commonly in organi c chemistry , we have to determine whether two structures are the same or different, and if they are different , what structural features are different . In order for two structures to be the same, al l bonding connect ions have to be identical , and in the case of doubl e bonds , the groups must be on the same side of the double bond in both structures . (A good exerc i se to do wi th your study group i s to draw two structures and ask i f they are the same ; o r draw one structure and a sk how to draw a different compound . ) (a) different compounds ; H and CH3 on one carbon of the double bond, and CH3 and CH2CH3 on the other carbon-same in both structures ; drawing a plane through the p orbitals shows the H and CH3 are on the same side of the double bond in the fi rst structure , and the H and the CH2CH3 are on the same side in the second structure , so they are DIFFERENT compounds

compare

H e' " CH 3 \ /

3 - - - - - - - e ::E-C- - - - - - - - - -

I \ H" " CH2CH3 .. ... .. ... ..

compare

these are DIFFERENT

(b) same compound; in the struc ture on the right, the right carbon has been rotated, but the bonding i s identical between the two structures (c) di fferent compounds; H and Br on one carbon , F and Ci on the other carbon in both structures ; H and CI on the same side of the plane through the C=C in the first structure , and H and F on the same side of the plane through the C=C i n the second structure , so they are DIFFERENT compounds (d) same compound: in the structure on the right, the right carbon has been rotated 1 200 2-9 (a) H H H

" , H - C - C = N - C - H

(b) ".. CH3 :N

" CH3 ",.. C ....... H

NOT INTER­CONVERTIBLE

CH3 - N : "

CH3 ",.. C ....... H / , � I , '-. 3 H \ H sp

sp2 two CH3's on opposite s ides of the C=N

two CH3's on the same s ide of the C=N

(c) the CH3 on the N is on the same side as another CH3 no matter how i t i s drawn-only one possible structure

2- 1 0 (a) and

".. CH3 : N

" ",.. C " CH3 CH3

H F " / C = C / "-

F H trans

. , . . two IdentIcal groups on one (b) no cis-trans i SOmeri Sm

}

. . (c) no cis-tram I somensm b f th d bl b d . . . c ar on 0 e ou e on (d) no cis-trans I somensm

(e) Q ... CH3 C = C ... ,

H H C I S

and QC � c,H

... "-H CH3 trans

29

"cis " and "trans " not defi ned for this example

Page 40: Solucionario de wade

2- 1 1 Models wi l l be helpful here . (a) c i s- trans i somers-the first i s trans , the second i s cis (b) constitutional i somers-the c arbon skeleton i s different (c) consti tutional i somers-the bromines are on different carbons in the first s tructure , on the same carbon in the second structure (d) same compound-just fl ipped over (e) same compound-just rotated (f) same compound-just rotated (g) not i somers-different molecu lar formulas (h) constitutional isomers-the double bond has changed position ( i ) same compound-just reversed (j) consti tutional i somers-the CH3 groups are in different relative positions (k) consti tutional i somers-the double bond i s in a di fferent posi ti on rel ati ve to the CH3

2- 1 2 (a) 2.4 D = 4 . 8 x 8 x 1 .2 1 A

8 = 0 .4 1 , or 4 1 % of a posi t ive charge on carbon and 4 1 % of a negat ive charge on oxygen (b) :0:

I .. .. ..... C .......

R ..... + R B

Resonance form A must be the major contributor. If B were the major contributor, the value of the charge separation would be between 0 . 5 and 1 .0 . Even though B is "minor" , it is qui te significant, explaining in part the high polari ty of the C=O .

2- 1 3 Both NH3 and NF3 have a pair of nonbonding electrons on the ni trogen . In NH3 , the direction of pol arization of the N-H bonds is toward the ni trogen ; thus , all three bond polarities and the lone pair polari ty reinforce each other. In NF3, on the other hand, the direction of polarization of the N-F bonds is away from the n i trogen ; the three bond polari ties cancel the lone pair polarity , so the net resul t is a very small molecular dipole moment .

polari ties reinforce ; l arge dipole moment 1 O·

� N� H .... \\ ' H

H

polarities oppose; small dipole moment

2- 1 4 Some magni tudes of dipole moments are difficult to predict ; however, the direction of the dipole should be straightforward, in most cases. Actual values of molecular dipole moments are given in paren theses. (Each halogen atom has three non bonded electron pairs , not shown below . ) The C-H is usual l y considered non-polar.

(a) LCi H it" / H � C� large dipole ( 1 . 54)

XX: CI I .. net

30

(b) H H �, +-­

C - F / H

I .. net

l arge dipole ( 1 . 8 1 )

Page 41: Solucionario de wade

2- 14 continued (c) LF � / " c � � � F

net dipole = 0

(e) Q1 0�,o.�0 T 'x�o" " o�

(d)

each end oxygen has one-half negative charge as i t i s the composite of two resonance forms; see ro or net solution to 1 -38 (b) smal l dipole (0 . 52)

(g) 0;13 (h)

� 1 1 l arge dipole (2 .72) C H / 'CH3

( i ) 1 \) (k)

/ N ,� ,/ \," ' CH H3C 3 CH3 F \\ � Il-F F

net dipole = 0

net

1 net

smal l dipole (0 .67)

(I) ----+ � C I - Be - C I

net dipole = 0

1 l arge dipole ( 1 .70)

net

(f) � � H - C N CD

I .. net

l arge dipole (2 .95)

\ l arge dipole

net

/ net

l arge dipole ( l .45)

(m) tt� � N " " H "'" \\ " H H net dipole = 0

In (k) through (m), the symmetry of the molecule al lows the indiv idual bond dipoles to cance l .

2- 1 5 With ch lorines on the same side of the double bond, the bond dipole moments reinforce each other, result ing in a l arge net dipole . With ch lorines on opposite s ides of the double bond, the bond dipole moments exact ly cancel each other, result ing in a zero net dipole .

large net dipole C� �CI

C == C / '" H H 1

31

net dipole = 0

Page 42: Solucionario de wade

(hydrogen bonds shown as wavy bond)

2- 1 7 (a) (CH3hCHCH2CH2CH(CH3h has less branching and boi ls at a h igher temperature than (CH3hCC(CH3h . (b) CH3(CH2)sCH20H can form hydrogen bonds and wi l l boi l at a much h igher temperature than CH3(CH2)6CH3 which cannot form hydrogen bonds. (c) HOCH2(CH2)4CH20H can form hydrogen bonds at both ends and has no branching; i t w i l l boi l at a much hi gher temperature than (CH3hCCH(OH)CH3 . Cd) (CH3CH2CH2hNH has an N-H bond and can form hydrogen bonds ; it wi l l boi l at a higher temperature than (CH3CH2hN which cannot form hydrogen bonds . (e) The second compound shown (B) has the higher boi l ing point for two reasons : B has a higher molecular weight than A ; and B , a primary amine with two N-H bonds, has more opportunity for forming hydrogen bonds than A, a secondary amine wi th only one N-H bond.

2- 1 8 (a) CH3CH20CH2CH3 can form hydrogen bonds with water and i s more soluble than CH3CH2CH2CH2CH3 which cannot form hydrogen bonds with water.

(b) CH3CH2NHCH3 is more water soluble because it can form hydrogen bonds ; CH3CH2CH2CH3 cannot form hydrogen bonds .

(c) CH3CH20H i s more soluble in water. The polar O-H group forms hydrogen bonds with water, overcoming the res is tance of the non-polar CH3CH2 group toward entering the water. In CH3CH2CH2CH20H, however, the hydrogen bonding from only one OH group cannot carry a four-carbon chain in to the water; thi s substance i s only sl ightly soluble in water .

(d) Both compounds form hydrogen bonds with water at the double-bonded oxygen , but only the smaller molecu le (CH3COCH3) disso lves . The cycl ic compound has too many non-polar CH2 groups to dissolve.

32

Page 43: Solucionario de wade

2- 19 H H H H H I I I I I

( a) H - C - C - C - C - C - H I I I I I

H H H H H alkane

(Usual l y , we use the tenn "alkane"

H H H I I I (b) H - C - C = C - C - C - H I I I I I

H H H H H alkene

only when no other groups are presen t . ) H H I I

(d) H - C - C = C - C - H

(g)

2-20

I I H - C C - H ' " /

H C - C H H ' I I ' H H H cycloalkyne

H H H H H

I I I I , C C - C - C - H H - C � " C � I I I I I H C H H - C .. .

� C H ' I

• H H C I H I H

H

cycloalkene

H H 0 I I " (a) H - C - C - C - H

(d)

(g)

I I

H H

aldehyde H H H H I I I I

H - C - C - O - C - C - H I I I I H H H H

ether

ketone

H H H H I I I I

(e) H - C - C - C - C - H I I I I

H - C - C H H I I

' H H H

cyc10alkane

H H H H " H I (h) , �C .. ,

H - C C - C :: C - C - H I I I

H - C C .. H H • " C � ... H H ' "

H C = C - H I I H H

alkyne , alkene , cycloalkane

H H H 0 H I I I I

(b) H - C - C - C - C - H I I I I H H H H

alcohol

H H 0 H " H" (e) , �

C .. ,

H - C C - C - O - H I I

H - C C - H , .. c � . H , . H

H H carboxyl ic acid

o H H C" - H H " (h) " C , /

H ' c; � . H . C - C .

H I I H H H

aldehyde

33

H H H H I I I I

(c ) H - C - C :: C - C - C - C - H I I I I H H H H

alkyne

H H H I I I (f) H , �

C .. � C = C - H C " C " I

H ,C " C ��

C' H I

H aromatic hydrocarbon and alkene

I;I H H ( i ) H C 'c ' , H

' C o:. " C � " C - H I I I I

H 'C "c �

C " C �_C

, H I I

H H

aromatic hydrocarbon and cycloalkene

H 0 H H I " I I (c) H - C - C - C - C - H I I I

H H H

ketone

(f) H 0 H H -' C � " C '- H

I I H - C C - H , ... c � . H , . H

H H ether

alcohol

Page 44: Solucionario de wade

2-2 1 H H 0 H

1 1 1 1 1

(a) H - C - C - C - N - C - H

(g)

1 1 1 1

H H H H amide

H I;I H H 'C ' H H 0

I I I I I I

H - C - C - C - C - C - O - H I I I I

H C H H H 'H ' H

H H H H 1 1 1 1

(b) H - C - C - N - C - C - H 1 1 1 1 1

H H H H H amine

H H H H I I I I

(e) H - C - C - O - C - C - H I I 1 I

H H H H ether

( i )

H H 0 H 1 1 1 1 1

�) H - C - C - C - O - C - H 1 1 1

H C H H'H ' H ester

H H H 1 1 1

(f) H - C - C - C - C :: N 1 1 1

H H H nitri Ie

H H , , H H U) " C . 1

H - C N - C - H I 1 ' H H - C C - H , .... C ... . H " H H H

carboxyl ic acid cyclic ester ketone and ether amine

(k)

(n)

o H 1 1 H

" ... C .. I

H - C N - C - H I 1

'

H - C C - H H , .... C ... ,

H " H H H

cyclic amide

I 1 H - C C - H , .... C ... ,

H " H H H

cyclic ester

H H , , H 0 H

( I ) C 1 1 1

H -'c ' . N - C - C - H I 1 1

H - C C - H H , .... C ... . H " H

H H

amide

H H H , . H (0) " C " H - C C - C :: N

I 1 H - C C - H , ' C ' ,

H " H H H

nitri le

(m)

a H C H

H -'C ' ' C '- H I 1

H - C C - H H ' . � ' ' H

H - C - H 1

H ketone and amine

ketone

2-22 When the identity of a func ti onal group depends on several atoms, all of those atoms should be circled . For example, an ether is an oxygen between two carbons , so the oxygen and both c arbons should be c irc led . A ketone is a carbony l group between two other carbons , so all those atoms should be circled.

(a) €2 §9H2CH3 (b) € 0Y3 (gIl (c) CH3 C - H

alkene ether

34 aldehyde

Page 45: Solucionario de wade

2-22 continued

I I g (d) H C - NH

amide (this also looks like an aldehyde, but an amide h as higher "priority " as you wil l see l ater)

(g)

aromatic

(j)

(e) CH6 c:3) H

amine ~ I I

(f) R C - O- H � carboxylic acid

S uggested by student Richard King: R i s the symbol that organic chemists use to represent alkyl and aryl groups . As you w i l l s ee in the course of your study, there are qui te a few ways that carbon and hydrogen atoms can go together to form alkyl and ary l groups . So when you see this symbo l , you should know that it represents ONl.. Y some combination of carbon and h ydrogen atoms-except when it includes other atoms .

(h) ( i)

8 alkene ketone

(k) (aromatic

CH3 CH3 �

R

2-23 Please refer to solution 1 -20, page 1 2 of this Solutions Manual .

2-24 The examples here are representative. Your examples may be di fferent and sti l l correct . What is important in thi s problem i s to have the same functional group.

(a) alkane : hydrocarbon wi th a l l s ingle bonds ; can be acyclic (no ring) or cycl ic

H H H I I I

H - C-C - C - H I I I H H H

(d) alcohol : contai ns an OH group on a carbon

H H I I

H - C - C - O- H I I H H

(b) alkene : contains a carbon-carbon double bond

H I

H - C - C == C - H I I I H H H

(e) ether: contains an oxygen between two carbons

H H I I

H - C - O- C - H I I H H

35

(c) alkyne: contains a carbon-carbon triple bond

H I

H - C-C-C - H I H

(f) ketone : conatins a carbonyl group between two carbons

H 0 H I I I I

H - C - C - C - H I I H H

Page 46: Solucionario de wade

2-24 continued

(g) aldehyde: contains a carbonyl group wi th a hydrogen on one s ide

H H a I I I I

H - C - C - C - H I I

H H

(j) ester: contains a carbonyl group with an O-C on one s ide

H a H I 1 / I

H - C - C - O - C - H I I

H H

(h) aromatic hydrocarbon : a cyclic hydrocarbon with alternating double and single bonds o (k) amine : contains a ni trogen bonded to one , two, or three carbons

H H or R group I I

H - C - N - H or R group I

H

(m) nitri le : contains the carbon-ni trogen triple bond: H3C - C N

(i ) carboxylic acid: contains a c arbonyl group with an OH group on one side

H 0 I I I

H - C - C - O - H I

H

(I ) amide : contains a carbonyl group with a ni trogen on one side

H a H I I I I

H - C - C - N - H I

H

2-25 Models show that the tetrahedral geometry of CH2CI2 precludes stereoisomers .

2-26 (a)

H H " /

C H -

CI_\

C- H

/ \ H H

(b) Cyclopropane must have 60° bond angles compared with the usual sp3

bond angle of 1 09 . 5 ° in an acyclic molecule .

(c) Like a bent spring, bonds that deviate from their normal angles or posit ions are highly strained. Cyclopropane i s reactive because breaking the ring relieves the strain .

2-27

(a) f) + ./ 0 " "

H ..... \ "

H

H

H (d) H '::, /' H beh i nd the pl ane

C of the paper

H ��) ""C

/ \� J '\. C,- H

H 'H J '\.

H H

a l l sp3, al l == 1 09 °

(b) 0_ ?:}'0 Sp3 , no bond angle because oxygen is bonded to only one atom

(c) (J �/H

" , N -- N H

" " I � H � both sp3 , al l == 1 09°

angles around sp3 atoms == 1 09° angles around sp2 carbon == 1 20°

36

Page 47: Solucionario de wade

2-27 continued

angles around sp3 atom � 1 09 ° angles around sp2 atoms � 1 20°

i n front o f the

(h) plane of the

behi nd the p l ane

of the pape; p aper

...-/ �(j {) a H / "'" ,,� H

H C I

H

( i )

2-28 For cl ari ty in these pictures , bonds between hydrogen and an sp3 atom are not l abeled; these bonds are

s-sp3 overlap .

(f)

(b) I ;;\'. . H - C - C - O - H A\(A/· ·

1 09°

(e)

(g)

2 2 7(�Yfp:� H /I-t C-t 1'-H

109° H H 1 09° sp3_sp2 sp2_sp3

37

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( i ) sp3_sp2 sp2-s � O J � I H �- f"'<; o,,�C � 120°

sp3_sp3 ___ I 1 09° I I--- sp2-sp2 + p-p

\ H /C � . /C � 120°

�/C,,\ �H H 'H\ Sp2-S

sp2_sp3

2-29 The second resonance form of formamide is a minor but s ignificant resonance contributor. It shows that the nitrogen-carbon bond h as some double bond character, requiring that the n i trogen be sp2

hybridized with bond angles approaching 1 20° .

2 -30

: 0 : I I

H - C - N - H I H

° ° 2 ° 0 ° sp ° 1 °

+ / H - C = N - H

I H

(a) The major resonance contributor shows a carbon-carbon double bond, suggesting that both carbons are sp2 hybridized with trigonal p lanar geometry . The CH3 carbon i s sp3 hybridized with tetrahedral geometry .

� :R : � :?� Sp2

H - C - C - C - H ..... I-----;.� H - C - C = C - H I I sp3 / 1 I H H H H

�nor m�or

(b) The major resonance contributor shows a carbon-nitrogen double bond, suggesting that al l three carbons and the n i trogen are sp2 hybridized wi th trigonal pl anar geometry .

+ • • + + H - N - C = C - C - H .. .. H - N - C - C = C - H .. • H - N = C - C = C - H

I I I I I I I I I I I I H H H H H H H H H H H H

minor minor major

2-3 1 In (c) and (d), the un shadowed p orbi tals are vertical and para l le l . The shadowed p orbitals are perpendicula; and horizontal . � (aJ H3C ,,,Q\2 (bJ hn. K NV

H3C"-M � (d)

CH3_QOYoQ?o O[Q[J riJJ

Page 49: Solucionario de wade

2-32 (a)

(c)

cis

H " "QD,, ,, , eH,CH] * �C - C , * trans

HJC M H

There are st i l l s ix coplanar atoms.

2-33 Coll inear atoms are marked with asteri sks .

2-34 (a) no cis-trans i someri sm (b)

(c) no cis-trans i somerism

(b) The cop lanar atoms in the structures to the left and below are marked w i th asteri sks .

(d)

(d) Theoretical ly , cyclopentene could show c i s-trans i somerism. In real i ty , the trans form is too unstable to exist because of the necessi ty of s tretched bonds and deformed bond angles . trans-CycJopentene has never been detected

c

·

l.s

0 If/ /J V "trans "--not possible because of ring strain

(e)

39

these are cis- trans isomers , but the des ignation of cis and trans to spec i fic structures is not defined because of four different groups on the double bond

Page 50: Solucionario de wade

2-35

(a) consti tutional i somers-the carbon skeletons are different (b) constitutional i somers-the position of the ch lorine atom has changed (c) c i s-trans i somers-the first is cis, the second i s trans (d) const i tutional i somers-the c arbon skeletons are different (e) c i s -trans i somers-the first is trans, the second is cis (f) same compound-rotation of the first structure gives the second (g) c i s-trans i somers-the fi rst is cis, the second is trans (h) consti tutional isomers-the posit ion of the double bond relat ive to the ketone h as changed (while i t i s true that the first double bond i s cis and the second is trans, in order to have cis-trans i somers, the rest of the structure must be i dentica l )

2 -36 CO2 i s l inear; i t s bond dipoles cance l , so i t has no net dipole . S02 i s bent, so i ts bond dipoles do not c ance l .

. . O == C == O net dipole moment = 0

t:o��"o: . . . . net dipole moment

..

2-37 Some magni tudes of dipole moments are di fficult to predict ; however, the direction of the dipole should be straightforward in most cases . Actual values of molecular dipole moments are given in parentheses. (The C-H bond is usual ly considered non-polar. )

(a) �'x/ '--J ............ CH3 N U 1 H / 'CH3 \

net l arge dipole moment

� � (b) CH3-C N CJ

I �

net

l arge dipole moment (3 .96)

net dipole moment = 0

or

(c )

/ net

l arge dipolemoment

� /Br Br Il l / I I I C B�� (d) 0;/0 1 R 1 Br

net dipole moment = 0 electron pairs on bromines are not shown

(0 �b 4N"",,� / I 2 I 2 net

CH2 CH2 " / CH2

moderate dipole moment

40

C CH3/ 'CH3 net

l arge dipole moment (2 .89)

(g) CH2 /CI CH; 'C / I 1 1 / CH2 /C"" 'CH H 2

net

moderate dipole moment

electron pairs on ch lorine are not shown

Page 51: Solucionario de wade

2-38 Diethyl ether and I -butanol each have one oxygen , so each can form hydrogen bonds with water (water supplies the H for hydrogen bonding with diethyl ether) ; their water solubi l i t ies should be s imi lar. The boi l ing point of I -butanol is much higher because these molecules can hydrogen bond with each other, thus requiring more energy to separate one molecule from another. Diethy l ether molecules cannot hydrogen bond wi th each other, so i t is re lati vely easy to separate them.

CH3CH2 - 0 - CH2CH3 CH3CH2CH2CH2 - OH

diethy l ether

can hydrogen bond with water cannot hydrogen bond with i tself

2-39 CN - CH]

N -meth y I pyrro Ii di ne

b .p . 8 1 °C

piperidine

b .p . 1 06DC

I -butanol

can hydrogen bond with water can hydrogen bond wi th i tself

tetrahydropyran

b .p . 8 8 DC

o- OH

cyc!opentanol

b.p. 14 1 °C (a) Piperidine has an N-H bond, so it can hydrogen bond with other molecules of i tself . N-Methylpyrrolidine h as no N-H, so it cannot hydrogen bond and will require less energy (lower boi l ing point) to separate one molecule from another.

(b) Two effects need to be explained: 1 ) Why does cyclopentanol have a higher boi l ing point than tetrahydropyran ? and 2) Why do the oxygen compounds have a greater difference in boi l ing points than the analogous nitrogen compounds?

The answer to the first question i s the same as in (a) : cyclopentanol can hydrogen bond with its neighbors whi le tetrahydropyran cannot .

The answer to the second questi on l i es in the text, Table 2- 1 , that shows the bond dipole moments for C-O and H-O are much greater than C-N and H-N; bonds to oxygen are more polarized, with greater charge separation than bonds to ni trogen .

How i s thi s reflected in the data? The boi l ing points of tetrahydropyran ( 8 8 DC) and N-methy lpyrrolidine (8 1 °C) are c lose ; tetrahydropyran molecules would have a s l ight ly stronger dipole­dipole attraction , and tetrahydropyran i s a l i ttle less "branched" than N-methy lpyrro l idine, so i t is reasonable that tetrahydropyran boi l s at a s l ightly higher temperature . The large difference comes when comparing the boi l ing points of cyclopentanol ( 1 4 1 DC) and piperidine ( 1 06°C) . The greater polari ty of O-H versus N-H is reflec ted in a more negati ve oxygen (more electronegati ve than ni trogen) and a more posi ti ve hydrogen , resul ti ng i n a much stronger intermolecular attraction . The conclusion i s that hydrogen bonding due to O-H i s much stronger than that due to N-H.

2 -40 (a) can hydrogen bond with i tse lf and with water (b) can hydrogen bond only with water (c) can hydrogen bond wi th i tself and with water (d) can hydrogen bond only with water (e) cannot hydrogen bond (f) cannot hydrogen bond

2-4 1 Higher-boi l ing compounds are l i sted .

(g) can hydrogen bond only with water (h) can hydrogen bond wi th i tself and wi th water ( i) can hydrogen bond only with water (j) can hydrogen bond onl y with water (k) can hydrogen bond only with water ( I ) can hydrogen bond with itself and with water

(a) CH3CH(OH)CH3 can form hydrogen bonds with other identical molecules (b) CH3CH2CH2CH2CH3 has a h igher molecular weight than CH3CH2CH2CH3 (c) CH3CH2CH2CH2CH3 has less branching than (CH3hCHCH2CH3 (d) CH3CH2CH2CH2CH2Cl has a higher molecular weight AND dipole-dipole interaction compared with

CH3CH2CH2CH2CH3

4 1

Page 52: Solucionario de wade

2-42

(a) ether

ether

(c ) aldehyde

(e) ester (cyc l ic )

alkene

(g)

(b)

(d)

(f)

(h)

a lkene

ketone

aromatic

amide (cycl ic)

amme ester

2-43 . .

: 0 : I I C

CH3/ t ' CH3

sp2-planar

: 0 : :0 : " I +

/ S ....... .... .. f----J.... / S ....... CH3/ • • CH3 CH / • • CH 3

t 3

sp3-tetrahedral

The key to thi s problem is understanding that sulfur has a lone pair of electrons . The second resonance form shows four pairs of e lectrons around the sulfur atom, an electroni c configuration requiring sp3

hybridization . Su lfur in DMSO cannot be sp2

l i ke carbon in acetone, so we would expect sulfur's geometry to be pyramidal (the four electron pairs around sulfur require tetrahedral geometry , but the three atoms around su lfur define i ts shape as pyramidal) .

42

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2-44

(a) pen ic i l l in G

(b) dopamine

(c) thyroxine

(d) testosterone

II \\ II sulfur replaces oxygen) CQ) ([)amide thioether ("thio" means

'1 _ \ CH 2 C -N __

aromatic

aromatic ) ©H2 � I amIne

CH2 CH2

(In l ater chapters , you wi l l learn that the OH group on a benzene ring is a speci al functional group cal led a "phenol ". For now, it fi ts the broad definition of an alcohol . )

aromatic

aromatic . ) amme

� CH2@e c arboxyl ic acid

43

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CHAPTER 3-STRUCTURE AND STEREOCHEMISTRY OF ALKANES

3- 1 (a) CnH2n+2 where n = 25 gives C2sHs2

Note to the student: The IUP AC system of nomencl ature has a wel l defined set of rules determining how structures are named. You w i l l fi nd a summary of these rules as Appendix 1 i n thi s Solutions Manual .

3-2 Use hyphens to separate letters from numbers. Longest chains may not a lways be written left to right.

(a) 3 -methylpentane (a lways find the longest chain ; it may not be written in a straight l ine) (b) 2-bromo-3-methylpentane (always find the longest chain) (c) 5-ethyl-2-methyl-4-propylheptane ("When there are two longest chains of equa l length, use the chain w i th the greater number of substituents . " ) (d) 4-isopropyl-2-methyldecane

3-3 Thi s Solutions Manual w i l l present l ine formulas where a question asks for an answer including a structure . If you use condensed structural formulas instead, be sure that you are able to " trans late" one structure type into the other.

(a) (b) (c)

(d)

3 -4 Separate numbers from numbers wi th commas .

(a) 2-methylbutane (b) 2 ,2 -dimethylpropane

(c) 3 -ethyl-2-methylhexane (d) 2,4-di methylhexane

(e) 3 -ethyl -2 ,2,4,5-tetramethylhexane (f) 4-t-butyl-3-methylheptane

3-5 (Hints: systematize your approach to these problems. For the i somers of a s ix carbon formula, for example, start wi th the i somer containing a l l s ix carbons in a straight chain , then the i somers containing a fi ve-carbon chain, then a four-carbon chain, etc. Careful ly check your answers to AVOID DUPLICATE STRUCTURES. )

(a)

n-hexane 2-methylpentane 3-methylpentane

2,2-dimethylbutane 2,3 -dimethylbutane

45

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3-5 continued

(b) �

n-heptane 2-methylhexane 3-methylhexane 2 ,2-dimethylpentane

T� �� 3 , 3 -di methylpentane 2,3-dimethylpentane 2,4-dimethylpentane 3 -ethylpentane 2 ,2 ,3 -trimethylbutane

3-6 For th i s problem, the carbon numbers in the substi tuents are indicated in i tal ics .

(a)

(d)

3-7

(a)

CH3 1

-CHCH3 1 2

I-methylethyl common name = isopropyl

CH3 .-!C �CH /1 3

3 0 CH3

l,l-di methylethyl

(b)

common name = r-butyl or rerr-butyl

(b)

CH3 I

-CH2CHCH3 1 2 3

2-methylpropyl common name = i sobutyl

(c)

3 -8 Once the number of carbons is determined, CnH2n+2 gi ves the formula . (a) CIOHn (b) C1sH32

3-9

(a) ( lowest b .p . ) hexane < octane < decane (highest b .p . ) -molecular weight

CH3 1

-CHCH2CH3 20/ 1 2 3

I-methylpropyl common name = sec-butyl

(b) (CH3hC-C(CH3h < CH3CH2C(CH3hCH2CH2CH3 < octane -branching

3 - 1 0

( lowest b .p . ) (highest b .p . )

(a ) ( lowest m.p . ) hexane < octane < decane (highest m.p . ) -molecu lar weight

(b) octane < CH3CH2C(CH3hCH2CH2CH3 < (CH3hC-C(CH3h -branching (lowest m .p . ) (highest m .p . )

46

Page 56: Solucionario de wade

3 - 1 1

3 - 1 2

- - - - - - - - - - - -- - - - - -- ---- -. ---- - - - - - - - - - - - - - ------- ---, 1 5 .0 kJ/mole : (3 .6 kcal/mole)

dihedral 0° 60° 1 20° angle 8

%��l H�H %��l H CH3 H

ec l ipsed staggered ec l ipsed

1 x 4.2 kJ/mole + 2 x 5 .4 kJ/mole = 1 5 .0 kJ/mole ( 3 . 6 kcaI/mole) H-H H-CH3

0° , , , H e 1 2.5

4 2 H&: H HCH3 ec l ipsed 5. 4

H�2H3

H'fJH

CH3 staggered

1 80° 240° , , , H e 1 2 .5

54 H&1

A l l energy values are per mole.

dihedral 300° 360° angle 8

4. 2

, H C 1 2.5 �3 CH3

H3C J!I eclipsed4. 2 ifH3

ec l ipsed 5.4

H3C�H

H'fJH

CH3 staggered

3.8 CH3 3.8

H3C�CH3

H'fJH

H staggered

(Note that the lowest energy conformers at 60° and 180° have at l east one CH3-CH3 interaction = 3.8 kJ (0 .9 kcal ) higher than ethane . )

Relative energies on the graph above were calculated using these values from the text: 3 .8 kJ/mole (0 .9 kcaI/mole) for a CHrCH3 gauche (staggered) interac t ion ; 4 . 2 kJ/mole (l.0 kcaI/mole) for a H-H ec l ipsed interaction ; 5 .4 kJ/mole ( 1 . 3 kcaI/mole) for a H -CH} ec l ipsed interaction ; 1 2 .5 kJ/mole (3 .0 kcaI/mole) for a CHrCH} eclipsed interact ion . These va lues in kJ/mole are noted on each structure and are summed to give the energy value on the graph . S l i gh t variation between th i s graph and the one i n the text are due to rounding.

47

Page 57: Solucionario de wade

H

3- 14

a l l bonds are staggered

- bold bonds are coming toward the reader from the plane of the paper

dashed bonds are coming toward the reader from the plane of the paper

(a) 3-sec-butyl-I,I-dimethylcyc lopentane (the 5-membered ring gi ves the base name) (b) 3-cyc lopropyl-I,I-dimethylcyc lohexane (the 6-membered ring gi ves the base name) (c ) 4-cyc lobutylnonane (the chain is l onger than the ring)

o 3- 1 6 (a) no cis-trans isomeri sm possible

(b)

and

cis

and

CIS

�H 3

� CH3 H trans

trans

(d) ----.Y CHJ ----.Y H

� 2 and � 2

c i s H CH3

trans 3- 1 7 In (a) and (b) , numbering of the ring is determined by the first group alphabetically being assigned to ring carbon 1 .

(a) cis-I-methyl-3-propylcyc lobutane ("m" comes before "p"-practice that alphabet!) (b) tral1s-I-t-butyl-3-ethylcyclohexane (the prefi xes t, s, and 11 are ignored in assigning alphabetical priority) (c) trans- l ,2-dimethylcyc lopropane (ei ther carbon wi th a CH3 could be carbon-I; the same name results)

3- 1 8 Combustion of the cis i somer gives off more energy, so cis- 1 ,2-dimethylcyclopropane must start at a higher energy than the trans i somer. The Newman projection of the c i s i somer shows the two methyls are ecl ipsed wi th each other; in the trans i somer, the methyls are sti l l ecl ipsed, but with hydrogens, not each other-a lower energy. "'\ more strain == �CHj ) higher energy

CH3

cis

HH

48

Page 58: Solucionario de wade

3-19 trans-1 ,2-Dimethylcyc lobutane i s more stable than cis because the two methy ls can be farther apart when trans , as shown in the Newman projections .

""\ more strain = �CH ; ) higher energy CH3

c i s

H H

In the 1 ,3 -dimethylcyc lobutanes , however, the cis al lows the methyls to be farther from other atoms and therefore more stable than the trans.

3-20

equatori al on ly

HACHl

CH3 � H

showing both axial and equatoria l

trans

axi al only

3 -2 1 The abbrevi ation for a methyl group, CH3 , is "Me". Ethyl i s "Et" , propyl is "Pr", and butyl is "B u".

� � H H (a) Me (b)

H

H Me

Me H H

Me

al l methy ls axial (a l l H's equatori al )

a l l methyls equatoria l (al l H's ax ia l )

Note that ax ia l groups al ternate up and down around the ring .

3 -22 Carbons 4 and 6 are beh ind the c irc les .

49

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3-23 The isopropyl group can rotate so that i ts hydrogen i s near the axial hydrogens on c arbons 3 and 5, s imilar to a methyl group's hydrogen, and therefore simi l ar to a methy l group in energy. The t-butyl group , however, must point a methy l group toward the hydrogens on carbons 3 and 5 , g iv ing severe diaxial interactions, causing the energy of thi s conformer to jump dramatical l y .

5=H3 }JJCHl

i sopropylcyclohexane t-buty Icyclohexane

3-24 The most stable conformers have substituents equatori al.

3-25

(a) c i s

(b) trans

equatori al , axial

H CH3 (b)�CH]

H --

EQUAL ENERGY

H

H axial , equatori al

Hp:J� � H3C �

H

CH 3

axi al , axial h igher energy

H equatorial , equatori al lower energy

(c) The trans i somer is more stable because BOTH substituents can be in the preferred equatori al positions .

3-26

Positions cis trans

1,2 (e,a) or (a,e) (e,e) or (a,a)

1,3 (e,e) or (a ,a) (e ,a) or (a,e)

1 ,4 (e,a) or (a,e) (e,e) or (a,a)

50

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3-27 The more stable confonner p laces the l arger group equatori al .

(a) CH2CH3 CH3 c;tH7 --- �CH2CH] ---

H H more stable

(b) 4�

H] H ---

--- �CH2CH' CH3

CH3 more stable H

(c) CH2CH3 H 3C , H C -CH

H ]H�CH2CH]

--(CH 3hCH ...

H H more stable

(d) H

more stable H

3-28 The key to detennining c i s and trans around a cyclohexane ring i s to see whether a substi tuent group is "up" or "down" relative to the H at the same carbon. Two "up" groups or two "down" groups wi l l be c i s ; one "up" and one "down" wi l l be trans. Th i s works independent of the confonnation the molecule i s in ! (a) cis- l ,3-dimethylcyclohexane (d) cis- 1 ,3-dimethylcyc lohexane (b) cis- l ,4-dimethylcyclohexane (e) cis- l ,3-dimethy lcyc lohexane (c ) trans-I,2-dimethylcyclohexane (f) trans-I,4-dimethy lcyc lohexane

H (b) ;------f-CCCH3h �CH3

H (c ) Bulky substi tuents l ike t-butyl adopt equatorial rather than axial posi t ions, even i f that means altering the conformation of the ring. The tw i st boat conformation al lows both bulky substi tuents to be "equatorial".

H

H CH 3 I

5 1

C -CH 2CH 3 I CH 3

Page 61: Solucionario de wade

3-30 The nomenclature of bicyc l ic alkanes is summarized in Appendix l .

(a) bicyclo[3 . 1 .0]hexane (b) bicyc lo[3 . 3 . 1 ]nonane (c) bicyclo[2 . 2 . 2]octane (d) bicyclo[3 . 1 . 1 ]heptane

3-3 1 Using models is essential for th i s problem.

H H

-----

from text Figure 3-29

rotate picture

H

3-32 Please refer to solution 1 -20, page 1 2 of this Solutions Manual .

3-33 (a) The third structure i s 2-methylpropane (i sobutane) . The other four structures are all n-butane. Remember that a compound's i denti ty is determined by how the atoms are connected, not by the position of the atoms when a structure is drawn on a page .

(b) The two structures at the top- left and bottom-left are both cis-2-butene . The two structures at the top­center and bottom-center are both I-butene. The unique structure at the upper right i s trans-2-butene . The unique structure at the lower right i s 2-methylpropene .

(c) The first two structures are both cis- l ,2-dimethylcyclopentane. The next two structures are both trans-l,2-dimethylcyc lopentane . The l ast structure i s different from all the others, cis- l ,3-dimethylcyc lopentane.

(d)

(e) Naming the structures shows that three of the structures are trans- l ,4-dimethylcyclohexane, two are the cis i somer, and one is cis-1,3-dimethylcyclohexane. Al though a structure may be shown in two different conformations , it s ti l l represents only one compound .

A B

('(.�:3

H3C � H

cis- l A-dimethy l

H H H 3 C�

CH3 cis- I ,3-dimethyl

D

52

Analysi s of the structures shows that some double bonds begin at carbon-2 and some at carbon-3 of the longest chain .

The three structures l abeled A are the same, with the double bond trans; B is a geometric i somer (cis) of A. C and Dare consti tutional i somers of the others .

��h � H CH3

trans- l ,4-di methy I

H

H-P-CH3

CH3 cis- l ,4-dimethy I

M H3C H

® CH3 H

trans-I,4-di methyl trans- l ,4-dimethyl

Page 62: Solucionario de wade

3-34 Line formul as are shown.

(Cl d (a) � (b)

(d) (Cl � (I) 8CH3

or 6 " , "CH2CH3 " " -

(g) (h) 01

(i ) ot-;or: H" "H CI

(j) � (k) 0-0 ( I ) Bxj'H

" H' 3-35 There are many possib le answers to each of these problems. The ones shown here are examples of correct answers. Your answers may be different AND correct . Check your answers in your study group .

(a)

2-methylheptane

(b)

4 ,5 -diethy ldecane

(c)

cis- 1 ,2-diethylcycloheptane

(d) only two possible answers CH3 O ,,,,CH3

or

3-methylheptane 4-methy lheptane

3 , 5 -diethyldecane

(Any combination is correct except using posi tion numbers 1 or 2 or 9 or 10 . Why won't these work?)

QCH2 CH3

CH2CH3 cis- 1 ,3 -diethylcycloheptane

[) " "CH3

6CH3

CH2CH3

NOT 1 ,5 NOT 1 ,6 NOT 1 ,7

cis- 1 ,4-diethylcycloheptane

trans-l,2-dimethy lcyc lopentane trans- 1 ,3 -dimethylcyclopentane

5 3

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3-35 continued (e)

Other ring sizes are possible, although they must have 6 or more carbons to be longer than the 5 c arbons of the substituent chain .

(2 ,3 -dimethy lpenty l )cyc loheptane (2 ,3-di methylpenty I )cyc looctane

(f)

bicyclo[ 4.4.0]decane

Any combination where the number of carbons in the bridges sums to 8 wil l work. (Two carbons are the bridgehead carbons . )

b icyc lo[3 . 3 .2]decane

3-36 HO - CH 2CH 2CH 3 HO - CH 2CH 2CH 2CH 2CH 3

HO - C H2CH 2CH 2CH 3 HO - CH 2CH 2CH 2CH 2CH 2CH 3

3-37

(a) 3 -ethyl-2 ,2 ,6-trimethylheptane (b) 3-ethyl-2 ,6 ,7-trimethy loctane (c) 3 ,7 -diethy l -2 ,2,8-trimethy ldecane (d) 2-ethy l-l, l -dimethylcyc lobutane

(e) bicyclo[4 . 1 .0]heptane (f) cis- l -ethy l -3-propylcyc lopentane (g) ( 1 , I-diethylpropyl)cyclohexane (h) cis-l-ethyl-4-isopropylcyclodecane

3 -38 There are ei ghteen i somers of C sH IS. Here are eight of them. Yours may be different from the ones shown . An easy way to compare is to name yours and see if the names match.

� Il-octane

� 2 ,2 ,4-trimethylpentane

3 -39 (a)

4 4 6

� 2-meth y Iheptane

~ 2 ,3 ,4-tri methy Ipentane

correct name : 3 -methylhexane ( longer chain)

(c) �I CI

correct name: 2-ch loro-3 -methy l hexane (begin numbering at end c losest to substi tuent)

54

� T 2 ,3 -dimethylhexane 3 -ethylhexane

T * 3-ethy l -3 -methy lpentane 2 ,2 ,3 ,3 - tetrameth y I butane

(b)

correct name : 3 -ethyl-2-methy lhexane (more branching with this numbering)

(d)

correct name: 2 ,2-dimethy lbutane (include a pos i t ion number for each substi tuent , regardless of redundancies)

Page 64: Solucionario de wade

3-39 continued

(e)

correct name: sec-buty lcyc lohexane or ( I-methylpropyl)cyclohexane (the l onger chain or ring is the base name)

3 -40

(f)

correct name: 1 ,2-diethylcyc lopentane (posit ion numbers are the l owest possible)

(a) n-Octane has a h i gher boi l i ng point than 2,2,3 -trimethy lpentane because l inear molecules boil h igher than branched molecules of the same molecular weight ( increased van der Waals interaction) .

(b) 2-Methylnonane has a h igher boi l ing point than n-heptane because it has a s ign ifi cantly higher molecular weight than n-heptane.

(c) n-Nonane boi l s h i gher than 2 ,2 ,5 -trimethylhexane for the same reason as in (a) .

3-4 1 The point of attachment i s shown by the bold bond at the left of each structure .

-CH2CH2CH2CH2CH3 1 ° ll-pentyl

-CH2CH2CHCH3 1 ° I

CH3 3-methy lbutyl

CH3 I

-CHCHCH 3 I

( i sopenty l )

2° CH3 1 ,2-dimethy lpropyl

-CHCH2CH 3 2° I

CH2CH 3 l -ethylpropyl

CH3 I

-CH - C -CH

-CH2CHCH2CH3 1 ° I

CH 2-methylbutyl 3

CH 3 I

-C -CH 2CH3 3 ° I

CH3 l , l -dimethylpropyl (t-pentyl )

2 I

3 1 ° CH3 2 ,2-dimethy lpropyJ

(neo-pentyl)

3 -42 In each case, put the l argest groups on adjacent carbons in anti positions to make the most stable conformations .

(a) 3 -methylpentane

2

� C-2 i s the front carbon with H, H , and CH3 C-3 i s the back carbon with H , CH 3 , and CH2CH 3

H

55

carbon-3 cannot b e seen; it is behind carbon-2

Page 65: Solucionario de wade

3-42 continued (b) 3,3-dimethylhexane

C-3 is the front carbon wi th CH3, CH3, and CH2CH3 C-4 is the back carbon w i th H, H, and CH2CH3

axial 3-43 (a) M:ax�al H

CH3 equatorial

CH3

equatorial I� H

(b)

equatori al

more stable (lower energy)

CH3 CH3 axi al axia l

less stable (higher energy)

carbon-4 cannot be seen; it is behind carbon-3

equatorial

(c) From Section 3- 1 4 of the text , each gauche interaction raises the energy 3.8 kJ/mole (0 .9 kcallmole) , and each axia l methyl has two gauche interactions, so the energy is : 2 methyls x 2 interactions per methyl x 3.8 kJ/mole per interaction = 1 5 .2 kJ/mole (3.6 kcallmole)

(d) The steric strain from the 1 ,3-diaxial interaction of the methyls must be the difference between the total energy and the energy due to gauche i nteractions:

3-44

(a)

(b)

(c)

23 kJ/mole - 1 5 .2 kJ/mole = 7.8 kJ/mole (5.4 kcallmole - 3.6 kcal/mole = 1 .8 kcallmole)

The more stable conformer p laces the l arger group equatoria l .

a CH(CH3)2 �7CH' --

H ��H3)2

--�

a CH2CH3

a CH 2CH3 fE:fH --� H

a CH2CH3 �CH(CH3h H e

H more stable

H �CH(CHJh CHli e

e I CH3

more stable

CH�CH2CHJ e e

H H more stable

56

Page 66: Solucionario de wade

3-44 continued

(d)

(e)

H

HH--�H2CH]

a CH3 H more stable

a CH3

H��H 2CH3

more stable H

3-45 (Using models is essential to this problem.)

In both cis- and trans-decal in, the cyclohexane rings can be in chair conformations. The re lati ve energies w i l l depend on the number of axial substi tuents.

trans no axial substi tuents MORE STABLE

3-46 chair form of glucose-al l substi tuents equatori al

HO HO

OH

H H

57

cis one axia l substi tuent

OH I

CH2

OH

(without ri ng H's shown)

Page 67: Solucionario de wade

4- 1 (a) H H H

I I I

CHAPTER 4-THE STUDY OF CHEMICAL REACTIONS

(b) (c) H H I I

(d) ..

H-C-C-C· H-C -C-C-H : I .

4-2

(a)

I I I H H H

H Q H CI - C Q + CI' --- CI-C·

IV

:� I

CI-C· + CI -CI HUUu ---

I H

H I

CI-C-C1 I

H

I I I H H H

+ H-CI

+ CI'

(b) Free-radical halogenation substitutes a halogen atom for a hydrogen . Even i f a molecule has only one type of hydrogen, substi tution of the fi rst of these hydrogens forms a new compound. Any remaining hydrogens in this product can compete with the in i ti al reactant for the avai l able ha logen . Thus, ch lorination of methane, CH4, produces a l l possible substi tution products: CH3C I , CH2C I2 , CHCI3 , and CCI4.

If a molecule has different types of hydrogens , the reaction can generate a mixture of the possible substitution products.

(c) Production of CCI4 or CH3CI can be control led by al tering the ratio of CH4 to C12. To produce CCI4, use an excess of CI2 and let the reaction proceed unti l a l l C-H bonds have been replaced with C-C1 bonds . Produc ing CH3CI is more chal lenging because the react ion tends to proceed past the first substi tution. By using a very l arge excess of CH4 to C12, perhaps 1 00 to 1 or even more, a chlorine atom i s more l ikel y to find a CH4 molecule than i t i s to find a CH3CI , so on ly a sma l l amount of CH4 i s transformed to CH3C1 by the t ime the C I2 runs out, with almost no CH2Cl2 being produced.

4-3 (a) Thi s mechani sm requires that one photon of l i ght be added for each CH3CI generated, a quantum yield of 1. The actual quantum yield is several hundred or thousand. The h igh quantum yield suggests a chain reaction, but th i s mechan i sm is not a chain; i t has no propagation steps .

(b ) Thi s mechan i sm confl ic ts w i th a t least two experimental observations . First , the energy of l ight requi red to break a H-CH3 bond i s 435 kJ/mole ( 1 04 kcallmole, from Table 4-2) ; the energy of l ight determined by experiment to in i t iate the reaction i s only 251 kJ/mole of photons (60 kcallmole of photons), much less than the energy needed to break th i s H-C bond. Second, as in (a) , each CH3Cl produced would require one photon of l i ght , a quantum yield of 1 , i nstead of the actual number of several hundred or thousand. As in (a), there i s no provi s ion for a chain process , s ince al l the radicals generated are also consumed in the mechan i sm .

4-4 (a) The twelve hydrogens of cyc lohexane are a l l on equivalent 2° c arbons. Replacement of any one of the twelve wi l l lead to the same product , ch lorocyclohexane. n-Hexane, however, h as hydrogens in three different posi tions: on carbon- l (equivalent to carbon-6) , carbon-2 (equi valent to c arbon-5) , and carbon-3 (equi valent to carbon-4). Monoch lorination of n-hexane wi l l produce a mixture of a l l three possible i somers : 1 - , 2-, and 3-chlorohexane.

59

Page 68: Solucionario de wade

4-4 continued (b) The best convers ion of cyclohexane to ch lorocyclohexane would require the ratio of cyclohexanel ch lorine to be a l arge number . If the ratio were smal l , as the concentration of ch lorocyc lohexane increased during the reaction, ch lorine would begin to substitute for a second hydrogen of chlorocyclohexane, generating unwanted products. The goal is to have chlorine attack a molecule of cyclohexane before it ever encounters a molecule of ch lorocyc lohexane, so the concentration of cyc lohexane should be kept hi gh.

4-5 (a)

(b)

-IlGOIRT - 2.1 kJ/mole =' - 2100 Ilmole e � e -(-2100 J/mole)/«8.314 J/K-mole)· (298 K»

= e 2100 I 2478 = e 0.847

Keq = 2 .3 =

[CH3SH] [HBr]

[CH3B r] [H2S]

[CH3Br]

=

[ill

K = degrees kelvin

initi al concentrations: final concentrations 1 - x 1 - x

o x

o x

x 0 x x 2

1 - 2x + x2 c::::> x 2 = 2.3 x2

- 4.6 x + 2.3 Keq = 2.3 = =

(l-x) (l-x)

o = 1 .3 x 2 - 4.6 x + 2.3 > x = 0.60 , 1 - x = 0 .40 (us ing quadratic equation)

4-6 2 acetone � diacetone Assume that the in i t ial concentration of acetone is 1 molar, and 5% of the acetone is converted to diacetone. NOTE THE MOLE RATIO.

ini t ial concentrations: final concentrat ions:

[di acetone]

[acetone] 2 =

[acetone]

1 M 0 .95 M

0.025

(0.95 )2

[diacetone]

o 0.025 M

6.Go = - 2 .303 RT 10glO Keq = - 2 .303 «8 .3 14 J/K-mole) 0 (298K)) o l og(0 .028)

= I + 8.9 kJ/mole (+ 2. 1 kcal/mole) I 4-7 6.5° wi l l be negative since two molecules are combined into one, a loss of freedom of motion . Since 6.so is negati ve, - T6.So is positi ve; but 6.Go is a l arge negative number since the reaction goes to completion . Therefore, /'li{0 must also be a large negative number, necessari ly l arger in absolute value than 6.Go. We can expla in th i s by formation of two strong C-H bonds (4 1 0 kJ/mole each ) after breaking a strong H-H bond (435 kJI mole) and a WEAKER C=C pi bond.

60

Page 69: Solucionario de wade

4-8 (a) 6.5° i s positive--one molecule became two smal ler molecules w i th greater freedom of motion (b) 6.5° is negative-two smal ler molecules combined into one l arger molecule wi th less freedom of motion (c ) 6.5° cannot be predic ted s ince the number of molecules in reactants and products i s the same

4-9 (a)

(b)

in i ti ation

propagati on

h� hv ( 1 ) Cl - C1 - 2 Cl·

i (2) "�,, Cl· + H-CH2CH3 - H-Cl + • CH2CH3

(3) h � h Cl -Cl + • CH2CH3 - Cl - CH2CH3 + Cl· i (4)

termination (5) (6)

Cl· + • Cl - Cl - CI Cl· + • CH2CH3 - Cl- CH2CH3 CH3CH2• + • CH2CH3 - CH3CH2CH2CH3

step ( 1 )

step (2)

step (3)

break Cl-Cl

break H-CH2CH3 make H-Cl

step (2)

break Cl-Cl make Cl-CH2CH3

step (3 )

!ili0 = + 242 kJ/mole (+ 58 kcal/mole)

!ili0 = + 4 1 0 kJ/mole (+ 98 kcal/mole) !ili0 = - 431 kJ/mole (- 1 03 kcal/mole) !ili0 = -2 1 kJ/mole (-5 kcaUmole)

!ili0 = +242 kJ/mole (+ 58 kcallmole) !ili0 = - 339 kJ/mole (- 81 kcal/mole) /�lr = -97 kJ/mole (-2 3 kcaUmole)

(c) 6.HO for the reaction is the sum of the !ili0 values of the indiv idual propagation steps:

- 2 1 kJ/mole + - 97 kJ/mole = - 1 1 8 kJ/mole

4- 1 0 (a)

(- 5 kcal/mole + - 23 kcallmole = - 28 kcaUmole)

in i t iation ( 1 ) h� hv Br - Br - 2 Br·

i

(2) propagation

(3)

,,�� Br· + H-CH3 - H-Br + • CH3

h � h Br- Br + • CH3 - Br-CH3 + Br·

step ( 1 )

step (2)

step (3)

break Br-Br

break H-CHl make H-Br -

step (2)

break Br-Br make Br-CH3

step (3)

!ili0 = + 192 kJ/mole (+ 46 kcallmole)

!ili0 = + 435 kJ/mole (+ 1 04 kcal/mole) !ili0 = - 368 kJ/mole (- 88 kcallmole) M-I0 = +67 kJ/mole (+16 kcaUmole)

!ili0 = + 192 kJ/mole (+ 46 kcal/mole) !ili0 = - 293 kJ/mole (- 70 kcal/mole) M-I0 = -1 0 1 kJ/mole (-24 kcaUmole)

(b) 6.HO for the reaction is the sum of the I'IJJO values of the indiv idual propagation steps :

+ 67 kJ/mole + - 101 kJ/mole = - 34 kJ/mole (+ 1 6 kcallmole + - 24 kcallmole = - 8 kcaUmole)

61

Page 70: Solucionario de wade

4- 1 1 (a) fi rst order: the exponent of [ (CH3hCCI ] in the rate l aw = 1 (b) zeroth order: [CH30H ] does not appear in the rate law ( i ts exponent is zero) (c) first order: the sum of the exponents in the rate law = 1 + 0 = 1

4- 1 2 (a) first order: the exponent of [cyclohexene] i n the rate l aw = 1 (b) second order: the exponent of [ Br2 ] in the rate law = 2 (c) third order: the sum of the exponents in the rate law = 1 + 2 = 3

4- 1 3

(a) the reaction rate depends o n neither [ethylene] nor [hydrogen] , s o i t i s zeroth order in both species. The overal l reaction must be zeroth order. (b) rate = kr (c) The rate law does not depend on the concentration of the reactants. It must depend, therefore, on the only other chemical present, the catalyst . Apparently, whatever is happening on the surface of the catalyst determines the rate, regardless of the concentrations of the two gases. Increasing the surface area of the catalyst , or simply adding more catalyst, would accelerate the reaction .

4- 1 4

(a)

t + 1 3 kJ/mole

HCI + • CH3 CH 4 + CI ·

....................... �-------

reaction coordinate

(b) Ea = + 1 3 kJ/mole ( + 3 kcal/mole) (c) !ili0 = - 4 kJ/mole ( - 1 kcal/mole)

4- 1 5

(a)

t Cl 2 + • CH3 .......... i 4

k II _____ ./...... . ....... .,. + ca mole

- 1 09 kJ/mole

reaction coordinate -----

(b) reverse: CH 3C I + CI · ----- C l2 + • CH 3

(c) reverse: Ea = + 1 09 kJ/mole + + 4 kJ/mole = + 1 13 kJ/mole ( + 26 kcallmole + + 1 kcallmole = + 27 kcaVmole)

62

Page 71: Solucionario de wade

4- 1 6

(a)

t numbers are kJ/mole

+ 1 92

reaction coordinate --(b) The step leading to the highest energy transit ion state i s rate- l imi ti ng . In th is mechanism, the first propagation step is rate- l imi ting :

B r· + CH4 -- HBr + • CH3

(c) (1) [t-------�; 1 +

"8· " means parti al radica l character on the atom )

(2) [H-�tt'---H------�; r

[H/8. 8.

]+

(3) H -�--------Br - - - -- - Br

(d ) WO for the reaction i s the sum of the WO values of the indiv idual propagation steps (refer to the solution to 4- 1 0 (a ) and (b»:

4- 1 7

(a) in i t iation ( 1 )

+ 67 kJ/mole + - 1 0 1 kJ/mole = - 34 kJ/mole ( + 1 6 kcallmole + - 24 kcallmole = - 8 kcaUmole)

hn I-I

hv - 2 I· r\�n I • + H -CH3 - H - I + . CH3

propagation 1

(2) h � h I � I +' • CH3 - I -CH3 + I·

step ( 1 )

s tep (2)

step (3 )

(3)

break I-I

break H-CH3 make H-I

step (2)

break I-I make I-CH3

step (3 )

Mfo = + 1 5 1 kJ/mole ( + 36 kcal/mole)

WO = + 435 kJ/mole (+ 1 04 kcal/mole) WO = - 297 kJ/mole ( - 71 kcallmole) MIo = + 138 kJ/mole ( + 3 3 kcaUmole)

WO = + 1 5 1 kJ/mole ( + 36 kcallmole) WO = - 234 kJ/mole (- 56 kcal/mole) MIo = -8 3 kJ/mole ( -20 kcaUmole)

63

Page 72: Solucionario de wade

4- 1 7 continued

(b) Mia for the reaction is the sum of the Mia values of the individual propagation steps :

+ 1 3 8 kllmole + - 8 3 kllmole = + 55 kJ/rnole (+ 33 kcallmole + - 20 kcallmole = + 1 3 kcaVrnole)

(c) Iodination of methane is unfavorable for both kinetic and thermodynamic reasons . Kinetical ly , the rate of the fi rst propagation step must be very s low because it is very endothermic ; the acti vation energy must be at least + 1 3 8 kllmole. Thermodynamical ly , the overal l reaction is endothermic, so an equi l ibri um would favor reactants , not products ; there i s no energy decrease to dri ve the reaction to products .

4-18 Propane has s ix primary hydrogens and two secondary hydrogens , a ratio of 3 : 1 . I f primary and secondary hydrogens were repl aced by ch lorine at equal rates , the ch loropropane i somers would reflect the same 3: 1 ratio, that is, 75% l -ch loropropane and 25% 2-chloropropane .

H H

(d) H H

H H

H H all are 20 H

4-20

3 a H abstraction H

I Cl- + CH -C-CH 3

I 3

break 30 H -C(CH3h make H-Cl

CH3

overall 3 a H abstraction

10 H abstraction H I

(b) H-- 30

OR H

I CH3-�-CH3

t CH3 t l o t 1 0

1 0

H H H

al l are 20 H

H

H -Cl +

H

(c ) H -- 30 I

CH3-C-CH 2CH3 t CH3 t t l o t 20 1 0

(e)

H

CH -C-CH 3 I

3 CH3

1 0

30

al l are 20 H except bridgehead H's ( labeled 30)

Mia = + 3 81 kllmole ( + 9 1 kcallmole) Mia = - 43 1 kllmole ( - 1 03 kcal/mole) Mr = -50 kJ/rnole ( -1 2 kcaVrnole)

H I

CI- + H -CH -C-CH 2 I 3 H -CI + -CH -C-CH 2 I

3 CH3

break 10 H-CH2CH(CH3h make H-CI overall 10 H abstraction

CH3

Mia = + 4 1 0 kllmole (+ 98 kcallmole) !J..Ho = - 43 1 kllmole ( - 1 03 kcallmole) /).H0 = -2 1 kJ/mole ( -5 kcaVrnole)

(Note: Mia for abstraction of a 1 0 H from both ethane and propane are + 4 10 kllmole (+ 98 kcallmole). It i s reasonable to use this same value for abstracti on of the 10 H in i sobutane . )

64

Page 73: Solucionario de wade

4-20 continued

t Cl o +

H I

H-CH2- � -CH3

CH3

reaction coordinate ..

1 ° radical; DJ{0 = - 21 kJ/mole (- 5 kcal/mole)

3° radical; I'1Ho = - 50 kJ/mole (- 12 kcaUmole)

Since DJ{0 for forming the 3° radical is more negative than Mlo for forming the 1 ° radical, it is reasonable to infer that the activation energy leading to the 3° radical is lower than the activation energy leading to the 1 ° radical.

4-21 2-Methylbutane can produce four mono-chloro isomers. To calculate the relative amount of each in the product mixture, multiply the numbers of hydrogens which could lead to that product times the reactivity for that type of hydrogen. Each relative amount divided by the sum of all the amounts will provide the percent of each in the product mixture.

CH3 I

CICH2 - CHCH2CH3 (6 1°H) x (reactivity 1.0) = 6.0 relative amount 6.0 x 100= 23.5

26%

4-22

CH3 I

CH3-C-CH2CH3 I

Cl (l 3°H) x (reactivity 5.5) = 5.5 relative amount

2� :� x 100 = 23%

CH3 I

CH3· CH· CHCH3 I

Cl (2 2°H) x (reactivity 4.5) = 9.0 relative amount

2� :� x 100 = 38%

CH3 I

CH3 - CHCH2CH2Cl (3 lOH) x (reactivity 1.0) = 3.0 relative amount

3.0 x 100 = 23.5 1 3%

(a) When n-heptane is burned, only 1 ° and 2° radicals can be formed (from either C-H or C-C bond cleavage). These are high energy, unstable radicals which rapidly form other products. When isooctane (2,2,4-trimethylpentane, below) is burned, 3° radicals can be formed from either C-H or C-C bond cleavage. The 3° radicals are lower in energy than 1 ° or 2°, relatively stable, with lowered reactivity. Slower combustion translates to less "knocking."

CH3 H I I

JV\ JV'

CH - �-C - �-CH - C -CH 3 ? + 2 I 3

CH3 CH3

isooctane (2,2,4-tnmethylpentane)

Any indicated bond cleavage will produce a 3° radical.

65

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4-22 continued

(b) CH3 CH3 I I

H C - C-O - H 3 I + o R -- H C-C-O-3 I

+ R-H an alkyl

CH3 radi cal CH3

When the alcohol hydrogen i s abstracted from t-buty l alcohol , a rel ati vely stab le t-butoxy radical ( (CH3)3C- O 0 ) is produced. Thi s low energy radical i s slower to react than alkyl radical s , moderating the reaction and producing less " knocking."

4-23

(a) 1 ° H abstraction

F 0 + CH3CH2CH3

break 1 ° H -CH2CH2CH3 make H-F overall 1 ° H abstraction

2° H abstraction

F 0 + CH3CH2CH3

break 2° H -CH(CH3)2 make H-F overal l 2° H abstraction

H - F +

M-[0 = + 4 1 0 kJ/mole ( + 98 kcaJ/mole) M-[0 = - 569 kJ/moJe (- 1 36 kcal/mole) Mlo = -159 kJ/mole ( -38 kcallmole)

H - F

MiO = + 397 kJ/mole ( + 95 kcaJ/mole) Mio = - 569 kJ/mole (- 1 36 kcal/mole) Mlo = - 172 kJ/mole ( -41 kcallmole)

(b) Fluorination is extremely exothermic and is l ikely to be i ndi scriminate in which hydrogens are abstracted. (In fact , C-C bonds are also broken during fluorination.)

(c) Free radical fl uorination is extremely exothermic . In exothermic reacti ons , the transi tion states resemble the starting material s more than the products, so whi le the 1 ° and 2° radica ls differ by about 1 3 kJ/mole (3 kcal/mole) , the trans i ti on states w i l l differ by only a tiny amount. For fluorination, then , the rate of abstraction for 1 ° and 2° hydrogens wil l be virtual l y identical. Product rati os wi l l depend on statistical factors only .

Fluorination is difficul t to contro l , but i f propane were monofluorinated, the product mixture would reflect the ratio of the types of hydrogens : six 1 ° H to two 2° H , or 3 : 1 rat io , giving 75% 1 -fluoropropane and 25% 2-fluoropropane .

4-24 H H H Br Br Br I I I I I I

CH -C- C-CH 3 I I

3 CH -C-C-CH 3 I I

3 CH - C-C-CH 3 I I

3 CH3CH3

Mechan ism CH3CH3 CH3CH3

in i ti ation

propagation

nn hv Br-Br -- 2 Br o

H I • ���-->� CH3- C- C-CH3 + Br o � HBr + CH - C-C-CH 3 I I

3 I I CH 3CH 3 CH3CH 3

H H Br I .� nn I I

CH -C-C- CH + Br-Br � CH -C-C- CH 3 I I

3 3 I I

3 C� C� C�C�

66

+ Br 0

Page 75: Solucionario de wade

4-25 (a) Mechanism nn hv

in i t iation B r-Br --- 2 Br 0 H H H IU� 6=: C( propagation a + B,·

-

.. ..

• H

H r-'� H Br � H \n A U H

+ B,-B,

- U + B,·

(b) Energy calculation uses the value for the al lylic C-H bond from Table 4-2.

First . i break al ly l ic H-CH[ri ng] WO = + 364 kJ/mole (+ 87 kcallmole) propagatIOn make H-Br WO - - 368 kJ/mole ( - 88 kcallmole) step overal l a l ly l ic H abstraction M-/0 = -4 kJ/mole ( -1 kcaUmole)

Second i break Br-Br propagation make 2° C-Br step overal l C-Br formation

The first propagation step is rate-limiting.

MfD = + 192 kJ/mole (+ 46 kcallmole) WO = - 285 kJ/mole ( - 68 kcallmole) M-/0 = - 93 kJ/mole ( -22 kcaUmole)

[ H 8 0 8 OJ t slmclu;c of Ihc transHi on slateo 0- - - - H - - - - - - - - - B,

+ HBr

(c) The Hammond postulate te l ls us that, in an exothermic reaction, the transi t ion state is c loser to the reactants in energy and in structure . Since the first propagation step is exothermic (although not by much), the transition state is c loser to cyclohexene + bromine radic al . This is indicated in the transi tion state structure by showing the H c loser to the C than to the Br. (d) A bromine radica l wi l l abstract the hydrogen with the lowest bond dissoc iation energy at the fastest rate . The al lyl ic hydrogen of cyc lohexene is more easi ly abstracted than a hydrogen of cyclohexane because the radical produced is stab i l ized by resonance. (Energy values below are per mole . )

a l iphatic: 397 kJ (95 kcal ) - H H H H - 364 kJ (87 kcal) : al lyl ic SLOWER 6 6 FASTER

4-26 . . . . . . . .

¢x ¢x ~ if ----- I • ----- ----- • I h- :::--..

OCH3 OCH3 OCH3 OCH3 B HA radical

67

Page 76: Solucionario de wade

4-27 () �-0J¢Ct R-O· J I h R

H3C 0 "TT

CH3 Vitamin E

.. R-O H ,

+

·0

II � I l-R H3C

CH3 stabilized by resonance­less reactive than RO·

4-28 The triphenylmethyl cati on is so stable because of the delocal ization of the charge . The more resonance forms a species has, the more stable it wi l l be.

+0 Oc b ----

/' O�

+Oc-b ---- o

Q=Cb � II

o 0 0 0 < }-c ---- < }-c + ---- o-c ---- < }-c b b b +U

� + + Q P +0 < }-c + ---- < }-c ---- < }-c b b b

(Note : these resonance forms do not inc lude the simple benzene resonance forms as shown below; they are signifi cant, but repeti ti ve, so for simpl ic i ty, they are not drawn here . )

4-29

(c )

4-30

(c)

Ph Ph

0-+/ 0-+/ 'I _ � C

, ---- � II C, ---- many more

Ph Ph

most stable (c) > (b) > (a) least stable + +

CH3 -CCH2CH3 I

(b) CH3-CHCHCH3 I

CH3 CH3

3 ° 2°

most stable (c) > (b) > (a) least stable . .

CH3 -CCH2CH3 I

(b) CH3-CHCHCH3 I

CH3 CH3

3 ° 2 ° 68

+ (a) CH3 -CHCH2CH2

I CH3

1 °

. (a) CH3 -CHCH2CH2

I CH3

1 °

Page 77: Solucionario de wade

4-3 1

:O:H� II ( I II - ..

H C-C-C-C-CH3 + :O-H 3 I • •

H base

:0 : :0: II II

-- H20 +

.. -:0 : :0:

.. -

: 0 : :0 : I I I I I I / C .... -• • /C ....

H C ..... ' c 'CH ___ ",C , ", C ____ C / C

H C ..... ' C ..... ' CH H C/ 'C/ 'CH 3 I 3

H

3 I

3 3 I 3

H H

-- H-Base +

4-34 Please refer to solution 1 -20, page 1 2 of this Solutions Manual .

4-35 transition states (a) and (c)

/�

(b) Mio is negative (decreases), so the reaction is exothermic .

t

4-36

t

--------- l E t - -------- - r Q

reaction coordinate --

� ..

(d) The fi rst transi t ion state detennines the rate since it is the hi ghest energy point. The structure of the fi rst transi ti on state resembles the structure of the intennedi ate since the energy of the transi tion state is c losest to the energy of the intennediate .

products

acti vation energy

tranSitIOn state

reactants

products

reaction coordinate --

69

Page 78: Solucionario de wade

4-37

t

4-38

� energy of h ighest transition state detennines rate

t t-Jl0 is posi tive --- --------------------

reaction coordinate -

The rate law is first order with respect to the concentrations of hydrogen ion and of t-butyl alcohol , zeroth order with respect to the concentration of ch loride ion, second order overal l . rate = kr [(CH3)3COH 1 [ H + 1 1 °

(d) H �--+--H

......t--'r-- H (e)

3 °

H

H

3° 1 °

� � CH3

(c) H H

H --- 3 °

H H

all are 2 ° H except for the 1 °

1 ° two types l abeled

� �

H al l are 2° H except for the two types l abeled a l l are 2° H except as l abeled

4-40

(a) break H-CH2CH3 and I-I , make I-CH2CH3 and H-I kllmole : ( + 4 1 0 + + 1 5 1 ) + ( - 222 + - 297) = + 42 kJ/mole

kcallmole : ( + 98 + + 36) + ( - 53 + - 7 1 ) = + 1 0 kcallmole

(b) break CH3CH2-CI and H-I, make CH3CHrI and H-CI kllmole: ( + 3 3 9 + + 297) + ( - 222 + - 43 1 ) = -17 kJ/mole

kcal/mole: ( + 8 1 + + 7 1 ) + ( - 53 + - 1 03) = - 4 kcallmole

7 0

Page 79: Solucionario de wade

4-40 continued (c) break (CH3hC-OH and H-C l , m ake (CH3hC-Cl and H-OH kJ/mole : ( + 381 + + 431) + ( - 331 + - 498) = -1 7 kJ/mole kcal/mole: ( + 9 1 + + 103) + (-79 + -119)= -4kcallmole

(d) break CH3CH2-CH3 and H-H, make CH3CHrH and H-CH3 kJ/mole: ( + 356 + + 435) + ( - 410 + -435) = -S4 kJ/mole kcallmole : ( + 85 + + 104) + ( - 98 + -104) = - 1 3 kcallmole

(e) break CH3CHrOH and H-Br, make CH3CHrBr and H-OH kllmole : ( + 381 + + 368) + ( -285 + - 498) = - 34 kJ/mole kcal/mole : ( + 91 + + 88) + ( - 68 + - 119) = - 8 kcallmole

4-41 Numbers are bond dissoc iat ion energies in kcallmole in the top l ine and kl/mole in the bottom l ine.

< >-CH2 > CH2 = CHCH2 > (CH3hC > (CH3)zCH > CH3CH2 > CH3

85 87 91 95 98 104 356 364 381 397 410 435

most stable least stable

4-42

(a) o _

eyCl Only one product; chlorination would work. B romination on a 20 carbon would not be predicted to be a h igh-yie lding process .

(b) eyCH3

_

eyCHP +

Ch�Hl + QCH1

Cl Chlorination would produce four constitutional isomers and would not be a good method to make only one of these. Monobromination at the 30 carbon would gi ve a reasonab le yield .

(c)

H H H CI H H I I

CH3-�-�-CH3 -

CH3 CH3

I I I I CH -C-C-CH + CH -C-C-CH Cl 3

I I 3 3

I I 2

CH3 CH3 CH3 CH3

Chlori nation would produce two consti tutional i somers and would not be a good method to make only one of these . Monobromination would be selective for the 30 carbon and would give an excellent yield.

(d)

4-43

CH3 CH3 I I

CH -C-C-CH 3 I I

3

CH3 CH3

ini t iation (1)

(2)

propagation

(3)

CH3 CH3 I I

- CH -C-C-CH CI 3 I I

2 Only one product ; ch lorination would gi ve a h igh yie ld . Monobromination would be very difficul t since al l hydrogens are on 1 ° carbons . CH3 CH3

�l � CI �� :6)

71

H

O ;C

Cl + C l·

Termination steps are any two radicals combining.

Page 80: Solucionario de wade

4-44 .

(a) CH2=CH-CH2 ....... f----l.�

. :0: :0:

I I I (c) CH -C-O· ....... f----l.� CH3-C=O 3

• • • •

(d) H H

OH I ....... f----l.�

�H UH H

4-45 (a) Mechanism

ini t iation nn hv

Br -Br --- 2 Br·

(e) H H �H U ....... f----l.�

H H

cYH

+ HBr

H3C � CH3 J H3C yy' CH3 H3C� CH3 l propagation H � '=Jj.J_ � � -1 WId·) � · W W f

The 3 ° a l lyl ic H is abstracted selecti vely (faster Br _

�j second propagation step than any other type In the molecule) , fonrung an intermediate represented by two non-equivalent resonance forms . Parti al radical character on two different carbons of the a l ly l ic radical leads to two different products . +

Br·

(b) There are two reasons why the H shown i s the one that is abstracted by bromine radical: the H is 3° and it is a l lyl ic , that i s , neighboring a double bond . Both of these factors stabi l i ze the radical that is created by removing the H atom.

4-46 Where mixtures are poss ib le , only the major product is shown .

(a) 0 � rY Br

� V only one product possible

� CH� (b) U �

CH3H I I

(c) CH -C-C-CH 3 I I

3 CH3 CH3

-

3 ° hydrogen abstracted selecti vely, faster than 2° or 1 °

CH3 Br I I CH -C-C-CH 3 I I

3 CH3 CH3

72

3 ° hydrogen abstracted selectively, faster than 1 °

Page 81: Solucionario de wade

4-46 continued

(d) 00 -

decal in

(e) �

(f)� -

Br

CO 3' hydrogen abstracted select ive ly. faster than 2'

Br � + 3 ° hydrogen abstracted selectively , faster than 2° or 1 °

� Br

both 2°-formed in equal amounts

UCIHCH3 I from resonance-stabi l i zed benzyl ic radical Br

h

(h) Q¢-OQ+ cb Al l hydrogens at the starred positions are equi valent and allylic. The H from the lower ri ght carbon has been removed to make an intermedi ate wi th two non-equivalent resonance forms, giv ing the two products shown . Drawing the resonance fonns is the key to answering this question correctly.

Br

4-47 (a) As CH3C\ is produced, it c an compete with CH4 for avai l able Cl ., generating CH2Cl2. This can generate CHCI3, etc.

RroRagation steRs CH4 + C l· .- HCl + • CH3 • CH3 + Cl2 .- ClCH3 + Cl· C lCH3 + C l • .- HCl + • CH2Cl • CH2C\ + Cl2 .. CH2C\2 + Cl· CH2Cl2 + Cl· .. HCl + • CHCl2 • CHCl2 + Cl2 .. CHCl3 + Cl· CHCl3 + Cl • .- HCl + • CCl 3 • CC l3 + Cl2 .. CCl4 + Cl·

(b) To maximize CH3C l and min imize formation of polychloromethanes, the rat io of methane to chlorine must be kept h igh (see problem 4-2).

To guarantee that a l l hydrogens are replaced with chlorine to produce CCI4, the ratio of chlorine to methane must be kept h igh .

73

Page 82: Solucionario de wade

4-48 (a) n-Pentane can produce three monochloro isomers. To calculate the relative amount of each in the product mixture , multiply the numbers of hydrogens which could lead to that product times the reactivity for that type of hydrogen . Each relative amount divided by the sum of al l the amounts will provide the percent of each in the product mixture.

(b)

4-49

CI- CH2CH2CH2CH2CH3

( 6 10 H) x (reactivity 1 .0) = 6.0 relative amount

6.0 x 1 00 = 1 8% 3 3 .0

CI I

CH3 -CHCH2CH2CH3

( 4 20 H) x (reactivity 4 . 5 ) = 1 8 .0 re lative amount

total amount = 3 3 .0

i�:� x 1 00 = 55%

C I I

CH3CH2 - CHCH2CH3

( 2 20 H) x (reactivity 4 . 5 ) = 9 .0 relative amount

9 .0 x 1 00 = 27% 33 .0

(a) The second propagation step in the chlorination of methane is highly exothermic (i1Ho = - 1 09 kJ/mole ( -26 kcal/mole». The transition state resembles the reactants, that is, the CI-CI bond will be slightly stretched and the CI-CH3 bond will j ust be starting to form.

8· 8· CI - - - - - - C I - - - - - - - - - - - - - - - CH 3 stronger weaker

(b) The second propagation step in the bromination of methane is highly exothermic (i1Ho = - 1 0 1 kJ/mo\e ( - 24 kcal/mole». The transition state resembles the reactants, that is, the B r-Br bond will be slightl y stretched and the Br-CH3 bond will j ust be starting to form.

B r - - - - - - B r - - - - - - - - - - •.. - CH 3 stronger weaker

74

Page 83: Solucionario de wade

4-50 Two mechanisms are possib l e depending on whether HO' reacts with ch lorine or with cyc!opentane.

Mechan ism 1

1 <1) HO-OH --- 2 HO' in i ti ation

(2) HO' + Cl -Cl ---- HO-Cl + Cl'

:>0

H

f) Cl· + ---- H-Cl + '0

propagation H H

(4) Cl -Cl + '0 ----

Cl>O + Cl'

Mechanism 2 f) HO-OH --- 2 HO' initi ation

:>0

H'O (2) HO' + ---- H-OH +

H H

f) Cl -Cl + '0 ---

CI>O + C l·

propagation

:>0

H (4) Cl' + ---- H-Cl + D

The energies of the propagation steps determine which mechanism is fol l owed . The bond dissociation energy of HO-Cl is about 210 kJ/mole (about 50 kcallmole) , making in i t iat ion step (2) in mechanism 1 endothennic by about 30 kJ/mole (about 8 kcallmole) . In mechanism 2, i n i ti ation step (2) is exothennic by about 10 1 kJ/mole (24 kcallmole ) ; mechanism 2 is preferred.

4-51 �� C� l_ C l

'c: Cl -C-H + :O-H --- H2O + Cl -C: ---I

U· I I

H H H a carbene

4-52 This cri t ical equation is the key to this problem: 11 G = 11 H - T 11 S

At 1400 K, the equi l ibrium constant is 1; therefore:

c=�> fj, G = 0 >

+

Assuming fj, H is about the same at 1400 K as i t is at calorimeter temperature :

-137 kJ/mole =

1400 K =

- 137,000 1400

= -98 J/K-mole (-23 callK-mole)

JIK-mole

. . :Cl :

This is a l arge decrease in entropy, consistent wi th two molecules combining into one .

75

Page 84: Solucionario de wade

4-53 Assume that ch lorine atoms (radicals) are sti l l generated in the in i ti ation reaction . Focus on the propagation steps. Bond dissoci ation energies are given below the bonds, in kJ/mole (kcal/mole) .

Cl o + H-CH3 - H-CI + 0 CH3 !1H = + 4 kJ/mole ( + 1 kcal/mole) 435 ( 1 04) 43 1 ( 1 03)

CI - C1 + 242 ( 58 )

o CH3 - CI - CH3 + Cl o 3 5 1 (84)

!1H = - 1 09 kJ/mole ( - 26 kcal/mole)

What happens when the different radical spec ies react wi th iodine?

Cl o + I - I 1 5 1 (36)

- I - CI + 0 1 2 1 1 (50)

!1H = - 60 kJ/mole ( - 14 kcal/mole)

I - I + 1 5 1 (36)

o CH3 - 1 - CH3 + I 0 234 (56)

!1H = - 83 kJ/mole ( - 20 kcal/mole)

Compare the second reaction in each pair: methyl radical reacting with ch lorine is more exothermic than methyl radical reacting wi th i odine ; this does not explain how iodine prevents the ch lorination reaction . Compare the first reaction in each pair: ch lorine atom reacting wi th i odine is very exothermic whereas ch lorine atom react ing with methane is sl ightly endothermic . Here is the key: ch lorine atoms wi l l be scavenged by iodine before they have a chance to react with methane. Without ch lorine atoms, the reaction comes to a dead stop.

4-54 (a) nn hv fl B r-Br - 2 Br 0 in i t iation f' � "

(2) Br 0 + H-SnB u3 ----- H-Br + o SnBu3

H � H

(3) � B' + 1 . snBUj - O· + Br - SnBu3

propagation H H

(4) O' --.. ":":\ n + H-SnBu3 - (j- H + o SnBu3

(b) A l l energies are in kJ/mole . The abbreviation "cy" stands for the cyc lohexane ring .

S tep 2 : break H-Sn, make H-Br: +3 1 0 + - 368 = - 58 kJ/mole

Step 3 : break cy-Br, make B r-Sn: +285 + - 552 = - 267 kJ/mole WOW !

S tep 4 : break H-Sn , make cy-H: + 3 1 0 + - 397 = - 87 kJ/mole

The sum of the two propagation steps are : - 267 + - 87 = - 354 kJ/mole -a hugely exothermic reaction .

76

Page 85: Solucionario de wade

4-55

Mechanism 1 :

(a) CI ° + 03 --- CIO ° +

(b) 2 CIO ° --- CI - O - O-Cl

hv (c) CI - O - O - Cl --- 02 + 2 CI -

Mechanism 2 :

(d ) 03 � 02 + °

(e) CI ° + 03 --- CIO ° + 02

(f) CIO ° + ° --- 02 + Cl o

The biggest problem in Mechanism 1 l ies in step (b). The concentration of CI atoms is very smal l, so at any given time, the concentration of CIO wi l l be very smal l . The probabi l i ty o f two CIO radica l s finding each other to form CIOOCI i s virtual ly zero . Even though thi s mechani sm shows a catalytic cyc le with Cl o (starting the mechanism and being regenerated at the end), the middle step makes it h ighly un l i ke ly .

S tep (d) i s the " l ight" reaction that occurs naturally in dayl ight . At n ight , the reaction reverses and regenerates ozone .

S tep (f) i s the crucial s tep . A l ow concentration of CIO will find a rel ati vely h igh concentration of ° atoms because the " l ight" reaction i s producing ° atoms in relative abundance. Clo i s regenerated and begins propagation step (e) , continu ing the catalytic cycle .

Mechanism 2 i s bel i eved to be the dominant mechani sm in ozone depletion . Mechanism 1 can be discounted because of the l ow probabi l i ty of step (b) occurring, because two spec ies i n very low, catalytic concentration are required to find each other in order for the step to occur.

4-56 (a) [ H _�I_

S

_

.

_ _ - H - _ _ _ _ _ �: r --- \ C2Hs-CI + DCI) + \C2H4DCI + HCI;

7�o 91%

D replacement: 7% 7 1 D = 7 (reacti v i ty factor) H replacement : 93% 7 5 H = 1 8 .6 (reactivity factor) re lative reacti v i ty of H : D abstraction = 1 8 .6 7 7 = 2 . 7

Each hydrogen i s abstracted 2 . 7 times faster than deuterium.

In each case, the bond from carbon to H (D) i s breaking and the bond from H (D) to CI i s forming .

(c ) In both reactions of ch lorine wi th ei ther methane or ethane, the first propagation step i s rate- l imiting. The reaction of ch lorine atom with methane is endothennic by 4 kJ/mole ( 1 kcal/mole), whi le for ethane th i s step is exothennic by 2 1 kJ/mole (5 kcallmole) . By the Hammond Postu late, differences in activation energy are most pronounced in endothennic reactions where the transition states most resemble the products . Therefore, a change in the methane molecule causes a greater change in its trans i tion state energy than the same change in the ethane molecule causes in its transi t ion state energy . Deuterium wi l l be abstracted more s lowly in both methane and ethane, but the rate effect wi l l be more pronounced in methane than in ethane .

77

Page 86: Solucionario de wade

CHAPTER 5-STEREOCHEMISTRY

Note to the student : S tereochemistry is the study of molecu lar structure and reactions in three dimensions . Molecu lar models wi l l be espec ia l ly helpful in th i s chapter.

5 - 1 The best test of whether a household object is ch iral is whether it would be used equal ly wel l by a left ­or ri ght-handed person . The ch i ral objects are the corkscrew, the writ ing desk , the screw-cap bottle (only

for refi l l ing , however; in use, i t would not be chiral ) , the ri fle and the knotted rope; the corkscrew, the bottle top, and the rope each have a twi st in one direction, and the rifle and desk are c learly made for right­handed users . A l l the other objects are achiral and would feel equi valent to ri ght- or left-handed users .

5-2

(a) cis

mirror

(b) trans

H3� H CH3

mirror

(c) cis first, then trans

mirror

mIrror

(d)

mIrror

ach iral-identical mirror images

chiral­enantiomers

ach iral-identical mirror i mages

ach iral-identical mirror i mages

CH3 chiral-I enanti omers

" , C H " � '-...

Br CH2CH3

79

Page 87: Solucionario de wade

5-2 continued

(e)

a a

mirror

(f) \J(j0" ° '" '"

mirror

5 -3 Asymmetric carbon atoms are starred.

Br (a) I

(b) OH OH I I

(c)

(d) OH OH I I

CI H (e)

6 80

chiral­enantiomers

chiral­enantiomers

no asymmetric carbons­same structure

no asymmetric carbons­same structure

enantiomers

no asymmetri c carbons­same structure

no asymmetri c carbon­same structure

Page 88: Solucionario de wade

5-3 continued COOH

(f)

(g)

(h)

(i )

U)

(k)

J / C ," " ' H H3C

NH2

~ * *

CI CI

~ * *

H CI

C I

jL7i CI

H

H

* " Q " '" C == C H / "

CH3

CH3

COOH I * " , C

H " � " H N

CH3 2

~ * *

CI CI � l * *

CI H

CI

� CI Q H

, .. ' / H C == C

/ " H3C H

H3C

enantiomers

same structure

enantiomers

no asymmetric c arbons-same structure

enantiomers

enantiomers

5-4 You may have chosen to in erchange two groups different from the ones shown here . The type of isomer produced wi l l sti l l be the same as l isted here .

Interchanging any two groups around a chira l i ty center wi l l create an enantiomer of the first structure .

Interchanging the Br and the H creates an enantiomer of the structure in Figure 5 - 5 .

Interchanging the ethyl and the isopropyl creates an en anti orner of the structure in Figure 5 - 5 .

On a double bond, interchanging the two groups on ONE of the stereocenters w i l l create the other geometric (cis-trans) i somer. However, interchanging the two groups on B OTH of the stereocenters w i l l gi ve the orig inal structure .

H " ", CH3 original C structure I I is cis C

H'" .......

CH 3

interchange H and CH3 on top stereocenter to produce trans

8 1

Page 89: Solucionario de wade

5-5 (a)

H " " " , H

plane of symmetry

(e) CHO 1

H " " C J ' CH OH

HO 2

chiral-no plane of symmetry

(b)

m Br Br

p lane of symmetry

(f) plane of symmetry

(c)

thi s view is from the right s ide of the s tructure as drawn in the text

H' Br H chiral-no plane of symmetry

p lane of symmetry

(d) CH2Cl 1

H " " C , J CH3

C l

ch iral-no plane of symmetry

(h) COOH 1

H " II C , J CH3 H2N chiral-no plane of symmetry

5-6 AL WAYS place the 4th priority group away from you. Then determine if the sequence 1 �2�3 i s c lockwise (R) or counter-c lockwi se (S) . (There i s a Problem-Solving Hint near the end of section 5-3 in the text that describes what to do when the 4th priority group is c losest to you . )

(a)

R

(d)

SU2 S - * : : * ....- S

(g)

H Cl

(b)

(h)

rotate CH3 up 3 CH3

* 1 1 C 2 Br � :: � CH2C

H 4

2

- -C l C l

82

3

H 4

S

R

only one asymmetric carbon

(c) �3h H 1

" 1 *C 3

(f)

C == C ..... : � CH / I -3

H H H 4

O � H

R

( i ) S �C " �� 3 /- CH2 I " ' " '/

(H3COhHC I , C -. H

CH(CH3h 4

see next page for an explanation of part ( i )

Page 90: Solucionario de wade

5 -6 continued

Part (i) deserves some explan ation . The di fference between groups 1 and 2 h i nge on what is on the "extra" oxygen .

CH(OCH3h � I

� - CH3

H - C - O - CH3 I t

H O� ,/ C � higher priority

5 -7 There are no asymmetric carbons in 5-3 (a) , (b) , (d), (e) , or ( i ) .

(c )

(f)

(g)

(h)

(j)

(k)

OH

s * 1 ./ C " , CH3CH2CH2....... " H

CH3

COOH

S J ./ C " , CH ./ �" H 3 NH 2

I;I � R * � S

Cl C l

C;: l� S * � S

H Cl

II R 0 , /( c == c H / "-H CH3

OH 1 * R " , C" H "� CH2CH2CH3 H3C

COOH 1 * R H " 'YC

" NH '" CH3 2

� I;I R *� * S

Cl Cl

� C;:l R � * R

Cl H

Q * S H " /

H'"

C == C" C�3 H

CH3 CH3

R

83

o - c I � .

H - C _ 0 / Imaginary

I lower priority

Page 91: Solucionario de wade

5-8

2 .0 g 1 1 0 mL = 0 .20 g/mL ; 1 00 mm = 1 dm

5-9

+ 1 .74° (0 .20) ( 1 )

= + 8 . 7 ° for (+ )-glyceraldehyde

0 .5 g I 10 mL = 0 .05 g/mL ; 20 cm = 2 dm

5- 1 0

- 5 .0° (0 .05) (2)

= - 50° for (-)-epinephrine

Measure using a solution of about one-fourth the concentration of the first . The value w i l l be ei ther + 45° or - 45°, which gives the s ign of the rotat ion.

5 - 1 1 Whether a sample i s dextrorotatory (abbrevi ated " ( +)" ) or levorotatory (abbreviated " ( -) ") i s determined experimental ly by a pol arimeter. Except for the molecule glyceraldehyde , there is no direct , universal correlation between direction of optical rotation (( +) and (-)) and designation of confi guration (R and S). In other words , one dextrorotatory compound might have R configuration whi le a different dextrorotatory compound might have S configuration . (a) Yes , both of these are determined experimental ly : the (+) or (-) by the polarimeter and the smel l by the nose . (b) No, R or S cannot be determined by either the polarimeter or the nose . (c ) The drawings show that (+)-carvone from caraway has the S configuration and (-)-carvone from spearmint has the R configuration .

o

(+)-carvone (caraway seed)

(R)-2-bromobutane

o

3

(For fun , ask your instructor if you can smell the two enantiomers of carvone. Some people are unable, presumably for genetic reasons , to di stinguish the fragrance of the two enantiomers . )

(-)-carvone (spearmint)

CH3 I / C " " " H CH3CH2 "

OH

(R)-2-butanol

one-third of mixture

+

CH3 I ./ C " ,

./ �" OH CH3CH2 H

(S)-2-butanol

two-thi rds of mixture Chapter 6 wi l l explain how these mixtures come about. For thi s problem, the S enantiomer accounts for 66 . 7% of the 2-butanol i n the mixture and the rest, 3 3 . 3% , is the R en anti orner. Therefore , the excess of one enantiomer over the racemic mixture must be 33 . 3% of the S, the enantiomeric exces s . (Al l of the R i s "canceled" by an equal amount of the S, algebraica l ly a s wel l a s in optical rotation . )

The optical rotation o f pure (S)-2-butanol i s + 1 3 . 5 ° . The optical rotation o f thi s mixture i s : 3 3 . 3% x ( + l 3 . 5 °) = + 4 . 5 °

84

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5- 1 3 The rotation of pure (+)-2-butanol i s + 1 3 . 5 ° .

observed rotat ion + 0.45 ° --------------------- = rotation of pure enantiomer

x 1 00% = 3 . 3 % optical purity + 1 3 . 5 ° = 3 . 3 % e .e . = excess o f (+) over (-)

To calculate percentages of (+) and (-) : (two equations in two unknowns)

(+) + (-) = 1 00%

(+) (-) = 3 . 3 %

(-) = 1 00% - ( + )

(+ ) ( 1 00% - (+)) = 3 . 3 %

2 (+) = 1 03 . 3 %

(+) = 5 1 .6% (rounded) (-) = 48 . 4%

5 - 1 4 Drawing Newman projections i s the c learest way to determine symmetry of conformations . (a) H

H ,,-h H

Br�CI H

(b) B'r' }Plane includes ',CI Br and CI ,

chiral­optical ly acti ve plane of symmetry containing

B r-C-C-CI ; not optical ly active

(d)

H�Ck2� H

Brl(:y�20 B, H : H

plane of symmetry-not optica l ly acti ve despite the presence of two asymmetric c�rbons

, ,

H H I I

Br - C - C - Cl I I

H H no asymmetric carbons

(e) Br H (f) H �CH2,,-h H

H �CH20 Br H H

no plane of symmetry­optical ly act ive (other chair form is equivalent-no plane of symmetry)

(c) H

H ,,-h H

Br�CI CI

chiral­optical ly acti ve

H I *

CICH2 - C - CI I Br

Br � - - - "' H 1� 4

H Br

plane of symmetry through C - l and C-4-not optica l ly acti ve

no asymmetric carbons

Part (2) Predictions of optic al acti v i ty based on asymmetri c centers give the same answers as predictions based on the most symmetric conformation .

85

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5- 1 5

(a) H '- " H

CI

" \,\ C == C == C ' / 'CI

no asymmetric carbons , but the molecule i s chiral (an allene) ; the drawing below i s a three­dimensional picture of the allene in (a) showing there is no p lane of symmetry because the substituents of an al lene are in different planes

(e)

(h)

H

�/ " '-j �J Br

p lane o f symmetry bi secting the molecule ; no asymmetric carbons ; not a ch iral molecule

two asymmetric carbons ; a chiral compound

(b)

no asymmetric carbons , but the molecule i s chiral (an allene)

(d) H H

"- /

no asymmetric carbons ; thi s al lene has a plane of symmetry between the two methyls (the p lane of the paper) , including all the other atoms because the two pi bonds of an al lene are perpendicular, the Cl i s i n the plane of the paper and the plane of symmetry goes through i t ; not a chiral molecule

C == C H / " /

CI C == C / "-

H H planar molecule-no asymmetric carbons ; not a chiral molecule

(f) (g) o 0

Br no asymmetric carbons , but the molecule is chiral due to restric ted rotat ion; the drawing below is a three­dimensional picture showing that the rings are perpendicular (hydrogens are not shown)

86

H H no asymmetric carbons , and the groups are not l arge enough to restrict rotation ; not a chiral compound

Page 94: Solucionario de wade

5- 1 6

(a) COOH

H+ OH

CH3

(c) CH3

HO+ H

CH2CH3 (R)-2-butanol

5 - 1 7

(a) CH20H

HO+ H

CH3

5- 1 8

(a)

(c )

COOH H

HO+ H CH3+ COOH

CH3 OH enantiomer enantiomer

CH3

Br+ H

CH2CH3

Br+ 1I

CH2CH3 CH3 same enantiomer

CH3

H+ OH

CH3

HO+ H

CH2CH3 CH2CH3 enanti omer same

CH3

HO+ H

COOH same

CH3

H+ Br

CH2CH3 enantiomer

CH2CH3

H+ OH

CH3 same

Rules for Fischer projections: 1 . Interchanging any two groups an odd number of times (once , three times, etc . ) makes an enantiomer. Interchanging any two groups an even number of times (e .g . twice) returns to the original stereoi somer. 2. Rotating the structure by 900 makes the enantiomer. Rotating by 1 800 returns to the original stereoisomer. (The second rule is an appl ication of the fi rst . Prove this to yourself . )

1 80 0 rotation of the right structure does not give left structure ; no plane of symmetry : chiral�nantiomers

CH20H

- - - - -Br H f' on the left ; also has plane of symmetry : same structure + 1 800 rotation of the right structure gi ves same structure as

CH20H

�� Br + Br plane of symmetcy

(:H , 3 , 87

same structure

Page 95: Solucionario de wade

5- 1 8 continued

CHO

(d) H + OH

H + OH

CH20H

(e)

(f)

Hi: H OH

CH20H

Hoi�

H OH

CH20H

mirror

CHO

HO + H

HO + H

CH20H

CH20H

1 80 0 rotation of the right structure does not give left structure ; no plane of symmetry : chiral--enantiomers

HO $ H n 1 800 rotation of the right structure gi ves same structure as - - - - - - - - - - - - - - - - - - - on the left ; also has p lane of symmetry : same structure

HO H

CH20H

H+�:

HO + H

CH20H

1 80 0 rotation of the right structure does not gi ve left structure ; no p lane of symmetry : chiral--enantiomers

5 - 1 9 If the Fischer projection is drawn correct ly , the most oxidi zed carbon wi l l be at the top ; thi s is the carbon with the greatest number of bonds to oxygen . Then the numbering goes from the top down.

(a) R (b) no chiral center (c) no chiral center

5 -20

(d) 2R, 3R (e) 2S, 3R (numbering down) (f) 2R, 3R

(a) enantiomers--configurations at both asymmetric carbons inverted

(g) R (h) S ( i ) S

(b) diastereomers--configuration at only one asymmetric carbon inverted (c) diastereomers--confi guration at only one asymmetric carbon inverted (the left carbon) (d) constitutional i somers-C=C shifted position (e) enantiomers--ch iral , mirror i mages (f) diastereomers--configuration at only one asymmetri c carbon inverted (the top one) (g) enantiomers--configuration at al l asymmetric carbons inverted (h) same compound-superimposable mirror images (hard question ! use a mode l ) ( i ) diastereomers--configuration a t on ly one chiral i ty center (the ni trogen) inverted

5 -2 1 CH3

(a) ' -- ':;ul " �:' U

CH3 meso structure

plane of symmetry

not optica l l y act ive

CIHB� l fr

Cl , -, , -* ' * , , ,

H3C : CH3

plane of symmetry

88

Page 96: Solucionario de wade

5-2 1 continued

(b)

CH3

Br+ Cl

Cl+ Br

CH3

Cl+ B

B,+ C:

50: 50 CH3

BrHC� lrC l

� - * *

CH3 H3C

C lHB� f I Br * - = *

H3C CH3

racemic mixture-not opti cal l y acti ve, although each enantiomer by i tself would be optical ly acti ve

: � planes of � i H tH3 ( symmetry

H 1 i f H H =t= H '---LJ/ (c)

/ : \ H H H3C : CH3

(d) HO * H � tCH20H

H B� H? H

Br * H �/ '\ CCH3 not ch iral , not opti cal l y acti ve

CHO

H � OH

H0 -T- H (e)

H � OH HO OH

5-22

(a)

H � OH

CH20H

optical l y acti ve

COOH

H+OH

H Cl

CH3 A

HO

COOH

HO+ H

Cl H

CH3 B

enantiomers : A and B ; C and D

CH20H HOH2C CH20H

optica l ly active

CH3

(f) * *

CH3 not optical ly active­superimposable on i t s mirror image­technical ly , thi s i s a meso structure (may require models ! )

(g) *

COOH

H + OH

Cl + H

CH3

COOH

HO + H

H + Cl

CH3 C D

plane of symmetry

CH3 1 CH3

*

meso-not optical ly act ive

diastereomers : A and C ; A and D ; B and C; B and D

89

Page 97: Solucionario de wade

5-22 continued COOH

(b) - � -$ �:- - =;��rY COOH

H+ OH

HO+ H

COOH F

COOH

HO + H

H + OH

COOH G

(c)

(d)

COOH E

enantiomers : F and G diastereomers : E and F; E and G

COOH

H ---If-- OH

H OH

H --I- Br

COOH H

COOH

HO -f-- H

HO H

H --I- Br

COOH L

COOH

HO -I-- H

HO -I-- H

Br --lf-- H

COOH I

COOH

H --If-- OH

H OH

Br --lf-- H

COOH M

COOH

HO --ll-- H

H ---If-- OH

H --II-- Br

COOH J

COOH

H --I- OH

HO --ll-- H

H --I- Br

COOH N

enantiomers : H and I ; J and K; L and M; N and 0 di astereomers : any pair which i s not enantiomeric

p Q enantiomers : P and Q diastereomers : P and T; P and U;

Q and T; Q and U; T and U

cm

rtiH3

H ' H

plane of symmetry MESO

T

90

COOH

H ---lf-- OH

HO ---lf-- H

Br --ll-- H

COOH K

COOH

HO r--I-- H

H --II-- OH

Br --lf-- H

COOH o

plane of symmetry

MESO U

Page 98: Solucionario de wade

5-23 Any diastereomeric pair could be separated by a physical process l i ke d is ti l l ation or crystal l ization . Di astereomers are found in parts (a) , (b) , and (d) . The structures in (c) are enantiomers ; they could not be separated by normal physical means .

5 -24 r enantiomers � 0 o+��

' H+�H' 0 CH3 CH3 H OH HO H HO H H OH COOH COOH ( (S)-2-butyl (R,R)-tartrate mirror (R)-2-butyl (S,S)-tartrate � diastereomers diastereomers

\... CH2CH3 H+O o CH3 H--If--- OH HO--l�H COOH (R)-2-butyl (R,R)-tartrate

mirror

o

HO--lf---H H ----i...-- 0 H COOH (S)-2-butyl (S,S)-tartrate

5 -25 Please refer to solution 1 -20, page 12 of th i s Solutions Manual .

5-26

(a) H I * " , C ",--C l " '/' ...... CH3 HO R chiral

(e) SHjB:, - - - ... - - - - - - - - _ .. . H * Br

R /CH2Br

meso ; ach iral

(b) CH20H H+OH CH3 R

ch i ral

(f) plane of symmetry

sHiB:, B/ * H

S CH2Br

chiral

91

(g )

sHi B '

/ * OH

R CH3 chiral

(h) S CH3 H'" * OH s � * OH H * OH

R / CH2CH3 chiral

Page 99: Solucionario de wade

5 -26 continued

(i) Br Br " "

( I )

Cn)

R

C = C = C ' C( �CI

chiral molecule , but no ch iral centers

chiral

5 -27

(j)(Y Br

I * "-chiral

s

plane of symmetry meso; achiral

(0) R

s

chiral

(k) o- Bf

achiral

(m)

chira l

(NH2 i s group 1 , CH3 i s group 4)

plane of symmetry

chiral

(a) <;:H3 en anti orner

<;:H3 * =

en anti orner * =

H - C - CI c::======�> CI - C - H >

(c)

CH2CH3 no plane of symmetry no diastereomer chiral structure

* -Br - C - H

* I Br - C - H

CH2CH2CH3

no p lane of symmetry ch iral structure

CH2CH3 chiral structure

en anti orner >

no p lane of symmetry no diastereomer chiral structure

<;:H3 * =

H - C - Br * I

H - C - Br

CH2CHzCH3 chiral s tructure

chiral structure

* -H - C - Br

* I Br - C - H

CH2CH2CH3 chiral structure

c::==============================�> ( inverting two groups on the diastereomer

92

bottom asymmetric carbon i nstead of the top one would also gi ve a diastereomer)

Page 100: Solucionario de wade

5-27 continued

(d) Br enantiomer c::=====�> H

H

no pl ane of symmetry chiral structure

diastereomer

* -

_� _':': � � �� _ _ _ _ plane of H .: C _ OH symmetry

CH1CH3 no enantiomer a meso structure, not chiml

diastereomer >

Br

chiral structure

* -HO - C- H * 1

H - C- OH

CH1CH3 chiral structure

> H not chiral ;

m eso structure

(inverting two groups on the right asymmetric carbon instead of the left one would also give a diastereomer)

(inverting two groups on the bottom asymmetric carbon instead of the top one wou ld also gi ve a diastereomer)

(f)

racemic mixture of enantiomers; each is chiml with no plane of symmetry

-------------�------------( �H2CH3 �H2CH�

5-28

* -H - C - OH * 1

HO - C - H +

* -HO - C - H * 1

H - C - OH �stereomer � : � �H2CH3 � CH2CH3 * :: diastereomer H - C- OH I f - - - - - � t - - - - - - - - - p ane 0

H .:. C _ OH symmetry

CH CH this meso structure is a diastereomer of each of the 2 3 enantiomers; it is not chiml

(a) CH70H (b) CHO (d�O+�{

1I

11+ 011

CH3

H+Br

CH3

93

H+ OH

CH3

Page 101: Solucionario de wade

5-29 Your drawings may look different from these and sti l l be correct . Check configuration by assigning R and S to be sure .

COOH (a) �" H CH3

NH

5-30

2

CHO (b) �" OH HOCH2 H

(a) same (meso)-plane of symmetry , superimposable (b) enantiomers--configuration inverted at both asymmetric carbon atoms (c) enantiomers--configuration inverted at both asymmetric carbon atoms

(d)

(d) enantiomers-solve thi s problem by switching two groups at a time to put the groups in the same positions as in the first structure ; i t takes three switches to make the i dentical compound, so they are enantiomers; an even number of switches would prove they are the same structure (e) enanti omers--configuration inverted at both asymmetric carbon atoms (f) diastereomers--configuration inverted at only one asymmetric carbon (g) enantiomers--configuration inverted at both asymmetric carbon atoms (h) same compound-rotate the right structure 1 80° around a hori zontal axi s and it becomes the left structure

5-3 1 Drawing the enantiomer of a ch iral structure i s as easy as drawing i ts mirror image.

(b) CHO

Br+ H

CH20H

(e ) CH3 " " , H

C == C == C '" / 'B

H r

(f) plane of symmetry-no en anti orner

5 -32

o

CHO

(c) HO H

HO H

HO H

(a) 1 .00 g / 20.0 mL = 0.050 g/mL ; 20.0 cm = 2 .00 dm

(0 .0500) (2 .00) = - 1 2 .5 °

(b ) 0 .050 g / 2 .0 mL = 0 .025 g/mL ; 2 .0 cm = 0 .20 dm

+ 0 .043° (0 .025) (0.20)

(d)

5 -33 The 32% of the mixture that is (-)-tartaric acid wi l l cancel the optical rotation of the 32% of the mix ture that i s (+)-tartaric acid , leaving only (68 - 32) = 36% of the mixture as excess (+)-tartaric acid to gi ve measurable optica l rotation . The specific rotation wi l l therefore be only 36% of the rotation of pure (+)-tartaric acid : (+ 1 2 . 0° ) x 36% = + 4 .3 °

94

Page 102: Solucionario de wade

5-34 (b) Rotation of the enantiomer will be equal in magnitude, opposite in sign: - 15.90°.

(c) The rotation -7.95° is what percent of - 15.900?

- 7 95° --' - x 100% = 50% e.e. - 15.90°

There is 50% excess of (R)-2-iodobutane over the racemic mixture; that is, another 25% must be R and 25% must be S. The total composition is 75% (R)-(-)-2-iodobutane and 25% (S)-(+)-2-iodobutane.

5-35 All structures in this problem are chiral.

Cal

H+

OH

H+ OH

CH20H A

CHO

HO+H

HO+H

CH20H B

enantiomers: A and B; C and D

CHO

H+ OH

HO+ H

CH20H C

diastereomers: A and C; A and D; Band C; Band D

(b)

E F G enantiomers: E and F; G and H diastereomers: E and G; E and H; F and G; F and H

95

CHO

HO+H

H+ OH

CH20H D

H

Page 103: Solucionario de wade

5-35 continued

(c) This structure is a challenge to visualize. A model helps. One way to approach this problem is to assign R and 5 configurations. Each arrow shows a change at one asymmetric carbon.

5-36

H CH3 H CH3 H3C CH3 H \ S

H .0.

H

H H

H H

CH3 H

� � L

H3C H3C

.0. � H H

H CH3

H3C H H3C

Summary R5S5 (K) is the enantiomer of RRR5 (M) 5555 (L) is the enantiomer of RRRR (P)

H

H H

CH3 H

The structures with two R carbons and two 5 carbons (J, N, and 0) have special symmetry. J is a meso structure; it has chirality centers and is superimposable on its mirror image-see Problem 5-20(h). Nand 0 are enantiomers, and are diastereomers of all of the other structures. Give yourself a gold star if you got this correct!

s� .Ix. �� S* A diastereomers B R*

meso

15,3R equivalent to 1R,35

meso

15,3R equivalent to 1R,35

C enantiomers D chiral chiral 1R,3R 15,35

C and Dare enantiomers. All other pairs are diastereomers.

*Structures A and B are both meso structures, but they are clearly dif ferent from each other. How can they be distinguished? One of the advanced rules of the Cahn-Ingold-Prelog system says: When two groups attached to an asymmetric carbon differ only in their absolute configuration, then the neighboring (R) stereocenter takes priority. Now the configuration of the central asymmetric carbon can be assigned: 5 for structure A, and R for structure B. This rule also applies to problem 5-37. (Thanks to Dr. Kantorowski for this explanation.)

96

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5-37 This problem is similar to 5-36.

(a) [CH3 CH3 CH3 CH3

H *

Br H *

Br Br *

H H *

Br 2 5 5

H 3 Br Br H H Br Br H

H *

Br H *

Br H * Br Br

* H

4 R R

sCH3 '-- � CH3 CH3 '-- �

CH3

A diastereomers B C enantiomers D meso meso chiral chiral

(b) 25,4R 25,4R 2R,4R 25,45 equivalent equivalent to 2R,45 to 2R,45

The configuration of carbon-3 in A and B can be assigned according the rule described above in the solution to problem 5-36: C-3 in A is 5, and C-3 in B is R.

(c) According to the IUPAC designation described in text Section 5-2B, a chirality center is "any atom holding a set of ligands in a spatial arrangement which is not superimposable on its mirror image." An asymmetric carbon must have four dif ferent groups on it, but in A and B, C-3 has two groups that are identical (except for their stereochemistry). C-3 holds its groups in a spatial arrangement that is superimposable on its mirror image, so it is not a chirality center. But it is stereogenic: in structure A, interchanging the Hand Br at C-3 gives structure B, a diastereomer of A; therefore, C-3 is stereogenic.

(d) In structure C or D, C-3 is not stereogenic. Inverting the H and the Br, then rotating the structure 180°, shows that the same structure is formed. Therefore, interchanging two atoms at C-3 does not give a stereoisomer, so C-3 does not fit the definition of a stereogenic center.

5-38 The Cahn-Ingold-Prelog priorities of the groups are the circled numbers in (a).

(a) H CH3 H CH3 10 (2)1 H2 H

" lev 0.1 H .......::C CH .. H C CH

"C '/" "c/ 'CH Pt "c/ "c/ 'CH I

''''' \ 3

H/" I /' \ 3

H H CH3 H H CH3

o 0 0 0 (R)-3,4-dimethyl-l-pentene (5)-2,3-dimethylpentane

(b) The reaction did not occur at the asymmetric carbon atom, so the configuration has not changed­the reaction went with retention of configuration at the asymmetric carbon.

(c) The name changed because the priority of groups in the Cahn-Ingold-Prelog system of nomenclature changed. When the alkene became an ethyl group, its priority changed from the highest priority group to priority 2. (We will revisit this anomaly in problem 6-21(c).)

(d) There is no general correlation between R and 5 designation and the physical property of optical rotation. Professor Wade's poetic couplet makes an important point: do not confuse an object and its properties with the name for that object. (Scholars of Shakespeare have come to believe that this quote from Juliet is a veiled reference to designation of R,5 configuration versus optical rotation of a chiral molecule. Shakespeare was way ahead of his time.)

97

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5-39 (a) The product has no asymmetric carbon atoms but it has three stereocenters: the carbon with the OH, plus both carbons of the double bond. Interchange of two bonds on any of these makes the enantiomer.

(b) The product is an example of a chiral compound with no asymmetric carbons. Like the allenes, it is classified as an "extended tetrahedron"; that is, it has four groups that extend from the rigid molecule in four different directions. (A model will help.) In this structure, the plane containing the COOH and carbons of the double bond is perpendicular to the plane bisecting the OH and H and carbon that they are on. Since the compound is chiral, it is capable of being optically active.

COOH

H

HO

H

(c) As shown in text Figure 5-16, Section 5-6, catalytic hydrogenation that creates a new chirality center creates a racemic mixture (both enantiomers in a 1: 1 ratio). A racemic mixture is not optically active. In contrast, by using a chiral enzyme to reduce the ketone to the alcohol (as in part (b)), an excess of one enantiomer was produced, so the product was optically active.

98

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CHAPTER 6-ALKYL HALIDES: NUCLEOPHILIC SUBSTITUTION AND ELIMINATION

6-1 In problems like part (a), draw out the whole structure to detect double bonds.

(a) vinyl halide (c) aryl halide (e) vinyl halide (b) alkyl halide (d) alkyl halide (f) aryl halide

6-2 (a) H (b) Br , , (d) I , J-C-I Br -C- Br

(Cl � I-C-H , , , H Br Br

(f)� (g) H (hl+ FX '-rH2F Br

6-3 IUPAC name; common name; degree of halogen-bearing carbon

(a) l-chloro-2-methylpropane; isobutyl chloride; 1 ° (b) diiodomethane; methylene iodide; methyl (c) 1,I-dichloroethane; no common name; 1 ° (d) 2-bromo-l,1,l-trichloroethane; no common name; all 1° (e) trichloromethane; chloroform; methyl (f) 2-bromo-2-methylpropane; t-butyl bromide; 3° (g) 2-bromobutane; sec-butyl bromide; 2° (h) l-chloro-2-methylbutane; no common name; 1 ° (i) cis-l-bromo-2-chlorocyclobutane; no common name; both 2° U) 3-bromo-4-methylhexane; no common name; 2° (k) 4-fluoro-l,1-dimethylcyclohexane; no common name; 2° (I) trans-l,3-dichlorocyclopentane; no common name; all 2°

6-4 CI Cl Cl

Cl

>

0

Kepone®

(e) Br

~ Cl or (CH3hCCI

CI CI

Cl

Cl

Page 107: Solucionario de wade

6-5 From the text, Section 2-9B, the bond dipole moment depends not only on bond length but also on charge separation, which in tum depends on the dif ference in electronegativities of the two atoms connected by the bond. Because chlorine's electronegativity (3.2) is significantly higher than iodine's (2.7), the C-Cl bond dipole is greater than that of C-I, despite C-I being a longer bond.

6-6

(a) n-Butyl bromide has a higher molecular weight and less branching, and boils at a higher temperature than isopropyl bromide. (b) t-Butyl bromide has a higher molecular weight and a larger halogen, and despite its greater branching, boils at a higher temperature than isopropyl chloride. (c) n-Butyl bromide has a higher molecular weight and a larger halogen, and boils at a higher temperature than n-butyl chloride.

6-7 From Table 3-2, the density of hexane is 0.66; it will float on the water layer (d 1.00). From Table 6-2, the density of chloroform is 1.50; water will float on the chloroform. Water is immiscible with many organic compounds; whether water is the top layer or bottom layer depends on whether the other material is more dense or less dense than water. (This is an important consideration to remember in lab procedures.)

6-8

(a) Step (1) is initiation; steps (2) and (3) are propagation.

nn hv (1) Br - Br -- 2 Br·

Cr\ (2) H2C = � - CH2 + Br • � HBr + { H2C = � - CH2 " .. Hi - � = CH2}

.� \n (3) H2C = C - CH2 + Br - Br � H2C = C - CH2 + Br·

H H I Br

(b) Step (2): break allyJic C-H, make H-Br: kJ/mole: + 364 - (+ 368) = - 4 kJ/mole

kcallmole: + 87 - (+ 88) = - 1 kcallmole

Step (3): break Br-Br, make allylic C-Br: kJ/mole: + 192 - (+ 280) = - 88 kJ/mole

kcallmole: + 46 - (+ 67) = - 2 1 kcal/mole

M! overall = - 4 + - 88 = - 92 kJ/mole (- 1 + - 2 1 = - 22 kcal/mole ) This is a very exothermic reaction; it is reasonable to expect a small activation energy in step ( 1), so this reaction should be very rapid.

6-9 (a) propagation steps

C�� H3C, /CH2 ) C=C + Bre / ,

H3C CH3

100

repeats chain mechanism

Br· + +

Page 108: Solucionario de wade

6-9 continued

The resonance-stabilized allylic radical intermediate has radical character on both the 1 ° and 3° carbons, so bromine can bond to either of these carbons producing two isomeric products.

(b) Allylic bromination of cyclohexene gives 3-bromocyclohex-l-ene regardless of whether there is an allylic shift. Either pathway leads to the same product. If one of the ring carbons were somehow marked or labeled, then the two products can be distinguished. (We will see in following chapters how labeling is done experimentally.)

6- 10

(a)

(b)

o NBS

CH3 I

CH3 -C-CH3 I hv

CH3

o- Br + y Br

CH3 I

CH3 -C -CH2CI I

CH3

the second structure from an allylic shift is identical to the first structure-only one compound is produced

This compound has only one type of hydrogen-only one monochlorine isomer can be produced.

CH3 I

CH3 I

CH3- � -H

CH2CH3 hv

CH3- �- Br

CH2CH3

Bromination has a strong preference for abstracting hydrogens (like 3°) that give stable radical intermediates.

hv

(or NBS)

Bromine atom will abstract the hydrogen giving the most stable radical; in this case, the radical intermediate will be stabilized by resonance with the benzene ring.

Br

(d) CO hv OJ +

the second structure from an allylic shift is identical to the first structure-only one compound is produced

(or NBS) Br

Bromine atom will abstract the hydrogen giving the most stable radical; in this case, the radical intermediate will be stabilized by resonance with the benzene ring.

6- 1 1 (a) substitution-Br is replaced (b) elimination-H and OH are lost (c) elimination-both Br atoms are lost

6- 12

(a) CH3 (CH2)4CH2 - OCH2CH3 (b) CH3 (CH2)4CH2 - eN

101

Page 109: Solucionario de wade

6- 13 The rate law is first order in both 1-bromobutane, C4H9Br, and methoxide ion. If the concentration of C4H9Br is lowered to one-fi fth the original value, the rate must decrease to one-fifth; i f the concentration of methoxide is doubled, the rate must also double. Thus, the rate must decrease to 0.02 molelL per second:

= ( 0.05 molelL per second ) x ( 0. 1 M ) ( 2.0 M )

0.02 molelL per second rate x =

original rate ( 0.5 M ) ( l.0 M )

new rate change in change in C4H9Br NaOCH3

A completely different way to answer this problem is to solve for the rate constant k, then put in new values for the concentrations.

rate = k [C4H9Br] [NaOCH3] � 0.05 mole L-I sec -I = k (0.5 mol L-I) (l.0 mol L-I) �

rate constant k = 0.1 L mol-I sec-I

rate = k [CH3IJ [NaOH] = (0. 1 L mol-I sec-I) (0.1 mol L-I) (2.0 mol L-I) = 0.02 mole L-I sec -I

6- 14 Organic and inorganic products are shown here for completeness.

(a) (CH3)3C - 0 - CH2CH3 + KEr

+ (c) (CH3hCHCH2 - NH3

-

Br

+ NaI

+ NaCI

(e) � I + NaCI

+ -

+ NH4 Br

(f) � F + KCI (lS-crown-6 is the catalyst and does not change; CH 3CN is the solvent)

6- 15 All reactions in this problem follow the same pattern; the only difference is the nucleophile (-:Nuc).

Only the nucleophile is listed below. (Cations like Na+ or K+ accompany the nucleophile but are simply

spectator ions and do not take part in the reaction; they are not shown here.)

� CI + -:Nuc ---- � Nuc + CI-

l-chlorobutane

(a) HO- (b) F - from KF/lS-crown-6 (d) -CN

(f) -OCH2CH3 (g) excess NH3 (or - NH2)

6- 16

(a) (CH3CH2)2NH is a better nucleophile-Iess hindered (b) (CH3hS is a better nucleophile-S is larger, more polarizable than 0 (c) PH:, is a better nucleophile-P is larger, more polarizable than N

(e) HC_C-

(d) CH3S- is a better nucleophile-anions are better than neutral atoms of the same element (e) (CH3)3N is a better nucleophile-Iess electronegative than oxygen, better able to donate an electron pair

(f) CH3S- is a better nucleophile-anions are generally better than neutral atoms, and S is larger and more polarizable than 0 (g) CH3CH2CH20- is a better nucleophile-Iess branching, less steric hindrance (h) 1- is a better nucleophile-Iarger, more polarizable

102

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6- 17 A mechanism must show electron movement. .. CH -O-CH + H+ 3

• • 3

H

+1", • • -

� CH3-0-CH3 + :Br: .. � Protonation converts OCH3 to a good leaving group.

---

H 1 CH3-�: + CH3-!3[:

6- 18 The type of carbon with the halide, and relative leaving group ability o f the halide, determine the reactivity.

methyl iodide> methyl chloride> ethyl chloride> isopropyl bromide» neopentyl bromide, } least. . b I' d'd reaCllve

most reactive t- uty IO 1 e

Predicting the relative order of neopentyl bromide and t-butyl iodide would be difficult because both would be extremely slow.

6- 19 In all cases, the less hindered structure is the better SN2 substrate.

(a) 2-methyl- 1-iodopropane ( 1 ° versus 3°)

(b) cyclohexyl bromide (2° versus 3°)

(c) isopropyl bromide (no substituent on neighboring carbon)

(d) 2-chlorobutane (even though this is a 2° halide, it is easier to attack than the 1 ° neopentyl type in 2,2-dimethyl-1-chlorobutane-see below) CH3

1 a neopentyl halide-(e) isopropyl iodide (same reason as in (d)) I /c"" hindered to backside attack

- C, �CCH

H3 by neighboring methyl groups , "\. Lt13 H H "'--- -:Nuc

6-20 All SN2 reactions occur with inversion of configuration at carbon.

(a)

HO: trans

(b) H -:CN

� R

8-Br : H

H--P<CH, H08-

transition state

inversion

103

+

:Br:

cis

.. -:Br:

Page 111: Solucionario de wade

6-20 continued .. -

H H + :Br:

_ H3CH 1 c::::::::>

CH3CH2 CH3

(d) cBr

H ",�I�

F�CH3 - • •

:SH ..

.. -:Br:

Br H ·B·· ;-

(e) H , H3CO , + . r. S �/� .

-,�,,� • •

CH3� :-� - � c::::::::>

(f)

6-2 1

H H � I

H-�: �� C� ID I C;CI

H -

R

+

H+ I

H + CH}

CH2CH}

:CJ :

(a) The best leaving groups are the weakest bases. Bromide ion is so weak it is not considered at all basic; it is an excellent leaving group. Fluoride is moderately basic, by far the most basic of the halides. It is a terrible leaving group. Bromide is many orders of magnitude better than fluoride in leaving group ability.

(b)

H" : �"'" CH . } F :

transition state

inverted, but still named S

H��, I CH}

---II"�

I I +

OCH3

.. -:Br:

(c) As noted on the structure above, the configuration is inverted even though the designations of the configuration for both the starting material and the product are S; the oxygen of the product has a lower priority than the bromine it replaces. Refer to the solution to problem 5-38, p. 97, for the caution about confusing absolute configuration with the designation of configuration.

Cd) The result is perfectly consistent with the SN2 mechanism. Even though both the reactant and the product have the S designation, the configuration has been inverted: the nomenclature priority of fluorine changes from second (after bromine) in the reactant to first (before oxygen) in the product. While the designation may be misleading, the structure shows with certainty that an inversion has occurred.

104

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6-22

slow � planar carbocation

.. -:Br:

+

+ CH3CH20H2 Be

(or simply HBr)

6-23 The structure that can form the more stable carbocation will undergo SN 1 faster.

(a) 2-bromopropane: will form a 2° carbocation (b) 2-bromo-2-methylbutane: will form a 3° carbocation (c) allyl bromide is faster than n-propyl bromide: allyl bromide can form a resonance-stabilized intermediate. (d) 2-bromopropane: will form a 2° carbocation (e) 2-iodo-2-methylbutane is faster than t-butyl chloride (iodide is a better leaving group than chloride) ( f) 2-bromo-2-methylbutane (3°) is faster than ethyl iodide (1°); although iodide is a somewhat better leaving group, the difference between 3° and 1 ° carbocation stability dominates

6-24 Ionization is the rate-determining step in SN l. Anything that stabilizes the intermediate will speed the reaction. Both of these compounds form resonance-stabilized intermediates.

H (x H I ---

H

H � H

V H allylic

benzylic +( )=CH2 --- OCHz � < }- �H2

6-25 H H I t:.. + I

+

CH3 - T - THCH3 • Br- + CH3 - C - CHCH3 + H -R- CH2CH3

(Br CH:1 H �

CH3 - � -CHCH3 ...... f'---- CH3 - � -CHCH3 / I I .. ..---..... ('I. I :?: CH3 CH3CH2- 9: H-+? CH3

CH2CH3 H CH2CH3 105

Page 113: Solucionario de wade

6-26 It is i mportant to analyze the structure of carbocations to consider if migration of any groups from adjacent carbons wi l l lead to a more stable carbocation . As a general rule, if rearrangement would lead to a more stable carbocation , a carbocation wi l l rearrange. (Beginning with this problem, only those unshared electons pairs involved in a particular step wi l l be shown. )

CH3 I

(a) CH3-C -CH-CH3 I 1\ CH3 I �

1- +

CH3 I +

CH3-y-y-CH3

CH3 H

c�� nucleophilic attack on unrearranged carbocation . .

CH3 :O-CH3 ! I

CH3-? - ?-CH3

CH3 H

nucleophil ic attack after carbocation rearrangement

CH3 I +

CH -C-C-CH � 3 I�I

3

CH3 H

Methyl shift to the 2° carbocation forms a more stable 3° carbocation.

cj: nucleophilic attack on unrearranged carbocation

H rl<-H I � ",--;::R - CH,CH;

CH3

106

CH3 :O-CH3 HRCH3 I I

-----l .. � CH3 -C - C -CH3 I I CH3 H

unrearranged product

H CH2CH3

d=�: CH 3

unrearranged product

Page 114: Solucionario de wade

6-26(b) continued

nuc leophi l ic attack after carbocation rearrangement

rf-H :b�CH2CH] ��.. � C))

HH ___

hydride shift

30 CH3

Hydride shift to the 2° carbocation forms a more stable 3° carbocation .

(c ) H H H

H

rearranged product

Note: braces are used to indicate

nucleophi lic attack on unrearranged carbocation

the ONE chemical species represented by multiple resonance forms. H H H :(jH OH

I �O·=C-CH3 • •

III H H : 0 :

0-.. I I ... O-C-CH I . . 3

: 0 :

o-H • • ?-H o-H • •

:?-H

I O=C-CH3 O-C-CH3 + ...... I----J .. � I . . +

unrearranged product

I I HO-C-CH • • 3

The most basic species in a mixture i s the most l ikely to remove a proton . In this reaction , acetic acid is more basic than iodide ion .

107

Page 115: Solucionario de wade

6-26(c) continued

nucleophilic attack after carbocation rearrangement

H H Hy)H

hydride

I + shift ---...... �

H H H

H H H H H H

:0: H II

: c;?: j allylic-resonance-stabilized HO -C -CH3 ..

: O - H I

:0 : H I I D�-C-CH3 HO -C-CH I . . 3

..... f--------

(removes H+ H as on p. 107) rearranged product

Comments on 6-26(c)

HDH �==C-CH3

plus two other I � resonance forms as shown on p. 107 H

(1) The hydride shift to a 20 carbocation generates an allylic, resonance-stabilized 20 carbocation.

(2) The double-bonded oxygen of acetic acid is more nucleophilic because of the resonance forms it can have after attack. (See Solved Problem 1-5 and Problem 1- 16 in the text.)

(3) Attack on only one carbon of the allylic carbocation is shown. In reality, both positive carbons would be attacked in equal amounts, but they would give the identical product in this case. In other compounds, however, attack on the dif ferent carbons might give different products. ALWAYS CONSIDER ALL POSSIBILITIES.

(d)ctH 0 CH2-I ---l"�

hydride shift followed by nucleophilic attack r+CH2 NCH3 � V ---l"� V '------' : � -CH2CH,;

108

The 1 0 carbocation initially formed is very unstable; some chemists believe that rearrangement occurs at the same time as the leaving group leaves. At most, the 10 carbocation has a very short lifetime.

Page 116: Solucionario de wade

6-26 (d) continued

alkyl migration (ring expansion) followed by nucleophilic attack

H 2° H �I + CH2 : 0 -CH2CH3

\ -.-.------�.�

6-27 CH3 0 I "

CH2

(a) CH3- �-0-C-CH3

CH2CH3

(b) CH3 -CHCH20CH3 (e) CHoCH2CHl

SN 1 , 3°

I

CH3

SN 1 , 3° SN2 , 1 °

6-28 CH3

I H20,�

CH CH /C"'H SN2-3 2 Br inversion

(R)-2-bromobutane� H20,�

S N l­racemization

\.. (R)-2-butanol

CH3

I /C""

CH CH ,'OH 3 2 H

(S)-2-butanol

CH3

I + /C"",

CH CH , OH

y

3 2 H

(S)-2-butano� 50 : 50 mixture-racemic

/ .. : 0-CH2CH3 I H

SN 1 , weak nucleophile

S N2, strong nucleophile

If SN 1 , which gives racemization, occurs exactly twice as fast as SN2, which gives inversion, then the racemic mixture (50 : 50 R + S) is 66.7% of the mixture and the rest, 3 3 . 3%, is the S enantiomer from SN2. Therefore, the excess o f one enantiomer over the racemic mixture must be 3 3 .3%, the enantiomeric excess.

(In the racemic mixture, the Rand S "cancel " each other algebraically as well as in optical rotation.)

The optical rotation of pure (S)-2-butanol is + l 3.5° . The optical rotation of this mixture is: 33 .3% x + l 3.5° = + 4.5°

109

Page 117: Solucionario de wade

6-29

(a) Methyl shift may occur simultaneously with ionization. The lifetime of 1 ° carbocations is exceedingly

short; some chemists believe that they are only a transition state to the rearranged product.

AgI +

CH3 1 ° methyl 30 CH3

I + shift I CH3-CJH2 • CH3- �-CHFHl �H3 I � H20:

CH3 I

CH3 - C-CH2CH3 ...

I :O-H

(b) Alkyl shift may occur simultaneously with ionization.

6-30 (a) H

CH2 \ I

� Ag+ ---I.� AgI +

:Br: UBf

..

alkyl shift­ring expansion

CH2 ..

+ CH30H2 +

The SN 1 mechanism begins with ionization to form a carbocation, attack of a nucleophile, and in the case of ROH nucleophiles, removal of a proton by a base to form a neutral product.

(b) In the E l reaction, the solvent (methanol, in this case) serves two functions: it aids the ionization process by solvating both the leaving group (bromide) and the carbocation; and second, it serves as a base to remove the proton from a carbon adjacent to the carbocation in order to form the carbon-carbon double bond. The SN1 mechanism adds a third function to the solvent: the first step is the same as in E l , ionization to form the carbocation; the second step has the solvent acting as a nucleophile-this step is dif ferent from E l; third, the solvent acts like a base and removes a proton, although from an oxygen (SN1) and not a carbon (El). Solvents are versatile!

110

Page 118: Solucionario de wade

6-3 1 This solution does not show complete mechanisms. Rather, i t shows the general sequence of a l l four pathways leading to five different products .

2-bromo-3-methylbutane

;on;z.;/

---------------- -----... --. . -- -- --- -- ---------------�

Five unique products shown in the boxes. One alkene can be produced from either the r or the 30 carbocation.

6-32

CH3

hydride shift ¥ _________ � H3C rearranged

+ CH3 30 carbocation

.......... -_ .............. -_ .............. .. . .

.. .. ___ ................................... __ I

two possible E 1 products

111

iH\ CH3

H3C .. � � T "'CH2

H : ----------------______ 1

H1C"CHl

_________ H ___________ _

two possible E1 products

Page 119: Solucionario de wade

6-33 Substi tution products are shown first, then e l imination products .

(a) � + ) (c )

without rearrangement with rearrangement

(d) o 6-34 Et is the abbreviation for ethy l , so EtOH is ethanol .

: Br : CH

EtOH 0·\3· CH3 Cf··�rH�Et

-----l.... • 0 J • • • + .., � HOEt I

• • Et

+

SN 1 product:

Cll3

�?: V Et + EtOH2

+

E I -MAJOR product­more substituted alkene

E l -MINOR product­more substituted alkene

6-35

+ EtOH2 +

without rearrangement

H" /

H �

crC-H • •

J HOEt . .

..

with rearrangement

112

+ EtOH2 +

Page 120: Solucionario de wade

6-36

yl NaOCH2CH3

..

Br

6-37 Br NaOCH3 � (a)

Br (b) �

Br

(c) ~ Br

H"'0CH] (d)

'" H

NaOMe

NaOH

NaOEt

H atoms that are removed are shown

Br

E2-minor

..

...

...

...

� E2-minor product-less substituted alkene (monosubsti tuted)

� +

minor alkene-monosubsti tuted

� +

major alkene-trisubsti tuted (cis + trans)

� +

minor alkene-monosubsti tuted

aCH]

minor alkene-disubstituted

113

+ � + yl E2-major OCH2CH3 product-morc substituted alkene SN2-probably (trisubsti tuted) very small amount

because of steric hindrance

OCR, � � +

major alkene- S N2 product-some disubsti tuted wi l l be formed on a (cis + trans) 2° carbon

� No SN2 product wi l l be formed; SN2 cannot

minor alkene- occur at a 3° carbon. disubsti tuted

OH

~ +

~ major alkene- SN2 product-very trisubstituted small amount because

2° carbon is hindered

OEt --O'CH]

+ (rCH]

major alkene- SN2 product showing trisubstituted inversion at the

carbon-some of this product wil l be formed on a 2° carbon

Page 121: Solucionario de wade

6-38 In systems where free rotation is possible, the H to be abstracted by the base and the leaving group (Br here) must be anti-coplanar. The E2 mechanism is a concerted, one-step mechanism, so the arrangement of the other groups around the carbons in the starting material is retained in the product; there is no intermediate to al low t ime for rotation of groups . (Models wi l l help .)

r:RCH3 Ph H' H � r/ C-C, -

/ ""CH Br� Ph

3

transition state

8-• • :t: :9CH3

Ph "" ••• , CH3 - "'C-C" . H� - 'Ph

trans

only geometric isomer H and Br anti -coplanar

�----------��----------� (- - - - - - - = stretched bond being broken or forme:0 possible from a concerted mechanism

orbital picture of this reaction

H H Ph

Br

H�:OCH I

• • 3 , ,

Br

transition state

...--pi bond

H ---.. � Ph --+---=--+--

trans

The other diastereomer has two groups interchanged on the back carbon of the New man projection , where it must give the cis-alkene.

H Ph H

Ph H • Ph CH3

Ph CH3

Br cis

114

Page 122: Solucionario de wade

6-39 lin this problem, a l l new internal alkenes fonn cis i somers as wel l as the trans i somers shown.

Section 1 Br � Br �

Section 2

�Br

C�OH.�H3

SN1

NaOCH3 OCH3

CH30H· �

SN2

NaOCH3 CH3

----j .. � H2C =< E2 CH30H CH3

AgN03

CH30H

+� El

+� E2

+� E2

�Br �

.. +� El

Section 3

CH3

H3C + Br

CH3

CH3

H3C + Br

C H3

Section 4

Br � Br �

NaOCH3 CH3 • H2C =< E2

CH30H CH3

CH30H CH3 CH3 • H3C + OCH3 + H2C =<

� CH3 SN 1 E l

CH3

NaOCH3 OCH3

CH30H" � +�

SN2 E2

CH30H OCH3 � . � +

SN1

115

El

+� El

+� E2

+� El

+� E2

+� E1

Page 123: Solucionario de wade

6-39 continued

Section 5 B r � Br

�aI I

-.� � � + �

E2 E2

6-40

(a) Ethoxide is a strong base/nucleophi le-second-order conditions. The I ° bromide favors substi tution over e l imination, so SN2 will predominate over E2.

� substitution-major el imination-minor

(b) Methoxide i s a strong base/nucleophi le-second-order conditions. The 2° chloride will undergo SN2 by backside attack as well as E2 to make a mixture of alkenes.

OCH3 � substi tution e l imination­

minor alkene e l imination­major alkene (cis + trans)

(c) Ethoxide is a strong base/nucleophile-second-order conditions. The 3° bromide is h indered and cannot undergo SN2 by backside attack. E2 is the only route possible.

CH3 - C=CH2 I

CH3

(d) Heating in ethanol i s conditions for solvolysis , an SN I reaction.

OCH2CH3 I

CH -C-CH 3 I

3

CH3

(e) Hydroxide is a strong base/nucieophile-second-order condi tions. The 1° iodide i s more l ikely to undergo SN2 than E2, but both products wi ll be observed.

CH3 - CHCH20H CH3 -C = CH2 I I

C� C� substitution-major el imination-minor

(f) S i l ver ni trate in ethanol/water is ionizing conditions for I ° alkyl hal ides that w i ll lead to rearrangement followed by substitution on the 3° carbocation.

OCH2CH3 from ethanol as I nucieophile

CH3 - ? - CH3

CH3 116

OH I

CH3-?-CH3

CH3

from water as n ucieophi le

Page 124: Solucionario de wade

6-40 continued

(g) S i lver nitrate in ethanoVwater is ionizing conditions for 1 ° alkyl hal ides that w i l l lead to rearrangement fol lowed by substi tution on the 30 carbocation.

OCH2CH3 from ethanol as I nucleophile

CH3 -C -CH2CH3 I

CH3

OH from water as I nucleophile

CH3-C-CH2CH3 I

CH3

(h) Heating a 30 hal ide in methanol is quintessential first-order conditions, ei ther E l or SN I (solvolysis). CH6OCH3 C 6 (!,"cc)

\. / substitution V

(SNl) el imination (E l )

( i ) Ethoxide in ethanol on a 30 hal ide wi l l lead to E2 elimination; there wi l l be no substitution.

CH3

A major alkene U tri substituted minor alkene disubstituted

6-41 Please refer to solution 1 -20, page 12 , of this Solutions Manual .

6-42 (a) � CI

(e) 0-I"" I

6-43

(f) CI I H - C - H I

CI

(g) CI I

C1-C - H I

CI

(h) I � Co '

(d) CI I CI-C - CH20H I

CI

( i ) CH2CH3 I

I- � -CH3 CH3

(a) 2-bromo-2-methylpentane (d) 4-(2-bromoeth y 1)-3-( fI uoromethy 1)-2-methy I heptane (e) 4,4-dichloro-5-cyc lopropyl-l- iodoheptane (b) l -chloro- l -methylcyclohexane

(c) 1 , I -dichloro-3-f1uorocycloheptane (f) cis- I ,2-dich loro- l -methylcyc lohexane

6-44 Ease of backside attack (less steric hindrance) decides which undergoes SN2 faster in all these examples except (b) .

(a) �CI

(b) �I

faster than � Primary R-X reacts faster than 2° R-X.

CI faster than �CI Iodide a better leaving group than chloride.

117

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6-44 continued Cl

(c) � faster than � Less branching on a neighboring carbon. Cl

Br (d) � Br faster than 00 Same neighboring branching, so 1 ° faster than 2°.

(e) OCH2Ci faster than OCI Primary R-X reacts faster than 2° R-X.

(f) �Br faster than XBr Less branching on a neighboring carbon.

6-45 Formation of the more stable carbocation decides which undergoes SN 1 faster in all these examples except (d). Cl Cl (a) 30+ faster than �

(b) � faster than � Cl 2° Cl 1°

(c) OBr faster than OCH2Br 2° 1°

(d) OJ faster than OCI (leaving group ability) 113

(e) Xr faster than ~

3° Br 2°

(t) Q faster than Q Br Br 2° allylic 2°

1 18

Page 126: Solucionario de wade

6-46 For SN2, reactions should be designed such that the nucleophile attacks the least highly substituted alkyl halide. ("X" stands for a halide: Cl, Br, or I.) (a)

(b)

(c)

(d)

(e)

(f)

(g)

6-47 (a)

Q-CH2X + HO- -

-

Q-CH20H 0 SCH2CHl

CL HO- 0 0 --� .. - � -

OH X O:.A X 0 some U OH OH

Q-CH2X + NH3 excess

HC:::C

( 1 ) CH3 cf \ -

-

-

-

/CH - 0 � __ ;."CH2CH3 CH3 1°

(2) �-CH3 \ I CH-X + CH2CH3 /

2°U CH3

also produced

Q-CH2NH2

H2C = CHCH2CN

CH3 \

� CH-O-CH CH / t 2 3

CH3 this bond formed CH3 \

.. CH-O-CH CH / t 2 3

CH3 this bond formed Synthesis ( 1 ) would give a better yield of the desired ether product. ( 1) uses SN2 attack of a nuc\eophilc on a 1 ° carbon, while (2) requires attack on a more hindered 2° carbon. Reaction (2) would give a lower yield of substitution, with more elimination. (b) CANNOT DO SN2 ON A 3° CARBON!

CH � bumps into H before it can find C CH2 I

3 '\ I I CH3-C-Cl + -OCH3 .. CH3-C I I C� C�

6-47 (b) continued on next page 119

elimination (E2) competes

Page 127: Solucionario de wade

6-47 (b) continued

Better to do SN2 on a methyl carbon:

6-48

CH3 I

CH3 - ? -O

CH3

� + CH - X 3

U

(a) SN2-second order: reaction rate doubles

CH3 I

CH3 - C-O - CH3 I CH3

(b) SN2-second order: reaction rate increases six times (c) Virtually all reaction rates, including this one, increase with a temperature increase.

6-49 This is an SN I reaction; the rate law depends only on the substrate concentration, not on the nucleophile concentration.

(a) no change in rate (b) the rate triples, dependent only on [ t-butyl bromide ] (c) Virtually all reaction rates, including this one, increase with a temperature increase.

6-50 The key to this problem is that iodide ion is both an excellent nucleophile AND leaving group. Substitution on chlorocyclohexane is faster with iodide than with cyanide (see Table 6-3 for relative nucleophilicities). Once iodocyclohexane is formed, substitution by cyanide is much faster on iodocyclohexane than on chlorocyclohexane because iodide is a better leaving group than chloride. So two fast reactions involving iodide replace a slower single reaction, resulting in an overall rate increase.

0- Cl + -CN moderate rate

.. 0-CN

o- Cl + -I

6-51

(a)

6-52

(a)

----� .. � 1 + -CN fast 0-

----� .. � CN + - 1 fast 0-

recycles-only a small amount needed (catalyst)

(c) (d)

rearrangement rearrangement

120

Page 128: Solucionario de wade

6-52 continued (b)

(i) · 0: H�( H H

equivalent resonance fonns H H }

.��" H H

(iv) �H2 H

equivalent resonance fonns i 1/��· .Jy�,,} loss of bromide gives unstahle 10 carhocation Ihat quickly rearranges to a 2° al lylic carbocation

equivalent resonance fonns-same as from part (iii) (c) Only one substitution product arises from equivalent resonance fonns. (i)

(iii) OCH2CH3

6-53

� cis + trans

most stable

al\ylic (+ on 3° and 2° C)

(ii)

(iv) OCH2CH3 � same as part (iii) cis + trans

+

>0>0> al\ylic (+ on 1 ° and 2° C)

121

least stable

Page 129: Solucionario de wade

6-54

H CHX-Z, H 020

I hydride t shift CH3

C(H

+ HXH2 I"

U OR I hydride t shift CH3

6

+

C( I alkyl shift-t nng expansIOn H

(j 6-55 Reactions would also give some elimination products; only the substitution products are shown here.

(a)

6-56

<;:H3

N::C-C-H

CH2CH3 SN2 gives inversion: only product:

(a) CH3CH20CH2CH3

C:::CH I (d) CH3(CH2)8CH2

(g) Q

(b) <;:H3 (c) <;:H2CH3 <;:H2CH3

H-C-OH I

H-<;:-CH3

CH3CH20 - <;: -CH3 + CH3 -<;: -OCH2CH3

CH2CH3 SN2 only

with inversion

(b) < }-CH2CH2CN

(e) < �N+- CH3 1-

(h) HO "'0"" CH3

122

- -- -- -

CH(CH3h CH(CH3h

solvolysis , SNl, racerruzation

(c) < }-SCH2CH3

(f) (CH3hC -CH2CH2NH2

Page 130: Solucionario de wade

6-57 substitution H+B,

Br + H CH 3

25,35

+ B'+ H H+Br

CH3 2R,3R racemic mixture

"--------- -----) y

HO+ H Br+ H

CH 3 2R,35

�KOH

+ H+ OH H+ Br

CH 3 25,3R

racemic mixture elimination H Br Br -:. ::: H '-Y rotate / '" ..

B� S=H3

H:MB, HO

' CH 3 H H'-')

H and Br � anti -coplanar

Regardless of which bromine is substituted on each molecule, the same mixture of products results. Each of the substitution products has one chiral center inverted from the starting material. The mechanism that accounts for inversion is SN2 . If an SN 1 process were occurring, the product mixture would also contain 2R,3R and 25,35 diastereomers. Their absence argues agai nst an SN 1 process occurring here.

trans

The other enantiomer gives the same product (you should prove this to yourself). The absence of cis product is evidence that only the E2 elimination is occurring in one step through an anti-coplanar transition state with no chance of rotation. If El had been occuring, rotation around the carbocation intermediate would have been possible, leading to both the cis and trans products.

6-58

(a) + 15.58° . . . x 100% = 98% of ongmal optIcal activity = 98% e.e. +15.90°

Thus, 98% of the 5 enantiomer and 2% racemic mixture gives an overall composition of 99% 5 and 1% R.

(b) The 1 % of radioactive iodide has produced exactly 1 % of the R enantiomer. Each substitution must occur with inversion, a classic S N2 mechanism. 6-59 (a) An SN2 mechanism with inversion will convert R to its enantiomer, 5. An accumulation of excess 5 does not occur because it can also react with bromide, regenerating R. The system approaches a racemic mixture at equilibrium.

�� �H3 f\ :: SN2 Br -C-H + Be .. H-C-Br + Br-inversion - -CH2CH3 CH2CH3

R 123 5

Page 131: Solucionario de wade

6-59 continued (b) In order to undergo substitution and therefore inversion, HO- wouid have to be the leaving group, but HO- is never a leaving group in SN2. No reaction can occur. (c) Once the OH is protonated, it can leave as H20. Racemization occurs in the SN1 mechanism because of the planar, achiral carbocation intermediate which "erases" all stereochemistry of the starting material. Racemization occurs in the SN2 mechanism by establishing an equilibrium of Rand S enantiomers, as explained in 6-59(a).

(b)

(c)

HO-C-H --.. :: H+ \:::) HO-C-H + H-C-OH

racemic mixture The carbocation produced in this E l elimination is 30 and will not rearrange. The product ratio follows the Zaitsev rule.

trace-from unrearranged 20 carbocation

minor trace amount

from either unrearranged 20 carbocation or rearranged 30 carbocation

from rearranged 30 carbocation

¥+�+Jy major minor trace-from unrearranged

20 carbocation from rearranged from rearranged 30 carbocation 30 carbocation

124

The carbocation produced in this E l elimination is 20 and can either eliminate to give the first two alkenes , or can rearrange by a hydride shift to a 30 carbocation which would produce the last two products. The amounts of the last two products are not predictable as they are both trisubstituted, but the first product will certainly be the least.

The carbocation produced in this El elimination is 20 and can either eliminate to give the first alkene, or can rearrange by a methyl shift to a 30 carbocation which would produce the last two products. The middle product is major as it is tetrasubstituted versus disubstituted for the last structure and monosubstituted for the first structure.

Page 132: Solucionario de wade

6-61 The allylic carbocation has two resonance forms showing that two carbons share the positive charge. The ethanol nucleophile can attack either of these carbons, giving the S N 1 products; or loss of an adjacent H will give the El product.

n O'CH2-B,

H SNI + H

1

- Br----

O'�H2 :O-CH2CH3 . . ..

H t • H

1

aCH2

+ H :�-CH2CH3

El

['tCH2 H �H :O-CH2CH3 H H "--./ • •

6-62

(a)

(b)

H

U··/'"W OH ... . .

H 0' /'"w SN I 9:H ...

H S 2 U··�H+

N OH -..

1 O'�H2 aCH2} I .. ..

H + H

CH2 - OCH2CH3 a- (1+ _ ... I H � H '---- : 0 -CH2CH3

O'CH2OCH2CHl

H ..

CH2 ('y'

OcH2CH3 .Z-CH2CH1 � I+ / ' .. H H�

('yCH2 � OCH2CHl H C;(H2

H

H C);,. -H2O: OH ... +1 H

H C);r:! -H2O: OH ... +1 H

OSHh "'� . . .!3r. +OH �

1 H

125

QH � ... H • •

H �:

"--./ .. (JH :ii,:

...

H

0': H

U B'

O B' + H20:

Page 133: Solucionario de wade

6-63 NBS generates bromine which produces bromine radical. Bromine radical abstracts an allylic hydrogen, resulting in a resonance-stabilized a\lylic radical. The a\lylic radical can bond to bromine at either of the two carbons with radical character. H I

H2C=C-C-CH3 I I H CH3

+ Br- -- HEr + 1H2C=C-C-CH3 I I H CH3

continues propagation Br I

• Br + H2C=C-C-CH3 I I H CH3

a\lylic } ...... f----I�- H2C -T = �� -CH,

H CH3 � Br2 Br I

+ H2C-C=C-CH3 I I H CH3

6-64 The bromine radical from NBS will abstract whichever hydrogen produces the most stable intermediate; in this structure, that is a benzylic hydrogen, giving the resonance-stabilized benzylic radical.

< }�HCH) � HBr + 1< }CHCH). � OCHCH)

1 Br � Br2 < }CHCH3 + Br·

...... f----I�_ • < >= CHCH3

(Even though three carbons of the ring have some radical character, these are minm resonance contributors. The product is most stable when the ring has all three double bonds intact, necessitating that the bromine bond to the benzylic carbon.) 6-65 Two related factors could explain this observation. First, as carbocation stability increases, the leaving group will be less tightly held by the carbocation for stabilization; the more stable carbocations are more "free" in solution, meaning more exposed. Second, more stable carbocations will have longer lifetimes, allowing the leaving group to drift off in the solvent, leading to more possibility for the incoming nucleophile to attack from the side that the leaving group just left. The less stable carbocations hold tightly to their leaving groups, preventing nucleophiles from attacking this side. Backside attack with inversion is the preferred stereochemical route in this case. 6-66 Br

aCH3

----.. ionization � Br +

H (t� CH3_--=------rearrangement

2° carbocation

126

3° carbocation mechanisms continued on next page

Page 134: Solucionario de wade

6-66 continued substitution on 20 carbocation

H�H 6CH3 :6-CH • • 3

..

substitution on rearranged 30 carbocation H I aCH3 :O-CH _ / • •

3 �- ....

elimination from rearranged 30 carbocation

6-67

(a) E2-one step

H

(�r CH3-CH-CH3 --

'-=:OH

CH3 :O-CH • • 3

-----I ... �

OH I CH3-CH-CH3 + Br-

(b) In the E2 reaction, a C-H bond is broken. When D is substituted for H, a C-D bond is broken, slowing the reaction. In the SN2 reaction, no C-H (C-D) bond is broken, so the rate is unchanged. (c) These are first-order reactions. (Blr slow

El H2C-CH-CH3 ... I H

SN I (BI r

The slow, rate-determining step is the first step in each mechanism. � H Explanation on next page .. - I, + fast :Br: H2C-CH-CH3 -- H2C=CH-CH3

slow H3C-CH-CH3 ... + �o: H C-CH-CH --3 3 fast + Br-

127

Page 135: Solucionario de wade

6-67 (c) continued The only mechanism of these two involving C-H bond c leavage is the E l , but the C-H c leavage does NOT occur in the slow, rate-determining step. Kinetic isotope effects are observed only when C-H (C--­D) bond c leavage occurs in the rate-determining step. Thus, we would expect to observe no change in rate for the deuterium-substituted molecules in the E l or SN I mechanisms. (In fact, this technique of measuring isotope effects is one of the most useful tools chemists have for determining what mechanism a reaction fol lows . ) 6-68 Both products are formed through E2 reactions. The difference is whether a D or an H is removed by the base. As explained in Problem 6-67, C-D cleavage can be up to 7 times slower than C-H cleavage, so the product from C-H c leavage should be formed about 7 times as fast. Thi s rate preference is reflected in the 7 : 1 product mixture . ("Ph" is the abbreviation for a benzene ring.)

D !f " .:-� H " , C - C

H " " Ph Br

- D, Br ..

H H " /

C = C

Ph/ '

H

requires C-D bond c leavage; s low; minor product

H ,H " .:-� H " C - C

Ph " 'I ' D B r

- H , Br ...

requires C-H bond c leavage ; 7 times faster; major product

6-69 The energy, and therefore the structure, of the transition state determines the rate of a reaction. Any factor which lowers the energy of the transition state wil l speed the reaction.

R \ � n " N: R' - X R " � R

---I"'�[ R�N�� - - - - R, - - -_ _ _ x8-l- R

�N+_ R' x-R" � R " � R R transition state-developing charge

This example of S N2 is unusual in that the nucleophi le is a neutral molecule-it is not negatively charged. The transition state is beginning to show the positive and negative charges of the products (ions), so the transition state is more charged than the reactants . The polar transition state wi l l be stabi l ized in a more polar solvent through dipole-dipole interactions, so the rate of reaction wi l l be enhanced in a polar solvent.

t , \, --- polar solvent­lower energy, faster reactIon

reaction coordinate ___

128

Page 136: Solucionario de wade

6-70 The problem is how to explain this reaction: facts Et2N : : NEt2

I HO- I H2C - CH - CH2CH3 � H2C - CH - CH2CH3 + Cl-I I

1) second order, but several thousand times faster than similar second order reactions without the NEt2 group

Solution Cl OH

1 2 2) NEt2 group migrates

Clearly, the NEt2 group is involved. The nitrogen is a nucleophile and can do an internal nucleophi l ic substitution (SNi), a very fast reaction for entropy reasons because two different molecules do not have to come together.

very fast 3

The slower step is attack of HO- on intermediate 3; the N is a good leaving group because it has a positive charge. Where wi ll HO- attack 3? On the less substituted carbon, in typical SN2 fashion. :NEt2

I H2C - CH - CH2CH3 I OH 2

This overall reaction is fast because of the neighboring group assistance in fonning 3. It is second order because the HO- group and 3 collide in the slow step (not the only step, however). And the NEt2 group "migrates", although in two steps. 6-71 The symmetry of this molecule is crucial. (a) Ph H Ph I I I CH3 - C - C - C - CH3 I I I H Br H

Regardless of which adjacent H is removed by t-butoxide, the product will be 2,4-diphenyl-2-pentene.

(b) Here are a Newman projection, a three-dimensional representation and a Fischer projection of the required diastereomer. On both carbons 2 and 4, the H has to be anti -coplanar with the bromine while leaving the other groups to give the same product. Not coincidentally, the correct diastereomer is a meso structure. Ph H Br H CH3 H 2 H3C Ph

� Ph Ph Br 3 Ph H3C Br CH3 Ph

H and Br are anti-coplanar CH3 129

H H

H

Page 137: Solucionario de wade

6-72 "Ph " = phenyl

(a) Ph I CH3CHCHCH3

I B r

NaOCH3 ...

E2

Ph Ph I I CH3-C = CHCH3 + CH3CHCH = CHz major rrunor

(b) H and Br must be anti -coplanar in the transition state

H ,H Kf CH3 Na OCH3 2R,3R \\

__

Ph \\\\\

CH3 Br

H

Ph I \\ H " 1 \\ IIC - C " CH ""- - 'CH 3

. 3 methyls CIS

H H�CH3

Ph�CH3 Br

CH3 � Ph ';( CH3

(c) H CH3 th H

Ph�CH3 CH3 ---{ V 'r H CH3 ------ Ph -+ r..

Br 2S,3R

Ph " \\CH3 '1 ,\ - IIC := C " CH3........... ' H

methyls trans

(d) The 2S, 3S is the mirror i mage of 2R, 3R; it would give the mirror image of the alkene that 2R, 3R produced (with two methyl groups cis). The alkene product i s planar, not chiral , so i ts mirror image is the same: the 2S, 3S and the 2R, 3R give the same alkene.

130

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6-73 All five products (boxed) come from rearranged carbocations . Rearrangement, which may occur simultaneously with ion ization, can occur by hydride shift to the 3° methylcyclopentyl cation, or by ri ng expansion to the cyclohexyl cation.

H C n 6H2 - Br 6H2

.. Br- +

hydride shift 6H ��H2 �H�� (y� H� · ·

___ H�CH3

H ..

&H CH3

A��CH3 U .. H

CH3 \6.b::\ X' H�CH3 U -alkyl shift (ring expansion) + (J'-- H H )q

H �CH3 U �

H -- .. O H

OH

� II (t.� + H�CH3

oqCH3 � •

.. HOCH

• • 3 -

CH2

6 CH6CH3

OCH3

6-74 Begin with a structure of (S)-2-bromo-2-tluorobutane . S ince there i s no H on C-2, the lowest priority group must be the CH3 . The Br has highest priority, then F, then CH2CH3, and CH3 i s fOUlth . Sodium methoxide i s a strong base and nucleophi le, so the reaction must be second order, E2 or SN2 .

(a) Br

H�CH3 NaOCH3 H ­----l .. � H3C ( \/ ,

H3C F H

131

Br H3C ffi H

H3C '<;Y- F H

NaOCH3 H3C -( '>/ ) ----- H3C

Page 139: Solucionario de wade

6-74 continued

In regular structural formulas, the reaction would give three products i nc luding the stereoisomers shown above .

� +

Br F

(5)

(b) 4 3 l� NaOCH3 4 3 l�

6-75

Br F

(5) F OCH3

(5)

In these structures, the numbers 1 to 4 indicate the group's priori ty in the Cahn-Ingold-Prelog system.

A cursory analysis of the designation of configuration would suggest to the uncritical mind that this reaction proceeded with retention of configuration-but that would be wrong ! You know by now that a careful analysis i s required. In the Cahn-Ingold-Prelog system, the F in the starting material was priority group 2, but in the product, because Br has left, F is now the first priority group. So even though the designation of configuration suggests retention of configuation, the molecule has actual ly undergone invers ion as would be expected with an SN2 reaction . (See the solution to problem 6-2 1 for a s imi lar example . )

(a) Only the propagation steps are shown . NBS provides a low concentratin o f Br2 which generates bromine radical in ultraviolet l ight . In the starting material , all 8 allyJic hydrogens are equ ivalent.

H H

Br · I Br - Br I '---I recycles

Br

+

Br H H

+ Br · + Br ·

(b) The first step in thi s first-order solvolysis is ionization , followed by rearrangement. H H 0?CrCH2 CH30H _ H -!CrCH2

---....... B r + + H II

mechanism continued on next page

132

hydride shift

allyl ic !

Page 140: Solucionario de wade

6-75 (b) continued H El H

--:H3�H +

H, � - CH2

� H� ..

H{yCH2 + CH30H2 +

SN I H3C H � �

.O - H � 1�·· CfCH2 I CH3�H. (j-CH2 CH3�H.

t H & �H2

H C (:YH 3 }� H � CH30H

• 1 + \

• ••

� CH2 CH OH 3 • •

X '3_ CH2 U+ CH \�H2

OCH3 1 � CH2

U + CH10H2 . +

(c) This first-order solvolys is generates an allylic carbocation in the ionization step.

Br H

EtOH ..

The El is shown on the next page.

• •

! Et�H Et , • • / H

• • ,.., Et H + ?) CD � H

Et�H �

H OEt l33

+ EtOH2 +

! Et�H OEt +

co These are the SN 1 products.

Et OH2 +

Page 141: Solucionario de wade

6-75 (c ) continued E l

H

EtOH

H

..

+ EtOH2 +

(d) The first step in this first-order solvolysis is ionization, fol lowed by rearrangement. Q<> _C:_30H __ � (0)

CH3�: Q(> CH3�H

' Q<> H Br � H : O- H � H OCH

j 'Y 1 Br - H3C SN 1 product from . unrearranged

rearrangement by alkyl shift carbocation

��Qj SN 1 on rearranged carbocation

� � � .. +��H

E l on rearranged carbocation

cp � CP +

CH3�H

:O� ) OCH3 H3C • • H �

G;1 CH3�H

� ((\ H H� �

+ CH30H2 +

H

134

CH30H2 +

Page 142: Solucionario de wade

CHAPTER 7-STRUCTURE AND SYNTHESIS OF ALKENES 7- 1 The number of elements of unsaturation in a hydrocarbon formula is given by:

2(#C) + 2 - (#H) 2

2(6) + 2 - (12) 2

= 1 element of un saturation

(b) Many examples are possible. Yours may not match these, but all must have either a double bond or a ring, that is , one element of unsaturation .

o 7-2 2(4) + 2 - (6)

2 = 2 elements of un saturation

H / C = C = CH2

H3C

1: 7 -3 Hundreds of examples of C4�NOCI are possible. Yours may not match these, but all must contain two elements of unsaturation .

CI �Oy NH2 � CI N = O

\d N - C = C - CI � U OH

CI�C N

OH

7-4 Many examples of these formulas are possible. Yours may not match these, but correct answers must have the same number of elements of unsaturation .

(a) C3H4CI2 => C3� = 1 �Cl CI �CI

Cl

(b) C4HgO => C4Hg = 1 OH

� o o �

13S

Page 143: Solucionario de wade

7 -4 continued (c) C4H402 => C4H4 = 3

° I I HC C - C - OCH3

0 , + ,.... 0 ' N (d) CsHsN02 => Cs sHs = 4

° [ r C C - NH,

° o N CUO

C - N (e) C6H3NClBr => C6 sHS = 5 Br N == C == CH2

)U A HC C - C == C - C CH I I Br NH

I Cl

Cl Br Cl

Note to the student: The IUP AC system of nomenclature is undergoing many changes, most notably in the placement of position numbers . The new system places the position number close to the functional group designation, which is what this Solutions Manual wi l l attempt to follow ; however, you should be able to use and recognize names in either the old or the new style. Ask your instructor which system to use.

7-5

(a) 4-methylpent- l -ene (b) 2-ethylhex- l -ene (c) penta- l ,4-diene (d) penta- l ,2 ,4-triene

(f) 4-vinylcyclohex- l -ene (" 1" is optional) (g) 3-phenylprop- l -ene (" 1 " i s optional) (h) trans-3 ,4-dimethy\cyclopent- l -ene (" 1 " is optional) ( i) 7-methy\enecyclohepta- l ,3 , 5-triene

(e) 2 ,5 -dimethy\cyclopenta- l ,3 -diene

7-6 (b), (e) , and (f) do not show cis, trans isomerism

(a) � (Z)-hex-3-ene (cis-hex -3 -ene)

(c)

\d\ (2Z,4Z)-hexa-2,4-diene cis, cis-hexa-2,4-diene

(d)

;=C (Z)-3-methylpent-2-ene

(E)-hCX�-enc \ (trans-hex-3-ene)

(2Z,4E)-hexa-2 ,4-diene cis, trans-hexa-2 ,4-diene

(E)-3-methylpent-2-ene

136

(2E,4E)-hexa-2 ,4-diene trans, trans-hexa-2,4-diene

"Cis" and "trans" are not clear for this example; "E' and "Z ' are unambiguous .

Page 144: Solucionario de wade

7-7 � (a) IL-

2,3 -dimethylpent-2-ene (neither cis nor trans)

CI (d) '0

5-chlorocyclohexa- l ,3 -diene (positions of double bonds need to be specified)

7-8

(a) 'x Br

(b) (c)

3 -ethylhexa- l ,4-diene I -methylcyclopentenc

(cis or trans not specified; the vinyl group is part of the main chain)

(e)

cis-3 ,4-dimethylcyclohexene (could also have drawn trans)

CI Br /=C

(f)

(E)-2 ,5-dibromo-3 -ethylpent-2 -ene (cis does not apply)

(E)-3-bromo-2-chloropent-2-ene (2)-3 -bromo-2-chloropen t -2-ene

(b)

(2E,4E)-3-ethylhexa-2,4-diene (2Z,4E)- (2E,42)- (2Z,42)-

(c) no geometric i somers

(d)

(2)-penta- l ,3 -diene (E)-penta- l ,3 -diene

(e)

(E)-4-t-butyl-5-methyloct-4-ene (2)-4-t -butyl-5 -meth yloct -4-ene

137

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7 -8 continued

(f)

CI CI CI CI

(2Z,5£)-3, 7 -dich loroocta-2 ,5 -diene (2E,5£)-3 ,7-dichloroocta-2 ,5-diene

CI CI

(2Z,5Z)-3,7 -dichloroocta-2 ,S -diene (2E,5Z)-3, 7 -dich loroocta-2 ,5-diene

(g) no geometric i somers (an E double bond would be too highly strained)

(h) W W (Z)-cyc Iodecene (£)-cyc lodecene

( i ) W W C) ( I E,S£)-cyc Iodeca- l ,S-diene ( 1 Z,S£)-cycIodeca- l ,5-diene ( I Z,5Z)-cyclodeca- l ,5 -diene

7-9 From Table 7- 1 , approximate heats of hydrogenation can be determined for similarly substituted a lkenes . The energy difference is approximately 6 kJ/mole ( 1 .4 kcal/mole), the more highly substituted alkene being more stable .

gem-disubstituted 1 1 7 kJ/mole (28.0 kcal/mole)

7- 1 0 Use the relative values in Figure 7-7 .

(a) 2 x (trans-disubstituted - cis-disubsti tuted) =

tetrasubstituted 1 1 1 kJ/mole (26.6 kcal/mole)

2 x (22 - 1 8) = 8 kJ/mole more stable for trans, trans (2 x (S . 2 - 4.2) = 2 kcal/mole)

(b) gem-disubstituted - monosubstituted = 20 - 1 1 = 9 kJ/mole (4 .8 - 2 .7 = 2. 1 kcallmole)

2-methylbut- l -ene i s more stable

138

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7- lO continued

(c) trisubstituted - gem-disubstituted = 25 - 20 = 5 kJ/mole (5 .9 - 4.8 = 1 . 1 kcallmole)

2-methylbut-2-ene i s more stable

(d) tetrasubsti tuted - gem-disubsti tuted = 26 - 20 = 6 kJ/mole

7- 1 1

(a)

2,3-dimethylbut-2-ene i s more stable o(CH3 strained but stable CH3

(6.2 - 4 .8 = 1 .4 kcallmole)

(b) could not exi st-ring s ize must be 8 atoms or greater to include trans double bond

(c )

(d)

Despite the ambiguity of this name, we must assume that the trans refers to the two methyls since thi s compound can exist, rather than the trans refeni ng to the alkene, a molecule which could not exist .

stable-trans in l O-membered ring

(e) unstable at room temperature�annot have trans alkene in 7-membered ring (possibly i solable at very low temperature-thi s type of experiment i s one of the chal lenges chemists attack with gusto) (f) stable-alkene not at bridgehead (g) unstable-violation of B redt's Rule (alkene at bridgehead in 6-membered ring) (h) stable-alkene at bridgehead in 8-membered ring ( i ) unstable-violation of Bredt's Rule (alkene at bridgehead in 7-membered ring)

7- 1 2 (a) The dibromo compound should boi l at a higher temperature because of its much larger molecular weight. (b) The cis should boi l at a h igher temperature than the trans as the trans has a zero dipole moment and therefore no dipole-dipole interactions . (c) 1 ,2-Dichlorocyclohexene should boi l a t a higher temperature because o f i ts much l arger molecular weight and larger dipole moment than cyclohexene.

7- 1 3

(a) O� 30 NaOH •

acetone

major E2

139

mmor E2

Page 147: Solucionario de wade

7 - 1 3 continued

(b)

minor E2 Cl

major E2 hindered base gives Hofmann product as major isomer

(c) 0 20 NaOCH3 ..

CH30H 6] + o major product i s difficult to predict for 2° halides

(d)

7 - 14

7- 1 5 (a)

=

SN2

NaOC(CH3h 0 ..

(CH3hCOH E2

-

Base: -,

H H "'� �Ph " C - C

H3C " / �B Ph r

E2 NaOC(CH3h i s a bulky base and a poor nuc leophi le, minimizing SN2

-H3C " " " " H " C === C "

Ph"'" ' Ph cis

E2 el imination

substitution product-since the substrate is a neopentyl halide and highly hindered, the SN2 substitution is slow, and elimination is favored

140

Page 148: Solucionario de wade

7- 1 5 continued (b)

(c)

(d)

(c)

H':r--f(jPhH

" Ph " "

Br Br

meso

B��� Ph B r one enantiomer of the d, l pair

Cl

H

Ph Ph (CH3CH2)3N : " /

C = C

NaOH ..

acetone

/ " Br H

only alkene isomer E2

B r Ph " /

+ minor substi tution products

C = C + minor substitution products

Ph/ "

H only alkene i somer

E2

HO H Cb SN2 i s the only mechanism possible because no H can get coplanar wi th Cl ; an el i mination product would violate B redfs Rule

H

See Appendix 1 i n thi s manual for numbering and naming bicyc l ic systems.

Models show that the H on C-3 cannot be anti-coplanar with the Ci on C-2. Thus, this E2 el imination must occur with a syn-coplanar orientation: the D must be removed as the Cl leaves .

7- 1 6 As shown in Solved Problem 7-3 , the H and the Br must have trans-diaxial orientation for the E2 reaction to occur. In part (a) , the c i s i somer has the methyl in equatorial posit ion, the NaOCH3 can remove a hydrogen from either C-2 or C-6, giving a mixture of alkenes where the most highly substituted i somer is the major product (Zaitsev) . In part (b), the trans i somer has the methyl in the axial posi tion at C-2 , so no el imination can occur to C-2. The only possible el imination orientation i s toward C-6.

(a)

B r �l I 6

� H eIther H .. H can be removed

(b) B r cj(. 1 . 6 H3C

only this H - H can be removed

c5 c5 NaOCH3 .. + Zaitsev orientation

C H30H

major minor

c5 NaOCH3 .. only alkene formed; stereochemistry of E2 precludes

CH30H other i somer from forming

14 1

Page 149: Solucionario de wade

7- 1 7 E2 el imination requires that the H and the leaving group be anti-coplanar; in a chair cyclohexane, this requires that the two groups be trans diaxial . However, when the bromine atom is in an axial position , there are no hydrogens in axial positions on adjacent carbons, so no el imination can occur.

7- 1 8

c� hydrogens lrans diaxial lo lhe Br

CH3

H

Br-- axial

(X)'" => � � ' Br � T -1 H H Br axial

(D B r H

C:::::> = Br � T �atOrial H

(a)

KOH �

H +

H

(b) Showing the chair form of the decalins makes the answer clear. The top i somer locks the H and the Or into a trans-diaxial conformation--optimum for E2 el imination. The bottom i somer has B r equatorial where it is exceedingly s low to el iminate .

7 - 1 9 Models are a b i g help for this problem.

(a) B r c{(; H

(b)

• CH3

the only H trans diaxial

H

NaOCH3 ..

both trans diaxial­gives two products

H

the only el imination product

D (from elimination of H and Br)

+ H

H

H = W- h : H H

H (from el imination of D and Br)

142

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7-20 o 0 (:.10 :

Ph H - Br �- �/ C - C , ---

/ "':'" / , � "

Ph B0 H

o 0 : I :

transition state

Ph " ," " " Ph --- " C-C " H """ - ' H

cis

only geometric i somer

both Br anti-coplanar

�----------��----------� (- - - - - - - = stretched bond being broken or forme:0 possible from a concerted mechanism

Br

H Ph

Ph

Br

7-2 1

(5, 5) ( :.10 : H

H C - B r 3 � �/ C-C / X" / , �" H

B0 CH3

Br - - - - - - - r

. B r

transition state

o oro 0

o •

H Br H3C� "

C ::..::..::..: C , "

, ," H 8 - , CH B r 3

transition state

.....-pi bond

H Ph --� .. � H -+----;;---1-- Ph

cis

H " , " , H ___ II , C-C '\\ H3C """ - 'CH3

cis

only geometric isomer

both Br anti -coplanar

------------��----------r. - - - - - - = stretched bond being broken or forme:0 possible from a concerted mechanism

Br

H3C H ..

H3C

Br

Br - - - - - - - I , H3C:Qx H

H3C , H

Br

transition state

143

:j: .....-

pi bond

H3C .. H3C

cis

Page 151: Solucionario de wade

7-22 (a)

(b)

(d)

(e)

B r � H B r

NaI

acetone

Br

NaI

acetone o cyclohcxene

Br r"f-CH3 yH CH2CH3

cis-hept-3-ene

The two bromine atoms must be anti ­coplanar in order to e l iminate . Only the bromines at C- l and C-2 fi t that requirement; the bromines at C-2 and C-3 cannot e l iminate .

(5)-3-bromo- l -ethyl-3-meth y Icyclohexene

rotation <D �· �H Br

acetone H NaI, 9 . �

H

In the first conformation shown , the bromine atoms are not coplanar and cannot el i minate. Rotation in this l arge ring can place the two bromines anti-coplanar, generating a trans alkene in the ring. (Use models ! )

b trans-cyclodecene

7-23 The stereochemical requirement of E2 el imination is anti-coplanar; in cyc lohexanes , thi s translates to trans-diaxial . Both dibromides are trans, but because the t-butyl group must be in an equatorial position , only the left molecule can have the bromines diaxial . The one on the right has both bromines locked into equatori al positions, from which they cannot undergo E2 el imination.

(CH3hC#� H B r --- trans-diaxial-can do E2

H (CH3hC� Br r HI

144 trans-diequatorial--cannot do E2

Page 152: Solucionario de wade

��U24 (Br C

_

H_3_ ..

..

03�

:�-CH2CH: o

CH

�?+-CH2CH3 . cj-

CH3

?: II H CH2CH3 � \ .

H2 H

r'��C" �.I � H ·R-CH2CH:

H H

�:�-CH2CH3 ..

• O-CH2CH3 I H

(j

CH2

()

CH]

(b) rYC:3 ��-H Br

Gf

CH37 +C�:�-CH2CH3

� , ...

H (t,H3 .. . . + O-CH2CH3

H (I d�: I 2° J .�-CH2CH3 �

CH3

H�

.7 + � .. c, � H H

H H I �:�-CH2CH3

..

H

� :O-CH2CH3 I H ()CH]

a

CH]

CH2CH}

carbocation rearrangement CH3 r:f;H

Vh<H 3°

_O/CH] 2° H I :O-CH2CH3

• • .. same product mixture as in PaIt (a)

145

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7 -25 In these mechanisms, the base removing the final proton is shown as HS04 -. It i s equal ly possible that water removes this proton.

H (a) � 1 +

(J0H

_ f_� __

__ (f0 .. -H ___

• • H -OS03H /). •

H20

G+ ...- H/-OS03H � f • H H

H /). 1 +

___ �CI) �H H,\ -OS03o/ ! }

/ OS03H

� cis + trans

major isomer minor isomer

H (c) • • � 1 + (XoH

_ t_� __

__ CfR-H ___ A • • H-OS03H Ll

CH3 CH3

H rearrange- ct H hydride shift

from unrearranged 2° carbocation H from rearranged 3° carbocation

greatest amount least amount

146

Page 154: Solucionario de wade

7-26

(a) f..Go = tJJ{0 - Tf..So

= + 116,000 Ilmol - 298 K ( Il7 1IK·mol)

= + 81,100 Ilmol =

+ 8l. l kl/mol (+19.3 kcaVmole)

f..Go is posi tive, the reaction is disfavored at 25°C

(b) f..GIOOO = + 116,000 Ilmol - 1273 K (117 lIK·mol)

= - 3 2,900 Ilmol = - 32 .9 klima I (- 8.0 kcal/mole)

f..G is negative , the reaction is favored at 1000°C

7-27 (a) basic and nucleophilic mechanism: B a(OH}z is a strong base (b) ac idic and electrophilic mechanism: the catalyst is H+ (c) free radical chain reaction: the catalyst is a peroxide that ini tiates free radical reactions (d) acidic and e lectrophi l ic mechanism: the catalyst BF3 i s a strong Lewis acid

7-28

(a)

H H H � I I I • • � H-OS03H

CH3 - C - C - C-OH ..

I I I • •

H H H

H H

H H H H I I I + I

CH3-C-C - C-OH I I I \.::;

H H H

+

H H H I I

CH3-C==C - CH3

H H H hydride I I I shift .. CH -C - C-C-H ...... I----

Hi) :\. 3

I...) + I I I 1+

CH3-C-C-C I I-.-A I E + Z

(b)

'----..- H H

H

H H H This 1 ° carbocation may or may not exist . It i s shown for clari ty.

(j-" "OCH3 + NaBr

147

Page 155: Solucionario de wade

7-29 (a)

(b)

H () .. �" 9.H H - OP03H2

------; .. �

H H

rl! 1 (J voH

H

GH (X H c ...... � " .. I H • •

H20: H H

+ H20

CH2-�H ("'1+ +

c5-�� �20· c5 6 \.. H00S03H

.

Two possible rearrangements

1. Hydride shift H + 1 (CH H-C-H�

H C f7H� 02 I) - CJ H H,o:.

+ • •

2. Alkyl shift-ring expansion + CH2 H

H- �J 1 H�

+ ...... C • • O C rl "\

- O'H H,o�

This 10 carbocation mayor may not exist . It is shown for clarity

o+c5 H

H

(c ) carbocation formation

HO: � � H0oS03� /'-..

� H H , T H ' �I�'-continued on next page

148

+ H20

Page 156: Solucionario de wade

7-29 (c) continued without rearrangement

H r:-. H I ��?: H2O: ..

H

H� �C1<

H H2O: + .. H

H

with hydride shift

H �

H H

H E+ Z

H� �H�

H

H H

~ H +( ..

+

1<H - ./"'--r� H H20: H H H --.-.-------

rearrangement by alkyl shift

El on rearranged carbocation

H �: H

H

149

H

H

El product from unrearranged carbocation

Page 157: Solucionario de wade

7-29 (d) continued E 1 on rearranged carbocation

�� <:;LJ H20:. 00 + H,o' +

7-30 Please refer to solution 1 -20, page 1 2 of this Solutions Manual.

7 -3 1

(a) � � Br (c) (d) 0 o (e)

Br

(f)

7-32

(g) Br�

(h) r; (a) 2-ethylpent-l-ene (number the longest chain containing the double bond) (b) 3-ethylpent-2-ene (c ) (3E,6E)-octa-l,3,6-triene (d) (E)-4-ethylhept-3-ene (e) l-cyc lohexy lcyclohexa-l,3-diene (f) (3Z, 5E)-6-ch loro-3-(chloromethyl)octa-l ,3,5 -triene

(") Z E 1 �

7-33 (a) E (b) nei ther-two methyl groups on one carbon (c) Z (d) Z

7-34

(a) � F

(Z)-l-fluoro­prop-l-ene

� F (E)-l-fluoro-prop-l-ene

F

A 2-fluoro-prop- l -ene

0 [>-F F 3-fluoro- fluorocyclopropane prop-l-ene

(b) C4H 7Br has one element of unsaturation, but no rings are permitted in the problem, so all the isomers must have one double bond. Only four i somers are possible with four carbons and one double bond, so

B r Br H Y ��� H Br

Br � /'o.... � �" y Br

B r - � '- �

Br � Br

�B r

Br � � Br

(b) Cholesterol , C27H460, has five elements of unsaturation . If only one of those is a pi bond, the other four must be rings"

150

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7-35

~ ~ ;=\' � (a) trans, trans cis,trans trans,cis cis,cis

(b) 2£,4£ 2Z,4£ 2£,4Z 2Z,4Z

This problem is i ntended to show the difficulty of using cis-trans nomenclature with any but the simplest alkenes. Cis and trans are ambiguous: the first alkene in part (a) is cis if the two similar substituents are considered, but trans if the chain is considered. The £-Z nomenclature i s unambiguous and i s preferred for a l l four of these i somers.

7-36 (a) and (d) have no geometric i somers

(b) (c) � trans-pent-2-ene

(E)-pent-2-ene cis-pent-2-ene (Z)-pent -2-ene

trans-hex -3-ene (E)-hex -3-ene

cis-hcx-3-ene (Z)-hex-3-ene

B r

(e) y Br

(f)

trans- 1 ,2-dibromopropene (E)- 1 ,2-dibromopropene

� Br cis- 1 ,2-dibromopropene 1- (Z)- 1 ,2-dibromopropene Br

trans,trans-hexa-2 ,4-diene (2£,4E)-hexa-2,4-diene

cis,trans-hexa-2,4-diene (2Z,4E)-hexa-2,4-diene

cis,cis-hexa-2,4-diene (2Z,4Z)-hexa-2 ,4-diene

7-37

(a)

(c)

7-38

(a)

F 1 F

>===< H H

Cl 1 Cl A Br Br

o

H F

>===< F H

dipole moment

Cl 1 Cl >===<

H H

= 0

larger dipole moment (no bromines opposing the dipole of the chlorines)

Br 1 B r (b) >===<

H H

(b) +� (c)

minor

Br CH3

>===< CH3 Br

dipole moment = 0

major major minor 15 1

Page 159: Solucionario de wade

7-38 continued

(d)

mmor major

7-39 Only major alkene i somers are shown. Minor alkene i somers would also be produced in parts (a), (b) and (d).

H H I I

H H I I

(a) CH -C-C-CH 3 I I

3 CH3-C=C-CH3 + H20 H OH E + Z

H (b) NaOC(CH3h

hindered base

- en + NaBr + HOC(CH3h

(c)

Br

H H I I CH -C-C-CH + Zn 3 I I 3 Br Br

CH3 CH3

H H CH3COOH I I ---...., .. � CH3-C=C-CH3

E + Z I I NaOH, tl (d)

7-40

(a)

CH -C-C-CH 3 I I 3 H Br

N aI, acetone ..

�Br

'"

Br 0 OR Zn, CH3COOH

(b) Q'0H H2S04, tl 0 ..

(c) Q' Br KOC(CH3h 0 ..

(d) 0 Br2 ' hv Q'Br KOC(CH3h 0 .. ..

7-41

(a) � only product

(b) � + 1\ + \ J �---�,,�---�

major

152

+ ZnBr2

minor

Page 160: Solucionario de wade

7-4 1 continued

(c) � + � (d) � )l U + U

(e) � � U + U major rrunor major rrunor

The E2 mechanism requires anti-coplanar orientation of R and Br.

7-42 The bromides are shown here . Chlorides or iodides would also work.

(a)

� Br

B r (b) Y (c) �

Br

(d) cSf Br

7-43

(a) There are two reasons why alcohols do not dehydrate with strong base. The potential leaving group, hydroxide, i s i tself a strong base and therefore a terrible leaving group . Second, the strong base deprotonates the -OR faster than any other reaction can occur, consuming the base and making the leaving group anionic and therefore even worse.

(b) A halide i s already a decent leaving group. S ince halides are extremely weak bases , the halogen atom is not easi ly protonated, and even if it were , the leaving group abi l i ty is not significantly enhanced. The hard step is to remove the adj acent R, something only a strong base can do-and strong bases will not be present under strong acid condi tions .

7-44

(a) � (b)

without rearrangement

(c)

153

(d) )==I rearrangement

�H maJor­Zaitsev

Page 161: Solucionario de wade

7 -45 continued with rearrangement

7-461 /' (a) /' .......... ,

major

hydride shift

+"y minor

(c) � +� major mInor

(b) +

major

(d)

major

ax

Cl

(CH3hC -.£:::::iCI eq

minor

minor

major­Zaitsev

only product from E2-anti -coplanar is possible only from carbon wi thout CH3 by remN��

3

) (e) y

D

The conformation must be considered because the leaving group must be in an axi al posi t ion. The t-butyl group is so large that it must be equatorial , locking the conformation into the chair shown. As a result , only the Cl at position 4 is axia l , so that is the one that must leave, not the one at position 3. Two isomers are possible but it is difficult to say which would be formed in greater amount.

+

7-47 H H H � . . � V �H ,

H' etc I ()H 0H

� +C ...

- H20 '\ H ' •

'- _ _ _ / " H20.

� H �

OSO H El works well because only one carbocation and

ct 3 only one alkene are possible. Substitution i s not a

H problem here. The only nucleophiles are water,

HSO - --� .. - which would s imply form starting material by a 4 .. reverse of the dehydration , and bisulfate anion. Bisulfate anion is an extremely weak base and poor nucleophi le; if it did attack the carbocation , the unstable product would quickly re-ionize, with no

154 net change, bac k to the carbocation.

Page 162: Solucionario de wade

7-48 CH3 CH3 I I

CH -C-C-CH ----3 I I

3 H +

HO : 9o� CH3 I

CH -C-C-CH 3 II I 3

: 0 : CH3

CH3 CH3 I I

CH -C-C-CH 3 I 1 +

3

HO "'9oH2

CH3 CH3 -H20 I�I --I"� CH -C-C-CH 3

I + 3

HO l methyl shift

CH3 + I

CH -C-C-CH 3 I I

3

: OH CH3

The driving force for thi s rearrangement i s the great stabi l ity of the resonance-stabi l ized, protonated carbonyl group.

7 -49 NBS generates bromine w hich produces bromine radical. Bromine radical abstracts an allylic hydrogen, resulting in a resonance-stabi l ized aUylic radical. The al lyl ic radical can bond to bromine at either of the two carbons with radical character. See the solution to problem 6-63.

NBS c::=> B r2 hv

2 Br 0 CCH2 1aCH2

+ Br 0 ____ \.-I �� <;:, H

H

reCYCI� 7-50

B ro aCH2

+ + H

Br

� \ NaOEt CH3 Ph

Ph HX� CHCH3

", H -----l.� '-r----/ + CH {" �

Ph B r Ph CH2CH3

25,3R B H -CHCH 3

HX! Ph

" , CH {"

Ph 25,35 Br

HBr

Ph

� + JYJ Ph Ph

A = E,Z mixture

Ph

� + Ph Ph

A = E,Z mi xture

E2 dehydrohalogenation requires anti-coplanar arrangement of H and Br, so specific c i s-trans I somers (8 or C) are generated depending on the stereochemistry of the starting material . Removing a hydrogen from C-4 (achiral) will give about the same mixture of cis and trans (A) from either diastereomer.

155

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7-51 � more stable than

~ more stable than

by 4 kJ/mote

by 16 kJ/mole

Steric crowding by the (-butyl group is responsible for the energy difference. In cis-but-2-ene, the two methyl groups have only s l ight in teraction. However, in the 4,4-dimethylpent-2-enes , the larger s ize of the t-butyl group crowds the methyl group in the cis i somer, increasing its strain and therefore i ts energy.

7-52 () C1 V ()' endocyclic exocyclic endocyclic exocyclic

\. tri substituted V

disubsti tuted ) \ tri substituted

V tri substituted)

9 kJ/mole 5 kJ/mole

A standard principle of science is to compare experiments which differ by only one variable. Changing more than one variable c louds the interpretation, possibly to the point of inval idating the experiment.

The first set of structures compares endo and exocyclic double bonds, but the degree of substi tution on the alkene i s also di fferent, so this comparison is not valid-we are not isolating simply the exo or endocyclic effect.

The second pai r is a much better measure of endo versus exocycl ic stabi l i ty because both alkenes are tri substituted, so the degree of substi tution plays no part in the energy values . Thus , 5 kJ/mole is a better value.

7-53 (a) CH3CH 2CH =CH2

(b) No reaction-the two bromines are not trans and therefore cannot be trans-diaxial .

(e) 0 (d) (X) H H

bromines are diequatorial­cannot undergo E2

No reaction-the bromines are trans , but they are diequatorial because of the locked conformation of the trans-decalin system. E2 can occur only when the bromines are trans diaxial .

156

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7-54 In E2, the two groups to be eliminated must be coplanar. In confonnationally mobile systems like acyclic molecules, or in cyclohexanes, anti-coplanar is the preferred orientation where the H and leaving group are 180° apart. In rigid systems like norbornanes, however, SYN-coplanar (angle 0°) is the only possible orientation and E2 will occur, although at a slower rate than anti-coplanar.

The structure having the H and the Cl syn-coplanar is the trans, which undergoes the E2 elimination. ( It is possible that the other H and Cl eliminate from the trans isomer; the results from this reaction cannot distinguish between these two possibilities.)

H

Cl extremely slow to eliminate­H and C\ not coplanar

trans

Cl

.

H

Cl '/syn­y �oplanar

Cl

H

7-55 It is interesting to note that even though three-membered rings are more strained than four membered rings, three-membered rings are far more common in nature than four-membered rings. Rearrangement from a four-membered ring to something else, especially a larger ring, will happen quickly.

without rearrangement

H� � + H20: y--

CH2 ..

with rearrangement-hydride shift H

(P�H2

3° carbocation, but still in a strained 4-membered ring

mmor

H20: •

157

unstable 1 ° carbocation-­short lifetime if it exists at all

minor

minor

Page 165: Solucionario de wade

7-55 continued with rearrangement-alkyl shift-ring expansion

H

��H2 C "......-A

-�. ("�-H�.

H C-C-H . .

O[ H MAJOR

H2

abstraction of adjacent H gives bridgehead alkene-violates Bredfs Rule

2 I H20: H H

2° carbocation on 5-membered ring­HOORAY!

H

('xCH/'RiI

.� � t - H20

1 ° carbocation-terrible!-will rearrange; can't do hydride shift, must do alkyl shift = ring expansion

t hydride shift

this is an unstable carbocation even though it is 3°; bridgehead carbons cannot be sp2 (planar-try to make a model), so this carbocation does a hydride shift to a 2°, more stable carbocation

this 2° carbocation loses an adjacent H to form an alkene; can't form at bridgehead (Bredt's Rule)­only one other choice

158

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CHAPTER 8-REACTIONS OF ALKENES 8-1 Major products are produced in greatest amount; they are not necessarily the only products produced.

(a) Br (b) Cl (c) OH (d) <rB'

+

Br �H +H ¢Til produced in about equal amounts (mixture of cis and trans)

8-2 H H H H allylic H H

H-t�H · .Q H+ H H --- H-C+ H

\ H H H H H H H

1 .. : Br:

H H

H-O-H

Br H H H

+

3-bromobut -l-ene 1-bromobut-2-ene Because the allylic carbocation has partial positive charge at two carbons, the bromide nuc\eophile can bond at either electrophilic carbon, giving two products.

8-3 (a) initiation steps

o ° ° I I II � I I L1

H3C-C-0-v' O-C-CH3 - 2 H C-C-O· 3 • •

° II

H C-C-O· + H-Br 3 "�JG -

° I I

H C-C-OH 3 • • +

(this radical has another resonance fonn but it is not gennane to this mechanism)

• Br

propagation steps

~ �

+ • Br C� - the 3° radical is more stable than the 1 ° radical

_)��� - kB' + • Br

1-bromo-2-methylpropane 159

Page 167: Solucionario de wade

8-3 continued

(b) initiation steps

H2 • • H C-C -0· + H-Br 3 • • ,,--JG

propagation steps

+ • Br

the 3° radical is more stable than the 2° radical

+ • Br

I-bromo-2-methylcyclopentane (cis + trans)

(c) initiation steps - 2

propagation steps -

8-4 HBr Br� (a) � • ROOR

(c) (yOH H2SO4 0 HBr •

OH � � H2SO4 HBr (d) • •

� ROOR 160

+ • Br

H I

/Cy Br Ph' ·

I the benzylic radical is stabilized by resonance, and more stable than the aliphatic radical

..

H � Br Ph H I + • Br

2-bromo-l-phenylpropane (recall that "Ph" is the abbreviation for "phenyl")

Br

(b) � HBr .. �

C)' Note: A good synthesis uses major products as intermediates, not minor products. Knowing

~ orientation of addition and elimination is critical to using reactions correctly.

Br

Page 168: Solucionario de wade

8-5 �CH3 H

• CH3 � H H-O-H CH 3 '-------"" "'"

I +

CH3 H

CH3�*H

CH31+ H

H 2°

methyl shift ..

CH3 + I CH3

CH3/C�CH3

3° I H H

nucleophilic attack by water

CH3 + I CH3

/C'tCH3 CH3 \[)oo ..

:O-H H I

H

proton removal CH3 I CH3

CH3 .. CHJ CH3

tfCH3CH3

°O-H CH3 + 0 0 O')H / I

1 2-butanol H

/ k� H 2,3-dimethy-

jfCH3CH3

HO H

.. 13

,CH3

H

+ H30+

;C\fCH3 CH3 4 ....

'-.. :O-H CH3/ "1' + H30+

H � I H

8-6

CH3 2,3-dimethyl-2-butene

(a) '6H (b) �H

U OH'

H

0d:J (c)

from 3° carbocation

8-7

(a)

CH>=<.H + Hg(OAc)

CH3 CH3

(b)

CH3

CH3trCH3

I H CH2CH3

NaBH4 ..

from 3° benzylic carbocation

..

CH3

CH3 AfgOAC CH3

CH3 �

CH2CH3

(H \i : H

ACO :/ I ° 0 0

II H3C-C 1fgOAC

CH3 CH3

o

"Ac" = acetyl

o II H3C-C'-O

"OAc" or "AcO" = acetate

161

I H CH2CH3

Page 169: Solucionario de wade

8-8 CH HO " 3

(f'::(OAC) HX (b)

U (c) C IOOCH3 :lo."

Hg(OAC)

III Hg(OAc) OCH3

(a)

(d)

8-9

(a) �

(b) &, (c)

8-10

(a) , (b) �

CI O-OCH J

Hg(OAc}z NaBH4 .- .-

CH30H

OCH3 � KOH,� 6 Hg(OAc}z NaBH4 OH

.-H2O

Hg(OAc}z

H BH3 0THF � BH2 ..

..

OH � H202

HO-.-

.-

Using an acid-catalyzed hydration In part (c) would initially form a 2° carbocation that would quickly rearrange to 3" on carbon-3. The desired product would not be synthesized.

� OH

(c), (d) � B H3 0T HF � H202 � .. .-

(e) , (f) 6 8-11

(a) �

(b) � (c) �

Br

HO-BH2

BH3 0TH F CHaB:, H202 ..

HO-

B H3 ° THF H202 -----;.-� .-

HO-

Hg(OAc}z NaBH4 .. ..

H2O

HO�

OH

.-

KOH,� � BH3 ° THF H202 .. .-HO-

162

OH

CHaO:

.-~ OH

Page 170: Solucionario de wade

8-12 Instead of borane attacking the bottom face of I-methylcycJopentene, it i s equally l ikely to attack the top face, leading to the enantiomer. s=<:. BH3 · THF

..

8-13

(a) CH3 6,'QH

8-14

(a)

(b)

Et Et

>=< Me H

Z

Et H

>=< Me Et

E

(b) V major

B H3 · THF

B H3 • THF

H

+ � OH

minor-steric hindrance to attack of BH 3

H

S S

Et � � Et K H 'M'

H"OH

e )

enantiomers

R S

Et � � H K Me'� H�Et

�---------.. V enantiomers

OH

-

CH3

The enantiomeric pair produced from the Z-a lkene i s diastereomeric with the other enantiomeric pair produced from the E-alkene. Hydroboration-oxidation is stereospeci fic , that i s, eac h a lkene gives J specific set of stereoi somers , not a random mixture.

8-15

(a) CO Hg(OAch NaBH 4

CO ... ... H2O

OH

(b) CO B H3 · THF H202

cp .. .. HO -

OH

163

Page 171: Solucionario de wade

8-15 continued

(c)

C

o

8-16

c5 empty p orbital in planar carbocation

CH3 CH3

\bottom .� � "'-- :Br: �

CH3 Br The planar carbocation is responsible for non-stereoselectivity. The bromide nuc\eophile can attack from the top or bottom, leading to a mixture of stereoisomers. The addition is therefore a mixture of syn and anti addition.

8-17 During bromine addition to either the cis- or trans-alkene, two new chiral centers are being formed. Neither alkene (nor bromine) is optically active, so the product cannot be optically active.

The cis-but-2-ene gives two chiral products, a racemic mixture. However, trans-but-2-ene, because of its symmetry, gives only one meso product which can never be chiral. The "optical inactivity" is built into this symmetric molecule.

This can be seen by following what happens to the configuration of the chiral centers from the intermediates to products, below. (The key lies in the symmetry of the intermediate and inversion of configuration when bromide attacks.)

IDENTICAL-SYMMETRIC

�------���--------� ( + "\

M�B

Gr

" +

H",:\ I

�··H

H"'�H3 H3�'H Br +

a Be nuc\eophile could attack at either carbon , : �.r: OR: Br: J

H-+ Br

Br+H

l H H CH3 - Br Br':r-(f CH3

'---<' + ", / �'H H'" Br CH3 CH3 Br

Br-+ H

H+ Br

CH3 2S,3S ENANTIOMERS 2R,3R

CH3

--------------------------------------------------______________________________________ u __________ u __ •

continued on next page

164

Page 172: Solucionario de wade

8-17 continued ENANTIOMERIC

TRANS �------���---------( B! CH \ U H

"",U�"CH3

H'/1V\� ( """'CH + ' \ / bromide attack on this CH3 11 3 B b r romonium ion gives the

+ same results a Be nucleophile could attack at either carbon :Br: OR :Br:

, •• J

Br+H

Br+H

t H H C CH3�Br

+ Br)1-J,H

/ �"'CH H\'\� Br H

3 CH3 Br

H+Br

H+Br

CH3 2R,3S IDENTICAL-MESO 2R,3S CH3

CONCLUSION: anti addition of a symmetric reagent to a symmetric cis-alkene gives racemic product, while anti addition to a trans-alkene gives meso product. (We will see shortly that syn addition to a cis­alkene gives meso product, and syn addition to a trans-alkene gives racemic product. Stay tuned.)

8-18 Enantiomers of chiral products are also produced but not shown.

(a)

(b)

(c)

Br .. -

Br-Br 0+ 0Br � + :Br: ___

�. "'Br

(Hex) + Br\ Br-Br --- ,U) �

"1 It"� :Br: H"

Et �'Hex ..

+

Bromide will attack the other carbon of the bromonium ion as well.

165

Br �e1-I

'>---f/ H"'/ '" Et Br

Page 173: Solucionario de wade

8-18 continued (d) Three new asymmetric carbons are produced in this reaction. All stereoisomers will be produced with the restriction that the two adjacent chlorines on the ring must be trans.

Cl - Cl + --G--I< + Cl-Cl ..

:Cl: j :Cl:

Cl CH� h rCI

��l Cl CH3

8-19 The trans product results from water attacking the bromonium ion from the face opposite the bromine. Equal amounts of the two enantiomers result from the equal probability that water will attack either C-l or C-2. Q B, - �r <:v

H H Q .. �

8-20

H20: water will do nucleophilic attack at either carbon

from Solved Problem 8-5 H

+

+

equal amounts of enantiomers

H enantiomer of product \ � in Solved Problem 8-5

the bromonium ion shown here is the enantiomer of the one shown in Solved Problem 8-5

a�OO-H "",CH3 0

0 0

� ..

°O+_ H \ H :OH �� :O-H � U>

CH3 ... U >

CH3 Br+

\ H \ Br \ Br H H from Solved Problem 8-6: the bromonium ion shown is meso as it has a plane of symmetry; attack by the nucleophile at the other carbon from what is shown in the text will create the enantiomer H - H

,Br .. a', + :Cl: a"'Br

//� �H H�;H

show products from I;I � both nucleophiles 0 I a"' Br attacking at this carbon oR - H

'"' H OH

these are the enantiomers of the structures shown in the text

166

Page 174: Solucionario de wade

8-21 The chiral products shown here will be racemic mixtures.

(e)

plus enantiomer

CH3 ,OH (y Rr +

(b) 1 _ -;;61 CH3 ,Cl (y Rr

Br

plus enantiomers

8-22 Cl

(a) � CI2 ~ .. H2O

CHOH 6 (c) H2SO4 .. !t,.

8-23

H (C)H3CHC1

HO C�

(b)

CI2 .. H2O

plus enantiomer

CI 6 KOH

!t,.

CH3 ,OH (yCl

..

(d)

0

l-! ' H3C,----<Cl

/ \"", HO H"'CH-�

plus enantiomer

CI

CI2 6 .... 0H .. H2O

(a) � (b) � (c) C):) (d) -0-< 8-24 Limonene, CIOH16, has three elements of unsaturation. Upon catalytic hydrogenation, the product, CIOH20, has one element of unsaturation. Two elements of un saturation have been removed by hydrogenation-these must have been pi bonds, either two double bonds or one triple bond. The one remaining unsaturation must be a ring. Thus, limonene must have one ring and either two double bonds or one triple bond. (The structure of limonene is shown in the text in Problem 8-23(d), and the hydrogenation product is shown above in the solution to 8-23(d).) 8-25 The BINAP ligand is an example of a conformationally hindered biphenyl as described in text section 5 -9A and Figure 5-17 . The groups are too large to permit rotation around the single bond connecting the rings, so the molecules are locked into one chiral twist or its mirror image.

"front" naphthalene nng

"rear"

"front" naphthalene ring

These simplified three-dimensional drawings of the enantiomers show that the two naphthalene rings arc twisted almost perpendicular to each other, and the large -P(Phh substituents prevent interconversion of these mirror images.

167

Page 175: Solucionario de wade

8-26 Methylene inserts into the circled bonds.

S (a) + c5

(b) �H + � :CH2

(c) ~ + d 8-27

(a) (b) (c)

8-28

(a) F CH212

R • Zn(Cu)

(b) 0 CH2Br2 �H • 50% NaOH (aq)

Br OH

(c) 6 H2SO4 0 CHCl3 • • L! 50% NaOH (aq)

8-29

BOTTOM VIEW

__ site of origi nal double bond

the original 6-membered ring is shown in bold bonds

SIDE VIEW

Br

Cl cjCl

(a) (b) (c) Co (d)CQ H

"

0 0 H ,

168 o H

Page 176: Solucionario de wade

8-30

(a) R 'c=o

+ :O :�

fa l....H V '0''\ H "'1 ( t \", H

• A H +

H "", " '"

,---

Et EtH " 'u):O(-�

" f\]\" H'�t E;H "C-C"

Et" - 'Et cis

ENANT IOMERS A r-------- ....... ----�\

EtJ-<�"

Ofl +

fl0j----(P Et 'I, ,\\ II/ H H \"

HO Et Et OH

R

-H + •

H-O: OR :O-H I I I H t H

Etf!

+ ,---<Ofl +

H O�" H 2;. Et

,!-I Et fl0'>----Y •

H"7 "OH2 Et • •

+ :O-H (b) , c=o fa l.... H v'o' H , ('\ "Et 'I, t "

:O :� A H + "

" "

,, ---

H' "

Et �" Et "

, U)( " " 1\(\,

H'�t �" Et "c-c" Et" - 'H

trans

IDENTICAL-MESO A r�-------- ------�\

Et�Ofl + floyfl

�"Et H"'/ � HO H Et OH

stereochemistry shown in Newman projections:

H Et v Et

x: H

trans

---H�

Et

-H + •

---H £� :

. . . . H-O : OR :O-H

I I H + H

Etf!

.,--<Ofl

+

H20/- "" Et

ElH flO�+ H"'/ "OH2 • • H

OH EtMH

H� Et OH

Et • •

rotate ..

OH Q)H H- Et

H Et

MESO

Remember the lesson from Problem 8-17: anti addition of a symmetric reagent to a symmetric cis-alkene gives racemic product, while anti addition to a trans-alkene gives meso product. This fits the definition of a stereospecific reaction, where different stereoisomers of the starting material (cis and trans) are converted into different stereoisomers of product (a dl-pair and meso form).

169

Page 177: Solucionario de wade

8-31 R

G'C=O 0, t...H

H 0' """ c t ) "" , /

----/' == C � �

I trans H

8-32 All chiral products are racemic mixtures.

R , C-O

II \ o H

o

R= < � \-C-O II

Mg2+ o

CH2CH3

(a) � H (b) CH3CH2 � OH '-/ �

HO/ ��CH2CH3

HotH anti addition to a trans-alkene gives

HO H meso product o

CH2CH3

o (c) cj''''CH1 (d) � rotate

. t�is is the (e) �H trans O

OH

� HO", CIS product �OH H " 'OH H � H

H OH 8-33 � H+ � - H+ � � --- � -- --- 1-Y +� H H H \. or) H CH 0 H • • 3 H OCH3

CH3-O: I enantiomers

H 8-34 All these reactions begin with achiral reagents; therefore, all the chiral products are racemic.

(a) ():OH OH

(b) o:�: H OH (c) CH3� ==:> / . � "'CH2CH3 HO H

HO+�Hl

HO+ H

CH3 H

(d) CH3HOH

'" H HO CH2CH3

H (e) CH1HOH

'" H HO CH2CH3

H (f) CH3 � OH

HOX�CH2CHl

H=f�� HO H

CH3

same as (d)!

H=f�� HO H

CH3 Refer to the observation in the solution to Problem 8-35 on the next page.

170

same as (c)!

H0=f�Hl

HO H

CH3

Page 178: Solucionario de wade

8-35

(a) � meso

HO OH

(b)� � racemic (d,l)

HO OH

(c) � CH3C03H rotate � .. .. meso H2 O

HO OH

(d) � Os04 rotate � .. .. racemic (d,l) H202 HO OH

Have you noticed yet? For symmetric alkenes and symmetric reagents (addition of two identical X groups):

cis-alkene + syn addition ----7 meso Assume that cis/synlmeso + 1 x + 1 = +1 cis-alkene + anti addition ----7 racemic are "same", and trans/anti! +1 x - 1 = -1

racemic are "opposite". trans-alkene + syn addition ----7 racemic Then any combination can -1 x +1 = -1

trans-alkene + anti addition ----7 meso be predicted, just like math! -1 x -1 = +1

8-36 Solve these ozonolysis problems by working backwards, that is, by "reattaching" the two carbons of the new carbonyl groups into alkenes. Here's a hint. When you cut a circular piece of string, you still have only one piece. When you cut a linear piece of string, you have two pieces. Same with molecules. If ozonolysis forms only one product with two carbonyls, the alkene had to have been in a ring. If ozonolysis gives two molecules, the alkene had to have been in a chain.

(a) two carbonyls from ozonolysis are in a chain, so alkene had to have been in a nng

8-37

(a) � (b)

(b) two carbonyls from ozonolysis are in two different products, so alkene had to have been in a chain, not a ring

(c) two carbonyls from ozonolysis are in two different products, so alkene had to have been in a chain, not a ring

E or Z of alkene cannot be determined from products

H

03�

Mc,� �o + o� OH

O�

17 1

Page 179: Solucionario de wade

8-37 continued

(c) 0=0 03 Me2S 00 0=0 • • +

(d) U 03 Me2S �H

• •

0 0

(e) U KMn04 �OH

• t!. 0 0

(f) U KMn04 d ·aH •

cold 1I110H

8-38 The representation for a generic acid wi l l be H-B, where B is the conjugate base.

:x:r+ +/ J.C

H I / B:-}

8-39 Catalytic BF3 reacts with trace amounts of water to form the probable catalyst:

F H - I +1 dimer

PTe\� __ /, -catalyst

etc . ...

tetramer trimer

172

Page 180: Solucionario de wade

8-40 H-B is used here to symbolize a generic acid.

alkenes

� polymers�

(colored)

-:B

Note : the dashed bond symbol is used here to indicate the continuation of a polymer chain .

8-41 H H H H I I \ /

------C-C· + C=C I I � n/)\...( \ H Ph � H

H H H H I I I I

------C-C-C-C· I I I I H Ph Ph H

10 radical , and not resonance-stabilized­this orientation is not observed

Orientation of addition always generates the more stable intermediate; the energy difference between a 10 radical (shown above) and a benzyl ic radical is huge. The phenyl substituents must necessarily be on alternating carbons because the orientation of attack is always the same-not a random process.

8-42 For c larity , the new bonds formed in this mechanism are shown in bold .

RO - OR

H H \ /

RO · + C=C � .!)\...(\ � H

2 RO· H H \ /

H H �'" ,C

/ C H H H H I I ( o"+-",, V \

---t.� RO-C-C. H H .. I I I I

RO-C-C-C-C· I I H H

etc. ----

173

I I I k) H H H) l\c/ / \

H H H H H H H H I I I I I I

RO-C-C-C-C-C-C· I I I I I I H H H H H H

Page 181: Solucionario de wade

8-43 o II

H COCH3 \ n /

. . -

:0 : : 0 : II I

H C-OCH3 H C-OCH3 I _I I II

HO : C=C --- HO-C-C : 11: . HO-C-C I I \ I I .. �/ \

8 -44

H Me H Me H Me

o II COCH3

\ n / C=C

/ \ H Me

. . · 0· · 0· COOMe

·11

· COOMe

· I·

H I H C-OCH3 H I H C-OCH3 I I _I I I II

HO - C-C-C-C : • • OO - C - C-C - C I I I I I I I I

H Me H Me H Me H Me ! ele.

COOMe COOMe COOMe COOMe COOMe

"

Me Me Me Me Me

Dashed bonds mean that the chain continues.

0 II

Plexiglas

aOH 0 H3C-C-00H Na .. -

H30+ <X'O " Br�

o;Y � OH

o

O F <X"'o � <X 4 MCPBA

OH

OH

174

B r2 - �

hv

Page 182: Solucionario de wade

8-45 In the spirit of this problem, all starting materials will have six carbons or fewer and only one carbon-carbon double bond. Reagents may have other atoms. major product from resonance

(a) stabilized radical on 10 and 30 C

(b)

0 Br2 •

hv (or NBS)

(c)

0 Br2 •

O:::J Na. O::�H-:yBC � � ::

anti-Markovnikov syn stereochemistry

o� o� 0: � CH,I, 0: CH3 Zn, CuCI CH3

ey Br eyC::N

BC-p KOH Br Br2 0- Br KOH o-0 H

hv· -only substit:on

-

(or NBS) is possible + Na MCPS: � \{ 0-0

-

OH ./ � ,A/ � H20

hi ° � T 8-46 Please refer to solution 1-20, page 12 of this Solutions Manual.

8-47

(a)

H 0

(d) Ao + 1 Br

(el e/' (c) q,

�:3

II, BH H 2

_aCH3 " " O H

intermediate

Cl

(f) V 175

peroxides do not affect HCl addition

L5F (g) r---

Page 183: Solucionario de wade

8-47 continued

O

CH3 (h) '" 'OH

""OH

(I) Q H (0)

q; 8-48

(a)

HgOAc intennediate

initiation r-,ri.

RO-OR ---- 2 RO·

(i) CH3

('f. IIOH

V ""OH

(n)

(p)

0:Jn RO. + H -Br .. ROH + Br· propagation

r�H Br· +�

(b) H

�H--C:S03;

H + L"H

QC� •

'---:9-H H

176

CO OH

q; Cl

Br -..� + • Br

Page 184: Solucionario de wade

8-48 continued

(c) H H H n�� n H-Br I

.I 1(1� 'fH ....... ,+ C H

OR fH hydride shift

• (fit:: (d)

� .. -:Br: .. Br 'Y'

----... Br Br , ...... ,.--.10..1 - I f"

:B�: I . . t Br 'J'

HO- + H-C-Br --- :C- Br --- Br- + :CBrz I I Br Br

00� V �

Brz .. aA B'

(e) H JY H--Q�

H

OR

H H�:C1: X f/ �

+C I H

XH�

�CH3

+C •

I H

;r:J _ (f) � B_r_-B" •

• • : � '--- .Br.

& . �� :Ci: + F�'

� CI

� HOC; � CH O-H • • 3 OCH 3.. �

3

Br

� Br

Br

� CI

� . . . Jy CH30 -H H�CH3 OCH

177

• • � 3

Page 185: Solucionario de wade

8-48 continued

(g) H � 0 J....< _H H-B H- T •

Ph

Recall that "Ph" is the abbreviation for phenyl.

8-49

(a) (f (b) (f (0) (f (d) (f (c) (f (f) (f (g) (f (h)

(i)

Hg(OAc)z NaBH4

HEr ROOR

.. ..

Hg(OAc)z NaBH4

OS04 H202

.. .

..

or cold, dilute KMn04

Zn(Cu)

..

50% NaOH (aq)

�OH

V OH

�CI

V OH

�Br V 'Br

178

H

H3C� H

Ph Ph

Page 186: Solucionario de wade

8-50

(a)

(d)

(g)

(j)

HQ "

OH CH3-Q--("

HO H�OH

, OH

CH3

-0-< Br

HO'8 Br "

OH CH3

(m) � CH3

(b)

(e)

(h)

(k)

CH3D--f0 ~ + CO2

BD--f Br

Cl

C I�-b-fCi "

Cl CH3

0

~ (c) H

HO

+ CH2=O

HD-f0H

(f) ", OH CH3

Br

( i ) -b-\ Br

CH,o� (I) OCH3

8-51 Each monomer has two c arbons in the backbone, so the substi tuents on the monomer wi l l repeat every two carbons in the polymer. Dashed bonds indicate continuation of the polymer chain .

Cl � F

Cl Cl Cl Cl I I I I

--- --- -CH2CHCH2CHCH2CHCH2CH -- - --

polyvinyl chloride, PVC

F F F F F F

F0-;--I I I I I I

----C -C -C-C -C-C----I I I I I I

F F F F F F F polytetrafluoroethylene, PTFE, Teflon

N N N N N III III III III III C C C C C � I I I I

--- ----CH2CHCH2CHCH2CHCH2CH -----

polyacry lonitrile , Orion

179

l ine formula representation

Cl Cl Cl Cl

F F F F F F F F F . . - . - .

F F F F F F F F F

N N N N III III III III C C C C . . . .

Page 187: Solucionario de wade

8-52 Without divinylbenzene, individual chains of polystyrene are able to s l ide past one another. Addition of divinylbenzene during polymerization forms bridges, or "cross l inks", between chains, adding strength and rigidity to the polymer. Divinylbenzene and s imilar molecu les are called crossl inkjng agents.

two polystyrene chains crosslinked by a divinylbenzene monomer shown in the dashed oval

8-53 A peroxy radical i s shown as the initiator. Newly formed bonds are shown in bold. H CH3 \ /

H CH�\.. ,C/ C I I r "-f--' v \

_-I.� RO-C-C. H CH� I I H CH3

8-54 0, 'C-OCH2CH3 I ethyl acrylate

H2C=CH

etc. ---

8-55 In each case, the compound (boxed) that produces the more stable carbocation is more reactive.

(a) (b) 0 3"6 al lylic

� (c) �

180

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8-56 Once the bromonium ion i s formed, i t can be attacked by either nuc\eophi le , bromide or chloride, leading to the mixture of products .

Q H H

B r-- B r � ~

H B r

��-� � :Cl : 'i---Y H �.. H Cl

8-57 Two possible orientations of attack of bromi ne radical are possible:

(A) anti -Markovnikov

>= \ . HBr hBr

+ • B r -- C, .- + • B r

/ B r 3 ° radical

(B) Markovnikov

>= 'l-r HBr y; + • B r -- �H2 � + • B r

H 1 ° radical

The first step in the mechanism i s endothermic and rate determining. The 3° radical produced in anti­Markovnikov attack (A) of bromine radical is several kl/mole more stable than the 1 ° radical generated by Markovnikov attack (B) . The Hammond Postulate tells us that it is reasonable to assume that the activation energy for anti -Markovnikov addition is lower than for Markovnikov addition. This defines the first half of the energy di agram.

The relative stabi l i ties of the final products are somewhat difficult to predict . (Remember that stabi l i ty of final products does not necessari ly reflect relative stabi l i ties of intermediates; this is why a thermodynamic product can be different from a kinetic product.) From bond dissoc iation energies (kJ/mole) in Table 4-2 :

anti -Markovnikov

H t0 3°C 3 81

Br to 1 ° C 285 666 kJ/mo\e

Markovnikov

H to 1 ° C 410

Br t0 3°C 272 682 kl/mole

If it takes more energy to break bonds in the Markovnikov product, i t must be lower in energy, therefore, more stable-OPPOSITE OF STABILITY OF THE INTERMEDIATES!

Now we are ready to construct the energy diagram; see the next page .

181

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8-57 continued

t Markovnikov , ,'/

, - . , . , , ,

anti-Markovnikov

reaction --

It is the anti-Markovnikov product that is the kinetic product, not the thermodynamic product; the anti­Markovnikov product is obtained since its rate-determining step has the lower act ivation energy.

8 -58 Recal l these facts about ozonolysis: each alkene c leaved by ozone produces two carbony l groups; an alkene in a chain produces two separate products; an alkene in a ring produces one product in which the two carbonyls are connected.

(a) GSO

H 0 H

o

0

H

0

8-59

+ CH2=O

CHO

¢ CHO

(b) + O� (d)

H

0

CH3

(a) CY CH3C03H U'OH

.. H2O

OH

(b) 0 OOH

OH

(or cold, dilute KMn04) cis + syn

(cis double bond plus syn stereochemistry of addition)

trans + anti (trans double bond plus anti stereochemistry of addition)

182

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8-59 continued

(c )

(d)

1 This structure shows a trans alkene in a to-membered ring, just the rotated view of the structure to the right.

o Cl2

B r2 ..

� B r

� Br

anti addition of Br2 requires trans alkene to give meso product

�tate around C-2

�"': Br2 ..

H � 2

Br

trans-cyclodecene (XC!

""OH

(e) CO �

B H3 - THF .. ~

OH

(f) � or CO or � _H_g _(O_A_ c-lh

.. � NaBHt, V-/ � CH30H

8-60

A) Unknown X, CSH9Br, has one element of unsaturation. X reacts with neither bromine nor KMn04, so the unsaturation in X cannot be an alkene; it must be a ring.

B ) Upon treatment with strong base (t-butoxide) , X loses H and Br to give Y, CSHg, which does react with bromine and KMn0 4 ; it must have an alkene and a ring. Only one isomer is formed.

C) Catalytic hydrogenation of Y gives methylcyc lobutane. This is a B IG c lue because i t gives the carbon skeleton of the unknown. Y must have a double bond in the methylcyclobutane skeleton, and X must have a Br on the methylc yc lobutane skeleton .

D) Ozonolysis of Y gives a dialdehyde Z, CSHg02' which contains all the original carbons, so the alkene c leaved in the ozonolysis had to be in the ring.

Let's consider the possible answers for X and see if each fits the information.

1)

if this is X

KO-t-Bu .---:f' I > U

ozonolysis

this must be Y; only one product

more possibi l i ties on the next page 183

° > d' + O=CH2

DOES NOT FIT OZONOLYSIS RESULTS so this cannot be X and Y

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8-60 continued

2)

3)

d KO-t-Bu B r 1 >

if thi s i s X

c( KO-t- B u > Br

i f this i s X

d major

Q major

0( Y would be a mixture of alkenes , but the e l imination + gives only one product. We already saw in example 1

minor that the exocycl ic double bond does not fi t the ozonolysis resul ts so this structure cannot be X.

0( Y would be a mixture of alkenes, but the e l imination + gives only one product, so this structure of X i s not

minor consistent with the information provided.

4) P KO -t-�u d + Q lozOnOIYSiS

SAME COMPOUND ;

> or< Br

if thi s i s X this must be Y ; on ly one product

The correct structures for X, Y, and Z are gi ven in the fourth possibi l i ty . The only structural feature of X that remains undetermined is whether i t is the cis or trans i somer.

Z, a dialdehyde

trans

8-6 1 The clue to the structure of a-pinene is the ozonolysis . Working backwards shows the alkene position .

o backwards

o >

� became carbony l carbons '(S)( a-pinene

After ozonolysis , the two carbonyls are sti l l connected; the alkene must have been in a ring, so reconnect the two carbony l carbons with a double bond.

~ B r2 ���T 'fitB T

'

, I I CH3 .. or

CH3 Br A A

~ Br {iXB T B r2 H2SO4 .. .. Zaitsev product H2O � 'OH � CH3

B C

� OH PhC03H �o H30+ {L:(OH

.. .. "

"

D E

184

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8-62 The two products from pennanganate oxidation must have been connected by a double bond at the carbonyl carbons. Whether the a lkene was E or Z cannot be determined by this experiment.

CH3(CH2) 12CH = CH(CH2hCH3

8-63

Unknown X Pt

Unknown X

shown here as Z which is the natural ly-occurring isomer

must have three alkenes in this skeleton

o 0

CH2 = O +H�

o

o 0

+ A/ H

There are several ways to attack a problem l ike this . One is the trial-and-error method, that is , put double bonds in al l possible positions unti l the ozonolysis products match. There are times when the trial-and-error method is useful (as in s imple problems where the number of possibi l it ies is few), but thi s is not one of them.

Let's try logic . Analyze the ozonolysis products careful ly-what do you see? There are only two methyl groups , so one of the three terminal carbons in the skeleton (C-8, C-9, or C- l O) has to be a =CH2 . Do we know which terminal carbon has the double bond? Yes, we can deduce that. If C- l O were double-bonded to C-4, then after ozonolys is , C-8 and C-9 must sti l l be attached to C-7 . However, in the ozonolysis products , there is no branched chain, that i s , no combination of C-8 + C-9 + C-7 + C- l . What if C-7 had a double bond to C- l ? Then we would have acetone, CH3COCH3 , as an ozonolysis product-we don 't . Thus, we can't have a double bond from C-4 to C- l O. One of the other terminal carbons (C-8) must have a double bond to C-7 . 8 � � 1 0

9 6 5

The other two double bonds have to be in the ring, but where? The products do not have branched chains , so double bonds must appear at both C- l and C-4. There are only two possibi l i ties for this requirement .

I H)--- or }{}- II

Ozonolysis of I would g ive fragments containing one carbon, two carbons, and seven carbons. Ozonolysis of II would gi ve fragments containing one carbon , four carbons, and five carbons. Aha! Our mystery structure must be II.

(Editorial comment: Sc ience i s more than a collection of facts . The application of observation and logic to solve problems by deduction and inference are critical scientific skil ls , ones that distinguish humans from algae . )

185

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8-64 In this type of problem, begin by detennining which bonds are broken and which are fonned. These wi l l always give c lues as to what is happening.

formed

H

H+ goes to most e lectronegati ve atom

q�+ .. \ 3 0 carbocation looks

HO for e lectrons , finds them at nearby alkene, forming a 6-membered ring (yes !)-leaves a 30 carbocation

1 protonated epoxide opens to gIve the most stable carbocation (3()

8-65 See the solution to Problem 8-35 for s impl ified examples of these reactions .

(a)

(b)

(c)

(d)

eOOH HOOCJ

fCOOH HOOe

(COOH eOOH

(COOH eOOH

OS04

HO OH � trans + syn � racemic

Hooe ,,1 "" H H eOOH Hooe � OH 7--<� " trans + anti � meso '" IleOOH HO H Hooe� OH '---< cis + anti � racemic / �/IH HO eOOH

HO OH � cis + syn � meso

Hooe "1 �/lleOOH H H

186

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8-66

u B03 - THF ..

H202 ..

HO- aCH3 1 1 1 0 I I I OH H

8-67 By now , these rearrangements should not be so "unexpected" .

� n U " "":

H � B�

H 20 I Cf.�C H

C+H�H 3 H

(YCH3

alkyl migration with ring expansion gives 3° carbocation in 6-membered ring--carbocation nirvana !

H H

H

+ : !lr:0'CH]

� B r

CH3

You must be asking yourself, "Why didn't the methyl group migrate?" To which you answered by drawmg the carbocation that would have been formed:

I;I 2° 3° C+ H O\} � -- QC<1''-CH3

CH I " H CH3 3 H The new carbocation is indeed 3° , but it is only in a 5-membered ring, not quite as stable as in a 6-membered ring. In all probabi l i ty , some of the product from methyl migration would be formed, but the 6-membered ring would be the major product .

8-68 Each alkene wi l l produce two carbonyls upon ozonolysis or permanganate oxidation. Oxidation of the unknown generated four c arbonyls , so the unknown must have had two alkenes . There is only one possibi l i ty for their positions .

H

CO H

the unknown

KMn04 ..

/j.

H a� COOH :: COOH H

187

COOH + I COOH

Page 195: Solucionario de wade

8-69 (a) Fumarase catalyzes the addition of H and OH, a hydration reaction . (b) Fumaric acid i s planar and cannot be chiral . Malic acid does have a chiral center and is chiral . The enzyme-catalyzed reaction produces only the S enantiomer, so the product must be optical ly active. (c) One of the fundamental rules of stereochemistry i s that optical! y inacti ve starting materials produce optical ly inactive products . Sulfuric-acid-catalyzed hydration would produce a racemic mixture of malic acid, that is, equal amounts of R and S. (d) If the product is optical ly active , then ei ther the starting materials or the catalyst were chiral . We know that water and fumaric acid are not chiral, so we must infer that fumarase is chiral . (e) The D and the OD are on the "same side" of the Fischer projection (sometimes cal led the "erythro" stereoisomer) . These are produced from either: (1) syn addition to cis alkenes, or (2) anti addition to tralls alkenes. We know that fumaric ac id is tralls, so the addition of D and OD must necessari ly be ant i . (f) Hydroboration i s a syn addition .

1 . B D3 • THF D OD DOOeH eORD

Hooe " , H " 'e == e '" H

....... 'eOOH

2. D202, DO- H >-1(, , " , .. ", " "

H +

DOOe H eOOD D OD

(note that OH exchanges

with D in DO-) eOOD eOOD

�+�D D�+� eOOD eOOD

As expected, tralls alkene plus syn addition puts the two groups on the "opposite" s ide of the Fischer projection (sometimes cal led " threo") .

8-70 Hg(OAc) Hg(OAc)

(a) �. 0H

+ Hg(OAc)

--

. . AcO :

mercunntum ion

Br Br

Br - Br ..

bromonium ion

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8-7 1 The addition of BH3 to an alkene i s reversible. Given heat and time, the borane wi l l eventually "walk" its way to the end of the chain through a series of addition-e l imination cyc les. The most stable alkylborane has the boron on the end carbon; eventual ly, the series of equ i l ibria lead to that product which is oxidized to the primary alcohol.

..

..

...

----

�H2 most stable H

189

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8-72 First, we explain how the mixture of stereoisomers results, then why.

We have seen many t imes that the bridged halonium ion permits attack of the nucIeophi le only from the opposi te side.

expected: Q Ci-CI ---

H Ph W CI� � �:S:1:

H Ph ~ H Cl

trans only

A mixture of cis and trans could result only if attack of chloride were possible from both top and hottom, something possible only if a carbocation existed at this carbon.

actual: '"

0� H Ph H --� H CI trans

+� H Ph cis

This picture of the p orbitals of benzene show resonance overlap with the p orbital of the carbocation. The chloride nucIeophile can form a bond to the positive carbon from either the top or the bottom.

Why does a carbocation exist here? Not only is it 3°, it is also next to a benzene ring (benzylic) and therefore resonance-stabilized. This resonance stabi l ization would be forfei ted in a halonium ion intermediate.

� 0":::: r'�/ // �

�CI

+

CI Ph :Ci : C a 'Ph CC 'Ci . . 0: + - � IIi H "'H CI CI CI

190

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CHAPTER 9-ALKYNES

9-1 Other structures are possible i n each case.

CH3CH2C::: CCH2CH3 CH3CH2CH2CH2C::: CH

0-C::: CH HC::: C - CH2CH = CHCH2CH2CH3

9-2 New IUPAC names are gi ven. The asterisk (* ) denotes acetylenic hydrogens of terminal alkynes.

(a)

(b)

* CH3CH2 -C:::C - H

but- l -yne

* CH3CH2CH2 - C:::C - H

pent- l -yne

CH3 - C:::C-CH3 but-2-yne

CH3CH2 - C::: C - CH3 pent-2-yne

CH3 I *

CH3CH - C:::C - H 3-methy lbut-l-yne

9-3 The decomposition reaction below is exothermic (/)J{0 = - 234 kJ/mole) as well as having an increase in entropy. Thermodynamical ly, at 1 500°C, an increase in entropy wi l l have a large effect on AG (remember AG = /)J{ - TAS). Kinetically, almost any activation energy barrier wi l l be overcome at 1 500°C. Acetylene would likely decompose into i ts elements:

HC::CH 1 500° � 2 C + H2

9-4 Adding sodium amide to the mixture will produce the sodium salt of hex- l -yne, leaving hex- l -ene untouched. Disti l lation wi l l remove the hex- l -ene, leaving the non-volati le salt behind.

CH� CHCH2CH2CH2CH3 } NaNH2.. J + _

�H� CHCH2CH2CH2CH3

H -C = C -CH2CH2CH2CH3 1 Na • C = C - CH2CH2CH2CH3 non-volatile salt

9-5 The key to this problem is to understand that a proton donor will react only with the conjugate base of a weaker acid. See Appendix 2 at the end of this Solutions Manual for an in-depth discussion of acidity.

-(a) H -C:::C-H + NaNH2 -- H -C:::C: Na+ + NH3

-(b) H-C:::C -H + CH3Li -- H-C:::C: Li+ + CH4

(c) no reaction: NaOCH3 i s not a strong enough base

(d) no reaction: NaOH is not a strong enough base

(e) H-C:::C:-

Na+ + CH30H -- H-C:::C - H + NaOCH3 (opposite of (c)) -

(0 H -C:::C: Na+ + H20 -- H -C:::C-H + NaOH (opposite of (d)) -

(g) no reaction: H - C:::C : Na+ is not a strong enough base

(h) no reaction: NaNH2 is not a strong enough base

191

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9-6 H-C:::C-H

NaNH2 ..

-H-C:::C :

CH3(CH2)S -C::: C -CH2CH3

9-7 1) NaNH2

CH3CH2Br Na+ • H-C:::C-CH2CH3 � NaNH2 CH3(CH2)SBr

.. Na+ :C:::C-CH2CH3

.. (a) H-C:::C-H 2) CH3CH2CH2CH2Br

H -C:::C-CH2CH2CH2CH3

1 ) NaNH2 1 ) NaNH2 (b) H-C:::C-H • H-C:::C-CH CH CH .. CH -C:::C-CH CH CH

2) CH3CH2CH2Br 2 2 3

2) CH3I 3 2 2 3

1 ) NaNH2 1 ) NaNH2 (c) H-C:::C-H .. H-C:::C-CH2CH3 • CH3CH2-C:::C-CH2CH3

2) CH3CH2Br 2) CH3CH2Br

(d) cannot be synthesized by an SN2 reaction-would require attack on a 20 alkyl halide

_��r/2°

CH3-C:::C : Na+ + CHCH2CH3 strong nucleophile; CH must be second order 3 fr CH3-C:::C-CHCH2CH3

I CH3

low yields; not practical

(e) H-C:::C-H 1 ) NaNH2 .. 2) CH3I

CH3-C:::C-H 1) NaNH2 .. 2) BrCH2CHCH3

I

CH3 -C::: C -CH2CHCH3 I

(f) H-C:::C-H 1) NaNH2

.. 2) Br(CH2)8Br

H2C -C::: C -CH2 ....... !----

CH3

H-C:::: C-?H Br�

2

� NaNH2 Na+ -,C-

r'. � . =C-CH

Br�2

CH3

Intramolecular cyclization of large rings must be carried out in dilute solution so the last SN2 displacement will be intramolecular and not intennolecular.

9-8

(a) H-C:::C-H 1 ) NaNH2

.. H20 ----l.� H-C:::C-CH20H

2) H2C=O

192

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9-8 continued

(b) H-C:::C-H

(c) H-C:::C-H

(d) H-C:::C-H

9-9 (Br H

1 1

1 ) NaNH2 2) PhyCH3

o

1) NaNH2 �

2) CH3I

1) NaNH2 �

2) CH3I

H -C � C -CH2CH2CH3 1 1 :OH Br H�oo

.. H20

9H ----l.-� H3C-C-C::C-H

I Ph

OH I 1 ) NaNH2 .-CH3-C:::C-H

2) HCCH2CH2CH3 II

CH3 -C::: C -CHCH2CH2CH3

o 3) H20

OH I 1) NaNH2

� CH3-C:::C-H 2) CH3CCH2CH3

CH3 -C::: C -CCH2CH3 I .

CH3

..

II o

3) H20

H H � -oo H \ ,1 :�H _ 1 C=C � H-C=C-C-C�C�

{) \ I ) Br CH2CH2CH3 Ii 1

-:oH/ t 0 0

H � H H } I ) _ 1 _ _ I H-C=(= C-CH2CH3 --- H-C=C=C-CH2CH3 � H-C=C-C-CH2CH3

1 HO-H 0 0 0 0

H U� 1- 0 0

t :OH 0 0 � H-OH H i - -0 0 - } U I -H-C=C=C-CH2CH3 � H-C-C=C-CH2CH3 � H-C-C=C-CH2CH3

1 0 0 1 I H H H

9-1 0 To determine the equilibrium constant in the reaction:

tenninal alkyne � internal alkyne !J.G = - RT In Keg !J.G = - 1 7.0 kJ/mole

(- 4.0 kcallmole)

R = 8.3 1 4 JIK-mole

( !1G) ( -(-17,000) ) Keg = e RT = e (8.314)(473)

[internal]

[terminal]

193

75 98.7% internal

1 .3% tenninal

= e4.32 = 75

Page 201: Solucionario de wade

9-11 (a) This i somerization i s the reverse of the mechanism in the solution to 9-9.

H�-•

• •

H-C-C C- CH2CH3 .. H-C-C C-CH2CH3 ---- H-C==C==C-CH2CHJ I ) . NH2 i - . . - } 1 1 1 • •

H H H I(

i H H } . H

HJH2

1- _ 1 1 H-C C-C-CH2CH3�H-C==C==C-CH2CH3 � H-(C==C==C-CH2CH3

• • • • • • I �

l � :� \""'H - NH

�H U

2 H 1

H-C C-C-CH2CH3 I H

(b) All steps in part (a) are reversible. Wi th a weaker base like KOH, an equi l ibrium mixture of pent-I­yne and pent-2-yne would result. With the strong base NaNH2, however, the final tenninal alkyne is deprotonated to give the acetylide ion:

H -C C -CH2CH2CH3 + NaNH2 - Na+ :C == C -CH2CH2CH3 + NH3

Because pent-l -yne is about l O pK units more acidic than ammonia, this deprotonation is not reversible. The acetyl ide ion is produced and can't go back. Le Chatelier's Principle tells us that the reaction wi l l try to replace the pent- l -yne that is being removed from the reaction mixture, so eventually all of the pent-2-yne will be drawn into the pent- l -yne anion "sink".

(c) Using the weaker base KOH at 200°C will restore the equilibrium between the two alkyne isomers wi th pent-2-yne predominating.

9- 12

(a)

( 1)

(2)

(b)

( 1 )

(2)

Br Br 1 1

H3C-T-T-CH3

H H

KOH --- H3C-C C-CH3

L1 Br Br

Br2' CCl4 I 1 CH3CH == CHCH3 .. CH3CH -CHCH3

Br Br I I

H-C-C-(CH2hCH3 1 1 H H

NaNH2 •

1500

Br2' CCl4 CH2 == CH(CH2) sCH3 •

H -C C -(CH2hCH3

Br Br 1 1

H2C -CH(CH2hCH3

194

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9- 12 continued

(c)

( 1 ) KOH

Br Br Br2' CCl4 "

(2) ----I .. � CH3CH -CH(CH2)4CH3

(d) Br

( 1 )

H

(2)

9- 1 3

Lindlar catalyst

Na

(c) Br

W NaNH2

cis H

One of the useful "tricks" of organic chemistry is the abi lity to convert one stereoisomer into another. Having a pair of reactions like the two reductions shown in parts (a) and (b) that give opposite stereochemistry is very useful , because a common intermediate (the alkyne) can be transformed into either stereoisomer. This principle is used again in part (d).

195

Br

H

KOH at 200°C could have been used instead of NaNH2

Page 203: Solucionario de wade

9- 13 continued Br (d)

� ~ trans

KOH -C C�

I Lindlar H2 , catalyst Br

H H \ I C == C cis

/ L 9-14 The goal is to add only one equivalent of bromine, always avoiding an excess of bromine, because two molecules of bromine could add to the triple bond if bromine was in excess. If the alkyne is added to the bromine, the first drops of alkyne wil l encounter a large excess of bromine. Instead, adding bromine to the alkyne wil l always ensure an excess of alkyne and should give a good yield of dibromo product.

9- 16

H + I

CH3CH2CH2 - C == C - H �

2° better than � :B�: 1 ° carbocation 0 0

2° carbocation and resonance-stabilized

CI H I I

H3C - C - C - (CH2)4CH3 I I

Cl H

H CI I I

+ H3C - C - C -(CH2)4CH3 I I

H Cl (b) The second addition occurs to make the carbocation intermediate at the carbon with the halogen because of resonance stabilization.

:C1: H I I - C==C -

0 0 y+ :C1: H

I I b·l· . -C-C- no sta IlzatlOn I +

H

H+ I I I I I resonance i :ci: H : �I: H } ---- -�-k- ...... t------t�� -- C-k- stabilization

196

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9-17

in i ti ation: RO - OR hv 2 RO·

RO • + H-Br --- RO-H + Br •

propagation: Br.�J j-CH2CH2CH3 H -C == C -CH2CH2CH3

I •

Br

The 2° radical is more stable than 1°. The anti-Markovnikov orientation occurs because the bromine radical attacks first to make the most stable radical, which is contrary to electrophi l ic addition where the H+ attacks first (see the solution to 9-15).

9- 18

(b) H - C C -(CH2hCH3 HBr ROOR

HBr

Lindlar catalyst

(or Na, NH3)

2HBr

H -C==C-(CH2hCH3 E + Z I I CI Cl

H -C==C-(CH2hCH3 E + Z I I Br H

H - C == C -(CH2hCH3 I I

H Br Br Br I I

H-C-C-(CH2) CH I I

3 3

Br Br

H Br I I

H-C-C-(CH2) CH I I

3 3

H Br

197

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9- 19

;)(� + CH3-r=c) CH2CII3

! H2O:

CH3 -C == C - CH2CH3 1 1+

H :OH I ) �H ! H2O:

CH3-'f ( 'f-CH2CH3

H :RH

t H+

H 1 +

CH3-C-C-CH2CH3 , '",\ H :O-H

.. � t Hi;:

H I

CH -C-C-CH2CH3 3 I I I

H 0

3-pentanone

both 2° vinyl carbocations

resonance-stabilized carbocations

+ CH3-(=A-CH2CH3

H20: ! CH3 -C == C -CH2CH3

+1 1 HO: H

I ) H)

H20: ! CH3-'f ) 'f-CH2CH3

H�: H

H+ t H

+ 1 CH3 - C - C -CH2CH3

r' 1 H-O: H

�. H20: t

H I

CH -C-C-CH CH 3 I I 1

2 3

0 H

2-pentanone

(resonance forms not shown)

The role of the mercury catalyst is not shown in thi s mechanism. As a Lewis acid, it may act like the proton in the first step, helping to form vinyl cations; the mercury is replaced when acid is added.

-H+ ---I.... CH3 -C == C - CH2CH3

I I +Hg OH

198

Page 206: Solucionario de wade

9-20

(a) But-2-yne is symmetric. Either orientation produces the same product.

° Sia2BH

CH3 CH3 \ / H202

CH3 CH3 \ / HO- I I

CH3-C C-CH3 .. C=C .. / \ HO-

C=C / \

---I"� CH3CHz - C - CH3

H BSia2 H OH

(b) Pent-2-yne is not symmetric. Different orientations of attack wi l l lead to different products on any unsymmetrical internal alkyne.

CH3 -C C -CH2CH3

�H S� CH3 CH2CH3

\ / C=C

/ \ H BSiaz � H202 , HO-

CH3 CH2CH3 \ / C=C

/ \ H OH � HO-

° I I

CH3CH2 -C -CH2CH3

9-2 1 ° I I

(a) ( 1 ) CH3CCH2CH2CH2CH3

° II

(b) ( 1 ) CH3CCH2CH2CH2CH3

(2) same mixture as in (b) ( 1) ° II

(c) ( 1) CH3CH2CCH2CH2CH3

�O (d) ( 1) VV

° II

CH3 CH2CH3 \ / C=C / \

Sia2B H � H202 , HO-

CH3 CH2CH3 \ / C=C / \

HO H

° I I

CH3 -C -CH2CH2CH3

(2) HCCH2CH2CH2CH2CH3

° II + CH3CH2CCH2CH2CH3

° II

(2) CH3CH2CCH2CH2CH3

�O (2)

VV 199

Page 207: Solucionario de wade

9-22

(a) CH3 H I I H CH3

\ / ----11 .. - H -C -C -BH) + C==C

I I -CH3 CH3

/ \ H3C CH3

CH3 H H H CH3 I I I I I

H-C-C-B-C-C- H I I I I

CH3 CH3 CH3 CH3

disiamylborane, Sia2BH

(b) There is too much steric hindrance in Sia2BH for the third B-H to add across another alkene. The reagent can add to alkynes because alkynes are l inear and attack is not hindered by bulky substituents.

9-23 o 0 II II

(a) (1) HO-C-C-(CH2hCH3 oxidation of a tenninal alkyne with neutral KMn04 produces the ketone and carboxylic acid without cleaving the carbon-carbon bond

o 0 II II

(b) (1) CH3 - C -C - CH2CH2CH3

o 0 II II

(c) (1) CH3CH2 -C -C -CH2CH3

o 0 11 II

(d) (1) CH CH-C-C-CH CH 31

2 3

CH3

(e) ( 1)

9-24

(a) CH3 -C:::C- (CH2)4 -C:::C-CH3

o II

(2) CO 2 + HO -C -(CH2hCH3 oxidation of a tenninal alkyne with wann, basic KMn04 cleaves the carbon-carbon bond, producing the carboxylic acid and carbon dioxide

o 0 II II

(2) CH3 - C -O H + HO -C -CH2CH2CH3

o II

(2) CH3CH2 -C -O H

(2)

(2)

200

o 0 II II

CH3CH-C-O H + HO -C -CH2CH3 1 CH3

o OH

HO

o

Page 208: Solucionario de wade

9-25 When proposing syntheses , begin by analyzing the target molecule, looking for smaller pieces that can be combined to make the desired compound. This i s especially true for targets that have more carbons than the starting materials ; immediately, you will know that a carbon-carbon bond forming reaction will be necessary.

People who succeed at synthesis know the reactions-there i s no shortcut. Practice the reactions for each functional group until they become automatic.

(a) analysis of target from ,,- -, -. acetylene

3° acetylenic alcohols made ( ./'""'-.... I\:'�\�i;��>: from alkylation

from acetylide plus ketones \/ r.f - �: ... ,:' of acetyhde , OH �'--' ' •• _---'

" • , : put on less reactive forward direction:

' . - 8 group first r� H-C=:C-H + NaNH2 ---- H-C=:C: Na+ .. H-C=C�

H-C=:C-H

t NaNH2

...... f------ Na+ -:C=:C�

� o

analysis of target: cyclopropanes are made by carbene insertion into alkencs; to get cis substitution around cyclopropane, stereochemistry of alkene must be cis; cis alkene comes from catalytic hydrogenation of an alkyne

NaNH2 CH31 NaNH2 ------I .. � ---- H3C -C =: C -H ..

CH3CH2Br .. H3C-C=:C-CH2CH}

H I Lindlar 2 , catalyst

H H >=<

H3C CH2CH3

CH212 ..

Zn(Cu)

(c) H CH2CH2CH3 )" /(

analysis of target: epoxides are made by direct epoxidation of alkenes; to get trans substitution around epoxide, stereochemistry of alkene must be trans; trans alkene comes from sodium/ammonia reduction of an alkyne CH3CH2 0 H

9-26 Please refer to solution 1-20, page 12 of this Solutions Manual.

20 1

Page 209: Solucionario de wade

9-27

(a) CH3CH2 -C:: C -(CH2)4CH3 (b) H3C-C::C-(CH2)4CH3 (c) ( )- C::C-H

(d) o- C::C-H

(f) �' Br

(e) CH3CH2 -C:: C - CHCH2CH2CH3 I

CH3

(g) CH3CH -C:: C - (CH2hCH3 I OH

(h) H3C C:: C -CHCH2CH3 \ I I C=C I \ CH2CH3

H H

CH3

(i) H - C:: C - CH2 - C:: C - CH2CH3 (j) H-C::C-CH=CH2 (k) H-C::C-("H

,C=CH2 H 9-28

(a) ethylmethylacetylene (b) phenyl acetylene

9-29

(a) 4-phenylpent-2-yne (b) 4,4-dibromopent-2-yne

9-30

(a) internal alkynes

CH3 - C:: C -CH2CH2CH3

hex-2-yne

CH3CH2 - C:: C - CH2CH3

hex-3-yne

CH3 -C:: C -CHCH3 I

CH3 4-methylpent-2-yne

(c) sec-butyl-n-propylacetylene (d) sec-butyl-t-butylacetylene

(c) 2,6,6-trimethylhept-3-yne (d) (E)-3-methylhept-2-en-4-yne

terminal alkynes

H - C:: C - (CH2hCH3

hex- l -yne

H - C:: C - CHCH2CH3 I CH3

3-methylpent- l -yne

H - C :: C -CH2CHCH3 I

CH3 4-methylpent- l -yne

CH3 I

H -C :: C - C - CH3 I CH3

NaNH2

(e) 3-methylhex-4-yn-3-ol (f) cycloheptylprop-I-yne

acetylide ions

C:: C - (CH2hCH3

- C:: C -CHCH2CH3 I .

CH3

C:: C -CH2CHCH3 I

CH3

CH3 I

C::C-C-CH3 I .

CH3 3,3-dimethylbut- l -yn

(b) All four terminal alkynes wi l l be deprotonated with sodium amide. � 9-3 1

(R�-: -CH3CH2-C .C==CH ---

Hb �aNH2 "

CHCl HC CH

H20 •

202

OH I

CH3CH2 - C-C-CH I

HC� "CHCI

ethchlorvynol

Page 210: Solucionario de wade

9-32

9-33

(a)

(d)

(g)

G) 9-34

(a)

(b)

(c)

(d)

(e)

H-C:::C-H NaNH2 CH3(CH2hBr

.. ..

CH3(CH2h (CH2)12CH3

CHiCH2h -C::: C -H NaNH2

"1 CH3(CH2)12Br

\ I H2 C = C ...... 1-------

/ \ Lindlar CH3(CH2h -C::: C - (CH2)12CH3

Cl I

CH2 = CCH2CH2CH3

H H catalyst muscalure

Cl I

(b) H3C -C -CH2CH2CH3 I Cl

H Br \ I

CH2 = CHCH2CH2CH3 (e) C=C I \

Br CH2CH2CH3

o 0 0 II II II

HO - C -C -CH2CH2CH3 (h) CO2 + HO - C - CH2CH2CH3

0 - II

Na+ : C ::: C -CH2CH2CH3 (k) H3C - C - CH2CH2CH3

Br H2O I

(c) CH3CH2CH2CH2CH3

Br Br I I

(f) H - <? - <? - CH2CH2CH3

Br Br

(i) H2C = CHCH2CH2CH1

0 II

(I) H - C - CH2CH2CH2CH3

NaNH2 H3C - C -CH2CH3 .. .. H-C:::C -CH2CH3

I 1500 Br

Br I KOH

H3C - C -CH2CH3 .. H3C -C::: C -CH3 I 2000 Br

_ NaNH2 _

+ CH3CH2-C=C-H .. CH3CH2-C=C: Na � CH3CH2CH2CH2Br

CH3CH2 -C:: C -CH2CH2CH2CH3

H3C H Br2 \ I

C=C ---I

H

H \

\ CCl4 CH2CH2CH3

H I Br2

C=C ---I

H3C \ CCl4 CH2CH2CH3

Br Br I I

CH3C -CCH2CH2CH3 • I

H H

Br Br I I

CH3C -CCH2CH2CH3 • I

H H 203

KOH .. H3C -C::: C -CH2CH2CH3

2000 this product could contain minor amounts of 3-hexyne from rearrangement

1 ) NaNH2

1 500 .. H - C == C - CH2CH2CH2CH3 2) H2O

Page 211: Solucionario de wade

9-34 continued

(f) Hz

W ..

Lindlar c is catalyst

Na ..

W NH3 trans (g)

(h) HC:: C - CHzCHzCHzCH3 _H_

z_0---i"� [ CH2 = ��H2CH2cH2cH3 1 - H3C -�CH2CH2CH2CH]

HZS04 unstable enol . HgS04

9-35

Br Br I I

CH3C -CCH2CH2CH3 • I

H H

KOH

this product could contain minor amounts of 3-hexyne from rearrangement

---J"� H3C - C:: C -CHzCH2CH3 200° major

Hz I Lindlar • catalyst

H3C CHzCH2CH3 \ I C=C I \

H H

KOH t Br� .. HC� + � MixtureA

200° from rearrangement-

Br

CompoundD b.p. 1 40- 1 50°C under vacuum

OH

\ could contain 3-hexyne J '--------.. / Y o

only the terminal alkyne reacts j with NaNH2 and acetone

OH 2�

204

+ � tMhture B

J� � CompoundC b.p. 80-84°C

Page 212: Solucionario de wade

9-36

(c) CH3CH2-C:::C - CH20H (after H20 workup)

(e)

(g)

9-37

(a)

(b)

(c)

(d)

(e)

(f)

OH I

CH3CH2 -C::: C - CHCH2CH2CH3 (after H20 workup)

OH I

CH3CH2 - C::: C - C -CH2CH3 I

CH3 (after H20 workup)

NaNH2 HC::C-H • HC::C:

NaNH2 HC::C-H • HC::C:

H2 H3C -C:: C - CH2CH2CH3 •

Lindlar synthesized in part (b) catalyst

H3C -C:: C - CH2CH2CH3

synthesized in part (b)

H3C -C:: C - CH2CH2CH3

synthesized in part (b)

HC = C - CH2CH2CH2CH3

synthesized in part (a)

Na

2 equiv. H2

Pt

2HBr

elimination on 3° halide

Na+ -0-0

CH3CH2CH2Br --=----"'--..::.......-... HC:: C -CH2CH2CH3 � NaNH2

..... f-----. CH3I

H3C CH2CH2CH3 \ I C=C

H H

H3C H \ I

205

C=C I \

H CH2CH2CH3

Br I

H3C -C -CH2CH2CH2CH3 I Bf

Page 213: Solucionario de wade

9-37 continued o 1 ) Sia2BH I I

(g) H -C:: C -CH2CH2CH3 from (b)

H -CCH2CH2CH2CH3

(h) H20

H-C::C-CH2CH2CH3 •

from (b) H2S04 HgS04

NaNH2

° I I

H3C -CCH2CH2CH3

(i) HC::C-H ---t .. � HC:: C:

alkene must be cis to produce the (±) product from anti addition

NaNH2

Review the stereochemistry in the solution to Problem 8-35, p. 171 of this Solutions Manual.

Lindlar catalyst

1) NaNH2 U) HC::C-H ---t .. � HC::C: Na+ ---l"� H C-C::C-CH

2) CH31 3 3

2 catalyst

Alternatively, trans-but-2-ene could be anti-hydroxylated with aqueous peracetic acid. OS04

H 1 Lindlar H3C CH3

\ I C=C

Review the stereochemistry in the solution to Problem 8-35, p. 171 of this Solutions Manual.

I \ H H

alkene must be cis to produce the meso product from syn addition

9-38 5 H2 Pt

meso

o-CH2CH2CH2CH3

� the fact that five equivalents of hydrogen are consumed says that X must have five pi bonds in the above carbon skeleton T�:2�dX

o O O � O O 0 II II I I I' 'I? 'I? \11 I I I I

H-C-CH CH -C-C-H + H-C-C-H + H-C-C-OH + H-C-OH 2 2 '-. � '"'-__ ----. ) � - y�-----

6 carbonyls � 3 alkenes

o-CH=CH-C::CH

Compound X

from C::C

2 carboxylic acids � 1 alkyne

Whether the alkene is cis or trans cannot be determined from these results.

206

Page 214: Solucionario de wade

9-39 Compound Z

ozonolysis � o o o o I I II II I I

CH3(CH2)4-C-H CH3-C-CH2-C-OH HO-C-H \ J \.'-__ ...... ,-__ -.;) y y from alkene from alkyne

Compound Z: CH3(CH2)4CH = C -CH2 -C == C -H Whether the alkene is E or Z cannot be determined from this information.

I CH3

9-40 All four syntheses in this problem begin with the same reaction of benzyl bromide with acetylide ion: N�� _ N��

HC::CH .. HC::C : � CH2Br .. PhCH2-C::CH .. PhCH2-C::C:

(b) PhCH2-C::C:

(c) PhCH2-C::C :

+ Br� allyl bromide

+ Br....../ ethyl bromide

+ Br....../ ethyl bromide

?' benzyl 6 use this helow

� I bromide

PhCH2-C::C� 6-phenylhex-l-en-4- yne

.. PhCH2-C::C� .. \ PdlBaS04

quinoline Lindlar catalyst

Na

Ph'>={ H H

cis-l-phenylpent- 2-ene

PhX trans-l-phen y lpent -2-ene

(d) The diol with the two OH groups on the same side in the Fischer projection is the equivalent of a meso structure, although this one is not meso because the top and bottom group are different. Still, it gives a clue as to its synthesis. The "meso" diol can be formed by either a syn addition to a cis double bond, or by an anti addition to a trans double bond. We saw the same thing in the solution to 9-37 (j).

Ph'>={ OS04

H H cis syn addition

product from part (b)

Ph

CH2Ph H+OH H+OH

CH2CH3 racemic

207

..

H30+ anti addition trans

product from part (c)

Page 215: Solucionario de wade

9-41

(a) 1 ) Sia2BH CH��-C�C-H •

2) H202, HO-

(b) f\ � .. reaction 1 •

� '" -.. / H-O-Et

R-C C-H + :O-Et R�C=\{ -==:-l

(c)

{ H +O�E' H :O�Et � O�Et

R - �-�/ ... • R-�-t � 'c=/

1 \ y\ reaction 2 R/ \H H H H H ! H20: H O-Et H :o� 1 1 1 1 H+ R-C-C-H .. R-C-C-H •

1 1 • • 1 1 H :O�H H20: H OH ilJ H 1

R-C-C-H 1 I I H °

H 1

-----< R -C -C -H ...... f----;.�

1 "+ H :OVH

H 1

H :O�Et 1 I)

R- C-C-H 1 1 H OH ! - HOEt

H } 1 +

R-?-?-H

H :O-H ..

alkyne

R-C CH

alkene R-C=CH2

1 H

RO-•

RO-•

R-C=CH 1 OR

sp2 carbanion

R- C-CH 1 I

2

H OR sp3 carbanion

The closer that electrons are to the nucleus, the more stable. An s orbital is closer to a nucleus than a p orbital i s, as p orbitals are elongated away from the nucleus. An sp2 carbanion is more stable than an sp3 carbanion because the sp2 carbanion has 33% s character and and the electron pair is closer to the positive nucleus than in an sp3 carbanion which is only 25% s character. The sp2 carbanion is easier to form because of its relative stabi l i ty.

208

Page 216: Solucionario de wade

9-42 Diols are made by two reactions from Chapter 8 and revisited in 9-41 (d): either syn-dihydroxylation with Os04' or anti-dihydroxylation via an epoxide using a peroxyacid and water. As this problem says to use inorganic reagents, the solution shown here wil l use OS04.

Recall the stereochemical requirements of syn addition as outlined in this Solutions Manual, p. 17 1, Problem 8-35:

cis-alkene + syn addition � meso cis-alkene + anti addition � racemic (±) trans-alkene + syn addition � racemic (±) trans-alkene + anti addition � meso

Part (a) asks for the synthesis of the meso isomer, so syn addition will have to occur on the cis-alkene. Part (b) will require syn addition to the trans-alkene to give the (±) product. (a)

(b)

HC::C-H HC::C : Na+

alkene must be cis to produce the meso product from syn addition, so reduction is done with Lindlar catalyst to produce the cis alkene

NaNH2

OS04 ..

HC::C-H .. HC::C : Na+

alkene must be trans to produce the (±) product from syn addition, so reduction is done with Na!NH3

.. OS04

f \ H H

209

.. Lindlar catalyst

Na .. NH3

NaNH2 "! Bfi -Zc::cJ-

NaNH2 "! Bfi -Zc::cJ-

Page 217: Solucionario de wade

9-43 This synthesis begins the same as the solution to problem 9-40:

NaNH2 _ � NaNH2 HC::CH .. HC::C: + CH2Br .. PhCH2-C:::CH .. PhCH2-C:::C : 01 benzyl

� bromide

The anion will add across the carbonyl group of the aldehyde:

°

lf H + H30+

PhCH2-C:::C : ---

MCPBA •

210

1 acid-catalyzed dehydration H2S04, Ll

Page 218: Solucionario de wade

CHAPTER 10-STRUCTURE AND SYNTHESIS OF A LCOHOLS

1 0- 1 Please see the note on p. 136 of this Solutions Manual regarding placement of position numbers. (a) 2-phenylpropan-2-ol (d) trans-2-methylcyclohexan-l-ol (" 1 " is optional) (b) 5-bromoheptan-2-ol (e) (E)-2-chloro-3-methy lpent-2-en-l-01 (c) 4-methylcyclohex-3-en-l-ol (" 1 " is optional) (f) (2R,3S)-2-bromohexan-3-ol 10-2 IUPAC name first, then common name. (a) cyclopropanol; cyclopropyl alcohol (b) 2-methylpropan-2-01; t-butyl alcohol

(c) l-cyclobutylpropan-2-01; no common name (d) 3-methylbutan-l-01; isopentyl alcohol

(also isoamyl alcohol)

1 0-3 Only constitutional isomers are requested, not stereoisomers, and only structures with an alcohol group. OH

(a) C3HgO �OH propan-l-ol A propan-2-01 �OH

� OH

�OH (b) C4HlOO � OH

butan-I-ol butan-2-01 2-methylpropan-I-ol 2-methylpropan-2-01

(c) C4HgO has one element of unsaturation, either a double bond or a ring. OH OH � HOi>-d [>-J

cyclobutanol cyclopropylmethanol l-methylcyclopropanol 2-methylcyclopropanol cis or trans

�OH � OH * � HO�

OH but-3-en-I-ol but-3-en-2-01 but-I-en-2-01 but-I-en-I-ol (E or Z)

* �OH

J:H �OH �

OH but-2-en-l-ol but-2-en-2-01 2-methyl- 2-methylprop-2-en-l-01 (E or Z) (E or Z) prop-l-en-I-ol

(d) C3H40 has two elements of unsaturation, so each structure must have either a triple bond, or two double bonds, or a three-membered ring and a double bond. All structures must contain an OH. (In the name, the "e" is dropped from "yne" because it follow by a vowel in "01". ) HO-C::C-CH3 * prop-l-yn-l-ol

HC::C - CH2 OH

H2C=C=C!I * OH

y.0H

'\ZOH prop-2-yn-I-ol propa-l,2-dien-I-ol cycloprop-l-en-I-ol cyc loprop-2-en-l-ol

*These compounds with the OH bonded directly to the carbon-carbon double bond are cal led "enols" or "vinyl alcohols." The structure with OH on a carbon-carbon triple bond is cal led an ynol. These are unstable, although the structures are legitimate.

211

Page 219: Solucionario de wade

1 0-4 (a) 8 , 8-dimethy lnonane-2,7 -diol (b) octane-l,8-diol (c) cis-cyclohex-2-ene-l,4-diol (d) 3 -cyclopentylheptane-2,4-diol (e) trans-cyclobutane-l,3 -diol

1 0-5 There are four structural features to consider when determining solubility in water: 1) molecules with fewer carbons will be more soluble in water (assuming other things being equal); 2) branched or otherwise compact structures are more soluble than l inear structures; 3 ) more hydrogen-bonding groups wil l increase solubi l i ty; 4) an ionic form of a compound will be more soluble in water than the nonionic form.

(a) Cyclohexanol i s more soluble because its alkyl group is more compact than in I-hexanol. (b) 4-Methylphenol is more soluble because its hydrocarbon portion is more compact than in I-heptanol , and phenols form particul arly strong hydrogen bonds with water. (c) 3 -Ethylhexan-3-ol is more soluble because its alkyl portion is more spherical than in octan-2-01. (d) Cyclooctane-l,4-diol is more soluble because it has two OH groups which can hydrogen bond with water, whereas hexan-2-ol has only one OH group. (The ratio of carbons to OH is 4 to 1 in the former compound and 6 to 1 in the latter; the smaller thi s ratio, the more soluble. ) (e) These are enantiomers and wil l have identical solubil ity.

10-6 Dimethylamine molecules can hydrogen bond among themselves so i t takes more energy (higher temperature) to separate them from each other. Trimethylamine has no N-H and cannot hydrogen bond, so it takes less energy to separate these molecules from each other, despite its higher molecular weight.

1 0-7 See Appendix 2 at the back of thi s Solutions Manual for a review of acidity and basicity. (a) Methanol i s more acidic than t-butyl alcohol. The greater the substi tution, the lower the acidi ty. (b) 2-Chloropropan-l-ol is more acidic because the electron-withdrawing chlorine atom is c loser to the OH group than in 3-chloropropan-l-ol . (c) 2,2-Dichloroethanol i s more acidic because two electron-withdrawing chlorine atoms increase acidity more than just the one chlorine in 2-chloroethanol. (d) 2 ,2-Difluoropropan-l-ol i s more acidic because fluorine is more electronegative than chlorine; the stronger the electron-withdrawing group, the more acidic the alcohol.

10-8

most acidic

sulfuric acid » 2-chloroethanol > water > ethanol > t-butyl alcohol

(CH3hCOH

least acidic

> ammonia > hexane

Sulfuric acid is one of the strongest acids known. On the other extreme, alkanes like hexane are the least acidic compounds. The N-H bond in ammonia is less acidic than any O-H bond. Among the four compounds with O-H bonds, the tertiary alcohols are the least acidic. Water i s more acidic than most alcohols inc luding ethanol. However, if a strong electron-withdrawing substituent l i ke chlorine is near the alcohol group, the acidity increases enough so that it is more acidic than water. (Determining exactly where water appears in this l i s t i s the most difficult part. )

10-9 Resonance forms of phenoxide anion show the negati ve charge delocalized onto the ring only at carbons 2, 4, and 6:

: 0:

66' 5::::-'" 3 4

..

: 0: HQ .. .

::::-... .. ..

: 0:

6 .. .. C-H

212

: 0:

OH . ... h-

: 0:

.. 6

Page 220: Solucionario de wade

10-9 continued

Nitro group at position 2

:0 : :0 :

Nitro group at position 3

:0:

:0 :

Nitro group at posi tion 4

:0:

.. ..

. .

:0 : :0 : Nitro at position 2 delocalizes negative charge.

Nitro at position 3 cannot delocalize negative charge at posi tion 2 or 4-no resonance stabil ization.

:0 :

Nitro at position 4 delocal izes negative charge.

- • • .."N, . • -

:0'" + 0 :

Only when the ni tro group i s at one of the negative carbons wi l l the ni tro have a stabi l izing effect (via resonance). Thus, 2-ni trophenol and 4-nitrophenol are substantial ly more acidic than phenol itself, but 3-nitrophenol is only s l ightl y more acidic than phenol (due to the inductive effect).

10-10 OR

A � � (a) Structure A is a phenol because the OH i s bonded to a benzene ring. As a phenol, it wi l l be acidic enough to react with sodium hydroxide to generate a phenoxide ion that w ill be fairly soluble in water. Structure B is a 2° benzyl ic alcohol, not a phenol, not acidic enough to react w ith NaOH. (b) Both of these organic compounds wi l l be soluble in an organic solvent l ike dichloromethane. Shaking this organic solut ion with aqueous sodium hydroxide wi l l ionize the phenol A, making it more polar and water soluble; i t w i l l be extracted from the organic layer i nto the water l ayer, while the alcohol will remain in the organic solvent. Separating these immiscible solvents w i l l separate the original compounds. The alcohol can be retrieved by evaporating the organic solvent. The phenol can be i solated by acidifying the basic aqueous solution and filtering if the phenol i s a solid, or separating the layers if the phenol is a l iquid.

2 13

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1 0- 1 1 The Grignard reaction needs a solvent containing an ether functional group: (b), (t), (g), and (h) are possible solvents. Dimethyl ether, (b), i s a gas at room temperature, however, so i t would have to be l iquefied at low temperature for it to be a useful solvent.

10- 1 2

(a) CH3CH2MgBr (b) �Li + LiI (c) F-D-MgBr

Li

(d) � + LiCI

1 0- 1 3 Any of three halides-chloride, bromide, iodide, but not fluoride-can be used. Ether is the typical solvent for Grignard reactions.

o MgCl

H \

(a) + C=O I H

�MgBr

H \

(b) + C=O I

H

o- MgI H

\ (c) + C=O I

H

Note: the alternative arrow symbolism could also be used, where the two steps are numbered around one arrow:

1 ) ether ...

2) H30+ OK

ether H30+ -- ..

ether H30+ -- ..

ether H30+ -- ...

OCH2OH

�OH

o-CH2OH

NO! Me BAD! This means that water is present with ether during the Grignard reaction.

, 'ether " , '

'ce ' •

}'fo+ , 3 ' , "

1 0- 1 4 Any of three halides-chloride, bromide, iodide, but not fluoride-can be used. Grignard reactions are always performed in ether solvent; ether is not shown here.

(a) two methods

IMgCl Hy H30+ Ib: Where two methods can be + -- ... used to form the target

0 compound, the newly

InH H30+ Ib: formed bond is shown in + CH3MgI bold. -- ...

0

(b) two methods OH

o MgB, 0 H30+ � H� 1.& + -- ..

0 OH �H H30+

~ + IMg-......./ -- ..

214

Page 222: Solucionario de wade

1 0- 1 4 continued

(c) o MgCI +

o

H� OH

dD 10- 1 5 Grignard reactions are always performed in ether. Here, the ether i s not shown.

(a) Any of the three bonds shown in bold can be formed by adding a Grignard reagent across a ketone.

(i) ( i i ) 0 (i i i ) CH3C:2MgBr � CH3C:2CH2MgBr (i ) ,\ ?:;-

01 '-("') O� + PhMgBr

Ph

� Ph H30+

�O

Ph

Ph I

( Ph 111

( i i)

(b) Ph :-MgBr + "-- 'c = 0 -- � Ph -C-OH

(d) two methods

�MgCI

I Ph

\ Cy H30+ + �>= O -- �

Cy

Cy = cyclohexy l

10- 16 :C!0 II :?) CH3 -C� MgBr --- CH -C-CI 3

I � Ph

(This is just a nucleophi lic substi tution where CI is the leaving group. The unusual feature is that it occurs at a carbonyl carbon.)

OH I

CH3-T-Ph

Ph

215

I Ph

+ Cy-MgCl

:0 - CI- I I

� �-Ph

Ph-MgBr l �:o:

work-up step

I CH -C-Ph

3 I Ph

ketone intermediate

Page 223: Solucionario de wade

1 0- 17 Acid chlorides or esters wi ll work as starting materials in these reactions. The typical solvent for Grignard reactions i s ether; i t i s not shown here.

0 H30+ I I (a) Ph-C-Cl + 2 PhMgBr � •

(b) JyOCH] + 2 CH3CH2MgI H30+

0

(c)

o I I

Ph-C-Cl + 2 o MgCI

10-18 :qD I I ./'0... . MgB r

(a) H-C� ---

o

OH I � C � I

H

. .

:?j H-C-OEt ��

II H30+ (b) ( i ) HC-OEt + 2 CH3CH2MgBr � •

0

II < }MgBr H30+

(i i ) HC-OEt + 2 � •

0

Ph I

Ph-C-OH I Ph

~ OH

� OH

aldehyde intermediate

II � MgBr � H30+

( i i i ) HC-OEt + 2 • � OR

10-19 Ether i s the typical solvent in Grignard reactions.

(a) < }MgBr + o

U �OR

V

216

Page 224: Solucionario de wade

10- 19 continued

(b) �MgCl

0

U +

�gl 0 (c) + U 10-20

_ ?'"O (a) HC C : �D (b)

H30+ �OH ---- �

H30+ �OH ---- �

10-2 1 There are more than one synthetic route to each structure; the ones shown here are representative. The new bonds formed are shown here in bold. Your answers may be different and still be correct.

(a)

(b)

(c)

�Br Li CuI (�CULi

�Br � - ---- �

2

� Br Li CuI ( �CULi

( ( U ----- �

2

�Br Li CuI (�CULi

()l CJ" ----- �

2

Alternatively , coupling l i th ium dicyclohexy!cuprate with I-bromobutane would also work. As this mechanism i s not a typical SN2, i t i s not as susceptible to steric hindrance l ike acetylide ion substi tution or a s imi lar SN2 reaction.

(d) �Br Li CuI

----- (�CULi 2

10-22 These reactions are acid-base reactions in which an acidic proton (or deuteron) is transferred to a basic carbon in either a Grignard reagent or an alkyllithium.

(a) CH3D + Mg(OD)I (b) CH3CH2CH2CH3 + LiOCH2CH3

(d) 0 217

o I I

+ CH3C-OLi

Page 225: Solucionario de wade

10-23 Grignard reagents are incompatible with acidic hydrogens and with electrophi l ic, polarized multiple bonds like C=O, N02, etc. (a) As Grignard reagent is formed, i t would instantaneously be protonated by the N-H present in other molecules of the same substance. (b) As Grignard reagent is formed, it would immediately attack the ester functional group present in other molecules of the same substance. (c) Care must be taken in how reagents are written above and below arrows. If reagents are numbered " 1. . .. 2. ... etc.", it means they are added in separate steps, the same as writing reagents over separate arrows. If reagents wri tten around an arrow are not numbered, it means they are added al l at once in the same mixture. In thi s problem, the ketone is added in the presence of aqueous acid. The acid will immediately proton ate and destroy the Grignard reagent before reaction with the ketone can occur. (d) The ethyl Grignard reagent wi l l be immediately protonated and consumed by the OH. Thi s reaction could be made to work, however, by adding two equivalents of ethyl Grignard reagent-the first to consume the OH proton, the second to add across the ketone. Aqueous acid wil l then protonate both oxygens.

OH 10-24 Sodium borohydride does not reduce esters.

(a) CH3(CH2) gCH20H (b) o

no reaction (c) no reaction (PhCOO- before acid work-up)

(I) c,tcc 1 °lfOH

(d) 6 (e) HO· Y Y �OCH3

OH

OH

6 10-25 Lithium aluminum hydride reduces esters as well as other carbonyl groups.

(a) CH3(CH2) 8CH20H (b) CH3CH2CH20H + HOCH3 (c) PhCH20H (d)

(e) HO�OH y + HOCH3

10-26 (a)

OH

� H

o OR �OH

o

OR OR �

o

(f)

NaBH4 •

CH30H

1) LiAIH4 •

2) H30+

1) LiAIH4 •

2) H30+

(b) � NaBH4 •

CH30H

HO�OH

HO�

�OH

OH

� 218

1) LiAlH4 OR 2) H30+

1) LiAIH4 OR •

2) H30+

Page 226: Solucionario de wade

1 0-26 continued

(c) ~ NaBH4 ~ 1) LiAIH4 .. OR ..

CH30H 2) H30+ 0 OH

(d) WO NaBH4 C�X:f LiAlH4 will NOT give the desired .. product. LiAIH4 will also reduce the CH30H

ester in addition to the ketone.

0 OH

10-27 Approximate pKa values are shown below each compound. Refer to text Tables 1-5, 9-2, and 1 0-3, and Appendix 5 at the back of the text. CH3S03H > CH3COOH > CH3SH > CH30H > CH3C:: CH > CH3NH2 > CH3CH3

< 0 4.74 "" 10.5 15.5 25 "" 35 50

most acidic

10-28 (a) 4-methylpentane-2-thiol (b) (Z)-2,3-dimethylpent-2-ene- l -thiol (" 1" is optional) (c) cyclohex-2-ene- l -thiol (" 1" is optional)

10-29 �

OR

HBr � Br ROOR

NBS ----l .. � � Br �oc (see Problem 8-2)

NaSH ..

NaSH ..

� SH

� SH

10-30 Please refer to solution 1-20, page 12 of this Solutions Manual.

1 0-3 1 (a) 5-methyl-4-n-propylheptan-2-01; 2° (b) 4-( I-bromoethyl)heptan-3-01; 2° (c) (E)-4,5-dimethylhex-3 -en-l-01; 1 ° (d) 3-bromocyclohex-3-en- l -01; 2° (" 1" is optional) (e) cis-4-chlorocyclohex-2-en-l-ol; 2° (" 1" is optional) (f) 6-chloro-3-phenyloctan-3-01; 3° (g) (l-cyclopentenyl)methanol; 1 °

1 0-32

(a) 4-chloro- l -phenylhexane- l ,5-diol (b) trans-cyclohexane-l,2-diol (c) 3-nitrophenol (d) 4-bromo-2-chlorophenol

219

least acidic

3-methylbutane-I-thiol

2-butene-l-thiol (but-2-ene-l-thiol)

Page 227: Solucionario de wade

1 0-33 Q (a) I 0 OC-OH

-6 OH

(f) &OH

10-34

OH (b)

OH

(g) ¢ (h)

I

OH OH OH (c) 6 (d)� (e) �H

HO H SH U) CH3S-SCH3

H� (i) 6 OH

(k)

~ (a) Hexan- l -01 will boil at a higher temperature as it is less branched than 3,3-dimethylbutan - l -01. (b) Hexan-2-01 will boi l at a higher temperature because its molecules hydrogen bond with each other, whereas molecules of hexan-2-one have no intermolecular hydrogen bonding. (c) Hexane- l ,5-diol wil l boil at a higher temperature as it has two OH groups for hydrogen bonding. Hexan-2-ol has on ly one group for hydrogen bonding. (d) Hexan-2-ol wi l l boil at a higher temperature because it has a higher molecular weight than pentan-2-01. All other structural features of the two molecules are the same, so they should have the same

intermolecular forces.

1 0-35 (a) 3-Chlorophenol is more acidic than cyclopentanol. In general, phenols are many orders of magnitude more acidic than alcohols. (b) 2-Chlorocyclohexanol i s s l ightly more acidic than cyclohexanol; the proximity of the e lectronegati ve chlorine to the OH increases i ts acidity. (c) CyciohexanecarboxyJic acid i s more acidic than cyclohexanol. In general , carboxyl ic acids are many orders of magnitude more acidic than alcohols. (d) 2,2-Dichlorobutan- l -ol is more acidic than butan-I-ol because of the two electron-withdrawing substituents near the acidic functional group.

10-36 (a) Propan-2-ol is the most soluble in water as it has the fewest carbons and the most branching. (b) Cyclohexane- l ,2-diol is the most soluble as it has two OH groups for hydrogen bonding. Cyclohexanol has only one OH group; chlorocyc lohexane cannot hydrogen bond and is the least soluble. (c) Cyclohexanol is the most soluble as it can hydrogen bond. Chlorocyclohexane cannot hydrogen bond, and 4-methylcyclohexanol has the added hydrophobic methyl group, decreasing its water solubil ity.

1 0-37 OH

(a) � Hg(OAch NaBH4 � • .-

H2O

(b) U BH3' THF H202 ({Nl • • HO- " "

OH

220

Page 228: Solucionario de wade

1 0-37 continued OH (c)� BH3 - THF H202 ~ � �

HO-OH

(d)� Hg(OAch NaBH4 � � H2O

..

10-38 OH (a)

VOH (b)� (e) JyPh

OH

(d) This problem confuses a lot of people. When a Grignard reagent is added to a compound that has an OH group, the first thing that happens is that the Grignard reacts by removing the H+ from the 0-.

o

00H+

o

ether 0 1 CH3Mg1 �

(o�ly one equivalent of 0- +MgI + CH4

Gngnard reagent added! � If a second equivalent (or excess) of Grignard reagent is added

If no more Grignard reagent H 0+ CH3Mg1 before hydrolysis, then it will add is added before acid 3 ether at the ketone. Acid hydrolysis hydrolysis, then the starting a will give the diol. material is recovered.

00H

(

Mg

' 03 Hp'.

HX

H3

0- +MgI UOH Ph Ph

(e) Q--'OH (f) Ph+OH Ph

Ph (g)Ph60H (h) crtu

OH HO�OCH3 HO�OH � (i) V 0 U) V (k)�

(m) (n) (0)

OH (l)d:) HO OH HO

221

Page 229: Solucionario de wade

10-39 All Grignard reactions are run in ether solvent. Two arrows are shown indicating that the Grignard reaction is allowed to proceed, and then in a second step, dilute aqueous acid is added.

o H30+ ?H

(a) �H + BrMg-..../ --- .. �

(b)

(c)

(d)

(e)

(f)

(g)

�Br Mg CH20 -- .. ether

0 BrMg-Q

)lH + ---

0

�OH

Hp+ OH ..

~ O

B' Mg U H30+ (YbH -- .. .. ether

OR' Mg CH20 H30+ V

OH ' h -- .. ..

ether o

Oil COCH2CH3 o �H

OH

~ (h) 0 �

1 0-40

(a)

(b)

+ XMgC C-CH3 �------���------� ( - '\ RMgX + H-C=C-CH3

any Grignard reagent where R is alkyl or alkenyl

BH3 - THF ..

Mg CH20 H30+

Technically, XMgC=CCH3 is a Grignard reagent because it is an organometallic compound of magnesium. However, it is not made in the usual fashion; it is made by deprotonating the terminal alkyne as shown.

Ph _ /'.... ........", 'CI -- .. ..

H Ph�OH

ether

222

Page 230: Solucionario de wade

1 0-40 continued

(c)

(e)

°

6 o �OEt o o

OH

6 o

(d) 6

OH �OEt o OH

(f) �OEt H30+ ..

�OH o 10-4 1

(a) VMgBr + CH,O _e_th_e--;r ..

(b) V

(d)

(e)

NaOH .. o

Mg � � ether

r-==\. + NaSH ---f '-- Br

HO�

�OH

V �

OH

V

�OH

�SH

1 eq. H2 .. Pt

°

6

(f) � Br� � (' 1 \ CULi + � Br I :"� j . � 1 0-42 The position of the equilibrium can be determined by the strength of the acids or the bases. The stronger acid and stronger base will always react to give the weaker acid and base, so the side of the equation with the weaker acid and base will be favored at equilibrium. See Appendix 2 in this Solutions Manual for a review of acidity.

(a) CH3CH20� + < '}-OH CH3CH20H + < '}-o-

stronger base

stronger acid

weaker acid

223

weaker base

products favored

Page 231: Solucionario de wade

1 0-42 continued (b) KOH + Cl-Q-OH H20 + Cl-Q-0- K'

Cl

(c)

(d)

(e)

Cl stronger stronger base ac id

OH

c6 + CH3O-h h stronger stronger acid base

aOH+ KOH

weaker weaker base acid reactants favored

(CH3hCO-stronger base

stronger base

+ CH3CH2OH stronger acid

stronger acid

(g) KOH + CH3CH20H weaker weaker base acid reactants favored

1 0-43

(a)�H

° OR �OH

OR °

�OR

o

weaker weaker acid base

products favored

weaker weaker base acid

products favored

H20 +

stronger acid stronger

base

(CH3hCOH + weaker ac id

CH3CH20-weaker base

products favored

(CH3hCOH + HO-weaker weaker acid base

products favored

stronger acid

1 ) LiAJH4 •

1) LiAJH4 •

2) H30+

stronger base

�OH

224

I ) LiAIH4 OR •

Page 232: Solucionario de wade

1 0-43 continued 0

(b) � 0

(C) � 0

OH NaBH4 � .. CH30H

OH NaBH, .

~ CH30H I ,&

OH

1 ) LiAlH4 OR ...

2) H30+

1 ) LiAlH4 OR ...

2) H30+

(d) OO NaBH4

00 1 ) LiAlH4

(e) ('

(f) /

... CH30H

OH r---<

'\ NaBH4 ...

CH30H OEt � OEt

0 0

OH " 1 ) LiAIH',. Q 2) H30+ OEt OH

II 0

OR ... 2) H30+

+ HOCH2CH3

10-44 The goal is to synthesize the target compound (boxed) from starting materials of six carbons or fewer. The product has 1 2 carbons, so the logical "disconnection" in working backwards is two six carbon fragments which could be joined in a Grignard reaction. The best way to make epoxides is from the double bond, and double bonds are made from alcohols which are the products of Grignard reactions.

I g ... I O Br M o MgBr

,& ether h

o +

H�

OH H30+ cro -- � I ,&

225

y H2S04 VD t MCPBA v<b target

Page 233: Solucionario de wade

10-45 All steps are reversible. H i · · �} HO . � ><H � HO � I +

XH HO· H-O • •

- H20 I · A H20 : :0 :

A .. ./C .. .. .. / + ...........

1 0-46 The symbol H-B represents a generic acid, where B- is the conjugate base.

(a)

(b)

� { H :OH OH n / H�

H-� H-t-C� H H

OH

H

" H- B .. r? ,

OH

�� H

lIfv0} B :- -HH 0: � HT\..

--- H

H H H H

+ H - B

• • + �

� cr� H r�cro:

H + H-B

(C)Q(�H_ I \.-:O- H H • • Q(0 :

. . I �

H

ao: >-

• • ...... H H C -

J �'H a:: 1 0-47

6' A

Mg .. ether

6MgBr 6° -...:::::

°2°

-...::::: ..

B c

I )� 1 2) H,o' OH

D

H2S04 ..

/).

E

226

ex�cess H2 Pt

2 Br2 ..

� H

. . + :O - H

isobutylcyclohexane

F

Page 234: Solucionario de wade

10-48 This mechanism is similar to cleavage of the epoxide in ethylene oxide by Grignard reagents . The driving force for the reaction is relief of ring strain in the 4-membered cyclic ether, which is why it will undergo a Grignard reaction whereas most other ethers will not.

R"MgX + .D . ---�U +

MgX

10-49 When mixtures of isomers can result, only the major product is shown.

(yO 1 ) CH3Mg1 CH3 HZS04 o-CH3 Hz .. UCH3 ... OOH A '" 2) H3O+ Pt

A B t KM�04 t Hz�04 (J=O �CH3

" "

G D OH

t Hz�04 t 1) CH3Mg1 2) H3O+ OH (J-O Q" " " 1 1 1 1 1 1 0

F C t I ) Mg, ether t PhC03H 2) cyclopentanone 3) H3O+

U Br HEr 0 ..

E

1 0-50 The most important reactant in the deskunking mixture is hydrogen peroxide. Thiols are oxidized to structures having one, two, or three oxygens on the sulfur; all of these functional groups are acidic. The sodium bicarbonate is basic enough to ionize these acids, making them water soluble where the soap can wash them away.

sulfenic acids

� SH HzOz � S

HzOz .. ...

3-methylbutane- l -thiol I OH

�SH HzOz �S HzOz .. ...

I

2-butene- l -thiol OH (but-2-ene- l -thiol) 227

sulfinic acids

�SPH HzOz

... I I °

OH �S' HzOz

... I I °

sulfonic acids �O I I S -OH I I °

° �I I

� S -OH I I °

Page 235: Solucionario de wade

1 0-5 1

Mg � ----.. Br ether t KOH, H20 �OH A +Na �O- Na+ B + /' Br

�O� C

�MgBr D

1 )�0

H ..

2) H30+ � E OH

H2S04, fj, + � 1) 03 F

2)� Br2 �

H Br

�O G � +

O� H

Br KOH, fj, �

H �C::C� plus other alkynes

1) NaNH2, 1 50°C + 2) H20 O� I)Sia2BH /"'.. /"'.. /"...

1 .. H -C :: C---"" ........., ........., ....... H I 7-

J Na+ -C =C�

1 o

� CH2CH3

HO-C - C = C� I CH3

K

228

non- l -yne

2 HBr + Br

� Br

L

Page 236: Solucionario de wade

CHAPTER l l-REACTIONS OF ALCOHOLS

1 1 - 1 (a) both reactions are oxidations (b) oxidation, oxidation, reduction, oxidation (c) one carbon is oxidized and one carbon is reduced-no net change (d) reduction: C-O is replaced by C-H (e) neither oxidation nor reduction-the C still has two bonds to 0 (f) oxidation (addition of X2) (g) neither oxidation nor reduction (addition of HX) (h) neither oxidation nor reduction (elimination of H20) (i) oxidation: adding an 0 to each carbon of the double bond (j ) the first reaction is oxidation as a new C-O bond is formed to each carbon of the alkene ; the second reaction is neither oxidation nor reduction, as H20 is added to the epoxide, and each carbon sti l l has one bond to oxygen (k) neither oxidation nor reduction: H-B is added in the first reaction, and B is replaced by 0 in the second reaction; overall, only H and OH are added, so there is no net oxidation nor reduction. (Note that in functional groups involving two carbons like alkenes or alkynes, both carbons have to be oxidized or reduced before the net change to the functional group is classified as oxidation or reduction. )

1 1-2 0 OH 0 (a) 6 H2Cr04 6 pcc 6 .. �

(b) H2Cr04 oH PCC no reaction .. � no reaction

0 OH 0

(c) cY H2Cr04 cY PCC cY .. �

0

(d) H2Cr04 6 pcc no reaction .. � no reaction

(e) H2Cr04 0 PCC no reaction .. � no reaction

H2Cr04 0 PCC (f) I I no reaction .. H3C -C -OH � no reaction

0 H2Cr04 PCC 0 I I I I (g) H3C -C - OH .. CH3CH2OH � H3C -C -H

0 H2Cr04 0 PCC I I I I (h) H3C -C -OH .. H3C -C -H � no reaction 229

Page 237: Solucionario de wade

1 1 -3 (a) A 1 0 alcohol loses two hydrogens when transformed to the aldehyde, and a 20 alcohol loses one hydrogen in fonning a ketone; each alcohol is oxidized. DMSO loses an oxygen from the sulfur; it is clearly reduced. (If you got those two, you did the problem correctly.) To be rigorous, oxalyl chloride undergoes a disproportionation reaction: one C is oxidized to CO2 and the other C is reduced to CO; however, the net effect on the elements in oxalyl chloride is "no change" . (b) Text section 8- 1 5B shows that dimethyl sulfide reduces an ozonide. In the process, dimethyl sulfide i s oxidized to DMSO. 1 1 -4 (a) Dehydrogenation does not occur at 250 C-either: 1) there is a high kinetic barrier (a high activation energy) for this reaction, or 2) it is thermodynamically unfavorable, with [)'G > O. The latter possibility is supported by the fact that the reverse reaction (catalytic hydrogenation of a carbonyl) is spontaneous at 250 C (see text section 10- 1 1 C) and therefore has [)'G < O. This makes the question of kinetics academic-a reaction that cannot proceed must be uselessly slow. (b) and (c) Kinetics will improve with increasing temperature for virtually all reactions , so both the hydrogenation and dehydrogenation reactions will go faster. In this case, however, the question is how to favor the dehydrogenation reaction. The answer is that thermodynamics will favor this reaction as the temperature is raised. The key is the fundamental thermodynamic equation [)'G = Mi - TflS. We can estimate that Mi > 0 since the product ketone plus hydrogen is less stable than the starting alcohol. Also. [)'5 > 0 since one molecule is converted to two: therefore, -T [)'5 < O. At low temperature (250 C), Mi dominates because T is so small, so [)'G > O. At a high enough temperature, the -T[)'5 term will begin to overwhelm Mi, and [)'G will become negative . For the reaction in question, this must be the case at 3000 C.

1 1 -5

(a)

PCC

AND HO-

(b) all three reagents give the same ketone product with a secondary alcohol

(c) OH - OH

--.....,.� CH3(CH2)4 Y"­

°

Na2Cr207 •

H2S04 AND PCC •

(d) all three reagents give no reaction with a tertiary alcohol ><H

� V --.- NR (no reaction)

230

Page 238: Solucionario de wade

Note to the student: For simplicity, this book wil l use these standard laboratory methods of oxidation: -PCC (pyridinium chlorochromate) to oxidize 1 ° alcohols to aldehydes; -H2Cr04 (chromic acid) to oxidize 1 ° alcohols to carboxylic acids; -cr03, H2S04, acetone (Jones reagent) to oxidize 2° alcohols to ketones.

Understand that other choices are legitimate ; for example, Swem oxidation works about as well as PCC in the preparation of aldehydes, and Collins reagent or PCC will oxidize a 2° alcohol to a ketone as well as chromic acid. If you have a question about the appropriateness of a reagent you choose, consult the table in the text before Problem 1 1 -2 .

1 1 -6 0

(a) �OH PCC �H •

0 (b) �OH

H2Cr04 �OH •

� cr03 � (c) • H2SO4

OH acetone 0

0 (d) �OH

PCC �H •

0

�OH H2Cr04 �OH (e) ..

(f) oH H2SO4 6 1 ) BH3 • THF (yoH cr03 &0 • • •

Ll 2) H202, HO- H2SO4 - H2O acetone

1 1 -7 A chronic alcoholic has induced more ADH enzyme to be present to handle large amounts of imbibed ethanol, so requires more ethanol "antidote" molecules to act as a competitive inhibitor to "tie up" the extra enzyme molecules.

1 1 -8 OH OH [ 0 ]

..

o 0 I I I I

CH3-C-CH pyruvaldehyde

[ 0 ] o 0 I I I I

CH3-C-COH pyruvic acid

pyruvic acid is a normal metabolite in the breakdown of glucose ("blood sugar")

231

Page 239: Solucionario de wade

1 1 -9 From this problem on , "Ts" will refer to the " tosyl" or "p-toluenesulfony l " group:

�� Ts � -��CH3 o

CH3 I

CH3 I

(a) CH3CH2 - OTs + KO - C - CH I

3 � CH3CH20 - C -CH3 I

+ KOTs

CH3 CH3 (E2 is also possible wi th this hindered base; the product would be ethylene, CH2=CH2)

(b) �OTS + NaI � �I + NaOTs

TsO H H

� + NaCN � � + NaOTs inversion--SN2 (c) "

R S

+ -cST, c5H3 0TS NH2 excess 6 + +NH, OTs (d)

NH3 NH3 .. ..

-

�OTs (e) + Na+ :C = CH

1 1 - 1 0

(a) �OH TsCI �OTs .. pyridine

(b) �OH TsCI �OTs • pyridine

(c) �OH TsCi �OTs .. pyridine

(d) �OH TsCI �OTs • pyridine

1 1 - 1 1

---l.� �C=CH + NaOTs

NaBr � Br •

excess NH3 � NH2 •

NaOCH2CH3 �O-...../ •

KCN �CN •

(a)Q- TsCl CH20H - d- "

pyn me Q-CH20TS

o OR Q-CH20- �-o-CH1

o 232

Page 240: Solucionario de wade

1 1 - 1 1 continued

(b) 0-CH20Ts

(c)

o� n �o o H - Br

V ·

1 1 - 1 3

major minor

(d)

Pt

H 6�OBO O � o r o + 0 0

----

o 0

:Br :

Br

o t

+ ��H2 --=-

V "-- :B�: � + H20 V Sr o 0

CH CH CH CH3 � n H C - t- OH H - C� 3

I 0 0

I �+ H C - C - OH 3

I 0 0 2

_ H20 + I 3 :C l : I

3 ----'l.� H3C - C _ +-- H3C - C - Cl

I � I CH3 CH3

1 1 - 1 4 The two standard quali tative tests are :

CH3 CH3

carbocation intermediate

1 ) chromic acid-distinguishes 3 ° alcohol from either 1 ° or 2°

R R +OH

R

H2Cr04 (orange)

• no reaction (stays orange)

R(or H) R +OH

H

233

H2Cr04 (orange)

R(or H) I

R - C = O + erJ+ blue-green

Page 241: Solucionario de wade

1 1 - 14 continued

2) Lucas test-<listinguishes 1 ° from 2° from 3° alcohol by the rate of reaction R R

+ ZnCl2 + R OH + HCl .. R Cl + H20 insoluble-" c loudy I I in < I minute R R

soluble R ZnCl2 R

2° R +OH + HCl .. R+CI + H2O insoluble-"cloudy" in 1 -5 minutes H H

soluble H ZnCl2 H

1 ° R +OH + HCl .. R +CI + H2O insoluble-"cloudy" in > 6 minutes H H (no observable reaction at room temp. )

soluble - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - . - - - - - - - - - - - - - - _ . .

(a)

(b)

(c)

(d)

(e)

Lucas: H2Cr04:

Lucas : H2Cr04:

Lucas: H2Cr04:

Lucas: H2Cr04:

Lucas : H2Cr04:

OH

A c loudy in 1 -5 min . i mmediate blue-green

OH

A cloudy in 1 - 5 min . immediate blue-green

�OH

cloudy in < 1 min . no reaction-stays orange

o

� no reaction no reaction-stays orange

no reaction cloudy in 1 -5 min . DOES NOT DISTINGUISH-immediate blue-green for both

�OH �OH

cloudy in < 1 min. * * no reaction DOES NOT DISTINGUISH-immediate blue-green for both � +OH

( **Remember that allylic catioIlS are reson�nce­stabilized and are about as stable as 1° cations. Thus, they will react as fast as 3u i n the Lucas test,

even though they may be 1 °. Be careful to notice

subtle but important structural features !)

no reaction cloudy in < 1 min. DOES NOT DISTINGUISH-stays orange for both

1 °

methyl 3° � • • -

shift + jI' ,CH3 :Br : ---1 .. - C I �

/ 1 .... CH2

Br � Even though 1 ° , the neopentyl carbon is hindered to backside attack, so SN2 cannot occur easi ly . Instead, an SN 1 mechanism occurs , with rearrangement.

234

Page 242: Solucionario de wade

1 1 - 1 7

CH3

0" o�

, • • ZnCl2 " H •

1 1 - 1 8

: C l : approach from above

H

H

a= CH3

C - CH + 3

This 3° carbocation is planar at the C+ so that the Cl- can approach from the top or bottom gi ving both the cis and trans isomers .

- approach : Cl : from

below

0= CH3

;: Cl CH3

�H CH (from Hel) H

3 3 ° I .� • • - CH CI G" · H + C ' :Cl : O'

C O " --- + 2° - •

hydride shift

3 �OH + PBrJ -� 3 � Br + P(OHh

1 1 - 1 9 OH

(a) 6 SOCl2

-CH3

OH

(b) 6 TsC I pyridine

-CH3

Cl

6 OTs

6 retention

NaCI

SN2-inversion

235

Cl

o Another possible answer would be to use PCl3 .

Page 243: Solucionario de wade

1 1 -20

(a) H D

1 Cl O O

O :�\S • •

• • I =0

I Cl · · --

H Cl D +01 - 1 ,

• • -

X S-O· 0

·(CI . .. -

H

+ I )�I D

OO- S=O. . . .

I � - HCI al

- . . I :o - s = o : · 0 . .

j + { r. Cl

A + � } D

O" -�- S=R:

�tH· � 0 - I allylic !

S02 +

:Cl :

D 00+ bel

. . + :C l :

(b) The key is that the intermediate carbocation is al ly lic, very stable and relatively long-lived. It can therefore escape the ion pair and become a "free carbocation" . The nucleophilic chloride can attack any carbon with positive charge, not just the one closest. Since two carbons have partial positive charge, two products result.

1 1 -2 1

(a) �OH HCl no reaction unless heated, then �CI ... ZnCI2 HEr

�Br �

PBr3 �Br �

P � I -

12 SOCI2

�CI �

236

Page 244: Solucionario de wade

1 1 -2 1 continued

(b) IOH HCl �

ZnCI2 HEr -

PBr3 -

P -12 SOCl2 -

(d) �OH

1 °, neopentyl

ICl (C)�OH HCl �Cl ..

ZnCl2

IBr � Br HBr �

I Br

PBr3 � Br .. II (poor reaction on 3°)

IC l P �I .. 12

(poor reaction on 3°)

SOCI2 �C l -

(poor reaction on }O)

CI HCl •

ZnC I2 no reaction unless heated, then

SN I -rearrangement � HBr

PBr3 -

P -12

�Br SN2-minor (hindered)

� Br

�I SOCIl. �Cl

237

Br

+ � SN l-rearrangemen t

Page 245: Solucionario de wade

1 1 -2 1 continued (e)

1 1 -22 OH

(a) � (b) �OH

OH (c) �

JxOH

(d) U (e)

HCI ..

ZnClz

HEr

p Iz

SOClz

'eYCI + SN 1 ; carbocation intermediate

can be attacked from either side by chloride

�" ,Br

U SN2 with inversion of configuration

� ,Br

U "

SN2 with inversion of configuration

� " I

U "

SN2 with inversion of configuration

�CI

U retention of configuration

+

major rrunor

+ major (cis + trans) minor

HzS04 , � + major (cis + trans) mmor

major minor

HZS04 , �

major minor trace

238

Page 246: Solucionario de wade

11-2 3

H I (y O:

V .. ° � II� P, '

Cl'l Cl C l

� H °

.. -CI+ II N

• 0�- f-Cl �-.�

Cl V ..

goo d leavi ng gro up � °

Olo-�-CI • • I ""i Cl +

H

E2 !O H I N+

o

cy clohex ene was fonn ed 0 without a car boca tion + i n ter medi a te

° • • II : O-P-CI • • I

Cl

+

H 1+ N

o 11 -24

this produc t ca n reac t with two more a lcohols to b eco me leavi ng groups in the

E2 el imi na tion

B oth mecha nis ms b egin with protona tion of the oxy gen.

H H �H+ H H H 1 1 1+ 1 1 • .

-H .. H-C-C-O. H-C-C-O-H 1 1 •

H H 1 1 • •

H H

O ne mecha nis m inv olv es a nother molecule of etha nol a cting as a bas e, giv ing elimi na tion .

H H H 1 1 1+ • H-T.-A T�·

-H i{qCH2CH3 H�

H H \ / C=C + H20 / \

H H

The other mecha nis m inv olv es a nother molec ule of etha nol a cting as a nuc leophi le, givi ng s ubs ti tution .

� H H H H • • 1 1 1+ + I) ItqCH2CH3 H-C-C-O-H .. CH CH -O-CH CH .. CH CH - O-CH CH

HOCH2CH3 1 �U· · · 3 2 • •

2 3 3 2 2 3 H H ··

11-2 5 A n equi mola r mi xtur e of metha nol a nd etha nol would produc e a ll th ree possib le ethers . T he diffi culty in s epa ra ting thes e compounds would pr eclude this method from b ein g a prac tica l route to a ny one of them. T his method is prac tica l o nly for s ymmetri c ethers , tha t is , where bo th a lky l groups ar e identica l .

H+ CH3CH20H + HOCH3 .. H20 + CH3CH20CH3 + CH30CH3 + CH3CHzOCH2CH3 tl.

239

Page 247: Solucionario de wade

11-2 6 � H H�' • • H+ ,"",1+ -H20

CH -O-H • CH -O-H • 3 • • �3 • • • • �I+ CH39.H

CH -O-CH • CH -O-CH 3 • • 3 3 3

11-27 H

CH -O-H 3 • •

• • � � I + (a) X�H '

H+ X�H �

VH • VH -H20

H + 1 y---..

H20: aC#"--.. H

• •

• &H (b) OCH3

H

--­H+

aO-H C/··

H �

H

+ • •

a:H H l H20: -

era (C�

OCH2-�H

HOCH2 '-"h!'

� H+ � ----;.� ..

-H20

H H

H 0. H-OH +

9.CH3 a:CHl H H20:

H H

d:O�OCH3 (cj:OH

OCH3 1+ • • -CH30H H H+ .. -

H H

1° +

H(5

H H

H H,I/.H C) hydride + b H shift_ �

H

� H20) +1 � � (1: �L)

H

240 �+6

Page 248: Solucionario de wade

11 -2 7 conti nued

(d) a�H,-w Ol+-H ___ il

H-

zO""'''

CH3 1 H(!j

3C CH�CH3 H3c;"CoCH3

'--C+ _ ) 7"

H20: .

H2C

11 -2 8 (a) n n -H20 n H3C++0CH, w· H3C-T-rf-CH3 · H3C-t-1-CH3

HO :OH HO �OH HO " " 1+

M ethy l s hift

H'C�CH� HO

A lky l s hift- ri ng contraction

H

Q-�H: Q-CH3 -QCHl H�: CH3 H

'" R + CH3 0 CH3

"- "" H20:

rmg contraction

-W----H3C C-CH +

3

HO

H,C, +Q 4 ·H'C� C CH3 CH3 1 '" -0

H3C H20: CH3

H-O: H-O+ " "

241

o

Page 249: Solucionario de wade

1 1-2 8 co ntinued (b)

CX;;

oH.f H+ �

OH ..

30 and doubly

cYf

oH H - H ° O<

0H benzylic carbocation 1+ 2 �

G:O-H \ + • • �C

/\ ���; Ph Ph Ph Ph Ph Ph �nS io n

a

o ··�a

+OH

P h

11 -2 9

H20: U

Ph � P h

Ph

.-------

O-H + / • •

a/�-H

- +

P h

Ph

+O-H ij� OH0

OH rIng � H+ f'-.-- / H expa ns io n � /H • V-C, + 1 •

C==C �C-C-H

[c � �-CH3 1 C [�-CH3 / " / 1 H H H H H 1

H S imi la r to the pina co l rea rra ngement, this mecha nis m invo lves a ca rbo ca tio n next to a n a lco ho l , with rea rra ngement to a p ro to na ted ca rbo ny l . Relief of

so me ri ng s tra in in the cyclo propa ne is a n a dded a dva nta ge of th e rea rran gement.

� HS�4-c:f

CH]

11 -30 ° II

(a ) 2 H3C",C'H (b)

° ° (c) if +H� (d)

11 -3 1 ° II

(a) Cl-C -CH2CH2CH3 + HOCH2CHzCH3

(c) H3C-<, }-OH ° II

+ CI-CCH(CH3h

242

(Y"'0 + CH2=0

�: °

° II

(b) CH3(CH2hOH + Cl-CCH2CH3

+ 0-0

II Cl-C � !J [>-OH (d)

Page 250: Solucionario de wade

11-32 .. -: 0 : : 0 : · 0· II I • II . : O-S- O CH ..... __ --i .. � 0==S- O CH3 ....... f---l .. � 0==S- O CH3 .. II 3 •• II •• I

: 0 : :0 : :0 : .. 11-33 Proton transfer ( ac id- base) reactions are much faster than almost any other reaction. M et ho xide wi l l act as a b ase and remove a proton from t he oxyg en much fast er than methoxide wil l act as a nuci eoph i le and displace water.

CH 3CH2 -O H + H+ ---

11-34

( a)

o- O H -- o- O CH2CH3

(b) T here are two problems with this att empted b imolecular dehydr at ion . F irst, all three possible ether combinations of cyci ohexanol and ethanol would be produced. S econd, hot sulfuric acid are t he co ndi tions for dehydrating secondary al cohols l ike cyci ohexanol, so el imination would comp ete with substituti on.

11-35

(a) W hat the student did:

N a+ -0, H X + TsO- CH2CH3

CH3 CH2- OX"" H ---� .. � ,

H3 C CH2CH3 sodium (S) -2- butoxide

H3 C C H2 CH3 (S)-2- ethoxybutane

T he product also has the S config uration , not the R. W hy? T he sub stitution is indeed an SN2 rea ct ion, b ut the substitution did not take place at the chiral center, so t he config uration of the starting material is retained, not inverted.

( b) T here are two ways to make (R) -2- ethoxybutane. S tart with (R)-2-b utanol , mak e th e anion , and substi tute on ethyl tosy late s imi lar to part ( a), or do an SN2 inversion at the chiral center of (S)-2- butanol .

SN2 works better at 1 ° carb ons so the for mer method would be preferred t o the latter.

( c) T his is not t he optimum method b ecause i t req uires SN2 at a 2° carbon, as discussed i n part (b ).

H OXH TsCI • T S- OXH

H3 C CH2 CH3 pyri dine H 3C CH2CH3 (S)-2-b utanol (S)-2- butyl tosylate

no inversion yet

243

low temperature ( high temp. favors el imi natio n)

HXO CH2 CH3

H3 C CH2CH3 (R)-2- ethox yb utan e

INVERSI O N!

Page 251: Solucionario de wade

11-36 o-� OH � o-� 0: Na+ + CH/�"'0-S03CH3 ---o-� OCH3 + :0-S03CH3 - - .. � - " 1 1-37

o SOCI2 II ° II 1) 2 CH3CH2MgBr CH2CH3

I CH3CH2-C-CI

2) H30+ • CH3CH2 -C -CH2CH3

I • (a) CH3CH2-C-OH

PCC ° II

CH2CH3 I 1) BH3• TIIP 2) H202, HO-

CH3CHCHCH2CH3 ....... I-----I OH

OH

OH � H2S04 CH2CH3 I

CH3CH = C -CH2CH3

(b) CH3CH2CH20H ---- CH3CH2-C-H 1) CH3CH2MgBr

.. � 2) H30+

cr03 ° H2S04

• � 11-38

(alo-0H

(blo-0H

HC03H UOH

" H+ o H30+

--- •

f!. 'OH

cr03 �O H2SO: U CH3CH2MgBr'"

acetone

any peroxy acid can b e used t o fOlm the epox ide whi ch is c l eaved to the trans­dia l in aqueous acid

OH H30+ _ Q--\ HCI

---

CI

Q--\ PBr3 �

CH3CH20H ___ CH3CH2Br /' Mg, ether

(c) OH Br ��� cr03

OR: H2S04

M MgBr �O Ho+ �H � g � U 3 H+ --- .. ---

ether from Dart (b ) T o-0H � o- Br Mg --­

ether

H,o+ � ) ��! ���A ------' .. � - Q-f;0H rn (d)

0 o-0H � O- Br ��. ether from (c)

� ) - �gBr _ O-(;;CHJ Q-f;0

CH3Br see the nex t pag e for an al tern ati ve endi ng

CH30H �r3

244

Page 252: Solucionario de wade

1 1- 38 continued (d) continued from the previous page

/"'yOH � /"'yBr � U U ether

� H30+ CH30H ° 0--* SNI

_ ____ OH _ OCH3 fmm eel H2S04 � .

(e) There are several possible combinations of Grignard reactions on aldehydes or ketones. This is one example. Your example may be different and still be correct.

o-0H

CH3CH2CH20H �C

PB,) Mg hO

H)O'� ____ /"'y Br ___ _ ____

U ether OH

cr03 /'.. '> H2S04

- �o

H+ Br2 Mg �OH ___ � ____ �Br ___ �MgBr L\ hv ether

H 0+ 3 � SNI CH3CH20H �';­�OH OCH2CH3" H2S04

1 1- 39 Please refer to solution 1-2 0, page 12 of this Solutions Manual.

1 1- 40

(a) �OH SOCl2

�CI -1°

PBr3 -�Br

P ---�I 12

3° oH 01 (c) HCl -

HBr 0' -

III 0 -

245

2° (b)Q-0H

Cd) OH (j

SOCl2 o-CI ..

PBr3 o-Br -

P 0-1 ..

12

&CI SOCl2 ..

PBr3 &Rr .. &1 P -

12

Page 253: Solucionario de wade

11-41

� (a) " (b) � R

(e) cJ °

(d) lfH R

(f ro m in version )

QCOOH QCl (e) (f) (g)�Br

CH20MgBr

(h) b (i)+OCH] U) � + CH30H (k) 0 (m) �O� (n) � + � + EtOH

11-42

major mInor

(I)

+ CH3CH3 H C;-H

(y0H (y0- Na+ (a) V � V

CH)CII,Br • (y0CH2CH3

V W illi amso n ether synthesis

(b) Br d NaOH.

(C) 0 B rM -E.. ether

OH S r d� �H ,V +

M gB r

(( Mg �

ether d H2S04

.. t:,.

°

y t t H30+

1 ) H� cr03

OMgBr II OH

2) Hp' • � H,SO;

a ceton e

H OH

° � (d)VOH � �o 1 ) CH3CH2MgBr

.. �

2) H30+ U 246

Page 254: Solucionario de wade

11-43 Major product for each reaction is shown.

(a)� (b)� cis + trans-rearranged cis + trans

(c)� cis + trans

(dlO (elO (t) 0 11-44

(a) CH3CH2CH2COOCH3

o I I

(d) CH3CH20-P-OH I OCH2CH3

11-45

o I I

Cl-�-CH3 o

methanesulfonyl chloride

11-46

HO",,�H _ . (a) /� SOCl2

..

S

HO',,-.;H _ . (b) /"-..../"-..../ TsCI

S pyridine

HO",,�H _ . (c) /"-..../� TsCI

..

S pyridine

rearranged

Note that (d), (e), and (f) , produce the same alkene.

VCOOCH2CH3 (c) I

h-

o H

pyridine II I +N CI-

..

(y0-�-CH3

V 0 + a C� S-retention

TSO� S

TSO� KEr

S could use NaOH if kept cold to avoid elimination

247

R-inversion

R-inversion

Alt�rnatlvely, PBr3 c()uld be used,

Page 255: Solucionario de wade

1 1- 47 continued (b) PBr3 converts alcohols to bromides without rearrangement because no carbocation intermediate is produced. OH Br

11-48

� PBr3 � U · U H

(a) UOH pec. �o

(b) VOH PBf3• V Bf

(e) 00H � 0°- Na' CH3I. (d) H OH

q

l

H QH q ' H

SOCI2 OR •

CH3 CH3 CH3 retention

�H

q

l

H

TsCI ct-.. • pyridine SN2

CH3 CH3 inversion

OH (e) ct

H2S04

CK. • I � Y 1 ) 03 HI04 .. ... q.oH 2) Me2S OH

(f) aCH20H H2Cr04

aCOOH

..

(g) -t-o-0H Cr03 +00 • H2SO4 acetone

(h) -t-o-0H H2SO4 -to ..

(i) �I

TsCI � (j) a

OH P,I2 .. aI ..

pyridine CH3 OH CH3 OTs cis cis

248

Page 256: Solucionario de wade

1 1 -49

(a)

(b)

(c)

(d)

yaH

yaH yaH yaH

PBr3

SOCl2

HCl ZnCl2

HBr

..

)J""Br inversion

yCi retention

QCI SNi

cis and trans

QBr SNi

cis and trans

(e) yaH 1) TsCl, pyridine

)J""Br inversion, SN2

1 1- 50

(a) Lucas:

(b)

(c)

(d)

(e)

Lucas:

2) NaBr

�OH no reaction

OH � cloudy in 1 -5 min. immediate blue-green Q-OH cloudy in 1 -5 min. immediate blue green

Q-OH cloudy in 1-5 min. immediate blue green

no reaction

OH � cloudy in 1- 5 min.

OH

� cloudy in < 1 min. no reaction-stays orange

o no reaction no reaction-stays orange

00

no reaction no reaction-stays orange

oc cloudy in < 1 min.

249

Page 257: Solucionario de wade

1 1-51

(a) 1 H H \ / p =c

\ ...

H : 0 : . . -

(b){ o:s ...

: 0:

cO ...

. . -

t : 0:

c6 ..

..

..

H H } \- / :c-c / \\

H 0 : . .

: 0: cO'� :::::-.... �

: 0:

:00 : 0: c6'�

� �

: 0 :

.. • cO . . t

: 0 :

.. • 00 : 0 :

... • cO � �

The las t three res onance for ms ar e s imi lar to the firs t three; the change is that the electr ons ar e shown in al tern ate pos itions in the benzene ri ng. To be r igorous ly cor rect, thes e three r es onance for ms sho uld be included, but mos t chemis ts would not wr ite them s ince they do not r eveal extr a char ge del ocalization; unders tand that they would s ti ll be s ignificant, even if not wri tten with the others.

(c)

1 1-52

: 0 : I I : O-S- CH • . I I 3

: 0 :

� A OH

I

: 0 : : 0 : I II ....... f---l.� 0==S- CH 3 • . I I ...... f---l.� 0==S- CH3 • . I

: 0 : :0 :

� M g •

D � ether C Br M gBr + ---

� B 0 E O M gBr

250

Page 258: Solucionario de wade

� Br _M_g--i.� U ether

0' M gBr _----;.�

X era w

v

1 1- 54 [��+ f)]J, 00 � 00 _-_H_2-i��

+ CH �5

HO O-D

alkyl shift-. .

n ng expansIO n recall that 10 carbocations probably do not exist; this could be considered a transition state

ri ng exp ansion fr om 1 ° car bocation to 2°, resonance­stabilized car bocation

N OTE :

T he mi gration directly above does N OT occur as the cation pr oduced is not r esonance- stabi l ized.

An alternative mechanism could be proposed: protonate the ring oxygen, open the ring to a 2° carbocation followed by a hydride shift to a resonance-stabilized cation, ring closure, and dehydration.

hydride shift to resonance stabilized cation r?\ OH (tOH j

:96 � II:� -

H:9b -2° car bocation

o 251

. .

H+ off o ne 0, H+

�90. H on the other ° H-O+ ..

Page 259: Solucionario de wade

1 1 -5 5 H Cl r ,., - oc/ :ci: 0 ( a) ("fOH • �oH -z nCI 2 - ��

V . . '----A Z nCl2 V \ I n thi s presentati on of the mec hanis m, th e HO Z- Cl } h

+ - n 2 Cl- + H20 t- ZnCI2 rearrangement i s sh own c onc urr ently wit ' . c leavage of th e C- O bond wi th no 1 0 H -Cl c arbocati on i nter medi ate.

( b) r'0 . r'0 �� �+ -V �OH H i" QS 03H V A .. _ V ;C ' H • • � '---'tf +9 H

H + H20 �

H • • Onc e thi s c arboc ati on is formed, Cf(+ �H2

removal of adjac ent protons C __ ----;.� produces the c ompounds sh own.

H OH2 ,-- .. (c ) A l l thr ee of th e produc ts go thr ough a c omm on carboc ati on i ntermediate .

• • � H ........ H \ H � � H dS 03H � '" �c� \. � � • � - � + H20: - �

OH OH OH �: O ... f .. H

� H dS 03H + i OU+r:(ll-_H�CO ) �

C�J +0) ! H20: H20 :Y H'··' H

00 + 00 ci? T h;s;s the pc oduc t fc om a p; nacol re arr ange ment.

252

Page 260: Solucionario de wade

11-56

(a) 00Hcr03 ..

H2S04

a o-CH3

.. a a CH3MgBr MgBr t Mg, ether +

CH3 -0-°1

OR:

acetone PBr3

CH30H --- CH3Br

CH3

" d" ("{-OH alternative en 109 V after hydrolysis of Grignard product

HOI H2S04 SN1

...

p� � HO I d�1 (b)0oH 0 Br PBr3

.. Mg o MgBr OH

e� H O+ if .. 3

a ..

H SO r03 PCC 2 4

CH3CH20H � H� � acetone a

(c) OH o MgBr if

I I Cr03 a � • II 0- MgBr'

H2S04 � . � °

acetone from (b) V I

CH3I. oE

(d) 2 �OH PBr3

.. 2 Mg

-� Br ether

or use the S 1 sho " N method wn 10 part (a)

2 �MgBr a

� cr03 • H2S04 acetone

� ! a EtOH, H+ II + II ..

...C .... H30 ... C .... OEt H OH .. H

H2Cr04 t CH.,OH

a OH

1 1) BrMg-..../ ___ � HO--..../ t 2 ) H30+ Mg PBr3

OH

(e) OMgBr a l)� ..

from (b) 2 ) H30+ VOH

253

two RMgX can add to an ester, making a 3° alcohol with two equivalent R groups; see solution to problem 10-18

Page 261: Solucionario de wade

1 1 -56 conti nued

(f) HO�

VOH

(g) HO"�

H

P B r3 M g B rM g� .. .. ether

H

P CC �o ..

TsCI TS O" � K CN

.. .. pyri di ne H'�

o OS 04

..

H

B r2 -hv

B r 6 N C

M g

ether

CH 3CH2B r

P B r3 t CH 3CH20H

PCC -

I I -57 F or a compli cated synthesi s li ke this , begi n by work ing backwar ds . Tr y to fi gur e out where the carbon framework came from; in thi s problem we are restricted to alcohols containing five or fewer carbons. The dashed boxes show the fragments that must be assembled. The most practical way of

formi ng car bon- carbon bonds i s by G ri gnard reacti ons. The epoxide must be f ormed from an alkene, and the alkene must have come from d ehydration of an alcohol produced in a Gri gnard reaction .

�o�<> BfM g�. Mg

,,l, ether

maj or i somer

1 . TsCl , p� 2. KO H,1\

avoid carbocation conditions to prevent rearrangement

, H ° ) y

t H30: \ cr 03 , H2S 04, aceto ne 0: MgB r

!WL� 254

2. Mgo-0H

Page 262: Solucionario de wade

1 1 -58

(a) Both of these pseudo-syntheses suffer from the misconception that incompatible reagents or conditions can co-exist. In the first example, the S N 1 conditions of ionization cannot exist with the SN2 conditions of sodium methoxide. The tertiary carbocation in the first step would not wait around long enough for the sodium methoxide to be added in the second step. (The irony is that the first step by itself, the solvolysis of t-butyl bromide in methanol, would give the desired product without the sodium methoxide.) I n the second reaction, the acidic conditions of the first step in which the alcohol is protonated are incompatible with the basic conditions of the second step. If basic sodium methoxide were added to the

sulfuric acid solution, the instantaneous acid-base neutralization would give methanol, sodium sulfate, and the starting alcohol. No reaction on the alcohol would occur. (b)

SN1 solvolysis conditions warm

Several synthetic sequences are possible for the second synthesis.

�OH Na �O-Na+ CH31 �O/ .. .-

OR �OH TsCI �OTs NaOCH3

�O/ .- .-pyridine

OR�OH PBr3

�Br NaOCH3

�O/ .- ..

11-59

Compound X: -must be a 1 ° or 2° alcohol with an alkene; no reaction with Lucas leads to a 1 0 alcohol; can't be allylic as this would give a positive Lucas test �OH

Compound Y :

6 -must be a cyclic ether, not an alcohol and not an alkene; other isomers of cyclic ethers possible

"-J5 ___ J5TS TsCI

.. pyridine

this reaction works fine, but wait.. .....

NaOCH3 ---l'-� NO REACTION!

cannot do an SN2 reaction

The Williamson ether synthesis is an SN2 displacement of a leaving group by an alkoxide ion. There are two reasons why this tosylate cannot unergo an SN2 reaction. First, backside attack cannot occur because the back side of the bridgehead carbon is blocked by the other bridgehead. Second, the bridgehead carbon cannot undergo inversion because of the constraints of the bridged ring system.

backside -§r attack is H OTs blocked

side view 255

� this carbon cannot invert H �OTS which is required in the SN2

\:::::t'J mechanism

side view

Page 263: Solucionario de wade

11-61 Let's begin by considering the facts. The axial alcohol is oxidized ten times as fast as the equatorial alcohol. (In the olden days, this observation was used as evidence suggesting the stereochemistry of a ring alcohol.)

Second, it is known that the oxidation occurs in two steps: 1) formation of the chromate ester; and 2) loss of H and chromate to form the C=O. Let's look at each mechanism.

AXIA� ;::7'.( H( � Step2.

0� FASTER

OH H ��� H 0, \.Cr03H EQUATORIAL HO�

H

So what do we know about these systems? We know that substituents are more stable in the equatorial position than in the axial position because any group at the axial position has 1 ,3-diaxial interactions. So what if Step 1 were the rate-limiting step? We would expect that the equatorial chromate ester would form faster than the axial chromate ester; since this is contrary to what the data show, Step 1 is not likely to be rate-limiting. How about Step 2? If the elimination is rate limiting, we would expect the approach of the base (probably water) to the equatorial hydrogen (axial chromate ester) would be faster than the approach of the base to the axial hydrogen (equatorial chromate ester). Moreover, the axial ester is more motivated to leave due to steric congestion associated with such a large group. This is consistent with the relative rates of reaction from experiment. Thus, it is reasonable to conclude that the second step of the mechanism is rate-limiting.

11-62 CI Cl � (a) H Cl 0 H \ / H Cl " 00 - Cl <001 ° o�'p __ Ooo (y'O-p/,OOOoo - + I") 1 00 :CI : H 00 1 - <00-P=O 00 QO-P=O o 0 Cl 'Cl

0 0 � 0 0 I 0 0 � 0 0 CI 0 0 � VO Cl 0 0 (Cl

Cl 1 o-p=o 1 CI

continued on next page 256

Ell H H Cl

(';(H __ �H + :o-f=o LiZ \.fin CI

H H H �:N� >

Page 264: Solucionario de wade

1-62 (a) co ntinued CI • • E2

I H • • ' - 0

• G5J�N) O(H H

+ 1 N

+ 0 + CI

(b)

H H � H

H3� h�h OH

POCI3 H3CK:;> •

pyrid ine .....-

H

1 O-P=O 1 Cl

H NOTH3C�

Z aitsev H

There m ust be a stereo chemi cal requirem ent in this el imi nation. I f the S ay tzeff alkene i s not p roduced because the m ethy l gro up i s trans to the l eaving group, then the H and the leaving gro up m ust be trans and the el imi nation m ust be anti- the characteri stic stereochem istry o f E2 e l imi natio n. This evidence differentiates between the two po ssibi l i ties in p ar t (a) .

11-63

(a) u · �H+

° RCH3 U -CH30H H

.. 1 +

° �QCH3 1Qtc5�H } H2� �

HOf) -H .. : 0 + -- O H

--- OH --0 • • H

1 H

• • H+ • • + 9.) :OH '--- � :OH H • • • •

H20 • • .. ° + \ • • r;O

'---"" H • • ' H

t HOVH ---.-

H00H °

.. .. ·9t H2� H

It is equal ly l ike ly for pro to nation to occur first on the ring o xy gen, fol lo wed by ri ng opening, th en replacement o f OCH3 by water.

D r. K anto rowski suggests this altern ative. H e and I wi l l arm wrestle to determi ne whi ch mec hanism is co rrec� ('l _

.. �-� H2� ��+ H H 'OO H � +k�

257

H 00H °

Page 265: Solucionario de wade

8SZ

(Il �qtJ ..

H� \O� I) H

(q) p;}nUllUO::l £9-11

Page 266: Solucionario de wade

CHAPTER 12-INFRARED SPECTROSCOPY AND MASS SPECTROMETRY

See p. 270 for some useful web sites with infrared and mass spectra.

12- 1 The tabl e i s com pleted by r eco gnizing that: (V) (A) = 1 0,000

v (cm·l) 4000 3300 3003 2198 1700 1640 1600 400 A (Jlm ) 2 .50 3.03 3. 33 4 .55 5.88 6.10 6.25 2 5 .0

12-2 I n general , o nly bo nds w i th di pol e mom ents wi l l have an IR abso rptio n .

H-C C-H H-C C-C� �C-C C-C�

t t t t t no y es yes

H I H3C-C-H k t

no y es 12- 3

y es

H3C / CH3 \ I

C=C I t \

H

y es (weak)

H

(a) al kene: C=C at 1 640 cm-I , =C-H at 3080 cm-l (b) al kane: no peaks indicating sp o r sp2 car bo ns pr esent

no y es

H CH3 \ I

C=C

HI t \

H y es

( c) This IR sho ws mo re than o ne gro up . There is a termi nal alky ne sho wn by: C=C at 2 100 cm-I , =C-H at 3300 cm-! . These signals i ndicate an aro matic hy dro carbo n as well: =C- H at 3050 cm-! , C=C at 16 00 cm-! .

12-4 (a) 2° ami ne, R- N H-R : o ne peak at 3300 cm-! indicates an N-H bo nd; this spectr um also sho ws a C=C at 1 640 cm-! (b) carbo xy lic acid: the extremely bro ad abso rptio n in the 25 00-35 00 cm-l range, w ith a "sho ul der" aro und 2 500-2 700 cm-! , and a C=O at 1 7 10 cm-! , ar e com pel l ing evidence for a carbo xy lic acid (c) al co hol: stro ng, bro ad O- H at 3330 cm-!

12 -5 ( a) co nj ugated k eto ne: the small p eak at 3030 cm-! suggests =C-H, and the stro ng peak at 1 685 cm-! IS co nsistent with a keto ne co nj ugated wi th the alkene. The C=C i s indicated by a very sm all peak aro und

1 600 cm-! . (b) ester: the C=O absor ptio n at 1 738 cm-! (higher than the keto ne's 1 7 1 0 cm-! ) , i n co nj unctio n with the stro ng C-O at 12 00 cm-! , poi nts to an ester (c) ami de: the two peaks at 3 160- 3360 cm-! ar e l ikely to be an NH2 gro up; the stro ng p eak at 1 640 cm-! i s too stro ng fo r an al kene, so i t m ust be a differ ent ty pe o f C=X, in this case a C=O , so lo w because i t is part

o f an a mi de

12-6

(a) The sm al l peak at 1 642 cm-! indicates a C=C, co nsistent with the =C- H at 3080 cm-! . This appears to be a simpl e al kene. (b) The stro ng absorp tio n at 1 69 1 cm-! i s unmi stakably a c=o. The smaller p eak at 1 62 6 cm-! indicates a

C=C, pro bably co nj ugated with the C= O . The two peaks at 2 7 12 cm-! and at 2 8 14 cm-! r epresent H-C= O co nfirm ing that thi s i s an al dehy de. 259

Page 267: Solucionario de wade

1 2- 6 conti nued (c) The str ong peak at 1 650 cm-' is C=C, probably conjugated as it is unusually str ong. The 1 703 em' peak appear s to be a conj ugated C=O, undeni ably a car boxy lic acid because of the str ong , br oad O-H absor pti on from 2 400-3 400 cm-' . (d) The C=O absor ption at 1 742 cm-' coupled with C-O at 122 0 cm-' suggest an ester. The small p eak at 1 604 cm-' , p eaks above 3 000 cm-', and peaks in the 600- 800 cm-' region i ndicate a benz ene r ing.

12 -7 (a) The M an d M +2 p eaks of equal inten sity i dentify the presen ce of bromi ne . The mass of M ( 1 56 ) mi nus the weight of the l ighter i sotope of br omi ne (79) gi ves the mass of the rest of the molecule: 1 56 - 79 = 77. The C6HS (p hen yl) group weighs 77; this compound is b romob enz ene, C6HSBr . (b) The mJz 12 7 peak shows that iodine i s presen t. The mol ecular ion mi nus iodine gives the r emainde r of th e molec ul e: 15 6 -12 7 = 2 9 . The C2HS (ethyl) group weighs 2 9; this compound i s iodoethane, C2HSI . (c ) The M an d M +2 peaks have relative intensities of about 3 : 1 , a sure sign of chlor ine. T he mass of M mi nus the mass of the lighter i sotope of chl orin e gi ves the mass of the r emainder of the molecul e : 90 - 35 =

55. A fragmen t of mass 55 is not on e of the com mon al ky l groups (IS, 2 9, 43 , 57, etc. , i ncreasi ng in in cremen ts of 1 4 mass units (CH2)), so the presence of an atom l ike oxy gen must be consider ed. In addit io n to the chlorin e atom, mass 55 coul d be C4H7 or C3H30 . P ossible molecular formulas ar e C4H7CI or C3H3ClO. (d) The odd-mass molec ul ar ion in dicates the p resenc e of an odd n umber of ni trogen atoms (alway s beg in by assumi ng one n itrogen). The rest of the mol ecule must be: 1 1 5 - 1 4 = 1 0 1; this i s most l ikely C7H 17 The formula C7H17N is the correct formula of a molecul e with n o e lements of un saturation. The seven carbon s probably incl ude alky l gr oups l ike ethyl or p ropy l or isopr opy l.

12- 8 Recall that radical s are not detected in mass spectr ometry; only positively- char ged ions ar e detected.

12 -9 [ CHJ. 57 CHj r CH3 -�H CH2 - �H

3

CH3 43 85

mJz 1 00

CH3 I + CH3-CH-CH2 + CH2-CH3

mass 2 9 r adicals

mJz 57

-

-

-

260

not detected

CH3 I CH3 I CH3-CH + CH2-CH-CH3

+ mJz 43 mass 57

CH3 CH3 I + I CH3-<;H + CH2-CH-CH3

mass 43

CH3 CH3 I I CH3-CH-CH2-CH

mJz 85 +

mJz 57

+ CH3 mass 1 5

Page 268: Solucionario de wade

1 2- 10 The molecular weig ht of each isomer is 1 16 g/ mole, so the molecular ion app ear s at rnJz 1 16. The left h al f of each str ucture is the same; loss of a three car bon radical g ives a stabi l ized cation , each with rnJz 73 :

iH2C:O� .. • H2C";�t iH2C: .. �'" · H2C,,+�} .. .. ° ° . . . .

rnlz73 rnlz73 W here the two stru ctures differ i s in the alpha-cleavage on th e right side of the oxy gen . A lpha-cleavage on the left structure loses two carbon s , whereas alpha- cleavage on the right structure loses on ly on e carbon.

[ �o0-/-r [ �o*- r I alpha- loss of CH CH I alpha- loss of CH3 t cleavage 2 3 t cleavag e

{�O .. �H2 ... �O�CH2} J� .. ,�, ... . �+��,}_ .. .. 1 R + R rnJz 87 rnJz 1 0 1

1 2- 1 1 2,6-D imethy lheptan -4-01 , C9H200, has molecular weight 144. The hig hest mass peak at 1 26 is not the molecular ion , but rather is the loss of water ( 1 8) from the molecular ion . [�r - H,o+ [�r

rnlz 1 44 rnlz 1 26

Th e peak at rnJz III i s loss of an other 1 5 (CH3) from the f ragmen t of rnJz 1 26 . This i s cal led ally lic cleavage; i t gen erates a 2°, al ly lic , reson an ce-stabi l iz ed carbocation.

[ � r-i�+-�}+ CH) rnJz 1 26 rnJz III

The peak at rnlz 87 results from frag men tation on on e side of the alcohol :

[�r �:����!C {JJH _ )ji

H

} + � rnlz 144 mlz 87

reson an ce-stabi l ized

1 2- 1 2 P lease ref er to solution 1 -20, p age 1 2 of this S olution s M an ual .

1 2- 1 3 D ivide the n umbers in to 1 0,000 to arrive at the an swer.

.

mass 57

(a) 1 603 cm-1 (b) 2959 cm-1 (c) 1 709 cm-1 (d) 1 739 cm-1 (e) 22 1 2 cm-1

1 2- 14

(a)

H CH2CH3 , / C==C or

/ t ' H H 1 660 cm-1

CH2CH3 /

O==C t 'H 1 7 10 cm-1

stronger absorption-larger dipole 261

(f) 3 300 cm-1

Page 269: Solucionario de wade

1 2- 14 continued

H CH2CH3 \ I

(b) C=C or I t \ H H 1660 cm-1

CH2CH3 I

(c ) "N=C or / t \ H H 1 660 cm-1

stronger absorption-l arg er dipole

H CH3 \ I

(d) C=C or I t \ H3C H 1 660 cm-1

(no dipole moment)

1 2- 1 5

H3C \

CH3 I

(a) C=C and I t \

H3C CH3 1 660 cm-1

weak or non- existent

(b) O� 1 620 cm-l � . conj ug ated

H OCH2CH3 \ I C=C

I t \ H H 1 640 cm-1

stronger absorption-larger dipole

H CH2CH3 \ I C=C

I t \ H H 1 660 cm-1

H CH2CH3 \ / C=C

I t \ H H 1 660 cm-1

stronger absorption-larg er dipole

and

H CH3 3000- <'c=cl 3 100 cm-1 I t \

H CH(CH3h 1 660 cm-1

moder ate i ntensity

1645 cm�1 0 � not conj ug ated

(c) both car bony ls show str ong absor ptions ar ound 1 7 10 cm-1

o o

(d)

II CH3(CH2)3 - C + H

2700-2800 cm-1 two small peaks

o y- 3300 cm-1 0\ . H br oad, str ong

1200 c m-1

and

and

II CH3(CH2h -C -CH3

a� 1 7 1O cm�1 str ong

262

Page 270: Solucionario de wade

12 - 1 5 c ontinu ed 1 C 3300 cm-

(e ) CH3(CH2h - C:: C - H and t CH3(CH2)6 -C:: N t 2 100-22 00 c m-1 22 00-2 300 c m-1 weak to moder at e intens it y moder ate to str ong int ens it y

(f) bot h c ar bonyls s how s tr ong abs or pt ions ar ou nd 1 7 10 c m-1 � 3300 c m-1 f H br oad, str ong o 0' 0 1\ I 1\

CH3CH2CH2 - C - OH and CH3 - CH - CH2 - C - H

2 �00 cm-1 t ver y br oad

0/ 1650 cm-1 II

2 700-2 800 cm-i t wo s mall peaks

1 7 1O cm-i �0 II

(g) CH3CH2CH2 -C - � - � o/ 3300 cm-1

two peaks

and CH3CH2 - C -CH2CH3

12 - 1 6

(a)

(c )

12 - 1 7

o � 1 700 cm-1 I I H C-OH

\ / ...........-C = C 2 400-3400 c m-1

/ t \ H CH3 1 640 c m-1

tV 3000- 3 100 cm-I �CH2 -C N V, 22to cm1 "-

1 600 cm-I

+ (a) CH3 •

�71 [ CH3 �

HjCH2CH2CH3 1 43

mlz 86

(b)

(d)

o � 1 7 1 5 c m-I II CH3-CH-C-CH3 I

CH3

H ./ 3000-3 1 00 cm-i

a/" H 3400 cm-i N ..... V "'=:: \ I ,CH2CH3, �, 2 900

T-3000 cm-I

CH3 I

1 600 c m-I

+ CH - CH2CH2CH3

CH3 I mlz 7 1

• CH3 -CH mlz 43 +

263

Page 271: Solucionario de wade

1 2- 1 7 c ontinued

(b)[CH3 -CH=b��9�CH2CH3r----l .. �i CH3_CH= bH3

<iH, - CH3_ �H_ bH3

CH, } rnJz 69

rnJz 98- al ly l ic c leavage [ �� CH3 ]t

(c ) cHJbHj-CH,- bH-CH1 45

alpha- 1 :OH CH3 + OH CH3 t c leavage I I II I ----I .. � + CH - CH2 -CH -CH3 ___ CH -CH2 - CH - CH3

rnJz 87 rnJz 102

[ j+ . CH3

CH3 -CH==CH - �H -CH3

rnJz 84

alpha­c leavage

i·· :OH

CH3-�H

(d) [&f-. ()�H'

benzyl c ation

rearr ange ..

rnJz 45

H C

o T he tropylium ion is a ch ar ac teris tic fragment fr om phen ylalkanes.

rnJz 1 3 4 � m1z 9 1 � /CH3 H'Cy r earr an ge HC+ .. rnJz 77 �H3 rnJz 43

H3C, + "CH3 C I CH3

rnJz 57

GCH rnJz 85

rnJz 1 44 • alpha-c leavage 1O-Fc�H3

• • O-�=c�H3} rnJz 1 29

264

Page 272: Solucionario de wade

1 2- 1 8

(a)

(b)

(c )

(d)

[�r -----

rnJz114

+ (5

. H

----- cJ + mJz 98 rnIz 83

+ �CH2 +

mJz 85

�CH2 + mJz 7 1 +

+ �CH2 +

mJz 5 7

o CH3 mass 1 5

CH3 I +

• CH3 H2C' mass 29

H2C......./ mass 43

Hi:� mass 57

CH3-C==CH-CH2 --

OH • :OH +OH

m1z 69 + oCH3 mass 15

[ j+ i .. . . }

�Hj CH, - CH,CH,CH, - �H' - gH, +. CH,CH,CH,CH, 31 rnIz 88 mJz 31 mass 57 ! - H20

[ CH, = CH - c::t C� CH, mJz 70 55

+ CH2 =CH -CH2 -CH2 + 0 CH3

mJz 5 5 mass 1 5

{ CH2=CH- �H2 � ---- �H2 -CH = CH2 } + 0 CH2CH3 mJz 4 1 mass 29

Oc+ ----- 1.0 rnJz 77

mJz 1 2 1

c ontinued o n nex t page 265

+ 0 NHCH2CH3 mass 44

Page 273: Solucionario de wade

12- 1 8 (e) c ontinu ed

[a�::rr mlz 121

12- 1 9 (a) The c har act er ist ic fr equ enc ies of t he OH abs or pt ion and t he C=C abs or ption wi l l i ndic ate the pr es enc e or abs enc e of t he gr ou ps. A s pectru m with an abs or pt ion ar ou nd 3300 c m-i wi l l have s ome c yc lohexanpl i n it; i f t hat s ame s pec tru m als o has a peak at 1 645 c m-i , t hen t he s ample wi l l als o c ont ain s ome c yc lo he xene. Pur e s amples wi l l have peaks r epr es ent at ive of only one of the c ompou nds and not the other . N ote that quantitation of the t wo c ompou nds wou ld be ver y diff icu lt by I R bec aus e the s tr ength of abs orpt ions ar e ver y diff er ent . Us ual ly , ot her met hods ar e us ed in pr ef er enc e t o I R f or qu antitat ive measur ements .

o y- 3300 c m- i 0\ 'H hr oad, str ong

12 00 c m-1

O -- 1 645 c m-1 moder ate

(b) M ass s pec trometr y c an be mis leading wit h alc ohols . Usu al ly , alc ohols dehydr ate in the inlet s ystem of a mass s pec tr omet er , and t he c har ac ter ist ic peaks obs er ved in the mass s pec tru m ar e thos e of t he alken e, not of t he par ent alc ohol. F or t his par ticu lar analys is , mass s pectr ometr y wou ld be u nr el iable and per haps mis leading.

12-2 0 (a) The "stu dent pr ep" c ompou nd mus t be I- br omobu tane. The mos t obvious featur e of the mass spec tr um is t he pair of peaks at M and M +2 of appr oximat ely equ al heights , c har act er is tic of a br omine atom. L oss of br omi ne (79) fr om the molecu lar ion at 1 36 gi ves a mass of 57 , C4H9, a butyl gr oup. W hic h of t he f our possi ble bu tyl gr ou ps? The peaks at 107 (loss of 29, C2Hs) and 93 (loss of 43 , C3H7) ar e c ons is tent w ith a l i near c hai n , not a br anc hed c hain . (b) The bas e peak at 57 is s o s tr ong bec aus e the c ar bon-halogen bond is the weakes t in the molecu le. Typic al ly , loss of halogen is the dominant fr agmentat ion in alkyl hal ides .

[4:r r � �{� rnJz 1 36

/'Br + rnJz 107

H2C. + Br

rnJz 93 +

/'-../ rnJz 57

+

mass 29

+ H2C-..../

mass 43

+ • Br mass 79

266

Page 274: Solucionario de wade

12 -2 1 (a) D euter iu m h as twic e th e mass of h ydr ogen, bu t s imi lar s pr ing c ons tant, k . Compar e th e fr equenc y of C-D vibr ation to C-H vibr ation b y s etting u p a r atio, ch an ging only th e mass (su bs ti tu te 2 m for m).

= = �lt2 = 0.707

Vo = 0.7 07 vH = 0.7 07 (3000 em-I) '" 2100 cm-1

(b) The func tional gr oup mos t l ik ely to be c onfus ed with a C- D str etc h is the alk yne (c ar bon- c ar bon tn ple bond) , whic h appears in th e s ame r egion and is of ten ver y weak.

12-22 ( � r-m/z 1 14

[+r �

m/z 1 14

r*r �

m/z 114

+ � m/z 57

1+ m/z 57

�. m/z 57

a 1° c ar boc at ion- not favor able

a 2° c ar boc at ion-r eas onable

a 3° c ar boc ation-t he best

The most l ikely fr agment ation of 2 ,2, 3 , 3- tetr amet hylbut ane wi l l give a 3° c ar boc at ion, th e mos t st able of the c ommon alkyl c at ions. The molecu l ar ion s hould be s mall or non- exis tent whi le m/z 57 is l ik ely t o be the bas e peak , wher eas t he molecu lar ion peaks wi l l be mor e pr omi nent for n -oc tane and for 3 ,4- dimethylh exane .

12 -2 3 (a) Th e infor mat ion t hat this mys tery c ompound is a h ydr oc ar bon mak es inter pr et in g the mass s pectru m much eas ier . ( It is r elat ively s imple to tel l i f a c ompound has c hlori ne, br omine , or n i tr ogen by a mass s pec tru m, but oxygen is di ffic ult to determ ine by mass s pec tr ometr y alone. ) A h ydr oc ar bon with molec ular ion of 1 1 0 c an

have only 8 c ar bons ( 8 x 12 = 96) and 1 4 hydr ogens . The for mula CgHl4 has t wo elements of uns atur at ion .

(b) The IR wi l l be us eful in detenni ning what t he elements of u n s atur ation ar e . Cyc loalk an es ar e gener al ly not dist i nguish able in th e IR. An alk en e sh ould h ave an abs or ption ar ound 1 600- 1 650 em-I; none is pr es ent in th is I R. A n alkyne s hould h ave a s mal l , shar p peak ar ound 22 00 em-I-PRESENT A T 2 12 0 em-I! A lso, a

sh ar p peak ar ound 3300 c m-1 indic ates a h ydr ogen on an alk yn e , s o th e alk yne is at on e end of t he molec ul e. B ot h elements of uns at ur at ion ar e acc ounted for by t he alk yne.

267

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12-2 3 (c ) T he onl y question i s how ar e the o ther carb ons arr anged. T he m ass spec tr um sho ws a pr ogr ess io n of

peak s fr om the mo lec ular ion at 1 10 to 95 ( lo ss of CH3), to 8 1 (los s o f C2Hs), to 67 (loss o f C3H7). The

mass spec tr um suggests i t i s a li near chain. T he extr a ev idence that hydro genation of the mystery com po und

giv es n-o ctane v eri fies that the chain is l inear . T he origi nal c ompound m ust b e o ct- 1- yne. ( d) T he base peak i s so str ong b ecause the ion pr oduc ed i s s tabil ized b y r esonanc e.

12 -2 4

m/z l lO

+

m/z 39 '-..... c:::=:::>

+

� m/z 95

� m/z 8 1 +

+

�_ m/z 67

(a) and (b) T he m as s spec is consistent with the for mul a of the alkyne, CSHJ4' mass 1 1 0. T he IR is no t c onsistent with the al kyne, howev er . O ften , symmetric all y substituted alk ynes hav e a mi nisc ule C=C peak, so the fac t that the I R does not sho w thi s peak does not prov e that the alkyne is ab sent. T he im pOlt ant ev idenc e in the I R is the signific ant peak at 1 62 0 c m-I; this abs orp tion is c har ac teri stic of a conjugat ed diene. I nstead of the al kyne b eing form ed, the reac tion must hav e b een a doubl e el im ination to the dic nc .

�*+ Br

12 -2 5

(a)

Q 0 � 17 60 cm-I

(c ) II � O- g-CH \ I

3 - T 12 00 cm-J ,

1 5 80 cm-J

(b )

268

o-H O , ...,... 1 72 0 cm-1 II \\ I 11--'/ '\ C-C-H - k t " 27 1 0,

1 600 c m-I 2 82 0 c m- 1

Page 276: Solucionario de wade

12 -2 6 (a)

o

A � Br M g /""-.... /""-....

ether .. ,,/ ............", - M gB r

H30+ .. �OH

2 -m eth y l hexan-2-01 C7HI60 m ol. wt. 116

(b) T he mol ec ul ar i on i s not vi sib le in the spec tr um. Alc ohols typic al l y dehydr ate i n the hot in let system of the mass spec tr om eter , especi all y true for 30 alc ohol s that ar e the easi est type to dehydr ate. T he two fragmentati ons that produc e a resonanc e-stabi l ized c arb oc ation give the m ajor peak s in the spec tr um at mlz 59 and 1 0 1 . [�OH l t ___

{ ;<�-H _ .. _� /<O+-H}

� H3C CH3 H3C CH3

mass 1 1 6 rnIz 59

.. .. } O-H

�c(+ CH3

rnIz 101

12 -27 T he unk nown c om pound has peak s in the mass spec at rnIz 1 98 , 1 55 , 127 ,7 1 , and 43 . I t i s helpf ul that the m asses of two of the fr agm ents, 1 5 5 and 43, sum to 198 , as do the other two fragm ent masses, 127 and 7 1 . W e c an say wi th c ertai nty that the mol ec ul ar i on i s at rnIz 1 98 , and that the unk nown i s a rel ativel y sim pl e m ol ec ul e wit h two m ai n fragm entati ons.

T hi s i s a fairl y hi gh m ass for a sim pl e c ompound; som e heavy group must b e pr esent. W hat is N O T pr esent i s N b ecause of the even mol ec ul ar ion m ass, nor CI nor B r b ec ause of the l ack of i sotope peak s , nor a pheny l group b ec ause of no peak at 77 . T he progression of alk yl gr oup m asses: 1 5 ,2 9,43 , 57 ,71 , 85 , 99- incl udes two of the peak s, so i t appear s that the unk nown c ontai ns a pr opyl gr oup and a pentyl group (the pr opyl c oul d b e part of the pentyl gr oup). T he 127 fragm ent i s k ey; the fragm ent C9HI9 has thi s m ass, b ut we would expec t m uc h m or e fr agm entati on fr om a nine c arb on pi ec e . T here must b e som e o ther expl anati on for thi s 127 pe ak.

A nd ther e is! T her e i s one pi ec e-m ore speci fic al ly , one atom- that has m ass 127: i odine! I n all prob abil ity, the iodi ne atom i s attac hed to a fr agment of mass 7 1 whic h i s CsHl1 ' a pentyl group. W e

c annot tell with c ertai nty what i som er i t is , so unl ess ther e i s som e other evidenc e, l et' s pr opose a str ai ght c hai n i somer, 1 -i odopentane.

155 127 I

T he 155 fr agment probabl y has this b ri dged struc ture

b ec ause iodi ne i s so bi g :

269

Page 277: Solucionario de wade

12-2 8

(a)

(b)

(c)

CH3 CH3 I I H C-C-C-CH3 3 I I

CH3 I HC-C-C-CH3

3 II I OH OH

T / s tr ong peak around 0 CH3

broad, s tr ong OH peak 17 1 5 cm- 1

around 3300 cm-I

o

o str ong peak ar ound 1 690 cm-I

two peak s at 27 00 and 2 800 cm-I

H 0/ t

ver y br oad 2 500- 3 500 cm-1 with a " shoulder " ar ound 27 00 cm-I

o

OH br oad, strong 011 pe ak ar ound 3 300 em-'

br oad COOH band mi ss ing

sti l l has a str ong OH a t 3300 cm-I but d iffer ent fr om COOH

I f you wish to find IR spectr a and mass spectr a of comm on compounds, ther e ar e two w eb sites that ar e ver y helpfuL E nter ing a name or molecular form ula wi l l giv e i som er s fr om which t o choose the desir ed s tr uctur e and the IR or MS if av ai lable in their d atabase.

http:// webbook .nisLg ov/ "NIS T " i s the U.S . N ational I ns ti tute of S tandards and T echnology.

http://www.aist.go.jpIRIODB/SDBS/menu-e.html T his is fr om the N ational I nstitute of A dvanced I ndustr ial S cience and T echnology of J apan.

270

Page 278: Solucionario de wade

CHAPTER 13-NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY

A b enzene r ing can b e wri tten with thr ee altern ating doub le b ond s or with a c ir cle in the r ing. A l l of the carb ons and hydr ogens in an un sub stituted b enzene r ing ar e equivalent, r egar dless of which symb olis m is used.

equi val ent to o The Japanese web site listed at the bottom of p. 270 also gives proton and carbon NMR spectra.

Reminder: The word "spectrum" is singular; the word "spectra" is plural.

1 3- 1

650 Hz -6 ( a) = 2. 17 x 1 0 = 2. 17 ppm dow nfield fr om TMS 300 x 1 06 Hz

(b) D iffer ence in magnetic field = 7 0,459 gauss x (2 . 17 x 10-6) = 0. 1 5 3 gauss

( c) T he chemical shift does not change w ith field str ength: «52 . 17 at b oth 60 MHz and 300 MHz.

( d) (2. 17 ppm) x ( 60 MHz) = (2 . 17 x lO-6) x ( 60 x 106 Hz) = 130Hz

1 3-2 N umb er s are chemi cal shi ft values, in ppm, derived fr om Table 1 3 -3 and the A ppendix in the text. Your predictions should be in the given range, or within 0.5 ppm o/the given value.

a ( a) b ( b) H CH3 b b * b 'c

.... CH3 � CH3 � CH3 CH3 ...... \ / a = «5 5-6 I a = «57 .2 b C = C b b = «5 0.9 h- b = «52 .3

/ \ CH CH3 CH3 H C....

3 b b

a CH3 ...... \ H b C� a

b

( c) � � ( d) c �H3 b a = «52- 5 b = /)2 . 5 c = /) 1-2 c�p*OC�J

a = /)7 .2 b = /) 3 .6 CH3-C-C::C-H

c em a

( e)

H H a a

c c H H

c � b � a H--u- CH2-C-O-H

H H c c

a = /) 10- 12 b "" «5 3 ( between tw o deshielding gr oups)

b

b �H3 (f)

CH -C-CH Br 3 I a

2

Br

c = /)7 .2 ( Th e h ydr ogens l ab el ed "c" ar e not equival ent. Th ey appear at r oughly the sam e chemi cal shift b ecause the sub stituent is neither stro ngly electr on-d onating nor withdr awing.) 271

a = «5 3- 4 b = «5 1-2

Page 279: Solucionario de wade

1 3 -3 c b a

(a) CH3CH2CH2Cl

three types of H

a

(b ) b a b

CH3CHCH3 I Cl

two types of H

(c) a

a �H3 b c CH3-C-CH2CH3

I CH3

a

b H a

(d) H*� C�3

H h B r c

H three types of H d 13 -4

(a) �*:3

t fiv e types of H

H h H c c

H d

four types of H

(b) T he three types of aromatic hydrogens appear in a relativ ely s mall s pace around () 7 .2 . The signal is complex b ecause all the peaks from the three types of hydrogens ov erlap.

N ote : N MR spectra drawn in this S olutions M anual w i l l repres ent peaks as single l ines. T hese l ines may not look l ike " real" peaks , b ut this av oids the problem of v ar iation am ong spectrometers and printers . I ndiv idual spectra may look different from the ones presented here , b ut al l of the impor tant information wi l l b e contained in thes e representational spectra.

1 3 -5

I 1 0

c CH3 a a b c I I I a I I

H C - C - O - C - CH - C - CH 3 I

2 3

CH3 c

I I I 9 8 7

I 6

� 9H / 12

3 H

� � 2H

TMS

-.J I I I I I I I 5 4 3 2 o 8 (ppm)

13 -6 The three spectra are identified with their str uctures. D ata are giv en as chem ical shift v alues, with the in tegration ratios of each peak giv en in parentheses .

S pectrum (a) b

OH C I a

CH3 - C - C::C - H I

CH3 c

S pectru m (c )

b b CH3

I a CH3-� - CH2Br

B r

a = cS 2. 4 ( 1 ) ( 1 H) b = cS 2. 6 ( 1 ) ( 1 H) c = I) 1 .5 (6) (6H)

S pectrum (b ) a a H H

c�,o*OC�J

H H a a

a = cS 6.8 (2 ) (4H) b = cS 3.7 (3 ) (6H)

a = I) 3 .9 ( 1 ) (2 H) (Thi s i s the compound in Problem 1 3-2 (f). You b = cS 1.9 (3) (6H) may wish to check your answer to that questio n

agains t the spectrum.) 272

Page 280: Solucionario de wade

1 3- 7 Chem ical shift values ar e appro ximate and may v ar y s li ghtly fro m yo ur s . T he splitti ng and integr atio n values sho uld m atch exactly, ho w ever.

(a)

I 10

(b)

I 1 0

(c)

I 1 0

b b H3C, /CH3

a HC - O - CH a / ,

H3C CH3 b b

I I 9 8

0 b a c II ClCH2CH2 - C - O CH3

I I 9 8

Note: the three types of

aromatic protons above

are accidentally equivalent because the

isopropyl group has little

effect on their chemical

shifts.

I I 9 8

I 7

I 7

.1! 5H

I 7

.1! 2 H

(To show the pattern of peaks, this multiplet " I is larger than it would appear on a real spectrum.)

I I I 6 5 4 (5 (ppm)

.1! 2 H

I I I I I I 6 5 4 (5 (ppm)

a a H H c * CH]

a - / H � /; C�b

CH3 H H c a a

I I , T 3

Q 3H

I 3

(To show the pattern of peaks, this m ultiplet

T 2

2 H

L I I

2

Q I H

is larger than it would , I appear on a real spectrum . )

" I I , I I I 1 I 6 5 4 3 2 (5 (ppm)

273

Q 12 H

I 1

I 1

� 6H

I

TMS

I T o

TMS

I I o

TMS

1 I o

Page 281: Solucionario de wade

1 3-7 co ntinued

(d) a H H a

c b * b e � l! CH3CH20 � Ii OCH2CH3 6H

4H

All four aromatic a H H a protons are equivalent .

12 4H TMS

I I I 1 I T I I I I I I I I I 1 0 9 8 7 6 5 4 3 2 o 8 ( ppm)

( e) 12 4H �

c a 0 b b 0 a c 6H I I I I CH3CH20 - C - CH2CH2 - C - OCH2CH3

l! N o splitting is ob serv ed when chemically 4H equivalent hydr ogens ar e adjacent to each other , as with hydr ogens "b " ab ov e .

I I 1 0 9

1 3 -8 0

I 8

( aJ 5 2 .4 H2?): H 8 5 . 8

8 1 . 8 H2C H 8 6 .7

\ H3C '(

CH3 J

8 1 . 2

I 7

T he signals at 1 . 8 and 2.4 ar e tr iplets b ecause each set o f pro tons has two neighbo ring hydro gens . T he N + 1 r ule corr ectly pr edicts each to b e a triplet.

TMS

I I I I T T I I I I T 6 5 4 3 2 o 8 ( ppm)

(b )

o

8 1 . 5 H2C I

8 1 .65 H2C , C H2

8 2. 1 allylic

H 8 6.2 CH3

8 1 .7 allylic

T he thr ee CH2 gro ups can b e distinguished b y chemical shift and b y splitting. T he al ly l ic CH2 w il l be the farthest downfield

o f the thr ee, at 2. 1 . T he signal at 1 . 5 is a tr iplet, so that must b e the one with o nly two neighbor ing pr oto ns. T he CH2 sho w ing the multiplet at 1 .65 must b e the o ne b etween the o ther two CH2 gr oups .

274

Page 282: Solucionario de wade

1 3 -9 (a)

1 0

(b)

I 1 0

(c )

1 0

i! 5H

T he thr ee types o f aro matic pro to ns 12 ar e accidentally IH equivalent. J ", 15 Hz

,........I-....

9 8 7

b c CH30 \ CH3

/ C = C / \

Cl H a

I I I 9 8 7

o e II c d (CH3hC

\ F- OCH2CH 3

C =C / \

H H a b

6

T 6

i! IH

I I I

I H I H

9 8 7 6

� I H

1 = 1 5 Hz ,........I-....

5 8 (ppm)

I 5 8 (ppm)

5 8 (ppm )

275

a H H a

�*� } H C = C a / \ H C(CH3h

4

12 3H

I 4

� 2H

4

b d

3 2

� 3H

I I 3 2

3 2

Q 3H

I

1

Q 9H

� 9H

TMS

I 0

TMS

I I o

TMS

o

Page 283: Solucionario de wade

1 3 -9 continued (x d Q

cx>= c 3H

H '-':: H

I b I .& b 1 1 The electron-withdrawing H H

The chemical shift Q � carbonyl group deshields the O�C

"'OH of -COOH is variable. It is 2H 2H protons adjacent to it, moving

TMS usually a broad them downfieid from their a peak. usual position at 8 7 .2. 1 I I I I I I I I I I I 10 9 8 7 6 5 4 3 2 o 8 (ppm)

1 3- 1 0 The formula C3H2NCI has three elements of unsaturation. The IR peak at 1 650 cm- I indicates an alkene, while the absorption at 2200 cm- I must be from a ni tri le (not enough carbons left for an alkyne). These two groups account for the three elements of unsaturation. So far, we have: \

IC N

C = C / \ + 2 H + Cl

H C = N

The NMR gives the coupling constant for the two protons as 1 4 Hz. This large J value shows the two protons as trans (cis, J = 10 Hz; geminal , J = 2 Hz). The structure must be the one in the box .

\ 1 -C = C

I \ Cl H

1 3 - 1 1 (a) C3H7CI-no elements of unsaturation; 3 types of protons in the ratio of 2 : 2 : 3 .

a b c Cl - CH2CH2CH3

a = <5 3 . 8 (triplet, 2H) ; b = <5 2. 1 (multiplet, 2H); c = <5 1 .3 (triplet, 3H)

(b) C9H 1002-5 elements of unsaturation; four protons in the aromatic region of the NMR indicate a disubsti tuted benzene ; the pair of doublets with J = 8 Hz indicate the substi tuents are on opposite s ides of the ring

The other NMR signals are two 3H singlets, two CH3 groups . One at <5 3 .9 must be a CH30 c=:====�> group. The other at <5 2 .4 is most l ikely a CH3 group on the benzene ring.

(para) <�

1> + 3 C + 6 H + 2 0 +

CH3-Q- + OCH3 + C + 0 + 1 element of unsaturation

1 element of un saturation

One way to assemble these pieces consistent with the NMR i s : b a H H

0

C�3*g- OC�3

H H b a

a = <5 7 .9 (doublet, 2H) b = <5 7 .2 (doublet, 2H) c = <5 3 .9 (singlet, 3H) d = <5 2.4 (singlet, 3H)

276

/ (Another plausible structure is to have the methoxy group directly on the ring and to put the carbonyl between the ring and the methyl. Thi s does not fit the chemical shi ft " al ues quite as well as the above structure, a s the methyl would appear around <5 2 . 1 or 2 . 2 instead o f 2.4 . )

Page 284: Solucionario de wade

13-13 (a)

, . , .

. . '

� 5 . 1

1 .. .. .. ... .. ..

1 , . , . , . Jbc = l .4 Hz � I I � Jbc = 1 .4 Hz

JJ------JU_ o

c I I a H C - OH

\ / c=c f e d / \

CH3CH2CH2 H b

a = 8 1 2. 1 (broad singlet, I H) b = 8 5 . 8 (doublet, I H) c = 8 7 . 1 (multiplet, I H) d = 8 2 .2 (quartet, 2H) e = 8 1 . 5 (sextet, 2H) f = 8 0.9 (triplet, 3H)

(b) The vinyl proton at 8 7 . 1 is He; it i s coupled with Hb and Hd, with two different coupling constants, Jbe and Jed' respectively . The value of Jbe can be measured most precise ly from the signal for Hb a t 8 5 . 8 ; the two peaks are separated by about 1 5 Hz, corresponding to 0.05 ppm in a 300 MHz spectrum. The value of Jed appears to be about the standard value 8 Hz, judging from the signal at 8 7 . 1 . The splitting tree would thus appear:

Jed = 8 Hz

� �

8 7 . 1

1 Jbe = I 5 Hz

. ... , " .. . .. ... ' ,, '

Jed = 8 Hz I I Jed = 8 Hz

over�n NMR

U

277

� , , ,

Jed = 8 Hz 'I

L

Page 285: Solucionario de wade

1 3 - 1 4

(a)

o c I I a H C - H

a = 8 9.7 (doublet, IH) b = 8 6.7 (doublet of doublets , I H) c = 8 7 .5 (doublet, I H) d \ I *H C C,

d - H H � Ii � b

H H d d

d = 8 7 .4 (multiple peaks , 5H)

The doublet for He at 8 7 .4 OVERLA PS the 5H peaks of Hd.

(b) Jab can be determined most accurately from Ha at 8 9 .7 : Jab "" 8 Hz, about the same as "normal " alkyl coupl ing.

Jbe can be measured from Hb at 8 6.7, as the distance between ei ther the first and third peaks or the second and fourth peaks (see diagram below) : Jbc "" 18 Hz, about double the "normal" alkyl coupl ing.

(c ) 0 6 .7

. . . . . . . . . . . . . . . 1 . . . . . . . . . . . . . . . 1 Jbc = 1 8 Hz 1 r;.:� �'�' 1 f " ��:� �'�' 1

J � � L 1 3- 1 5 0 (a) a = 0 1 .7

b "" 0 6 .8 c = 0 5 -6 d = 0 2. 1

(b) a = doublet b I I d H C - CH3

\ I c= c a I , H3C H

c

b = mu ltip let (two overlapping quartets-see part (c» c = doublet d = singlet

(c) using Jbc = 1 5 Hz and Jab = 7 Hz: o 6.8

J . . r.

· · · . . . ;: = 1 :·;:· · . . · · 1

. . . . . . . 'I;� -;. � . · ;::::·1- · · · · · · · · · · · '1';:'-;' � . . ' � � : : 1-· · · · · · (j:�'= 7 Hz ( • • • � • • • • • , • • • • • • ,

� � L J � L 278

Page 286: Solucionario de wade

1 3 - 1 6

1 3 - 1 7 (a)

(b)

replace He

replace Hd

replace Ha ..

replace Hb �

replace He on the left �

replace He on the right ..

(c) The Hd protons are also enantiotopic.

1 3 - 1 8 b

(a) BXyr "H

e a � CH3

CH3 -::., H H c d

This compound has five types of protons. � and Hd are diastereotopic . a = � 1 . 5; b = � 3.6; c ,d = � 1 .7; e = 8 1.0

279

cis diastereomer

trans diastereomer

non-superimposable mirror images =

nantiomers; therefore, Ha and Hb arc enantiotopic and are not distingui shable by NMR

(b)

enantiomers

e H

H b

e H

This compound has s ix types of protons He and Bct are diastereotopic , as are � and Hr. a = 8 2-5 ; b = 8 3 .9 ; c ,d = 8 1 .6 ; e ,f = 8 1 .3

Page 287: Solucionario de wade

1 3- 1 8 continued

(c) b H c H

H b

d H

Br

This compound has s ix types of protons . He and Hf are diastereotopic . a ,b ,c = & 7.2; d == & 5.0; e,f = & 3 .6

1 3 - 1 9

(a) .. � H

a (d) H Cl

\ / C=C / \

H H b c

This compound has three types of protons. Ha and Hb are diastereotopic. a,b = & 5-6 ; c = & 7-8

H 1

CH CH - O-H 3 2 • •

"" + H-B --- + I') ". -CH3CH2 - �-H + .B --- CH3CH2 - �: + H-8

(b) . . ') " -

CH CH - O-H +:B 3 2 • •

.. � "" --- CH3CH2-R: + H-B H

1 --- CH3CH2- O: + :B . .

1 3 -20 The protons from the OH in ethanol exchange with the deuteriums in D20. Thus, the OH in ethanol is replaced with OD which does not absorb in the NMR. What happens to the H? It becomes H OD which can usually be seen as a broad singlet around & 5 . 25. (If the solvent is CDCI3, the i mmiscible H OD will float on top of the solvent, out of the spectrometer beam, and its signal wi l l be missing.)

R OH + D 20 - R OD + HOD

CH3CH2 OH + D20 3H

2H

HOD Peaks from OH are usual ly broad

TMS because of rapid proton transfer. This is especial ly true in water, II I or HOD. I ,

I I I J I I I I I I I 1 0 9 8 7 6 5 4 3 2 o

& (ppm)

280

Page 288: Solucionario de wade

13-2 1

(a) The fonnula C4HlO02 has no e lements of unsaturation , so the oxygens must be alcohol or ether functional groups. The doublet at 8 1 .2 represents 3H and must be a CH3 next to a CH. The peaks centered at 8 1 .65 integrating to 2H appear to be an uneven quartet and signify a CH2 between two sets of non-equivalent protons; apparently the coupl ing constants between the non-equivalent protons are not equal , leading to a complicated pattern of overlapping peaks. The remaining five hydrogens appear in four groups of IH, IH, IH, and 2H, between 8 3.7 and 4.2 . The 2H multiplet at 8 3 .75 i s a CH2 next to 0, with complex spl i tting (doublet of doublets) due to diastereotopic neighbors . The IH multiplet at 8 4.0 is a CH between many neighbors . The two IH signals at 8 4. 1 -4.2 appear to be OH peaks in different environments, partial ly split by adj acent CH groups .

HO H H H

I :: :: H V:: = (H3-r - + -CH2- + -OH + -OH + -CH2 � H3C . % OH

Of the two possible structures:y--r====�=o======;:: '/ The 3-di�en�onal view H 1.2 I 1.65 375 shows c learly that the two Hs

I

CH -C-CH OH 3 I

2 OR CH3-C-CH2CH20H on a CH2 are not equi valent

I ) CH20H

this one has no chiral centers and would not display such complex spl i tt ing

OH-- 4.1-42

CORRECT!

If the possibility of intramolecular H -bonding is H

considered, the differences in the environments � for each H become even more obvious.

H

H

(b) The fonnula C2H7NO has no elements of unsaturation . The N must be an amine and the ° must be an alcohol or ether. Two triplets, each with 2H, are certain to be -CH2CH2 - . Since there are no carbons left, the Nand 0, with enough hydrogens to fi l l their valences, must go on the ends of thi s chain . The rapidly exchanging OH and NH2 protons appear as a broad, 3H hump at 8 2.8 .

13-22

I 10

° II

I 9

I 8

CH3 -C - OCH2CH3

8 2.05 8 4. 1

HOCH2CH2NH2

8 3.7 � '-- 82.9

I 7

I 6

I 5

8 (ppm)

I 4

!! 3H

I 3

� 3H

12 2H

TMS

I I I I I I 2 1 0

The key is what protons are adjacent to the oxygen . The CH30 in methyl propanoate, above, absorbs at 8 3.9, whereas the CH2 next to the carbonyl absorbs at 8 2 .2 . Thi s i s in contrast to ethyl acetate, at the left.

281

Page 289: Solucionario de wade

1 3-23

Ha = [) 2.4 (singlet, IH) Hb = [) 3.4 (doublet, 2H) He = [) 1 .8 (multiplet, 1 H) Hd = [) 0.9 (doublet, 6H)

1 3 -24

d CH3

H2 I H-O-C -C-H I Q c b

(a) The formula C4H802 has one e lement of un saturation . The IH singlet at [) 1 2 . 1 indicates carboxylic acid. The IH multiplet and the 6H doublet scream isopropyl group.

° CH3• II

CHi CH-C-OH

(b) The formula C9HlOO has five e lements of unsaturation . The 5H pattern between [) 7.2-7 .4 indicates monosubsti tuted benzene. The peak at [) 9.85 is unmistakably an aldehyde, trying to be a triplet because it i s weakly coupled to an adjacent CH2. The two triplets a t [) 2 .7-3 .0 are adjacent CH2 groups .

H2 °

Vc,cJl H I H2

h-(c) The formula CSH802 has two elements of unsaturation. A 3H singlet at [) 2.3 is probably a CH3 next to carbonyl . The 2H quartet and the 3H triplet are certain to be ethyl; with the CH2 at [) 2.7 , this also appears to be next to carbonyl.

° ° I I II CH3CH2 -C-C-CH3

(d) The formula C4H80 has one element of unsaturation, and the signals from [) 5 .0-6.0 indicate a vinyl pattern (CH2=CH-). The complex quartet for IH at [) 4.3 is a CH bonded to an alcohol, next to CH3. The OH appears as a IH singlet at [) 2 . 5 , and the CH3 next to CH is a doublet at [) 1 .3. Put together:

OH I

CH2=CH-CH-CH3 but-3-en-2-ol

(e) The formula C7H160 i s saturated; the oxygen must be an alcohol or an ether, and the broad IH peak at [) 1 . 2 is probably an OH. Let's analyze the spectrum from left to right. The expansion of the IH multiplet at () 1 .7 shows seven peaks-an i sopropy l group! Six of the nine H in the pattern at.5 0.9 must be the doublet from the two methyls from the i sopropyl. The 2H quartet at .5 1 . 5 must be part of an ethy l pattern, from which the methyl triplet must be the other 3H of the pattern at.5 0.9. This leaves only the 3H singlet at <5 1.1 which must be a CH3 with no neighboring Hs.

H I

-OH + -CH2CH3 + CH3- � - + -CH3 + 1 C

CH3

These pieces can be assembled in only one way : CH3 I

HO -C -CH(CH3}z 2,3-dimethylpentan-3-ol I CH2CH3

282

Page 290: Solucionario de wade

1 3 -25 Chemical shift values are estimates from Figure 1 3-4 1 and from Appendix lC, except in (c) and (d), where the values are exact.

(a) estimated

a a C � C I \

C C b '0/ b

I I I 200 1 80 1 60

(b) estimated

b a /C c C

'c II I c, /C a C c

b

I I I 200 1 80 1 60

(c) exact values

Q 8 1 45

l! 8 200

I I I 200 1 80 1 60

I 1 40

I 1 40

I 1 40

l! 8 1 30

I I I 1 20 100 80

Carbon - 1 3 8 (ppm)

i! 8 1 25

I I I 120 1 00 80

Carbon - 1 3 8 (ppm)

Q 8 70

I 60

I 60

0 c

s;:. d C b,..-:CH2 CH;' a'C /" 8 1 26 3

I CH3

I I I 1 20 1 00 80

Carbon - 1 3 8 (ppm)

283

e

I 60

TMS

I I I I

40 20 0

Q s;:. 8 30 8 22

TMS

I I I I

40 20 o

g 8 26 �

<5 1 7

TMS

l I I I

40 20 o

Page 291: Solucionario de wade

1 3-25 continued

(d) exact values

12 8 136

f! 8 192

I I I I 200 1 80 1 60 1 40

1 3-26

(a) '637 0

'68-- � t -- '630

I 1 0

1 3-27

(a)

I 200

'6209

I I 9 8

a a C=C / \

C C b '0/ b

I I 1 80 1 60

I 7

f! '6 1 30

I 1 40

0 II � H C-H

8 135 \ / a C=C

I c H

I I I 1 20 100 80

Carbon - 1 3 & (ppm)

(b) 30 -0- 2 = 15 37 -0- 2 .7 = 14 8 -0- 1 = 8

As a general rule,

b\ H

the 1 5-20 rule works fairly well .

0 II

H3C-C-CH2-CH3 b a c

I I I 6 5 4

& (ppm)

12

I 60

I 3

'670

I I I 1 20 1 00 80 Carbon - 13 & (ppm)

284

I 60

f! 2H

II

TMS

I I I I

40 20 o

12 3H

� 3H

TMS

I I I I 2 1 o

TMS

1II1 I I I

40 20 o

Page 292: Solucionario de wade

1 3-27 continued

(b)

I 200

(c)

b a,..,...C, c C C II I

C, ,..,...C a C e

b

I 1 80

8 200

I I 200 180

(d)

� 8 1 92

I I 200 1 80

I I 1 60 1 40

h 8 14 5

I I 1 60 1 40

h 8 1 36

I I 1 60 1 40

� 8 1 2 5

I I I I 1 20 100 80 60

Carbon - 1 3 8 (ppm)

0 c d C b/.CH2

CH;' a 'c /'"

£ 8 1 26

I I

3 I

CH3 e

I 1 20 1 00 80

Carbon - 1 3 8 (ppm)

H \

I 60

0 II C - H

I a C=C

Ie b\ H H

£ � 1 3 5

I I I I 1 20 1 00 80 60

Carbon - 1 3 8 (ppm)

285

h £ 8 30 8 22

TMS

,II, I I I

40 20 o

Q � 8 26 8 1 7

TMS

J !t I I JII, I I I

40 20 o

TMS

,11, I I I

40 20 0

Page 293: Solucionario de wade

1 3-28 The ful l carbon spectrum of phenyl propanoate is presented below.

� (5 1 73

C

I I I I

200 1 80 1 60

g/'6 1 26/CH

f · · · ·

'6 1 29 · ·

CH · · ·

.d '6 15 1

C

J I

140

� '6 1 2 2 CH

f e 0 gOO-g -CH2CH1

_ c b a f e

I I I 1 20 100 80

Carbon - 1 3 8 (ppm)

I I 60 40

The DEPT-90 wi l l show only the methine carbons, i.e . , CH. All other peaks disappear.

DEPT-90

g/'6 1 26/CH

f �

'6 1 29 (5 1 22 CH CH

12 � '628 8 9 CH2 CH3

TMS

1 I I

20 o

The DEPT- 1 3 5 wi l l show the methyl , CH3, and methine, CH, peaks pointed up, and the methylene, CH2• peaks pointed down.

DEPT-135

g/'6 1 26/CH

f �

(5 1 29 '6 1 22 CH CH

By analyzing the original spectrum, the DEPT-90, and the DEPT- 1 35, each peak in the carbon NMR can be assigned as to which type of carbon i t represent.

286

CH,

TMS

I

Page 294: Solucionario de wade

13 -29 S ince allyl bromide was the s tarting material , i t i s reasonable to expect the al lyl group to be present in the impurity : the triplet at 1 1 5 is a =CH2 ' the doublet at 1 38 is =CH- , and the triplet at 63 is a deshielded al iphatic CH2 ; assembling the pieces forms an allyl group. The formula has changed from C3HSBr to C3H60, so OH has replaced the Br. H2C=CH-CH20H

8 1 1 5 ---' 8 Its "- 8 63 tnplet doublet triplet

Allyl bromide is easi ly hydrolyzed by water, probably an SN 1 process .

1 3 -30

1 3 -3 1

H2C=CH-CH2Br

° II

c /C, C a ° \ / C-C

d b

b a /C, c C C II I C C

a 'C/ c

b

13 -32 Compound 2

H20 H2C=CH -CH20H + HBr

a = 8 180, singlet b = 8 70, triplet c = 8 28 , triplet d = 8 22, triplet

a = 8 1 28 , doublet b = 8 25, triplet c = 8 23 , triplet

Two elements of un saturation in C4�02' one of which i s a carbonyl , and no evidence of a C=C, prove that a ring must be present.

Two elements of unsaturation in C6HIO must be a C=C and a ring . Only three peaks indicates symmetry.

Using PBr3 instead of H2S04iNaBr would give a higher yield of bromocyclohexane.

Mass spectrum: the molecular ion at mJz 1 36 shows a peak at 1 3 8 of about equal height, indicating a bromine atom is present: 1 36 - 79 = 57 . The fragment at mJz 57 is the base peak; this fragment is most l ikely a butyl group, C4�' so a l i ke ly molecular formula is C4H9Br. Infrared spectrum: Notable for the absence of functional groups : no O-H, no N-H, no =C-H, no C=C, no C=O =} most l ikely an alkyl bromide. NMR spectrum: The 6H doublet at 8 1.0 suggests two CH3's spl i t by an adj acent H-an isopropyl group. The 2H doublet at 8 3 . 2 i s a CH2 between a CH and the Br.

Putting the pieces together gives isobutyl bromide.

1 3 -33 The formula C9H\\Br i ndicates four elements of unsaturation , just enough for a benzene ring. Here is the most accurate method for determining the number of protons per signal from integration

values when the total number a/protons is known. Add the integration heights : 4.4 cm + 1 3 .0 cm + 6.7 cm = 24 . 1 cm. Divide by the total number of hydrogens : 24. 1 cm.;. l lH = 2 . 2 cmIH. Each 2.2 cm of integration height = I H, so the ratio of hydrogens is 2 : 6 : 3.

The 2H singlet at I) 7.1 means that only two hydrogens remain on the benzene ring, that is, it has 4 substi tuents. The 6H singlet at 82 .3 must be two CH3's on the benzene ring i n identical environments. The 3H singlet at <') 2 .2 is another CH3 in a slightly different environment from the first two. Substi tution of the three CH3's and the Br in the most symmetric way leads to the structures on the next page.

287

Page 295: Solucionario de wade

13-33 continued a second structure is also possible although it is less likely because the Br would probably deshield the Hs labeled "a" to about 7.3-7.4

13-34

(a)

(c)

a b H CH3

C�3* B'

H CH3 a b

a = 07 . 1 (singlet , 2H) b = 0 2 .3 (singlet, 6H) c = 0 2 .2 (singlet, 3H)

b H3C H a

C�3* B'

b H3C H a

The numbers in i talics i ndicate the number of peaks in each signal.

CH3 - CH2 -CCI2 - CH3 3 4 1

10 CH -CH - CH 3

I 2 3

CH3

7 (b) CH -CH - OH 3

I 1 CH3

2 H H

(assume OH exchanging rapidly-no spl itti ng)

(d) H*C73 (e) Q-CH1 o 2

H H 1

(all of these benzene H's are accidentally equivalent and do not split each other)

13-35 Consult Appendix 1 in the text for chemical shift values. Your predictions should be in the given range, or within 0.5 ppm oflhe given value.

(a)

(c)

(e)

(g)

< }H al l at 0 7 . 2

01.6 CH3 - 0 -CH2 - CH2 - CH2Cl

03 .4 03 .8 03 . 2

o II

CH3 - CH2 - C - CH3 (j 1.0 0 2 . 5 0 2 .0

o H �g - H

08.0 >=<- 09-10 H H

07 . 5 08 .0

The C=O has i ts strongest deshielding effect at the adjacent H (ortho) and across the ring (para). The remaining H (meta) i s less deshielded. 288

(b)

all at 0 1.3

(d) CH3-CH2-C-C-H 8 1 .2 0 2 .2 0 2 .5

CH\ 0 4.3 <5 2-5 (f) CH-O-CH - CH -OH � 2 2

CH3 03 .8 0 3.8 00.9

(h ) o 06-7 I I CH3-CH=CH -C- H

o 1 .7 8 5-6 8 9-10

Page 296: Solucionario de wade

1 3-35 continued

(i) o CH3 II /

HOOC- CH2- CH2- C-O - CH - 8 4.0 8 1 0- 1 2 02.3 02 . 3

"CH3 0 1 . 1

(k) H�H l& l .4 HY:7(H f � H �

07.2 8 2. 5

1 3-36

benzyl ic

c a c CH - CH - CH a = 04.0 (septet, 1 H)

U) H H HH i H ,�-�/ H }

, / , I 01 .3 /C, F=C\ 04.5

H C - C H

(I)

/ I I' �� 8 1 . 3 0 1 .7

H

al lyl ic

H� 06 . 0 � - H\

al ly l ic and benzyl ic

3 I

3

OH b = 02.5 (broad singlet, 1 H) (rapidly exchanging)

c = 0 1 .2 (doublet, 6H) b

1 3-37 (a) The chemical shift in ppm would not change: 04.00.

(b) Coupling constants do not change with field strength: J = 7 Hz, regardless of field strength.

(c) At 60 MHz, 04.00 = 4.00 ppm = (4 .00 x 10-6) x (60 x 1 06 Hz) = 240 Hz The signal is 240 Hz downfield from TMS in a 60 MHz spectrum.

At 300 MHz, (4.00 x 1 0-6) x (300 x 1 06 Hz) = 1 200 Hz The signal is 1 200 Hz downfield from TMS in a 300 MHz spectrum. Necessarily , 1 200 Hz i s exactly 5 times 240 Hz because 300 MHz is exactly 5 t imes 60 MHz. They are directly proportional.

1 3-3 8 H H * c b R d H � Ii CH2- CH2-O - C - CH3

H H '------y--' a

289

a = 07 .2-7 . 3 (multiplet, 5H) b = 04.3 (trip let, 2H) c = 02.9 (triplet, 2H) d = 02.0 (singlet, 3H)

Page 297: Solucionario de wade

13-39

Q (a) b a c

3H £ CH3 -O-CH2CH3 3H

g 2H

TMS

III 1 10 9 8 7 6 5 4 3 2 1 o

8 (ppm)

(b) 0 b Q £ c a II (CH3hCH -C-CH3 3H 6H

f! (To show the pattern IH of peaks. this multiplet TMS is larger than it would j I, I appear on a real spectrum.)

1 0 9 8 7 6 5 4 3 2 o 8 (ppm)

I a b a a

(c) CICH2CH2CH 2Cl

411 I Q

2H

I

TMS

II ..--

10 9 8 7 6 5 4 3 2 o 8 (ppm)

290

Page 298: Solucionario de wade

13-39 continued

(d)

12 2H

f! I H

I I I I 1 0 9 8 7

(e)

f! 12 2H 2H

� IH

I 6

b d H NH2

���� H NH2 b

The NH2 group shields nearby protons on benzene.

I I 5 4

8 (ppm)

a H H b - /

I I 3 2

d * CH1

02N � ;) O-C�C

CH3 a H H b d

� IH

(To show the pattern of peaks, this multiplet is larger than it would

g 4H very broad and variable

I

g 6H

,I I I I appear on a real spectrum.)

I I I I I I I I 1 0 9 8 7 6 5 4 3

8 (ppm)

(f) S ignal (a) is split into a quartet because of the adjacent CH3 with J = 7 Hz. Each of those peaks is then split into a doublet because of the coupling with the trans H, J = 15 Hz. This is called a doublet of quartets, and it is drawn here as two quartets. In a real spectrum, these peaks would overlap and would not be a clean doublet of quartets. See the spl itting tree for problem 13-15.

f! 12 I H

I . ,

I III I I I I I 10 9 8 7 6

V

3

�I d � C

H3C C H3

b H

I I I 5 4 3

8 (ppm)

291

I I 2 1

g 3H

I I 2

TMS

I I o

TMS

I I o

TMS

I I o

Page 299: Solucionario de wade

1 3-40 (a) The NMR of I-bromopropane would have three sets of signals , whereas the NMR of 2-bromopropane would have only two sets (a septet and a doublet, the typical isopropyl pattern).

(b) Each spectrum would have a methyl singlet at 8 2. The left structure would show an ethyl pattern (a 2H quartet and a 3H triplet) , whereas the right structure would exhibit an isopropyl pattern (a IH septet and a 6H doublet) .

(c) The most obvious difference i s the chemical shift of the CH3 singlet. In the compound on the left , the CH3 singlet would appear at 8 2. 1 , whi le the compound on the right would show the CH3 singlet at 8 3.8. Refer to the solution of 13-22 for the spectrum of the second compound.

(d) The spl itting and integration for the peaks in these two compounds would be identical, so the chemical shift must make the difference. As described in text section 1 3-SB, the alkyne is not nearly as deshielding as a carbony l , so the protons in pent-2-yne would be farther upfield than the protons in butan-2-one, by about O.S ppm. For example, the methyl on the carbonyl would appear near 8 2 . 1 whi le the methyl on the alkyne would appear about 8 1 .7.

13-4 1 The multipl icity in the off-resonance decoupled spectrum is given below each chemical shift: s =

singlet; d = doublet; t = triplet; q = quartet. It is often difficult to predict exact chemical shift values; your predictions should be in the right vicinity . There should be no question about the multipl ic ity and DEPT spectra, however.

(a) estimated

q s

.!! 8 1 70

c

I I I I

200 1 80 1 60

t

I 1 40

I I I 1 20 1 00 80

Carbon - 1 3 8 (ppm)

I 60

I 40

I 20

I o

The DEPT-90 spectrum for ethyl acetate would have no peaks because there are no CH groups.

DEPT-135

292

Page 300: Solucionario de wade

1 3-4 1 continued

(b) actual values b a c

H2C = CH - CH2Cl 1 1 8 1 34 45 t d t

g 0 1 34 CH

I I I I 200 1 80 1 60 1 40

CH

DEPT-90

CH

DEPT-135

12 0 1 1 8 CH2

I r I 120 100 80

Carbon -13 8 (ppm)

c b 1 39/s

I 60

(c) actual v&lues Carbons b and c are accidentally equivalent.

M 0 1 29

CH

< >/H2 H2 d '/ '\ C - C - Br

1 27 e f d c-b 39 33

Q 1 29 129 t t

o 1 27 d d 0 1 39 CH C

I I I I I I I I

200 1 80 1 60 1 40 1 20 1 00 80 Carbon - 1 3 8 (ppm)

DEPT-90 and DEPT- 1 35 on next page 293

I 60

� 045 CH2

I 40

I 40

I 20

f 033 CH2

I 20

(CHJ)4Si

TMS

I r o

(CHJ)4Si

TMS

I T 0

Page 301: Solucionario de wade

1 3-4 1 (c) continued

CH DEPT-90

CH DEPT-135 CH

CH2

13-42 The multiplicity of the peaks in this off-resonance decoupled spectrum show two different CR's and a CH3. There is only one way to assemble these pieces with three chlorines.

Cl CI I I

020-- CH3- CH - CH q t I � CI 075

060 d d

13-43 There is no evidence for vinyl hydrogens, so the double bond is gone. Integration gives eight hydrogens, so the formula must be C4HsBr2' and the four carbons must be in a straight chain because the starting material was but-2-ene. From the integration, the four carbons must be present as one CH3, one CH, and two CH2 groups. The methyl is split into a doublet, so it must be adjacent to the CH. The two CH2 groups must follow in succession, with two bromine atoms filling the remaining valences. (The spectrum is complex because the asymmetric carbon atom causes the neighboring protons to be diastereotopic.)

H B r H Br I I I I

H- C- C- C- C-H I I I I

H H H H � �

d a c b

a = 8 4.3 (sextet, IH) b = 8 3.6 (triplet, 2H) c = 8 2.3 (multiplet, 2H) d = 0 l.7 (doublet, 3H)

1 3 -44 There is no evidence for vinyl hydrogens, so the compound must be a small, saturated, oxygen­containing molecule. Starting upfield (toward TMS), the first signal is a 3H triplet; this must be a CH 3 next to a CH2. The CH2 could be the signal at 8 l.5 , but it has six peaks: it must have five neighboring hydrogens, a CH3 on one side and a CH2 on the other side. The third carbon must therefore be a CH2; its signal is a quartet at 0 3 .6 , split by a CH2 and an OH. To be so far downfield, the final CH2 must be bonded to oxygen. The remaining IH signal must be from an OH. The compound must be propan-l-ol.

a = 03 .6 (quartet, 2H) b = 0 3.2 (triplet, IH) c = 0 1 .5 (6 peaks, 2H) d = 00.9 (triplet, 3H)

294

Page 302: Solucionario de wade

13-45

(a)

CH3 CH3 \ / C=C

/ \ CH3 H

Isomer A

b d CH3 CH3

\ / C=C

/ \ CH3 H c a

Isomer A

a = 85.2 (quartet, 1H) b = 8 1.7 (singlet, 3H) c = 8 1.6 (singlet, 3H) d = 8 1.5 (doublet, 3H)

+ H CH3

\ / C=C / \

H CH2CH3 Isomer B

d f H CH3

\ / C=C

/ \ H CH2CH3 d e g Isomer B

d = 84.7 (singlet, 2H) e = 82.0 (quartet, 2H) f = 8 1.7 (singlet, 3H) g = 8 1.0 (triplet, 3H)

(b) With NaOH as base, the more highly substituted alkene, Isomer A, would be expected to predominate-­the Zaitsev Rule. With KO-t-Bu as a hindered, bulky base, the less substituted alkene, Isomer B , would predominate (the Hofmann product).

13-46 "Nuclear waste" is comprised of radioactive products from either nuclear reactions, for example, from electrical generating stations powered by nuclear reactors, or residue from medical or scientific studies using radioactive nuclides as therapeutic agents (like iodine for thyroid treatment) or as molecular tracers (carbon-14, tritium H-3, phosphorus-32, nitrogen-I 5, and many others). The physical technique of nuclear magnetic resonance neither uses nor generates any radioactive elements, and does not generate "nuclear waste". (Some people assume that the medical application of NMR, medical resonance imaging or MRl, purposely dropped the word "nuclear" from the technique to avoid the confusion between "nuclear" and "radioactive" .)

13-47

Mass spectrum: The molecular ion of m/z 117 suggests the presence of an odd number of nitrogens. Infrared spectrum: No NH or OH appears. Hydrogens bonded to both sp2 and sp3 carbon are indicated around 3000 em-I. The characteristic C=N peak appears at 2250 em-I and aromatic C=C is suggested by the peak at 1600 cm-I. NMR spectrum: Five aromatic protons are shown in the NMR at 8 7.3. A CH2 singlet appears at 8 3.7. Assemble the pieces:

H H

H* H H mass 77

+ CH2 + C:::N

mass 14 mass 26

295

H o-� C -C:::N - I

H

mass 117

Page 303: Solucionario de wade

13-48 This is a challenging problem, despite the molecule being relatively small. Mass spectrum: The molecular ion at 96 suggests no Cl, Br, or N. The molecule must have seven carbons or fewer. Infrared spectrum: The dominant functional group peak is at 1685 cm-I, a carbonyl that is conjugated with C=C (lower wavenumber than normal, very intense peak). The presence of an oxygen and a molecular ion of 96 lead to a formula of C6HsO, with three elements of unsaturation, a c=o and one or two c=c. Carbon NMR spectrum: The six peaks show, by chemical shift, one carbonyl carbon (196), two alkene carbons (129, 151), and three aliphatic carbons (23, 26, 36). By off-resonance decoupling multiplicity, the groups are: three CH2 groups, two alkene CH groups, and carbonyl.

Since the structure has one carbonyl and only two alkene carbons, the third element of unsaturation must be a ring.

C=O C=CI CH2 + CH2 + CH2 + 1 ring I ,

H H Since the structure has no methyl group, and no H2C=, all of the carbons must be included in the ring. The only way these pieces can fit together is in cyclohex-2-enone. Notice that the proton NMR was unnecessary to determine the structure, fortunately, since the HNMR was not easily interpreted except for the two alkene hydrogens; the hydrogen on carbon-2 appears as the doublet at 6.0.

13-49

cyclohex -2-enone

>

The mystery mass spec peak at mlz 68 comes from a fragmentation that will be discussed later; it is called a retro-Diels-Alder fragmentation.

The key to the carbon NMR lies in the symmetry of these structures.

b ccra CH3

h c a CH3 b ortho-xylene

erC

:3

,

/ c �/

J " , a d "

,/ hb CH . c 3 meta-xylene " ,

" ,

3

(a) In each molecule, the methyl carbons are equivalent, giving one signal in the CNMR. Considering the ring carbons, the symmetry of the structures shows that ortho-xylene would have 3 carbon signals from the ring (total of 4 peaks), meta-xylene would have 4 carbon signals from the ring (total of 5 peaks), and para-xylene would have only 2 carbon signals from the ring (total of 3 peaks). These compounds would be instantly identifiable simply by the number of peaks in the carbon NMR . (b) The proton NMR would be a completely different problem. Unless the substituent on the benzene ring is moderately electron-withdrawing or donating, the ring protons absorb at roughly the same position. A methyl group has essentially no electronic effect on the ring hydrogens, so while the para isomer would give a clean singlet because all its ring protons are equivalent, the ortho and meta isomers would have only slightly broadened singlets for their proton signals. (Only a very high field NMR, 500 MHz or higher, would be able to distinguish these isomers in the proton NMR.)

296

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13-50 (a), (b) and (c) The six isomers are drawn here. Below each structure is the number of proton signals and the number of carbon signals. (Note that splitting patterns would give even more clues in the proton NMR.) H C CH

H CH

3 'rI" 3 H CH3 H H H _ l-. /'

13 ;L >=< >=< y� D

IxH IxC

6 H H H3C H H3C CH3 H 3xH* 3xC

2xH 3xC

2xH 2xC

2xH 2xC

*Rings always present challenges in stereochemistry. When viewed in three dimensions, it becomes apparent that the two hydrogens on a CHz are not equivalent: on each CHz, one H is cis to the methyl and one H is trans to the methyl. These are diastereotopic protons. A more correct answer to part (b) would be four types of protons; whether all four could be distinguished in the NMR is a harder question to answer. For the purpose of this problem, whether it is 3 or 4 types of H does not matter because either one, in combination with three types of carbon, will distinguish it from the other 5 structures.

5xH 4xC

(d) Two types of H and three types of C can be only one isomer: 2-methylpropene (isobutylene). only isomers that would not be distinguished from each other would be cis- and trans-but-2-ene. )

(The

13-51 For DEPT to be useful in distinguishing isomers, there should be different numbers of CH3, CHz, CH, and C ,ignal, in �NMR' �

~ 6 C signals 5 C signals 5 C signals one CH3 peak four CHz peaks one CH peak

one CH3 peak two CHz peaks two CH peaks

one CH3 peak two CHz peaks one CH peak one C peak

A carbon NMR could not distinguish isomers Band C, and if there is any overlap in the signals, it might have a difficult time distinguishing A. Fortunately, the patterns that would appear in the DEPT-90 would distinguish B (two CH peaks up) from the other two (l CH peak up), and the DEPT -135 would instantly distinguish A (4 CHz peaks down) from C (2 CH2 peaks down). 13-52 (a) MS or IR could not easily distinguish these isomers: same molecular weight and same functional group. They would give dramaticaIIy different proton and carbon NMRs however.

+OCH]

2 singlets in HNMR 3 s inglets in CNMR

�OCHZCH3

4 signals, all with splitting, in HNMR 4 singlets in CNMR

297

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13-52 continued (b) The only technique that would not readily distinguish these isomers would be MS because they have the same molecular weight and would have similar, though not identical, fragmentation patterns.

o

)lo� IR: C=O about 1 730 cm-i HNMR: methyl singlet and ethyl pattern CNMR: ester C=O about 8 1 70

IR: C=O about 1 7 1 0 cm-i HNMR: 3 singlets CNMR: ketone C=O about 8 200

(c) The big winner here is MS : they have different molecular weights, plus the Cl has the two isotope peaks that make a Cl atom easily distinguished. The other techniques would have minor differences and would require having a detailed table of frequencies or chemical shifts to determine which is which. F-o-OH C'-o-OH

M+ = mlz 1 12 M+ = mlz 128 , 1 30

298

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CHAPTER 14-ETHERS. EPOXIDES AND SULFIDES

1 4- 1 The four solvents decrease i n polarity in this order: water, ethanol , ethyl ether, and dichloromethane. The three solutes decrease in polarity in this order: sodium acetate , 2-naphthol , and naphthalene. The guiding principle i n determining solubi l i ty is, "Like dissolves l ike." Compounds of s imilar polari ty wi l l dissolve (in) each other. Thus , sodium acetate wi l l dissolve in water, wi l l dissolve only s l ightly in ethanol , and wi l l be virtual ly insoluble in ethyl ether and dichloromethane. 2-Naphthol wi l l be insoluble in water, somewhat soluble i n ethanol , and soluble i n ether and dichloromethane . Naphthalene w i l l be insoluble i n water, partial ly soluble in ethano l , and soluble in ethy l ether and dichloromethane. (Actual solubi l i ties are difficult to predict, but you should be able to predict trends.)

14-2 CI I CH2CH3

- + / Cl-Al-O : I \

Oxygen shares one of its electron pairs with aluminum; oxygen is the Lewis base, and aluminum is the Lewis acid. An oxygen atom with three bonds and one unshared pair has a positive formal charge . An aluminum atom with four bonds has a negative formal charge .

CI CH2CH3

14-3 The crown ether has two effects on KMn04: first, it makes KMn04 much more soluble in benzene; second, i t holds the potassium ion tightly , making the permanganate more avai l able for reaction. Chemists cal l this a "naked an ion" because it is not complexed with solvent molecules.

IS-crown-6 a "crown ether"

benzene ..

o ,0 '� -/ Mn

/' '\:, 0' 0

Please see the note on p. 13 of this Solutions Manual regarding placement of position numbers.

1 4-4 IUPAC name first; then common name (see Appendix I in this Solutions Manual for a summary of IUPAC nomenclature)

(a) methoxycyclopropane ; cyc lopropyl methyl ether (b) 2-ethoxypropane; ethyl i sopropyl ether (c) l-chloro-2-methoxyethane; 2-ch loroethyl methyl ether (d) 2-ethoxy-2,3-dimethylpentane; no common name (e) 2-t-butoxybutane; sec-butyl t-butyl ether (f) trans-2-methoxycyc lohexan-l-ol ; no common name

14-5

(a) + () o + 2 H20

The alcohol is ethane-l,2-dio l ; the common name is ethylene glycol. 299

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1 4-5 continued

(b) H

H O :o� H O :60 I I H+ I IJ H2C -- CH2 ----l .. � H2C -- CH2

\.

() = o

H O : OH I I H2C- CH2

H2C-CH2 I \

:0: 0 + H30+ -I I H2C--CH2

H20: HO ) H2�-CH2

+ I � HO :O �H I I

H2C-CH2

-The mechanism shows that the acid catalyst is regenerated at the end of the reaction .

1 4-6

.. � H O: H+

I H2C-CH2

I HO :0: I I

H2C-CH2

+ � H OH (I

H2C-CH2 . � I

H O: 0 I I H2C- CH2

(a) dihydropyran (b) 2-chloro-l,4-dioxane (c) 3-isopropylpyran

(d) trans-2,3-diethyloxirane; trans-3,4-epoxyhexane; trans-3-hexene oxide (e) 3-bromo-2-ethoxyfuran

14-7

+ CH3CH2CH2CH2

mJz 57

(f) 3-bromo-2,2-dimethyloxetane

mJz 73

300

CH3 I

+CH , CH3

mJz 43

mJz 1 01

Page 308: Solucionario de wade

1 4-8 SN2 reactions , including the Williamson ether synthesis , work best when the nucleophile attacks a 1 ° or methyl carbon . Instead of attempting to form the bond from oxygen to the 2° carbon on the ring, form the bond from oxygen to the 1 ° carbon of the butyl group.

The OH must first be transformed into a good leaving group: either a tosylate , or one of the hal ides (not fluoride) .

� OH TsCI � OTs ..

pyridine

OH 0- Na+ O�

Q �Q � OTs

Q ..

1 4-9

(a) (f0H Na CH3CH2CH2Br (f0�

... ..

OH Na CH3I OCH3 (b) A ... .. A

OH

Na OH CH3I (c)

¢

... ..

¢Hl N0 2

(d) CH3CH2CH20H

(e)

1 4- 1 0

CH3CH20H CH3 I

CH -C- OH 3 I CH3

(a) ( 1 ) �

or

� OH

(2) �

N0 2

Na CH3CH2Br ... ... CH3CH2CH20CH2CH3

Na CH3CH2CH2Br �

Na •

... CH3CH 2CH20CH2CH3 CH3 PhCH2Br

CH -C - OCH2Ph --_..

3 I CH3

Hg(OAc)2 OCH3 NaBH� � • CH30H

Na CH31 ... ...

OCH3 � 30 1

3-butoxy-l,l-di meth y Icyclohex:me

Page 309: Solucionario de wade

14-10 continued

(b) (1)

(2)

(c) (1) (2)

(d) (1)

(2)

(e) (1)

o Hg(OAch ..

Alkoxymercuration i s not practical here ; the product does not have Markovnikov orientation . (yoH CH31 Na -- ..

6 Hg(OAch NaB H4 .. ..

CH30H

Hg(OAch NaBH4 ------� .. � .. y

OH

(yOCH1

OCH1

(2) Wi l l iamson ether synthesis would give a poor yield of product as the hal ide i s on a 2° carbon.

OH

Na K� --

(f) (1) ==< Hg(OAch NaBH4 ( }-o+ .. .. o-OH

(2) Wi l l iamson ether synthesis is not feasible here . SN2 does not work on either a benzene or a 3° halide.

14-11 An important principle of synthesis is to avoid mixtures of i somers wherever possible; minimizing separations increases recovery of products. Bimolecular dehydration is a random process . Heating a mixture of ethanol and methanol with acid wi l l produce all possible combinations: dimethyl ether, ethyl methyl ether, and diethy l ether. Thi s mixture would be troublesome to separate .

302

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1 4- 12

Ether formation

.. r"'.. H+ CH3CH2CH2 -�.H

Dehydration

H nl+ CH3CH2CH2 - O -H 4 '--..

..

H� -CH2CH2CH3

.. � H20 H

• • ( 1 + CH3CH2CH2 - � - CH2CH2CH3

H nl+ CH3CHCH2 - ° -H I..)

.. H�

H2�

-

Remember I:!.G = M/- TM? Thermodynamics of a reaction depend on the s ign and magnitude of AG. As temperature increases, the entropy term grows i n importance. In ether formation, the I:!.S is smal l because two molecules of alcohol give one molecule of ether plus one molecule of water-no net change in the number of molecules. In dehydration , however, one molecule of alcohol generates one molecule of alkene plus one molecule of water-a large increase in entropy. So TI:!.S is more important for dehydration than for ether formation. As temperature increases, the competition wi l l shift toward more dehydration .

14- 1 3

(a) This symmetrical ether a t 1 ° carbons could b e produced in good yield b y bimolecular dehydration.

(b) This unsymmetrical ether could not be produced in high yield by bimolecular dehydration . Wil l iamson synthesis would be preferred .

/'-..../OH

/"'- OH

(c) Even though thi s ether i s symmetrical , both carbons are 2°, so bimolecular dehydration would give low yields. Unimolecular dehydration to give alkenes would be the dominant pathway. Alkoxymercuration­demercuration is the preferred route.

or } Hg( OAc)z � �

OH

303

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14- 14

0 -• • H- Br R� "-.; (Q • • - I

Br £LoH ---­• • H-Br

'4 "-.; :Br: H

(b) o HI

HBr _ < }- OH

(d) � � O) HI

(e)

� � OH� I

HBr -

BrCH2CH2 -CH -CH2 - Br I

14- 1 6 � �r CH3

� • • / + B o Br' 'Br

Br r-\+ 1_ --- � :'0- B- Br

I BUOHI + HOBBr 2

H • • '0 Bu 1 ,+ -� : O-B-Br

I � I H Br ! 2 H20

B( OHh + 2 HBr

I 1 H3C Br

H .. H 'O� Bu, 1_ .. .. :O-B-Br two proton transfers happen

· • -Sr here: one H+ comes off of oxygen, another H+ goes on the other oxygen; several d ifferent scenarios of how this happens could be proposed

304

Br JL Br

�+ ,Br • • . OJ-B + .Br.

I \ • •

H3C� .. � � H20: Bu, "' Br Ea

' , .. :0-B + CH3Br . . , Br

nucleophilic attach on CH3 faster than on I 0 carbon

Page 312: Solucionario de wade

14-17 Begin by transforming the alcohols into good leaving groups like halides or tosylates: PBr3 NaSH /'.... /'.... NaOH /'.... /'....

�OH - �Br • /' .......", 'SH • /' .......", 'S- Na+

OH TsCI OTs 1 A ·d·- A pyn me I

�S �

14- 1 8 The sulfur at the center of mustard gas is an excellent nucleophile, and chloride is a decent leaving group. Sulfur can do in internal nucleophilic subsitution to make a reactive sulfonium salt and the sulfur equivalent of an epoxide. (a) Cl� • • �Cl C '-- .

S.

very reactive alkylating agent �

HN�S�CI

inactivated enzyme

(b) NaOCI is a powerful oxidizing agent. It oxidizes sulfur to a sulfoxide or more likely a sulfone, either of which is no longer nucleophilic, preventing formation of the cyclic sulfonium salt.

Cl�·S�Cl NaOCI

14- 1 9

Cl� • • �Cl S+ I

o sulfoxide

o + CI�I�CI

S++ I

o sulfone

Generally, chemists prefer the peroxyacid method of epoxide formation to the halohydrin method. Reactions (a) and (b) show the peroxyacid method, but the halohydrin method could also be used.

(a) ==< o

D( MCPBA ..

HO H2SO4 0 (b) 'r- MCPBA P - / • t1 CH2CI2 Ph Ph Ph

CI CIl) 0 (c) U BH3 · THF H202 NaOH .. - .. HO-

CIU Hg(OAc}z NaBH4 CI

U (d) OH NaOH � .. .. .. H2O

(e) �OH NaOH � ..

Cl 305

Page 313: Solucionario de wade

14-20 (a) 1 ) tert-Butyl hydroperoxide is the oxidizing agent. The (CH3hCOOH contains the 0-0 bond just l ike a peroxyacid. 2) Diethyl tartrate has two asymmetric carbons and is the source of asymmetry; i ts function is to create a chiral transition state that is of lowest energy , leading to only one en anti orner of product. Thi s process is called chirality transfer. 3) The function of the titanium (IV) isopropoxide is to act as the g lue that holds al l of the reagents together. The titanium holds an oxygen from each reactant­geraniol , t-BuOOH, and diethyl tartrate-and tethers them so that they react together, rather than just having them in solution and hoping that they wi l l eventually col lide . (b) Al l three reactants are required to make Sharpless epoxidation work, but the key to enantioselective epoxidation is the chiral molecule, diethyl tartrate . When it complexes (or chelates) with titanium, it forms a large structure that is also chiral . As the t-B uOOH and geraniol approach the complex , the steric requirements of the complex allow the approach in one preferred orientation. When the reaction between the alkene and t-B uOOH occurs , it occurs preferentially from one face of the alkene , leading to one major stereosiomer of the epoxide. Without the chiral diethyl tartrate in the complex, the alkene could approach from one side j ust as easily as the other, and a racemic mixture would be formed. (c) Using the enantiomer of diethyl L-tartrate, cal led diethyl D-tartrate, would give exactly the opposite stereochemical results.

t-B uOOH

Ti(Oi-Pr)4 •

O:O

+�H2CHJ

H+OH

COOCHzCH3

I � �O

�" " " " '-...OH

diethyl D-tartrate , the enantiomer of diethy l L-tartrate

1 4-2 1 R , C=O

+

'

OO' )

: O -�

: O : � · A rO � H V 'O '

H " '" t '). " " , Me "" H30+

" " ---.

H' ""

H'�e

�'Me 'C==C'

Me' 'H

trans-but -2-ene

Me HMe

Me�OH

+ Ho0H ........ '--_

/ �" Me H "7 '" HO H Me OH

� ) y IDENTICAL-MESO

stereochemistry shown i n Newman projections:

HCH31 V ):H CH3 --HIi

trans 306

. . . . H - O : OR : O - H I I I

H t H

Mel---<.�-:;

OH HoKle H

'I, + \\'

+ "'Me H '\' +

�.:� OH

H t<::>rCH3

H3Cy H OH

rotate •

OH Q)H

H3C H H3C H

MESO

Page 314: Solucionario de wade

14-2 1 continued H" "H

I, \' "C==C" Me' 'Me

> Mex� OH HO HMe

'" + � '" H H "'/ '" HO Me Me OH �---------- ----------/ Y

cis-2-butene

stereochemistry shown in Newman projections:

H CH3 H --+---;;;.....-+-- CH3 -- --

ENANTIOMERS

rotate ..

OH

� C1h H 0: Cli 3 2 • •

H�CHI H3C H

CIS

14-22

14-23

RC0 3H ----I .. � H2C -CH2 \/

o

anhydrous HBr--only Bc nucleophiles present

H C CH H C - C� 2 ,,/ 2 ..

2" / 2 ..

·0· ·0 � :Br: • •

'-----A H 0r • 1 +

H

CH3CH20CH2CH20H Cellosolve®

CH2CH2Br 1 ---l"�

: O-H H \ Br .. � --...."

CHIRAL­RACEMIC MIXTURE

/ : B r : t

.. CH2CH2Br �I+ H-O-H

BrCH2CH2Br + H20

aqueous HBr-many more H20 nucleophiles than Bc nucleophiles

H 2C, /CH2 _ H2C --- c0. H2C-CH2 , - " Q :O-H .. I 1+ ..

:0: '-----A

H 0r :?+ k HO :?,H H29. H H�

307

H2C-CH2 I 1

HO OH

Page 315: Solucionario de wade

14-24 The cyclization of squalene via the epoxide is an exce l lent (and extraordinary) example o f how Nature uses organic chemistry to i ts advantage . In one enzymatic step, Nature forms four rings and eight chira l centers ! Out of 256 possible stereoisomers , only one is formed !

1 4-25 � � •

:\. - . . � ,,------":0 -CH3 H • • H

14-26 (a)

� O� O- Na+

H (d ) (J N � � I OH �

1 4-27

(a ) ""- / HI 80 � OH

o (c) ,A' '" ,,'" " , " Me Et Me H

o (d)

"A' ""

, ... ,,,, " Me Et Me H

(b)

(e )

CH30 -•

CH30H

H+ •

CH30H

HO

\ = bonds formed

H �CH3. � + :�- 'CH1

H OCH3

H2N �0_ Na+

N C � O- K+

(b) HoX3oH

Me H H0'>---f/ Et " '1 "'-OCH3 Me

Me�OH

; �" Me CH30 H

308

H OCH3

(c) s � �O- Na+

(f)

o :N == N ==N � O_' Na+ +

Page 316: Solucionario de wade

1 4-28 Newly fonned bonds are shown in bold. OH (a ) I

� OH (b) �OH (c )

1 4-29 Please refer to solution 1 -20, page 1 2 of this Solutions Manual .

1 4-30 0 ""-""" (a ) /" O� (b) � O� (c) � (d) � O� (e) � O ....... (

f) 00 ;AH � HACH3 (g) (h )

''''

, ( i )

H3C H

14-3 1 (a ) sec-butyl isopropyl ether (b) t-butyl isobutyl ether (c) ethyl phenyl ether (d) chloromethyl n-propyl ether

14-32 (a ) 2-methoxypropan- l -01 (b) ethoxybenzene or phenoxyethane (c) methox ycyclopentane (d) 2,2-dimethoxycyclopentan- l -01 (e ) trans- l -methoxy-2-methylcyclohexane

1 4-33 (a ) � + y + H2 O

Br Br

HO H

(e) trans-cyclohexene glycol (f) cyclopentyl methyl ether (g) propylene o xide (h) cyclopentene oxide

(f)

trans-3-chloro- l ,2-epoxycycloheptane (g) trans- l -methoxy- l ,2-epoxybutane ; or ,

trans-2-eth y 1 -3 -methox yoxirane (h ) 3-bromooxetane ( i ) 1 ,3-dioxane

(b) � + � Br + H2 O

Br

(c) and (d) no reaction (e ) < )-- OH + CH3CH21 (f)

� OH OCH3

� OH -+o� (h) �NHCH3 (g) "

( i )

HO'" H

309

Page 317: Solucionario de wade

14-33 continued H O

(j )

1 4-34

o (k) �

OCH3 (n) ry."CH3 �1 I 0H

H

Br

( I ) � OH

(a ) On long-term exposure to air , ethers form peroxides . Peroxides are explosive when concentrated or heated . (For exactly this reason, ethers should never be distilled to dryness .) (b) Peroxide formation can be prevented by excluding oxygen . Ethers can be checked for the presence of peroxides , and peroxides can be destroyed safely by treatment with reducing agents . 14-35 (a ) Beginning with (R)-butan-2-o l and producing the (R) sulfide requires two inversions of configuration .

HO H � H Br � R 5

An a lternative approach would be to make the tosylate , displace with chloride or bromide (SN2 with inversion ) , then do a second inversion with NaSCH3.

(b) Synthesis of the (5) isomer directly requires only one inversion .

1 4-36

(a )

H O H � R

TsCI Ts O H � R

pyridine

molecular ion mlz 1 02

.-. � ))� } ./ ............,., " CH2

mlz 73

3 10

Page 318: Solucionario de wade

14-36 continued (b) 3 1 0CH3

+ . H H

\ • • + \+ C - O :

/ \ ----- C = O

/ \ mJz 3 1

molecular ion mJz 1 02

+ / �?,

H mJz 7 1

H H

� . . :OCH3 + 1

CH -"

CH3

H H

+ • •

OCH3 I I CH

" CH3

OCH3

(after H migration)

mJz 59

i :OCH3 + 1

�CH .. •

I I + . . }

�CH mJz 87

14-37

14-3 8

o: � 0 · + � �O - H

H+ I

_ ( CH2

-:;:::.CH2 C I CH3

(a�

MCPBA �

CH2Cl2

(b� MCPBA •

CH2Cl2

(c� MCPBA ..

CH2Cl2

OH I

OH I

CH20H

OCH2_ - CH2

• •

��20: OCH2 _

CH2 • •

/ H H20 : C +/J

CH3 I

CH3

� � �

311

1 ) PhMgBr �

2) H2O

NaOCH3 •

CH30H

H+ �

CH30H

-0 I • • \" CH3 H OH

+- H30+

� Ph OH

�OCH3 OH

�OH OCH3

Page 319: Solucionario de wade

o A

G

D

(X0H Na

" "'0 �

- + (X0 Na " " '0 (X0CH] CH3Br

C � ''' 0 "

G H

1 4-40 The student turned in the wrong product ! Three pieces of information are consistent with the desired product: molecular formula C4HIOO; O-H stretch in the IR at 3300 cm-I (although it should be strong, not weak) ; and mass spectrum fragment at rnJz 59 (loss of CH3). The NMR of the product should have a 9H singlet at 8 1 .0 and a IH singlet between 8 2 and 8 5. Instead, the NMR shows CH3CH2 bonded to oxygen. The student isolated diethyl ether, the typical solvent used in Grignard reactions.

1 4-4 1

(a)

(b)

Predicted product 59 CH3

CH]H- OH

CH3 C4HIOO

Isolated product

C4HIOO O-H at 3300 cm-I O-H at 3300 cm-I due to water contamination

UOH Na U

O- Na+}_

-UO�

/"-.../OH PBr3 � Br �

\ }- OH NaOH \ }- O- Na+ � � \ }-o�

�OH TsCI

�OTs � pyridine

312

Page 320: Solucionario de wade

1 4-4 1 continued

(c)

(d) t HBr, ROOR

(e)

Hg(OAc)z NaBH4 -----l .. � ..

1 ) BH3 • THF ..

Br NaOCH3

..

OH

1 4-42 In the first sequence, no bond i s broken to the chiral center, so the configuration of the product i s the same as the configuration of the starting material . �H

[a]o == - 8 .24°

(Assume the enantiomer shown is levorotatory . )

Na �

�CH2CH [a]o == - 1 5 .6°

In the second reaction sequence , however, bonds to the chiral carbon are broken twice, so the stereochemistry of each process must be considered.

� 0-Q-• • II

H OH CI - S . . . - -',,;;. � II � !J � 0 ..

o

, � � H,;�\-�-Q-"-/ "-/ "-/ \. H 0

t :t� �s 0- _ ) ---- H, o - s-Q-undOUbtedly, some E2 1 � 0 produ

.cts will als�

CH CH 0·- Na+ form In thiS reacHon. 3 2 • • • RETENTION OF SN2-INVERSION CONFIGURATION

The second sequence involves retention fol lowed by inversion , thereby producing the enantiomer of the 2-ethoxyoctane generated by the first sequence . The optical rotation of the final product w i l l have equal magni tude but opposite s ign, [a]o == + 1 5 .6° .

313

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14-43 � CH3 CH3 /""\ CH3 CH3 CH3 . . '- '\ I I . '. - t I I I . .

HO:

OH2C

\ FH -- HO - CH2 - CH - �:

OH2C

\ FH -- HO - CH2 - CH - 0 - CH2�- CH - �:

:0 : :0 : CH3 I -

CH3 CH3 CH3 H2C - CH a I I I . . - � HO - CH - CH - O - CH - CH - O - CH - CH - O : 2 2 2

• • + HOu

H� C� C� C� I I I

HO - CH2 - CH - 0 - CH2 - CH - 0 - CH2 - CH - OH + HO-

14-44 The text uses HA to indicate an acid; A- is the conjugate base .

.. OH

+ H- A

+ : A- I Note that all three carbocation intermediates are 3° ! t

• A­·

"-.H " I '-.... {...C

C+ .. OH

. . O - H . .

This process resembles the cyclization of squalene oxide to lanostero l . (See the solution to problem L 4-24 . ) I n fact , pharmaceutical synthesis o f steroids uses the same type o f reaction cal led a "biomimetic cyclization" .

314

Page 322: Solucionario de wade

(b)

(c) :0 Attack of water gave inversion of configuration at the chiral center; R became S.

" , / \CH2 H

" I H C '-- • • -3

R HO: R

No bond to the chiral center was broken. Configurat ion is retained; R stays as R.

(d) The difference in these mechanisms lies in where the nuc leophi le attacks. Attack at the chiral carbon gives inversion; attack at the ach iral carbon retains the configuration at the chiral carbon. These products are enantiomers and must necessari l y have optical rotations of opposite sign.

1 4-46 methyl cellosolve CH30CH2CH20H

To begin , what can be said about methy l cellosolve? Its molecular weight is 76; i ts IR would show C - -O i n the 1 000- 1 200 cm-I region and a strong O-H around 3300 em- I ; and i t s NMR would show four sets of signals in the ratio of 3 : 2 :2 : 1 .

The unknown has molecular weight 1 34 ; this i s double the weight of methy l cel losolve, minus 1 8 (water) . The IR shows no OH, only ether C-O. The NMR shows no OH, only H-C-O i n the ratio of 3 :2 :2 . Apparently, two molecules of methyl ce l loso lve have combined in an acid-catalyzed, bimolecular dehydration .

CH30CH2CH2 - ° - CH2CH20CH3 1- CH30CH2CH2 �O - CH2CH20CH3 "-________ ... . . I )

(th is compound i s ca lled "diethylene glycol dimethy l ether" , or a shortened, common name is "dig lyme" )

H20 : � H

315

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14-47 The fonnula CgHgO has five elements of unsaturation (enough (4) for a benzene ring). The IR i s usefu l for what is does not show. There i s neither OH nor C=O, so the oxygen must be an ether functional group.

The NMR shows a 5H signal at 8 7 .2, a monosubsti tuted benzene. No peaks in the 8 4.5 -6.0 range indicate the absence of an alkene, so the remaining element of un saturation must be a ring. The three protons are non-equivalent, with complex splitt ing.

< > + 2 C + 0 + 3 H + ring ether

These pieces can be assembled in only one manner consistent with the data.

H H �H

(Note that the CH2 hydrogens are not equivalent (one is cis and one is trans to the phenyl) and therefore have distinct chemical shifts . )

14-48 The key concept is that reagents always go to the less h indered side o f a molecule first. I n this case, the "underneath" side is less hindered; the " top" side has a CH3 hovering over the double bond and approach from the top wi l l be much more difficult, and therefore slower, than approach from underneath .

hindered

Br2 side

less J I MCPBA

MCPBA approaches from the bottom; the epoxide is fanned from MCPBA without the participation of any other reagent. The epoxide fonns on the less hindered side.

fonnation of the bromohydrin begins with fonnation of the bromonium ion on the less hindered side

water must attack from the side opposite the bromonium ion , from the TOP

Br

B

2,6-lutidine

c

Epoxides B and C are diastereomers ; they wil l have different chemical and physical properties. Nucle�philes can react with C much faster than with B for precisely the same reason that explained their fonnatlOn: approach from underneath i s less hindered and is faster than approach from the top.

316

o

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CHAPTER IS-CONJUGATED SYSTEMS, ORBITAL SYMMETRY, AND

ULTRAVIOLET SPECTROSCOPY

1 5 - 1 Look for: 1 ) the number of double bonds to be hydrogenated-the fewer C=C, the smaller the MI; 2) conj ugation-the more conjugated, the more stable , the lower the �; 3) degree of substitution of the alkenes-the more substituted, the more stable, the lower the !1H.

(a) � < � < � < smallest !1H

� < ==C� < � biggest !1H

< < < < smal lest !1H biggest !1H

1 5 -2 Reminder: H-B i s used to symbolize the general fonn for an acid, that i s , a protonated base; B- is the conj ugate base . H

H H H H

H H H Hf"B

• H H H The key step i s hydride shift from a 2° carbocation to a 2° allyiic, resonance-stabil ized carbocation, which can subsequently lose a proton to fonn a conjugated diene.

317

H H H

H H H allylic ! B:

H H H

H H H

Page 325: Solucionario de wade

1 5-3 (You may wish to refer to problem 2-6.) orbital picture

(a) H H " " 1 2 3 / "

C == C == C H

" t "H

sp

The central carbon atom makes two n bonds with two p orbitals . These p orbitals must necessari ly be perpendicular to each other, thereby forcing the groups on the ends of the allene system perpendicular.

(b) H " /

" " c == C == C Cl

'" "

Cl

H

mirror Cl

H '" \\\

H

C == C == C '"

/ 'CI

non-superimposable mirror images = enantiomers

1 5-4 Carbocation stabi lity depends on conjugation (benzylic, al lyl ic) , then on degree of carbon carrying the positive charge.

i H H \ I C - C + II \

H2C CH3 2°

H H � H

:�� Jl H C H

I H

1 ° less significant contributor

more significant contributor

H H

H

equivalent to the first resonance form

318

Page 326: Solucionario de wade

The most basic species in the reaction mixture wil l remove this proton . The oxygen of ethanol i s more basic than bromide Ion .

1 5-6

+

allyJic H3C Y'� ...... H H3C , + � H () C � e

two carbon:-===:/·

� o . fv electron deficient I

so nucleophi le C H CH can attack ei ther 2 3

AgC I +

6� �I IYI iC (i � H H

CH:CH'0.H3

'----f- OCH2CH3 � H

H

H - O - CH C� • • 2 13 �

H - O : I

two carbons are electron defic ient so nucleophi le can attack ei ther

CH2CH3 I

.. CH2CH3 • •

+ H �0.HJ

'----+-+ ?,- CH,CH3 '-l( H H

319

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1 5 -7

'"' � H - Br

( ..

(b) � '"' 0 0 \ / H - Br O - C - C ...

H / o 0 / \\ H3C CH2

same carbocation { as in (a)

H3C H \ / C == C H / \ ("' 1 +

H C CH - O - H 3 2 0 0 � - H20

H3C H \ / + C - C / \\

electron deficient 0B° :- so nucleophile 0 0 r 0 can attack either

..

� - H20

H3C CH2 two carbons are

320

_ electron deficient

°B� o so nucleophile 0 0 0 0 can attack either

Page 328: Solucionario de wade

1 5 -7 con� n I +

&{ B r (c) / ..... Br - Br

H2C -�? - ? H H H H

. . :Br :

Br Br I I

H2C - C - C == CH2 I I

H H

al lyl ic

...........

� +

Br } I +

H2C -? == ?)CH2

H H

_ two carbons are :B� : e lectron deficient so nucleophi le can attack ei ther

Br Br I I

H2C - C == C - CH2 I I

H H

While the bromonium ion mechanism i s typical for isolated alkenes, the greater stabi l i ty of the resonance­stabi l ized carbocation wi l l make it the lower energy intermediate for conj ugated systems.

(d) C� I Ag+ i +

H3C - C == C - CH2 .. H3C - C == C - CH2 I I - AgCl I I �

H H H H � H20 . .

• • +

H - ?� )H

H3C - C == C - CH2 I I

H H

H29.

OH I

H3C - C == C - CH2 I I

H H

321

al lyl ic -

� +

! +

H3(f-r CH2} . .

H O 2. •

two carbons are e lectron defic ient • •

+ so nucleophile

O

H v? - H _ can attack either

H3C - C - C - CH2 I I

H H

H20 . .

OH I

H3C - C - C == CH2 I I

H H

Page 329: Solucionario de wade

1 5-7 continued

same carbocation as in (d) (e) ?�A + � + allylic +

} H3C - ? - ? == CH2 _:

g; H3(C ? - ? == CH2 -- H3C - ? == ? - fH2

H H H H H H , H2R I H2R

1 5 -8

(a)

two carbons are + electron deficient so nucleophi le can attack either

OH

+

OH I I

H3C - C - C == CH2 + H3C - C == C - CH2 I I

Br Br I I

H2C - C - C == CH2 + I I

A H H

I I H H

Br Br I I

H2C - C == C - CH2 I I

B H H

H H

(b) � n Br 20 / ..... B r - Br I +

H H H H

Br Br I I

. . -:Br :

H2C - C - C == CH2 I I

A H H

--

allyl ic

• • - two carbons are :Br : electron deficient

• •

so nucleophi le Br Br can attack either

I I + H2C - C == C - CH2

I I B H H

(c) The resonance form A + , which eventually leads to product A , has positi ve charge on a 2° carbon and is a more significant resonance contributor than structure B+ . With greater positive charge on the 2° carbon than on the 1 ° carbon, we would expect bromide ion attack on the 2° carbon to have lower activation energy. Therefore, A must be the kinetic product . At higher temperature , however, the last step becomes reversible, and the stabi lity of the products becomes the dominant factor in determining product ratios. As B has a disubstituted alkene whereas A i s only monosubstituted, it is reasonable that B is the major, thermodynamic product at 60° C .

322

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1 5 -8 continued (d) At 60° C, ionization of A would lead to the same allyl ic carbocation as shown in (b), which would give the same product ratio as formation of A and B from butadiene.

Br (Br I I

H2C - C - C == CH2

A I I

H H

- Br- I +

B r Br I I

al lyl ic

B r 1 0} I +

H2C - T == T)CH2

H H

_ two carbons are :B� : electron defic ient

so nucleophile

Br B r I I

can attack eIther

H2C - C - C == CH2 + H2C - C == C - CH2

1 5-9

(a)

(b) nn hv Br - Br --- 2 B r -

A I I I I

H H 10 % 90 % B H H

HBr + �Q H H ) eX allyl ic 0:

H___ I H

. H H { � n Br - Br

H H

�-------------------- Br - + recycles in chain mechanism

Br

1 5 - 1 0 ("Pr" i s the abbreviation for n-propyl , used below. )

NBS generates a low concentration of Br2 a � CCl4 �N - Br + H - Br -

a

+ Br - Br

323

Page 331: Solucionario de wade

1 5 - 1 0 continued

initiation n n hv Br - Br --- 2 B r ·

propagation

Br · + pr -�-�== CH2 --.� HB, + { P, -�J= CH' ...... t---l.� P,JJ- �HJ �ri? ��O �� ) I -hexene

H H H H I I recycles in chain mechanism I I

L--__ � __________ Br . + Pr - C - C == CH2 + Pr - C == C - CH2 I I

B r B r

The HBr generated in the propagation s tep combines with NBS t o produce more Br2 ' continuing the chain mechanism.

1 5- 1 1

(a) U B, (b) Br � + These are the major products from abstraction of a 2° al lyl ic H.

1 5 - 1 2 Both halides generate the same al lyl ic carbanion.

Br + I MgBr

• �

H3C - C == C - CH2 H3C - C = C - CH2 � � M� � �

1 5 - 1 3

(a)

1

MgBr Mg I

---l�� CH3CHCH3 + CH3CH == CHCH2Br ether

324

(e) < ) elI,Br

H

benzylic radicals are even more stable than al lyl ic

+

I H3C - C - C == CH2

I I H H

Page 332: Solucionario de wade

1 5- 1 3 continued

Thi s synthesis could also be performed sequential ly .

1 5- 1 4

o

1 5 - 1 5

+

(d) O +

1 5 - 1 6

(a)

C O OCH3

C O OCH3

OMe

OMe

+

Br - CH2 add one-half equivalent o

o

(b) CH) O:( CH30

(e)

CH,o,( (b) 0

+

O! OEt

+

NCXCN

NC CN

325

dec-5-ene

H2C - CH2CH2CH,

F

C O OCH3

CN

(c) Co + � eN

0 (f) 0 <0 +

0

(C)6CH3

I . . . . . . C O OCH3 . . . . . . CHO

CH3 racemic mixture; wedge and dashed bonds show relative stereochemistry

Page 333: Solucionario de wade

1 5- 1 7 These structures show the alignment of diene and dienophi le in the Diels-Alder transition state, leading to 1 ,4-orientation in (a) and 1 ,2-orientation in (b) .

(a) : 0 :

this left structure is a VERY minor resonance contributor; however, it explains the orientation for the diene as carbon- l is more negative and carbon-2, a 3°C, is s l ightly more positi ve because of methyl group stab i l ization

(b)

1 5 - 1 8 For clarity, the bonds formed in the Diels-Alder reaction are shown in bold.

COOCH3 (b)

1 5 - 1 9 For a photochemically allowed process, one molecule must use an excited state in which an electron has been promoted to the first anti bonding orbital . All orbital interactions between the excited molecule 's HOMO* and the other molecule's LUMO must be bonding for the interaction to be al lowed; otherwise, i t i s a forbidden process.

HOMO* excited state of diene HOMO

bonding interaction J (constructive overlap�

LUMO of dienophi le

anti bonding interaction (destructive overlap)

In the Diels-Alder cycloaddition, the LUMO of the dienophi le and the excited state of the HOMO of the diene ( labeled HOMO*) produce one bonding interaction and one anti bonding interaction. Thus, this is a photochemically forbidden process.

326

Page 334: Solucionario de wade

15-20 For a [4 + 4] cyc\oaddition :

(a) Thermal

HOMO

(

anti bonding in teraction

)

bonding interaction

(

Photochemical

) HOMO* 11:3

one interaction is anti bonding

� forbidden

both interactions are bonding

� allowed

(b) A [4 + 4] cycloaddition is not thermal ly al lowed, but a [4 + 2] (Diels-Alder) i s !

1 5 -2 1

HOMO

bonding interaction�

'--

LUMO

A

bonding J interaction �

11:3

A = E C l E = - l = I cm A = 0.50 c l

convert mass to moles : 1 mg x 1 g 1 mole x -- = 6 x 1 0--6 moles 1 60 g

C =

E =

6 x 1 0--6 moles I O mL

0.50

( 6 X 1 0-4 ) ( 1 )

1 000 mg

1 000 mL 6 0--4 M x

= x l l L

= 833 "" 8 327

Page 335: Solucionario de wade

1 5-22 (a) 353 nm: a conjugated tetraene-must have highest absorption maximum among these compounds ; (b) 3 1 3 nm: c losest to the bicycl ic conjugated triene in Table 1 5-2; the diene i s in a more substituted ring, so i t is not surprising for the maximum to be sl ightly higher than 304 nm; (c) 232 nm: s imi lar to 3 -methylenecyclohexene in Table 1 5-2; (d) 273 nm: 1 ,3 cyclohexadiene (256 nm) + 2 alkyl substituents (2 x 5 nm) = predicted value of 266 nm; (e) 237 nm: l ike 3 -methylenecyclohexene (232 nm) + 1 alkyl group (5 nm) = predicted value of 237 n m

1 5-23 Please refer to solution 1 -20, page 12 of th is Solutions Manual .

1 5-24

(a) isolated (b) conjugated (c) cumulated (d) conjugated

b 4 0 3 1

W 70 3

5 5

1 5 -25

(a) � (b) o-Cl one equivalent of HCl

(e) conjugated 0 (f) cumulated ( 1 ,2) and conjugated (2,4) H

(c)

OH

/ C � H2C = C = C " CH H 2

" Br

(d) � + � (e) Br� + Br � Br Br

Br Br

(f) �

+ � + � B r

Br Br minor- Br not conjugated

(g) (h)

(i)

328

OH

Page 336: Solucionario de wade

15-26 Grignard reactions are performed in ether solvent. For c larity, the bonds formed are shown in bold.

(a) � Br + � BrMg I

(b) N' + BrMgr •

15-27

(a)

(b)

(c)

(d)

5

4

+ CH2 CH2 C H2 �CH2""'''I--·· �?�CH2 ....... 1--•• ��H2

: 0 :

H H H

·6 : 0 : : 0 : : 0 :

-........ C ..... :0 H

:0 :

'iC, o H

a: O- H

H

329

+ I H

H

:0 :

Page 337: Solucionario de wade

15-27 continued

(e) { &�H2_ (f) H H (!=.yO: __ (

C yO: V HC �

(g)

15-28 A

(a) A = Ee l E= - l = 1 cm A = 0.74 c l

convert mass to moles: 0.00 10 g x 1 mole

3.9 x 10-6 moles = 255 g

3.9 x 10-6 moles 1000mL = 3.9 x 10-5 M c = x

100mL lL

0.74 El E = '" ( 3.9 x 10-5 ) ( 1 )

(b) This large value of E could only come from a conjugated system, el iminating the first structure. The absorption maximum at 235 nm is most l ikely a diene rather than a triene. The most reasonable structure is:

compare with: co "max = 235 nm

Solved Problem 15-3

330

"max = 232 nm

Table 15-2

Page 338: Solucionario de wade

15-29

(a) � + � +

Br B r trans

(b) CPr" is the abbreviation for n-propyl , used below. )

NBS generates a low concentration of Br2 o �N- Br + I1-Br --

o initiation

nn hv Br-Br --- 2 B r·

propagation

o �N-H

o

Br� cis

+ Br-Br

first step is abstraction of al ly lic hydrogen to generate allylic radical

H H I I

Br· + Pr -C -C ==CH2 ",,--,r I-hexene

�--------- Br· + recycles in chain mechanism

15-30

(a) 0 � COOH

(b)

---I"� HBr +

H H I I

Pr - C -C ==CH2 I Br

COOCH3

331

+

radical wi l l be a mixture of c is and trans

H H t I I Pr- C ==C- �H2 n�) Br - Br

H H I I

Pr -C == C -CH2 I Br

CIS + trans

(c ) ,-k � COOH

Page 339: Solucionario de wade

15-30 continued

(e) (f)

'" "'eN 0= ",eN

",

I _ "" " eN

",CN ",

(Note: Yield of the product in (f) would be smal l or zero due to severe steric interaction in the s-cis conformation of the diene. See Figure 15- 16 . )

15-3 1 f �+/ (a) 1 H2C � k

(b)

(c)

(d)

(e)

more significant contributor: 2°

i .. -:0:

I + CH3-C=NH2

{ H�C � H

{ H2C�

H

0( ..

0(

0(

0

H

..

..

:0:

� H

H

H

most sign ificant contributor-

more significant contributor: 2°

:0: t I more significant contributor--H2C == C - CH3 negative charge on more electronegatIve atom

II • • more significant contributor-:0: t CH3-C-NH2 no charge separation

0

0( .. H

H H

:0:

.. .. H

H H

negative charge on more e lectronegative atom

332

Page 340: Solucionario de wade

15-31 continued (f)

most significant contributor---all atoms have ful l octets

.. ..

more significant contributor­negative charge can be stabilized by electronegative ch lorines

15-32

(a) The absorption at 1630 cm- i suggests a conjugated alkene_ The higher temperature al lowed for migration of the double bond_

(b) UCH,CH,CH] desired

OCH,CH2CH] actual

(c) expected:

doubly al lyl ic mlz 122

mlz 79

[ ]+--{o } � . � � actual : + � OCH,:f-CH2CH] .6 CH, �CH'

mlz 122 mlz 93

15-33

(aj ( + �H (b) 0 C 'lL Cl +

o

o (dj ( ¢ +

(e) o + o

333

mass 43

+ • CH2CH3 mass 29

+

Page 341: Solucionario de wade

15-33 continued 0

(h) ( i ) d' CI +tb : S�O (g) CI*CI CI +Q-CI CI*CI ::;:,- CI fCN CI ::;:,- CI + I CI CI CI CI

15-34 0 NC H

(a)

0 7 Co+ I 90° C NH

� 0

exo H

(b) The endo i somer is usually preferred because of secondary p orbital overlap of C=O with the diene in the transition state . (c) The reasoning in (b) applies to stabi l ization of the transition state of the reaction , not the stabi l i ty of the product. Arguments based on transition state stabi l ity apply to the rate of reaction, inferring that the endo product is the kinetic product. (d) At 25° C, the reaction cannot easily reverse, or at least not very rapidly . The endo product i s formed faster and is the major product because its transition state is lower i n energy-the reaction is under kinetic control . At 90° C, the reverse reaction is not as slow and equi l ibrium is achieved. The exo product is less crowded and therefore more stable-equil ibrium control gives the exo as the major product.

t �

." -'. exo transition state-no secondary orbital overlap , , . . , . ,."

. �'. . .

,.,., ... endo transition state-

,/ secondary orbital overlap stabil ization

, . . , . ...... .. ...

endo product

exo product-more stable, less crowded �-------------------------------------------reaction -

334

Page 342: Solucionario de wade

15-35 Nodes are (a) represented by dashed l ines.

(b) - 1[6*

- 1[5*

H ::' _H '" H 1[ 1

(C)

1t6*

1ts*

1t4*

1t3

1t2

1tl

( ) ( ) ( )

0

0

0

335

Page 343: Solucionario de wade

15-35 continued

(d) Whether the triene i s the HOM O and the alkene is the LUMO, or vice versa, the answer wi ll be the same.

(e)

15-36

Thermal

LUMO

one anti bonding interaction

� forbidden

o

+

o

HOM O*

Photochemical

LUM O two bonding interactions

� allowed

(a) J H2C== ?- ? == ? - CH2 ....... f---� .. H2C== ? - ?- ?==CH2 .. 1 H H H H H H

(b) Five atomic orbitals wil l generate five molecular orbitals .

(c) The lowest energy molecular orbital has no nodes . Each higher molecular orbital wil l have one more node , so the fifth molecular orbital wil l have four nodes .

336

Page 344: Solucionario de wade

15-36 continued

(d) Nodes are represented by dashed l ines.

(e) __ 11:5*

11:4*

_1 11:3 1L n,

JL n,

11:5*

11:4*

11:3 (non bonding)

11:2

11:1

(f) The HOMO, 11:3, contains an unpaired electron giving thi s species its radical

character . The HOMO is a non-bonding orbital with lobes only on carbons 1, 3 , and 5, consistent with the resonance picture. i�-�-�}

337

Page 345: Solucionario de wade

15-36 continued (g) --

---

H _H

(h) __

H H

_H

1£5*

1£4*

1£3

1£2

1£]

1£5*

Mg

ether ..

Again , it is 1£3 that determines the character of this species. When the single e lectron in 1£3 of the neutral radical is removed, positive charge appears only in

the posi tion(s) which that electron occupied. That i s , the posi tive charge depends

on the now empty 1£3' with empty lobes (posit ive charge) on carbons 1, 3, and 5, consistent wi th the resonance description .

J � ___ � ___ � l 1 + + + f

Again, it is 1£3 that determines the character of this spec ies. The negative

charge depends on the filled 1£3' with lobes (negative charge) on carbons 1, 3,

and 5, consistent with the resonance description. i-·�-�-�;t a A

MgBr H3C-C-OH I

CH3 (b) Use Appendix 3 to predict Amax values . Alkyl substi tuents are c irc led .

transoid cyclic diene = 2 17 nm 4 alkyl groups = 20 nm exocyclic alkene = 5 nm TOTAL = 242 nm

desired product

c isoid cyclic diene = 253 nm 3 alkyl groups = 15 nm TOTAL = 268 nm

338

c isoid cyclic diene = 253 11m 4 alkyl groups = 20 nm TOTAL = 273 nm--AHA!

actual product B

Page 346: Solucionario de wade

15-37 continued

(c)

hydride shift

B is produced in preference to the other cisoid diene above because Bls diene system is more highly substi tuted and therefore more stable

Hi): � B Q H3C- ? -H

CH3

15-38 It is stunningly c lever reactions l ike this that earned E. J. Corey his Nobel Prize. Diels-Alder

�o COOCH3

DCH3

+ 1 ---o imagine the diene on top of the dienophile in the transition state, leading to the endo adduct

retro-Diels-Alder

o

new o �

0= C = 0 dienophi le

same as

o 339

endo product

COOCH3

amazing!

o

Page 347: Solucionario de wade

16- 1

CHAPTER 16-AROMA TIC COMPOUNDS

Note: The representation of benzene w ith a c ircle to represent the 1t system is fine for questions of nomenclature, properties , i somers, and reactions. For questions of mechanism or reactivity, however, the representation with three alternating double bonds (the Kekule picture) is more informative. For c larity and consistency , thi s Solutions Manual will use the Kekule form exc lusively . o � equivalent representations of henzene c::::::> ©

Kekule form used in the Solutions Manual

H H

H • • ' C . • H • C·- • C . H . • C,, . H • C · -. C .

• C -. • C • H ' '. C · • H

• C . • C • H ' • C

'·' . H

H H

16-2 All values are per mole .

(a)

(b)

(c)

16-3

(a)

benzene -208 kJ

1 ,4-cyclohexadiene -240 kJ MI = + 32 kJ

benzene -208 kJ

- cyclohexene - 1 20 kJ

f.,}{ = - 88 kJ

1 ,3-cyc lohexadiene -232 kJ - cyc lohexene

H I

H C H .... C,.. .::::-C'

- 1 20 kJ

f.,}{ = - 1 1 2 kJ

H I

H C H 'c-:::' 'c'

II 1---- I II ..... C, -:::.C ....

H C H I

H

H H "-C=C "

..... C.::::- ,..C .... H C H

I H

H H "-C-C " { I I ---- II II } C=C C-C / , / ,

H H H H

- 49. 8 kcal

- 57.4 kcal

+ 7.6 kcal

- 49.8 kcal

- 28 .6 kcal

- 2 1 .2 kcal

- 55.4 kcal

- 28.6 kcal

-26 .8 kcal

341

Page 348: Solucionario de wade

1 6-3 continued

(b)

16-5 Figure 1 6-8 shows that the fi rst 3 pairs of electrons are in three bonding molecular orbitals of cyclooctatetraene. Electrons 7 and 8, however, are located in two different nonbonding orbitals . As in cyclobutadiene, a planar cyc looctatetraene is predicted to be a diradical , a particular ly unstable electron configuration.

Models show that the angles between p orbitals on adjacent 1l: bonds approach 90°.

1 6-7 (a) nonaromatic: internal hydrogens prevent planarity (b) nonaromatic: not al l atoms in the ring have a p orbital (c) aromatic: [l4]annulene (d) aromatic: also a [ 1 4]annulene in the outer ring: the internal alkene is not part of the aromatic system

16-8 Azulene satisfies all the criteria for aromaticity, and it has a Huckel number of 1l: electrons: 10. Both heptalene ( 121[ electrons) and pentalene (8 1l: electrons) are anti aromatic .

1 6-9

(a) (b) 1l:s* 1l:6* 1[7* 1l:6* 1 1 1l:7*

nonbonding

1l:4 1[5 �!1�!g.x 1l:4 1l:s

1l:2 H H H

1[3 1[2 1l:3 1l:1

This electronic configuration is antiaromatic.

342

Page 349: Solucionario de wade

16-9 continued

(c)

These are TOP views. Nodes are shown by dashed lines.

16- 10

(aj ____ �y--. ____ _ ______ _

n2* �-----iII[

(b)

(c)

--- n2* n3* - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - non bonding

nI

nI i s bonding; n2* and n3* are antibonding.

n, nI ��-------- �------�) \�------- �------�/ Y Y cation-aromatic anion-antiaromatic

343

Page 350: Solucionario de wade

16- 1 1

(a)

(b) --- n5* antibonding

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - _ . non bonding n2 n3

(c) R" j (" 7l:z H n3

\. n1 )

y cation-antiaromatic

16-12 (a) antiaromatic: 8 n electrons, not a Huckel number (b) aromatic : 10 n electrons, a Huckel number (c) aromatic: 18 n electrons, a Huckel number (d) antiaromatic: 20 n electrons, not a Huckel number (e) nonaromatic: no cyclic n system (f) aromatic : 18 n electrons, a Huckel number

7l:z

\.

344

bonding

R" j � H j r" n1

y anion-aromatic

n3

)

Page 351: Solucionario de wade

1 6- 1 3 The reason for the dipole can be seen in a resonance form distributing the electrons to give each ring 61t electrons. This resonance picture gives one ring a negative charge and the other ring a positive charge.

aromatic tropylium ion 6 1t electrons

�---------��----------( several resonance forms delocaJizing the ___ positive charge around the seven-membered

aromatic cycIopentadienyl anion 6 1t electrons

�--------���--------� "\

several resonance forms ____ de localizing the negative charge around the five­membered ring

nng "'---- ----�

16- 1 4

+ AgBF4 -- AgCl +

CI

The crystalline material soluble / in polar organic solvents is cycIopropenium tetrafluoroborate.

composite resonance picture shows that both rings are aromatic with charge separation

aromatic cyc\opropenium ion

!KCI

Cl-<Jl

1 6- 1 5 Draw resonance forms showing the carbonyl polarization, leaving a positive charge on the carbonyl carbon. 0 0 0 & 0

t t t .. . . :0: :0: :0:

I I I +C 0 +C B 0

2 1t electrons-4 1t electrons-AROMATIC

The three-and seven-membered rings are aromatic; the five-membered ring is antiaromatic and, nut surprisingly, very reactive.

6 1t electrons- ANTI AROMATIC AROMATIC

345

Page 352: Solucionario de wade

l6-l6 o 0 1 7N:(: 00 'I I N �oo )

9 N N I 0 0 3 H l6-l7

The structure of purine shows two types of nitrogens. One type (N-l, N-3, and N-7) has an electronic structure l ike the nitrogen in pyridine; the pair of electrons is in an sp2 orbital planar with the ring. These electrons are avai lable for bonding, and these three nitrogens are basic . The other type of nitrogen at N-9 has an electronic structure l ike the nitrogen of pyrrole; i ts e lectron pair is in a p orbital , perpendicular to the ring system, and more importantly, an essential part of the aromatic pi system. With this pair of electrons, the pi system is aromatic and has 10 electrons, a Huckel number, so the electron pair is not avai lable for bonding and N-9 is not a basic nitrogen .

(a) The proton NMR of benzene shows a single peak a t 8 7 .2; alkene hydrogens absorb a t 84.5-6. The chemical shifts of 2-pyridone are more s imi lar to benzene's absorptions than they are to alkenes . It would be correct to infer that 2-pyridone is aromatic .

(b) H H � x):� ---.. ---..

H�H J:� .. H N + 0:

The lone pair of electrons on the nitrogen in the first resonance form is part of the cycl ic pi system. The second resonance form shows three alternating double bonds with six electrons in the cyclic pi system, consistent with an aromatic electronic system. H H

H These resonance forms show that the hydrogens at positions with greater electron density are shielded, decreasing chemical shift .

"" 7 &7.26 Hx�.i�. H N 0 : � ... -..... Hn�. H N 0 :

These resonance forms show that the hydrogens at positions with greater positive charge are deshielded, increasing chemical shift. H 0 0 _

:0 :

87.3l H

346

This resonance form of thymine shows a cyclic pi system with 6 pi electrons , consistent with an aromatic system. Four of these electrons came from the two lone pairs of electrons on nitrogens in the first resonance form.

Page 353: Solucionario de wade

16- 1 8 (a) (b)

( .. ;�: / \ :N �.S.:

0

aromatic: 4 n aromatic: 4 n electrons in e lectrons in double bonds double bonds

(c)

d 0

not aromatic : no cyclic n system

(d) H H

(�)�6 o 0+

. . . .

plus one pair plus one pair because of sp3

aromatic : cation on carbon-4 indicates an empty p orbital ; two n bonds plus a pair of electrons from oxygen makes a 6 n electron system

from oxygen from sulfur carbon

. . -(e)6

� 6 o 0 +

(f) O I . . sp3 N H

. . . . aromatic : resonance fonn shows "push-pul l " of electrons from one 0 to the other , making a cyclic n system with 6 electrons

not aromatic : no cyclic n system because of sp3

carbon

16- 19

16-20

H H 1 1

H � • • ,B ........ � H H +,B ...... + H �N N' ..... N' N'

1 I ...... f-. -I.� _ 1 1 1_ ,B ....... . ,B...... ,B.;:::. ,B ......

H N H H N H 1 1+

H H

(a) 9

aromatic: resonance form shows electron pair from N making a cyclic n system with 6 electrons

Borazole is a non-carbon equivalent of benzene. Each boron is hybridized in Its

nonnal sp2. Each nitrogen is also sp2 with Its pair of electrons in its p orbital. The system has six n electrons in 6 p orbitals--aromatic!

� � OCOI � � --- � --- � � � h- h- � � 1 0

anthracene

347

Page 354: Solucionario de wade

16-20 continued

(b) anthracene H

_B_r�n

----,B

.. �r

0 ©-c00 H Br �cQo� I�

.... � �

.Br. H Br H

H

H H plus many other resonance forms

H bromide attack at C-9 leaves two aromatic rings

bromide attack at C-lO leaves two aromatic rings

cis and trans

(c) A typical addition of bromine occurs with a bromonium ion intermediate which can give only anti addition. Addition of bromine to phenanthrene, however, generates a free carbocation because the carbocation is benzylic, stabi l ized by resonance over two rings. In the second step of the mechanism, bromide nucleophi le can attack either side of the carbocation giving a mixture of cis and trans products.

(d)

L1 ---

---

E 1

16-2 1 C l 6

chlorobenzene

C l �Cl �Cl 1 ,2 ,3-trichlorobenzene

Br + , ·B • C �'" • r.

H H

--../ resonance-stabi l ized

Cl �Cl

V

H

Cl o �Cl

a-dichlorobenzene m-dichlorobenzene ( 1 ,2-dichlorobenzene) ( 1 ,3-dichlorobenzene)

Cl C l qCl o Cl �Cl

Cl 1 ,2 ,4-trichlorobenzene 1 ,3 ,5-trichlorobenzene

348

+ HBr

Cl

¢ Cl

p-dichlorobenzenc ( 1 ,4-dichlorobenzene)

Cl (yCI

YCI

CI 1 ,2 ,3 ,4-tetrachloro­benzene

Page 355: Solucionario de wade

16-2 1 continued CI �CI

CIVCI

1 ,2 ,3 ,5-tetrachloro­benzene

16-22 (a) fluorobenzene (b) 4-phenylbut-I-yne (c) m-methylphenol , or

3-methylphenol

CI �CI

CIY CI

1 ,2,4 ,5-tetrachloro­benzene

CI �CI

CI¥CI CI

1 ,2 ,3 ,4 ,5-pentachloro­benzene

CI CI�CI

CIYCI CI

1 ,2 ,3,4,5,6-hexachloro­benzene

(d) o-ni trostyrene (g) 3 ,4-dinitrophenol (e) p-bromobenzoic acid, or

4-bromobenzoic acid

(common name: m-cresol) (f) i sopropoxybenzene, or

isopropyl phenyl ether

(h) benzyl ethyl ether, or benzoxyethane, or (ethoxymethyl)benzene, or a-ethoxytoluene

16-23 These examples are representative. Your examples may be different from these and sti l l be correct.

(a) < }-O-CH3

methyl phenyl ether or methoxybenzene or anisole �

(e) o-CH-OH

I� .0 diphenylmethanol

16-24 + CH2

H H

..

H H

H

(b) CH3 --< }- S03H

p-toluenesulfonic acid

Br

(0 < f < >

(c) < }- Li (d) 02N-{ }-OH

phenyl l ithium These are phenols. Thi s is 4-nitrophenol .

CH3CH20

(g) � >-CH,OH

Br 1 ,3-dibromo-2-phenylbenzene

3-ethoxybenzyl alcohol or 3-ethoxyphenylmethanol

HiiH H + H H ...... C

H I � I H

.. .. .. .. ..

H H H I

H H

H

H

349 H

Page 356: Solucionario de wade

16-25 :O� 0-1 H+ Ij_� CH -CH == CH2 ..

Amax 220 nm (strong) 258 nm (weak)

< )+¥3 CH-CH _ ==CH � -H20 2

i O-CH=CH-t � • < > �H-CH=CH2 }

plus resonance fOnTIS with positive charge on the benzene ring

o-� CH == CH -CH2 - 1 OH

Amax 250 nm (strong) 290 nm (weak)

electronic systems with extended conjugation absorb at longer wavelength

H20 : � -- o-� CH==CH- ?�2

- crH H20 :

16-26 Please refer to solution 1 -20, page 12 of this Solutions Manual .

1 6-27 OCH3 OH C O OH NH2 CH3 (a) ON02 (b) � OCH3 (e) 9 (d) 9 (e) 6 1.0 1.0 1.0

1.0 .0

Cl OCH3

HC=CH2 HC=CH2 (0 9 (g) 9

1.0

HC=CH2 Br

(k) � CH2 OCH3 (I) 6 1.0

HO

NH2 N02

CH O H H (j) I (h� (; ) C. Cl-

Ih 0 CH30 OCH3 -

c-0 (m)¢H

I� .0

CH3

350

CH3 (n) OCH3 (0)

I � II .0

Na+

Page 357: Solucionario de wade

16-28 (a) 1 ,2-dichlorobenzene (ortho) (b) 4-nitroanisole (para) (c) 2 ,3-dibromobenzoic acid (d) 2,7 -dimethoxynaphthalene

16-29 (5 toluene

CH3 �CH l

llACH3

1 ,2,3-trimethylbenzene

o-xylene qCHl

CH3 1 ,2,4-trimethylbenzene

(e) 3-chlorobenzoic acid (meta) (0 2,4,6-trichlorophenol (g) 2-sec-butylbenzaldehyde (ortho) (h) cycIopropenium tetrafluoroborate

£ CH3 .& CH3

1 ,3,5-trimethylbenzene (common name: mesitylene)

¢ CHJ

p-xylene

1 6-30 Aromaticity is one of the strongest stabil izing forces in organic molecules . The cycIopentadienyl system is stabi l ized in the anion form where it has 6 'It electrons, a Huckel number. The question then becomes: which of the four structures can lose a proton to become aromatic?

Whi le the first , third, and fourth structures can lose protons from sp3

carbons to give resonance­stabilized anions, only the second structure can make a cycIopentadienide anion. It wi l l lose a proton most easi ly of these four structures which, by definition, means it is the strongest acid.

16-31

(a)

H Hyly H HVH

H

�H H

H H � �

H H

H

351

AROMATIC

H H

H H H H

H H

H

Page 358: Solucionario de wade

16-3 1 continued

(b) Br

6

(c) Br

6

u B r

o +

Jl Br U +

Br

c::::::=> ¢ Br

6 +

Br

Br

Cl � Br

+

flZ); Br +

Br Br a B' B'L6 I + I � � \. J

Y different positions of double bonds

+

Br

a:;; Br

(ignoring enantiomers)

(d) The only structure consistent with three isomers of dibromobenzene is the prism structure, called Ladenburg benzene. It also gives no test for alkenes, consistent with the behavior of benzene. (Kekule defended his structure by claiming that the "two" structures of ortho-dibromobenzene were rapidly interconverted, equil ibrating so quickly that they could never be separated. )

(e) We now know that three- and four-membered rings are the least stable, but this fact was unknown to chemists during the mid- 1 800s when the benzene controversy was raging. Ladenburg benzene has two three-membered rings and three four-membered rings (of which only four of the rings are independent), which we would predict to be unstable. (In fact , the structure has been synthesized. Called prismane, it i s NOT aromatic, but rather, is very reactive toward addition reactions. )

352

Page 359: Solucionario de wade

16-32 H H

(a)

(b)

(c)

(d)

(e)

(t)

(g)

HA H nonaromatic

0 U aromatic-6 11: electrons

H H 0 nonaromatic

N [ ) N aromatic-6 11: electrons

° ° 0 0

nonaromatic

aromatic-6 11: electrons

ill W

H I

anti aromatic-4 11: electrons

o

aromatic-2 11: electrons

N o o N o H H +"N' o

aromatic-6 11: elec trons

nonaromatic aromatic-6 11: electrons

nonaromatic

H 1 -C o

aromatic-6 11: e lectrons

aromatic-6 11: electrons

o 0 o (::] - c

I H

H 1+ C o anti aromatic-4 11: electrons

o N I

H nonaromatic

if oxygen is sp2 => anti aromatic (8 11:

H I

B (B is Sp2 but donates o no e lectrons to the 11:

. . system) antJaromatlc-4 11: electrons

o +N I

H aromatic-6 11: electrons

o 0 o (00] +c

I H

e lectrons) ; if oxygen is sp3 => nonaromatic

=> not aromatic in either case

aromatic-6 11: electrons

H � oo H -N/' N-+ \ I aromatic-6 11: electrons

nonaromatic

H

anti aromatic-12 11: electrons

aromatic-10 11: electrons

this i s a tough call-it has 10 11: electrons so it could be aromatic , but internal H's might force it out of planarity

353

Page 360: Solucionario de wade

1 6-32 continued H H (h) H H CH3

I I I (B is sp2, but 0 + C C- B donates no

0 0 0 electrons to the 1t system)

nonaromatic aromatic- anti aromatic- aromatic-6 1t electrons 8 1t electrons 6 1t electrons

16-33 The c lue to azulene i s recognition of the five- and seven-membered rings. To attain aromaticity, a seven-membered carbon ring must have a positive charge; a five-membered carbon ring must have a negati ve charge . Drawing a resonance form of azulene shows thi s : OJ --.. ---. C» --.. ---� ri:�

sance } <0@

composite picture

The composite picture shows that the negative charge is concentrated in the five-membered ring, giving rise to the dipole.

16-34 Whether a n itrogen i s strongly basic or weakly basic depends on the location of i ts electron pair. If the electron pair i s needed for an aromatic 1t system, the nitrogen wi l l not be basic (shown here as "weak base") . If the electron pair i s i n ei ther an sp2 or sp3 orbital, i t i s avai lable for bonding, and the nitrogen is a

"strong base" . H (a)

(d)

� HN: " N : I \ I' strong

weak base base

H weak N / base ce·' I � �\

/N strong base strong base

(b)

(e)

I N 0

strong base

H I () 0

strong base

(c)

(f)

O�

N : \ I

strong base

/ strong • • base N

g1 weak base

16-35 Where a resonance form demonstrates aromaticity, the resonance form is shown.

(al O (b) : 0 :

1"-': 6 /.

+R aromatic + �

aromatic

.. (c) : 0 : :6

aromatic

+ (d) :0 : :0 :

BAD 0 NOT aromatic-although it can be drawn , the resonance form on the left i s NOT a significant contributor because the oxygen does not have a ful l octet; the form on the right shows the correct polarization of the carbonyl , but it's still not aromatic because of only 4 1t electrons 354

(e) : 0 :

0 aromatic

Page 361: Solucionario de wade

16-35 continued

Nitrogens whose e lectrons are needed to complete the aromatic 1t system wi l l not be basic . • • ,..... basic (D � + (g) � + (h) + �,..... not basic (i ) �,..... basic

(j) NH2

o 9 c;):;iC C�"�

ic

tf.:a,iC

aromatic , •• 0.. sp3 + N O :

: 0 : H not basic • • - • • -aromatic; N i s not basic but the 0- i s a weak base

aromatic NOT aromatic because of sp3 carbon ; both N are basic , although the top one

(k) +� [j

B­H

aromatic; not basic

H N

(I) (::] o . .

NOT aromatic : i f both N and 0 are sp2 , then the pi system has 8 e- ; N will be basic

16-36 �H o --.. �-

(a)

..

6 _ .. ---- ..

-

C CH2

.. · -"6

..

(b) init iation

nn hv

-

B r - Br - 2 B r-

propagation steps on the next page

is conjugated, lowering its basicity

o (mlO

NOT aromatic: if 0 is sp2, then the pi system has 8 e-

..

+

.. ..

CH2 CH2 6 .. -

6·:·

355

-

-

aromatic ; bottom N is not basic because it has donated its electron pair to make the ring aromalic

+

23

(5

Page 362: Solucionario de wade

1 6-36 continued

propagation

� r-·CH2

n! 6 Br - Br + 1.&

HEr +

----l .. � B r· +

resonance-stabilized

(c) Both reactions are SN2 on primary carbons , but the one at the benzyl ic carbon occurs faster. In the transition state of SN2, as the nuc leophi le is approaching the carbon and the leaving group is departmg, the electron density resembles that of a p orbital . As such, it can be stabi l ized through overlap with the 1'[ system of the benzene ring. • • 8-

: Br :

H

stabi l ization through overlap

:O-CH3 8- • • 16-37

(a)� VlCH� +

3 isomers

only 1 isomer

(c ) The original compound had to have been meta-dibromobenzene as this is the only dibromo isomer that gives three mononitrated products .

Br Br � � �N02 + VlBr Vl Br

Br o y Br

Page 363: Solucionario de wade

16-38

(a) The fonnula CgH70CI has five e lements of unsaturation, probably a benzene ring (4) plus ei ther a double bond or a ring. The IR suggests a conj ugated carbony l at 1 690 cm-1 and an aromatic ring at 1602 cm-I. The NMR shows a total of five aromatic protons, indicating a monosubstituted benzene. A 2H singlet at 8 4.7 is a deshielded methylene.

< > o I I

+ C + CI } o o-� C - CH I

2 -

CI

(b) The mass spectral evidence of molecular ion peaks of 1 : 1 intensity at 1 84 and 1 86 shows the presence of a bromine atom. The rnlz 1 84 minus 79 for bromine gives a mass of 1 05 for the rest of the molecule, which i s about a benzene ring plus two carbons and a few hydrogens . The NMR shows four aromatic hydrogens in a typical para pattern (two doublets), indicating a para-disubstituted benzene. The 2H quartet and 3H triplet are characteristic of an ethyl group.

-0- + Br +

16-39 (a) � l ike the ends of a � conjugated diene � �

} Br-o-CH2CH3

o

Diels·Al der product

16 -40 new sigma bonds shown in bold

(a) No, biphenyl is not fused. The rings must share two atoms to be labeled "fused". (b) There are 1 2 n electrons in biphenyl compared with 10 for naphthalene. (c) Biphenyl has 6 "double bonds". An i solated alkene releases 1 20 kJ/mole upon hydrogenation .

predicted: 6 x 1 20 kJ/mole (28 .6 kcaVmole) "" 720 kJ/mole ( 172 kcaUmole)

observed: 4 1 8 kJ/mole ( 1 00 kcal/mole)

resonance energy: 302 kJ/mole (72 kcaVmole)

(d) On a "per ring" bas is , biphenyl i s 302 -:- 2 = 1 5 1 kJ/mole, the same as the value for benzene. Naphthalene's resonance energy is 252 kJ/mole (60 kcaVmole) ; on a "per ring" basis , naphthalene has only 1 26 kJ/mole of stabi l ization per ring . This i s consistent with the greater reactiv i ty of naphthalene compared with benzene. In fact, the more fused rings, the lower the resonance energy per ring, and the more reactive the compound. (Refer to Problem 1 6-20. )

16-4 1 Two protons are removed from sp3 carbons to make sp2 c arbons and to generate a n system with to n electrons.

H

H 357

Page 364: Solucionario de wade

16-42

(a) � H l:J" H

�-0C - H

aromatic OCH

aromatic

""-- Ag+ r>--+ (c) V- Cl .. �CH

aromatic

(d) C�::: NaOH. r:}

- H

H H H 1-

C

Ocomatk

(e) C4H9Li (f) ..

0

Y Ag+ ..

Cl

aromatic

• • • • + o 0 ()-o H aromatic

16-43 These four bases can be aromatic , partially aromatic, or aromatic in a tautomeric form. In other words, aromaticity plays an i mportant role in the chemistry of all four structures. (Only electron pairs involved in the important resonance are shown . ) For part (b) , ni trogens that are basic are denoted by B. Those that are not basic are shown as NB. Note that some nitrogens change depending on the tautomeric form. (a) and (c)

cytosine

: 0 :

{�B/ H

�N�O . • NB I • ••

H uraci l

continued on the next page

aromatic to the extent that this resonance form contributes

.. : 0 : r7�:_ l_� .. N O·

+ I NB • •• H

aromatic to the extent that this resonance form contributes

358

aromatic

OH

r NB

� N�OH B tautomer­aromatic

Page 365: Solucionario de wade

16-43 continued

16-44

.. · 0 · ·. 0 ·. • • aroratic

H '�� N) '

__ ...

_. HN��.�

H,N � N Jl-N NB H,N"/� N Jl-N NB B B \ B B \ H H

guamne aromatic to the extent that this resonance form contributes

BN

B:):NH 2

NB

I I '\\ ful ly aromatic

..

/ N N NB

B \ H adenine

(a) Antiaromatic-only 4 n electrons.

tautomer­ful ly aromatic

(b) This molecule is electronical ly equivalent to cyc lobutadiene. Cyclobutadiene is unstable and undergoes a Diels-Alder reaction with another molecule of itself. The t-butyl groups prevent dimerization by blocking approach of any other molecule .

(c) Yes , the n itrogen should be basic. The pair of electrons on the nitrogen i s in an sp2 orbital and is not part of the n system.

b' a'

...

I 2

Analysis of structure I shows the three t-butyl groups in unique environments in relation to the nitrogen. We would expect three different s ignals in the NMR, as is observed at _1 1 00 C. Why do signals coalesce as the temperature is increased? Two of the t-butyl groups become equivalent-which two? Most l ikely, they are a and c that become equivalent as they are symmetric around the n itrogen. But they are not equivalent in structure I-what i s happening here?

What must happen is an equilibration between structures I and 2, very slow at -1100 C, but very fast at room temperature, faster than the NMR can differentiate. So the signal which has coalesced is an average of a and a' and c and c'. (Thi s type of low temperature NMR experiment is also used to differentiate axial and equatorial hydrogens on a cyclohexane. )

The NMR data prove that I i s not aromatic , and that I and 2 are i somers, not resonance forms. I f I were aromatic , then a and c would have i dentical NMR signals at a l l temperatures.

359

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1 6-45

Mass spectrum: Molecular ion at ISO; base peak at 135, M - 15, is loss of methyl.

Infrared spectrum: The broad peak at 3 500 cm-l is OH; thymol must be an alcohol. The peak at 1620 cm-l suggests an aromatic compound.

NMR spectrum: The singlet at (') 4.8 is OH; it disappears upon shaking with D20. The 6H doublet at (') 1 .2

and the IH multiplet at (') 3 .2 are an isopropyl group, apparently on the benzene ring. A 3H singlet at (') 2.3

is a methyl group, also on the benzene ring. Analysis of the aromatic protons suggests the substitution pattern. The three aromatic hydrogens

confinn that there are three substituents. The singlet at (5 6.5 is a proton between two substituents (no neighboring H's). The doublets at (5 6.75 and (5 7. 1 are ortho hydrogens, splitting each other.

X �Y + CH]

Z

Several isomeric combinations are consistent with the spectra (although the single H giving 8 6.5 suggests that either Y or Z is the OH group-an OH on a benzene ring shields hydrogens ortho to it, moving them upfield). The structure of thymol is:

¢r:;� thymol

CH3 The final question is how the molecule fragments in the mass spectrometer:

OH ..

CH3

H CH3 I

'C+

OH I h- ..

CH3 mlz 135

H CH3 I

'c

..

¢r

+ OH I .. h- +

CH3

• CH3

resonance stabilization of this benzylic cation includes fonns with positive charge on three ring carbons and on oxygen (shown)

360

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16-46

Mass spectrum: Molecular ion at 1 70 ; two prominent peaks are M - 1 5 (loss of methyl) and M - 43 (as we shall see, most likely the loss of acetyl, CH3CO). Infrared spectrum: The two most significant peaks are at 1680 cm-1 (conjugated carbonyl) and 1600 cm-I (aromatic C=C). NMR spectrum: A 3H singlet at 8 2.7 is methyl next to a carbonyl, shifted slightly downfield by an aromatic ring. The other signals are seven aromatic protons. The IH at d 8 . 7 is a deshielded proton next to a carbonyl. Since there is only one, the carbonyl can have only one neighboring hydrogen. Conclusions:

+ rnJz 1 27 including 7H =} mass 120 for carbons =} 1 0 C

The fragment C1oH7 is almost certainly a naphthalene. The correct isomer (box) is indicated by the NMR.

16-47

o II C-CH3

This isomer would would have two deshielded protons in the NMR.

Although all carbons in hexahelicene are sp2, the molecule is not flat. Because of the curvature of the nng system, one end of the molecule has to sit on top of the other end-the carbons and the hydrogens would bump into each other if they tried to occupy the same plane. In other words, the molecule is the beginmng of a spiral. An "upward" spiral is the nonsuperimposable mirror image of a "downward" spiral, so the molecule is chiral and therefore optically active.

The magnitude of the optical rotation is extraordinary: it is one of the largest rotations ever recorded. In general, alkanes have small rotations and aromatic compounds have large rotations, so it is reasonable to expect that it is the interaction of plane-polarized light (electromagnetic radiation) with the electrons in the twisted pi system (which can also be considered as having wave properties) that causes this enormous rotation.

361

three-dimensional picture of hexahelicene showing the twist in the system of six rings

Page 368: Solucionario de wade

16-48 The key concept in parts (a)-(c) is that an aromatic product is created. (a) +

(b)

5.:\ U + H+

A

:O-H

� o· -----.. ..

:O-H I + -.0·· 0

:O-H I

c a .. ..

The protonated carbonyl gives a resonance-stabilized cation. Protonation of the singly bonded oxygen does not generate a resonance-stabilized product.

+

6� :O-H :O-H :O-H

6· I I

c c o· 0 I . + H+ ---- .. .. .. .. h-

B AROMATIC

The last resonance form shows that the cation produced is aromatic and therefore more stable than the corresponding nonaromatic ion. In this second reaction, the products are more favored than in the first reaction which is interpreted as the reactant B being more basic than A.

erCI o + +0 • • ".......-A OH OH

CI- + .. ..

C (rCI o + +0 • • ".......-A OH OH Ih- CI- + .. .. Ih-D

AROMATIC

The product from ionization of C is stabilized by resonance. The ionization product of D is not only resonance-stabilized but is also aromatic and therefore more stable. A reaction that produces a more stable product will usually happen faster under milder conditions because the transition state leading to that product will be stabilized, leading to a lower activation energy.

362

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16-48 continued (e) uGH

H+ catalyst o Dehydration of F produces an aromatic product that is more stable than the product from E. A reaction that produces a more stable product will usually happen faster under milder conditions because the transition state leading to that product will be stabilized, leading to a lower activation energy.

(d)

E

H+ catalyst H20 + o

F AROMATIC . . ..

HO� V base ---·0 ... 1)- :0 :0

phenol Resonance stabilization of the phenoxide anion shows the negative charge distributed over the one para and two ortho carbons .

HO

base ---

umbelliferone (Umbelliferone is one of about 1000 coumarins isolated from plants, primarily the families of Angiosperms: Fabaceae, Asteraceae, Apiaceae, and Rutaceae.)

Resonance stabilization of the anion of umbelliferone gives not only the same three forms as the phenoxide anion, but in addition, gives an extra resonance form with (-) charge on a carbon, and the most significant resonance contributor, another form with the (-) charge on the other carbonyl oxygen. This anion is much more stable than phenoxide which we interpret as enhanced acidity of the starting material, umbelliferone. In fact, the pKa of umbelliferone is 7 .7 while phenol is about 1 0 .

. . :0

:0

363

Page 370: Solucionario de wade

16-49 Humulon (sometimes spelled humulone), even though highly resonance stabilized, cannot be aromatic because the carbon shown at the bottom of the ring is tetrahedral and must be sp3 hybridized. A ring is aromatic only when all of the atoms in the ring have a p orbital which the sp3 carbon does not.

16-50 Are you familiar with the concept "tough love", that is, sometimes you have to be stem with someone you love for their own good? The concept is that sometimes to demonstrate one emotion, the behavior has to appear exactly the opposite. Granted, this is a stretch to apply it to atoms, but the point is that sometimes atoms like oxygen and nitrogen can have conflicting effects: they can withdraw electron density by their strong electronegativity (an inductive effect) but at the same time, they can donate electron density through their resonance effect. This phenomenon will prove important in the reactivity of substituted benzenes described in Chapter 1 7 . • •

:0 :

pyridine I '" 2

""":::::3

yrN 98.60

4 9 7.25

9 7.64

yr+� 98.19

I '" 2

""":::::3 4 9 7.40

9 7.32

pyridine N-oxide

Nitrogen is electronegative, so it exerts a deshielding effect on H-2 in pyridine. The effect diminishes with distance as expected with an inductive effect, although it is harder to explain why H-4 is deshielded more than H-3. In pyridine N-oxide,with an even more electronegative oxygen attached to the N, it would be reasonable to expect that the hydrogens would be deshielded. This is the case with H-3; we can infer that the effect on H-3 is purely an inductive effect.

Even more interesting is the shielding effect on H-2 and H-4; these chemical shifts are shifted upfield. This must reflect the other side of oxygen's personality, the donation of electron density through a resonance effect. Drawing the resonance forms clearly shows that the electron density at H-2 and H-4 (and presumably H-6) increases through this donation by resonance, in perfect agreement with the NMR results. As we will see in Chapter 1 7 , in most cases, resonance effect trumps inductive effect.

. . :0 : :0 :

¢en + 11 QtH

I '" --- I ..

C

......::::: ......::::: 9 9 9 9

---

:0 : + II

(�J(H _C 9

I 9

:0 :

�¢:H HC� --- . I � 9

9

Note: The practice of applying human emotions to inanimate objects is called "anthropomorphizing". For example, we say that an atom is "happy" when it has a full octet of elecctrons. In casual conversation, this gets a point across, but it is not appropriate in rigorous scientific terms, for example, on exams. Under more formal conditons, we are expected to use the specific terms of science because they are well defined and do not permit sloppy or fuzzy concepts.

As our equipment technician says: "Don't anthropomorphize computers. They hate that."

364

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CHAPTER 17-REACTIONS OF AROMATIC COMPOUNDS

The representation of benzene with a circle to represent the 1t system is fine for questions of nomenclature, properties, isomers, and reactions. For questions of mechanism or reactivity, however, the representation with three alternating double bonds (the Kekule picture) is more informative For clarity and consistency, this Solutions Manual will use the Kekule form exclusively.

1 7- 1

1 7-2

Sigma complex

Oc H

I OH

:!=!H2� � C I

/ ,,+ ) / " H :� H � H OH

H addition product­NOT AROMATIC

While the addition of water to the sigma complex can be shown in a reasonable mechanism, the product is not aromatic. Thus, it has lost the 1 5 2 kllmol (36 kcal/mole) of resonance stabilization energy. The addition reaction is not favorable energetically, and substitution prevails.

CI CI � I . . + • • 1-

: CI-Cl: + AI-Cl --- : CI -CI - Al - CI I 1 CI CI

H H

H H CI CI H I - I-. . + • • + : CI -CI - Al -Cl -----.. :Cl- AI-CI + .. v· I I

Cl C

H H CI CI H +'H

H H t H H

H CI . . H :CI: HCI + .. CI .. .. Cl

H H + AICI3

H H t H H H CI 1-

:CI-AI-C) .:U I CI

365

Page 372: Solucionario de wade

1 7-3 Like most heavy metals, thallium is highly toxic and should not be used on breakfast cereal. +

�==== .. � Tl(OCOCF3h + CF3COO-

1 7-4 CH3 CH3

:0: I I N+ ----

:gD CH3

CH3 I N02

H --.. --... VH CH3

Benzene's sigma complex has positive charge on three 2° carbons. The sigma complex above shows positive charge in one resonance form on a 3° carbon, lending greater stabilization to this sigma complex. The more stable the intermediate, the lower the activation energy required to reach it, and the faster the reaction will be.

1 7 -5 delocalization of the positive charge on the ring

delocalization of the negative charge on the sulfonate group . .

: 0 : :0: : 0: • • I I • • I I • • I :O- S- Ar ....... I--J�- O==S-Ar ---- O== S- Ar • . I I I • • I I

:0: :0: :0: . .

366

("Ar" is the general abbreviation for an "aromatic" or "aryl" group, in this case, benzene; "R" is the general abbreviation for an "aliphatic" or "alkyl" group.)

Page 373: Solucionario de wade

1 7-6 (a) The key to electrophi l ic aromatic substi tution lies in the stabi l ity of the sigma complex . When the electrophile bonds at ortho or para positions of ethylbenzene, the positive charge is shared by the 3D carbon with the ethyl group. Bonding of the e lectrophile at the meta posi tion lends no particular advantage because the positive charge in the sigma complex is never adjacent to, and therefore never stabi l ized by, the ethyl group.

ortho CH2CH3 { CH2CH3 �I B q- ;r - FeBr3 -- a: � � � CH

+

CH2CH3 I

+ C Q Br H

3 D-good

(b) Electrophil ic attack on p-xylene gives an intermediate in which only one of the three resonance forms IS stabi l ized by a substituent (see the solution to Problem 17 -4) . m-Xylene, however, i s stabi l ized in two of i ts three resonance forms. A more stable intermediate gives a faster reaction .

1 7-7 For ortho and para attack , the positive charge in the s igma complex can be shared by resonance with the vinyl group. Thi s cannot happen with meta attack because the positive charge i s never adjacent to the vinyl group. (Ortho attack is shown; para attack gives an intermediate with positive charge on the same carbons . )

367

&:} "extra " resonance form

Page 374: Solucionario de wade

1 7-8 Attack at only ortho and para positions (not meta) places the positive charge on the carbon with the ethoxy group, where the ethoxy group can stabi l ize the positive charge by resonance donation of a lone pair of electrons . (Ortho attack is shown; para attack gives a simi lar intermediate . )

1 7-9 OIthOcB, 0B'

: �H

H ?'

� H � H� � NH2

NH2 1 + C Br V H NH2

"extra " resonance form

+ �B' � ll) - H

"extra " resonance form mCla& Br H6 � Br � QB'

� H

para NH2 HQ Br H

1 7 - 1 0 OCH(CH3h

6 + 8"

� H NH2 + 1

C Q �

Br H

o + Br2 ___ (XBr '" Br

H H

+

6 Br H "extra "

resonance form

+ HBr (g)

¢H Br H

S ubsti tution generates HBr whereas the addition does not. If the reaction is performed in an organic solvent, bubbles of HBr can be observed, and HBr gas escaping into moist air will generate a c loud. If the reaction is performed in water, then adding moi st li tmus paper to test for acid wi l l differentiate the results of the two compounds.

368

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1 7- 1 1 (a) Nitration i s performed with n itric acid and a sulfuric acid catalyst. In strong acid, amines in general, including ani l ine, are protonated.

+

o (b) The NH2 group i s a strongly activating ortho,para-director. In acid, however, i t exists as the protonated ammonium ion-a strongly deactivating meta-director. The strongly acidic nitrating mixture itself forces the reaction to be slower. (c) The acetyl group removes some of the electron density from the nitrogen, making it much less basic; the nitrogen of this amide is not protonated under the reaction conditions . The N retains enough electron density to share with the benzene ring , so the NHCOCH3 group is sti l l an activating ortho,para-director, though weaker than NH2 .

: 0 : • • I I Ph - � - C - CH3 ..

H

: 0 : + I

Ph - N = C - CH3 I

H

1 7- 1 2 Ni tronium ion attack at the ortho and para positions places positive charge on the carbon adjacent to the bromine, allowing resonance stabi l ization by an unshared electron pair from the bromine. Meta attack does not give a stabi l ized intermediate.

ortho Br Br aNOl

aN02

H .. .. H .. ::::-... CH HC h

+ +

meta Br Br

G Ht) N02 .. .. ::::-... ::::-...

H H . . para Br : B r :

HQ I + C

.. .. Q .. H N02 H N02

N02

..

. . :Br :

I + C N02 (J H . •

Br

. . + Br:

C(02

I H h

"extra " resonance form

.. .. e N02

H H

+ Br : Br

Q .. .. QH H N02 H N02

"extra " resonance form

369

Page 376: Solucionario de wade

17 - 13 Cl (a) O=!r H (b) O; HOcda�:-a�:} : �)� d!

r

1 H H H resonance-stabilized

(c) A bromine atom can stabilize positive charge by sharing a pair of electrons. Bromine can do this in any cationic species, whether from electrophilic addition or electrophilic aromatic substitution.

1 7- 14

(a)

(b)

(c)

(e)

1 7- 1 5 (a)

<i � Cl N02 COOH

(x

Br

� N02 OH ~ � CH3

N02

CH3 02N'Q

+ I :::::-.. Cl +

+

COOH O

Br +

O N :::::-..

2 OH 02N'(\

I + :::::-.. CH3 h ��"g o,p-d'''''"'

V CH3 weak o,p-director

CH3 (rN02

� Cl trace COOH

(d)

~ � N02 OCH3 OH (xN02

� CH3 trace

370

Page 377: Solucionario de wade

1 7- 1 5 continued (b) Cl C l

02Nn � Y N02

+

N02

I + � N02

(c) OH strong o,p-director

(d)

(e)

(f)

1 7- 16

<r NO'

Cl weak o,p-director

q-N02 0 'I � NH - C - CH3 +

- strong o,p-director

CH3 weak o,p-director

o °h2N

0 I I • • 1/ \\ "

CH3 - C - NH I \ C - NH2 strong o,p-director - strong m-director

'------y---' � activating deactivating

Cl weak o,p-director Jy N02

V N02 strong m-director trace

OCH3 strong o,p-dlrector Jy N02

V N02 strong m-director trace

(a) S igma complex of ortho attack-the phenyl substituent stabil izes positive charge by resonance : H C

I + I E

H ____ H ____

Para attack gives s imilar stabi l ization. Meta attack does not permit delocalization of the positive charge on the phenyl substituent.

371

E

H

Page 378: Solucionario de wade

1 7 - 1 6 continued (b)

(i)

< ) < }-N02 +

°2 N

\ > � > OH OH ( i i ) 02 N OH

\ > � ) trace

+ \ > \ �N02 + \ > ) > °2N ( i i i )

o 02N-{ > < }-C - CH3 +

Q-o-N02 ° I I

� II � II C - CH3

(i v) \ f \ }-N02 °2 N

\ f � > +

N02 N02 minor

(v) N02 major (minor amounts of ni tration on the outer rings)

1 7 - 1 7

Cc�-n (a) �H + AlCl3 ��): f �

AlCI4- +

-:?

:¢� .. • ..

372

}

Page 379: Solucionario de wade

1 7- 1 7 continued

+ • • CH3 - Cl - AlCI3

v· OCH3 1

. . : OCH3

I + c CH3

· O H BC1I3 ��l: a:] ....... ;...-...,.�H5�H3

....... ----l�

para I somer also formed by s imi l ar mechanism

..

(c ) CH3 :ci : � I I AICl3

CH AlCI4-

I� methyl 3° CH + I

3 shift

. . 5CH]

• I H �

CH - C - CHCH • 3 I

3 CH - C - CHCH 3 I + 3 ---l.� CH3 - C - CHCH3 I

CH3

CH3 + I + CH3 - C - CHCH3

I CH3

:CI : /

CH3 2° CH3

CH(CH3h

\ � � _ CH Only a small amount of the ortho isomer might be produced as steric

� I 3 interactions wi l l discourage thi s approach path of the electrophi le .

1 7- 1 8

(aJ ex: (b) CH3

I CH3 - � - OH

CH3

CH3

H

+ HF -- Ct H

C . + H

+ BF3 �

o o CHu CH - c If � 3

I CH3 -

373

Page 380: Solucionario de wade

1 7 - 1 8 continued (c)

+ (d) HO - CHCH3 + BF3 -- CHCH3

I I CH3 CH3

17 - 1 9 In (a), (b), and (d), the electrophi le has rearranged.

(a) CHU CH -c r; � 3 I _ CH3

(b) gives desired product

(b) (c) No reaction : nitrobenzene is too deactivated for the Friedel Crafts reaction to succeed.

~ plus over-alkylation products

(c) gives desired product, plus ortho i somer; use excess bromobenzene to avoid overalkylation (d) gives desired product, plus ortho i somer (e) gi ves desired product

1 7-2 1 C(CH3h C(CH3h (a) 0 (CH3hCCI 6 HN03 ¢ • • (separate from ortho, although

AICI3 H2SO4 steric hindrance would prevent much ortho substitution)

(excess) N02

(b) 0 6 ¢ CH3I S03 (separate from ortho) • •

AICI3 H2SO4

(excess) S03H

374

Page 381: Solucionario de wade

1 7-2 1 continued

(c) 2) Cl2 ¢ 0 CH3I (separate from artha) • •

AlCl3 AlCl3

(excess) Cl

1 7-22 ° I I

(a) 0 ° 1 ) AlCl3 aC - CH2CH(CH)), I I

� I + Cl - C - CH2CH(CH3h .. � I 2) H2O ° I I �C - C(CH3h

o (b) 0 °

I I + Cl - C - C(CH3h

1 ) AlCl3 •

2) H20 ° I I

0-0 I I

+ Cl - C � II 1 ) AlCl3

• 2) H20

ac'Q (e) 0 (d) Q CO, HCl

... AlCliCuCl

H3C0-o-CHO

OCH3

(e) ° I I

?' I

I I 1 ) AlCl3 ?' 3 3 Zn(Hg) HCl o ° aC - C(CH )

+ Cl - C - C(CH3h • I ' • � 2) H20 �

(f) ?' I I 0 ° � I + Cl - C - CH2CH2CH3

° I I

I ) AICl) ac - CH,Cll,CH) .. I •

2) H20 � Zn(Hg), HCl

375

aCH2c(CH)))

� I

aCIl2CH2CH2CH3

� I

Page 382: Solucionario de wade

1 7 - 23 Another way of asking this question i s thi s : from B (ei ther by SN I or SN2)?

Why is fluoride ion a good leaving group from A but not -:�N: A ��:NU C

B N02 Formation of the anionic sigma complex A is the rate-determining (slow) step in nucleophil ic aromatic substi tution . The loss of fluoride ion occurs in a subsequent fast step where the nature of the leaving group does not affect the overal l reaction rate . In the SN 1 or SN2 mechanisms, however, the carbon-fluorine bond is breaking in the rate-determining step, so the poor leaving group abi l i ty of fluoride does indeed affect the rate .

t nucleophi l ic aromatic substitution

1 7 -24

C l

SN I

376

t ;>, � <!) f=: <!)

SN2

H

Page 383: Solucionario de wade

17-25

(�2N�:�-CH]

y ;R: �IX;CH]

:�'N�y �

- . . '0' : I • Cl OCH3 0 Cl _ • • N�

+/1 DOCH3 .0.... - N . 0 .... .. �I

... � I I

N02

(b)

Cl

CH3

OH

..

.... N ..... O-0 .... + OCH3 02N'¢

4 -Cl-

N02

CH3

.... N ..... _ 0 .... + 0 C_ I

.... N • • -:0 .... + ..... 0: . .

R CU OCH3 R Cl t oell, -o:

N�1 {: .. �C ...

H -o:N�1 I y ... � y

.... N ..... _ 0 .... + 0 - . • N .. -:0

.... + ..... 0:

CH3 CH3

.. I � -o o � � //C;,- A4�:9.n�,,)lc�o

..

Cl

CH3

... JL �c -H H'L \CH{ f oj

377

CH3

H

o H

Page 384: Solucionario de wade

17-25 continued (c) Br

I � .....

+ Br NH2CH3 B

+ + Br NH2CH3

CH3NH2 HC . .

O

r NH2CH3

I I .. N02

(d) CI 02N

N02

NHNH2 °2N

N02

..

...... N 0/+ .......

0-

.. .. . .

..

C I

...... N ·0""" ....... . . -

•• • + 0 : -........ N ....... . . _ :0 + 0 :

NHCH3 "'+

(H-�HCH3

N02

NH2NH2 ..

.. CH3NH2

+ NH2NH2

..

"N, _ 0"+ 0

+

- Cl-

N02

N02

..

:0:

-.. ; N :0

o II +N -0"

- Br--

...... N ....... 0"""+ 0-

+ + 0 Cl NH2NH2 NH2NH2 1115 "N, _

0 ..... + 0

..

+N -0" I I ----..-

. . C­I

N, .. -• � + 0: .0 ..

l R Cl

+ NH2NH2

+N -0" ..

"N, _ 0"+ 0

-.. ", N ........... _ :0 + 0: . . . .

378

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17-26 Assume an acidic workup to each of these reactions to produce the phenol, not the phenoxidc ion. Cl Cl OH

(a) 6 0 Clz HN03 .. ..

AICl3 HZS04 ¢

¢ from addition-NaOIt, '7 1 elimina�ion

1'1. � mechanism; only this isomer

(b)

(c)

(d)

(e)

NOz NOz + ortho

Cl OH OH

0 Clz 6 NaOH 6 3 Brz B'q Bf .. .. ..

�I

AlCl3 3500 C

Br CI Cl Cl

0 2 Clz

¢

NaOH

¢ Q .. .. + AlCl3 1'1. � OH

Cl OH via benzyne mechanism + ortho

6 ¢ & ¢ Clz NaOH .. .. + AlCI3 1'1.

� OH Cl via benzyne mechanism OH

+ ortho o

'7 I CH3CHzCHz-C -C� O I I

� AICl3

0= CCHzCHzCH3 61 Zn(Hgl � HCl

CHzCHzCHzCH3 CHzCHzCHzCH3 Jr + ¢I .. NaOH VOH � 1'1. via benzyne mechanism OH

379

6CH2CH2CH'

I Clz t AICl3 CHzCHzCHzCH3

¢ + ortho

Cl

Page 386: Solucionario de wade

17-27

from chlorobenzene + NaOH + heat

17-28 (a) First, the carboxylic acid proton is neutralized. o

(b)

"

0r1 � C"'OH

V + NH3 -- NH/ +

+ Na+ plus other resonance forms

o

Q:H _I(� n H H H H

r?' OC� r(�yOCH3 H-O-t-B� �OCH3

� I Li. tL.) ------.. lL�) Li7 I .. I c / C G" / -C � I + Li+ I� I H H H H

the negati ve charge avoids the carbon bearing the electron­donating methoxy group

plus resonance forms

plus resonance forms

380

+ 0 -NH3 II U ryC"'O-.. H

.... C h H · ( pius resonance

Na. � forms o

resonance forms

plus resonance forms

Page 387: Solucionario de wade

1 7-29

more C12 , �, pressure

..

first product formed from benzylic substitution

(c)

cis + trans

1 7-30

VCOOH (a) I h-

17-3 1 r-,,, hv

CIXt l CIV'CI

CI

COOH

(b) ¢

COOH

CI-Cl .. 2 CI. initiation

propagation step 1

CH3 I

· C-C1

6

CH3 I

CI-C-C1

addition of chlorine to the 1t system of the ring, giving a mixture of stereoisomers

crCOOH (c) I h- COOH

CH3 I

· C-C1

r\r CI-CI

(b)

CI· +

6 ..

6 ..-..-

propagation step 2

381

o (f) II

OC.

h- H

CH3 I C-CI

6 C H

t

NH2

Page 388: Solucionario de wade

1 7-32 r'\ r\ hv Br -Br � 2 Br - initiation

CH3 CH3 I

· C-H

CH3 I C-H

6

�-�-H (A Br- + V �

6 � propagation step 1

Br- +

CH3 I

Br-C-H

6 nr '/'" I Br-Br ..

� propagation step 2

CH3 I

· C-H

6

C H

t

1 7-33 A statistical mixture would give 2 : 3 or 40% : 60% a to � . To calculate the relative reactivities, the percents must be corrected for the numbers of each type of hydrogen.

a: 56% 2H

= 28 relative reactivity �: 44% 3H

= 1 4 .7 relative reactivity

The reactivity of a to � is 28 14 .7

1 .9 to 1

1 7 -34 Replacement of aliphatic hydrogens with bromine can be done under free radical substitution conditions, but reaction at aromatic carbons is unfavorable because of the very high energy of the aryl radical. Benzylic substitution is usual ly the only product observed. Br Br �H3 �Br3 cc)Br � (a) (1 ) Br6-C-CH3 (2) Br6-C-CBr3 (b) ( l )

I "'-': (2) �

:-- I :-- I h Br Br ...... "' ...... " all four benzylic

hydrogens replaced

382

I excess Br2 • hv, time

��, �Br Br Br Br

Page 389: Solucionario de wade

1 7-35 + + n

CH2-Br

6-� 6�,t5 - 6 -

C 6�o

17-36

H H �H ! CH3CH2-A: + 1 ) CH3CH20-CH2 I CH3CH2-R-CH2

6 CH3CH2-0: 6 I ..

.. I � �

(a) Benzylic cations are stabilized by resonance and are much more stable than regular alkyl cations. The product is l-bromo- I-phenylpropane.

(b) H + 1

(]HC [Hl

,.,

I H-Br �

..

HC-CHCH3

6 moee stable than

• • - resonance-

V

2° benzylic-:Br: stabilized . .

Br H 1 1

HC-CHCH3

6 I-bromo-I-phenylpropane

383

H 1 +

6CHCH3

Page 390: Solucionario de wade

17-37 (a) The combination of HEr with a free-radical initiator generates bromine radicals and leads to anti­Markovnikov orientation. (Recall that whatever species adds first to an alkene determines orientation.) The product will be 2-bromo- l -phenylpropane. (b) Assume the free-radical initiator is a peroxide. R�R --- 2 RO·

r'lr' RO· + H -Br --- ROH + Br ·

6HCVC: I . Br �

.. more stable than

17-38

(Br • recycles in chain mechanism)

Br I

Br· +

MgBr ,

Br I 6CHCH3

2-bromo- l­phenyl propane

CH2CH20H I

(a) HC=CH2 6� 6ClI3

Mg --­ether

CHCH3 6 L\. H+ ---6C

H3

(b)

9 OCH3

(C)

6

CH3CH2Br •

AICl3

HN03 •

H2S04

¢C�2CH3 �

+ ortho OCH3

¢ N02

+ ortho

hv •

Br2 or NBS

hv •

Br2 or NBS

¢B'

N02

384

Br I ¢]CH3 OCH3

NaCN •

CH30H •

1'1

¢CN

N02

OCH3 I ¢1I3 OCH1

Page 391: Solucionario de wade

17-39 OH 0 enzyme

¢ ¢

HO-OH catalysts

+ H-O + O-H + ..

OH

17-40 OCH2CH3 (a) Q �I

� CH3

(d) *OH Br Br �

� I CBr3

Br

17-4 1 (a) , o

cQ o 17-42

OH 6 3 Br2 -

0

0 I I

(b) &CH]

� CH3

(e) 0

¢t CH3 °

(b)

OH Brh Br

y Br2 ---

Br C6H30Br3

(c)

(f)

I I H H

OH OH

I + I � Br'(\ � CH3 � CH3 Br

OH (CHJhC*

� CH 3 C(CH3h

(c) o o

o o

Br'(y Br

Br Br

C6H20Br4

17 -43 Please refer to solution 1 -20, page 12 of this Solutions Manual .

17-44 (a) C(CH3h (b) CH3CH2CHJ (c) C(CH3h (d) Br (e) C(CH3h

(

08

H 6 6 6 6 rearrangement rearrangement

385

Page 392: Solucionario de wade

rearrangement 17 -45 Products from substitution at the ortho position will be minor because of the steric bulk of the isopropyl group. (a) Br (d) COOH

I (

QCH3h (Q;CH3)2 (

QCH3h

('QCHlh 6 CH

O

CHl

71 � Br S03H ,.C CH(CH3h 17-46 (a)

0

(c)

0 °0 (d)

0 (e)

0

0" .... CH 3

° ° Br

�Cl ~ .. AlCl3

Zn(Hg) � HCl

" � 1 NBS � -- 1 �

Br

~ CH30H

... t:.

from (a)

CI OH I

U � CH2=CH-CH2 1) BH3• THF ... �I ..

AICI3 2) H202, HO-o

Br

Mg Br2' hv ---- � I � or NBS

CI

Cl2 6 NaOH ... ... AlCI3 350°C

Cl

Cl2 6 HN03 ... ..

AlCl3 H2SO4

° OH

U H,o+�

---- I ... �

OH 62CH3 6 NaOH .. CH3CH2Br

Cl Cl

¢

Cl2 ¢rCl .. �I AICI3

N02 386 N02

Page 393: Solucionario de wade

1 7 -46 continued (0 three methods

� Hg(OAch NaBH, � Ih .. .. I h OH H2O from (c) ---;;- �V � °OCH2MgBr H 0+ .. .. 3

I : from (c) CH3-CH

(g)

2) 9 COOH COOH

RN03 KMn04

¢

Fe or Sn or Zn

¢

.. .. H2SO4 f1

N02 (h) two possible methods

2)

Br2 ' hv ..

or NBS

OR

07 I _B_ r-l2 .. � ,:, AlCl3

8

BrMg

� I ether"

Br 6 Mg ether

major

9 Br

.. HCl(aq)

N02 NH2

8

Mg:r � H2O

-- .. °

MgBr 6+� major

COOH COOH

¢

RN03

~ .. H2SO4 �

N02 Br Br

387

OR

VCI Mg I h ether from (d)

OH

new bond in bold

OH

Page 394: Solucionario de wade

1 7 -46 continued CI OH o OH OH

(I) 0 CI2 � I FeCi; 6 6 KOH

--IJ,.

17-47 (a;;;N01

y (bl¢

N02 C(CH3h

�CI --. AICI3

(el J) US03H

Zn(Hg) --HCI

(d) no reaction: Friedel-Crafts acylation does not occur on rings with strong deactivating groups like N02

(e) CH30 0 qC'CH

l

(¢Hl OCH3

Br*Br or I

(gl�CI

Y (h) CH3 ¢ basic workup

I to Isolate

� free amine

CH3 �

CH2Br CH2Br N02 o (i) � -o-� CCH2CH3 Ph-C -� _ H

U)�S03H

� I H 0

N02

(koCH]

(m) I II

N-C-CH CH2CH3

I 3�

CH � ... CH3 stronger o,p-director than CH

3 C 3

II o 1 7-48 Major products are shown. Other isomers are possible.

388

NH2 (I)

a

COOH

� I COOH

H (same as E)

Page 395: Solucionario de wade

1 7-50 < }CH2CH2COOH

starting material molecular weight 1 50

product

molecular weight 132 = loss of 18 = loss of H20

IR spectrum: The dominant peak is the carbonyl at 1 7 1 0 cm-I. No COOH stretch . NMR spectrum: The splitting is complicated but the integration is helpful. In the region of 8 2.6-3 .2, there are two signals each with integration value of 2H; these must be the two adjacent methylenes, CH2CH2 . The aromatic region from 8 7 .3 -7.8 has integration of 4H, so the ring must be disubstituted. Carbon NMR: Four of the aromatic signals are tall, indicating C-H; two are short, with no attached hydrogens, also showing a disubstituted benzene. Also indicated are carbons in a carbonyl and two methylenes. � ___ I�S\ (F-� CH2CH2- C-8 c:=:::>

. , � , retain lose 1 H

o or cO

molecular weight 1 32

The product must be the cyclized ketone, formed in an intramolecular Friedel-Crafts acylation.

1 7-5 1 (a)

(jl .H �H Br

"w.� H + � A H

..

H

389

Page 396: Solucionario de wade

17 -51 continued (b) Assume the free-radical initiator i s a peroxide.

R�rdR --- 2 RO - } r r'J C' initiation

RO - + H-Br --- ROH + Br-

� 0 -Br· �I H ) H propagation

step 1

propagation step 2

1 7-52 CI

..

�Hm HO: + , � .. �

rol H ---.. -.- (9,C H � . � � C

I Br Br H H

+ Br-

a substituted benzyne Attack A

� 2 .. - CO, HO: + . . �

HOmC�n HomH ____ /'" I H -0l! '7 I � �

product A Attack B OH

----

H� lJlJ product B

390

Page 397: Solucionario de wade

1 7-53 The electron-withdrawing carbonyl group stabilizes the adjacent negative charge. plus

��\ • � ��::o:

17-54 (a) �

� (e)

1 7-55

()

+ Na+

(b)� �

(f)�H

:0: •

H'OO plus other resonance forms

°

resonance forms °

H-OEt dJ U r -... C h-H . \ Na' �

HcO H 0

. :� H .�A) H U + Na·f.

plus resonance forms

(d)�H3h

Q < }-C + + 2 HS04- + H30+ 6 o-�y -

6

0H + 2 H2S04 ---

H20 colorless conjugated­

yellow Concentrated sulfuric acid "dehydrates" the alcohol, producing a highly conjugated, colored carbocation, and protonates the water to prevent the reverse reaction. Upon adding more water, however, there are too many water molecules for the acid to protonate, and triphenylmethanol i s regenerated.

391

Page 398: Solucionario de wade

1 7-56

o 1 7 -57 (a) bromination at C-2

bromination at C-3

Cl Cl

l;�C:OH Y Cl

(why is this the major isomer?)

(why is this the major isomer?)

o Br (�-i izf �'b H'C� � � �<H�"" - HH� - HH • • _ three resonance forms :Br: +

,-0y Br

� H

��C/H _

Br

(,.0yH � Br

H H �:�!: j only two resonance forms

O(H Br

(b) Attack at C-2 gives an intermediate stabilized by three resonance forms, as opposed to only two resonance forms stabilizing attack at C-3. Bromination at C-2 wil l occur more readily .

392

Page 399: Solucionario de wade

1 7-58 NH2 I

abbreviate PhCH2 - CHCOOH phenylalanine

as

NH2 I R

F ·0 · • • F I I 02N

...

N02

-.. ;N, .. :0 C

/N, _ 0 /+ a

R � NH2R +N� -/H -0/ . 1. .C

..

+

+��H2R

-0/ I I .. ..

O-/�'O- - .. /N, .. ; :0 + O.

...

..

CH2Ph I

HN-CHCOOH °2N °2N

NH2R ...

N02 N02

393

. . :0:

I +N -.. / :0

o-:/�'o-

1 + 1(jH2R

-0"'- I .. I C _ I

/ N, • • -/+ O· :0 .

Page 400: Solucionario de wade

17-59 i + H ·0· ·0/ . . � � . )l H-C� )l

H3C CH3 H3C CH3

:Cl :

OH I

OH

CH3 CH3

OH I

H 'tl � -� HOXCH, '!?� CH, CH3 H CH3

� H>b�:H} �OH 1 +OH OH

HoiCH3 HoiCH3 HoiCH3 CH3 CH3

r OH I

��[J .. ..

L C-CH3 (�H3

OH I

["1 I :CH "

C-CH I 3 CH3

OH

OH

CH3

plus other

.. resonance forms

CH3 Ho-O-FQ-OH

CH3 .. .. HO�C-CH3 H20: �:� I � CH3

bisphenol A plus other resonance forms

394

Page 401: Solucionario de wade

1 7-60 (a) This is an example of kinetic versus thermodynamic control of a reaction. At low temperature, the kinetic product predominates: in this case, almost a 1 : 1 mixture of ortho and para. These two isomers must be formed at approximately equal rates at 00 C. At 1000 C, however, enough energy is provided for the desulfonation to occur rapidly; the large excess of the para isomer indicates the para is more stable, even though it is formed initially at the same rate as the ortho. (b) The product from the 00 C reaction will equilibrate as it is warmed, and at 1000 C will produce the same ratio of products as the reaction which was run initially at 1 000 C.

17-6 1

6

Br2 ---FeBr3 Q Q

0 II

Mg CH3CCH2CH3 H+ -- • --ether

Br MgBr CH3 - C - CH2CH3 I

(possible alternative syntheses include acylation followed by Grignard) OH

17-62

reactive sites

+

from chi oro­benzene + NaOH at 350°C

As we saw in Chapter 16, the carbons of the center ring of anthracene are susceptible to electrophilic addition, leaving two isolated benzene rings on the ends. Benzyne is such a reactive dienophile that the reluctant anthracene is forced into a Diels-Alder reaction. 1 7-63 Cl 0- Na+ (aJ -<rCI

NaOH -<rCI CICH2COO- Na+

• •

CI � 11 Cl � w. ;Y��H CI�

CI Cl Cl 2,4,5-T

cla, 0:CC� Cl 2 Cl- +

CI � 0 � CI TCDD

Two nucleophilic aromatic substitutions form a new six-membered ring. (Though not shown here, this reaction would follow the standard addition-elimination mechanism.)

395

(This is the compound used to poison Ukrainian political leader, Boris Yushchenko.)

Page 402: Solucionario de wade

1 7 -63 continued (c) To minimize fOnTIation of TCDD during synthesis: 1 ) keep the solutions dilute; 2) avoid high temperature; 3 ) replace chloroacetate with a more reactive molecule like bromoacetate or iodoacetate; 4) add an excess of the haloacetate.

To separate TCDD from 2,4,5-T at the end of the synthesis, take advantage of the acidic properties of 2,4,5-T. The 2,4,5-T will dissolve in an aqueous solution of a weak base like NaHC03. The TCDD wil l remain insoluble and can be filtered or extracted into an organic solvent like ether or dichloromethane. The 2,4,S-T can be precipitated from aqueous solution by adding acid.

+ 1 7-64 :O:� :OH (a) � H+ � �O -----. .. =-'\ �O �

o 0

OH

OH

o :OH

OH

OH

plus 3 resonance fOnTIS with positive charge on the benzene ring

plus resonance fOnTIS with positive charge on the ring and on the oxygen

¢ ('i�---.J...-J �O:

o

HO OH

o

o plus resonance fOnTIS on both benzene rings, the oxygen of the phenol, and the oxygen in the ring

396

OH

o plus resonance fOnTIS with positive charge on the ring and on the oxygen

Page 403: Solucionario de wade

..

° red dianion

(c)

1 7-65

6 ¢ CH3 CH3 fuming

H2SO4 Br2 Brq Br H2O BrO Br .. .. I� .. I� AICl3 A

S03H S03H para position blocked

1 7-66 A benzyne must have been generated from the Grignard reagent . + ((I Br � �S MgBr

. 011 + 0 � F � \--�-----V�----�)

° t

397

Page 404: Solucionario de wade

1 7 -67 The intermediate anion forces the loss of hydroxide. �H OH CH3

axH OH

For simplicity, abbreviate I I >1 R

.0 NHCH3 .0

Li+ H H OH \ /OH ffR �c .. c2 I • r :· R

H � L/ H 'C:Y � plus

resonance forms

---.-

H H , .

+ OH-H H

I I

R___ I �R a....-;c. Os:, H"S: .0 H .o

plus other resonance I forms t Li·

H I �C

yCH3

H � NHCH3 '"cH3CH,o - H-DJ'-R u �

1 7-68 OH OH

¢

NaOH •

lcquiv. ¢ CH3I_ I � .0

OH OCH3

1 equiv. Cl-C(CH3h

AlCl3

plus resonance forms

OH �C(CH3h

Y BHA, butyl.ted hydroxy.niso\c

OCH3 h ydroq uinone plus other isomer

OH 6 OH

CH3I A AICI; Y

CH3

OH 2 equiv. H C) C

V

c(CH3h Cl- C(CH3: ( 3 3

I:

BIrr, butyl.ted hydroxytoluenc AlCl3

CH3

398

Page 405: Solucionario de wade

1 7-69 Solve the problem by writing the mechanism. (See the solution to problem 17-2S(a) for an identical mechanism.)

Br � J):I �B " • • -

rHO: � 2 • • I � attack C-;

02N N02

Br

a

OH

02N N02 only product formed

..

Hydroxide attack on C- l puts the negative charge on carbons that do not have the N02 groups, so these anions are not stabi lized.

Hydroxide attack on C-2 puts the negati ve charge on carbons with nitro groups, thereby increasing the stabilization by delocalizing the negative charge. This intermediate is formed preferential ly .

399

Br

I I ..

l.5(Br OH

02N C- N02 H

also stabilized by resonance onto the nitro group

Br

J3tr �H � :C-

02N ...... N02 also stabilized by resonance onto the nitro group

Page 406: Solucionario de wade

CHAPTERl�KETONESANDALDEHYDES 1 8- 1 (a) 5-hydroxyhexan-3-one; ethylj3-hydroxypropyl ketone (b) 3-phenylbutanal; j3-phenylbutyraldehyde (c) trans-2-methoxycyclohexanecarbaldehyde (or (R,R) if you named this enantiomer); no common name (d) 6,6-dimethylcyclohexa-2 ,4-dienone; no common name 18-2 (a) C9HlOO =:} 5 elements of unsaturation

IH doublet (very small coupling constant) at 0 9.7 =:} aldehyde hydrogen, next to CH 5H mUltiple peaks at 0 7.2-7.4 =:} monosubstituted benzene I H multiplet at 0 3.6 and 3H doublet at 0 1 .4 =:} CHCH3

o II < � + -CH- + -CH �

I CH3

The splitting of the hydrogen on carbon-2 , next to the aldehyde, is worth examining . In its overall sh:lpe, it looks like a quartet due to the splitting from the adjacent CH3. A closer examination of the peaks sho\NS that each peak of the quartet is split into two peaks: this is due to the spli tting from the aldehyde hydrogen. The aldehyde hydrogen and the methyl hydrogens are not equivalent, so it is to be expected that the coupling constants wil l not be equal . If a hydrogen is coupled to different neighboring hydrogens by different coupling constants, they must be considered separately, just as you would by drawing a splitting tree for each type of adjacent hydrogen.

(b) CgHgO =:} 5 elements of un saturation cluster of 4 peaks at 0 1 28- 1 45 =:} mono- or para-substituted benzene ring peak at 0 1 97 =:} carbonyl carbon (the small peak height suggests a ketone rather than an aldehyde) peak at 0 26 =:} methyl next to carbonyl or benzene

o < }-C-CH3 o or CH3-o-C-H

more likely also possible

18-3 A compound has to have a hydrogen on a y carbon (or other atom) in order for the McLafferty rearrangement to occur. Butan-2-one has no y-hydrogen.

y position-no carbon

t o II

CH3 -C -CH2 -CH3 a a J3

401

Page 407: Solucionario de wade

18-4

[ It ] t j : R: :I�; L CH3 �H2CH2CH2CH2CH2CH3 --- 1 +C-CH3 --- C-CH3 r

43 85 rnIz 43

�+ CH2CH2CH2CH2CH2CH3 rnIz 128

rnIz 85

McLafferty rearrangement

18-5 The first value is the n to n*; the second value is the n to n*. The values are approximate.

(a) < 200 nm; 280 nm; this simple ketone should have values similar to acetone

(b) 230 nm; 3 10 nm; conjugated system (2 10) plus 2 alkyl groups (20) = 230; the value of 310 nm is similar to Figure 18-7 : ketone (280 nm base value) plus 30 nm for the conjugated double bond = 3 10 nm

(c) 280 nm; 360 nm; conjugated system (2 10) plus 1 extra double bond (30) plus 4 alkyl groups (40) = 280; similar reasoning for the other transition, starting with an average base value of 290

(d) 270 nm; 350 nm; same as in (c) except only 3 alkyl groups instead of 4

REMINDERS ABOUT SYNTHESIS PROBLEMS: 1. There may be more than one legitimate approach to a synthesis, especially as the list of reactions gets longer . 2. Begin your analysis by comparing the target to the starting material. If the product has more carbons than the reactant, you will need to use one of the small number of reactions that form carbon­carbon bonds. 3. Where possible, work backwards from the target back to the starting material. 4. KNOW THE REACTIONS. There is no better test of whether you know the reactions than attempting synthesis problems.

402

Page 408: Solucionario de wade

1 8-6 All three target molecules in this problem have more than six carbons, so all answers will include carbon-carbon-bond-forming reactions. So far, there are three types of reactions that form carbon-carbon bonds: the Grignard reaction, SN2 substitution by an acetyl ide ion, and the Friedel-Crafts reactions (alkylation and acylation) on benzene.

0 OH 0 (a) 0 Bf� 0MgBf l ) H� � H2C�; � .. ether 2) H30+

(b) 0

U Bf identical sequence as in part (a) ~ I � ..

OR o 0

o Cl� � AlCl3

... U this method using Friedel-Crafts acylation is more efficient as it is only one step

(c) a synthesi s as in part (a) could also be used here

OR 0 � Br Mg �MgBr 1) � � L-.l � L-.l 2) H30+

.. L-.l 6H OR

�C�4

(:j'Br Na+ -C=CH (:j'C=CH .. Hg2+ H 0 , 2

.. H2SO4 og

1 8-7 (a) (J (J BuLi

... SX S S • • S ::-C'" H H I

H

(b) (J CH3I (J

S • • S SX S ::-C" ..

I CH3 H H from (a)

(J BrCH2CH2Ph ..

SX S PhCH2CH2 H

BuLi ... ...

Ph Br�

403

0 H30+ I I ... PhCH2CH2 -CH

HgCI2

(J Ph H30+ � �

HgCI2

0 Ph �

Page 409: Solucionario de wade

1 8-7 continued

(c) rl S • . S 'C" I H

from (a)

PhCH2Br • rl SXS

PhCH2 H

BuLi PhCH2Br • • rl SXS

PhCH2 CH2Ph

H30+ HgCl2

o Ph� Ph

(d) rl Q-CH2B' � rl BuLi

-PhCH2CH2Br

• ('j() � Ph S:-C"S o-:xS S I H H

from (a) ('j � � � H30+ , HgCl2 Ph

18-8 0 0

(a) if (b) � (c) 0 II

I h CH3(CH2h -C -CH2CH3

18-9 0 \l-(J II (a) CH3(CH2h -C-CH2CH3 (b) PhCH2CN (c) PhCH2C

(simple SN2) 18-10 o

II

(a) I � � O Br

h ether OMgB' CH3CH2C=� H30+ -- OCCH2CH3

(b) CH3CH2C::N + BrMgCH2CH2CH2CH3 H30+

--- --o �

(c) CH3(CH2hCOOH + 2 CH3CH2Li H30+ -----

o � (d

6

CH2Br NBS. 6 NaCN.

6

CN MgBr

6 H,o' ::::'--1�---

404

o 6CH2C-o I� h

Page 410: Solucionario de wade

IS-II o II

(a) VCH,oH (b) VCH OH (C) � o

(d) � IS-12 Review the reminders on p. 402. There are often more than one correct way to do syntheses, but a more direct route with fewer steps is usually better.

o (a)� -..;::: H I h

� VV (c) 0

� VV

PhBr .Mg PhMgBr

PhBr .Mg PhMgBr

LiAIH4

..

H30+ •

H30+ •

OH 0 � Cr03,H20 � V V H2S04

• V V

OH

H,o'. � VV OR NaBH4 •

CH30H (solvent)

HC::CCH2CH3 (d) 0 _ t NaNH2 �H Na+ C::CCH2CH3 H30+

OH

�C�CCH2CHJ I • • h V IS- 13 The triacetoxyborohydride ion is similar to borohydride, BH4 -, where three acetoxy groups have replaced three hydrides. (a)

Na+

o II C

0" '- 0 -I -1/ H - � - O \ 0y

o

(b) o 0 II II .... C... ....C... • •

0 ' 0 0) 0 ' 0 :0 : �O-B-=-H Na+ + )!.:: -- �O - B + 1 / I \.- • . - . R� H / I R--1'H yO � yO H

o 0 1 H3CCOO-H O-H

R+H 405 H

Page 411: Solucionario de wade

18-14 Trimethylphosphine has a-hydrogens that could be removed by butyllithium, generating undesired ylides .

CH3 1+ BuLi CH3 1 + -

CH3 1+ CH3-P-CH2R �

1 CH -P-CHR + 3 1 CH -P-CH R 2 1 2

CH3 CH3 desired ylide

CH3 wrong ylide

rotation ..

• • �+ :�hPPh3

\\\ I" H'" IIIH Me Me

:O---! PPh) /

H IIIIIIC _ C ""

" H + Ph3PO ....... f--- " , 1.:1:1 . III Me"""" - 'Me H',r�/ IH

Me Me cis-2-butene

The stereochemistry is inverted. The nucleophile triphenylphosphine must attack the epoxide in an anti fashion, yet the triphenylphosphine oxide must eliminate with syn geometry.

(b)

o 18-16

O'H " o .. . fl17 ,: PPh3

CIS �

. . :0:

rotate and eliminate ..

trans

1 ) Ph3P + PhCHO (a) CH2=CHCH2Br .. CH2=CHCH-PPh3 � PhCH=CH-CH=CH2 2) BuLi

OR 1 ) Ph3P - + O=CH -CH=CH2

... PhCH -PPh3 .. PhCH = CH -CH = CH2 2) BuLi

(b) PhCH=CH -CH2Br 1 ) Ph3P +

... PhCH=CH -CH -PPh3 2) BuLi

+ PhCH=CH-CH=O CH2-PPh3 �

406

PhCH = CH -CH = CH2

Page 412: Solucionario de wade

1 8- 1 7 Many alkenes can be synthesized by two different Wittig reactions (as in the previous problem). The ones shown here form the phosphonium salt from the less hindered alkyl halide.

(a)

(b)

(c)

1 ) Ph3P + PhCHzBr .. PhCH -PPh3

2) BuLi

+

2) BuLi CHz-PPh3

o + )l - PhCH=C(CH3)z

H3C CH3

PhCH = CH -CH = 0 1) Ph3P + PhCHzBr .. PhCH -PPh3

2) BuLi .. PhCH = CH -CH = CHPh

18- 18

(a)

1 8- 19

least amount of hydrate

+ �O _ �CHz CHz-PPh3 + U U

OH I

Cl C-C-H 3 I

. . :O: � "

I H-OH CH -C-CH 3

I 3

OH

407

OH

OH I

CH -C-CH 3 I

3

OH

greatest amount of hydrate

+ HO-

Page 413: Solucionario de wade

18-20

(c) err t-Bu-C-t-Bu :::::: .. ;:::::=�

�-:C N

18-21 0

(a)�

V HCN

. . :0: � '"""

I H-CN CH CH -C-H .. 3 2

I CN

:o:� '""" I H-CN t-Bu-C-t-Bu

I CN

OH

vi' (b) OH 0

. . I pcc II �H-·�H H

(c)

OH I

CH CH -C-H 3 2 I

OH I

CN

t-Bu-C-t-Bu I CN

OH

HCN �H CN

OH I

u

CHCOOH

Note: Mechanisms of nucleophilic attack at carbonyl carbon frequently include species with both positive and negative charges. These species are very short-lived as each charge is quickly neutralized by a rapid proton transfer; in fact, these steps are the fastest of the whole mechanism. In most cases, this Solutions Manual will show these two steps as occurring at the same time, even though you have been admonished to show all steps of a mechanism separately. The practice of showing these proton transfers in one step is legitimate as long as it is understood that these are two fast steps.

408

Page 414: Solucionario de wade

18-22

(a) 0: (J<-:NH; H H H

(Jh: �20:JW� O)'H

:0:\ :o-� d,+ HO+ C} HO+ (NHPh 3 .

NHPh 3 _ I • • • • H H20: '---./

two fast proton transfers

:o:� :O� H-O-H ( 1+ 1 �� 1 �� Ph-C-H ---- Ph-C-H - Ph-C-H

1 • • 1 (r� NHCH, 1 NHCH3

two fast proton transfers

Ph-C-H I I

:NCH3

409

+ - H20 .. iPh-C-H � Ph-�-H} H 0: I I I 2 �U�CH3 H -NCH3

plus three resonance forms with positive charge on the benzene ring

Page 415: Solucionario de wade

18-23 Whenever a double bond is formed, stereochemistry must be considered. The two compounds are the Z and E isomers.

Z

18-25 This mechanism is the reverse of the one shown in 18-22(c) on the previous page .

Ph-C- H • Ph-C-H � Ph-C-H .. 1 +�t H20 : :0: II

. 1 1 H30+ II I

.NCH3 j H -NCH3 H -NCH3 � + • •

plus three resonance forms with ') (Positive charge on the benzene ring . . H-O: . . H-O: -CH3NH2 I

. � H-O-H 1+ Ph-C-H I

NHCH3

. . H-O: 1

Ph-C-H ...... f----+1 Ph-C-H .. Ph-C-H .. Ph-C-H t H-O+ VII

Ph-C-H

o 18-26 02N

NH2-NH-O-N02 \ )

A ..

Y. abbreviate "Z" Q: :o:� CH3-C-CH3 -­'-- : NH2Z

I H30+ CH3-C-CH3 ---+� H20: (k�

(I H 0+ I 3 H-NHCH3

+ � NHCH3

:o� I H30+ CH -C-CH • 3 I 3

NHZ

H-O-H (1+ CH -C-CH 3 I 3

NHZ

two fast proton transfers

• CH -�-CH t 3 I

3

H-NZ ..

Page 416: Solucionario de wade

N-NH2 (b) oJ

I� .b

18-28 o II

(a) PhCHO +

",NHPh (d) N Ph)l Ph

H2NNH - C - NH2 (b) ho + H2NOH

(c) 0

oJ + H2N-NHPh

(d) 0

0+

02N

NH2-NH� }-N02

(e) C(fNH2 CH3

I 0 o

18-29

. . � II H+ ·0· �

Ph-CH ---

plus resonance forms with (+) on benzene ring

: O-H :O-H O-H + •• } II +1 I

Ph -CH ---- Ph -CH -. -. ----l.� Ph -CH '" J H�CH3 1+ "---.:. : OCH 3

� k) H.qCH3 �

plus resonance forms ,------- : O-H j Ph _ CH ....... ,...----jW��ili (+�:��:ne rin}g � H20

1 +�CH3 (�H3

H-O-H +1) Ph-CH

I OCH3

OCH3 I

H+ I --- Ph-CH

I OCH3

hemiacetal

H..2:0CH3 VI Ph-CH Ph -CH acetal

I OCH3

I OCH3

411

Page 417: Solucionario de wade

1 8-30 CH3 CH3 1 1 • • • . t :00

0:""\ H 0+ �

oo/ H :0

6 ..

1 8-3 1

(c) 0 0

CH3 H 1 1 �

:00

0: f . . • . H30+

..

t -CH30H + H :O�

6

H2O: • •

r\ � + 2 HO OH

(e)0H +

HOD

412

CH3 CH3 1 1 :0, + :� .� . . V

H20: ! CH3 H 1 +1" . . . .

...... f---

:OoO-H

H20:

:0:

6

(b) 0 I I

CH3 - CH + 2 (CH3hCHOH

(d)OO +

HO)

HO

(f) )(l HO II OH o

Page 418: Solucionario de wade

�';led lX�U uo p�nu!luo:> WSlUnlj:>�W S!SAI0.1pAlj

:9zHt o

H . ..... J� :0 '0

�� HO

o-J, + • • J, • • H-O '0: H-O: + 0: '----1 '----1

0- Q J, + • • J, • • H -0 '0: H - 0: + 0: '----1 '----1

-Q�Q +0 H

H-O� 0 �:o 0 + '---1 '----1 (q)

Q-Q FHQ � : 0: (n)

H H

+6 ·o/� :0 :0 + -------

Page 419: Solucionario de wade

1 8-33 continued (b) hydrolysis mechanism continued

OH OH r + r :0, .... OH H-O� .... OH ((O· � C, ____ ·0· C � rOH +

� H30+ HO

.. H + HI\ 6-6 �6

(c) The mechanisms of fonnation of an acetal and hydrolysis of an acetal are identical, just the reverse order. This has to be true because this process is an equilibrium: if the forward steps fol low a minimum energy path, then the reverse steps have to fol low the identical minimum energy path. This is the famous Principle of Microscopic Reversibility, text section 8-4A.

(d)

CO-;;;o:CO o RJ 0 �9+ 00-00 o �O·· +0 :0

18-34 0

(a) A

H • • • • • I � H

H20: � CJ:::'t �+ CCl �CCl

+O� OH OH '--- R RH OH \ � +OH

1 � OH /h·· '---0;-0-WOH � CO

.. : OH

.. �� 1 equivalent NaBH4

VCHO CH30H

Aldehydes are reduced faster than ketones (keeping this reaction cold will increase selectivity) . Alternatively, sodium triacetoxyborohydride could be used for this reduction; see problem 18-13.

1 equivalent 1\

HO OH ..

414

Page 420: Solucionario de wade

18-34 continued 1 equivalent

0 (\ (c) 6

CH2B' HO OH ..

H+

The last step protonates the oxygen, dehydrates the alcohol, and hydrolyzes the acetal.

18-35 (a) o COO

: AgO HO after adding H+

(d) HO

V

COOH

HO

(\ (\ 0 o 0 o 0 H� a

CH2R' Mg

aCH2MgB' + ..

ether \.. )

0

..

(b)

oD

COOH

415

V t

[MgBr]+

t H30+

a II

HC:: C -CH2 -CH2 - C - CH3

(c)

y

COO: AgO

0 after adding H+

Page 421: Solucionario de wade

18-36 hydrazone formation -•• •• �

j: 0,00 0 + " H20 cJ'- .. (NHNH2_ " ",-- :NH2NH2 I :OH H--./ ..

reduction of the hydrazone

two fast proton transfers

H-N II

C •• .. O ....... N -.. ·N H � �II � H"OH OC� ('i2:N

V H " H -- V H

416

.. :OH

NJfNH2

Page 422: Solucionario de wade

1 8-37 1\

(a) H H

CO (b) � V H' 'H

(d) (c) H H �

1 8-38 Please refer to solution 1 -20, page 12 of this Solutions Manual.

18-39 IUP AC names first; then common names. Please see the note on p. 1 36 of this Solutions Manual regarding placement of position numbers. (a) heptan-2-one; methyl n-pentyl ketone (b) heptan-4-one; di-n-propyl ketone (c) heptanal; no simple common name (d) benzophenone; diphenyl ketone (e) butanal; butyraldehyde (f) propanone; acetone (lUPAC accepts "acetone")

(g) 4-bromo-2-methylhexanal; no common name (h) 3-phenylprop-2-enal; cinnamaldehyde (i) hexa-2,4-dienal; no common name (j) 3-oxopentanal; no common name (k) 3-oxocyclopentanecarbaldehyde; no common name (I) cis-2,4-dimethylcyclopentanone; no common name

18-40 In order of increasing equilibrium constant for hydration: 0 0 0 o o II

CH3-C-CH3 < I I I I

ClCH2 - C-H < H-C-H least amount of hydration

18-4 1

18-42 o

o b CH3 } II - I

H-C-CH -C-CH c 2 I

3 -f! CH 3

A By comparison with similarly substituted molecules shown in the text:

U 1t--t1t* base value (210) plus 3 alkyl groups (30) = 240 nm CH3 n--t1t* 300-320 nm

417

greatest amount of hydration

� 9H

Page 423: Solucionario de wade

18-43

C6HJ002 indicates two elements of unsaturation.

The IR absorption at 1708 em·1 suggests a ketone, or possibly two ketones since there are two oxygens and two elements of un saturation. The NMR singlets in the ratio of 2 : 3 indicate a highly symmetric molecule. The singlet at (5 2.15 is probably methyl next to carbonyl, and the singlet at (5 2.67 integrating to two is likely to be CH2 on the other side of the carbonyl.

> o o II II

H3C -C -CH2 -CH2 -C -CH3

Since the molecular formula is double this fragment, the molecule must be twice the fragment.

Two questions arise. Why is the integration 2 : 3 and not 4 : 6? Integration provides a ratio, not absolute numbers, of hydrogens. Why don't the two methylenes show splitting? Adjacent, identical hydrogens, with identical chemical shifts, do not split each other; the signals for ethane or cyclohexane appear as singlets.

18-44 The formula CJOHI2O indicates 5 elements of unsaturation. A solid 2,4-DNP derivative suggests an aldehyde or a ketone, but a negative Tollens test precludes the possibility of an aldehyde; therefore, the unknown must be a ketone.

The NMR shows the typical ethyl pattern at 8 1 .0 (3H, triplet) and (5 2.5 (2H, quartet), and a monosubstituted benzene at 8 7.3 (5H, multiplet). The singlet at (5 3.7 is a CH2, but quite far downfield, apparently deshielded by two groups. Assemble the pieces:

18-45

(a)

(b)

o II 0- + CH2 + C

H" ,CH3 0) (CH II I

/C,�CH2 H C

H2

rn/z 86

)H>-. �CH3 o �CH II I

/C,�,CH2 CH3 CH

I CH3

rn/z 1 14

+ .

+ .

..

..

[ H

r HJ:CH2 rn/z44

0 ....... H

I C,

CH/ 'CH 3 I CH3

rn/z 72

418

CHCH3 + II CH2

mass 42

+ . /CH3 CH

+ II CH2

mass 42

Page 424: Solucionario de wade

1 8-45 continued

(c) H O} (CH

I I I

. CH3

C .Jr-.... CH H3C"""" 'c"""" 'CH3 H2

mJz 1 14

18-46

�O 0+

1 8-47

+ . ..

H °X

O

[ r H • 0 .......

I ....... C"

H3C -""':CH2 mJz 58

XO 0

+ CHCH3 I I CHCH3

mass 56

0 OH A molecular ion of mJz 70 means a fairly small molecule. A solid semicarbazone derivative and a negative Tollens test indicate a ketone. The carbonyl (CO) has mass 28, so 70 - 28 = 42, enough mass for only 3 more carbons. The molecular formula is probably C4H60 (mass 70); with two elements of unsaturation, we can infer the presence of a double bond or a ring in addition to the carbonyl.

The IR shows a strong peak at 1790 cm-I, indicative of a ketone in a small ring. No peak in the 1600- 1 650 cm-I region shows the absence of an alkene. The only possibilities for a small ring ketone containing four carbons are these: 0 0

A 4<)' �

B 3

CH3

The HNMR can distinguish these. No methyl doublet appears in the NMR spectrum, ruling out B. The NMR does show a 4H triplet at & 3. 1 ; this signal comes from the two methylenes (C-2 and C-4) adjacent to the carbonyl, split by the two hydrogens on C-3. The signal for the methylene at C-3 appears at 8 2.0, roughly a quintet because of splitting by four neighboring protons.

The unknown is cyclobutanone, A. The symmetry indicated by the carbon NMR rules out structure B. The IR absorption of the carbonyl at 1790 cm-I is characteristic of small ring ketones; ring strain strengthens the carbon-oxygen double bond, increasing its frequency of vibration. (See Section 12-9 in the text. )

18-48 (a) The conjugated diene has a maximum at 235 nm (see Solved Problem 15-3) and the ketone has a maximum at about 237 nm, so the n to n* transition cannot be used to differentiate the compounds. (b) The ketone has an n to n* transition around 3 1 5 nm that the diene cannot have.

419

Page 425: Solucionario de wade

18-49 o I I (a) (�l���etal) CH3CH2CH2 -C -CH3 + 2 CH30H

o 6 + CH3CH,oH (b) hemiacetal (old: hemiketal)

(c) acetal

(d) acetal (old: ketal) 1\

HO OH

18-50 (a) G.o: :0: �

II I H30+

(X0H+

H (e) acetal O =(

OH H

(f) diether, inert to hydrolysis

(H2 0 .... .... CH (g) imine(s) + I

",CH NH2 0"

(h) imine 0=0 + H2N -Q .. :o� H-O-H ( 1+

H C-C-H .. H C-C-H 3 ,-- : NH2NHPh 3 I H O.

I H30+ .. H3C-?-H .. HC-C-H 3

I

OCH3 I

CH3-CH I

OCH3 acetal

+NHNHPh 2 • (� J NHNHPh NHNHPh

two fast proton transfers

H3C-C-H .. iH C-C-H --- H3C- � -H } I I H 0: 3

II I : NNHPh 2 l l � - NNHPh H -NNHPh "----/' U + • •

420

O-H I

CH3-CH • 1+ HOCH3 :kV· :O-H

I CH3-CH

I OCH3

hemiacetal

Page 426: Solucionario de wade

Ph3PO + .. �';Ph' � ,Ph) 18-50 continued (c) C. 0: U"---- : CH2 - tPh, �V C�2 � V rCH2 �

crCH2 /\ +/\ /\ 0 +

(d) °o�:� °o�-H

H30+ . ----

'O-H :90�� ..

:O� O-H o

(e)

OH

H-+/0 �OH o l

OH � HO

• • � H :0

' I

o +

OH � '0 OH r""c/

,.H30+ 0 ...

Hi): � OH <---J H / �/+ o

.. H20: O·'H

+ H

-6 r�:. 6 H �

H H,O+

.. �H � cmo: N / + • 021+ 'O-H �

.. H

�� +N 1 H

1 H

O 'b�H 0 'O-H � + / .' .. C, 4) .. <;:, r:J2-�';-H H3EJD2+ :�.;-H

-- + H ... H N� N /N, H /N, H H H H H

/ , I H H H

� '/ • '0 O· ;0 . 0 . • • C, •

� .' /N� 'H H30+ NH

C,

H 'H� 3 H

421

Page 427: Solucionario de wade

18-51 o II (a) CH3-CH KCN ..

HCN

Ph3P (b) PhCH2Br ..

OH I

CH3-CH-C::N H30+ ..

OH I

CH3-CH-COOH 0= ,Ph o HC BuLi + 6 .. PhCH -PPh3 ..

(c) NaBH4 Alternatively, using sodium triacetoxyborohydride 00 1 equivalent 00

CH30H .. selectively reduces the aldehyde; see Problem 18-13.

CHO CH20H o 1 equivalent o H OH

(d) 0 1I0/\01! � NaBH4 � ,H

CHPH" C Q '

1130+.

Q�1I0 C 0'" '0

LJ C

0'" '0 '--1

CH3CH2CH2Br + PPh3

t o 1 equivalent (e) 0 /\

tBuLi o +

HO OH o CII3CH2CH -PPh� C

(

f)

roo (g) roo (h) roo

.. H

C 0'" '0

LJ 1 equivalent roo H2

.. Pt

OH Raney Ni co-H .. H2

NaBH4 OH

CII,oIl· roll 422

CHCH2CH3 o ,II 1130+.

,..C, o 0 LJ

5112:113

C II o

Page 428: Solucionario de wade

1 8-52 All of these reactions would be acid-catalyzed. (a) q + H2NOH

o

(c) � O� �NH2 + LJ

(c) ONH2 +

1 8-53

,CH3 O =C

\ CH3

(a) °2N

VN/NH� }NO,

(c) NOH (d)

L

(b) 0 + H2N'O

�CHO

(d)ro 1\ I + HO OH � 0

o (f) 6

(b) 0 I I

+ 2 CH30H

cfN-r

C -NH2

1\ (e)

� OCH3

I CH3 -CH

I OCH}

(f) OH I

(g) N ... CH2CH3 < }-CCH2CH3

(h) rl H -CH I OCH3

423

SX S CH3CH2 H

Page 429: Solucionario de wade

1 8-54

(a) (] BuLi (] CH31 (] .. .. SX S S, . ...... S SX S

_ C I H H H CH3 H

C /eH, t

/CH, • • C " c H2O: ..

+ ' H .. .. ' H :S :S · ·' H · ·' H

HS�S - H

0: I I

..... C .... H 'CH3

(c) Mercuric ion, Hg2+, assists the hydrolysis in two ways. First, mercuric ion is a Lewis acid of moderate strength, performing the same function as a proton from a protic acid.

The effectiveness of Hg2+ as a Lewis acid is partly due to its charge: even complexed with a sulfur, the mercury atom sti l l has a positive charge, attracting the sulfur's electrons.

A second explanation for mercuric ion's effectiveness in hydrolysis of thioacetals lies in the complex formed between the ion and the two sulfur atoms. This stable complex effectively removes HSCH2CH2SH from the equi l ibrium, shifting the equil ibrium to product. (An example of whose principle? His initials are "Le Chatelier" . )

S ............ Hg2+

t S

Thiols are often referred to as "mercaptans" because of their abi l ity to CAPTure MERcury.

424

Page 430: Solucionario de wade

1 8-55 The key to this problem is understanding that the relative proximity of the two oxygens can dramatically affect their chemistry .

1 ,2-di oxane The "second" isomer described: two oxygens connected by a sigma bond are a peroxide. The 0-0 bond is easi ly cleaved to give radicals. In the presence of organic compounds, radical reactions can be explosive.

Mechanism of acetal hydrolysis

HO�O - H ---

• • / H :0 ....... f---i .. � I +

/C, H H

1 ,3-dioxane The "third" isomer described: two oxygens bonded to the same sp3 carbon constitute an acetal which is hydrolyzed in aqueous acid. See the mechanism below.

. . 0 CCH + 2

.. :0 " ' H

}

425

.. ..-

(0) o

l ,4-dioxane The "first" isomer described: an excellent solvent (although toxic), these oxygens are far enough apart to act indepen­dently . It is a simple ether.

(:

/H

" C H2O: • ' H ' 0 " ' H

Page 431: Solucionario de wade

18-56 (a)

6

] (b) CH]6CH] (c) NOH (d) 1\

6 0 H /

.;; C (e) N - N - Ph (t) PoH (g) no reaction (h) Her 6H (i) 0 0) c5 (k) Na+ -(j O N (I) HO COOH 0 1 8-57 The new bond to carbon comes from the NaBH4 or NaBD4, shown in bold below . The new bond to oxygen comes from the protic solvent.

(a) 0 NaBD4 H2O OH I I I

CH3 - C - CH2CH3 .. .. CH3 - C - CH2CH3 I D

(b) 0 NaBD4 D20 OD " I

CH3 - C - CH2CH3 .. .. CH3 - C - CH2CH3 I D

(c) 0 NaBH4 D20 OD " I

CH3 - C - CH2CH3 .. .. CH3 - C - CH2CH3 I

H

1 8-58 While hydride is a smal l group, the actual chemical species supplying it, AlH4-, is fairly large , so it prefers to approach from the less hindered side of the molecule, that is, the side opposite the methyl . This forces the oxygen to go to the same side as the methyl, producing the cis isomer as the major product.

H H " /

H/C�fd

H � H less "- I -

_-, .. �

H3C�0

H H

hindered H · AI · H face H

426

H+ ---.

OH Q CIS CH3 major product

Page 432: Solucionario de wade

18-59 (a)

- + OO O CH An exocyclic double bond is less stable than an endocyclic double bond.

BuLi CHz - PPh3 • 2

methy lenecyc lohexane

(b) The difficulty in synthesizing methylenecyclohexane from cyclohexanone without using the Wittig reaction rests in the stability of the double bond inside the ring (endocyclic) versus outside the ring (exocyclic ) .

6 + CH]MgBr

� H20.

HOoCH] :

+.

A dehydration following the E l mechanism passing through a carbocation intermediate will give the more substituted, endocyclic double bond as the major product. The only chance of making the exocyclic double bond as the major product is to do an E2 elimination using a bulky base to give Hofmann orientation (Chapter 7).

t HBr cold

Br CH3 o K+ -O-I-Bu • E2

°

+

major minor

c5 + 6

minor major

18-60 ° � I I

(a) 0 CI - CCHzCH2CH3 zn(Hg� � HCI V

(b) VC ' N

(e) 0 CI2 • AICI3

(d)

0:> I � �

CH3CHzMgBr H3O+ • �

ether

Cl OH 6 NaOH 1 .0 3500 C 6 ° I I

VC - CH2CHl 1 .0

OCH3

I) Na0,!;l 6 2) CH31 .0

¢l co HCl • AlCI3 CuCI

CHO

HZCr207 •

0::50 I : conc. H2SO:

° c6 Alternatively, the acid chloride could be made with SOCIz, then cyclized by Friedel-Crafts acylation with AICI3.

427

Page 433: Solucionario de wade

18-61

(a) OH I

(yC ,

V PhH

(d) OCH2CH3 I

()C, IH OCH2CH3

(b)

(e)

0 (c) II

()C. 0-

+ AgO ('I (f)

dc:�

H 0 I II

N-N-C-NH II

2 ()C-H

()CH1

18-62 H (a) � CH3Mg1 H30

+

.. � � cr03 , H20 � o

(b)�C=CH

ether I OH

HgS04 H20

H2S04

.. � 1) B uLi /"'.-..

.. I I

o

.. H2S04

H 0+ 3 .. 0

o

(c) ('I SXS H H

1) BuLi ('I --� .. �

S 2) CH31 Sx

CH3 H 2) CH3

(CH2hBr S�

CH3 HgCI2�

(d) � cr03, H20

.. H2S04 �

(e) �OH 1) excess CH�Li

o 2) H30+

(f) � _ CH3Mg1 H30+

C=N .. � ether

o

� o

� o

(g) 1) 03

.. 2) Me2S � +

o

428

CH3 , O=C

\ CH3

Page 434: Solucionario de wade

18-63

(a) �OH

(b) �CH2

(c) �C=CH

PCC ---.-

o �H

o �H

o 1) BuLi ..

HgCl2 �H Cl

(e) �Cl KOH

---.-H20

o �H

(This reaction needs a solvent like THF to keep all reactants in solution.)

o II 1) LiAIH4 (f)�

OH ..

o PCC II

�OH -- � H 2) H30+

ft SOCl, ft �(Ot-BuhH

OR �OH " �Cl

18-64

(a) � o

18-65 (a) ketone: no reaction (b) aldehyde: positive

(b) � + � o 0

(e) +

(c) enol of an aldehyde-tautomerizes to aldehyde in base: pOSitive (d) hemiacetal of an aldehyde in equilibrium with the aldehyde in base: positive (e) acetal-stable in base: no reaction (f) hemiacetal of an aldehyde in equilibrium with the aldehyde in base: positive

429

(c) � o

Page 435: Solucionario de wade

18-66 The structure of A can be deduced from its reaction with J and K. What is common to both products of these reactions is the heptan-2-01 part; the reactions must be Grignard reactions with heptan-2-one, so A must

be heptan-2-one.

(l SXS �C=CH H CH3 1) BuLi

D �H�)4B' S S

�CH3 E

�MgBr Hg2+ H30+

H2S04 HgS04 F H20

1 1) CH3CHO B 2) H30

+

OH 0 � Na2Cr207•

� c H2S04 A

�N� 1) PhMgBr

J !H3:' �OOH

2) H30+

� I Ph

430

~ G

MgBr

1) 6� K 0

OH

o

Page 436: Solucionario de wade

18-67 The very strong 11: to 11:* absorption at 225 nm in the UV spectrum suggests a conjugated ketone or

aldehyde. The IR confirms this: strong, conjugated carbonyl at 1690 em-

I and small alkene at 16 10 em-I.

The absence of peaks at 2700-2800 em-

I shows that the unknown is not an aldehyde.

The molecular ion at 96 leads to the molecular formula: o II

C=C-C-C 96

-64 , I 32 mass units => add 2 car

bons and 8 hydrogens '( mass 64

molecular formula = C6H80 = 3 elements of un saturation

Two elements of unsaturation are accounted for in the enone. The other one is likely a ring.

The NMR shows two vinyl hydrogens. The doublet at 8 6.0 says that the two hydrogens are on

neighboring car

bons

(two peaks = one neigh

boring H

).

doublet: 1 neigh

boring H

..-I--. H H 0 I I II

C-C=C-C-C + 1 C + 6 H + 1 ring

No methyls are apparent in the NMR, so the 6H group of peaks at 8 2.0-2.4 is most likely 3 CH2 groups. Com

bining the pieces:

The mass spectral fragmentation can be explained

by a "retro" or reverse Diels-Alder fragmentation:

0

H�TH2 �

CHz H C

Hz rnIz 96

+ .

--

0 II

HxC

H C Hz

rnIz 68

+ .

+ CHz II CHz

loss of 28

In the HNMR , one of the vinyl hydrogens appears at 8 7.0. This is typical of an a,j3-unsaturated carbonyl b

ecause of the resonance form that shows deshielding of the I3-hydrogen.

H H 0 I I II

C-C=C-C-C 13 a

87.0 --....... H H :0: +1 I I

....... f---t .. � C-C-C=C-C 13 a

431

Page 437: Solucionario de wade

18-68 Building a model will help visualize this problem. (a)

H • • I :OyO CH20H -"":CH

HO OH

HO

OH open-chain fonn

OH

OH

w --

same as

H H I I

-

:OyO CH20H +-"":CH

HOy�

.yCH20H

HO�OH

OH cyclic fonn-hemiacetal

-

H H I I :OyC:O CH20H

• •

"CH

··

+

(b) Yes, the cyclic fonn of glucose will give a positive Tollens test. In the basic solution of the Tollens test,

the hemiacetal is in equilibrium with the open-chain aldehyde with the cyclic fonn in much larger concentration. However, it is the open-chain aldehyde that reacts with silver ion, so even though there is only a small amount of open-chain fonn present at any given time, as more of the open-chain fonn is oxidized by silver ion, more cyclic fonn will open to replace the consumed open-chain fonn. Eventually all of the cyclic fonn will be dragged kicking and screaming through the open-chain fonn to be oxidized to the carboxylate. Le Chatelier's Principle strikes again!

cyclic fonn """"" open-chain fonn Ag+

NOT larger concentration at equilibrium

smaller concentration reversible at equilibrium

432

• oxidized to carboxylate

Page 438: Solucionario de wade

18-69

OH H

Any carbon with two oxygens bonded to it with single bonds belongs to the acetal family. If one of the oxygen groups is an OR, then the functional group is a hemiacetal. Thus, the functional group at C-2 is a hemiacetal. (The old name for this g;rollp is hemiketal as it came from a ketone.

)

(b) Models will help. Ignore stereochemistry for the mechanism.

HO

HO

H2�

1

HO

HO

OH

1 oof' H+

OH

o *'

OH 00,... H o

OH

OH

OH

HO

same as

OH

HO

HO

HO

HOH2C

OH

433

H O� o 0

OH

\ H I OH

\. (00,... H ---"C-O

OH

+ 0 0 OH

H

Page 439: Solucionario de wade

18-70 Recall that "dilute acid" means an aqueous solution, and aqueous acid will remove acetals.

&H 1) __

X

_

M

_

gJ

_�-=�._ � 2)H30

+ V U I

H+ t II

H C OH H C OH 0 3()'OH �aBH4 3� H

H G

OH 0 0 cY excess� OH PCC H

--

B

PhNHN NNHPh �H

c

�)CH3MgI

2) H30

+ (removes acetal

)

1 equiv. (\

HO OH ..

H+

434

(P� A j H2Cr04 H20

o 0

crOH

E j Zn (Hg) HCI

o (i'oH

F

Ag+

-- no reaction D is identical to A

Page 440: Solucionario de wade

18-7 1 0 II

I H ()C,

18-72

fIfi HS SH S, /S -----. .. - C W Ph

/" '

H

H

1) BuLi fI 0+ 0 H3 II S, /S ..

2) PhCH fu /"C, HgCI2 Ph/"C'

CH Ph 2 Ph CH2Ph 2

H H H (a

)�w

l........ _� __ � Cl: dH

--- (1-: o H C+ 0"""" �'

H

ROH �H

HR

0=: OR

/ ..

(b) This is not an ether, but rather an acetal, stable to base but reactive with aqueous acid. (C)Q- ..

H W O:RU 9-: - ROH. { QtC:oH}

H2� ! O-H / t/O-H -Ct- .. Q w .. H H

O H

o 2.. 0 '--- • •

:O-H \ ('O+-H .. '-.... H /· ·

� + .. . l........ _ �-H o I I :O-H H • •

0-H J H O" �- � 2 • • o IIY"\ \L) I :0-

H +

435

U-H o II I :0: H

Page 441: Solucionario de wade

18-73

(a) First, deduce what functional groups are present in A and B. The IR of A shows no alkene and no carbonyl: the strongest peak is at 1060 cm-1, possible a C-O bond. After acid hydrolysis of A, the IR of B shows a carbonyl at 17 15 cm-1: a ketone. (If it were an aldehyde, it would have aldehyde C-H around 2700-2800 cm-1, absent in the spectrum of B.) What functional group has C-O bonds and is hydrolyzed to a ketone? An acetal (ketal)!

R'O OR' '\.. /

° II

C R-C-R / , R R

A mol. wt. 1 16 =>

C6H 1 ZOZ There is only one ketone of formula C4HgO: butan-2-one.

° II

H3C - C - CHzCH3 ..... t---B

B

A must have the same alkyl groups as B. A has one element of unsaturation and is missing only CZH4 from the partial structure above. The most likely structure is the ethylene ketal. Is this consistent with the NMR?

03.9 (singlet, 4H) � HzC-CH2

I \ 0XO A

o 1.3 (singlet, 3H) i H3C � - CH3 } 00.9 (triplet, 3H)

0 1.6

(quartet, 2H)

What about the peaks in the MS at rnJz 87 and WI? The 87 peak is the loss of 29 from the molecular ion at 1 16.

'W H3C CHz-CH3

rnJz116

+ .

The peak at rnJz 101 comes from loss of CH3 from the molecular ion at 1 16.

rnJz 101

436

plus two resonance forms with positive charge on the oxygen atoms

plus two resonance ..... _---I .. � forms with positive

charge on the oxygen atoms

Page 442: Solucionario de wade

18-73 continued

1

:1-H

C + ...... 1--___ • / "-Me Et

:3V

Hp . II H20 C

• •

/ "-Me Et

:0: II

� C / "-Me Et

B

. :or:"OH} "c

/ "-Me Et

18-74 The strong UV absorption at 220 nm indicates a conjugated aldehyde or ketone. The IR shows a strong carbonyl at 1690 em-I, alkene at 1625 em-I, and two peaks at 2720 em-I and 28 10 em-I -aldehyde!

o II

C==C-C-H The NMR shows the aldehyde proton at 8 9.5 split into a doublet, so it has one neighboring H. There are only two vinyl protons, so there must be an alkyl group coming off the B carbon:

H H 0 ,..-------------,. I I II

R-C==C-C- H

H H 0 I I II CH3-C==C-C- H

The only other NMR signal is a 3H doublet: R must be methyl. "crotonaldehyde"

18-75 .r-" ·0· EtO l.O EtO· • (a) 0 � 0

• D " � : . ; "I ... II H .B

- II � : O . EtO-P .0 . EtO -P � O. P,- I) R · P,- • � ..... R + " II I� I 1'\'-- 1

EtO""'" I ...... C -- EtO""'" I ...... C + C --

R-C-C - R' --R-C ..!.. C -R' OEt 1 OEt I - R'./ "' R' I I I I

R R R R' I R R'

R R' , /

(EtOhP02- + C==C 437 � '

R'

Page 443: Solucionario de wade

18-75 continued OEt (b) R (X + :

-OEt � /1 � OEt

(c) (i)(EtOhP + Br _ � .... COOMe .. (Et0}z0P, � .... COOMe

� � -EtBr � � �COOMe I MeO­(CH3}zCHCHO t

..... (-------°

(ii) (EtOhP + BrCH2COOMe .. - EtBr

(EtOhOP�COOMe + "- �) ---------'

-----y COO�

MeO-

18-76 (a) 0

H+ UNMe2 ..

H 1 ('::�( )�.-H

\!!.)C+

(b) Aminoacetal linkage

�� in the dashed bOXCS'�2

-olD .----1 N�lo HOW:

N : N

/.: HOW

: :

'0 : '0 : I I 1 I

�------' .�---- -- )

OH OH deoxyadenosine deoxycytidine 438

(c) The first step in the mechanism in part (a) is protonation of the amine's electron pair. The nitrogens of the DNA nucleosides, however, are part of aromatic rings, and the electron pairs are required for the aromaticity of the ring. (See the solution to problem 16-42 for a description of the aromaticity of these nucleoside bases.) Protonation of the nitrogen will not occur unless the acid is extremely strong; dilute acids will not protonate the N and therefore the nucleoside will be stable.

Page 444: Solucionario de wade

CHAPTER 19-AMlNES 19-1 These compounds satisfy the criteria for aromaticity (planar, cyclic 1t system, and the Huckel number of 4n + 2 1t electrons): pyrrole, imidazole, indole, pyridine, 2-methylpyridine, pyrimidine, and purine. The systems with 6 1t electrons are: pyrrole, imidazole, pyridine, 2-methylpyridine, and pyrimidine. Th� systems with 10 1t electrons are: indole and purine. The other nitrogen heterocycles shown are not aromatic because they do not have cyclic 1t systems.

19-2

(a) CH3 I

H3C- �-NH2 CH3

(d)

19-3 (a) pentan-2-amine (c) 3-aminophenol (or meta-)

(b)

(e)

NH2 I

CH3-CHCHO

(b) N-methylbutan-2-amine (d) 3-methylpyrrole

(c) Q N

H C/ ........ CH 3 3

(f)

(e) trans-cyclopentane-l,2-diamine (f) cis-3-aminocyclohexanecarbaldehyde

19-4 (a) resolvable: there are two asymmetric carbons; carbon does not invert (b) not resolvable: the nitrogen is free to invert (c) not resolvable: it is symmetric (d) not resolvable: even though the nitrogen is quaternary, one of the groups is a proton which can

exchange rapidly, allowing for inversion (e) resolvable: the nitrogen is quaternary and cannot invert when bonded to carbons

19-5 In order of increasing boiling point (increasing intermolecular hydrogen bonding): (a) triethylamine and n-propyl ether have the same b.p. < di-n-propylamine (b) dimethyl ether < dimethyl amine < ethanol (c) trimethylamine < diethylamine < diisopropylamine

19-6 Listed in order of increasing basicity. (See Appendix 2 for a discussion of acidity and basicity.) (a) PhNH2 < NH3 < CH3NH2 < NaOH (b) p-nitroaniline < aniline < p-methylaniline (p-toluidine) (c) pyrrole < aniline < pyridine (d) 3-nitropyrro\e < pyrrole < imidazole

19-7

(a) secondary amine: one peak in the 3200-3400 cm-i region, indicating NH (b) primary amine: two peaks in the 3200-3400 cm-i region, indicating NH2 (c) alcohol: strong, broad peak around 3400 cm-i

439

Page 445: Solucionario de wade

19-8 A compound with formula C4H"N has no elements of unsaturation. The proton NMR shows five

lypes of H, with the NH2 appearing as a broad peak at 8 l.15, meaning that there are four different groups

of hydrogens on the four carbons. The carbon NMR also shows four carbons, so there is no symmetry in

this structure; that is, it does not contain a t-butyl group or an isopropyl group.

The multiplet farthest downfield is a CH deshielded by the nitrogen; integration shows it to be one H. There is a 2H multiplet at 8 l.35, the broad NH2 peak at 8 1.15, a 3H doublet at 8 l.05, and a 3H triplet at 80.90. The latter two signals must represent methyl groups next to a CH and a CH2 respectively. So far:

NH2 I

-C­H HC-C-3 H The pieces shown above have one carbon too many, so there must be one carbon that is duplicated: the

only possible one is the CH, and the structure reveals itself.

19-10 (a)

80.90 82.8 (multiplet)

(triplet) H, H / H3C-C� -C -CH 8 l.05 (doublet)

I 3 8 l.35 / NH� (multiplet)

"'---r""'" 8 l.15 (singlet)

(b) 4l.0 -- CH3 I

CH3CH2 -N -CH2CH3

(c) 44.7 + 0

II �51.1/ CH3CH2-CH 12.4 (An older printing of the text used values of 13.8,47.5, and 58.2. The values shown here are taken from a spectrum.)

t 7.9 t 20l.9 (d) 25.8

+ CH3CH2CH2OH

t t 10.0 63.6

[H)C+;�2-r��1-CH2CHf � iCH)-CH2-t-�H2 - CH)-CH2-�=CH2}

mlz 87 j . mlz 58 + CH3CH2

mass 29 + • CH3

mass 15 (c) The fragmentation in (a) occurs more often than the one in (b) because of stability of the radicals produced

. along with the iminium ions. Ethyl radical is much more stable than methyl radical, so pathway

(a) tS preferred. 440

Page 446: Solucionario de wade

19- 1 1 Nitration at the 4-position of pyridine is not observed for the same reason that nitration at the 2-position is not observed: the intermediate puts some positive character on an electron-deficient nitrogen, and electronegative nitrogen hates that. (It is important to distinguish this type of positive nitrogen without a complete octet of electrons, from the quaternary nitrogen, also positively charged but with a full octet. It is the number of electrons around atoms that is most important; the charge itself is less important.)

1+ • • + GOOD: - N- VERY BAD: -N

1 mechanism

o �II � _J fD+-

N 0

I H NO H NO H� N02

a�� · 0 -----'. :�l) � 6 N + N N N

VERY BAD­N does not have an octet of electrons

not produced because of unfavorable in termedi ate

19- 12 Any e lectrophilic attack, including sulfonation, is preferred at the 3-position of pyridine because the intermediate is more stable than the intermediate from attack at either the 2-position or the 4-position. (Resonance forms of the sulfonate group are not shown, but remember that they are important!)

The N of pyridine is basic, and in the strong acid mixture, it will be protonated as shown here. That is part of the reason that pyridine is so sluggish to react: the ring already has a positive charge, so attack of an electrophile is slowed.

441

Page 447: Solucionario de wade

19- 13

Cl

�-oo � :OCH3 I _

0_

0

__ ..

N

19- 14

(a)

� / -0 0

� _jlC :NH� N Br

cj;�-

o�H�5Hl N _N N GOOD

stabilized by induction from nitrogen �NH2 (jib:, NH:

N - �

GOOD

-Cl� �' N

This is a benzyne-type mechanism. (For simplicity above, two steps of benzyne generation are shown as one step: first, a proton is abstracted by amide anion, fol lowed by loss of bromide.) Amide ion is a strong enough base to remove a proton from 3-bromopyridine as it does from a halobenzene. Once a benzyne is generated (two possibilities), the amide ion reacts quickly, forming a mixture of products.

Why does the 3-bromo follow this extreme mechanism while the 2-bromo reacts smoothly by the addition­e limination mechanism? Stability of the intermediate! Negative charge on the electronegative nitrogen makes for a more stable intermediate in the 2-bromo substitution. No such stabilization is possible in the 3-bromo case.

442

Page 448: Solucionario de wade

19-15 n H � n ..--...... CH3 - I + I), HC03- ..--...... CH3 - I Pr- N-H .. Pr- N-H � Pr- N - H ..

I H 19- 16

(b) excess NH3 + Br� --- H2N�

(c) excess NH3 + PhCH2Br

19- 17 ° II

(a) CH3C -NHCH2CH3

19- 18 If the amino group were not protected, it would do a nucleophilic substitution on chlorosulfonic acid. Later in the sequence, this group could not be removed without cleaving the other sulfonamide group.

19-19

continued on next page

NH-S03H NH-� 6� I AI � both / / c. ====::> Y sulfonamide,

�2NH2 J

sulfathiazole

443

Page 449: Solucionario de wade

19-19 continued ¢OCHl +

S02CI

19-20

(a)�

(d) H3C, ,CH3

CXJ

NH2

6

(b)

NHCOCH3

9'-':: HCI •

h- H20 !1

S02 I

NH

6

NH2

9 S02 I NH

6 sulfapyridine

lllt? H3C, ,..CH3 N

+ aCH3

(e) ,CH3 V\H1

CH3 ,

H1C-N�

CH3 I

(c) N o + H2C=CH2

(I) �N-CH] 19-2 1 Orientation of the Cope elimination is similar to Hofmann elimination: the less substituted alkene is the major product. (a)�

+ (CH3hNOH

(b) H2C=CH2

major

HO'N� +� +� minor

+ (CH3CH2)2NOH

(c) 0 + (CH,hNOH (d) H2C=CH2 + 0 N OH

+0 r N �OH

nunor

19-22 The key to this problem is to understand that Hofmann elimination occurs via an E2 mechanism requiring anti coplanar stereochemistry, whereas Cope elimination requires syn coplanar stereochemistry.

(a) H,

FH(CH3h

/ � .:-...;CH3

� C

H2C \ H

Hofmann orientation loses a hydrogen from the CH3 and the N(CH3h group to make the less substituted double bond

444

Page 450: Solucionario de wade

19-22 continued (b) Hofmann elimination

(CH ) N+ 5=H(CH3h 3 3 \� ,) f�CH3 H " ",CH(CH3h "C --- C ..

H "I "--\ HO: H3C H �

'I, ,,\

IIC::::: C" H3C

� "'CH3

E (Saytzeff product-more highly substituted)

Cope elimination

.. 19-23

H " ",CH3 '1/ ,\\ IIC-C"

H3C� - "'CH(CH3h

Z (Saytzeff product-more highly substituted)

+ o Aliphatic diazonium ions are very unstable, rapidly decomposing to carbocations.

(b) /'N" NO

� (c) NO

I

o + (d) N2

6 Cl-

Aryl diazomum IOns are relati vely stable If kept cold.

19-24 The diazonium ion can do aromatic substitution like any other electrophile.

� H i-03S-o-� N=NO--- -03S-o-� N=NV-I+ � } -H

_ �(CH3h � _ N(CH3h

. .. . plus two other resonance forms most slgmflcant resonance contnbutor j .. _ with positive charge on the ring :Cl: (or some other base) ..

- 03S -< >-N = N -< >-N(CH3h

methyl orange 445

Page 451: Solucionario de wade

19-25

(a)

8 + NaN02

• HCI

+ N2 CI-

a (b) N2 CI- CI

6 a from (a)

(c)

8

CuCI ..

° I I � (SCCHl

CH3C-;I : I

F

HBF4 -t::.. 6

° I I

HN-CCH3

3 CH31 CH3'¢rCH3 - I �

CH3

H

NH H30+ CH3 *

2

..

9' CH3

t::.. � I

NaN%2 +

CH"

CH3

'¢r CH3 H3P02 I • �

3 9' CH3 CH qN2 CI-

HCI .

�I CH3 CH3

+ + (d) N2 CI- Br (e) N2 CI- I

a CuBr 6 a KI 6 .. ..

from (a)

+ + (f) N2 CI- CN (g) N2 CI- OH

a CuCN 6 a H2O 6 .. .. H2SO4

t::.. +

(h) N2 CI-

a + � }-OH .. < }-N'Ni }-OH

HO HO 446

Page 452: Solucionario de wade

19-26 General guidelines for choice of reagent for reductive amination: use LiAlH4 when the imine or oxime is isolated. Use Na(CH3COOhBH in solution when the imine or iminium ion is not isolated. Alternatively, catalytic hydrogenation works in most cases. (a) 0 (i' H

(b) 0 II

PhCH2 - C - CH3

(c) H I

H2NOH ..

if

0

NOH 1) LiAlH4 NH2

II I PhCH2 - C - CH3 .. PhCH2 - C - CH3

2) H2O I H

CH2Ph I

0 Ph)l H Na(CH3COOhBH 0 + ..

(d) 0

2)

N,Ph

NHPh

6 H+ 6 1) LiAlH4 6 + -. .. 2) H2O

(e) 0 NOH

c5 6 H2NOH 6 1) LiAlH4 .. ..

W 2) H2O

19-27 0-r (a) H 0 r" I

1) LiAlH4 0 + Cl� --- 0 .. 0 (b)

6

a

+ CI'l) ---

2) H20

0 HN� HNJ) 6 A I)UAIH� 6 A :::::..... 2) H20 :::::.....

447

OR H

c� H

Page 453: Solucionario de wade

19-28 Use a large excess of ammonia to avoid multiple alkylations of each nitrogen.

�Br + NH3 ____ �NH2

19-29 o

(a)�N: o o

BrCH2Ph ..

excess

o C<N-CII2Ph

o o

NH2NH2 ..

!l H2NCH2Ph

(b)� v--t Br(CH2hCH3 c<

NH2NH2 - I N - (CH2)sCH3 • H2N(CH2)sCH3 :::-... !l

o o (c)�

�N:

o

o o

Br(CH2hCOO- � -----4 .. _ I N - (CH2hCOO-must use anion :::-... to avoid protonating 0 the phthalimide anion

1) NH2NH2 ' !l ...

2) H+

19-30 Assume that LiAIH4 or Hz/catalyst can be used interchangeably.

PhCH2Br + NaN3 -- PhCH2N3 H2

PhCH2NH2 (a) Pt ..

(b) � Br � �C= N 1) LiAlH4

�NH2

V V 2) H20 V (c) 0 �OH

1) LiAIH4 •

2) H20 � TsCl

.. OH pyridine �OTs

OR 0 �OH SOCl2

..

�NH2

o �Cl

� -�OTs��C=N

from (c)

1) LiAIH4 •

2) H20

NH3 -

+ NaN3 �N3

o �NH2

1) LiAIH4 + 2) H20

�NH2

1) LiAIH4 �NH2 .. 2) H20

448

H2N(CHlhCOOH

Page 454: Solucionario de wade

19-30 continued

(e)

(D

Br H \� � R

Br H \� � R

NaN3 .. SN2-inversion

NaCN .. SN2-inversion

H N3 � 1) LiAIH4 •

1) LiAIH4 ..

� NH2 � S �H2NH2 S

(g) 0 � KCN HO CN 1) LiAIH4 H� .. .. � HCN

19-3 1 To reduce nitroaromatics, the reducing reagents (H2 plus a metal catalyst, or a metal plus HCI) can be used virtually interchangeably. Assume a workup in base to give the free amine final product. (a)

o Sn • HCl

(b) Br Br

¢

Br

o 6 f; U Br

Sn HCI

from (a) (d) 2) COOH

6

¢ N02

+ ortho

Fe HCI

)) U Br

COOH Jr U NO 2

Fe .. HCI

19-32 CH30 �.. :0 : :0: } B, �' 0 , I I .. y-:�H I I .. - , U I I ..

PhCH2 - T-C- �,"-, H R-C-� : ---- R-C=N

H,: R-C �N

H' ) Br

CH3 H H ( , � , 449 mechan;,m cont;nued on next page-:�H!

Page 455: Solucionario de wade

19-32 continued

i :0:- - .. � i (:0:- :0: t .. I :OH .. � • • - Br- '-,1 .. II .. -R-N=C-OH � R-N=C=O --- R-C=N-Br --- R-C-N-Br U· �U .. t :0: t :0: -.. I I .. .. I I � �H

""OH .. - ..

R-N-C-OH H""OH

R-7�C-?� :OH2 R-N: I H

+ CO2 -----l .. � R -N - H I H � H HJ

Stereochemistry at a chiral carbon is lost if the carbon goes through a planar intermediate, e.g., either a carbocation or a free radical. Configuration at a chiral carbon can be inverted during substitution by a nucleophile from the side opposite the leaving group. However, when the carbon retains all four pairs of electrons, as in this Hofmann rearrangement, it retains its configuration.

19-34

'" �NH2

H CH3 R

(a) The acyl azide of the Curti us rearrangement is similar to the N-bromoamide of the Hofmann rearrangement in that both have an amide nitrogen with a good leaving group attached. Subsequent alkyl migration to the isocyanate and hydrolysis through the carbamic acid to the amine are identical in both mechanisms. (b) The leaving group in the Curtius rearrangement is N2 gas, one of the best leaving groups known "to man or beast", as we used to say. (c) ,----------------, ,

, .-----------------abbreviate as "R"

Page 456: Solucionario de wade

19-35 Please refer to solution 1-20, page 12 of this Solutions Manual.

19-36 (a) primary amine; 2,2-dimethylpropan-l-amine, or neopentylamine (b) secondary amine; N-methylpropan-2-amine, or isopropylmethylamine (c) tertiary heterocyclic amine and a nitro group; 3-nitropyridine (d) quaternary heterocyclic ammonium ion; N,N-dimethylpiperidinium iodide (e) tertiary aromatic amine oxide; N-ethyl-N-methylaniline oxide (f) tertiary aromatic amine; N-ethyl-N-methylaniline (g) tertiary heterocyclic ammonium ion; pyridinium chloride (h) secondary amine; N,4-diethylhexan-3-amine

19-37 Shown in order of increasing basicity. In sets a-c, the aliphatic amine is the strongest base. H

I (a) Ph-N-Ph

H I N 0 < (b)

pyrrole

(c) CNH < pyrrole

< PhNH2

N 0 < � HN "N \ I

< H

<

o NH2 aliphatic amine is the strongest base aliphatic amine is the strongest base; pyrrole's aromaticity would be lost if protonated

aliphatic amine is the strongest base; pyrrole's aromaticity would be lost if protonated

ONH2 (d) I �

02N h < ONH2 < I

H C h 3

Basicity is a measure of the ability to donate the pair of electrons on the amine. Electron-withdrawing groups like N02 decrease basicity, while electron-donating groups like CH3 increase basicity. °

(e)�NH I 2

h

ONH2 crCH2NH2 < I < I aliphatic amine is the strongest base h h (amides are not basic)

Congratulations if you got this one right! CH3 is only slightly electron-donating by induction so the first structure puts the least electron density on the N. The second structure is more basic (by about 1 pK unit) because NH2 is electron-donating by resonance. The last structure is the strongest base for a completely different reason: steric inhibition of resonance. ( See the last topic in Appendix Two of this Manual.) The ortho methyl interferes with the isopropyl substituents on N and it forces the NR2 out of planarity with the ring, so that the electron pair on the N is not conjugated with the aromatic pi system. Essentially, this N becomes an aliphatic amine. The pKb of 2-methyl-N,N-diethylaniline is 6.9, about 2.5 pK units more basic than aniline. It is not as strong a base as a tertiary aliphatic amine because the aromatic ring is somewhat electron withdrawing by induction, yet the ortho methyl reduces the pi overlap of the nitrogen's electrons to make the N significantly more basic.

451

Page 457: Solucionario de wade

19-38 (a) not resolvable: planar (c) not resolvable: symmetric (e) not resolvable: symmetric (g) not resolvable in conditions where the proton on N can exchange

(b) resolvable: asymmetric carbon (d) resolvable: nitrogen inversion is very slow (f) resolvable: asymmetric nitrogen, unable to invert (h) resolvable: asymmetric nitrogen, unable to invert

19-39 The values of pKb of amines or pKa of the conjugate acids can be obtained from Table 19-3. The side of the reaction with the weaker acid and base will be favored at equilibrium. (See Appendix 2 for a discussion of acidity and basicity.) (a)

(b) I[)

(c)

N H

+

(d) ONH3 I +

� pKa 4.60

19-40

(a) PhCH2CH2NH2

19-4 1

.. --

o H

Q H

�I � JI + CH3COO-N+

I H pKa 8.75 I[J + CH3COO­.N,

H H pKa "'- 1

o N +

pKa 1 1. 12

O NH2 n I + N � .... +, H H pKa 1 1.27

products are favored

reactants are favored

products are favored

products are favored

(e) cr(\� NH,

retention of configuration

(a) PhCH2CH2CH2NH2 (b) �NH2

(e) d��- CHJ

(d) eN,

o OH

(e)(D �N

I CH3

452

(h) O NH2 I after workup

� with base

Page 458: Solucionario de wade

19-4 1 continued (i) ogNHCH1 U) 0CH2CH2NHCHl (k) NHCH3

I CH3(CH2)3CHCH2CH3

(I) CH2NH2 I

PhCH2CHCH3

(m) /"'N� (n) � Q (0) qH OCH2CH3 N02

(p) � HO CN

LiAIH4 ..

H20

- � HO CH2NH2

19-42 +

(a) CH3 � NH2 NaN02 CH3 � N2

o HCl- 0 Cuctt / I

Cl- CH30CN

(b) CH10CN

from (a) + N (c) CH1'O 2

from (a) + N (d) CH1'O 2

from (a)

Cl-

Cl-

1) LiAIH4 ..

2) H20

Kl ---

H2S04 ..

H20

°

CH3 �CH2NH2

o CH10 I

CH3 �OH o NH II (e) CH3'O 2 CH3C-Cl I •

CH3 '0 NHCOCH3 HN03 CH3 0 NHCOCH3

� I -� I (+ isomer)

H2S04 °2N

CH30NH2 JH O+ after workup ?' I J 3

with base � !1 °2N

(f) CH3 '0 NH2 =<J Na(CH3COOhBH I + ° -�

CH1 � �--o o

453

Page 459: Solucionario de wade

19-43 This fragmentation is favorable because the iminium ion produced is stabilized by resonance. Also, there are three possible cleavages that give the same ion. Both factors combine to make the cleavage facile, at the expense of the molecular ion.

CH3 I

CH3t?-NH2

CH3 58

mlz 73

19-44

(a) 0 � I

(b) ONH2

I + �

(e) C N-CH3

(I) C N-CH3

H202

H202

+ .

--- oCH3 + mass 15

.. C/o N�

CH3

.. C /o N� CH3

(g) CH3 Y'1(COOH SOCl2 o ..

1) LiAlH4 ..

t:. ----

454

CH3 I +

....... f---l .. � C = NH2 I CH3

mlz 58

�-o Cl-C ..

(N.oH

"'CH 3

Page 460: Solucionario de wade

19-45 The problem restricts the starting materials to six carbons or fewer. Always choose starting materials with as many of the necessary functional groups as possible.

(a) HO-Q-NH2

(c) HO

Mg

HO-< )

19-46

HO

HOOCH2CH2NH2

dopamine

NBS ..

H2, Pt ..

455

phenacetin

HO

HOOCH2B'

methamphetamine

Page 461: Solucionario de wade

19-46 continued (b) �

H2

�n C N� o 00=7

H H

H2 --­Pt COH

R: 1 2

� H+) NH,

r'Y:l -­��! I NH2 TI NH2 l bas

:

:

) H :. H • •

W-HO: 1 } co co .......... + N

- H20 .. .. �

co 19-47

(a) 0 "

H 1 H

H2 ---Pt

H+O: 1 1 H H

I H t

(X)� � � base:��N)

N + 1 )

o I I

.. H

o I I aC'OH SOCl2

• aC ...

I Cl :::::-...

NH3 .. r?'Y

C, o NH2 1) LiAIH4 r?'YCH2NH2

2)H20 .. 0

(b) o I I ac'

I H :::::-...

+ NH3

(c) o II

CI-C-CH3 •

C

NH

Na(CH3COOhBH aCH2NH2

� I ::::: :::::-...

C � 1) LiAIH4 C N-C-CH3 2)H20 "

N -CH2CH3

456

Page 462: Solucionario de wade

19-47 continued (d) 0 r-, Na(CH3COOhBH 0- r-,

o + HN0 .. N0 (e) HO�OH SOCI2 ...

NH CI�CI � H2N�NH2

o 0 _ / 0 0 o 0 1) LiAlly 2) H20

H2N�NH2 19-48 0 (a)

�Br -N� NH2NH2 �NH2 + -- ..

L1 0

(b) CH2CH3

6 HNO] I .. � H2SO4 QCH3

Sn � I HCI

'" QCH]

(c) �OH

Ph 0

(d) �Br

Ph

N02 NH2

SOCI2 ...

�CI NH3

____ �NH2

+ -� o

Ph 0 Nao� Ph 0

� N Ph H2

--NH2NH2 ...

�NH2

L1 Ph

(e)�Br

Ph

KCN �CN 1) LiAlH4 ... I 2)H O

"' � NH2 Ph

2 Ph

457

Page 463: Solucionario de wade

19-49 (a) When guanidine is protonated, the cation is greatly stabilized by resonance, distributing the positive charge over all atoms (except H):

NH� 1 NH2 NH2 NH2 .. II .. w .. II .. .. I .. + I • •

H,N-C-NH, � H,N-C-NH, �H,N- �-NH, �H'N=� NH, NH, } • • I +

H2N-C=NH2.

(b) The unprotonated molecule has a resonance form shown below that the protonated molecule cannot have. Therefore, the unprotonated form is stabilized relative to the protonated form. This greater stabilization of the unprotonated fonn is reflected in weaker basicity.

o· .. -o� + //. H2N N

- \ :0:

..

(c) Anilines are weaker bases than aliphatic amines because the electron pair on the nitrogen is shared with the ring, stabilizing the system. There is a steric requirement, however: the p orbital on the N must be paral lel with the p orbitals on the benzene ring in order for the electrons on N to be distributed into the TC

system of the ring.

If the orbital on the nitrogen is forced out of this orientation (by substitution on C-2 and C-6, for example),

the electrons are no longer shared with the ring. The nitrogen is hybridized sp3

(no longer any reason to be

sp2), and the electron pair is readily available for bonding � increased basicity.

H H \/ C ___

H

458

As surprising as it sounds, this aniline is about as basic as a tertiary a liphatic amine, except that the aromatic ring substituent is electron-withdrawing by induction, decreasing the basicity slightly. This phenomenon is cal led "steric inhibition of resonance". We will see more examples in future chapters. Also, it is the last topic in Appendix 2. (See another example in the solution to 19-37(f).)

Page 464: Solucionario de wade

19-50 In this problem, sodium triacetoxyborohydride will be represented as Na(AcOhBH.

(a) /'.... /'.... TsCI /'.... /'.... KCN /'.... /'.... 1) LiAIH4 /'.... /'.... _ NH2 / ........, -OH --- / ........, -OTs --- / ........, -CN .. / ........, ........,

pyridine 2) H20

H (b)

�OH PCC --- �O

CH3NH2 � NHCH3 ..

(c) �OH

� OH

TsCI -pyridine cr03 ..

H2S04 H20

Na(AcOhBH

�OTs NH3 • �

NH2

./"--./ N a(AcOhBH /"'--.../ II + �NH2 • I o HN� �(ACOhBH 0 /"'--.../ I I I CH3-CH

�N� tpcc CH3CH20H

o (d) CH3 CH2Br (r � a Na0't

" �CH20H �C-

H

(e)

aCH2NHCH2CH2CH]

(rCH3

I • � KMn04

0

�COOH

o ·

0

V PctV Na(AcOhBH

�NH2 from (c)

o " I ---?" I

NH2 (rCOCI ac-

� NH3 � jBr2 NaOH

NaN02 �NH2

.. HCI 0 0

(I) 0 �CI I

� •

.& AICI3

�HNO] I •

.& H2SO4

� zna[gJ y .& HCI .&

�OH

° H2Cr04 SOCI2 �

.. --- CI

Clemmensen N02 reduces both NH2

459

Page 465: Solucionario de wade

19-50 continued °

(g) o� y�l � Zn(Hg� ' / AICl3 U I HCI

°

YOH H2C,?' 5 YCl

19-5 1 ° I I

uY HN03 Fe, HCI f)I I h H SO

� • I h 2 4 H2N after workup with base

(a)o �I CI- CCH2CH2CH3 Zn(Hg)

(b)O (c)

Ph-Q-� N02 � HCI

19-52 S H �H I

�'''''t) coniine

� • HCI

'--CN-CH1

Hofmann elimination

Ph-Q-NH2

460

Page 466: Solucionario de wade

19-53 unknown X

(a) -fishy odor � amine

-molecular weight 101 � odd number of nitrogens

� if one nitrogen and no oxygen, the remainder is C6H)S Mass spectrum:

-fragment at 86 = M - 15 = loss of methyl � the compound is likely to

have this structural piece: CH3 � - N a-cleavage

IR spectrum:

-no OH, no NH � must be a 3° amine

-no C=O or C=C or C=N or N02

NMR spectrum:

--only a triplet and quartet, integration about 3 : 2 � ethyl group(s) only

assemble the evidence: C6H)SN, 3° amine, only ethyl in the NMR:

(b) React the triethylamine with HCI. The pure salt is solid and odorless. +

(CH3CH2hN + HCI -. (CH3CH2hNH CI -salt

(c) Washing her clothing in dilute acid like vinegar (dilute acetic acid) or dilute HCI would form a water­soluble salt as shown in (b). Normal washing will remove the water-soluble salt.

---

yQ o H

H

(b) Both answers can be found in the resonance forms of the intermediate, in particular, the resonance form that shows the positive charge on the N. This is the major resonance contributor; what is special about it is that every atom has a full octet, the best of all possible conditions. That does not arise in the benzene intermediate, so it must be easier to form the intermediate from pyrrole than from benzene. Also, acylation at the 3-position puts positive charge at the 2-position and on the N, but never on the other side of the ring, so this substitution has only two resonance forms. The intermediate from acylation at the 3-position is therefore not as stable as the intermediate from acylation at the 2-position.

461

Page 467: Solucionario de wade

19-55

(a) F

OZN'Q�oo

I NHzR ..

N02

000 + 0 0-; + 0 + olio F NH2R 0 1 �o F NHzR II1jF NH2R _

0 0

N:"

Y

- 0 0 N+ N+ :0'" 'c 0 I :0'"

I -0'" I I .. . �.. �

� � 00 C-

+ + 1

+ ... N.... _ ... N.... _ ... N .... 0 0-

0'" 0 0'" 0 :0'" 0:

(b) Why is fluoride ion a good leaving group from A but not from B (either by SNI or SN2)?

_yF Nu

N02 HC� I

A NOz

A �� :NU

C B

Formation of the anionic sigma complex A is the rate-determining (slow) step in nucleophilic aromatic substitution. The loss of fluoride ion occurs in a subsequent fast step where the nature of the Jeaving group does not affect the overall reaction rate. In the SN 1 or SN2 mechanisms, however, the carbon-fluorine hond is breaking in the rate-determining step, so the poor leaving group ability of fluoride does indeed affect the rate.

t >-.

lrl� ______________ __

nucleophilic aromatic substitution

t

(c) Amines can act as nucleophiles as long as the electron pair on the N is available for bonding. The initial reactant, methylamine, CH3NHZ' is a very reactive nucleophile. However, once the N is bonded to the benzene ring, the electron pair is delocalized onto the ring, especially with such strong electron­withdrawing groups like N02 in the ortho and para positions. The electrons on N are no longer available for bonding so there is no danger of it acting as a nucleophile in another reaction.

462

delocalized ,

N02

Page 468: Solucionario de wade

19-56 Compound A Mass spectrum:

-molecular ion at 73 = odd mass = odd number of nitrogens;

if one nitrogen and no oxygen present � molecular formula C4H11N

-base peak at 44 is M - 29 � this fragment must be present: -N -CH3- � I -,tCH2CH3

EITHER OR H-C-CH2CH3

19-57

44 � I H 2

a-cleavage

IR spectrum:

-two peaks around 3300 cm-I indicate a 1 ° amine; no indication of oxygen

NMR spectrum:

-two exchangeable protons suggest NH2 present

-I H multiplet at 8 2.8 means a CH-NH2 A

The structure of A must be the same as 1 above:

Compound B an isomer of A, so its molecular formula must also be C4H11N

IR spectrum:

-only one peak at 3300 cm-I � 2° amine

NMR spectrum:

-one exchangeable proton � NH -two ethyls present B The structure of B must be:

+ - CH2 = NCH2CH3

I H resonance-stabilized

rnJz 58

(a) The acid-catalyzed condensation of P2P (a controlled substance) with methylamine hydrochloride gives an imine which can be reduced to methamphetamine. The suspect was probably planning to use zinc in muriatic acid (dilute HCl) for the reduction.

0Jl + -Cl

phenyl-2-propanone, P2P methamphetamine (phenylacetone)

(b) The jury acquitted the defendant on the charge of attempted manufacture of methamphetamine. There were legal problems with possible entrapment, plus the fact that he had never opened the bottle of the starting material. The defendant was convicted on several possession charges, however, and was awarded four years of institutional time to study organic chemistry.

463

Page 469: Solucionario de wade

1 9-58

Mass spectrum:

-molecular ion at 87 = odd mass = odd number of nitrogens present

-if one nitrogen and no oxygens � molecular formula CSH13N R tCH2NH2

-base peak at mJz 30 � structure must include this fragment 30

IR spectrum:

-two peaks in the 3300-3400 cm-1 region � 1 ° amine

NMR spectrum:

-singlet at 8 0.9 for 9H must be a t-butyl group

-2H signal at 8 1 .0 exchanges with D20 � must be protons on N or °

8 1.0

80.9{ CIl3_ �H3

CH2-l, �H3 t 82.4

H H \ I

-.. C = N+ / \

H H mJz 30

Note that the base peak in the MS arises from cleavage to give these two, relatively stable fragments:

19-59 (a tough problem)

CH3 H H I \ /

CH3 -C· + C = N + 1 / \

CH3 H H

mJz 30

molecular formula CIIHl6N2 has 5 elements of unsaturation, enough for a benzene ring; no oxygt:ns precludes N02 and amide; if C:::::N is present, there are not enough elements of un saturation left for a

benzene ring, so benzene and C=N are mutually exclusive

IR spectrum:

--one spike around 3300 cm-1 suggests a 2° amine

-no C=N

-CH and C=C regions suggest an aromatic ring

Proton NMR spectrum:

-5H multiplet at 8 7.3 indicates a monosubstituted benzene ring (the fact that all the peaks are huddled

around 7.3 precludes N being bonded to the ring)

-1 H singlet at 8 2.0 is exchangeable � NH of secondary amine

-2H singlet at 8 3.5 is CH2; the fact that it is so strongly deshielded and un split suggests that it is between

a nitrogen and the benzene ring

fragments so far: < >-CH, -N + NH + 4 C + 8 H + I clement of unsaturation

continued on next page 464

Page 470: Solucionario de wade

19-59 continued Carbon NMR spectrum: -four signals around 5 125-138 are the aromatic carbons -the signal at 5 65 is the CH2 bonded to the benzene -the other 4 carbons come as two signals at () 46 and () 55; each is a triplet, so there are two sets of two

equivalent CH2 groups, each bonded to N to shift it downfield fragments so far:

< } CH2-N + NH + CH2 + CH2 + 1 element of unsaturation CH2 CH2

There is no evidence for an alkene in any of the spectra, so the remaining element of unsaturation must be a ring. The simplicity of the NMR spectra indicates a fairly symmetric compound.

Assemble the pieces:

19-60 (a) H

0 Ph�N�

Jl �H 11 CH3 �a(AcOhBH

( � N �

V

465

(b)

( HN�

Page 471: Solucionario de wade

19-61 Not only is substitution at C-2 and C-4 the major products, but substitution occurs under surprisingly mild conditions.

Begin by drawing the resonance forms of pyridine N-oxide: . . -:0: 1+

0 .. ..

:0: 11+ N -OR . .. �

:0: 11+

(�) ..

C-H

.. ..

:0: 11+

- N RO Resonance forms show that the electron density from the oxygen is distributed at C-2 and C-4; these positions would be the likely places for an electrophile to attack.

Resonance forms from electrophilic attack at C-2 . . -:0: :0: ··0··­. . ··0· .... . . 1++ E II + E 1+ E 1+ E aN H " _ aN

H " .. (:fN

H " - cN� H

.0 0 HC 0 � CH + + N does not have an octet; not a significant resonance contributor

GOOD! All atoms have octets.

Resonance forms from electrophilic attack at C-3

··0··­. . ··0··­. . . . -:0: 1+ 1+ 1+

(N\ H�Q

N,,= aN'CH

H .. - H " - H + � � C H E E E These two resonance forms place two positive charges on adjacent atoms-not good.

Resonance forms from electrophilic attack at C-4 . . -:0: 1+

OR ..

H E

. . -:0: 1++

- 0 .. H E N does not have an octet; not a significant resonance contributor

..

continued on next page 466

:0: 11+

0 ..

H E GOOD! All atoms have octets.

. . -:0: 1+

-Q HC + H E

Page 472: Solucionario de wade

19-61 continued The resonance forms from electrophilic attack at C-3 are bad; only one of the three is a significant contributor, which means that there is not much resonance stabilization. When the electrophile attacks at C-2 or C-4, however, there are two forms that are good plus one great one that has all atoms with full octets. Clearly, attack at C-2 and C-4 give the most stable intermediates and will be the preferred sites of attack.

19-62 (aJ (f0: �C-!l

continued on next page

H H + I� � CH3

I I H-N-H

O<>?: ....... f--l.� 0/9.: H�NCI!3 O- Q.H

B- is the conjugate base of the acid HB CH3 I

H-N:

1� B:

CH3 I

H - N:

d·· C-OH '-""1+

H

,...... d .. � C -9.H

� � called an aminal

467

Page 473: Solucionario de wade

19-62 continued

(b)crO:'� H -B .. .�

0

+C - C-OH q • •

11 B: B- is the conjugate base

of the acid HB

o 0

aD O�:·· � o�: .. C /' .. -H20 C - OH H - B C - OH -' ��+ �. / �H�_ 0::0 � (X/O

I enamme H

To this point, everything is the same in the two mechanisms. But now, there is no H on the N to remove to form the imine. The only H that can be removed to form a neutral intermediate is the H on C next to the carbocation.

(c) A secondary amine has only one H to give which it loses in the first half of the mechanism to form the neutral intermediate called an aminal, equivalent to a hemiacetal. In the second half of the mechanism, the H on an adjacent carbon is removed to form the neutral product, the enamine. The type of product depends entirely on whether the amine begins with one or two hydrogen atoms.

468

Page 474: Solucionario de wade

CHAPTER 20-CARBOXYLIC ACIDS

20-1 CH3 I

(a) CH3CH2CHCOOH

(d)

(

)

'>

tOH

H H

(g) COOH

Cl � CI

(j) HOOC�COOH Cl

(c) cf:�:� H

(h) NCOOH

�COOH CH3

o

(k) �COOH

20-2 IUPAC name first; then common name.

(a) 2-iodo-3-methylpentanoic acid; a-iodo-p-methylvaleric acid (b) (Z)-3,4-dimethylhex-3-enoic acid (c) 2,3-dinitrobenzoic acid; no common name (d) trans-cyclohexane-1,2-dicarboxylic acid; (trans-hexahydrophthalic acid) (e) 2-chlorobenzene-\ ,4-dicarbox ylic acid; 2-chloroterephthalic acid (f) 3-mcthylhexanedioic acid; p-methyladipic acid

(f)

. NH2 (1) I - COOH HOOC �

(I) 0(y0H V a

20-3 Listed in order of increasing acid strength (weakest acid first). (See Appendix 2 for a review of acidity.) Br

I (a) CH3CH2COOH < CH3 - CHCOOH < CH3 - CCOOH

I I Br Br

The greater the number of electron-withdrawing substituents, the greater the stabilization of the carboxylate ion. (b) CH3CHCH2CH2COOH < CH3CH2CCH2COOH

I I Br Br

< CH3CH2CH2CHCOOH I Br

The closer the electron-withdrawing group, the greater the stabilization of the carboxylate ion. (c) CH3CH2COOH < CH3-CHCOOH < CH3 - CHCOOH <

I I o C�N CH3-CHCOOH

I N02

The stronger the electron-withdrawing effect of the substituent, the greater the stabilization of the carboxylate ion.

469

Page 475: Solucionario de wade

20-4

�COOH �CHO

"-�

CH20H H2S04

Y shake with ether and water

ether water

�COOH soluble in ether �CH20H

�CHO shake with NaOH (aq)

ether

{ NaOH (aq) unchanged �CH OH t

ionized form by NaOH /"'-.. /"'-.. /"'-..

2 �

COON of ca�boxylic ./ ........",- ........",- -CHO

a aCId is soluble acidify with HCl (aq) and ether in water

ether

�COOH evaporate ether t

�COOH

470

HCl (aq) NaCl

Page 476: Solucionario de wade

20-5 The principle used to separate a carboxylic acid (a stronger acid) from a phenol (a weaker acid) is to neutralize with a weak base (NaHC03), a base strong enough to ionize the stronger acid but not strong enough to ionize the weaker acid.

�-o-OH 0°

y < }- co,

shake with ether and NaHC03 (aq)

ether

A ,-------- -------�\ CH3-Q-� OH + 00 \.. - ) y

shake with NaOH (aq)

ether

00

! evaporate

Q�

471

NaHC03 (aq)

< }-cooNa

2. filter or extract ! 1. add HCl

�COOH �- pure

NaOH (aq)

CH3-Q-oNa

2. fil ter or extract ! 1. add HCl

CH3-Q-0H pure

Page 477: Solucionario de wade

20-6 The reaction mixture includes the initial reactant, reagent, desired product, and the overoxidation product-not unusual for an organic reaction mixture.

(a) �eHO < :N �COOH � er03 ,, ____________ �eH2_0- H-----------) y

shake with ether and water

ether water

�eOOH have appreciable (some compoundS)

N ether and water C

" solubility in both

o

�eH20H b.p. 137° e �eHO

b.p. 102° e (b) Pentan-l-ol cannot be removed from pentanal by acid-base extraction. These two remaining products can be separated by distillation, the alcohol having the higher boiling point because of hydrogen bonding.

20-7 The eOOH has a characteristic IR absorption: a broad peak from 3400-2400 cm-I, with a "shoulder" around 2700 cm-I. The carbonyl stretch at 1695 cm-I is a little lower than the standard 1710 cm-I, suggesting conjugation. The strong alkene absorption at 1650 cm-I also suggests it is conjugated.

472

Page 478: Solucionario de wade

20-8 (a) The ethyl pattern is obvious: a 3H triplet at b 1.15 and a 2H quartet at b 2.4. The only other peak is the COOH at b 11.9 (a 2 .1 b unit offset added to 9 .8).

(b)

.i! IH

1 :1 I I I 10 9 8

o II

CH3CH2-C-OH

o .i! II Q £ H-C-CH2CH3

I I I 7 6 5 b (ppm)

£ 3H Q

2H TMS

II II I I I I I I 4 3 2 1 0

The multiplet between <5 2 and <5 3 is drawn as a pentet as though it were split equally by the aldehyde proton and the CH2 group. These coupl ing constants are probably unequal, in which case the actual splitting pattern wil l be a complex multiplet. (c) The chemical shift of the aldehyde proton is between b 9-10, not as far downfield as the carboxylic acid proton. Also, the aldehyde proton is split into a triplet by the CH2, unl ike the COOH proton which always appears as a singlet . Final ly, the CH2 is split by an extra proton, so it will give a multiplet with complex splitting, instead of the quartet shown in the acid.

20-9 • • :OH + 1 /C ............ H2C==C OH \ H

t .. +OH II

/C ............ H2C==C OH \

H

.. ..

.. ..

. . :OH I

+ hC ............ H C-C:I" OH 2 \ H . . :OH I C�+

H2C==C/ ""'OH \

H

473

Page 479: Solucionario de wade

20-10 :OH [ 0 ] t

CH3CH2tCH'iHg-OH - CH3CH,· +

87 CH3 mass 29

+ I /C, .. H2C==C 9,H \ CH3

m/z 1l6 plus resonance forms as shown in 20-9

m/z 87 McLafferty rearrangement

20-11

H H3CI1�

� :ft' �C'OH

CH3 m/z 1l6

t

--

H3C, CH II + CH2

mass 42

H, .. 0: I ,/.C, HC/' OH

I CH3 mlz 74

t

(a) �C C� conc. KMn04 • �COOH

or 1 ) 03, 2) H20 (h) C-:::) conc. KMnO� C:COOH

t, H 0+ COOH , 3

(c)

< }-Br

(d) �

(e) Q CH3

(f) �I

OR �I

o CH2CH20H CH2COOH Mg

-­ether < }-MgBr D H30+ 6 H2Cr04 6 � --- I • I � �

Br PBr3��

conc. KMn04 •

t"H30+

Mg CO2 -- ---ether

Mg CO2 H30+ -- --- � ether COOH ¢ COOH H30+ �COOH

COOH �

• �CN H30+ t, • �COOH KCN 474

Page 480: Solucionario de wade

20-12 (a) first intennediate

:O-H i + II • • RC-�H

:O-H

.. :O-H 1 + • •

� RC-OH �

.. } :O-H 1 + RC==�H

second intennediat RC( � i·· :O-H / RC � \\

iY-H } RC \ :OR :OR +OR .. . .

(b) The mechanism of acid-catalyzed nucleophi lic acyl substitution may seem daunting, but it is simpl)' a succession of steps that are already very famil iar to you.

Typically, these mechanisms have six steps: four proton transfers (two on, two off), a nucleophi lic attack, and a leaving group leaving, with a little resonance stabi lization thrown in that makes the whole thing work. The six steps are labeled in the mechanism below: Step A proton on (resonance stabi l ization) Step B nucleophile attacks Step C proton off Step D proton on Step E leaving group leaves (resonance stabilization) Step F proton off

StepD

Step A

i + • • • • :O-H :O-H :O-H II • • 1+ • • 1 + CH)C-OH _ CH)C-OH _ CH3C=OH L . . � .. .. f

.. � 1 o / ............ H CH2CH3 Step B

O-H StepC 1 CH3C-OH /R.......... A I + H �H2CH3 l ""0 .....

H'" ....... CH CH 2 3

- H20 / / //'0 1 :O-H :O-H +O-H }

---I�� CH3C+ � CH3C � CH3C . Step E \ \\ \ :RCH2CH3 + RCH2CH3 :�CH2CH3

475

Page 481: Solucionario de wade

20-12 (c) All steps are reversible, which is the reason the Principle of Microscopic Reversibi l i ty applies. Applying the steps as outlined on the previous page: (abbreviating OCH2CH3 as OEt) Step A proton on (resonance stabi lization) Step B nucIeophile attacks H -B is the acid catalyst

:B- is the conjugate base, although in hydrolysis reactions, water usual ly removes H+

Step C proton off Step D proton on Step E leaving group leaves (resonance stabi l ization) Step F proton off

: O-H +1

cr C

: O-H 1+ O/'OEt

: O - H : O-H 1 I

\ : O-H

H I a

c�aC

�O:t � CH H C � + +

C+ C (f 'OEt

Step B

:O- H H � : O-H 1 + I) . ·O·H I • • C-O: • 2 C-O: aAEt '

H St.pC aAEt \

�11

H-O

)

Step D � +k�

: OH

same series ____ - cr� I �

Q

(

O�

H of resonance -- -forms as above

Step F II : OH,

- EtOH .. Step E

: O-H I • • C-O : Oil +1). \H · OEt � ·1 H

476

OH I

(JC=O

Page 482: Solucionario de wade

20- 13 For the sake of space in this problem, resonance forms wil l not be drawn, but remember that they are critical!

'0'

PhXJ

� H+ + H ·0 .... ).? plus 5 resonance forms Ph �H

+H�CH 3

O .... H

Ph�H 10+ ;»H .... • "CH 3

H� CH 3+ • • H c:htH 'CH 3

H+ + H . . H ... 0 .... + ,-

Ph+H o 'CH 3 + -H20

p

c+ Ph /O '" H plus 4 resonance forms

'CH 3, HERE'S THE DIFFEREN CE! HR CH 3 + CANNOT LOSE H+ TO

H • • CH MAKE CARBON YL ( �O"'" 3

Ph+H O'CH 3 H� CH 3+

..... CH 3 o

Ph�H o 'CH 3

477

'0'

PhX='

+ H�'!J + H ·0 .... ).? plus 6 resonance forms

Ph �OH +H� CH 3

O .... H

Ph�OH 10+ ;»H .... • ·'CH 3

H� CH 3 + • • H ·0 .... 0h-t-OH 'CH 3

H+ � H . . H ...0 ....

+ ,-Ph+OH

o 'CH 3 t -H20

Ph yO� plus 5 resonance forms

o .\ CAN LOSE H+ TO 'C� MAKECARBONYL

+H� CH 3 o II CH Ph�O"'" 3

Page 483: Solucionario de wade

20-14 (a)

° { :0 : :0:- } II II . . I . • R-C-OH � R-C-OH � R-C-OH • • H+ + I + + I '- H H

BAD-two adjacent positi ve charges

(b) Protonation on the OH gives only two resonance forms, one of which is bad because of adjacent positive charges. Protonation on the C=O is good because of three resonance forms distributing the positive charge over three atoms, with no additional charge separation.

i + • • • • t :0: � :O-H :O-H :O- H II H+ II . . I + • • I + RC-OH � RC-RH �RC-RH � RC=9.H

(c) The carbonyl oxygen is more "basic" because, by definition, it reacts with a proton more readily. It does so because the intermediate it produces is more stable than the intermediate from protonation of the OH. 20-15 (a)

(b)

(c)

OH JvCOOH

� ° I I

+ CH30H use CH30H as solvent

HC -OH + CH30H .. use CH30H as solvent

..

° I I

OH JvCOOCH3

� + H20 � remove water with

molecular sieves

HC-OCH3 remove by distillation b.p.32°C

�COOCH2CH3

� + H20 � remove water with molecular sieves or by distillation

478

Page 484: Solucionario de wade

20-16 The asterisk (" * ") denotes the 180 isotope. (a) and (b)

i + • • • • :O-H : O-H : O-H " • • 1+ • • I +

CH3C-OII � CH3C-OH � CH3C=OH 1 .. � .. .. f

,..----:O-H 1 CH3C-OH

1 0 * CH3

0* .. � j

H/' " CH3

O-H 1

.. CH3C-OH 0* 1+

H/" 'CH �O* � H/" 'CH3

(c) The 180 has two more neutrons, and therefore two more mass units, than 160. The instrument ideally suited to analyze compounds of different mass is the mass spectrometer. 20-17 .. + O-Et : O-Et :o-Et (a) I I II first intermediate: H-C+ ---- H-C ---- H-C

\ \\ + \ : O-Et O-Et : O-Et

.. . . + • • : O-H : O-H O-H I I II second intermediate: H -C + ---- H-C ---- H-C \ \\ + \ : O-Et O-Et : O-Et

The more resonance forms that can be drawn to represent an intermediate, the more stable the intermediate. The more stable the intermediate, the more easily it can be formed, that is, under milder conditions. These intermediates are highly stabilized due to delocalization of the positive charge over the carbon and both oxygens. A trace of acid is all that is required to initiate this process.

479

Page 485: Solucionario de wade

20-17 continued (b) +

:O :� II H+ HC-OEt .. i

:O - H II

HC-OEt

� . . IF :O - Et I HC-OH

I OH

I) - EtOH / H-)5-Et i :O- H

HC -OH .. HC+ ----

I \ OH :RH

20-18 0 I I (a) H 3C'O C ....

I OH �

20-19 (a)

o

o II

o

(b) arCH -C - Cl ?' 2 �I LiAl(O-t-BuhH

..

. . :O-H 1+ • • r

-O�t

H/R'H t

. . :O-H / HC ----

\\ +OH

o II

. . } :O- H I +

HC=OEt

r:-':O-H +O-H P-//� H

HC \ :OH . .

o " H-C

\ OH

arCH -C-H ?' 2 �I

480

B214 selectively reduces a carboxylic acid in the presence of a ketone. Alternatively, protecting the ketone as an acetal, reducing the COOH, and removing the protecting group would also be possible but longer.

Page 486: Solucionario de wade

20-20

20-21

. � :0: Li+ '\ \ R-C-R' + H+ :6: Li+ / .

o \ I

R - C - R'

HO \

R-C-R' \ H+ HR:---./

a hydrate

· ·Y+ O-H H20: I IV

.. R-C-R' ..

(a) 0 CH3CH2 -C - OH + 2 Li-{ � _ _

H_2-10 .. �

(b)

O

R C -OH

20-22 .. :0

HO o: �Jt? �Cl Y + Cl - If Ph 0

HO I

R -C - R' +1) H-�-H 1- H20

:O-H l R-�- R'f +

:0 : HO, ;'0iYCI • • C • • + I Cl

Ph 0

t .. _ t :0 : • • + • • �. • plus three other

- � H(-O�� Cl resonance forms ........ f----' .. �

'0' -{ r-Cl ...." 1 with positive charge

HOyO�CI + • • ( Ph 0 on the benzene ring

plus resonance forms '0' '0'

:o�oj( �Cl -:O�R�� �Cl I'" I( ..' ',/\0 IrJ P� Cl ph °

481

o .. )l + Ph Cl

Cl Ph 0

O=C=O +

+ Cl

Page 487: Solucionario de wade

20-23 (a) Co: I I

Ph-C - Cl + H-O-CH2CH3 � ..

(b) CO: H II I

.. :?j

:0 : II

Ph - C ",1+ H-OCH2CH3 �t� R

Ph - C -�CH2CH3

:0 :

CH -C-C1 + H-N-CH3 3 � • • CH -C-C1 3 +1 U

H-N-CH

20-24 (a) 0 d'oH

(b) 0

�OH

o �Cl

I 3

H

20-25 Please refer to solution 1-20, page 12 of this Solution Manual.

20-26 (a) 3-phenylpropanoic acid (c) 2-bromo-3-methylbutanoic acid (e) sodium 2-methylbutanoate (g) trans-2-methylcyclopentanecarboxylic acid (i) 7,7-dimethyl-4-oxooctanoic acid

20-27 (a) f3-phenylpropionic acid (c) a-bromo-f3-methylbutyric acid,

or a-bromoisovaleric acid (e) sodium f3-methylbutyrate (g) o-bromobenzoic acid (i) 4-methoxyphthalic acid

(b) 2-methylbutanoic acid (d) 2-methylbutanedioic acid (f) 3-methylbut-2-enoic acid (h) 2,4,6-trinitrobenzoic acid

(b) a-methyl butyric acid (d) a-methylsuccinic acid

(f) /3-aminobutyric acid (h) magnesium oxalate

482

Page 488: Solucionario de wade

20-28 (a) 0 (b) (c)

II CH3 - C -OH

(XCOOH (H-tO- ) Mg2+ � COOH 2

(e) 0 (f) 0 II II

ClCH2-C - OH CH3-C - Ci

0-) o 2

(h) 0 < }-C - O- Na+

(i) 0 II

FCH2-C-0- Na+

20-29 Weaker base listed first. (Weaker bases come from stronger conjugate acids.) (a) ClCH2COO- < CH3COO- < PhO­

(c) PhCOO- Na+

20-30 (a)

(b) (XCOOH 7"1 + 2 NaOH

(XCOO- Na+

�I � COOH (c)

CH3-Q-COOH

(d) CH3 - CHCOOH + I Br

(e) < }-COOH +

CH3CH2COO- Na+

Na+ -0-1 �

COO- Na+

.-

..

no reaction

CH3- CHCOO- Na+ I Br

< }-COO- Na+ + HO-1 �

20-3 1 0 o II

CH3C-OCH2CH3 < lowest b.p. (n°C) �O�OH

(b.p. 143°C) II

< CH3CH2CH2 -C - OH highest b.p. ( l62°C)

The ester cannot form hydrogen bonds and will be the lowest boiling. The alcohol can form hydrogen bonds. The carboxylic acid forms two hydrogen bonds and boils as the dimer, the highest boiling among these three compounds . 20-32 Listed in order of increasing acidity (weakest acid first): (a) ethanol < phenol < acetic acid EWG = electron-withdrawing group (b) acetic acid < chloroacetic acid < p-toluenesulfonic acid (c) benzoic acid < m-nitrobenzoic acid < o-nitrobenzoic acid } Acidity increa�es. with: (d) butyric acid < (3-bromobutyric acid < a-bromobutyric acid 1. closer proxImIty of EWG

Cl 2. great number of EWG Brn- 0= [yF 3. increasing strength (e) COOH < COOH < COOH (electronegativity) of EWG

483

Page 489: Solucionario de wade

20-33 Acetic acid derivatives are often used as a test of electronic effects of a series of substituents: they are fairly easily synthesized (or are commercially available), and pKa values are easi ly measured by titration.

Substituents on carbon-2 of acetic acid can express only an inductive effect; no resonance effect is possible because the CH2 is sp3 hybridized and no pi overlap is possible.

Two conclusions can be drawn from the given pKa values. First, all four substi tuents are electron­withdrawing because all four substituted acids are stronger than acetic acid. Second, the magnitude of the electron-withdrawing effect increases in the order: OH < Cl < CN < N02 . (It is always a safe assumption that nitro is the strongest electron-withdrawing group of all the common substituents .) 20-34 (a) Ascorbic acid is not a carboxylic acid. It is an example of a structure called an ene-diol where one of the OH groups is unusually acidic because of the adjacent carbonyl group. See part (c) . (b) Ascorbic acid, pKa 4.71, is almost identical to the acidity of acetic acid, pKa 4.74.

H OH �" 0 0 HOCH2 _ -ene

, HO OH

'( diol I

(c) The more acidic H will be the one that, when removed, gives the more stable conjugate base. start here

R-<;l(

0 _R

-}:(O}��-}:(

O� { R-}:(

O_R1�O}

:0 OH :0: OH HO OH HO :0 : HO 0 : . . _ . . t . . _ . . t THREE more only two resonance fOnTIS

!. • • �� (d) In the slightly basic pH of physiological fluid, ascorbic acid {

R »(0 0:- eresonance fOnTIS acidic

'\..J. i s present as the conjugate base, ascorbate, whose structure can :R OH be represented by any of the three resonance fOnTIS on the left

20-35 (a)Q)

r; � - CH20H

\ COOH (e)o-

( i) COOH �CHO

of part (c).

� CH2COOH (b) 0-

(f) Ph I CH3CH2 -CHCH20H

484

(C) � � o

(d) 2�COOH

(h)�

COOH

V--COOH

(g)C�H

(k) �o COOH

Page 490: Solucionario de wade

20-36 (a)�

Mg CO2 H30+ .. --- .. Br ether

(b)� conc. KMn04 2 ..

�, H30+

(c) 0 0

�H Ag+ �OH

.. NH3 (aq)

(d) 0 SOCI2 0

�OH �C I

11) LiAlH4 .. 2) H30+

OH �

�COOH OR

,./"-COOH

OR H2Cr04 .. LiAI(Ot-BuhH 0 .. �

PCC J

(e) CH30H �COOH H+ ... �COOCH3 OR CH2N2 ...

KCN Hp+ ... .�

H

SOCI2 OR .. �COOH

C H30H �COCI .. �COOCH3

(f) O'COOH LiAlH4 H30+ O'CH2OH .. ...

H 1 � 3 ... 2 I CH3 (g)

U

CH2COOH SOCI CH NH QgN' h h 0

OR

(h) CI ( (XCI H2 (XCI I + --I --(COOH � COOH PI COOH

Diels-Alder

485

B2H6 ...

Page 491: Solucionario de wade

20-37 �_}-COOH < }-OH o-CH,oH 0-,

----V shake with ether and HCl (aq)

('hCOOH, PhOH, PhCH20H) V

shake with NaHC03 (aq)

+ PhNH3 Cl-

shake with NaOH (aq) and ether

NaOH (aq) ether

shake with NaOH (aq)

ether

PhCH20H ! evaporate

< }-CH20H pure

PhCOONa

shake with ether and HCl (aq)

NaCI

ether

PhNH2 evaporate!

< }-NH2 pure

evaporate � NaCI PhCOOH •

�_ COOH pure

NaOH (aq) PhONa

NaCI PhOH

486

evaporate .. pure

Page 492: Solucionario de wade

20-38 (a)

1;(H "OH N

OH "", OH + CH3 t racemic S 0 (R + S)

(b) Isomers which are R,S and S,S are diastereomers.

20-39 (a) PhCH2CH2OH TsCI KCN

.. .. PhCH2CH2CN pyridine !PBr3

Mg

H30+ .. �

CO2 H30+ PhCH2CH2Br .. .. .. PhCH2CH2COOH ether

(b)crCH (fCH 2 HBr 3 ---t .. � Br

(d) Br 1\ c;cr� HO O� o h W� ,

(e) Li ()COO: 2 6 _ _

H_30_+ ...

Mg .. ether

o dD

487

PhCH2CH2COOH

('-r(CH3

V COOH

�COOH

yV o (l

HO OH ..

W, �

Page 493: Solucionario de wade

20-40 (a) Mass spectrum:

-mlz 152 ==> molecular ion ==> molecular weight 15 2 -m1z 107 => M - 45 => loss of COOH

K -m1z 77 => monosubstituted benzene ring, H -Q

+

H H IR spectrum: -3400-2400 cm-i , broad ==> O-H stretch of COOH -1700 cm-i ==> C=O -1240 cm-i ==> C - O -1600 cm-i ==> aromatic C=C NMR spectrum: � 6.8-7.3, two signals in the ratio of 2H to 3H ==> monosubstituted benzene ring � 4.6, 2H singlet ==> CH2, deshielded

Carbon NMR spectrum: � 170, small peak ==> carbonyl � 115-157 , four peaks ==> monosubstituted benzene ring; deshielded peak indicates oxygen substitution on the ring

(b) Fragments indicated in the spectra:

< > mlz 14

COOH mlz 45 and from IR mlz77

This appears deceptively simple. The problem is that the mass of these fragments adds to 136, not 152-we are missing 16 mass units � oxygen! Where can the oxygen be? There are only two possibilities:

< }O -CH2COOH < }CH2-O- COOH

How can we differentiate? Mass spectrometry! 8157 ( }CHjO - COOH ( YatCH2COOH

91 93 J phenoxyacetic acid

This structure is consistent with the peak at 8 157 in the carbonNMR.

The mlz 93 peak in the MS confirms the structure is phenoxyacetic acid. The CH2 is so far downfield in the NMR because it is between two electron-withdrawing groups, the 0 and the COOH.

(c) The COOH proton is missing from the proton NMR. Either it is beyond 10 and the NMR was not scanned (unlikely), or the peak was broadened beyond detection because of hydrogen bonding with DMSO.

488

Page 494: Solucionario de wade

20-41 (a) 0 6 a-valemlactone

(b) :O� ro�

\

V

OH�

. . H-O : tt='�: .. H-O :

O

+t........ O· o· 0 : 0+ 0 : --- ---

o AD:

6 20-42

Co: CH3 II � • . / C O=C

Ph/' ......... Cl \ O - H

o II Ph-C-O - C-CH

I I 3

:0:

..

.. :0 : I + Ph-C- O=C-CH � I • • I

3 Cl :O- H

:0: II Ph -C-O-C- CH ---• • I 3

:cl : � I · · � H -? : plus two other resonance forms

. . :0: I +

Ph -C-O-C-CH I • • I 3 Cl :O- H . .

(s: Ph-C-O -C-CH I • . I I

3 Ccl :O-H

+

20-43 (a)

CH3 acetal � H30+

ester (an ester in a ring is called a lactone)

--qCH3 CH2CO'l

carboxylic Compound 1 acid 489

CH3 OH alcohol

Compound 2

Page 495: Solucionario de wade

20-43 continued (b) Compound 1 has 8 carbons, and Compound 2 has 6 carbons. Two carbons have been lost: the two carbons of the acetal have been cleaved. (This is the best way to figure out reactions and mechanisms: find out which atoms of the reactant have become which atoms of the product, then determine what bonds have been broken and formed.) (c) Acetals are stable to base, so the acetal must have been cleaved when acid was added.

CH3

oAo H30+ 5 �CH3

4 CH2COOH 2 1

(d) The carbons have been numbered above to help you visualize which atoms in the reactant become which atoms in the product. The overal l process requires cleavage of the acetal to expose two alcohols. The 3° alcohol at carbon-3 can be found in the product, so it is the primary alcohol at carbon-5 that reacts with the carboxylic acid to form the lactone.

+ 0 -H resonance stabilized; H3C --'( leaves the reaction

H +

.. : OH OH

� �.+ _ +0

H ,+

• :0 •. - '----{O- H

�:.O

QO-H

CH3 two fast

H-.aC\�-H

CH3

. . :0 ·0 ·0 • CH3

OH

:O -H

� H-O:

c5-o",-,,-H ,

.. H :0

....... 11--- CH3 OR 490

..

OR H+ transfers

. . :O-H

.Q +0 • CR3"

OH

OR ! -H20

:O-H ' + .OQ· ... C •

CH

OR 1

Page 496: Solucionario de wade

20-44 (A more complete discussion of acidity and electronic effects can be found is Appendix 2 .) A few words about the two types of electronic effects: induction and resonance. Inductive effects are a result of polarized (J bonds, usually because of electronegative atom substituents. Resonance effects work through 11:

systems, requiring overlap of p orbitals to delocalize electrons.

All substituents have an inductive effect compared to hydrogen (the reference). Many groups also have a resonance effect; all that is required to have a resonance effect is that the atom or group have at least one p orbital for overlap.

The most interesting groups have both inductive and resonance effects. In such groups, how can we tell the direction of electron movement, that is, whether a group is electron-donating or electron-withdrawing? And do the resonance and inductive effects reinforce or conflict with each other? We can never "tum off' an inductive effect from a resonance effect; that is, any time a substituent is expressing its resonance effect, it is also expressing its inductive effect. We can minimize a group's inductive effect by moving it farther away; inductive effects decrease with distance. The other side of the coin is more accessible to the experimenter: we can "tum off' a resonance effect in order to isolate an inductive effect. We can do this by interrupting a conjugated 11: system by inserting an sp3-hybridized atom, or by making resonance overlap impossible for steric reasons (steric inhibition of resonance).

These three problems are examples of separating inductive effects from resonance effects.

(a) and (b) In electrophilic aromatic substitution, the phenyl substituent is an ortho,para-director because it can stabilize the intermediate from electrophilic attack at the ortho and para positions. The phenyl substituent is electron-donating by resonance.

+

BUT: H E

..

o < }-CH2-C-O-H

H E

..

is a stronger acid than

+

H E

plus other resonance forms

o II

H -CH2 -C -0 - H

The greater acidity of phenylacetic acid shows that the phenyl substituent is electron-withdrawing, thereby stabilizing the product carboxylate's negative charge. Does this contradict what was said above? Yes and no. What is different is that, since there is no p-orbital overlap between the phenyl group and the carboxyl group because of the CH2 group in between, the increased acidity must be from a pure inductive effect. This structure isolates the inductive effect (which can't be "turned off') from the resonance effect of the phenyl group.

We can conclude three things: (1) phenyl is electron-withdrawing by induction; (2) phenyl is (in this case) electron-donating by resonance; (3) for phenyl, the resonance effect is stronger than the inductive effect (since it is an ortho,para-director).

491

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20-44 continued (c) The simpler case first-induction only: o

II o II

CH30-CH2 -C-O-H is a stronger acid than H-CH2 -C-O-H

There is no resonance overlap between the methoxy group and the carboxyl group, so this is a pure inductive effect. The methoxy substituent increases the acidity, so methoxy must be electron-withdrawing by induction. This should come as no surprise as oxygen is the second most electronegative element. The anomaly comes in the decreased acidity of 4-methoxybenzoic acid: o 0

CH30 -< >- C -0 - H is a weaker acid than H -< >- C - 0 - H

Through resonance, a pair of electrons from the methoxy oxygen can be donated through the benzene ring to the carboxyl group-a stabilizing effect. However, this electron donation destabilizes the carboxylate anion as there is already a negative charge on the carboxyl group; the resonance donation intensifies the negative charge. Since the product of the equilibrium would be destabilized relative to the starting material, the proton donation would be less favorable, which we define as a weaker acid . . .

:0: CH OV C-O-H 3 • • + -

Methoxy is another example of a group which is electron-withdrawing by induction but electron-donating by resonance.

(d) This problem gives three pieces of data to interpret: o o II II

(1) CH3-CH2-C-O-H is a weaker acid than H-CH2-C-O-H

Interpretation: the methyl group is electron-donating by induction. (2) CH3

< � 0 - H is a weaker aeid than < >-0 -H

CH3

Interpretation: the methyl group is electron-donating by induction. This interpretation is consistent with (1), as expected, since methyl cannot have any resonance effect. (3) CH3

� ct is a stronger acid than \4- O-H CH3

o-� it - O-H

Interpretation: this is the anomaly. Contradictory to the data in (1) and (2), by putting on two methyl groups, the substituent seems to have become electron-withdrawing instead of electron-donating. How?

Quick! Turn the page!

492

Page 498: Solucionario de wade

20-44 continued Steric inhibition of resonance! In benzoic acid, the phenyl ring and the carboxyl group are all in the same plane, and benzene is able to donate electrons by resonance overlap through parallel p orbitals. This stabilizes the starting acid (and destabilizes the carboxylate anion) and makes the acid weaker than it would be without resonance.

:0: < � C - �-H �

. .

0:9:

.. + C-O-H .. plus other resonance forms

Putting substituents at the 2- and 6-positions prevents the carboxyl or carboxylate from coplanarity with the ring. Resonance is interrupted, and now the carboxyl group sees a phenyl substituent which cannot stahdizc the acid through resonance; the stabilization of the acid is lost. At the same time, the electron-withdrawing inductive effect of the benzene ring stabilizes the carboxylate anion. These two effects work together to make this acid unusually strong. (Apparently, the slight electron-donating inductive effect of the methyls is overpowered by the stronger electron-withdrawing inductive effect of the benzene ring. )

H ""CH3 """ \\\\\\' 0 •

• •

�---�!�o-H H CH3 • --t··-���.-----..---­. "

COOH group is perpendicular to the plane of the benzene ring­no resonance interaction.

20-45 (a)

� H o stock bottle

o this three-dimensional view down the C-C bond between the COOH and the benzene ring shows that COOH is twisted out of the benzene plane

�OH o students' samples

(b) The spectrum of the students' samples shows the carboxylic acid present. Contact with oxygen from the air oxidized the sensitive aldehyde group to the acid. (c) Storing the aldehyde in an inert atmosphere like nitrogen or argon prevents oxidation. Freshly prepared unknowns wil l avoid the problem.

20-46 H h �COOH Ph' I' H

CH2I2 A Zn, cuci H"'" IIIIICOOH Ph H

SOCl2 .. H"A",COCI

Ph H (Simmons-Smith reaction, Sec. 8-9A)

(Hofmann

A

rearrangement)

A

Br2 ' HO-H ,,,,, IIIII NH2"

H20 H "',' II'''CONH2 � H � H

493

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20-47 Products are boxed. (a) All steps are reversible in an acid-catalyzed ester hydrolysis. (abbreviating O CH2CH3 as OEt) Step A proton on (resonance stabilization) Step B nucleophile attacks H -B is the acid catalyst

:B- is the conjugate base, although in hydrolysis reactions, water usually removes H+

Step C proton off Step 0 proton on Step E leaving group leaves (resonance stabilization) Step F proton off

! :O-H 1 ..-----....

.. :O-H 1 +

vC'OEt

:O-H :O-H 1 1

ac�aC�O�t

� CH HC 0 + +

:O-HH�

\ :O-H

H 1 C + C (f 'OEt

:O-H O/��Et :OH2

Step B

1 + / \ .OH C-O:� • 2 1 • • C-O· � I \

V:REt H " � I " StepC

V : REt A

�11

A-a) StepD � +k�

:OH cr� t;) 1-EtOH 1 same series "/ + \ of resonance ....... E----;.� , I ( H Step E forms as above ....... ""-

Step F II :OA,

494

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20-47 continued (b)

+ H ..

~ ·0 (c) • \\ OH �/ HI'B.

·0/

. . . �/�H

OH

:O-H I

OC"

:0:

I

.. : O - H I

:OH

+O-H II ....., c

O�: � ·

c

00:

.. -

• • /H :0 ..

\ OH

s.� ---

0: II c

00:

: B-•

495

..

:0:

+ I HC)h]

• • /H :0 +

\ ,...-:OH �� .. :OH

A cyclic ester is called a lactone. Lactones form when the OH nucleophile is just a few carbons away from the carbonyl electrophile.

Page 501: Solucionario de wade

20 -47 continued (d) � /�, ___ HO C �H I I � :OH

o ..

.. -O·

HO�C/'" II o

Esters can be fonned only in acid, not in base. 20-48 Cyanide substitution is an SN2 reaction and requ�res a 1 o,or 2° ��rbon with � leaving �roup., The Grignard reaction is less particular about the type of hahde, but IS sensitive to, and IncompatIble with, acidic functional groups and other reactive groups. (a) Both methods wil l work.

< }-CH2Br Mg CO2 H30+

---- ---- ---­ether

OR NaCN .. < }-CH2COOH

(b) Only Grignard wil l work. The SN2 reaction does not work on unactivated benzene rings. 0-,' Mg CO2 H30+ 0-" 1 _ \ Br ____ ____ ____

1 \ COOH ether _

(c) Grignard will fail because of the OH group. The cyanide reaction wil l work, although an excess of cyanide wil l need to be added because the first equivalent wil l deprotonate the phenol. *

HO 1 \ CH2Br .. 3 .. -0-" NaCN H 0+ - tJ.

HO-o-CH2COOH

(d) Grignard will fail because of the OH group. The cyanide reaction will fail because SN2 does not work on unactivated sp2 carbons. In this case, NEITHER method will work.

Ho-o-Br

(e) Both methods will work, although cyanide substitution on 2°C will be accompanied by elimination. 0- Br Mg CO2 H30+

---- ---- ---­ether

(f) Grignard will fail because of the OH group. The cyanide reaction wil l work. Since alcohols are much less acidic than phenols, there is no problem with cyanide deprotonating the alcohol. *

H0-o-CH2Br NaCN.. H�O+ .. H0-o-CH2COOH

* The pKa of HCN is 9.1, and the pKa of phenol is 10.0. Thus cyanide is strong enough to pull off some of the H from the phenol, although the equilibrium would favor cyanide ion and phenol. The pK;l of secondary alcohols is 16-18, so there is no chance that cyanide would deprotonate an alcohol.

CN + HO-{ ) � H-CN + -o-{ ) The side with the pKa 9.1 weaker acid is favored. pKa 10.0

496

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20-49 (a) H , .......... 0,

° H

° H ........ 'H

.. -

... -

H + + ° -0

""""" 'H pKa11.6

H+ + _ ........ H ° pKa 15.7

Hydrogen peroxide is four pK uni ts (104 times) stronger acid than water, so the hydroperoxide anion, HOO -, must be stabi lized relative to hydroxide. This is from the inductive effect of the electronegative oxygen bonded to the 0- ; by induction, the negative charge is distributed over both oxygens. The oxygen in hydroxide has to support the ful l negative charge with no delocalization .

(b) °

H ) C)lO .--H

°

H C)lO .......... O ....... H )

.. -

.. -

w +1 Jl-H ) C ° ...

W + 1 Jl . . /o-. H 3 C ° . .

..

H ) C�O } pKa 4.74

-

H)C.--l�/O- } pKa 8.2

The reason that carboxylic acids are so acidic (over 10 pK units more acidic than alcohols) is because of the resonance stabi l ization of the carboxylate anion with two equivalent resonance forms in which all atoms have octets and the negative charge is on the more electronegative atom-the best of all possible resonance worlds. The peroxyacetate anion, however, cannot deloca lize the negative charge onto the carbonyl oxygen; that negative charge is stuck out on the end oxygen like a wet nose on a frigid morning. There is some delocalization of the electron density onto the carbonyl, but with all the charge separation, this second form is a minor resonance contributor. This resonance does explain, however, why peroxyacetic acid is more acidic than hydrogen peroxide. It does not come close to acetic acid, though.

(c) � ........ OyC H3

o °

H ) C)lOJ

Carboxylic acids boi l as the dimer, that is, two molecules are held tightly by hydrogen bonding. The dimer i s an 8-membered ring with two hydrogen bonds as shown with dashed lines in the diagram. This works because the carbonyl oxygen has significant negative charge, and the H-O bond is weak because it is a relatively strong acid. The b.p. is 118°C.

497

Do peroxyacids boi l as the dimer? The author does not know, but there are three reasons to suspect that they do not. First, the b.p. is lower (l05°C) instead of higher suggesting that they do not boi l as a team but rather individual ly. Second, the dimer shown is a lO-membered ring-sti l l possible but less likely than 8-membered. Third and most important, the electronic nature of the carbonyl group, as implied in the resonance forms in part (b), places less negative charge on the carbonyl oxygen, and the H is less acidic, suggesting that the hydrogren bonding is much less strong.

Page 503: Solucionario de wade

20-50 Spectrum A: C 9H 1002 � 5 elements of unsaturation 811.8, IH � COO H 87.3, 5H � monosubstituted benzene ring 83.8, IH quartet � C HC H3 8 1.6, 3 H doublet � C HC H3

Spectrum B: C4H602 � 2 elements of unsaturat ion 812.1, IH � COO H 86.2, IHsinglet � H-C==C 85.7, IH singlet � H -C==C

(' )-�H-COOH CH3

H COO H \ / C==C 8 1.9, 3 H singlet � vinyl C H3 with no H neighbors CH 3 -C =C / \ H CH3

Spectrum C: C 6H 1002 � 2 elements of unsaturation 812.0, IH � COOH 87.0, IHmultiplet � H-C =C -COOH

H I 85.7, IH doublet � C =C -COOH

8 2.2-0.8 � C H2C H2C H3

498

must be trans due to large coupling constant in doublet at 8 5.7

Page 504: Solucionario de wade

CHAPTER 21-CARBOXYLIC ACID DERIVATIVES

21-1 IUPAC name first; then common name (a) isobutyl benzoate (both IUPAC and common) (b) phenyl methanoate; phenyl fonnate (c) methyl 2-phenylpropanoate; methyl a-phenyl propionate (d) N-phenyl-3-methylbutanamide; l3-methylbutyranilide (e) N-benzylethanamide; N-benzylacetamide (f) 3-hydroxybutanenitrile; j3-hydroxybutyronitrile (g) 3-methylbutanoyl bromide; isovaleryl bromide (h) dichloroethanoyl chloride; dichloroacetyl chloride (i) 2-methylpropanoic methanoic anhydride; isobutyric formic anhydride (j) cycIopentyl cycIobutanecarboxylate (both IUP AC and common) (k) 5-hydroxyhexanoic acid lactone; 8-caprolactone (I) N-cycIopentylbenzamide (both IUPAC and common) (m) propanedioic anhydride; malonic anhydride (n) I-hydroxycyclopentanecarbonitrile; cycIopentanone cyanohydrin (0) cis-4-cyanocyclohexanecarboxylic acid; no common name (p) 3-bromobenzoyl chloride; m-bromobenzoyl chloride (q) N-methyl-5-aminoheptanoic acid lactam; no common name (r) N-ethanoylpiperidine; N-acetylpiperidine 21-2 An aldehyde has a C-H absorption (usually 2 peaks) at 2700-2800 cm-I. A carboxylic acid has a strong, broad absorption between 2400-3400 cm-I. The spectrum of methyl benzoate has no peaks in this region. 21-3 The C-O single bond stretch in ethyl octanoate appears at 1170 cm-I, while methyl benzoate shows this absorption at 1120 and 1280 cm-I. 21-4 (a) acid chloride: single C=O peak at 1800 cm-I; no other carbonyl comes so high (b) primary amide: C=O at 1650 cm-I and two N-H peaks between 3200-3400 cm-I (c) anhydride: two C=O absorptions at 1750 and 1820 cm-I

21-5

(a) The formula C3HSNO has two elements of unsaturation. The IR spectrum shows two peaks between 3200-3400 cm-I, an NH2 group. The strong peak at 1670 cm-I is a C=O, and the peak at 1 610 cm-I is a

C=C. This accounts for all of the atoms. The HNMR corroborates the assignment. The IH multiplet at 0 5.8 is the vinyl H next to the carbonyl. The 2H multiplet at 0 6.3 is the vinyl hydrogen pair on carbon-3. The 2H singlet at 0 4.8 is the amide hydrogens. The CNMR confinns the structure: two vinyl carbons and a carbonyl. (b) The formula CSHS02 has two elements of unsaturation. The IR spectrum shows no significant OH, so this compound is neither an alcohol nor a carboxylic acid. The strong peak at 1 730 cm-I is likely an ester carbonyl. The C-O appears between 1 050-1250 cm-I. The IR shows no C=C absorption, so the other element of unsaturation is likely a ring. The carbon NMR spectrum shows the carbonyl carbon at 0 171,

the C-O carbon at 0 69, and three more carbons in the aliphatic region, but no carbons in the vinyl region between 0 100-150, so there can be no C=c. The proton NMR shows multiplets of 2H at 0 4.3 and 2.5, most likely CH2 groups next to oxygen and carbonyl respectively.

The only structure with an ester, four CH2 groups, and a ring, is o-valerolactone:

499

Page 505: Solucionario de wade

(b)

:0 : :0 : II � II) Ph -C-O. .......C,--I Cl CH3 H

. . :0: I ....... CH3

+ C :0/ '0 II Ph-C-O: ---­

I H

:0 : II -- PhCH -O-C-CH 2 +I} 3

. . :0 : I ....... CH3 . ...... C .... :0'" 'Cl

H�A Cl :0 : II PhCH -O-C-CH 2 • • 3

. . ·0 · ·1�CH3 . ...... C ....

+1 Ph-C-O : ---­I

:0'" eCI I + Ph-C=O :

I H H plus three resonance forms with positive charge delocalized on the benzene ring

the carbonyl oxygen is more nucleophilic than the single-bonded oxygen because the product is resonance stabilized

. . (c) :0: :0 : 0 ·0 · II . . �II) I I . . ·I� Ph-C-O : ........ C, ---Ph-C+0-C-CH3 I Cl CH3 I 1"\ H H Cl �

nucleophilic attack by this oxygen does not generate a resonance-stabilized intermediate

500

:0 : I' ....... CH3 . ...... C :0'" I + .. Ph-C=O :

cl-�(A plus all the resonance forms as above

o :0 : II II --- Ph -C-0 -C -CH + I}

3 H�� CC\

o : 0 : II II Ph-C-O-C-CH • • 3

Page 506: Solucionario de wade

2 1 -6 continued 0 (d) :0 : 0

Ph_�J oA H

---

---

The leaving group is ethoxide ion, CH3CH20-, a very strong base. Ethoxide would never be a leaving group in an SN2 reaction as it is too strong a base .

---

:0 : 0 I I +

.... Ph _ 0 0 I I � N :O� / � 0 0 I / p us H H� resonance

: 0 :

�N .... Ph

+

I H

form

:0 : leaving group

� �--... + .... Ph - o .� N � : 0

/ '" o . H H�

:0 :

�N .... Ph I H

+ HO� 21-7 Figure 2 1 ·9 is critical! Reactions which go from a more reactive functional group to a less reactive functional group (" downhil l reactions") will occur readily. (a) amide to acid chloride wi l l NOT occur-it is an "uphil l " transformation (b) acid chloride to amide wil l occur rapidly (c) amide to ester wil l NOT occur-another "uphil l " transformation (d) acid chloride to anhydride will occur rapidly (e) anhydride to amide wil l occur rapidly 21-8 o o

I I + HOCH2CH3 ---- CH3CH2 - C - OCH2CH3 + HCl

o + HO-{ > ---- �O �

+ HCl � (c) 0

< �C - Cl + HOCH2 --< > ----

(d) 0 -<J o-C - C l + HO ----

501

o < �C - OCH2� >

+ HCl

+ HCl

Page 507: Solucionario de wade

21-8 continued (e) 0

II CH3 -C-Cl + CH3 I HO-C - CH I

3 CH3

---

CH3 O=< �H3

O-C-CH I

3 CH3

+ HCl

This reaction would have to be kept cold to aVOid elimination . Esters of t ­butyl alcohol are hard to make.

(f) o o Cl�Cl o

+ 2 �OH --- �0�0� o

21-9 o o II II (a) H3C - C - Cl + HN(CH3h --- H3C -C - N(CH3h + HCl

(b) 0 II H3C - C -Cl

o + H2N-{ .> --- H3C - C - NH-{ .> + HCl

o (e) OC -CI ��

+ NH3 --- �C-NH2 + HCl

(d) 0 < }-C -Cl + HNJ

o --- < }-C - NJ + HCl

21-10 II I I II

+ 2 HCl

o II

(a) (i) 0 0 -0 0 -Q H3C - C - 0 - C -CH3 + HOCH2 Ij_� � H3C -C - OCH2 � II + HO -C-CH3

(ii) 0 0 II II r-- � r-- o

II H C-C - 0 -C -CH3 + HN 3 '--

� H3C - C -N,-- + HO - C - CH3

(b) (i) 0.0 : 0 I I I I • • H3C-C-0-CCH3 + HOCH2Ph � .. �

. . :0 : 0 I'.;) • • I I H3C-C ( �-CCH3 + I """

• • H-O-CH Ph • • 2

+ :0 : 0 :0 : 0 I I I I I I - . . I I H3C-C-0-CH2Ph + H -0-CCH3 " H3C-C + :0-CCH3 . . . . I · .

H.0o�CH,Ph / � 502

Page 508: Solucionario de wade

21-10 (b) continued (ii)

c�: R . . :0 : 0 I� . . I I H3C-C-0-CCH3 + HNEt2 --� .. H C-C-O-CCH 3 + I l,..� . 3

:0 : \ I H3C-C-NEt2

H- NEt2 t

o :0 : 0 I I \ I . . \ I

+ H-0-CCH3 .. H3C-C + :0-CCH3 . . I · .

21-11 � + O

· :0 : I I . . CH3C-0-CH2Ph + H-,NCH3 �.

--

:0 : . . \ I CH3C- NHCH3 + H -R -CH2Ph ...... f---

T.S.+ �

8-:0 : I i 8-CH C - - - - O-CH Ph 3 • • 2

I H-NHCH3 +

503

:OJ I • • CH37l,.."q -CH2Ph H-NHCH3

+

� T.S.+

:0 : \ I • • C�H37 + :R-t CH2Ph

H-NHCH3 + I leaving group

Page 509: Solucionario de wade

21-12

I I OCH2CH3 -·O- CCH3 + • • . . � .

:f)

)t��H'CH3 o I I

+ HO - CCH3 �O:

.• � -. . . nCH,CH3 �O

- ·ti Co

0

�'H� o I I � ;--.....

CH3C - 0 - H _ \ . . HOCH2CH3 + o • • I I

:R- CCH3 • :P.CH2CH3 +

21-13 .....--........ i + • • :0 : � :O - H :O - H

I I H+ I I . . I . . Ph - C - OEt - Ph - C - OEt --- Ph - C - OEt ___ . . +) . .

OH H n I "" I + H+ I • .

Ph - C - O - Et --- Ph - C - OEt I I O - C� O - C�

CH3P.H �

OH I .. Ph - C - OEt

CH30H "" I + ·'-..H - �- CH3

. . }

:O - H I • •

Ph - C = �Et

� - EtOH

. . :O - H :O - H

I + O- H�

Ph _ gV

I CH30H o I

Ph - C + ---I

:R- CH3 Ph - C ---

I I + R- CH3

504

. .

:0 - CH3 . . •

I I Ph - C - OCH3

Page 510: Solucionario de wade

2 1 -14

H 1 �.,pH � • • OH

;9.� ; rR� o OCH3 -Ii' 0 OCH3

\

505

Page 511: Solucionario de wade

2 1 - 1 5

i + .. H • • H • • H } • • 0 ·0 0 ·0" 0 ·0" o .0 ., . . . I I I I HC6s03H I I • • e l l • • c+ I I + ., c /'0-""- � /'H' '- -<->- /'H' �H' '-

OH I • •

CCO-C-O CH � • • I • • , + ... 3 I CH3 � �

� COOH :R,

t

OH • • I • •

H

CCO-C - O+ CH � · · 1 \...0( " '" 3 I CH3 � � COOH :R, H

OH I • • CCO-C-O CH � · · I · ·' ''' 3 I CH3 �

� COOH :<], H

• • H 0 -

CC�OOH

�H OH / +�t c - o HSO; cc · · I 'C .. CH3 ... I CH3 I I H SO

� ·0 · 2 4 COOH • •

� two rapid proton transfers

H • • "' 0 : H • • - 0 : • • I • • I o I I ----. C + H3C " .... OH CC

O-C + CCO=C � • • I � + I I CH3 � I CH3 � COOH � COOH

:0 : I I • • C 0 - ,

CC

• • CH3 aspirin I � COOH

506

-HS04

....

t H�· � O +

I I • • C

CCR-

I I CH3 � COOH

Page 512: Solucionario de wade

21-16 The asterisk (*) wil l denote \ 80.

(a) :0101 : :0.) I . .

H C - C - O*R � H C-C - O*R 3 �-. . 3 I V . .

: O • • H :R - H

The alcohol product contains the \ 80 label, with none in the carboxylate. The bond between \ 80 and the tetrahedral carbon with (R) configuration did not break, so the configuration is retained. (b) The products are identical regardless of mechanism.

:0: "' : O - H : O- H : O-H � i + • • • • t II H+ I I . • I • • I • • H3C - C - O*R --- CH3 C -O*R �CH3 C - O*R � CH3 C = O*R . . ( + . . +

OH H I ,, 1+

CH C - O*R 3 • • I O - H

� H20 : � f 9H , OH

H+ I . . I -- CH3 C - R*R .... CH3 C - O*R

1 H20 : ,, 1 + O- H '-..."... H- � - H

�I - HO*R I . . �

. . . . : O - H : O - H + O- H :0:

I I H3C - C + ........ f----II.� H3C -C

I I I I IV H20:

� H3C - C ---I

I I H3C-C- OH

: O- H + O - H : O-H acetic acid

(c) The \ 80 has 2 more neutrons in its nucleus than \ 60. Mass spectra of these products would show the molecular ion of acetic acid at its standard value of mlz 60, whereas the molecular ion of 2-butanol would appear at mlz 76 instead of mlz 74, proving that the heavy isotope of oxygen went with the alcohol Th i S demonstrates that the bond between oxygen and the carbonyl carbon is broken, not the bond between the oxygen and the alkyl carbon.

To show if the alcohol was chiral or racemic, measuring its optical activity in a polarimeter and comparing to known values would prove its configuration . (The heavy oxygen i sotope has a negligible effect on optical rotation . )

507

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21 -17 (a) A catalyst is defined as a chemical species that speeds a reaction but is not consumed in the reaction. In the acidic hydrolysis, acid is used in the first and fourth steps of the mechanism but is regenerated in the third and last steps . Acid is not consumed; the final concentration of acid is the same as the initial concentration . In the basic hydrolysis, however, the hydroxide that initial ly attacks the carbonyl is never regenerated. An alkoxide leaves from the carbonyl, but it quickly neutralizes the carboxylic acid. For every molecule of ester, one molecule of hydroxide is consumed; the base promotes the reaction but does not catalyze the reaction. (b) Basic hydrolysis is not reversible. Once an ester molecule is hydrolyzed in base, the carboxylate cannot form an ester. Acid catalysis , however, is an equi librium: the mixture wil l always contain some ester, and the yield will never be as high as in basic hydrolysis . Second, long chain fatty acids are not soluble in water unti l they are ionized; they are soluble only as their sodium salts (soap). Basic hydrolysis is preferred for higher yield and greater solubi lity of the product. 21-18

. . . .

:0 : :0 : . .

:0 :

0 : 0 :

°G:O

-

. .

O :OH

. .

.. O · O - H :ttr C·� • ---l�� �:� ----t�� C·· C . .

. .

�' H �' H 21-19

:O - H i + I I • • R - C-NH2

i + :O - H I I • • R -C-RH

21-20 (a) :��

H3C - C�NMe2 _ • • �

:OH

. .

.. �

. .

:O - H I

R -C-NH + 2

. .

:O - H I R - C-OH

+

. .

508

+

J 0 -1 H3C- b==0 ----

Page 514: Solucionario de wade

2 1 -20 continued

i + (b) :0 : � :O - H I I H+ I I • •

H3C - C - NMez -- MeC - NMez

. . :O - H I

----- MeC - NMez

(+

. . r :O - H ----- Meb = �ez

. . : OH H I '" I

MeC - NMez I + O - H

HzO : �

� :OH H+ I . . I

� MeC -- NMez .. MeC - NMez I HzO : '" I + O - H H - O - H � . .

! - IJN(CH,h

. . :O - H I

H3C - C + ----­I

:O - H

. . :O - H I

H3C - C -----I I

+ O - H

+ O- H�l I IV • • H3C - C

HN(CH ) I

3 Z .. :O - H

:0 : I I

H3C - C - OH +

+ HzN(CH3h

2 1 -2 1 In the basic hydrolysis (2 1 -20(a» , the step that drives the reaction to completion is the final step, the deprotonation of the carboxylic acid by the amide anion . In the acidic hydrolysis (2 1 -20(b» , protonation of the amine by acid is exothermic and it prevents the reverse reaction by tying up the pair of electrons on the nitrogen so that the amine is no longer nucleophilic.

n � " 2 1 -22 Ph - C N : - Ph - C = N:- H - OH

I· .. Ph - C = N - H

I Y'\ :O - Hj \ :OH :O - H � . . . .

:0 : I I

Ph - C - N - H ) h :

OH l . .

. . (?: Ph - C - NHz

I U :O - H

! :OH

HO':l H � .. i . . -

Ph- C - N - H � I I • • :0 :

- Ph - T= N - Hr :0 : . . -

:0 : I I - . . I I • • - I • • i :0 : :0:- r

----l .. � Ph - C + :NHz .. Ph- C - O : --- Ph- C = O I Y'\ J • • • •

:O - H � + NH3

509

Page 515: Solucionario de wade

21-23

Ph - C= N- H Hi):� H��+- H

H + I I II Ph - C- N- H ...... E---l .. � Ph - C- N - H ----Ph - C- NH H � 0

I I . . . . 2 I :O - H Note: species with positive charge on carbon adjacent to benzene also have resonance forms (not shown) with the positive charge distributed over the ring. 21-24

. . Jr\ H20 : :<./- H

U

(a) Reduction occurs when a new C-H bond is formed. In ester reduction , a new C-H bond IS formed in the first step and in the third step. This can also be seen in this mechanism where the steps are simi larly l abeled. (b) Co:

Step 4-workup; � alkoxide protonation OH

H H 2 1-25 H

I (a) «(5CH2CH' (C

O �NH 2

Step 1-reduction

Step 3-reduction

(d) (' N · H O� 5 1 0

H (e) Q (t) ex>

. N , . H3C CH2CH2CH3 CH2NH2

Page 516: Solucionario de wade

21-26

Ph MgI Ph "MgI + \ I C = N . •

\ .... / - MgI C = N+ .. / • "--.".. H+ H3C

/ \ H3C H """----""'H Ph H

H20: 1 "\ I I :O� C- N :

I 1 + I \ � H CH3 H

Ph H I / Ph H Ph

I { • • I (':1 + - NH3 :O- C-N : .. :O - C- N- H • :O - C + I I \ \--... H+ I H CH3 H H

21-27

Q:

I I I I CH3 H H CH3

Note : species with positive Charge! on carbon adjacent to benzene also have resonance forms (not shown) with- the posit ive charge distributed over the ring,

• • + :?j MgBr CH3CH2 - C - Cl + Ph - MgBr ----I .. �

� CH CH - C- Cl 3 2 I V

Ph � MgBrCl

..

+ NH4

OH � • • - + G JI' - :0 : MgBr :0 :

H+ I I I

I CH3CH2 - C - Ph I .. CH3CH2 - C- Ph .. CH3CH2 - C- Ph workup I Ph -:::. MgBr - -./ Ph Ph �

511

Ph I + O= C ( I I H CH'1

H) �

:0 : I I + Ph '- C -- CH3

Page 517: Solucionario de wade

21-28 (a) 0

Ph)lOR + 2 � MgBr H30+ -- ..

these alcohols can also be synthesized from ketones: OR 0

� Ph OR 0

+ �MgBr H30+ -- ..

� H30+ + PhMgBr --- ..

(b) 0 HAOR

+ 2 �MgBr __

H30+ ..

(c) �MgBr __ H30+ CH3C N + ..

OR N ::-.... �C + CH3MgBr -- H30+ � ..

21-29

OH � Ph

OH � Ph OH � Ph OH � H

0 � 0 �

:0 : i

:0 : I I I I Et - C �AlCI3 --- AlCI4- + Et - C ' .. .. +

·0+ L Et� l� r «f0CHJ

I (0 AICI4-

o

o I I C- Et

------

:OCH3 . .

H o II C- Et

------

+ OCH3 . .

-o-� I I CH30 _ CCH2CH3 + HCl + AlCl3

512

o

H II

6-� C :OCH3 . . :OCH3 . .

Page 518: Solucionario de wade

21-30 (a)

0 Ij-� o 0 0-\\ I I AlCl� < '}-cl l -1 � ' _ \ C - Cl -o o 21-31

I I AlCl3 + �Cl

r Zn(Hg)

HCl

(a) (i) 0 0 0 HC -0 -CCH3 + H2N -\ .J � HC -NH -\ .J

o (b) (i)

C�: ft H - C-0-CCH3 + H2N Ph � . .

( i i )

00. 0 · 1 1 · I I

H-C-0-CCH3 + HOCH2Ph �

� ..

. . :0: 0 I::,) . . I I

I I + HO -CCH3 o

II + HO - CCH3

:0: 0 I I . . I I

H-C-O-CCH �

+ I l,� . 3 H - C + :O -CCH I . . 3

H- NHPh

. .

H�

� :0: 0 I I . . I I

H-C- NHPh + H -R-CCH�

·0· 0 :0: 0 ·1::,) • • I I I I - • • I I H-C ( 9-CCH3 � H - C + :0- CCH3

+ 1 ",,· · ,, 1 +· ·

H-O-CH Ph l{0Q�CH?Ph ) • • 2 � �

513

Page 519: Solucionario de wade

21-32

(a) �OH

o I I

+ o 0 I I I I HC-O-CCH3

o I I

�O",C ' H

HC -Cl does not exist, so acetic fonnic anhydride is the most practical way to fonnylate the alcohol . (b) �OH +

o 0 I I I I

CH3C -O -CCH3

o I I

�O",C'CH3

Acetic anhydride is more convenient and less expensive than acetyl chloride. (c) 0=$ o

(VlNH2 VyOH

o

+ NH3

o The acid chloride would tend to react at both carbonyls instead of just one; only the anhydride wi l l give this product.

(d) c$ o

o

�OCH3 �OH

o The acid chloride would tend to react at both carbonyls instead of just one; only the anhydride will give this product.

514

Page 520: Solucionario de wade

H H I I

: 0 : 0

+/ ------

SIS

.......... J� • •

: 0 + 0 . .

.. HO

• • , + II��� HO� ..

HO

o

.. .. HO +H • •

HO

r

Page 521: Solucionario de wade

21-34

(a) Q:oH I +

h COOH

a 0 II II CH3C -O-CCH3

General ly, acetic anhydride is the optimum reagent for the preparation of acetate esters. Acetyl chloride would also react with the carboxylic acid to form a mixed anhydride.

(b)Q:0H I + CH30H

h COOH

Q:0H I +

h COOCH3 Fischer esterification works well to prepare simple carboxylic esters. The diazomethane method would also react with the phenol, making the phenyl ether. (c) rh CH2N2 rh

�COOH ..

�COOCH3 Fischer esterification would make the ester, but in the process, the acidic conditions would risk migrating the double bond into conjugation with the carbonyl group. The diazomethane reaction is run under neutral conditions where double bond migration will not occur.

21-35 Syntheses may have more than one correct approach .

(a)

(b)

(c)

(d)

(e)

(f)

0 I I ether Ph-C-OCH3 + 2 PhMgBr ..

0 II H-C-OCH2CH3 + 2 PhCH2MgBr

o II

H30+ ---

ether ..

OH I

Ph -C-Ph I Ph

H30+ OH I --- H-C-CH Ph

I 2

CH2Ph o II

Ph -C-OCH3 + H2NCH2CH3 � Ph -C- NHCH2CH3

0 H30+ OH II ether I

H -C-OCH2CH3 + 2 PhMgBr .. --- H-C- Ph I Ph

0 H30+ II ether Ph -C-OCH3 + LiAIH4 .. --- PhCH20H

a H30+ a II II Ph-C-OCH3 .. Ph -C-OH + CH30H

516

Page 522: Solucionario de wade

21-35 continued 1) TsCl, pyridine (g) PhCH20H • PhCHzC:: N 2) KCN from (e)

(h) a OH I II ether PhCHz-C-OCH(CH3h + 2 CH3CHzMgBr • PhCH2-C-CHzCH3 I from (g)

(i) How to make an 8-carbon diol from an ester that has no more than 8 carbons? Make the ester a lactone!

a 01

0 3 8

4 7 5 6

+ LiAIH4

21-36 There may be additional correct approaches to these problems.

(b) H I H a a C=O CH3 I II II I I

6 HC-0-CCH3 6 LiAIH4 �6 • •

21-37 Prepare the amide, then dehydrate.

CHzCH3

H O+ vlOH ether. � 3 8 OH

4 7 5 6

(a) a SOClz � NH3 � POCI3 �OH • �CI • �NHz .�C::N

(b) a II PhC-OH

a

21-38

(a)

o a SOCl2 II NH3 II POCI3 ----l.� PhC-CI • PhC-NHz ..

o a

517

PhCH2 -C::N

Page 523: Solucionario de wade

2l-38 continued (b) TsCI NaCN

----l.� PhCH2CH 2CN pyridine

Fe •

HCI

2l-39

(a) �OH

Cl

NaN02 CuCN -� .. � .. HCl

CN

¢ Cl

(c) �OH �Br

�C ...

CH3 II o

21-40 0 . . . . . .-CH3-N=C=O ..

iCH3-N=C-O : �•

Ar-OH 1 + • •

• • H-O- Ar ..

! Mg

ether

.. - .. } ---- CH3-N-C=O

• • I • •

((\J,'-Af

Ar= + I) 1 Ar -OH two rapid

Ar 0

9 .H • • proton transfers

518

Sevin (carbaryl)

Page 524: Solucionario de wade

21-41 (a)(X0 >=0 o

(i) carbonate ester (iii) not aromatic

(b) ao (i) thiolactone (iii) not aromatic

(c) (l Sy S o (i) thiocarbonate ester (iii) not aromatic

("y0H

+ CO2 �OH

(l SH SH

(i) a substituted urea enediamine (iii) AROMATIC-more easily seen in the resonance form shown

imine

(The enediamine product would not be stable in aqueous acid. It would probably tautomerize to an imine, hydrolyze to ammonia and 2-aminoethanal, then polymerize.)

(e) At first glance, this AROMATIC compound does not appear to be an acid derivative. Like any enol, however, its tautomer must be considered.

0-N OH I H

... U H30+ [['COOH C'COOH H O + �

.. ---3

.. H COOH N 0 NH2 N 0 I enamine I H H

lactam imine + NH3

o 0)( NH -.. f----'l ... �

\=..! HO NH2 --- polymer \=..!

(i) a carbamate Dr urethane see part (d)

(iii) AROMATIC-more easily seen in the resonance form

519

Page 525: Solucionario de wade

21-42 (a) Carbamoyl phosphate is a mixed anhydride between carbamic acid and phosphoric acid. It would react easily with an amine to form an amide bond (technically, a urea), with phosphate as the leaving group. (b) N-Carbamoylaspartate has a carbonyl with two nitrogens on either side; this group is a urea derivative. (c) The NH2 group on one end replaces the OH of a carboxylic acid on the other end; this reaction is a nucleophilic acyl substitution. (d) Orotate is aromatic as can be seen readily in the tautomer. It is called a "pyrimidine base" hecause of its structural similarity to the pyrimidine ring.

o OH HoN\ N�

��h �HJl O�N COO- HO�N COO-I H

21-43 Please refer to solution 1-20, page 12 of this Solutions Manual .

21-44 (g) benzonitrile

N� � _ _ � pyrimidine N

(a) 3-methylpentanoyl chloride (b) benzoic formic anhydride (h) 4-phenylbutane nitrile; y-phenylbutyronitrile (c) acetanilide; N-phenylethanamide (d) N-methylbenzamide (e) phenyl acetate; phenyl ethanoate (f) methyl benzoate

21-45 (a) 0

I I (b) 0 0

I I I I

(i) dimethyl isophthalate, or dimethyl benzene-l,3-dicarboxylale (j) N,N-diethyl-3-methylbenzamide (k) 4-hydroxypentanoic acid lactone; y-valerolactone (I) 3-aminopentanoic acid lactam; �-valerolactam

(c) a _ PhC-OCH2CH3 PhC-O-CCH3

II -./\ PhC-��

(e) OH I Ph -C- Ph

I Ph

H (f) a

II PhC-H

2 1-46 When a carboxylic acid is treated with a basic reagent, the base removes the acidic proton rather than attacking at the carbonyl (proton transfers are much faster than formation or cleavage of other types of bonds). Once the carboxylate anion is formed, the carbonyl is no longer susceptible to nucleophilic attack: nucleophiles do not attack sites of negative charge. By contrast, in acidic conditions, the protonated carbonyl has a positive charge and is activated to nucleophilic attack.

basic conditions o a I I II

R-C-OH + -OR' --- R-C-O- + HOR'

acidic conditions o I I

R -C - OH + H +

anion-not susceptible to nucleophilic attack

OH I R -C-OH + rapidly attacked by R'OH nucleophile

520

Page 526: Solucionario de wade

21-47 o o (a)o- II r;

_� O-C-CH3

o (h) ( )-O-C-H (C)c(NH-{ > I OH h

anhydrides react only once

o

(d) C ... OH

II o

OH

I II

o CH3 I

(f) O .... C=O

H3CO

(e) if'� ... c ... CH3

h 0

��rCH) amines are more nucleophilic than alcohols

21-48 (a) o-

r;_

� NH2 + o 0 II II

H ... C ... O ....

C ... CH 3

o

---� (_)-NH-C - H

o II + C

HO .... "'CH 3 (b) 0 o o 0 II II II II (}C

"'OH SOCl2 I � h

(}C'CI Na + -OOCCH3 • (}"-':: C ... O ....

C ... CH3

h

(c) H (toH +

� OH H

(d) 0

ClxO

CI 0

0

a�°Y'° :: oAo H

o �OH w.d;o h OH /:!,. h

-H2O

HOCH(CH3h �OCH(CH3h - �OH

(e)

0 0

0 0 C Ag> C OH

NH3 (aq) �

OH

o

o OH 0

H+ 6 H+ ) NaBH4 �H /:!,. � ...

/:!,. �COOH�H30H �COOH \... - H20 ) -H20

Y (f)� (g) 0 1\ 0 (f) 0 A HolloH. X I) LiAlH, • A Ac,o

• A Y H+,-H20 X 2) H30+ Y Y MeOOC COOMe MeOOC COOMe HOH2C CH20H AcOH2C CH20Ac any ester where ethylene glycol displaces aqueous acid workup methanol will be reduced with LiAIH4 removes ketal

521 protecting group

Page 527: Solucionario de wade

21-48 continued OH OH 0

(h) 0 MCPBA d KCN 6·,,·CN KOH 6 ,,"COOH 00] Q-COOH .. ... ... ...

H2O H2O,/). H2SO4

OH 0 f\. dH,o. OR Q- Bf f\. 0-0 Bf, 6 ... Bf CtD] HO OH Br 1. Mg .. ... ..

H20 H2SO4 H+, - H2O 2. CO2

21-49 (a) :0 :0:-I I • • � I� Ph-C�R\ - Ph-:fJ< :0: HCl + 0 �

II � Ph -C -- "-':: 0 "1 0 :ci: I H-O ..

. . -:0.) I ..

Ph-C-OEt --.. � I v··

:O-H

:O :� :O-H (c) i +

II H+ II .. Ph-C-OEt - PhC-REt

:O-H 1

------ PhC-OEt (+ .. H20 : !

OH H � OH

\=�. h

I "" 1 + H+ I . • 1 PhC-O-Et -- PhC-OEt .. PhC-OEt 1 1 H20: "" 1+ O-H O-H H-O-H � - EtOH . . :O-H I Ph -C + ------

I :O-H

:O-H 1 Ph-C I I

+ O-H

Note: species with positive charge on carbon adjacent to benzene also have resonance forms (not shown) with the positive charge distributed over the ring.

'-.......".. ..

+O-H� II� H20 : Ph-C ... 1

:O-H

522

Page 528: Solucionario de wade

(g) 0 H • • -

:0: 0 I :0: 0 II II ' 0 ' I::> • • II H C-C-O -CCH + X;' . H -- H C-C-O-CCH 3 � '-' '--''--'H3 T / " " ,

3 I l.,,�' 3

� + H t H-°Xv'

R configuration

No bond to the chiral center is broken, so the configuration is retained.

21-50

(a) 0 11 -0

Ph-C-O (b) 0

Oil C-NHCH3

(g) /j C'OCH3 OH

523

. . ... ......

(c) 0 II I"'J Ph-C- N0

(h) H>::: �OH

:0:

(d) 0 H-o II I

CC-N

COOH

(i) 0 0- 11 C-CH3

Page 529: Solucionario de wade

21-50 continued

(j) o II C'O- Na+ NHz

(k) CH3 I

PhCHzCHCHzNHz

(I) (m) (!�OH

OH

21-51 Products after adding dilute acid in the workup: (a) 0

II HC-OH + HO-Ph

(b) 0

�OH + HOCHzCH3

(c) �COOH (d) (H

+ HO

XO

�OH 21-52

(a) CHz-OH I CH-OH

I CHz-OH

glycerol o II

(b) CHz -O-C-(CHz),zCH3 I � CH - 0 -C - (CHZ)I2CH3

I � CHz -0 -C -(CHZ)I2CH3

21-53 OH (a) 6 CO,HCI

.. AICl3/CuCI

� Gatterman-Koch

OH OH

I) LiAIH4 ..

2) H20

OH 0

&" "-': C,

I H

h + para

OH HO 0 0 II

CH2 - 0 -C -(CH2)12CH3 I � CH-O-C-(CHz)12CH3

I � CH2 -0 -C -(CHz)12CH3

trimyristin

CHz-OH I CH-OH

I CHz-OH

glycerol

+ 3 CH3(CHzhzCHzOH

tetradecan-1-ol

OH 0

0 II

&" H2Cr04

I

: C'OH ...

AczO

CH3-C&\i "-':

C, .. I

0

h H

OH OH (b)6 ¢ ¢ 1 equivalent ¢ ,

"-': HN03 Fe AczO NHz is more ... .. .. nucleophilic than OH h HZS04 HCl

NOz NHz CH -C-NH + ortho

3 II

524 0

Page 530: Solucionario de wade

21-54

(a) I

�OH

o 0 II II

HC-O-CCH3

H30+ + 2 CH3CH2MgBr -- ..

OH

� (e) See the solution to 21.36(b) for one method. Here is another: reductive alkylation.

o +

HJlH

(f) 0-NH2

(g) 0-Br

(h) CH3O

XO

CH30 0

Mg ether

+

... o-MgBr

two equivalents

CNH2 ..

NH2

0 II

+ H -C-OEt

H I

(XO

N 0 I

H

o

H,o + 0-9H-o -- -- CH

(yCONH2

t POCI3 (yC::N

II

+ CH30H � + ... �C"'OCH3 + HoD

large '--1 excess

525

Page 531: Solucionario de wade

21-55 Diethyl carbonate has two leaving groups on the carbonyl. It can undergo two nucleophilic acyl substitutions, followed by one nucleophilic addition.

(a) 0 II

EtO-C-OEt PhMgBr [ <i? 1 PhMgBr

----�.� Ph -C-OEt ----�.� subst. 1 subst. 2

Mg ---'-I .. � CH3CH2MgBr ether o

I I 3 CH3CH2MgBr + EtO-C-OEt

o PhM B H30+ II g r Ph - C- Ph • ..

addition

OH I

CH3CH2 -C -CH2CH3 I CH2CH3

OH I

Ph-C-Ph I Ph

21-56 Triethylamine is nucleophilic, but it has no H on nitrogen to lose, so it forms a salt instead of a stable amide.

G: CH3C-CI + :NEt3

. . :�j --- CH3C-CI

IV +NEt3

:0:

When ethanol is added, it attacks the carbonyl of the salt, with triethylamine as the leaving group.

C�: + :0:

CH3C-NEt3 + HO-CH2CH3 � .. ---

An alternate mechanism explains the same products, and is more likely with hindered bases: o � .. I I � like an E2 + 1\ -I H�-CH2CH3

H2C)""C-CI .. Et3NH CI- + H2C=C=O

� I • NEt a ketene H... .---. 3

526

Page 532: Solucionario de wade

21-57 (a) 1) 03

H

H2C�OH 2) Me2S

o

6 .. w !'i

H

O�OH

! Ag+ , NH3 (aq)

0-

O�OH

(b) -o� CH30 _

Br2 --- CH30-O--Br Mg CO2 H30+

---I"� � ___ ether

CH30-O--COOH

SOCI2 + CH30-O--COCI

(c)

CHP-O--CONH2 • NHJ

�CH2Br KCN �CH2CN 1) LiAIH4 �CH2CH2NH2

o .. 0 2) H20" 0

(d) *COOH 1) NaOH

*COOH *CH20H

7 1 excess..

7 1 1) LiAIH! 7 1 HO

::::-"" OH 2) CH3I CH 0

::::-"" OCH 2) H20 CH 0::::-""

OCH excess 3 3 3 3 OH OCH3 OCH3 ! TsCl

pyridine

CH2CH2NH2 CH2CN CH20Ts *1 ..,1) LiAIH4 *1 ..,KCN *1

::::-.... 2) H20 ::::-"" ::::-.... CH30 OCH3 CH30 OCH3 CH30 OCH3 OCH3 OCH3 OCH3

527

Page 533: Solucionario de wade

21-58 o II (a) CH3CH2 0-C-OCH2CH3

(d) 0 II C

0'" ..... 0 '---1

(b)

(e) CH3 I

CH3-� -OH + CH3

o II Cl-C-Cl

(c) 0 _ II � CH30-C- � �

H

�H3 � CH3 -� - 0 -C -Cl

CH3

21-59 (a) The rate of these nucleophilic acyl substitution reactions is controlled by two factors: stability of the starting material as detennined by the amount of resonance donation from the leaving group into the carbonyl, and the leaving group ability which is determined by the basicity of the leaving group, the least basic being the best leaving group.

LEAST STABLE: no significant sharing of electrons from chlorine

electrons from oxygen are also distributed through the ring and nitro; very little resonance stabilization

leaving group ability and basicity:

Cl- > 02N-o- 0-

weakest base; negati ve charge best leaving group delocalized through

ring and nitro

electrons from oxygen are also distributed through the ring; small resonance stabilization

o .. )l .. :0 0:

do > � }-O-

negative charge delocalized through ring

>

MOST STABLE: most significant resonance donation of electrons from oxygen

CH3O-

strongest base; worst leaving group

The least stable starting material with the best leaving group will be fastest to react. The most stable starting material with the poorest leaving group will be slowest to react. (b) ( 0: .. � II�NUC

:O�O: .. / \

··0·· Nuc C ·

:oX-o: / �\

Cl-C C-Cl I 'Cl Cl/ I Cl-C C-Cl

I 'Cl Cl/ I \ Cl Cl triphosgene

Cl Cl;'

:0: .. )l " :0 Nuc

/ Cl-C

I'Cl Cl + Cl-

The first step is the standard attack of a nucleophile like CH3 0H at the carbonyl carbon to make the tetrahedral intennediate. The second step is key: the collapse of the tetrahedral intennediate produces one equivalent of phosgene. Attack of a second nucleophile of the other side of triphosgene would release a latent (hidden or trapped) equivalent of phosgene from that side too. Thus, the equivalent of three molecules of phosgene are locked into the trip has gene molecule. Eventually, all six positions would be substituted with methanol producing three molecules of dimethyl carbonate for each molecule of triphosgene.

528

Page 534: Solucionario de wade

o H II 21-60

(a) CH3-N=C=O + H20 � H3C-N-C-OH � CH3NH2(g) + CO2 (g) methyl isocyanate a carbamic acid-unstable

Both of these reactions are exothermic. In a closed vessel like an industrial reactor, the production of gaseous products causes a large pressure increase, risking an explosion.

(b) 0 CH3-N=C=O � CH3-N=C-O: .. 1·· ··-�•

H-OH 1+ " • • H-O-H ..

(c)

o I I

CI-C-C1 + phosgene

21-61

o I I

OH CI-C-O �-� � VJ

--- CH _�r=-c=o } 3 • • I • • (\!'-H

+ I) ! H - OH two rapid H -RH • • proton transfers

.. � H-OH H II ..

.. .. H C-N-C-O-H 3 .,.../ V decomposition could be proposed as either acid or base catalyzed

(a) (i) The repeating functional group is an ester, so the polymer is a polyester (named Kodel®). (ii) hydrolysis products:

o 0 HO-C-o-C-OH + HOCH2-o-CH20H

(iii) The monomers could be the same as the hydrolysis products, or else some reactive derivative of the dicarboxylic acid, like an acid chloride or an ester derivative.

(b) (i) The repeating functional group is an amide, so the polymer is a polyamide (named Nylon 6). (ii) hydrolysis product: 0

H2N II � OH

(iii) The monomer could be the same as the hydrolysis product, but in the polymer industry, the actual monomer used is the lactam shown at the right.

529

Page 535: Solucionario de wade

21-61 continued (c) (i) The repeating functional group is a carbonate, so the polymer is a polycarbonate (named Lexan®).

(ii) hydrolysis products: -o-CH3� HO r;_� ��OH

CH3 (iii) The phenol monomer would be the same as the hydrolysis product; phosgene or a carbonate ester would be the other monomer.

(d) (i) The repeating functional group is an amide, so the polymer is a polyamide. (ii) hydrolysis product:

H2N -o-COOH p-aminobenzoic acid, PABA, used in sunscreens

(iii) The monomer could be the same as the hydrolysis product; a reactive derivative of the acid like an ester could also be used.

(e) (i) The repeating functional group is a urethane, so the polymer is a polyurethane. (ii) hydrolysis products: H2N'CtNH2

HOCH2CH20H + CO2 + I h CH3 (iii) monomers:

+ O=C=N�

I

� N=C=O HOCH2CH20H � CH3

(f) (i) The repeating functional group is a urea, so the polymer is a polyurea. (ii) hydrolysis products:

(iii) monomers:

21-62

CO2 + NH2(CH2)9NH2

° )l or CI CI

or an equivalent carbonate ester

(a) Both structures are 13-lactam antibiotics, a penicillin and a cephalosporin. (b) "Cephalosporin N" has a 5-membered, sulfur-containing ring. This belongs in the penicillin class of antibiotics.

530

Page 536: Solucionario de wade

21-63 The rate of a reaction depends on its activation energy, that is, the difference in energy between starting material and the transition state. The transition state in saponification is similar in structure, and therefore in energy, to the tetrahedral intermediate :

·0· ��� OCH3 �_ I OH tetrahedral intermediate

The tetrahedral carbon has no resonance overlap with the benzene ring, so any resonance effect of a substituent on the ring will have very little influence on the energy of the transition state.

What will have a big influence on the activation energy is whether a substituent stabilizes or destabilizes the starting material. Anything that stabilizes the starting material will therefore increase the activation energy, slowing the reaction; anything that destabilizes the starting material will decrease the activation energy, speeding the reaction.

smaller Ea because of destabilized starting material

energy

__ ::--- similar transition state energy

larger Ea because of stabilized starting material

'------------------ reaction

(a) One of the resonance forms of methyl p-nitrobenzoate has a positive charge on the benzene carbon adjacent to the positive carbonyl carbon. This resonance form destabilizes the starting material, lowering the activation energy, speeding the reaction. -'0· :0: /8+ . ,. =C + 11/ +r C-C-OCH3

:0: - -_ . .

poor resonance contributor, destabilizing the starting material; no effect in the transition state

(b) One of the resonance forms of methyl p-methoxybenzoate has all atoms with full octets, and negative charge on the most electronegative atom. This resonance form stabilizes the starting material, increasing the activation energy, slowing the reaction.

21-64

�COOH + NH3 ___ �COO-

531

good resonance contributor, stabilizing the starting material; no effect in the transition state

A

Page 537: Solucionario de wade

21-65 A singlet at cS 2.15 is H on carbon next to carbonyl, the only type of proton in the compound. The IR spectrum shows no OH, and shows two carbonyl absorptions at high frequency, characteristic of an anhydride. Mass of the molecular ion at 102 proves that the anhydride must be acetic anhydride, a reagent commonly used in aspirin synthesis. ° °

II II C C

H C""'" '

0/ '

CH 3 3

Acetic anhydride can be disposed of by hydrolyzing (careful ly! exothermic!) and neutralizing in aqueous base.

21-66 IR spectrum: -sharp spike at 2250 cm-I ===> C N

NMR spectrum: -triplet and quartet ===> CH3CH2

-this quartet at 8 4.3 ===> CH3CH20 -1750 cm-I ===> C = ° } maybe an ester -1200 cm-i ===> C - ° -2H singlet at 8 3.5 ===> isolated CH2

° I I

CH3CH20 + C + CH2 + C N

sum of the masses is 113, consistent with the MS The fragments can be combined in only two possible ways:

° ° II

CH3CH20 - C - CH2 C N

A

I I CH3CH20CH2 - C - C N

B

The NMR proves the structure to be A. If the structure were B, the CH2 between oxygen and the carbonyl would come farther downfield than the CH2 of the ethyl (deshielded by oxygen and carbonyl instead of by oxygen alone). As this is not the case, the structure cannot be B.

The peak in the mass spectrum at mlz 68 is due to a-cleavage of the ester: [ R ]t J :R: :I?t L

CH3CH20 t - CH2 C N .. 1 + CCH2CN " .. CCH2CN r 68 mlz 68

mlz 113

532

Page 538: Solucionario de wade

21-67

The formula C5�NO has 2 elements of unsaturation. IR spectrum: The strongest peak at 1670 cm-1 comes low in the carbonyl region; in the absence of conjugation (no alkene peak observed), a carbonyl this low is almost certainly an amide. There is one hroad peak in the NHlOH region, hinting at the likelihood of a secondary amide.

o H I I I

-C-N-

HNMR spectrum: The broad peak at 8 7.55 is exchangeable with D20; this is an amide proton. A broad, 2H peak at 8 3.3 is a CH2 next to nitrogen. A broad, 2H peak at 8 2.4 is a CH2 next to carbonyl. The 4H peak at 8 1.8 is probably two more CH2 groups. There appears to be coupling among these protons bUl it is not resolved enough to be useful for interpretation. This is often the case when the compound is cyclic, with restricted rotation around carbon-carbon bonds, giving non-equivalent (axial and equatorial) hydrogens on

the same carbon. o H I I I

CH2-C- N-CH2 + CH2 + CH2

CNMR spectrum: The peak at 8 175 is the C=O of the amide. All of the peaks between 8 25 and 8 50 are aliphatic sp3 carbons, no sp2 carbons, so the remaining element of unsaturation cannot be a C=C; it must be a ring. The carbon peak farthest downfield is the carbon adjacent to N.

The most consistent structure: o II

N/ OC, H

8-valerolactam

533

Page 539: Solucionario de wade

21-68

IR spectrum: A strong carbonyl peak at 1720 cm-I, in conjunction with the C-O peak at 1200 cm-I, suggests the presence of an ester. An alkene peak appears at 1660 cm-I.

o I I

C=C C-O-C HNMR spectrum: The typical ethyl pattern stands out: 3H triplet at � 1.25 and 2H quartet at � 4.2. The chemical shift of the CH2 suggests it is bonded to an oxygen. The other groups are: a 3H doublet at () 1.8, likely to be a CH3 next to one H; a IH doublet at 0 5.8, a vinyl hydrogen with one neighboring H; and a IH multiplet at 0 6.9, another vinyl H with many neighbors, the far downfield chemical shift suggesting that It is beta to the carbonyl. The large coupling constant in the doublet at 05.8 shows that the two vinyl hydlTlgens are trans. 0 /H I I

C=C + CH C + OCH2CH3 + C-O-C / 3-1 If H

There is only one possible way to assemble these pieces: o 1.8, d H3C H 0 5.8, d

\ / C=C

/ \ 06.9, m H C -OCH2CH3 0 1.25, t

I I + o 04.2, q

CNMR spectrum: The six unique carbons are unmistakeable: the c=o of the ester at 0 166; the two vinyl carbons at 0144 (beta to C=O) and at 0 123 (alpha to C=O); the CH2-O of the ester at 0 60; and the two methyls at 0 18 and at 0 14.

Mass spectrum: This structure has mass 114, consistent with the molecular ion. Major fragmentations:

+ H3C H \ / C=C

/ \ +

+ . ---J"� CH3CH20. +

mass 45

H3C H \ / C=C

/ \ H C+

I I :0:

plus two other resonance forms

rnIz 69

H C-O=CH2 rnIz 99 g .. plus one other resonance form

534

Page 540: Solucionario de wade

21-69 If you solved this problem, put a gold star on your forehead. The formula C6Hg03 indicates 3 elements of unsaturation. IR spectrum: The absence of strong OH peaks shows that the compound is neither an alcohol nor a carboxylic acid. There are two carbonyl absorptions: the one about 1770 cm-

I is likely a strained cyclic ester (reinforced with the C--O peak around 1150 em-I), while the one at 1720 em- I is probably a ketone.

(An anhydride also has two peaks, but they are of higher frequency than the ones in this spectrum.) o II C-O-C

o II

C-C-C

Proton NMR spctrum: The NMR shows four types of protons. The 2H triplet at 8 4.3 is a CH2 group next to an oxygen on one side, with a CH2 on the other. The IH multiplet at 83.7 is also strongly deshielded (probably by two carbonyls) , a CH next to a CH2. The 3H singlet at 8 2.45 is a CH3 on one of the carbonyls. The remaining two hydrogens are highly coupled, a CH2 where the two hydrogens arc not equivalent. There are no vinyl hydrogens (and no alkene carbon in the carbon NMR), so the remaining element of un saturation must be a ring. Assemble the pieces:

o I I

o

\: - 0 -CH2CH2CH + II

C-C-CH3 + 1 rinV

Carbon NMR:

y o 0

A� o C...... CH3

HKH 1 On each carbon, the "up" hydrogen is cis to the acety l group, H H r while the "down" hydrogen is trans. Thus, the two hydrogens on each of these carbons are not equivalent, leading to complex splitting.

o 8 53 0 8 200

8173 --g � g / 0/ '

C/ 'C--- 830 \ /

868 ---C-C •

824

535

Page 541: Solucionario de wade

CHAPTER 22-CONDENSATIONS AND ALPHA SUBSTITUTIONS OF CARBONYL COMPOUNDS

22-1 (a) OH O� CH2 -�==CH2

- 2

(b) Enol 1 will predominate at equilibrium as its double bond is conjugated with the benzene ring, making it more stable than 2.

(c)

basic conditions fonning enol 1

H 0 1 H :0: I II _ I II

Ph -( C -C -CH3 :::: ... ;:::=:=:::: Ph -� -C -CH3 I -: OH H� ··

acidic conditions forming enol 1

H :O :� n I II H-B I II I I Ph-C-C-CH3 .. 1 H :�-H H :O-H }

Ph-C-C-CH ---- Ph-C-C-CH I

3 IJ +

3 I

H H H � ri' H :O-H

I I

basic conditions forming enol 2

o H�. 1 \I (I :�I-!. PhCH2 -C -CH2 :::: ... ;::::::::=::::

:0: II

PhCH2 -C -�.H2 ...... t----l.�

537

Ph -C==C-CH3 1

: O-H I

PhCH2 -C == CH2 2

Page 542: Solucionario de wade

22-1 (c) continued

acidic conditions forming enol 2

:O: � � I I H-B PhCH2-C-?H2 ..

H

22-2

:O:� 0"" H I') 'CH .. 3 ....

R

+ :O- H I I

PhCH2 -C -CH2 ---­I H

. . : O - H I

PhCH -C- CH 2 .. 2 + "-I

H � B: , . .

:O-H I

B:

PhCH2 -C = CH2 2

planar enol This planar enol intermediate has lost all chirality . Protonation can occur with equal probability at either face of the pi bond leading to racemic product.

• • /H :0

(plus one other resonance form for each) 22-3

o

cis

0

mixture of diastereomers

CIS

... ..

. . :0:

unaffected _ , by base "" H

H201 1 0 CH3

+

H

CH3 CH3

538

trans

racemic mixture

o �CH3

U" " H

Page 543: Solucionario de wade

22-4

(a)

(b)

. . :0: :0:

.. - I I I H2C-C-CH3 .. .. H2C=C-CH3

.. :0: :0:

CtH ... ..

(yH

. . .. :0: :0: :0: :0: (c) :0: :0: V V )l .. )l .. - C � CH" -H C 0 CH3 H3C C- CH3 H3 3 3 I

H H H

22-5 . . o :0: :0: o &'''�H • -

"'" -:OH '- ..

+ CI-0C� Cf_&H 22-6

o H � :0: :0: I I rl -..

Ph-C�C. :OH

-H • •

I .. i .. - t

I I - I Ph-C- S:.

H2" - Ph-C=CH2

H

o CI

Ph -ell I - C

� l iCI

_ .. ....----. !l :OH � ..

:0: CI :0: CI I I _ I I I

... _____ j :R: j:l

C:1 I CI 1 Ph -C-CH '"--" � ..

o CI I I I

o CI I I I

---- Ph-C-C-H I)

rH

-:RH! CI01

.. - } :0: CI

... .. Ph-�= �H

1 .. - } Ph-C-C-Cl .. - Ph-C==C-Cl - Ph-C-C-Cl • • --- Cl

\....(CI

539

I CI

Page 544: Solucionario de wade

22-7 For this problem, the cyC\ohexyl group is abbreviated "Cy".

o H� :0: :0:

Q4 = Cy

o Br II rl -, Cy-C�C

:OH -H

. • i .. - }

" - I Cy-C-CH2 .. .. Cy-C=CH2 •

• • Br0r

I I I Cy-C-C-H I) I

H

o Br " �- Br Cy-C-I) �J

1 ·0· Br ."

.

-I .. Cy-C-�H b-B

(H

-:OH I .. l :0: Br } I I

Cy-C=CH

.. i :0: Br :0: Br } " - I I I Cy-C-C-Br'" • Cy-C=C-Br •

• • --- ---A Br 0r

·0· na· Bf (;:1 r �II· I Cy-C-CBr3 C-C-Br_--::-

I CY-� :RH rH

22-8

(a) �COO- Na+

U + CHCl3

o :0: " . .

Cy-C-R : ,, � -+ HCBr3 ---- Cy-C-O- H + :CBr3 bromoform

(b) Coo- Na + � + CHI3 U (precipitate)

(c)

'--../

o Br " I

Ph-C-C-CH3 I Br

22-9 Methyl ketones, and alcohols which are oxidized to methyl ketones, will give a positive iodoform test. Al l of the compounds in this problem except pentan-3-one (part (d)) will give a positive iodoform test.

540

Page 545: Solucionario de wade

22-10

22-11 0

+ H :0""'" I I

o 0 ....... H ....... H :0 0 I I �cy" · �?V • �ci

\ I AcO : j o

0ly Br

0

Cl2 •

H

�H 0 0- II :Br: �cy

Br

0

KOH •

H� BGBr

..

6 CH3COOH 6Cl

L1 6 from Solved Problem 22-2 E2 elimination

22-12

(a) Br

�COOH

22-13

(a) 0 �

(b) no reaction: no a-hydrogen

(b) 0

033

+

(c) 0 Jl ............. . OH HO ' 'I I(

541

Br 0

(d) no reaction: no a-hydrogen

Page 546: Solucionario de wade

Ul � N

VO � .. II n ::c � ::c N ""0

::s-

t :s: .. rQ zj J.. , \

\.

Q .. n (�+ � ::c

� ::c

t t Q-ZO

Q+ .. n-o: I

::c � ::c

� ::c

��: 0 y J n

+ \'" Ql ·o-::c GZ-� ��;Y

t �:q: � IJ "

n n \

\ 0 o ::c

::c ::c � ::c

N N

I ..... VI

q-<J ::c

( A \

Q={J ::c ::c

t Q+-ZO ::c ::c

\ )

r� y N ::c o I C<5�

z 0 t��

.::c • C�O. z 0

0 . . \l o ..... .. ...

+'iJ , J.. \

o .. n=�+ t

� G ..

+n- � :

::c

::c C��� 0

Page 547: Solucionario de wade

22-16

(a) N ... CH3 II

Ph -C-CH3

(b) H3C'N ... CH3 I

Ph -C=CH2 Cd) o-iJ

22-17 Any 2° aliphatic amines can be used for this problem.

(a) 0

6 +

(b) 0 � (c) 0 "

PhC-CH 3

22-18

H n I N 0

6 H + 6 1) �Br

~ .. .. - H2O 2) H30+

o � Ph

H 0 I 0 " 0 0 0 H + I 1) PhC-CI " " PhC=CH2 • PhC-CH2 -CPh + •

- H2O 2) H30+

Y\ HO-H ...

In general, the equilibrium in aldol condensations of ketones favors reactants rather than products. There is significant steric hindrance at both carbons with new bonds, so it is reasonable to conclude that this reaction of cyclohexanone would also favor reactants at equilibrium.

543

Page 548: Solucionario de wade

22-19

(a) OH o (b)

H

Ph 22-20 All the steps in the aldol condensation are reversible . Adding base to diacetone alcohol promoted the reverse aldol reaction. The equilibrium greatly favors acetone.

o CH3 H � II I ("I CO 2-

CH -C-CH -C-O : 3 .. 3 2

I ••

CH3

o H

o CH3 II () I •• -

CH -C-CH -C-O : 3 2 I U· CH3

+

II I CH3 -C-CH2

22-21 + . . :O :�Bi II H-B

CH -C-CH 3 I

2 CH3 -g -CH2 ....... t---i .. � CH3 -� -CH2 __

:O-H :O-H } I � + "-I B :-

·O-H • I

CH3-C==CH2

H

a CH3 H I I I I

CH3-C-CH2-?-O

CH3

-B :

---

H H�

:O-H I

CH3-C-CH3 +

+ ••

:R�H

?H3

� :?-H ?H3� CH3 -C-CH2-C

I-O ....... If---il .. � CH3 -C-CH2-C-O :

+ I··

544

CH3 CH3 carbons of the eiectrophi Ie are shown in bold just to keep track of which carbons come from which molecule

Page 549: Solucionario de wade

22-22

(a) acidic conditions o CH3 H I I I I �W

H C-C-CH -- C-O : • 3 I I · . H CH3

(b) basic conditions . .

o CH3 H :0: CH3 H :0: CH3 H II I I I I I II - • • � 1,,1

H C-C-CH-C - O ----l--< H3C-C==CH-C-0 ....... '---;.�H C -C-CH-C-O: 3 ( I I -: OH I 3 I • .

22-23

H� .. CH3 CH3

o � CH]

I II _I II

I I / H3C-C-CH=C • •

\ + :RH CH3

H :0: I I

H O i H :0:

H3C-C-C-H • H C-C-C-H ... •

(I -.. 3

• •

.. - } H3C-C==C-H

H :OH \ � .. � CH]CH2�fQ j

HO CH3 0 I I I I

CH CH - C-C-C - H 3 2 I I)

t -:R0 n �:O: CH3 0

HO- H I I I I ... CH3CH2-T- T-C-H

H H

.. -{ H-O)

CH3 :0: HO CH3 :0: }- CH3 0 I I I I I I I I I I CH3CH2 -C -C - C -H ....... '---;.�CH3CH2 -C -C == C -H CH3CH2 -C == C - C -- H I U· I I

H H H

545

Page 550: Solucionario de wade

22-24

(a)

22-25 (a)

o

H

(b) M Ph Ph

(c)

Step 1: carbon skeletons

H3C� H

,0

� H3 H H H

Step 2: nucleophile generation

H 0

> CH3 H

H3C� CH3 0

+

H 0

H--H H H

H :0: H :0: _I II I I H C-H --&-11 C -:OH

i ·

· t H-�-C-H • • H-C�C-H

H�·· (2,2-Dimethylpropanal has no a-hydrogen.)

Step 3: nucleophilic attack CH3 H

HC�� • 3 �� H:0: CH 3 •

• q. _I II H -C-C-H . .

Step 4: conversion to final product

:0: H a ++11 (CH3hC C-H

H H \""­H-OH

H-O: H 0 � II (CH3hC

�C-H

H H\ I-:OH ,

. . i H-O H :0: H-O: H :0: } I I ! (I I II (CH) C-C-C�C-H... • (CH) C-C-C-C-H 33 I 33 I U·

H H j H 0 I "

(CH3hC-C�C-C-H Step 5: combine Steps 2, 3, and 4 to complete the mechanism �

546

Page 551: Solucionario de wade

22-25 continued

(b) Step 1: carbon skeletons

H

Ph�O

H3C H Step 2: nucleophile generation

Step 3: nucleophilic attack

Step 4: conversion to final product

> H

Ph-{ +

o

CH3 0 I I I

Ph-C==C - C-H I H

Step 5: combine Steps 2, 3, and 4 to complete the mechanism

547

Page 552: Solucionario de wade

22-26 This solution presents the sequence of reactions leading to the product, following the fonnat of the Problem-Solving feature. This is not a complete mechanism.

Step 2: generation of the nucleophile

H 0 _I II H :0: I II H-C-C-CH3 -(I :OH

i H :0:

H-C-C-CH .. • • •

3 .. � H-b= b-c�

H� ··

Step 3: nucleophilic attack �: -� :�: Ph-C-H + H-C-C-CH �.. 3

Step 4: dehydration

H-O) H 0 I 1 II Ph-C-C-C-CH3 • I '--- I :OH

H H� ··

H- O : H 0 1 I II __ H-OH • Ph-C-C-C-CH3 1 I

H 0 1 II Ph-C=C-C-CH3 1

H

H H

The same sequence of steps occurs on the other side. H 0 I I I

Ph-C=C-C-CH3 I H

o -.. H 0 H I I :RH I " I + Ph-C-H .. Ph-C=C-C-C=C-Ph

548

I I H H

final product

Page 553: Solucionario de wade

A

. . :0: :0:

:�� l ' HC-Ph �

EtO-� CH3 ..

B There are three problems with the reaction as shown:

1. Hydrogen on a 3° carbon (structure A) is less acidic than hydrogen on a 2° carbon. The 3° hydrogen will be removed at a slower rate than the 2° hydrogen.

2. Nucleophilic attack by the 3° carbon will be more hindered, and therefore slower, than attack by the 2° carbon. Structure B is quite hindered.

3. Once a normal aldol product is formed, dehydration gives a conjugated system which has great stability. The aldol product C cannot dehydrate because no a-hydrogen remains. Some C will form, but eventually the reverse-aldol process wil l return C to starting materials which, in tum, wil l react at the other a-carbon to produce the conjugated system. (This reason is the Kiss of Death for C.) 22-28

(a)

22-29 H H

(b) >Vy Ph

o

Ph yY H CI ==�> Ph i1n H CI ==�> Ph Y H

H 0 H 0 0

22-30 0 o

- --

o

The formation of a seven-membered ring is unfavorable for entropy reasons: the farther apart the nucleophile and the electrophile, the harder time they wi l l have finding each other. If the molecule has a possibility of forming a 5- or a 7-membered ring, it wi l l almost always prefer to form the 5-membered ring.

549

Page 554: Solucionario de wade

22-32 (a)

:0:

�) a a

:0: a

not feasible: requires condensation of two different aldehydes, each with a-hydrogens

II � II PhC -CH2CH3 + CH3CH2C - Ph feasible: self-condensation

(c)

a

H

feasible: only one reactant has a-hydrogen; cannot use excess benzaldehyde as acetone has two reactive sites

feasible; however, the cyclization from the carbon a to the aldehyde to the ketone carbonyl is also possible

550

a

CH3

feasible: symmetric reagent will give the same product in either direction of cyclization

Page 555: Solucionario de wade

22-33 (a) and (b)

starting diketone

22-34

- .. o 0 :RH

( CkJ

o 0

o

(a) The side reaction with sodium methoxide is transesterification. The starting material, and therefore the product, would be a mixture of methyl and ethyl esters.

o 0 II II H3C -C -OCH2CH3 + NaOCH3 ::: ... ;:::=="� H3C -C -OCH 3 + NaOCH2CH3

(b) Sodium hydroxide would irreversibly saponify the ester, completely stopping the Claisen condensation as the carbonyl no longer has a leaving group attached to it.

22-35

o II H3C -C -OCH2CH3 + NaOH

CH3 0 I II H C-C-C-OEt .. 3

"( I -:OEt H • •

'----.-/

o CH 0 II I 3 II (CH) CH-C-C-C-OEt 3 2

1 CH3

explanation on next page 551

Page 556: Solucionario de wade

22-35 continued There are two reasons why this reaction gives a poor yield. The nucleophilic carbon in the enolate is 3° and attack is hindered. More important, the final product has no hydrogen on the a-carbon, so the deprotonation by base which is the driving force in other Claisen condensations cannot occur here. What is produced is an equilibrium mixture of product and starting materials; the conversion to product is low. 22-36

o 0 (a) �OCHJ

o 0 (b) Ph _ Jl Jl

..........." '( 'OCHzCH3 Ph

22-37

H O i H :0: H :0: } PhCHz �-g-OMe 00 � PhCHz �-=-g-oMe III � PhCHz �== b-OMe

I) :OMe 00 H

,-----/ .. �:D 1 PhCHzCHz -C - OMe

o H 0 o 0 (?: f R " I " - MeO-

PhCH2CHz -C -C -C -OMe ...... 1--- PhCHzCHz -C -C -C -OMe I) 1

o

Ph

Ph

Ph

1 CHzPh

III �o :OMe o 0 OMe

o 0 :0: :0:

OMe III � Ph

o o ! W (workup)

MeO CH2Ph

:0: :0:

o 0 :0: :0:

OMe

o H 0 " 1 "

OMe ==== PhCHzCHz -C -C -C -OMe 1 CH2Ph

552

Page 557: Solucionario de wade

o II o I I (b) PhCH2C -OCH3 (c) CH3CHCH2 -C -OCH2CH 3

22-39

:0:

I CH3

This one would be difficult because the alphiJ-carnon IS hindered.

. .

:0: H� (I I I - . . G- I I C I (CH - C-OEt _:9_.E_t._�

�.H - C-O� CH=C-OEt

C-OEt {C-OEt C-OEt g :g� g This is the final 0 j product, after II removal of the a- C CH - C -OEt hydrogen by I ethoxide, followed C :-... by reprotonation "" 0

- EtO-0(

o I I CCH - C-OEt I r'4 C-OEt

:�) during the workup . H� :0:

. . :0:

CI) I I -• .

CH - C-OMe :RMe ---II.� C-OMe II

0- I I C I CH - C-OMe CH=C-OMe

{..

C-OMe C-OMe :0 g o

This is the final product, after removal of the a­hydrogen by ethoxide, followed by reprotonation during the workup.

o

C II

yH - C-OMe

C::--... ""0

-MeO-..

553

j o

C II CH- C-OMe I" C-OMe

:6)

Page 558: Solucionario de wade

22-40

(a) not possible by Dieckmann-not a !3-keto ester (b) �COOCH3

�COOCH3 + NaOCH3 (mixture of products results)

COOCH2CH3

(c) C(COOCH3 + NaOCH3 COOCH3

O� + NaOCH2CH3 (d) C COOCH2CH3

o

22-41

H 0 I "

H-C-C-OEt -I ) -:OEt H'---.-/

..

_I II_OEt �

H-C-C

The protecting group is necessary to prevent aldol condensation. Aqueous acid workup removes the protecting group.

I I H-C=C-OEt

H :0: t i H :0:

r :0 \ �g-OEt

o H 0 (:0: H 0 II I II Ph-C-C-C-OEt ..

-EtO- �I I II Ph-C-C-C-OEt

# o �. � :OEt ) �

Ph OEt

H

o 0

Pho/OEt

H

I H

:0:

I) I EtO H

. . :0: :0: :0: :0: :0:

-Jl �Jl � --- � Jl � OEt Ph- T 'OEt Ph- T 'OEt

H H H wj o H 0 II I II

Ph-C-C-C-OEt I H

554

Page 559: Solucionario de wade

22-42

(a) 0 0

Ph�OCH3

Ph

(c) 0 0

EtO�OEt

o

22-43

(a) 0 o II II Ph -C-OEt + CH3CH2 -C-OEt '---.--/

(c) o o II II EtO-C-CH2Ph + EtO-C-OEt �

22-44 o 0 0 HO

(b) 0 0

CH3�OCH3 + Ph

o 0

Ph �OCH3

plus 2 self-condensation products-a poor choice because both esters have a-hydrogens

(d) 0 0

EtOVOEt

CH3

(b) o o 0 II II II PhCH -C - OMe + MeO-C-C-OMe �

(d) o o II II (CH3hC-C-OMe + CH3CH2CH2CH2"C-OMe

� H ---- H

(a) (f if (b) � (c) 0V'0 -----actually present in the enol form:

22-45

(a) two ways: 60 0 II �C - Ph OR

(b) 0 � 0 0 II /' �II II

o �OCHl

VyPh

o

CH3CH2-C-CH2CH3 + CH3CH20-C-C-OCH2CH3

(c) 0

�CH2CH3

(d) two ways: 00 0 II �C-OCH2CHl

555

o 0

OR �OCH2CHl

OCH2CH3

Page 560: Solucionario de wade

22-46 :0: :0:

(a) Jl )l H3C C OEt

(b)

(c)

(d)

I H

:0: :0:

H CJlC-JlCH 3

I 3

H .. :0: N� )l "c,. ;-

C OEt I H

:0: :0: +

" )l - N -:0/ 'c CH

• • I 3

H

· .

:0: :0:

----- H3C¥OEt

H · .

:0 : :0:

----- H3C¥CH3

H . . - :0: :N

� 0 "C ----- OEt

H · .

:0 : :0: +10 - • • N ----- :0/ CH

• • 3 H

(other resonance forms of the nitro group are not shown)

· .

:0: :0:

----- H3CyOEt

H · .

:0: :0:

----- H3CyCHl

H · .

:0: N� yl �C h-

----- OEt

H · .

:0: :0: + l Iyl

- • • N h------ :0 / CH

• • 3

H

22-47 In the products, the wavy lines cross the bonds that must be made by alkylation, before hydrolysis and decarboxylation produce the substituted acetic acid.

0 0 0 (a) 0 1 ) NaOEt .Jl.Jl

EtO�OEt 2) PhCH-,B; EtO' I' �OEt CH2Ph

H30+ ..

o �OH + CO2 + 2 EtOH

(b) 0 0 � 1) NaOEt EtO OEt 2) CH)

..

0

II

0 0 I) NaOEt y 2) CH3I

.. EtO OEt

(c) 0 0 0 � I) NaOEt V EtO OEt 2) Ph(CH2);Br EtO OEt H30+

.. II

CH2CH2Ph

(d) 0 0 0 0 H30+

CH2Ph

H30+ 0 �OH .. + CO2 + 2 EtOH

II

0 Ph�OH

+ CO2 + 2 EtOH

COOH EtO OEt 2) Br(CH2)4Br" EtO OEt

� I) 2 NaOEt � .. 6 + CO2 ''' 2 EtOH

II

556

Page 561: Solucionario de wade

22-48 (a) Only two substituent groups plus a hydrogen atom can appear on the alpha carbon after decarboxylation at the end of the malonic ester synthesis. The product shown has three alkyl groups, so it cannot be made hy malonic ester synthesis. desired product

0 0 0 0 0 H 0

EtO�OEI ==> EIOYOEI ==> H00H � �

C�OH

R\ R2 R \ R2 H3C CH3

(b) H� - '>-- '>-- 1 =,i+ :0: Li+ :0:- }

� Jl OCH + :N : • HN : + 'CAOCH CAOCH x ' 3 \-- \-- '" 3 --- '" 1 H3C CH3 Li+ / / H3C CH3 H3C CH3

pKa 24 LDA pKa 40

With such a large difference in pKa values, products are favored » 99%. (c) 0 :0: H Jl OCH LDA -'C AOCH X ' 3 • '" 3

rBe r �

VOCH3 H30+

� (to

H3C CH3 H3C CH3

22-49 0 "

plus resonance form as shown in part (b)

SN2 o ..

/). OH H3C CH3 H3C CH3 o

(a) PhCH2CH2 - C - CH3 (b) cr + CO2 + EtOH (c) 6 + co, + EtOH + CO2 + EtOH

22-50 In the products, the wavy lines indicate the bonds that must be made by alkylation, before hydrolysis and decarboxylation produce the substituted acetone. (a) 0 0

1) NaOEt 0 0 •

EtO� 2) PhCH2Br EtO�

(b) 0 0

EtO�

(c) 0 0

EtO�

1) 2 NaOEt 2) Br(CH2)4Br

1) NaOEt .. 2) PhCH2Br

CH2Ph 0 0

EtO()' o 0

EtO� CH2Ph

0

H�O

+

. � + CO2 + EtOH

CH2Ph

H30+ 0

\3!' + CO2 + EtOH � /).

1) NaOEt .. 2) Br�

o 0

EIO�Ph

) A�:'�+

0

CO2 + EtOH +

557 �

Page 562: Solucionario de wade

22-5 1

(b) 0

(a) There are two problems with attempting to make this compound by acetoacetic ester synthesis. The acetone "core" of the product is shown in the box. This product would require alkylation at BOTH carbons of the acetone "core" of acetoacetic ester; in reality, only one carbon undergoes alkyl ation i n the acetoacetic ester synthesis. Second, it is not possible to do an SN2 t ype reaction on an unsubstituted benzene ring, so neither benzene could be attached by acetoacetic ester synthesis.

o o

Ph � Ph LDA

Ph II � . Ph � Br

Ph�

(c) 0 Ph � Ph

22-52 � y 0

.. �C'

H plus resonance fonns showing delocalization of e- into ring and C=O

n N

Ph � Ph � Br H30+

----l .. � ..

l-"

Ph - C1 H �H2 -C - Ph

"

came from > Ph - CH 0 I I "

Ph - C-C- CH3 +

Ph - CH - C - CH3 a

forward direction o " NaOEt

Ph - C - CH2 ( COOEt ) temporary ester group

Michael acceptor

� �CH O Ph - C - CH + I I) I I � ( COOEt) Ph - C-C -CH3

resonance­stabilized

o

o

Ph�

o - . .. I I CH2 - C - Ph

Michael donor

Ph - CH t-CH2 - C - Ph I �

( IOOEt�

Ph - CH -CH -C - Ph CO2 + EtOH + I �

Ph - CH - C -CH3 Ph - CH - C - CH3 22-53 First, you might wonder why this sequence does not make the desired product:

CP

O H �H

0

0

.�� ft 0 0

H --- - + � .. .. MVK poor yield

resonance­stabilized

The poor yield in this conjugate addition is due primarily to the numerous competing reactions the ketone enol ate can self-condense (aldol), can condense with the ketone of MVK (aldol), or can deprotonate the methyl of MVK to generate a new nucleophile. The complex mixture of products makes this route practically useless. (continued on next page)

558

Page 563: Solucionario de wade

22-53 continued What pennits enamines (or other stabil ized enolates) to work are: a) the certainty of which atom is the nucleophile, and b) the lack of self-condensation. Enamines can also do conjugate addition:

H I o

o o

high yield

22-54 The enolate of acetoacetic ester can be used in a Michael addition to an a,l3-unsaturated ketone l i k� MVK. 0 0 0 0 � NaOEt. •

EtO �� EtO

o

O /:: o

CO2 + EtOH + � 8 y � a

22-55 J + 1 : N -' C - CH = CH2 ----- : N =C-CH= CH2 -

• • + l ----- : N == C == CH - CH2 r acrylonitrile ( ! Nuc :

N C - CH2 - CH2 I Nuc

n � Nuc - H i - . . .. : N -C-CH-?H2

Nuc - . . } ----- : N == C == CH - ?H1 . Nuc

- • • II - • • I + :0: i :0: :o: }

:O - N - CH = CH ----- :O - N=CH-CH • :O - N = CH - CH • • + 2 • • + �2 - . . + I 2

Nuc : + Nuc nitroethylene + (some resonance fOnTIS of the nitro group are not h :0: s own) I I

:0- N - CH2 - CH2 .. :0- N - CH-CH2 • • + I • •

+ - I Nuc Nuc

559

Page 564: Solucionario de wade

22-56 (a) o I I

PhCH == CH -C - OEt

EtO�OEt

o 0

(c) two ways �0

:

t

H2=CH - C N

OR

(d)

(f)

followed by hydrolysis and decarboxylation

H3C...... /CH3 0 � II � CH2==CH-C- Ph

UCH3

followed by hydrolysis

o I I PhCH == CH -C -OEt

EtO�OEt

o 0 followed by hydrolysis and decarboxylation

(b) CH2=CH-C-�

EtO�OEt

o 0 followed by hydrolysis and decarboxylation

o o CH2=CH-C N

560

followed by hydrolysis

� �aOEt

0 (1\ �OEt + �

o o

(could also be synthesized by the Stork enamine reactions)

Page 565: Solucionario de wade

22-57

Step 1: carbon skeleton

comes from CH3 >

o

Step 2: nucleophile generation

o

These hydrogen atoms more acidic than the other alpha hydrogens because the anion can be stabilized by the benzene ring.

plus resonance forms with negative charge on the benzene ring

Step 3: nucleophilic attack (Michael addition)

H 0

continued on next page 561

. . 0: . . -H

Page 566: Solucionario de wade

22-57 continued

Step 4: conversion to final product (nucleophile formation)

plus one other (enolate) resonance form

(nucleophilic I o attack) t

(base-catalyzed dehydration)

: 0:

o

o· 0 .. plus one other (enolate) resonance form

Step 5: The complete mechanism is the combination of Steps 2, 3, and 4. Notice that this mechanism is simply described by: 1) Enolate formation, followed by Michael addition; 2) Aldol condensation, followed by dehydration.

22-58

Step 1: carbon skeleton o II

Uo'" CH=CH-C-OH

=====� " comes from

>

Step 2: n�leophile generation i : 0:

II - II CH2 - C -OCOCH3 • : CH2 - C -OCOCH3 ...... I----!.� D CH3COO-H

562

o 0 II II

CH3-C-O-C-CH3

· 0 · .. - } CH2 • � � OCOCH3

Page 567: Solucionario de wade

22-58 continued OH 0

Step 3: nucleophilic attack Ph -�H -CH2 - g -OCOCH3

0·0: 0 :O:-�H -OAC

II _ II I " '-..-Ph-C -H + : CH2-C-OCOCH3 ----Ph-CH-CH2-C-OCOCH3 �

Step 4: conversion to final product

OH 0 I II Ph-CH-CH-C-OCOCH3 ..

(OH :0: I _ II Ph -CH - CH -C -OCOCH I)

H CH3COO- U.

3 plus one other (enola!e)

hydrolysis

� 0

resonance fonn

II Ph -CH = CH - C -OCOCH3

II II H+ II II :O: � i :O -H 0

PhCH==CH-C-0-C-CH3 .. PhCH==CH - C-0-C-CH3

t �H+ • •

OH :0: :O -H 0 t I II I II

PhCH==CH-C-0-C-CH3 .. PhCH==CH-C-0-C-CH3 I • • � +

'"" + H20: . / H -0 -H plus one other resonance fOlm

( • • with positive on the other oxygen .. I two fast H20: t proton transfers

OH Q.O+-H O -H

I II I PhCH==CH-C-0-C-CH3 ... O==C

IV I OH C�

plus two other resonance fonns

o

i ·· H -O:

+ PhCH==CH - ? +

: OH

t � 0�

H20: II II PhCH==CH - C-OH

• • / -H -O. }

.. PhCH==CH - ? :OH ..

plus one other resonance fonn with positive on the other oxygen

Step 5: The complete mechanism is the combination of Steps 2, 3, and 4.

563

Page 568: Solucionario de wade

22-59 The Robinson annulation consists of a Michael addition followed by aldol cyclization with dehydration. In the retrosynthetic direction, disconnect the alkene formed in the aldoVdehydration, then disconnect the Michael addition to discover the reactants.

(a) aldol and dehydration forms the a,� double bond:

> o o

Michael addition forms a bond to the �' carbon:

;> o

(b) aldol and dehydration forms the a,� double bond:

o o

;> o o

Michael addition forms a bond to the W carbon:

o

o

22-60 Please refer to solution 1-20, page 12 of this Solutions Manual.

S64

Page 569: Solucionario de wade

22-61 The most acidic hydrogens are shown in boldface. (See Appendix 2 for a review of acidity.) (Braces around resonance forms omitted here.) (a)

H

o

/H o

:0:

H I

:0:

. .

R: --. .

:0 . . --:0:

H . .

(d)

o H I o :Oyf::yo:

�V--:0 . .

(e) 0 0

�" U COCH�

(f)

Because of the stereochemistry of the double bond, the H atoms on the left side of the ring are not equivalent to those on the right side. However, the two enolates will be equivalent except for the double bond geometry.

same enolate as in (b)

:0:

565

:0:

. . ·9

H . .

0:

--

. . 0:

: 0:

H . .

0: . .

. .

:0:

H . .

0: . .

:0: :0:

Page 570: Solucionario de wade

22-61 continued

(g) H a I "

H ...... C� ...... CH 'c "c I \ I H H H

(h) H a I H "

'iC ........ 1 ...... CH H2C C

I H

---

---

H : 0: I "

- ........ C� ...... CH H C "C 2

I H

H I

:0: "

- ........ C� ...... CH H C "C 2

I H

....--

....--

same enolate as in (g)

H : 0: I "

'iC ......... ;-...... CH .. H2C C

I H

H : 0: I "

'iC ......... ;-...... CH .. H2C C

I H

.. H :0: I I

.. 'iC ............... CH H2C C/'

I H

..

H :0: I I

.. 'iC ............... CH H2C C /'

I H

22-62 In order of increasing acidity. The most acidic protons are shown in boldface. (The approximate pKa values are shown for comparison.) See Appendix 2 for a review of acidity.

H uCOOCH3

<

(g) pKa 25

least acidic

,.----

a

.6COOCH3

<

(a) pKa 13

, 1_ .......... .... .. .... _ ........ ............ ....

fully deprotonated by ethoxide ion

a 11011 H H

(b) pKa20

a

ex: <

(c) pKa 11

566

< 0°11 (f)

pKa 17-18

0COOII

<

(d) pKa5

<

a H

coaCH3

(e) CN pKa 2

most acidic

Page 571: Solucionario de wade

22-63 � H. o 0

cHMcH3

H H ---

O� '0

�Jl CH; T .... CH3

The enol fonn is stable because of the conjugation and because of intramolecular hydrogen-bonding in a six-membered nng.

keto

In dicarbonyl compounds in general, the weaker the electron-donating ability of the group G, the more it will exist in the enol fonn: aldehydes (G = H) are almost completely enolized, then ketones, esters, and finally amides which have virtually no enol content.

H enol

o 0

NG H H

keto

22-64 The wavy line lies across the bond fonned in the aldol condensation. �) OH 0

H

0 (b<i) 0 -H20 ..

OH

(c) OH 0 OH _ 2 H20 0

Ph�Ph .. Ph�Ph

(e)

o o

22-65 The wavy line lies across the bond formed in the Claisen condensation. o 0 0 0

(b) �O (c) ..... A� OCH3 � ,., D

o

Cxj o 567

H enol

o

c6

Page 572: Solucionario de wade

89S

H .. ..

H

H H .. H -OH

o :.Q: � u

t -DH-

1 HO: .. )

H

(e)t9-ll w�lqOJd U! uO!1esu�puo::lloPle)o ws!ue4::l�w (e)

99-ll

Page 573: Solucionario de wade

:0:

69S

-----OZH-0 0

t�) :O-H O-H

:0:

----

:a \Jy"z: H

+ :O-H

H-O o w� IH H-O: H-O·

+ •

-----.....

a

H t

H-O a

a-H V�:O:H (q)W-ll W�IqOld U! u0!lusu�puo::> 10PIU)0 ws!ump�w (q)

p�nu!luo::> 99-ll

Page 574: Solucionario de wade

22-66 continued (c) mechanism of Claisen condensation in problem 22-65(a) �o

.. OMe -:OMe ..

H H � I _ I 4 . 1 J:

OMe A\ OMc /T" �Me

-MeO-•

OMe

(this product will be deprotonated by methoxide, but regenerated upon acidic workup)

(d) mechanism of Cia is en condensation in problem 22-65(b) o o

OMe H ..

'" -:OMe '---..-- ..

o

:0:

~ -MeO-

(this product will be deprotonated by methoxide, but regenerated upon acidic workup)

570

OMe

..

o

:0:

:0: ..

Page 575: Solucionario de wade

22-67 All products shown are after acidic workup.

(a) aldol self-condensation \YTa

H +

(b) Claisen self-condensation

OllaEt+

a (c) aldol cyclization

a

CHa

� COaEt

� a

� Ha- OC ..

a (d) mixed Claisen

a a

CHO

Ha--

caaEt Eta--

a a lfOEl + H3C� Eta- � ..-

a a a a

OR lfCH3 + E1O� Eta- � ..

(e) mixed aldol a a a lfCH3 + HJl ph Ha- �Ph ..

(f) enamine acylation-attempt at aldol would give self-condensation

Q 6 a

+ CI.Jl ph

a a

_--1"-_ HP·. � Ph

571

Page 576: Solucionario de wade

22-68 0 o o II

(a) CH3CH2 -C-CH2CH3 (b)A 0 + CO2 �+ CH3CH20H (c) CH3CH2CH2 hCH3 + CO2 + CH30H Ph ..... V .... CH3

(d) 0 � (e) Ph U

(f) 0 CrYH2CH2CH2CH3

OCH3 o o (g) 6CH2CH2CH2CH)

22-69

(a) reagents: Br2' H+ (b) reagents: Br2' PBr3, followed by H20

(c) reagents: excess 12 (or Br2 or Cl;», NaOH

(d) 0 II

Ph-C-H +

(e) Oyo + H

+ -Ph3P-CHCH3 ..

O

<-Ph NaOH

..

PhCH=CHCH3 + Ph3P=O o Ph W o CH3 H

22-70 In the products, the wavy lines indicate the bonds that must be made by alkylation, before hydrolysis and decarboxylation produce the substituted acetic acid.

(a) 0 0

EtO

� 1 ) NaOEt

OEt ..

(b) 0 0 II II 1 ) NaOEt

EtO�OEt ..

o 0 H30+ 0

EtO

V

OEt !1 " �

OH CH2Ph CH2Ph o 0

EIOY

OEI 1 ) NaOEt

2� \.-J Br

CO2 + 2 EtOH + O

u(0H

572

+ CO2 + 2 EtOH o 0

OEt

Page 577: Solucionario de wade

22-70 continued ( ) a a 1 ) 2 NaOEt c

• II II 2) Br(CH2)sBr EtO�OEt

COOH 6 + cO2 + 2 EtOH

22-71 In the products, the wavy lines indicate the bonds that must be made by alkylation, before hydrolysis and decarboxylation produce the substituted acetone.

1) NaOEt •

(a) a a 1 ) NaOEt 0 0

EtO� 2) EtB: EtO� � 2 2()) CH Br Eta

Ih

(b) a a

EtO� 1 ) 2 NaOEt a a

EtO()'

a o + CO2 + EtOH

o ~

+ cO2 + EtOH

(c) The acetoacetic ester synthesis makes substituted acetone, so where is the acetone in this product?

substituted acetone

The single bond to this substituted acetone can be made by the acetoacetic ester synthesis. How can we make the a,13 double bond? Aldol condensation!

a

make by conjugate--- : addition

a a

EtO� + � o NaOEt .. Eta

make by aldol cyclizationldehydration

a a

573

a a NaOH A .�

Page 578: Solucionario de wade

22-72 These compounds are made by aldol condensations followed by other reactions. The key is to find the skeleton make by the aldol.

(a) Where is the possible a,\3-unsaturated carbonyl in this skeleton? OH 0 reverse 0 0

I L I I aldol I I II

PhCH2CH2-CHPh ==>PhCHjCH-C-Ph ===> Ph-C-H + CH3-C-Ph

forward synthesis o o II I I

Ph-C-H + CH3-C-Ph o

NaOH I I .. PhCH = CH -C - Ph

(b) The aldol skeleton is not immediately apparent in this formidable product. What can we see from i t? Most obvious is the \3-dicarbonyl (I3-ketoester) which we know to be a good nucleophile, capable of substitution or Michael addition. In this case, Michael addition is most likely as the site of attack is \3 to anotho. carbonyl. 0 !�"""'\. QO O� Ph

forward synthesis 0

¢o NaOH ..

Ph '. I + \. > � OCH3

, OCH3 j ' 0

\ �-ketoester/ AHA! The aldol product reveals itself.

\ ",/ (See the solution to 22-71 (c).) .........

0 0 0 Q � Ph 1 ) NaOCH3 Ph + ..

OCH3 2) H+ OCH3 0 0

(c) The key in this product is the a-nitroketone, the equivalent of a \3-dicarbonyl system, capable of doing Michael addition to the \3-carbon of the other carbonyl. ,-.

(¥�> Jly-AJl \'" NO;'., N 02 aldol

". product

forward synthesis o �o+ A NaO� AJl 574

reverse � R aldol �� ==�> 0 +

Page 579: Solucionario de wade

SLS

: 0: ..

/ ..

0R:; 4d - ??[( WJ 4d - :)H

HO: 0 . .

H�H t d_Hi?

-1(--:R: 0

(I ·0· HO: . .

: 0: ..

: 0:

+H .. Hy 4d-:) I I o o

(q)

H£;J I( HO: H::J .. � o (n) : 0:

(L-ZZ

Page 580: Solucionario de wade

22-73 continued (c) Robinson annulations are explained most easily by remembering that the first step is a Michael addition, followed by aldol cyclization with dehydration.

OMe

o

OMe

:0 ..

OMe

.c-

plus other resonance forms

·1 j;H �MiChael addition)

� OMe

o

two rapid .l'mton transrc,�HJ JO n

n H -OH

r- H��C. H -OH

-.. 0 I·�

:RH H

plus one other resonance form

OMe

..

(dehydration) o

576

OMe

plus one other resonance form

Page 581: Solucionario de wade

22-73 continued

(d) the negative charge in this structure is stabilized by resonance with the carbonyl

H

H20:

H

H H

� cr-�HP' N+

�O H

+ OH N�\ �;�") H H30+ � �.

H20:

°

two rapid proton transfers

H

�I 'OH

ON--+ H

� H I H I H30+

two rapid proton transfers

o + o l

�o: + H

2 /' .. R) two rapid

H - R: � proton transfers

plus one other H30+ resonance fonn

plus one other resonance fonn H20:

o

+ -H20

H20: � c;r. o � H

o H

577

°

Page 582: Solucionario de wade

22-74� (a)��

22-75 o

o

NaOCH3 --- ..

Dieckmann

-o

H 0+ 3 ..

H 0+ 3 •

578

o

aldol, then dehydration

o

� Br •

alkylation

o � hydrolysis,

. decarboxylatIOn

o

Page 583: Solucionario de wade

22-75 continuued (b) 0

+ o

Ph

22-76

(a) :o:� '" H-B ..

o o

NaOH Robinson reduction

o HO

Claisen

doing this reaction first blocks this side of the ketone

:O- H I rlJC�

�OH plus one other resonance form

Ph

anion formed in Claisen used without isolation in the next step O-H

---I.... + ..... H C ......... .

" H-B

�H plus one oth9r

· •

resonance form

! B:- .. (O-H ..

o plus one other resonance form

o 579

Page 584: Solucionario de wade

22-76 continued .. ...

:O -H

(;) (\ R I r)·

b�H

I � H '1 H B • i --.-

c-1+

OH OH :OH

. . :O -H

0- ..

1 OH t

+ - "'\

plus one other resonance fonn

:O -H

plus one other resonance fonn

� ! : O -H H 1

B:- c:�<> H

..

1+ c -

1+ :OH :OH

plus other resonance fonns

:0: 0

--

" H-B :��

! o 0

Q. GH H20:) 580

Page 585: Solucionario de wade

22-76 continued

(c) 0

o n H .. ,-

C_ ____ I • C_N.

! HO -.: H l r -C_ H "" 1 ____

o

.. :0: &H

+ �CN

22-77 All of these Robinson annulations are catalyzed by NaOH. (a) CH'l + � (b) (l

+ � Ph 0 Ao �o �o

CH3 (e) Uo

+ o

581

H

Page 586: Solucionario de wade

22-78

a a :0: :0: . .

:0: :0:

E'O�OE' -:1iE'· E'OYOE;---H H� H

22-79

CH20P032-I C==O I

CO2 +

HO-CH .. I .. .OH HC-O-H •• • I • • � �

HC-OH I CH20P032-

To identify the retro­aldol, we must first locate the HO that is beta to the C=O.

plus two other resonance forms

Ph, /H C II C

H./ 'COOH

CH OPO 2-I 2

3

C==O I

t : 0: :0:

+ • • � C_ OEt

Ph, /H C II

I H

C EtOOC./ 'COOEt

[ Ph, /H 1 C

--- II HOOC./

C, COOH

CH20P032-I C == a (plus one other

_I resonance form) HO{CH - HO-CH

I • • -HCi 0:

I "\......Ie. HC-OH

I CH OPO 2-2 3

CH OPO 2-I 2 3

C==O I

HO-CH2 dihydroxyacetone phosphate

582

HC==O I

HC-OH I CH20P032-glycer­aldehyde 3-phosphate

Page 587: Solucionario de wade

22-80 This is an aldol condensation. P stands for a protein chain in this problem .

H :O:� 1 II H+ P -CH2CH2 C - CH ---

1 H . .

H :O-H

. . H :O-H 1 1

P -CH2CH2 C J CH 1 ..7 +

plus one other resonance form

H;) . . . H20. H :0-H this is another molecule

of protonated aldehyde 1 1 � P-CH2CH2 C-CH ! .. 1 1.....1 enol serving as a nucleophile

P-CH2CH2 C=CH + 1 +

H

H • • I + two fast H H

+ 1 I

:OH C==O-H proton transfers ..

:OH C==O 1 1

P-CH2CH2CH2 -CH· CHCH2CH2-P I) 1

P-CH2CH2CH2 -CH· CHCH2CH2-P plus one other resonance form ! - H2 0

CHO CHO 1

P-CH2CH2CH2 . CH = CCH2CH2-P ..

+ 1 P-CH2CH2CH2 - CH � CCH7.CHr-P

...... 1 .

22-81 (a) aldol followed by Michael

o NaOEt )l + CH2(COOEth -----l ... �

H � aldol

�-----H

EtOOC CH� Y 2 + CH2(COOEth EtOOC

NaOEI ! M;chael

.. EtOOC yyCOOEt

EtOOC COOEt

(b) Michael followed by Claisen condensation; hydrolysis and decarboxylation o

EtOOC J OEt NaOEt � I I Michaei

�O

o 0

EtoociSOEt NaOEt ...

EtOOCp Claisen

o 0

Ao�

+ CO2 + EtOH + �

583

Page 588: Solucionario de wade

22-81 continued (c) aldol, Michael, aldol cyclization, decarboxylation

o 0 EtOOe � NaOEt EtOOe ....... A (0 �d� • IT �

H 0l. H \.� eOOEt

e 02 + EtOH +

NaOEt • Michael

eOOH

584

o Etooe h r.

� EtOOe

NaOEt ! aldol

a EtOOC�

eOOEt

Page 589: Solucionario de wade

CHAPTER 23-CARBOHYDRA TES AND NUCLEIC ACIDS

Reminder about Fischer projections, first introduced in Chapter 5, section 5- 10: vertical bonds are equivalent to dashed bonds, going behind the plane of the paper, and horizontal bonds are equivalent to wedge bonds, coming toward the viewer.

Fischer projection:

CHO

HO +H

HO +H

CH2 0H

view X from X

left A' �'.'T side X B_ � "4 X

X

23-1 CHO CHO

H-t--OH HO H

HO H H OH

H OH HO H

H OH HO H

CH2 0H CH2 0H glucose mirror

image

A

B

o� H "C,..

HO-C-H I

HO-C-H

CH20H

view Y from

right y side "'r.'�

Y > �

y

CH2 0H

°

HO H

H OH

H OH

CH2 0H fructose

All four of these compounds are chiral and optically active.

23-2

(a) CHO

H +OH

H +OH

CH2 0H

CHO

HO +H

HO +H

CH2 0H

CHO

H +OH

HO +H

CH20H

'A .B

CH2 0H

:iOH

H 0=t=H

HO H

CH2 0H mirror image

CHO

HO + H

H+OH

CH20H two asymmetric carbons => four stereoisomers (two pairs of enantiomers) if none are meso

585

Page 590: Solucionario de wade

23-2 continued (b)

one chiral center � two stereoisomers (enantiomers)

(c) An aldohexose has four chiral carbons and sixteen stereoisomers. A ketohexose has three chiral carbons and eight stereoisomers. 23-3 ��H (a)

CH20H

23-4 CHO

HO H

H OH H

HO H HO

HO H HO

CH20H

(b)

CHO

OH

H

H

CH20H

CHO

H+OH

CH20H

CHO

HotH

HO H

CH20H

CHO

HO+H

CH20H

CHO

HO+H

CH20H L-( -)-glucose L- ( + )-arabinose L-( + )-erythrose L-( -)-gJyceraldehyde

23-5

23-6

2 CHO -

4H-C-OH 1 -CH20H 3

D =R

Hi

:CH1

HO H

CH20H

23-7 (a) Ph

H+OH

H NHCH3

1 CH3

2 CHO -

1 HO-C-H 4 -CH20H 3

L =S

Hoi

�H2CHJ

H OH

CH20H

Ph

Hot H

H3CHN H

2 CH3

586

Ph

H+ OH

H3CHN H

3 CH3

Ph

HO+H

H+NHCH3

4 CH3

Page 591: Solucionario de wade

23-7 (a) continued Ph

H = OH H+ NHCHl

1

Ph

HO = H H3CHN+H

2

Ph

H = OH H3CHN+H

3

Ph

HO = H H+ NHCH)

4

(b) Ephedrine is the erythro diastereomer, represented by structures 1 and 2. Structures 3 and 4 represent pseudoephedrine, the threo diastereomer. (c) The Fischer-Rosanoff convention assigns D and L to the configuration of the asymmetric carbon at the bottom of the Fischer projection. Structure 1 is D-ephedrine; structure 2 is L-ephedrine; structure 3 is L· pseudoephedrine; and structure 4 is D-pseudoephedrine. (d) It is not possible to determine which is the (+) or (-) isomer of any compound just by looking at the structure. The only compound that has a direct correlation between the direction of optical rotation and its D or L designation is glyceraldehyde, about which the D and L system was designed. For all other compounds, optical rotation can only be determined by measurement in a polarimeter.

23-8

(a) CHO (b) CHO (c) CHO H OH HO 2 H HO H H

3 OH HO H H 3 OH H OH HO H HO H H OH H OH H OH

CH20H CH20H CH20H D-allose D-talose D-idose

(d) CHO CHO H OH H�t--OH

HO H H 4 OH

r:=::�> HO H � inversion at bottom chiral center � HO --t=;;... H L-series sugar

23-9

CH20H CH20H L-arabinose D-xylose

D-mannose 1

CHO HO 2

HO ---,3,+-_: \ rotare

H ----'i4 t-- 0 H C:===:::::::>

H 5 OH 6CH20H 587

Page 592: Solucionario de wade

23-10 D-allose OH

HO

23-11 D-talopyranose

23-12 5

(a) HOH2C

4

OH H

1 OH

OH at C-3 is axial

H

OH (b)

H

OH

H OH

OH groups at C-2 and C-4 are axial

5 HOH2C

4

OH OH

OH

H

D-arabinofuranose D-ribofuranose

23-13

23-14

6 HOH2C

5

H H

(a) ex-D-mannopyranose

OH

HO H

H axial = ex

(d) ex-D-arabinofuranose

HOH2C

OH H

OH

CH20H 1

(b) �-D-galactopyranose

OH

H

equatorial = �

OH

(c) �-D-allopyranose

OH

HO equatorial = �

OH

(e) �-D-ribofuranose

H HOH2C OH cis to CH20H

=�

OH trans to

CHzOH = ex OH OH

588

Page 593: Solucionario de wade

23-15 a = fraction of galactose as the a anomer; b = fraction of galactose as the � anomer

a (+ 150.7°) + b (+ 52.8°) = + 80.2°

a + b = 1; b = 1 - a solve for "a"

a (+ 150.7°) + (1 - a) (+ 52.8°) = + 80.2° c::====�> a = 0.28; b = o.n The equilibrium mixture contains 28% of the a anomer and 72% of the � anomer

23-16 H, ....... 0

C/

• • - � 1 HO: H-C-OH .. �+ H OH

CHzOH

erythrose

The planar enolate can reprotonate from either side, producing a mixture of erythrose and threose.

23-17 CHzOH 1 C=O I Y"\

H:+ :� :OH

H+ OH

CHzOH

fructose

.. H� ....... 0: 'c ..

II • • C-OH

H , ....... 0 C/

H+ OH

CHzOH

H, ....... 0 C/

I 1 H-C-OH + HO-C-H

H+OH

CHzOH erythrose

CHzOH 1

C=O _I

HO-C:

H+ OH

H+ OH

CHzOH

CH,oH � I C=O 1

HO-C-H

H+ OH

H+ OH

CHzOH 589

H+ OH

CHzOH threose

H r'OH

CHzOH 1

C=O I

+ H-C-OH :+�: CHzOH

Page 594: Solucionario de wade

23-18 1 CH20H 1 2C==O

31Y\ H:+ :� :OH --

H +OH

6CH20H fructose

CH20H 1 -

:C-OH 1

O==C ----

�+OH

H +OH

CH20H

H"OH ! 1 CH20H

21 H-C-OH

31 O==C +

H + OH

H +OH

6CH20H

CH20H 1 C==O

- 1 • •

CH20H I • . -C-O:

HO-C: ----I I

HO-C

H +OH

H +OH

CH20H

CH20H 1 C-OH I I

:O-C �+ OH

H +OH

CH20H

CH20H 1

HO-C-H 1

O==C

H+OH

H+OH

CH20H

590

H+OH

H +OH

CH20H �H

CH20H I

�H C-OH HO: I) II

... :O-C �+OH

H +OH

CH20H

Page 595: Solucionario de wade

23-19

23-20

CHO

HXOH

HO H

HO-+---H

H OH

CH20H D-galactose

CHO

NaBH4 ..

CH20H

H OH

HO H ----_ ...... ........ _------ .. - plane of symmetry HO H

H OH The reduction product of D-galactose still has four chiral centers but also has a plane

CH20H of symmetry; it is a meso compound and is therefore optically inactive.

CH20H CHO

HiOH H OH

HO H NaBH4 HO H

H

NaBH4 HO

CH20H

OH

H

HOXH

HO H

23-21 (a)

HO

HO

H

H

H1=0H

H OH

CH20H D-glucose

COOH

H

H

OH

OH

CH20H

D-mannonic acid

23-22 (a) COOH

HO H

HO H

H OH

Hi OH

COOH

mannaric acid

..

(b)

(b)

..--

H OH

H OH

CH20H D-glucitol

COOH

H -+- OH

HO H

HO H

H OH

CH20H

D-galactonic acid

COOH

H OH

HO H

HO H

H OH

COOH

-H OH

H OH

CHO L-gulose

(c) no reaction-D-fructose is a

H1=0H

HO H

CH20H L-gulose

ketohexose; only aldoses react

galactaric acid (meso)

591

Page 596: Solucionario de wade

23-23 CHO COOH

H OH H OH

HO H HN03 HO H ..

HO H HO H

H OH H OH

CH20H COOH galactose meso-

optically A inactive

23-24

(a) not reducing: an acetal ending in "oside" (b) reducing: a hemiacetal ending in "ose" (c) reducing: a hemiacetal ending in "ose" (d) not reducing: an acetal ending in "oside" (e) reducing: one of the rings has a hemiacetal

CHO

H OH

HO H HN03 ..

H OH

H OH

CH20H glucose

B

(f) not reducing: all anomeric carbons are in acetal form

23-25

(c) (d) HOH2C

HO

H

COOH

H OH

HO H H OH

H OH

COOH optically active

OH OH axial = a axial = a

23-26 H OH H OH

HO��O �W HO��O H - H20 \ .. .. \ �I+ .. HO C�OH HO C_O -H It HO I • • It HO I

• • H H H H

H OH • • � • •

HO��O CH39

.H

HO��O H CH39.H

HO I

\_OCH� HO I

\5A�H3·

H HO I H HO I

H H H H

592

HO

OCH2CH3 cis to

CH20H= B

OH

H H

Page 597: Solucionario de wade

23-27 OH

HO

H

H30+ ..

CN I

O-CHPh

H °

2

OH

HO

H

CN

+ I ./ HO-CHPh

/ a cyanohydrin

II Ph-CH + 8

OH

HCN is released from amygdalin. HCN is a potent cytotoxic (cell-killing) agent, particularly toxic to nerve cells.

23-28

HOH2C

OH H

Ct and B-D-fructofuranose

OH CH3CH20H HOH2C

W CH20H _ H20

..

OH H

ethyl B-D-fructofuranoside

+

The aglycone in each product is circled. HOH2C CH20H

23-29 OH

HO ..

H H

OH H

ethyl Ct-D-fructofuranoside

OH

HO ..

.0 H • : I CH -0 �. � 'V SO,eH,

593

OH

H

+ -OS03CH3

H

Page 598: Solucionario de wade

23-30

(a) CH)OH2C CH20CH)

OCH) H

23-31

(a) OAc

H

23-32

(a) CHO

H-J.- OH r -------- l --------- ', HO --l-- H :

H i oH

H -t- OH

CH20H

,

H

OAc

D-g\ucose

~ PhNHNH2

W

CHO

HO H r--- .. .. --- - ... -- .. ---- . HO H

H OH

H OH

CH20H

D-mannose

l H I

C=NNHPh I C=NNHPh

r--------·---------

HO i H

H -t- OH

H--i--OH

CH20H

594

(b) OCH)

OCH)

H

(b) AcOH2C OAc

AcO OAc

CH20H

° r-------- .. -_ .. -----

: HO , H , , , H OH , H OH

CH20H

/ D-fructose

Page 599: Solucionario de wade

23-32 continued

(b)

CHO PhNHNH2

H---+-OH W ..

r--------HO---+-H

HO i H

H + OH

CH20H

D-galactose

H I

C=NNHPh PhNHNH2 CHO I ...... 1----

C=NNHPh W HO H r--------HOIH

HO H

H OH

CH20H

r--------HOI H

HO H

H OH

CH20H

D-talose

D-Talose must be the C-2 epimer of D-galactose.

23-33 Reagents for the Ruff degradation are: 1 . Br2, H20; 2. H202, Fe2(S04h .

CHO CHO

H---+-OH r-.. --.......... HO---+-H

H=t=0H

H OH

CH20H .. - _ ... - .. - -_ ... --- -_ .... _ . D-glucose

Ruff •

CHO r--------HO---+-H

H=t=0H

H OH

CH20H ------------------D-arabinose

Ruff •

HO H r--------HO H

H=t=0H

H OH

CH20H --------- _ ... ------_ . D-mannose

23-34 Reagents for the Ruff degradation are: 1 . Br2' H20; 2. H202, Fe2(S04h .

CHO

HIOH

HO H

HO�H

H�OH

CH20H

D-galactose

Ruff �

CHO

HOtH

HO H

H�OH

CH20H

D-Iyxose

595

Ruff •

CHO

HotH

H OH

CH20H

D-threose

Page 600: Solucionario de wade

23-35 Reagents for the Ruff degradation are: 1 . Br2' H20; 2. H202, FeiS04h .

H

H

H

H

23-36

HO

H

H

CHO CHO

OH

OH

OH

OH

CH20H

D-allose

CHO

H

OH Br2

H2O OH

CH20H

Ruff H ..

H

H

H202 .. .. FeiS04h

CHO

OH

OH

OH

CH20H

CHO

Ht OH

H OH

CH20H

HO H

Ruff H OH ...

H OH

H OH

CH20H

D-altrose

CHO

l) HCN HO H 2) H3O+ .. H OH 3) Na(Hg)

H OH

CH20H

D-arabinose D-erythrose D-arabinose

23-37 H CHO

HO H HO H2NOH· HCI

H OH .. H

H OH H

CH20H

D-arabinose

'C=NOH

H AC20

OH ..

OH

CH20H

oxime

596

HO

H

H

C N

H

OH

OH

CH20H

cyanohydrin

HO-..

H2O

CRO

H OH

+ H OH

H OH

CHzOH

D--ribose

CN- +

CHO

HtOH

H OH

CH20H D-erythrose

Page 601: Solucionario de wade

23-38 Solve this problem by working backward from (+)-glyceraldehyde.

CHO

CHO

H+ OH <�: ==::1

CH20H

H�* :H

CH20H

D-threose D

D-( + )-glyceraldehyde

COOH

H--if--OH HN03

HO--if--H ..

H---If--OH

COOH not optically active

COOH

H---tt-OH

HO ---If-- H HN03

HO---ll--H

H�I--OH

COOH not optically active

..

u CHO CHO

H ---If-- OH HO ---tl-- H OR

HO H HO H

H ---tl-- ° H H ---tl-- ° H

CH20H cannot be C

/ CH20H

must be C

D-Iyxose

�------���--------�

( CHO CHO \ H ---tl-- OH HO ---tt- H

HO -I-- H AND HO ---II-- H

HO�f--H

H---tl--OH

CH20H A

D-galactose

'"

HO�I--H

H OH

CH20H B

D-talose

/ r-----",..

597

COOH

HO+ H

H+ OH

COOH

optically active tartaric acid (not meso)

HN03

COOH

HO---tl--H

� HO---lf--H

H�f--OH

COOH optically active

COOH

HO---tl--H

HN03 HO_f--H •

HO---tl--H

H---If--OH

COOH optically active

Page 602: Solucionario de wade

23-39 Solve this problem by working backward from (+)-glyceraldehyde.

CHO

H + OH <¢:: ===::J

CH20H

D-( + )-glyceraldehyde

23-40 (a) HOH2C

(b)

OH H D-fructose

OH

CH30H +

CHO :=t=:: CH20H

D-erythrose F

CHO

H-t-OH

H OH

H OH

CH20H D-ribose

E

iCH2OCH3

2 ° CH30 3

H

H 4 OCH3

H OH

6 CH2OCH3

: r: CH20H

optically inactive erythritol

CH20H

H-t--OH

H OH

H OH

CH20H optically inactive ribitol

(c) Determining that the open chain form of fructose is a ketone at C-2 with a free OR at C-5 show", that fructose exists as a furanose hemiacetal.

598

Page 603: Solucionario de wade

23-41

(a) CH20H

OH H methyll3-D-fructofuranoside

HC=O I C=O I CH20H

+

D-glyceraldehyde

(b) H OH

HO�O 2H5IO� H OCH3

H OH CH20H

methyl I3-D-fructopyranoside

HC=O H30+ I

___ C=O I

CH20H

+

Production of one equivalent of formic acid, and two fragments containing two and three carbons respectively, proves that the glycoside was in a six-membered ring.

(c) In periodic acid oxidation of an aldohexose glycoside, glyceraldehyde is generated from carbons 4,5, and 6. If configuration at the middle carbon is D, that means that carbon-5 of the aldohexose must have had the D configuration. On the other hand, if the isolated glyceraldehyde had the L configuration, then the original aldohexose must have been an L sugar.

OH 23-42

OH

HO HO

H a anomer of maltose 13 anomer of maltose

H H H

599

Page 604: Solucionario de wade

H 0

B anomer of maltose H

OH

OH

H

OH

H 0

OH H

�O

H OH

o�ioH 0

_HO n-

� open chain fonn of maltose H H

OH

H

o + AgO

(mirror)

23-44 Lactose is a hemiacetal. Therefore, it can mutarotate and is a reducing sugar.

OH OH

H H

o

OH

H H H

o

a anomer of lactose H OH B anomer of lactose H

OH

OH

H

23-45 Gentiobiose is a hemiacetal; in water, the hemiacetal is in equilibrium with the open-chain fonn and can react as an aldehyde. Gentiobiose can mutarotate and is a reducing sugar.

23-46 Trehalose must be two glucose molecules connected by an a-I, I '-glycoside. OH

a-D-gl ucopyranos y l-a-D-gl ucopyranos i de

H

H H

600

Page 605: Solucionario de wade

23-47 raffinose OH

galactos

glucose

HOH2C

fructose

H

OH H

invertase ..

melibiose = 6-0-( o.-galactopyranosyl)-OH D-glucopyranose

H

0.-1,6'

HO

H H

+

HOH2C OH

The lower glycoside linkage is a-I ,2'

from the glucose point of view, but f}-2,1' from the fructose point of view.

OH H

fructose (D-fructofuranose)

23-48 cellulose acetate

� O� /0

"C I H

CH3

23-49 cytosine: NH2 eta I

H uracil: 0

C ,H

I N "

N AO

..

I H

guanine:

0 II 0

0-C-CH3 II 0 0-C-CH3 II

0-C-CH3

0

0 I H

.yCCH3

0 0 0 I C H I H

.y , o CH3 �C C 0/ '

CH // 30

I H H

" C CH3 // 0 2:: I N�OH

OH N� � JL _ _ I � N NH2

601 H

o� H

" CH3

Page 606: Solucionario de wade

23-50

(a)

Aminoglycosides, including nucleosides, are similar to acetals: stable to base, cleaved by acid.

H +

R • N

1 • R • R2NH2 H +

OH H OH • • '---"" I R,I,R I + �RA) CH2 0 + HO-CH2 0:::-...

• H H H ----I.� ��\ - �� 3° aliphatic

OH OH OH OH ) OH OH

H20: ! amine­strong base

OH � I • • H CH2 0 :0'

/ iU H� H H20: C�H2 R :0'

hen:iacetal form � • • H

of nbose H H

OH OH

(b) NH2 NH2

OH C1 ?H Cl .. -tH2 0 N 0 CH2 0 N+ R:

.. H H H H

OH � O

�NylN

hH .. N-t ) H H

H H � . . � I OH OH OH OH site of

protonation; weak

�2 � N

OH OH

cytidine base adenosine NucJeosides are less rapidly hydrolyzed in aqueous acid because the site of protonation (the N in adenosme, and in cytidine, the oxygen shown with the negative charge in the second resonance form) is much less basic than the aliphatic amine in an aminoglycoside. Nucleosides require stronger acid, or longer time and higher temperature, to be hydrolyzed.

This is important in living systems as it would cause genetic damage or even death of an organism if its DNA or RNA were too easily decomposed. Organisms go to great length and expend considerable energy to maintain the structural integrity of their DNA.

23-5\ H . . .

N :O:--- -H-N

t�N-H -- - -:N� Iribos�1 N ={+ )= N +

N-H ----·:O: � guanine Ii . . �

cytosine

H.

_ • • y\CH3

N-H-- -:O � .. �

t I '\ ..... N ..-: N N f N: ----H + I -r--, � N=.! • • � nbose .0. . .

adenine thymine

The polar resonance forms show how the hydrogen bonds are particularly strong. Each oxygen has significant negative charge, and in each pair, one H -N is polarized more strongly because the N has positive charge.

602

Page 607: Solucionario de wade

23-52 Please refer to solution 1-20, page 12 of this Solutions Manual.

23-53 CHO

(a) H OH

(b)

HO H HO

OH H OH

H H OH

CH20H

23-54 OH

(a) (b) HO

OH

H H

(c) OH (d)

OH

H

23-55

(a) O-aldohexose (D configuration, aldehyde, 6 carbons) (b) O-aldopentose (0 configuration, aldehyde, 5 carbons) (c) L-ketohexose (L configuration, ketone, 6 carbons) (d) L-aldohexose (L configuration, aldehyde, 6 carbons) (e) O-ketopentose (D configuration, ketone, 5 carbons) (f) L-aldotetrose (L configuration, aldehyde, 4 carbons)

(c) CH20H

H

OH

HO

HO

H

H

OH

(g) 2-acetarnido O-aldohexose (0 configuration, aldehyde, 6 carbons in chain, with acetarnido group at C-2)

603

Page 608: Solucionario de wade

23-56 H

, 'l0 C 0.0- H

, ,.-::0: c/ H 0' "D ,c/

o•o t1 (a) I tr\ HR: I-

HO-C- H -- HO-C:

HO H HO

H OH H

H OH H

.. ..

H

OH

OH

II HO-H HO-C

HO H

H OH

H OH

CH20H CH20H CH20H D-mannose

CH20H i) I HO-H

o==c ..

HOIH

H OH

H OH

CH20H D-fructose

H,

-/OH oC

• . . I

o==C

H0-j H

H ----+-OH

H OH

CH20H

H, /OH

C II

--- :o-c

HO H

H OH

H OH

CH20H

Cl H'C/OH • '.- H II HO. �'o. C .. 0-..

.. HOIH

H OH

H OH

CH20H

enediol

(b) The a isomer has the anomeric OH trans to the CH20H off of C-5. The �-anomer has these groups cis.

HOH2C CH20H

a-D-fructofuranose

c� HOH2C A OH

604

OH H �-0-fructofuranose

Page 609: Solucionario de wade

23-57

(a)

HCN CHO �

H+OH �

(b)

CH20H

COOH

HO+H

H OH

COOH

CN

HO+H

H OH

CH20H A

+ CN

H+OH

H OH

CH20H B

Ba(OHh ..

H2O

Ba(OHh ..

H2O

COOH

H+OH

HO H

COOH

COOH

HO+H

H OH

CH20H C

COOH

H+OH

H OH

CH20H D

COOH

� HO+H

H OR

COOH (-)-tartaric acid

HN03 ---

COOH

H+OH

H Oll

COOH meso-tartaric acid

COOH

H+OH

H OH

COOH (S,S)-(-)-tartaric acid (R,R)-( + )-tartaric acid (R,S)-meso-tartaric acid

23-58

(a) D-(-)-ribose

23-59

(a)

(c) CH30H2C

H

(b) D-( + )-altrose (c) L-( + )-erythrose

(b)

OCH3

OH (d)

CH20CH3

OH H

605

(d) L-(-)-galactose (e) L-(+)-idose

OH

H H

OCH3

OCH3

H

Page 610: Solucionario de wade

23-60

(a)

HO

H

23-61

H

H

OH

H OH

H OH

°

I

HO HO�

H

(a) methyll3-D-fructofuranoside

OH

H

OH

H

(b) 3,6-di-O-methyl-I3-D-mannopyranose

(b) HOH2C

OH OH H

HO

H

(c) 4-0-( CI.-D-fructofuranosy I )-I3-D-galactopyranose

CH20H

°

(d) /3-D-N-acetylgalactopyranosamine, or 2-acetamido-2-deoxy-/3-D-galactopyranose

23-62 These are reducing sugars and would undergo mutarotation:

-in problem 23-59: (b) and (c); -in problem 23-60: (a) and (c); -in problem 23-61: (b), (c), and (d)

23-63

(a) COOH (b) CHO CH20H H OH HO+ H 6:0

HO H HO H HO H + + others

HO H HO H HO H

H OH H OH H OH

CH20H CH20H CH20H

606

(c) OH

OB

H-\4-i- \ . --H

,.OCH3

H

Page 611: Solucionario de wade

23-63 continued (d) COO-

H-t-0H

HO--f--H

HO--f--H

HioH

CHzOH

(h) CHzOH

H--+-OH

HOIH

HO H

H OH

CHzOH

(k) CHO

H� OH

(e) CHzOH

H OH

HO H

HO+H

H OH

CHzOH

(i) CHO

HO H

HO H

H OH

CHzOH

(f) OAc OAc

H� OAC AcO �

H H

(j) CHO

H OH

H OH

HO H

HO H

H OH

CHzOH

H ,0 OCH

19}CH�

", CHOIO 3

+

H

CHO

HO±H

H OH

HO H HO H

H OH

CHzOH

H

excess ° ° HO--f--H HI04 I I I I

.. 5 H-C-OH + 1 H-C-H HO----t=-H from C-l from C-6

H � OH through C-5

CHzOH

23-64 Use the milder reagent, CH3VAgzO, when the sugar is in the hemiacetal form; the mild conditions prevent isomerization. When the carbohydrate is present as an acetal (a glycoside), use the more basic reagent, NaOHl(CH3)zS04; an acetal is stable to basic conditions.

(a) CH20CH3

°

CH3°I H

H OCH3

H OH

CH20CH3

using CH3I, AgzO

(b) CHO

H OCH3

CH30 H

H OCH3

H OH

CHzOCH3

using (CH3)2S04' NaOH

607

(c)

H

CH30

H

H

CHO

OCH3 + H

OCH3

OH

CHzOCH3

CHzOCH3 bo

CH3O-+--H

HIoCH3

H OH

CH2OCH3 using (CH3hS04' NaOH

Page 612: Solucionario de wade

23-64 continued

(d) CHO

H �OCH3

CH30� H

CH30 H

H ioH

CH2OCH3

CHO

+ H OCH3

CH30 H

H OH

H OH

CH2OCH3

(e) CHO

H=COCH3

CH30 H

HrOCH]

H OH

CH2OCH3

CHO

+ HXOCH3

CH30 H

H+OCH]

H OH

CH20H

using CH3I, Ag20 using CH31, Ag20

(f) CHO

H NHCOCH3 using (CH3hS04' NaOH

CH30 H

H OH

H OH

CH2OCH3

23-65

(a) These D-aldopentoses will give optically active aldaric acids.

CHO CHO

HO H HO H D-arabinose D-Iyxose

H OH HO H

H OH H OH

CH20H CH20H

(b) Only D-threose (of the aldotetroses) will give an optically active aldaric acid.

CHO

D-threose HO + H

H+ OH

CH20H

608

Page 613: Solucionario de wade

23-65 continued

(c) X is D-galactose.

X:

CHO

H OH

HOIH

HO H

H OH

CHzOH ! Ruff

CHO

HO H

HO H

H OH

CHzOH

HN03 ..

HN03 •

COOH

H OH

HO H

HO H

H OH

COOH

COOH

HO--!- H

optically inactive

HO -f-- H optically active

Hi OH

COOH

The other aldohexose that gives an optically inactive aldaric acid is D-allose, with all OH groups on the right side of the Fischer projection. Ruff degradation followed by nitric acid gives an optically inactive aldaric acid, however, so X cannot be D-allose.

(d) The optically active, five-carbon aldaric acid comes from the optically active pentose, not from the optically inactive, six-carbon aldaric acid. The principle is not violated.

(e) CHO

HO--!- H

HO-f--H

H-f--OH

CHzOH

Ruff ---

CHO

HotH

H OH

CHzOH D-threose

HN03 •

609

COOH

HotH

H OH

COOH (S,S)-tartaric acid optically active

Page 614: Solucionario de wade

23-66 (a)

CHO

H+OH HCN

..

CH20H

CN

HotH

H OH

CH20H

CN

+ :=+=:: CH20H

(b) The products are diastereomers with different physical properties. They could be separated by crystallization, distillation, or chromatography. (c) Both products are optically active. Each has two chiral centers and no plane of symmetry.

23-67

(a) The Tollens reaction is run in aqueous base which isomerizes carbohydrates.

H� '¢f=9 \ u .. -

HO H HO:

H--II--OH

H--II--OH

CH20H

HO-f--H

H--II--OH

H--II--OH

CH20H D-glucose

+

H-t-OH

H-t-OH

CH20H plus one other resonance fonn

HO-t--H

H---1r--OH

H---1r--OH

CH20H D-mannose

H fc]\-H ....... C/··

(II C-OH

HO-I--H

H-It-OH

H--I--

H ....... /0: C/

1-r : C- OH

HO'l H HO H ..

H--ll--OH

H OH

CH20H

plus one other resonance fonn

(b) Bromine water is acidic, not basic like the Tollens reagent. Carbohydrates isomerize quickly in base, but only very slowly in acid, so bromine water can oxidize without isomerization.

610

Page 615: Solucionario de wade

23-68 Tagatose is a monosaccharide, a ketohexose, that is found in the pyranose form.

61 1

Page 616: Solucionario de wade

Z19

HO

HO

H

-o o --- -'d'i°

H \ H 0

ZH:J/

o Z I H:J-O-d-O

II -

o

0--

Z I 0 0-

H:J-O-d-I I

O-d-\I II

O-d-O 0 0 \1

o

(q)

Page 617: Solucionario de wade

23-72 Bonds from C to H

are omitted for simplicity.

NH2 A N V\ 5' �N �O o?

c

23-73

o \ 0 NH2 P N " N

� t -0/ \ � I ,) A o? N

o \ �O {O H P:.--' -0/ \ CHJ I to o?

T

0\ ...,0 :cO

,H p"-N N G

-0""'" \ � I A NH o? N 2

°

3' � (a) No, there is no relation between the amount of G and A. (b) Yes, this must be true mathematically. (c) Chargaffs rule must apply only to double-stranded DNA. For each G in one strand, there is a complementary C in the opposing strand, but there is no correlation between G and C in the same strand.

613

Page 618: Solucionario de wade

23-74 Bonds from C to H are omitted for simplicity.

HO 5'

HO 5'

a

CH3�;: H

lN� 0 a

OH

a

CH3�;: H

lN� 0

N II N+

a

2'-deoxythymidine (abbreviated dT)

3'-azido-2',3'-dideoxythymidine (AZT)

�'" No phosphate can attach to the azide group, so synthesis of the DNA chain is terminated.

a a a - II II II O -P -O-P-O-p -O

I I I a a a

- - -

AZT 5'-triphosphate

5'

a

CH3�;: H

lN� 0

N II N+ II N

61 4

a

Page 619: Solucionario de wade

23-75 Recall from section 19-17 that nitrous acid is unstable, generating nitrosonium ion. +

(a) H-O-N=O + H+ - H20 + N=O

AI:�" l A + N=R N 0 H

cytosine (C)

+

t

N -""'::::N

I N Ao H �

if this species exists, its lifetime is very short as water will attack quickly

two rapid proton transfers

H20: ! +

:N=N

t�' I N Ao H

- H2O ..

H H+ H -""'::::N -""'::::N I I .. I I N�O N�O H H

H • • H � ':tO� \ OH f� H20:� r�� ... ===:: N O N 0 H H

enol tautomer of uracil

(b) In base pairing, cytosine pairs with guanine. If cytosine is converted to uracil, however, each replication will not carry the complement of cytosine (guanine) but instead will carry the complement n1' uracil (adenine). This is the definition of a mutation, where the wrong base is inserted in a nucleic acid chain.

(c) In RNA, the transformation of cytosine (C) to uracil (U) is not detected as a problem because U is a base normally found in RNA so it goes unrepaired. In DNA, however, thymine (with an extra methyl group) is used instead of uracil. If cytosine is diazotized to uracil, the DNA repair enzymes detect it as a mutation and correct it.

615

Page 620: Solucionario de wade

23-76 Q (a) I h < }-C - O 6qse

OH OH

(b) Trityl groups are specific for 1 ° alcohols for steric reasons: the trityl group is so big that even a 2° alcohol is too crowded to react at the central carbon. It is possible for a trityl to go on a 2° alcohol, but the reaction is exceedingly slow and in the presence of a 1 ° alcohol, the reaction is done at the 1 ° alcohol long before the 2° alcohol gets started.

(c) Reactions happen faster when the product or intermediate is stabilized. Each OCH3 group stabilizes the carbocation by resonance as shown here; two OCH3 groups stabilize more than just one, increasing the rate of removal. The color comes from the extended conjugation through all three rings and out onto the OCH3 groups. Compare the DMT structure with that of phenolphthalein, the most common acid base indicator, that turns pink in its ring open form shown here. Phenolphthalein is simply another trityl group with different substituents.

+ CH3 :0'" + H

:o� one of the major resonance contributors ¢ showing the delocalization I I of the positive charge on the oxygen 0- -0-

Ij_� C �-!J OCH3 ¢ < �c� }OH

23-77 (a) acetal �

(c)

H

Ph-\-O

o

6

3 OH

OH

OH

H

o phenolphthalein

o

(b) Ph�O >

benzaldehyde

OH �-D-glucose

OH

The 4- and 6-0H groups of glucose reacted with benzaldehyde to form the acetal.

(c) and (d) The chiral center is marked with (*). This stereoisomer is a diastereomer since only one chiral center is inverted. This

OH diastereomer puts the phenyl group in an axial position , definitely less stable than the structure shown in part (a). Only the product shown in (a) will isolated from this reaction.

Only the 2'- and 3'-OH groups are close enough to form this cyclic acetal (ketal) from acetone, called an acetonide.

616

Page 621: Solucionario de wade

24-1

(a)

(c)

24-2

CHAPTER 24-AMINO ACIDS, PEPTIDES, AND PROTEINS COOH

I "'''CH Ph H2N �H 2

COOH

H2N���,oH

COOH (b)

I H N � " CH2CH2CH2NH- C -NH2 2 H I I

(d) COOH

I '''''CH H2N �H 2

NH

(a) The configurations around the asymmetric carbons of (R)-cysteine and (5)-alanine are the same. The designation of configuration changes because sulfur changes the priorities of the side chain and the COOH.

2 3 (5)-alanine with COOH COOH group priorities shown � 3 � 2

1 " '" CH 1 '"'' CH SH H2N 3 H2N 2 4H 4H

(R)-cysteine with group priorities shown; note that the COOH and the CH2SH priorities are reversed compared with (5)­alanine

(b) Fischer projections show that both (5)-alanine and (R)-cysteine are L-amino acids.

COOH

H2N+ H

CH3

(5 )-alanine L-alanine

COOH

H2N+H

CH2SH

( R)-cysteine L-cysteine

24-3 In their evolution, plants have needed to be more resourceful than animals in developing biochemical mechanisms for survival. Thus, plants make more of their own required compounds than animals do. The amino acid phenylalanine is produced by plants but required in the diet of mammals. To interfere with a plant's production of phenylalanine is fatal to the plant, but since humans do not produce phenylalanine, glyphosate is virtually non-toxic to us.

24-4 Here is a simple way of determining if a group will be protonated: at solution pH below the group's pKa value, the group will be protonated; at pH higher than the group's pKa value, it will not be protonated.

H H 1 1 + (a) H2N - ?-COO- (b) H-r C\-COOH

CH(CH3h � (c) (d)

617

Page 622: Solucionario de wade

24-4 continued alanine H + I

lysine H + I

aspartic acid H

(e) (i) pH 6

(ii) pH 11

(iii) pH 2

H N-C-COO-3 I CH3

H I H N-C-COO-2

I CH3

H + I H N-C-COOH 3

I CH3

H N -C-COO-3 I

(CH2)4NH3 + H I

H2N-?-COO-

(CH2)4NH2

H + I H3N -?-COOH

(CH2)4NH3 +

+ I H3N-?-COO-

CH2COO-

H I H N-C-COO-2

I CH2COO-

H + I H N-C-COOH 3

I CH2COOH

24-5 . . . . H -N-H H -N -H H -N -H + I I I +

H -N==C-N -H ....... t--I.� H -N -C -N -H ....... 1-. --I"� H -N-C==N--H I I I + I I I

H H H H H H t + H -N -H

II H-N -C-N -H

I I H H

Protonation of the guanidino group gives a resonance-stabilized cation with all octets filled and the positive charge delocalized over three nitrogen atoms. Arginine's strongly basic isoelectric point reflects the unusual basicity of the guanidino group due to this resonance stabilization in the protonated form. (See Problems 1-39 and 19-49(a).)

24-6 tryptophan histidine

H I H2N-? -COOH

CH2

not bask -ro (like pyrrole) I

H

H I H2N-?-COOH

CH2

G� basIc --.. N not basic (like pyridine) • • (like pyrrole)

The basicity of any nitrogen depends on its electron pair's availability for bonding with a proton. In tryptophan, the nitrogen's electron pair is part of the aromatic 1t system; without this electron pair in the 1t system, the molecule would not be aromatic. Using this electron pair for bonding to a proton would therefore destroy the aromaticity-not a favorable process.

In the imidazole ring of histidine, the electron pair of one nitrogen is also part of the aromatic 1t system and is unavailable for bonding; this nitrogen is not basic. The electron pair on the other nitrogen, however, is in an sp2 orbital available for bonding, and is about as basic as pyridine.

618

Page 623: Solucionario de wade

24-7 At pH 9.7, alanine (isoelectric point (IEP) 6.0) has a charge of -1 and will migrate to the anode. Lysine (IEP 9.7) is at its isoelectric point and will not move. Aspartic acid (IEP 2.8) has a charge of -2 and will also migrate to the anode, faster than alanine.

caili� � � � �nOde

t t t lysine

o alanine

-1 aspartic acid

-2 24-8 At pH 6.0, tryptophan (IEP 5.9) has a charge of zero and will not migrate. Cysteine (lEP S.O) has a partial negative charge and will move toward the anode. Histidine (IEP 7.6) has a partial positive charge and will move toward the cathode.

catho� � � � �nooc

t t t histidine tryptophan cysteine

H H H + I + I H3N -?-COO- H3N -C-COO-

I

histidine (partially protonated imidazole ring)

24-9

(a) O=C -COOH I

CH3

(b) O=C-COOH

(c)

I CH2CH(CH3h

O=C-COOH I

CH20H

(d) NH3

O=C-COOH .. I H2, Pd CH2CH2CONH2

� \N� I

H tryptophan

H2N -CH -COOH I

CH2CH(CH3h

H2N -CH -COOH I

CH20H

H2N -CH -COOH

CH2S-

cysteine (partially deprotonated sulfur)

(The SH is more acidic than the NH3 + group.)

I CH2CH2CONH2

619

Page 624: Solucionario de wade

24-10 All of these reactions use: first arrow: (1 ) Br2IPBr3' followed by H20 workup; second arrow: (2) excess NH3, followed by neutralizing workup.

(a) H2C -COOH .. Br - CH-COOH I (1) I

H H

(b) H2C -COOH .. Br -CH-COOH I (1 ) I

CH2CH(CH3h CH2CH(CH3h

(c) H2C-COOH .. Br - CH-COOH I (1 ) I

CH(CH3h CH(CH3h

(d) H2C -COOH .. Br -CH-COOH I (1) I

CH2CH2COOH CH2CH2COOH

.. H2N - CH-COOH (2) I

H

.. H2N -CH -COOH (2) I

CH2CH(CH3h

.. H2N - CH -COOH (2) I

CH(CH3h

.. H2N -CH -COOH (2) I

CH2CH2COOH

In part (d), care must be taken to avoid reaction a to the other COOH. In practice, this would be accomplished by using less than one-half mole of bromine per mole of the diacid.

o o COOEt I

N - CH 1 ) NaOEt .. 2) BrCH(CH3h

COOEt I

N - C -CH(CH3h I

o

I COOEt COOEt

o �abbreviate COOEt GI � ! H30+

(b) COOEt

(}-?H COOEt

(c) COOEt CN - ?H

COOEt

N-CH

�OOEt

COOEt

H2N - CH -COOH I

CH(CH3h

1 ) NaOEt G I H30+

2) BrCH Ph- N - C-CH2Ph - H2N -CH-COOH 2 I � I

1 ) NaOEt ..

2) BrCH2CH2COO-\ .,. salt, not acid-why?

COOEt CH2Ph

H30+ -

� H2N - CH -COOH I CH2CH2COOH

620

Page 625: Solucionario de wade

24-1 1 continued

(d) COOEt 0-�H 1) NaOEt H30+

----,t.� H2N - CH -COOH

COOEt

24-12

COOEt

tJ. I CH2CH(CH3h

COOEt I

AcNH -CH 1) NaOEt I

• AcNH - C - CH2Ph H30+

----,t .. � H2N -CH - COOH I

COOEt 2) BrCH2Ph I

COOEt tJ. I

CH2Ph acetamidomalonic ester

24-13

(a)

o I I

PhCH2-CH NH3, HCN

..

NH2 I

PhCH2-CH I

C N

H2N -CH -COOH I

CH2Ph

(b) While the solvent for the Strecker synthesis is water, the proton acceptor is ammonia and the proton donor is ammonium ion.

+ PhCH2-CH

t I

:NH2

PhCH2-CH I I

+ NH2

this protonated imine is rapidly attacked by cyanide nucIeophile

� �+

:0: H - NH3

PhCH, -�H two fast pMon tmnSfl

+NH2 I) 0NH H ° 3

'--../

H-O - H +Jr\ �:O- H -H20 + 1) H3N -H I ...... f--- PhCH2 -CH .. PhCH2 - CH I I

NH2 NH2

C N I

PhCH2-CH I

NH2

(abbreviate as

R -C N on the next page)

621

Page 626: Solucionario de wade

24-13 (b) continued

mechanism of acid hydrolysis of the nitrile � • •

_ .�H+ J _ + + JI' • • l} H H20: R-C==N. .. l R-C==N -H .. .. R - C==N -H - R - C==N-H

�I+ H -O -H \. ..

+ I I H+"'--- • • 1+ R - T==N - H 1 H H H } H20: +

R -C -N - H '" .. R -C-N - H � R - C==N - H

:O - H . . � H20:

�20:

I II • • I :O - H +O -H :O - H . . . .

H -OH H two rapid OH H :O -H :O-H VI I I proton transfers I � I + - NH3 I • •

+.. i .. R - C -N -H .. R - C -N-H .. R-C - OH

I + --- R - C==OH

I I I + • •

HO � H+ HO H

24-1 4

(a) R NH3, HC� C - H

H20 I CH2CH(CH3h

(b) 0 I I NH3, HCN C-H .. I H2O

H

(c) 0 I I NH3, HCN C -H .. I H2O CH(CH3h

H20: H2N - CH - COOH ....... 1--­

I

H30+ H2N - CH-CN ..

I !l CH2CH(CH3h

H30+ H N -CH-CN 2

I .. !l

H

H30+ H2N - TH-CN .. !l

CH(CH3h

622

CH2Ph

H2N -CH -COOH I

CH2CH(CH3h

H2N -CH -COOH I H

H2N - CH - COOH I

CH(CH3h

+ :O� I I • •

R - C -OH

Page 627: Solucionario de wade

24-1 5 In acid solution, the free amino acid will be protonated, with a positive charge, and probably soluble in water as are other organic ions. The acylated amino acid, however, is not basic since the nitrogen is present as an amide. In acid solution, the acylated amino acid is neutral and not soluble in water. Water extraction or ion-exchange chromatography (Figure 24-1 1 ) would be practical techniques to separate these compounds.

24-1 6 0 o + I I

abbreviate H3N -C H - COEt as I

I I R-COEt

CH2Ph

:O :� 'OH � OH I I H+

' I I' . . H20: I R-C -OEt - RC-OEt • RC-OEt .. + • • • •

24-1 7 +

plus two other + I '" H20 : resonance forms H - � - H �

o + I I

H3N-?H - COH

CH2Ph

(protonated form in acid solution)

. �H H20 : I I - EtOH

• RC �.�---I

:O -H

plus two other resonance forms

PhCH20H +

OH I

RC -OEt

t:J OH H I r\1+

RC-O -Et l

O-H

H2, Pd + H3N -?H -COO- -----i .. � H3N -CH - COOCH2Ph

H+ I -- H3N -CH -COO-

CH2CH2CONH2

24-1 8 ° I I

PhCH20 - C -Cl ..

+ H N-CH-COO-3

I CH2CH2SCH3

CH2CH2CONH2

o H I I I

PhCH20C -N -CH -COOH

+

I CH2CH2SCH3 ! H2, Pd

PhCH3 + CO2 + H3N -?H -COO-

CH2CH2SCH3

623

I CH2CH2CONH2

+ CH3Ph

Page 628: Solucionario de wade

24-19 :0 : ..

:0 : :0 : :0 :

Ni N " :0 : :0: :0: : 0: 1 1 :0 : :0 : :0 : :0 :

N� N� :0: :0: :0 : :0: .. - - ..

These are the most significant resonance contributors in which the electronegative oxygens carry the negative charge. There are also two other forms in which the negative charge is on the carbons bonded to the nitrogen, plus the usual resonance forms involving the alternate Kekule structures of the benzene rings.

24-20 o : 0 : 0

(a)

(b)

(c)

(d)

+ I I : I I : i I H N - CH·C�NH·CH·C�NH·CH·C-O-3

I : I : I HO -CHCH3: CH2Ph : CH2CH2SCH3

Thr Phe ' Met

o : 0: 0: 0 + I I : I I : I I : I I , . ,

H N -CH . C � NH . CH . C -+- NH . CH . C -7- NH . CH . C - 0-3 I : I : I : I

CH20H: (CH2h : H : CH2Ph : I : :

seryl HN - C - NH2 glycyl phenylalanine II arginyl NH

0 : 0: 0: 0: 0 + I I: I I : II : I I : I I I I I I

H N -CH . C --f NH . CH . C -:- NH . CH . C -7" NH . CH . C -:- NH . CH . C -0-3 I : I : I : I : I CH3CH2CHCH3 : (CH2h : (CH2h : CH2 : (CH2)4

. I : I : I : I I (isoleucine) SCH3 CONH2 COOH NH2 M (methionine) Q (glutamine) D (aspartic acid) K (lysine)

0: 0: 0: 0 : 0 + I I: I I : I I : II : II I • I I

H N - CH . C --f NH . CH . C -:- NH . CH . C -7" NH . CH . C � NH -CH - C - 0-3 I : I : I : I : I (CH2h : CH : CHMe : CHCH: CH I ' 2 , . 2' 3' 2 . I ' ' I ' I COOH '

CHMe2' .

CH2CH3'

OH E (glutamic acid) L (leucine) V (valine) I (isoleucine) S (serine)

624

Try spelling your name in peptides!

Page 629: Solucionario de wade

24-21 S I I

(a) /C, HN NPh

\ I HC -C I �

CH3 0

24-22

H2N-Ile--7Gln--7peptide

Step 4

H2N-Gln--7peptide

S I I

(b) /C, HN NPh

\ I HC-C=O I

CH(CH3h

1 ) PhNCS ..

S I I (c) /C,

S I I /C,

HN NPh \ I

HC-C=O I (CH2)4NH2

S I I

(d) /C,

C NPh \ I C -C I � H 0

HN NPh \ I + H2N-Gln--7peptide

HC-C=O I

CH3 -CHCH2CH3

1) PhNCS

S I I /C,

HN NPh + H2N-peptide \ I HC-C=O

I CH2CH2CONH2

24-23 Abbreviate the N-terrninus of the peptide chain as NH2R .

(a) This is a nucleophilic aromatic substitution by the addition-elimination mechanism. The presence of two nitro groups makes this reaction feasible under mild conditions.

F · 0 · + · 0 · + 0 +

02N'Q;-...... . .

I NH2R�

. ". yF NH2R • I�. F NH2R 11+

F NH2R

- • • �- - • • N ND :0" C . I :0" I -0" I I .. . �.. � � � ..

C -1+ N02 ,. N+ _ ,. N+ _ ,. N • • -

0" '0 0" '0 :0" '0 :

I � I +:r::--� �

N02 N02

\

(b) The main drawback of the Sanger method is that only one amino acid is analyzed per sample of protein. The Edman degradation can usually analyze more than 20 amino acids per sample of protein.

625

Page 630: Solucionario de wade

24-24 I trypsin I N-tenninus � � 1

Tyr-Ile----Gln-Arg-Leu----Gly-Phe-Lys-Asn-Trp--Phe----Gly-Ala-Lys�ly-GIn-GlnoNHz

24-25

l t r C tcnrunu, (amide fonn) I chymotrypsin

Phe-Gln-Asn Pro-Arg-Gly·NH2

Cys-Tyr-Phe Asn-Cys-Pro-Arg

\ Tyr-Phe-Gln-Asn ) �---------- ---------_/ Y Cys-Tyr-Phe-Gln-Asn-Cys-Pro-Arg-Gly· NH2

24-26 abbreviations used in this problem:

H2N f-CH - COOH I Rl CH 3

o ° II II PhCH20C - NH - CH COH I

CH3 R2

mechanism of fonnation of Z-Ala

c�: PhCH O-C-CI + H N-R i � 2 �o.

° II ° II R2 - COH ¢:::=l PhCH20 - C-NH-R i ---

626

o H II 1+ :Cl : + PhCH O - C-N-R i

• • 2 ( �k

Page 631: Solucionario de wade

24-26 continued

two possible mechanisms of ethyl chlorofonnate activation

mechanism 1

R :�� R2 -C -OH + CI -C -OEt -00,,--/ �:?j

CI-C -OEt 1 + :O -H I

R2-C=0

o 0 II 0 0 II R2 -C -0 -C -OEt +1)

H"",\ � :9!: o 0 II II

R2 -C -0 -C -OEt

�:?j ___ - Cl- J C)-C-OEt 1 I

H-O : 0 I + II R2 - C=0 -C-OEt

mechanism of the coupling with valine Goo 0 oil 0 II

:0+

II R2-C -OH

plus two other resonance fonns

o 0 II II R2 - C -0-C -OEt

R2 -C -0 -C-OEt + H N -R3 ----�200

t H-O : 0 I 0 0 II R2 - C -0 -C -OEt

+

� � 1 00- i H-O : 0 :Cl: +11 II . . . .

.. R2 - C-0 -C -OEt o 0

:0 : H 0 0 0 II II II PhCH20C - NH - CH - C -NH - CH - C -OH ---

" I + R2 -C -N -R3 + CO2 (I I I

CH3 CH(CH3h H�-:�Et

627

Page 632: Solucionario de wade

24-27 1 \��� o 0 0 ° I I I I I I I I

PhCH20C -NH -CH - C -NH -CH - C -NH -CH -C -OH I I I

CH3 CH(CH3h CH2Ph

! o I I

Cl-C-OEt o 0 0 0 I I I I I I I I

Z -NH -CH -C -NH -CH -C -NH -CH -C -0 -C -OEt I I I

CH3 CH(CH3h CH2Ph t H2NCH2COOH glycine

o 0 0 ° I I I I I I I I

Z -NH -CH - C -NH -CH -C -NH -CH -C -NH -CH -C -OH I I I I

CH3 CH(CH3h CH2Ph H

j 1 H2N-CH-COOH

bH2CH(CH3h

° 0 0 ° I I I I I I I I

leucine

Z -NH -CH -C -NH -CH - C -NH -CH -C -NH - CH -C -NH -CH - COOH I I I I I

CH3 CH(CH3h CH2Ph H CH2CH(CH3h

. H2, Pd

o 0 0 ° / I I I I I I I

H N -CH - C -NH -CH -C -NH -CH -C -NH -CH -C -NH -CH - COOH 2 I I I I I

CH3 CH(CH3h CH2Ph H CH2CH(CH3h

628

Page 633: Solucionario de wade

24-28 ° II ° II

PhCH20C - Cl + H2N -CH -COOH -­I Z - NH-CH -C-OH I CH3 -CHCH2CH3

isoleucine CH3 - CHCH2CH3

I ° II Cl-C-OEt

o ° II II Z -NH -CH - C -0 -C -OEt I

CH3 - CHCH2CH3 t H2NCH2COOH

o ° II II Z -NH -CH -C - NH -CH -C -OH I I

CH3 - CHCH2CH3 H

° II 1 CI-C-OEt 0 0 0 I I II II

glycine

Z -NH - CH - C -NH -CH - C -0 -C -OEt I I CH3 - CHCH2CH3 H 1 H2N -?H -COOH asparagine

CH2CONH2

o ° II II Z -NH -CH - C -NH -CH - C -NH -CH -COOH I I I

CH3 - CHCH2CH3 H CH2CONH2

t H2' Pd o ° I I II

H2N -CH -C -NH -CH -C -NH -CH -COOH I I I CH3 - CHCH2CH3 H CH2CONH2

629

Page 634: Solucionario de wade

24-29 In this problem, "Cy" stands for "cyc1ohexyl".

o-N=C=N-Q c=:::::::> Cy -N = C = N -Cy DCC

mechanism

a :N --Cy II I CH3C-O-C"

a II + CH3C-�:\ H3N -Ph 1 Cy-N=C J-Cy

a N-Cy II II ....... f----I .. � CH C-O-C 3 , • • _ N --Cy N-Cy

plus other resonance forms

a a II + • • II CH3C-NHPh + Cy-N-C - NH-Cy (I ..

H � resonance­stabilized �

+

630

a II Cy - NH -C - NH - Cy

DCU

Page 635: Solucionario de wade

24-30 ���� 0 0 0 0 I I II I I " -0 Me3COC - NH - CH - C -NH -CH - C -NH -CH - C -0 P

I I I CH3 CH(CH3h CH2Ph

+ CF3COOH

0 0 0 + I I I I " -0 H3N - CH - C-NH-CH - C-NH - CH - C -O p

I I I CH3 CH(CH3h CH2Ph

DCC I � Boc-glycine • Me3COC - NHCH2COOH

o 0 0 0 0 I I I I I I I I " -0

Me3COC -NH - CH - C -NH -CH -C - NH - CH - C - NH - CH - C -0 P I I I I

H CH3 CH(CH3h CH2Ph

+ CF3COOH

0 0 0 0 + I I I I I I " -0

H3N -CH - C -NH -CH - C -NH . CH - C - NH - CH - C -0 P I I I I

H CH3 CH(CH3h CH2Ph

DCC 1 Mc,cci1-NH . erH . COOH Boc-leuc;nc

CH2CHMe2

o 0 0 0 0 I I I I I I I I " -0

Boc NH -CH -C -NH - CH - C - NH -CH - C -NH -CH - C -NH - CH - C -0 P I I I I I

CH2CHMe2 H CH3 CH(CH3h CH2Ph

tHF o 0 0 0 0

+ I I I I I I I I I I H N - CH - C - NH - CH -C - NH - CH - C - NH -CH -C - NH -CH - C -OH

3 I I I I I CH2CHMe2 H CH3 CH(CH3h CH2Ph

631

Page 636: Solucionario de wade

24-31 a I I -

Me3COC - NH -CH - COO + I

CH2CONH2

CH2Cl a a

I I 11 --0 --- Me COC-NH-CH-C-O P 3

I � CH2CONH2

Boc-asparagine

o 1 CF3CO:H

II

+ 11 --0 H N-CH -C-O P 3

I a I

CH2CONH2

Boc-glycine Me3COC - NHCH2COOH DCC

a a a

I I I I 11 --0 Me COC-NH-CH - C-NH-CH -C-O P 3

I I H CH2CONH2 l CF3COOH

a a + I I 11 --0

H N - CH -C - NH-CH - C-O P 3 I I

H CH2CONH2 a 1 I I

Boc-isoleucine Me3COC - NH - ?H - COOH DCC

CH3 - CHCH2CH3

a a a a I I I I I I 11 --0

Me3COC -NH -CH - C -NH - CH -C - NH -CH -C -a P I I I

CH3 - CHCH2CH3 H CH2CONH2 .HF a a a

+ I I I I II H N-CH -C-NH-CH -C - NH-CH-C-OH 3

I I I CH3 - CHCH2CH3 H CH2CONH2

632

Page 637: Solucionario de wade

24-32 Please refer to solution 1-20, page 12 of this Solutions Manual.

24-34

(a)

(c) 0 II

:0 : :0 :

N

:0 : :0 :

CH3C -NH -?H - COOH

+ CO2 + I � CHO

(d) C'x,COOH +

HOOCy:) N ' , N

(CH2)4NHCOCH3 I H H I H O==C

L-proline N-acetyl- I D-proline CH3

the enzyme does not recognize D amino acids o II

(e) H2N-CH - CN I

(f) H2N -CH - COOH I (g) Br-CH -C-Br I

CH2CH(CH3h H3C -CHCH2CH3

(h)

24-35

(a)

H2N -CH - COOH OR I CH2CH(CH3h

leucine

o NH3

H3C -CHCH2CH3 isoleucine

o II H2N-CH - C -NH2 I

CH2CH(CH3h

with a water workup, the acid bromide would become COOH

another possible answer, depending on whether the acid bromide is hydrolyzed before adding ammonia

II C-COOH .. H2N -CH - COOH I valine

(b)

I CH(CH3h CH(CH3h

excess H20 NH3

H2C-COOH ---i ... _ --- Br-CH-COOH ---I PBr3 I

H3C -CHCH2CH3 H3C -CHCH2CH3

633

H2N -CH - COOH I H3C -CHCH2CH3

isoleucine

Page 638: Solucionario de wade

24-35 continued o

(c) I I NH3, HCN C-H .. I H20

CH2CH(CH3h

o COOEt I

N-CH 1 ) NaOEt

.. I

COOEt 2) B�H2Ph

o

24-36

(a) H2N -CH -COOH + HOCH(CH3h I CH3

H2N -CH -COOH I

CH2CH(CH3h

o COOEt I

N-? -CH2Ph

COOEt

o

o

leucine

H H2N -T -COOH

CH2Ph

phenylalanine

o I I

(b) H2N-CH -COOH + PhC-CI I

(pyridine) I I ----1 .. � PhC -NH - CH -COOH

CH3

o I I

o I I

I CH3

(c) H2N-CH -COOH + PhCH20C-CI --­I

PhCH20C - NH - CH -COOH

CH3

(d) H2N -CH -COOH I

CH3

24-37 COOH

+

- TsCI ... HO-C-H

pyridine -CH3

o 0 II II

Me3COC - 0 - COCMe3

COOH excess - NH3

TsO-C-H -CH3

634

I CH3

o II

-- Me3COC - NH - CH -COOH I

CH3

COOH -

... H-<;:-NH2 -CH3

D-alanine

Page 639: Solucionario de wade

24-38

24-39

24-40

(a)

(b)

(c)

o COOEt I

N-CH

o

I COOEt

1) NaOEt

o H

COOEt N

N-? -CH, ,--1 COOEt

o L\! H30+

H2N -CH - COOH I

racemic histidine l'

NH

N ==.i 0 II NH3, HCN H C-H .. H2N -C -CN

H2O I I H2C H2C

to N 0 to N 0 H

o 0 II II

H

H N-CH - C-NH-CH -C-OH 2 I I

neutral

CH2 CHCH3 I I

CH2SCH3 OH

o 0 II II

H N-CH - C-NH-CH - C-OH neutral 2 I I

CHCH3 CH2 I I

OH CH2SCH3

0 0 0 II II II

H30+ H .. H2N -C-COOH L\ I

H2C

to N 0 H

racemic tryptophan

H2N -CH - C -NH -CH . C -NH - CH - C -OH basic: two basic side chains and one I I I acidic side chain

(CH2h CH2 (CH2)4NH2 I I

HN-CNH2 COOH II NH

635

Page 640: Solucionario de wade

24-40 continued

(d)

24-41

acidic: carboxylic acid side chain, and the SH is weakly acidic

(a), (b), (c) isoleucine glutamine

� ____ �A �� ________ � ( CH3 \( 0 0 \ I * II II

CH3CH2 -CH - CI H - NH - C -CH - CH2CH2- C -NH2

I * C-terminus --- CONH2 NH - CO - CH2NH2 --

\ J

Peptide bonds are denoted with asterisks (* ) .

'( glycine

N-terminus

(d) glycylglutamylisoleucinamide; Gly-GIn-He • NH2

24-42 o 0

Aspartame: I I I I } H2N -CH . C - NH - CH . C - OCH3 methyl ester

I I CH2COOH CH2Ph

Ic,-_� ,--___ ) �'-_--..,--_�I Y '( aspartic acid phenylalanine (from Edman degradation)

no free COOH => no reaction with carboxypeptidase

Aspartame is aspartylphenylalanine methyl ester.

24-43

o 0 0 0 0 II II II II I I

H2N -CH . C - N H - CH . C - N H - CH . C - NH - CH . C - NH - CH . C -OH I I I I I

CH2Ph CH3 H CH2 CH3 I

CH2SCH3

'-----r--' '-----r--''-----r--' '----r-" '-----r--' phenylalanine alanine glycine � ) y

from Edman degradation

methionine alanine

� from carboxy-peptidase

636

Page 641: Solucionario de wade

24-44 (a) ° II

protect the N-tenninus of the first amino acid 0 II PhCH20C - Cl + H2N -CH -COOH I

CH2CHMe2

� Z-NH - CH -C-OH I CH2CHMe2

leucine 1 ° II CI-C -OEt

activate the C-terrninus

o ° II II Z -NH - CH -C -0 -C -OEt I

CH2CHMe2

add the next amino acid 1 H2N -?H -COOH

CH3

o ° II II Z - NH -CH - C -NH - CH -C -OH I I

CH2CHMe2 CH3

activate the C-terrninus ! � Cl-C-OEt

0 0 0 II II II

alanine

Z -NH -CH -C -NH - CH -C -0 -C -OEt I I CH2CHMe2 CH3

add the next amino acid j H2N -?H - COOH phenylalanine

CH2Ph

o ° II II Z -NH -CH -C -NH -CH - C -NH - CH -COOH I I I

CH2CHMe2 CH3 CH2Ph

deprotect the N-tenninus t H2, Pd

o ° II II H2N -CH -C -NH - CH - C - NH -CH - COOH I I I

CH2CHMe2 CH3 CH2Ph

637

Page 642: Solucionario de wade

24-44 continued (b) II 1\

r: CH2Cl 0 °

� --- Me3COC-NH-TH - C-0-0 ° II

Me3COC - NH - CH - COO- + I

V U CH2Ph

CH2Ph

Boc-phenylalanine CF3COOH N-terminus

attach C-terminus of N-protected amino acid to polymer support

1 deprotect

o + 1\ -0

H N -CH - C-O P o

3 I

II CH2Ph Me3COC - NH -CH - COOH I dd . Boc-alanine

I DCC a . next ammo CH3 aCId and couple

0 0 0 II II " -0 Me COC - NH- CH-C -NH-CH-C-O P 3

I I CH3 CH2Ph

I CF COOH deprote.ct t 3 N-termmus

o 0 + II

" -0 H N-CH-C -NH-CH - C-O P 3 I I CH3 CH2Ph

Boc-leucine

o II 1 add next amino Me3COC -NH -?H - COOH DCC acid and couple

CH2CHMe2

0 0 0 ° \I II II 1\ -0 Me3COC - NH -CH - C -NH· CH . C -NH . CH . C - 0 P

I I I CH2CHMe2 CH3 CH2Ph

+ HF de protect and remove from polymer

0 0 0 + II II II

H3N -CH . C -NH· CH . C -NH . CH . C -OH I I I CH2CHMe2 CH3 CH2Ph

638

Page 643: Solucionario de wade

24-45 o II

protect N-terminus o

PhCH20C - Cl + H2N -CH -COOH I

---II

Z - NH-CH-C-OH I

o 0 II II

CH3 alanine

CH3

activate the C-tenninus

Z -NH - CH - C -0 -C -OEt I CH3 1 H2N -TH - COOH valine

CH(CH3h add the next amino acid

o 0 II II

Z -NH - CH - C -NH -CH -C -OH I I CH3 CH(CH3h ! H2• Pd dcprolect thc N-terminus

o 0 II II

H N -CH - C-NH-CH - C -OH 2 I I

react the N-terminus of the dipeptide at the left with the N-protected, C-activated tripeptide below CH3 CH(CH3h

o 0 0 0 1 II II II II Z -NH -CH - C -NH - CH - C -NH -CH - C -0 -COEt

I I I H3C -CHCH2CH3 CH2CHMe2 CH2Ph

o 0 0 0 II II II II

Z -NH -CH -C -NH -CH -C -NH - CH -C -NH -CH -C -NH . CH -COOH I I I I I

H3C -CHCH2CH3 CH2CHMe2 CH2Ph CH3 CH(CH3h

� H2, Pd deprotect the N-terminus

o 0 0 0 II II II II

H N -CH - C -NH . CH . C -NH . CH - C -NH . CH - C -NH . CH -COOH 2 I I I I I

H3C -CHCH2CH3 CH2CHMe2 CH2Ph CH3 CH(CH3h lie Leu Phe Ala Val

639

Page 644: Solucionario de wade

24-46

(a) There are two possible sources of ammonia in the hydrolysate. The C-terminus could have been present as the amide instead of the carboxyl, or the glutamic acid could have been present as its amide, glutamine.

(b) The C-terminus is present as the amide. The N-terminus is present as the lactam (cyclic amide) combining the amino group with the carboxyl group of the glutamic acid side chain.

(c) The fact that hydrolysis does not release ammonia implies that the C-terminus is not an amide. Yet, carboxypeptidase treatment gives no reaction, showing that the C-terminus is not a free carboxyl group. Also, treatment with phenyl isothiocyanate gives no reaction, suggesting no free amine at the N-terminus. The most plausible explanation is that the N-terminus has reacted with the C-terminus to produce a cyclic amide, a lactam. (These large rings, called macrocycIes, are often found in nature as hormones or antibiotics.) 24-47

(a) Lipoic acid is a mild oxidizing agent. In the process of oxidizing another reactant, lipoic acid is reduced.

S-S oxidized form

(b)

O==C

COOH

� C==O I I

HC -CH2SH I

HSCH2-CH

HN �

(c) 0

S-S �R I

NH �

COOH

SH SH reduced form

� � O==C C==O

I I ---l.� HC-CH2S -- SCH2-CH

I I HN NH � SH SH � �R

o II

H20 + RC-H + � (CH2)4COOH I I \ / ---l.� RC-OH + N (CHZ)4COOH

S-S

24-48

(a) histidine: H2N -CH -COOH I CH2

f NH

basic --- N--.J\not basic

SH SH

640

Page 645: Solucionario de wade

24-48 continued

(b) t N -H I N-H l � -H .. .. .. �

+ N� H :N � H :N Jl H I I I

H H H In the protonated imidazole, the two N's are similar in structure, and both NH groups are acidic.

(c) tj-H tj-H

(J H+ -H+ --- ----

N +N I I

H H resonance-stabilized

We usually think of protonation-deprotonation reactions occurring in solution where protons can move with solvent molecules. In an enzyme active-site, there is no "solvent", so there must be another mechanism for movement of protons. Often, conformational changes in the protein will move atoms closer or farther. Histidine serves the function of moving a proton toward or away from a particular site by using its different nitrogens in concert as a proton acceptor and a proton donor.

24-49 The high isoelectric point suggests a strongly basic side chain as in lysine. The N-CH2 bond in the side chain of arginine is likely to have remained intact during the metabolism. (Can you propose a likely mechanism for this reaction?)

H2N -CH - COOH I

(CH2h I

HN � C-NH2 II

NH arginine

H2N -CH - COOH I

(CH2h I

NH2 + ornithine

641

o II

H2N--C-NH2 urea

Page 646: Solucionario de wade

24-50

(a) glutathione:

0 0 0 II II II

H N - CH -C -NH -CH -C -NH -CH - C -OH 2 I I I

HOOCCH2CH2 CH2SH H "" ..... __ -.. ,.-__ -,,1 '-----y---" \. ..... _----.. ,--_-,,'

gluta�c acid cysteine glJcine (from Edman (from carboxy-degradation) peptidase)

(b) reaction: 2 glutathione + H202 --- glutathione disulfide + 2 H20

structure of glutathione disulfide: 0 0 0 II II II H2N -CH - C -NH -CH -C - NH -CH - C -OH

I I I HOOCCH2CH2 CH2S H I < I new S-S bond

24-51

HOOCCH2CH2 CH2S H I I I

H2N -CH - C -NH -CH -C -NH -CH -C-OH II II II 0 0 0

end groups: N-tenninus Ala --------- lie C-tenninus chymotrypsin fragments

A Glu-Gly-Tyr (middle) B Ala-Lys-Phe

N-terminus

�------c Arg-(Ser? Leu?)-lIe

c-tenninuy y

Ala-Lys-Phe-Glu-Gly-Tyr-Arg-(Ser? Leu?)-Ile

trypsin fragments D Ala-Lys

E Phe-Glu-Gly-Tyr-Arg

� F 'l -

se-r

--

-Le-

u'}lIe

y Ala-Lys-Phe-Glu-Gly-Tyr-Arg-Ser -Leu-Ile

642

)

I

Page 647: Solucionario de wade

24-52

(a) One important factor in ester react iv ity is the ab il ity of the alkoxy group to leave , and the ma in f actor in determining leav ing group ab ility is the stab ilization of the anion . An NHS ester is more reactive than an alkyl ester because the an ion R2NO- has an electron-w ithdrawing group on the 0-, t hereby distributing the negative charge over two atoms instead of just one. A simple alkyl ester has the full negative c harg� on the oxygen with nowhere to go, RO-.

o (b»

)l +

R OH

o

\ abbre\iate J

O-Succ

:0) - F3C�-SUCC

O=C""'"

+

I R

tetrahedral intermed iate

:0:

NHS ester

o

o

----

re act ive anhydride o 0

Rt.�)lCP3

:O-Succ . .

� :0) 0

R�o)l

CF] o I Succ tetrahedral intermediate

NHS trifluoroacetate is an amaz ing re agent. Not only does is act ivate the carboxylic ac id through a m ixed an hydr ide to form an ester under mild conditions , but t he NHS leaving group is also the nucleoph ile that forms an ester that is bot h stable enough to work wit h , yet eas ily reactive w hen it needs to be. Perfect!

(c) :� )l ./ Succ + H2NR R � ----

643

: 0:

)l + "R + -:O-Succ R N

J / �

H H

:0: )l "R R N

H + HO-Succ

Page 648: Solucionario de wade

24-53 NaN0 2 + HCI , ) (a) COOH Y

H 2N �' R

• HONO

.. H nitrous acid

L makes diazonium ions, Sec. 19-17

COOH

+� N N "

R H

intermediate 1 L

NaN) •

azide as nucleophile, Sec . 19-21B

COOH

" R" N H

)

intermediate 2 D

H 2 -

Pd

COOH

" R" NH H 2

D

(b) The product has the opposite configuration from the starting because of the stereochemistry of the reactions. Diazotization does not break a bond to the chiral center so the L configuration is retained in Reaction 1. It is wel l documented that azide substi tution is an SN2 process that proceeds with inversion of configuration; this is why this process works. The third reaction does not break or form a bond to the chiral center, so the D configuration is retained.

644

Page 649: Solucionario de wade

25-1 0 I I

CHAPTER 25-LIPIDS

CH2 -0 - C - (CH2)12CH3 I � trimyristin CH - 0 - C - (CH2)12CH3

25-2 0

I � CH2 -0 - C - (CH2h2CH3

II all c i s o I I

CH2 - 0 -C -(CH2hCH == CH(CH2hCH3 CH2 - 0 - C - (CH2)16CHl I � excess H2 I �

CH - 0 -C -(CH2hCH == CH(CH2hCH3 Ni

CH - 0 - C - (CH2)16CH3 I � I � CH2 - 0 -C -(CH2hCH == CH(CH2hCH3 CH2 -0 - C - (CH2)16CH3

25-3

(a)

2

(b)

2

(c)

3

25-4

triolein, m.p . -40 C tristearin, m.p. 72° C (liquid at room temperature) (solid at room temperature)

0 II

CH3(CH2)16 -C -0- Na+ +

0 II

CH3(CH2)16 -C - 0- Na+ +

0 II

CH3(CH2h6-C-0- Na+ +

[ a 1 2+ II

Ca -- CH3(CH2)16 -C - 0 2 Ca [ a 1 2+ II

Mg -- CH3(CH2)16 - C - 0 2 Mg [ a 1 3+ II

Fe -- CH3(CH2)16-C-0 3 Fe

+ 2 Na+

+ 2 Na+

+ 3 Na+

(a) Both sodium carbonate (its old name is "washing soda") and sodium phosphate wi ll increase lhe pH above 6, so that the carboxyl group of the soap molecule will remain ionized, thus preventing precipitation.

(b) In the presence of calcium, magnesium, and ferric ions, the carboxylate group of soap will form precipitates called "hard-water scum", or as scientists label it , "bathtub ring". Both carbonate and phosphate ions will form complexes or precipitates with these cations, thereby preventing precipitation of the soap from solution.

645

Page 650: Solucionario de wade

25-5

abbreviate S03 - as .lVV\l\l\l\lVVVo-S03-

-03S S03-

S03-

25-6 In each structure, the hydrophilic portion is circled. The uncircled part is hydrophobic.

,'-I

-', , ,

0(' � � , N ' I ' + , , I ' , ,

h- " ' -- ',' Cl-............ -

benzalkonium chloride _ .. _-----_ ............ __ ... __ ... __ ... __ ... _----------------------(�?_���O�O-../'O�O-../'O�O�?-�(��I ----- - -------------- ---_ .... _--------------_ .... _------_ ............ ---�---- ... - ... -

" " ,

, :'

Na+ -O�N " II I '." 0 " ............... _--_ ...

, , 0" ,

, , , ,

Nonoxynol-9

Gardol®

646

Page 651: Solucionario de wade

25-8

(a)

(b)

pentamer

o " CH2 -0 - C - (CH2)nCH 3 I �

� NaOH

CH -O-C-(CH ) CH ... .

I :�: 2" ]

CH -O - P - O: 2 I • •

:0: . .

..

electrophile

o I I

CH2 -0 -C -(CH 2)nCH 3 I � ------ CH - 0 -C - (CH2)nCH 3

I :?: CH -O - P=O 2

I •• OH

:0: . . - :0:

25-9 Estradiol i s a phenol and can be ionized with aqueous NaOH. Testosterone does not have any hydrogens acidic enough to react with NaOH. Treatment of a solution of estradiol and testosterone in organic solvent with aqueous base will extract the phenoxide form of estradiol into the aqueous layer, leaving testosterone in the organic layer. Acidification of the aqueous base will precipitate estradiol which can be filtered. Evaporation of the organic solvent will leave testosterone.

647

Page 652: Solucionario de wade

25- 1 0 Models may help. Abbreviations: "ax" = axial; "eq" = equatorial . ring junctures are axial to one ring and equatorial to another.

(a)

(c)

H

eq

HO

OH ax

H

H ax

ax CH3

CH3 ax

ax

ax CH3

ax HO

(b)

ax to A eq to B

H

Note that substituents at cis-fused

ax to B

H

............ - .....

eq to A CH3

ax OH o

25- 1 1 -------"" 0 " -' " " ...

. ,

, .

. . .

, , .

, . .

. . ,

, ,

, , , ,

..............

... - -- ........ --

, . , ,

. . ,

, .

, ,

H: ,

geranial

camphor

o

, ,

. , , ,

. .

, . .

. ,

.

,

, , , ,

OR ..

o

, .

. . , ,

........ - ..... - .......... --

, .

. .

, , . . .

, , . .

,

, , , .

OR': , . . ,

, . .

. . . . . , ,

:' OR , .

, ,

.

, , .

, , , , ,

... ......... ",

o

camphor shown in top view

648

. . , .. '

, .

, , , ,

, , ,

..... - ...........

menthol

. .

>-':-ii-+-- COO R :

"

abietic acid

, , ,

, , . . ,

Page 653: Solucionario de wade

25- 1 2 B-carotene

25-13

. . , , . .

, - "- .. _---' . .

a-famesene sesqui terpene

- , , • I

� , , --- .. _ .. _ .. --, , .

,� . , ,

.. - .. _- ... _ -..... ,

�--"--"-- .. ..............

. .

,

, . . .

, -

. .

,

"'''''''' ..... , , .

, .

, . .............. .. ..... ' l imonene monoterpene

(Limonene can also be circled in the other direction around the ring-see menthol in 25- 1 1. )

OR

. . . .

, , . '

'- . , .

. .

-.... - .. -'

, . .

. � ---,'--:...::...:;--.:....-- --' --

V zingiberene sesquiterpene

. I

'--, ,

... - ..... - ..... , .

.................. " \..------Y

a-pinene monoterpene

.. � .. ......

25- 14 Please refer to problem 1 -20, page 1 2 of this Solutions Manual.

25- 1 5

(a) triglyceride (b) synthetic detergent (c) wax (d) sesquiterpene (e) steroId

649

, .

Page 654: Solucionario de wade

25- 1 6

(a) 3 CH3(CH2hCH = CH(CH2hCOO- Na+ + HOCH(CH20Hh

(b)

(c)

o

soap glycerol

-g - (CH2)16CH3 CH -0 20

. I II

H tristeann -O-C-(CH2116C 3

II H Bf

CH 0

B f)l--1<' � II

CH

"""

_ I

0-C-(CH2h6 3

0

) CH

CH2 -

II (CH27 ]

0-C - (CH2h CH -H H Br

I 2

H H Br

Br)U(, ---- - 0 B f)l--1<' �

' --II

"" " _

O - CH 0

(CH2);C - I II (CH2hCH3

CH3(CH2h

0 _ C _ (CH2h (mixture 0 CH -f diastereomers) 2

(d) 3 CH3(CH2hCHO + o

(e) 3 CH3(CH2hCOOH + o

(f)

II II CH2 -0 -C - (CH2hCHO CH2 -0 - C - (CH2hCOOH I R I R CH -0 -C - (CH2hCHO I R CH -0 -C - (CH2hCOOH I R CH2 -0 -C - (CH2hCHO CH2 -0 -C - (CH2hCOOH

H 0 II

CH2 -0 -C - (CH2h' (CH2hCH3 o I H H I I H ,, /

CH3(CH2h (CH2h C -0 - CH "'" C

I R "�'

(mixture of diastereomers) CH2 -0 -C - (CH2h (CH2hCH3

(g) 3 CH3(CH2hCH = CH(CH2hCH20H + HOCH(CH20Hh glycerol

650

Page 655: Solucionario de wade

25-17 H2 1 ) LiAIH4

(a) CH3(CH2hCH = CH(CH2)7COOH ----:-- - CH3(CH2)16CH20H Nl 2) H30+

H2 (b) CH3(CH2hCH = CH(CH2hCOOH � CH3(CH2)16COOH

(c) CH3(CH2)16COOH + HOCH2 (CH2)16CH3 H+ ----

t1 CH3(CH2) 16COOCH2 (CH2) 16CH3

from (b) from (a)

1 ) 03 (d) CH3(CH2hCH = CH(CH2hCOOH - CH3(CH2hCH = 0 + 0 = CH(CH2hCOOH

2) Me2S nonanal

KMn04 (e) CH3 (CH2hCH = CH(CH2hCOOH

H20, t1- CH3(CH2)7COOH + HOOC(CH2hCOOH nonanedioic acid

(f) CH3(CH2hCH = CH(CH2hCOOH

25-18 (a) o

II

excess Br2

PBr3

H20 •

H2C -0 -C - (CH2)16CH3 NaOH H2C -OH o ..

I II t1 I HC -0 -C - (CH2hCH = CH(CH2hCH3 HC -OH I R + I

H2C -0 - P -OCH2CH2NH3 H2C -OH 6- glycer ol

(b) o II

H2C -0 -C -(CH2)14CH3 NaOH

I R HC -0 -C - (CH2)14CH3 I R +

H2C -0 - P -OCH2CH2N(CH3h I 0-

..

t1

H2C -OH

I HC -OH I

H2C -OH glycerol

651

H H Br Bk (CH2)6b:COOH CH3(CH2h

Na+

Na+

o II

o II

0-C - (CH2)16CH3

0 -C - (CH2hCH = CH(CH2hCH3 o II

Na+ 0-P-OCH2CH2NH2 I

Na+ 0

o II 2 Na+ 0 -C - (CH2)14CH3

o II +

Na+ 0 -P -OCH2CH2N(CH3h I

. Na+ -0

Page 656: Solucionario de wade

3

25-20 Reagents in parts (a), (b), and (d) would react with alkenes. If both samples contained alkenes, these reagents could not distinguish the samples. Saponification (part (c)) , however, is a reaction of an ester, so only the vegetable oil would react, not the hydrocarbon oil mixture.

25-2 1 (a) Add an aqueous solution of calcium ion or magnesium ion. Sodium stearate wi l l produce a precipitate , while the sulfonate wi l l not precipitate . (b) Beeswax , an ester, can be saponified with NaOH. Paraffin wax is a solid mixture of alkanes and will not react. (c) Myristic acid wi l l dissolve (or be emulsified) in di lute aqueous base . Trimyristin will remain unaffected. (d) Triolein (an unsaturated oil) will decolorize bromine in CCI4, but trimyristin (a saturated fat) will not.

25-22

(a) o I I

CH2 - 0 - C - (CH2)12CH3

(b)

asymmetric * I � _ carbon CH - 0 - C - (CH2hCH - CH(CH2hCH3 I �

CH2 - 0 - C - (CH2hCH == CH(CH2hCH3

o I I

CH2 - 0 - C - (CH2hCH == CH(CH2)7CH3 I � CH - 0 - C - (CH2)12CH3 I � CH2 - 0 - C - (CH2hCH == CH(CH2hCH3

25-23 0 I I

CH2 - 0 - C - (CH2)16CH3

asymmetric * I � _ carbon CH - 0 - C - (CH2hCH - CH(CH2hCH3 I �

CH2 - 0 - C - (CH2hCH == CH(CH2hCH3

652

optically active

not optically active ; symmetric

optically active

Page 657: Solucionario de wade

25-23 continued

(a) o

0-g - (CH2) 16CH3 CH2-0 I _ g _ (CH2)16CH3 CH-O 0

not optical ly active

I -g-(CH2)16CH3

0 CH2-O

I I H O-C(CH2)16C 3

(b) Br � H

"""Br

0 C1 li2 -

Br.... H � Bf

'--1",.,., I I --'

CH ••••••• .L.J/ / "'

(CH2);C-O I � / '\CH2),CH,

CH,(CH2)7

_ C _ (CH2), optlca

f d' stereomers , l ly active

) CH2 _ 0 (mixture 0 la

(c) products are not optically acti ve

HOCH(CH20Hh + Na+ -OOC(CH2)16CH3 + 2 Na+ -OOC(CH2hCH == CH(CH2hCH3

(d)

g lycerol

o I I

CH2 -0 -C -(CH2)16CH3 J � CH - 0 -C - (CH2hCHO I � CH2 -0 -C - (CH2)7CHO

optically active

+ 2 0 == CH(CH2hCH3 not optically active

25-24 The products in (b) , (c) and (f) are mixtures of stereoisomers,

(a) (b) (c)

Br Br

H

Br + CH20 0

B r Br Br

(f)

HO

653

(e)

+ CO2

Page 658: Solucionario de wade

25-25 Is it any wonder that Olestra cannot be digested !

oleic

� 0 0 L

654

o 0

#

#" l inolenic

Page 659: Solucionario de wade

(a)

H

(b)

OH eq

HO" ",

25-27

(a) sesquiterpene

, ,

. .

, ,

,

\ , (b) monoterpene

, ,

.

,

, ,

.. ... __ ... (c) sesquiterpene

, ,

, ,

.

, , ,

. .

,

"" H --

\ . , ... .,. ... - .......

,

:'--- -... " o \

. ,

, ,

,

\ \ . . .

... - ..... - ..... ---

---

COOH

NHCH2COOH I

Like a soap or detergent, this cholic acid-glycine combination has a very polar "head" and a slightly polar "tai l ". The "tai l " can dissolve non-polar molecules and the polar "head" can carry the complex into polar medi a.

........... .. ... ... ---- ..

OR ,

, ,

,

OR

, ..........

,

, ,

, ,

. .

\ "

.

, .

............. ,

. . \ \ \ , , ,

OR

655

\ , --- ---

, , .

............

, . \

..... -- ..

,

, ,

, ,

, ,

, ,

, , H

,

, , \ \ ,

,

I

.. - .. - ......

.......... _----_ ........

--.

, ,

\ , . .

.

Page 660: Solucionario de wade

25-27

(d) sesquiterpene , , · · · · · · ,

_ ....... ,

,

, ,

, .. -_ .... - ...........

, , , .

CH3 J OR " --'

, ,

" .... --- ............. : CH3 · · · · · · ,

. . ,

, , , , , CH3 \

. .

... : : ........................... '

CH3 CH3': " I .............. -

25-28 The fonnula ClsH340 2 has two elements of unsaturation; one is the carbonyl , so the other must be an alkene or a ring. Catalytic hydrogenation gives stearic acid, so the carbon cannot include a ring; it must contain an alkene. The products from KMn04 oxidation detennine the location of the alkene:

HOOC(CH2)4COOH + HOOC(CH 2)IOCH3 :::::> HOOC(CH2)4CH == CH(CH 2)IOCH3

If the alkene were trans , the coupling constant for the vinyl protons would be about 1 5 Hz; a 10 Hz coupling constant indicates a cis alkene. The 7 Hz coupling is from the vinyl H's to the neighboring CH2·s.

25-29

COOH linolenic acid

COOH linoleic acid-known

COOH NEW

COOH NEW

COOH NEW

COOH NEW

COOH oleic acid-known

In addition to the new isomers with different positions of the double bonds , there has been growing concern over the trans fatty acids created by isomerizing the naturally occuring cis double bonds. More manufacturers are now listing the percent of "trans fat" on their food labels . 25-30 The cetyl glycoside would be a good emulsifying agent. It has a polar end (glucose) and a non­polar end (the hydrocarbon chain) , so it can dissolve non-polar molecules, then carry them through polar media in micelles. This is found in Nature where non-polar molecules such as steroid honnones, antibiotics , and other physiologically-active compounds are carried through the bloodstream (aqueous) by attaching saccharides (usually mono, di , or tri) , making the non-polar group water soluble.

656

Page 661: Solucionario de wade

25-3 1 (a) Four of these components in Vicks Vapo-Rub® are terpenes . The only one that is not is decane; it is not comprised of i soprene units despite having the correct number of carbons for a monterpene.

Three of the four terpenes have had their isoprene units indicated in previous problems; in each of those three, there are two possible ways to assign the isoprene units , so those pictures wi l l not be duplicated here.

a-pinene;

see 25- 1 3

decane

. . . . .

cineole; eucalyptol ( .

.......

.. __ ........

. . . . .

menthol; see 25- 1 1

OH

camphor; see 25- 1 1

o camphor

(b) Vicks Vapo-Rub® must be optically active as it contains four optically active terpenes.

25-32

(a) Of the two, only nepetalactone is a terpene. The other has only 8 carbons and terpenes must be in multiples of 5 carbons.

(b) Of the two, only the second is aromatic as can be readily seen in one of the resonance forms.

, · · · · " , .............. _- ........ '1 .. , " .

. . . , ......... --_ .... -

, .

. , ,

" 0 __ ... ___._ ° I 00/ 0:

CI:?0 o+�o

::::-... :-. ::::-... // . 0

(c) Each compound i s c leaved with NaOH (aq) to give an enolate that tautomerizes to the more stable keto form. o o o

NaOH - o

o a(coo-NaOH

657

Page 662: Solucionario de wade

25-33

(a) High temperature and the diradical O 2 molecule strongly suggest a radical mechanism.

(b) Radical stability follows the order: benzylic> allylic > 3° > 2° > 1°. A radical at C-ll would be doubly allylic making it a prime site for radical reaction.

(c) • • • • · R - R · �,---H

}

initiation • Ll

H

�C; ..J j'\ .. . .

propagation step 1 • R - R .

COOH

COOH

•• • • • ��H

} . 0 - 0 · _ _ _ I • •

COOH

....... f---l�� plus two allylic resonance forms

�T H ! propagation step 2 . . . .

:O-OH I • •

�T-""'=� t � � H

foul-smelling aldehydes, ketones , and carboxylic acids = "rancidity"

COOH recycles to begin propagation step J

COOH

COOH

(d) Antioxidants are molecules that stop the free radical chain mechanism. In each of the two cases here, abstraction of the phenolic H makes an oxygen radical that is highly stabi lized by resonance, so stable that it does not continue the free radical chain process. It only takes a small amount of antioxidant to prevent the chain mechanism. Interestingly, in breakfast cereals , the antioxidant is usually put in the plastic bag that the cereal is packaged in , rather than in the food itself. BRA and BHT can also be used directly in food as there is no evidence that they are harmful to humans. • •

:0: :0:

(H3ChC � Y

(CH3hC �c(CH3h Y

OCH 3 CH 3

BRA radical BHT radical 658

Page 663: Solucionario de wade

CHAPTER 26-SYNTHETIC POLYMERS

Note: In this chapter, the " wavy bond" symbol means the continuation of a polymer chain . .JVVVVVVVVV"I..

H H H H I I I I

IVVVV'C-C - C - C· I I I I

H Ph Ph H

10 radical, and not resonance-stabilized­this orientation is not observed

Orientation of addition always generates the more stable intermediate; the energy difference between a 10 radical (shown above) and a benzylic radical is huge. Moreover, this energy difference accumulates with each repetition of the propagation step. The phenyl substituents must necessarily be on alternating carbons because the orientation of attack is always the same-not a random process.

26-2 PhCOO - OOCPh ----I .. � 2 Ph· + 2 CO 2

H CH3 \ /

Ph· + C=C , T!)� \ � H

H CH3H CH3H CH3 I I I I I I

Ph-C-C - C-C - C - C. etc.

I I I I I I H H H H H H

659

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26-3 The benzylic hydrogen will be abstracted in preference to a 2° hydrogen because the benzylic radical is both 3° and resonance-stabilized, and the 2° radical is neither.

benzylic

H H \)H 1 1 ( I 1

I'V'VVV' c -C -C -C ,JVV\J\A

1 1 1 1 Ph H Ph H

middle of a polystyrene chain

H H Ph H 1 I 1 I

.JVVVV C -C -C -C ,JVV\J\A

1 I (k Ph H new benzylic radical

1

H ( H \ rU c==c /

H H H Ph H I 1 1 1

\ Ph

I'V'VVV' C -C -C - C ,JVV\J\A

1 I I I Ph H H

H2C, /H ? o

Ph

+

+

H H H H 1 1 1 1

oC-C-C-CJVVVV\ I 1 I 1 Ph H Ph H

growing polystyrene chain

H H H H I I I I

H-C-C-C-CJVVVV\ 1 1 1 1 Ph H Ph H

tenninated chain

> Ph Ph Ph

Ph Ph

Ph

26-4 Addition occurs with the orientation giving the more stable intennediate. In the case of isobutylene, the growing chain will bond at the less substituted carbon to generate the more highly substitited carbocation.

H Me I 1+

I'V'VVV' C -C

26-5

I 1 H Me

H Me \ /

+ C==C / \

H Me

H Me H Me 1 I 1 1+

----I .. � .JVVVV C -C -C -C 1 I 1 1

H Me H Me

3° carbocation-favored

(a) chlorine can stabilize a carbocation intennediate by resonance H H H H 1 1+ 1 I

.JVVVV C - C .. .. IVVV\I' C -c 1 1 1 I I

H :Cl: H :Cl+

660

OR

H Me Me H I 1 I 1+

IVVV\I'C-C-C-C 1 1 1 1

H Me Me H 1 ° carbocation---disfavored (also more steric hindrance)

Page 665: Solucionario de wade

26-5 continued (b) CH3 can stabilize the carbocation intermediate by induction

H H 1 1 +

IVVVV" C -C 2° (not the best case imaginable, but stil l possible) 1 1 H CH3

(c) terrible for cationic polymerization : both substituents are electron-withdrawing and would destabilize the carbocation intermediate

26-6 benzylic \.

H H H 1 1 1 1

IVVVV" C -C - C - C .JVVVV\ 1 1 1 1 Ph H Ph H

middle of a polystyrene chain

Ph Ph

Ph Ph Ph

H CaaCH3 1 1+

.JVVVV C-C destabilized carbocation

+

1 1 H C N

H 1

H H H H hydride 1 1 1 1 transfer

C-C-C-C.JVVVV\ ---+ 1 1 1 1 Ph H Ph H

growing polystyrene chain

H Ph H 1 1 1

IVVVV" C -C -C -C .JVVVV\ ..

1 1 I 1 H H Ph H H \ I

/ H 2C, H C==C

C/ I \

1 + H Ph

Ph

H H H H 1 1 1 1

H-C - C-C-C.JVVVV\ 1 1 1 1 Ph H Ph H terminated chain

+ H H Ph 1 1 I

H 1

.JVVVV C - C - C -C.JVVVV\ 1 1 + 1 Ph H H

new henzylic cation

Polystyrene is particularly susceptible to branching because the 3° benzylic cation produced by a hydride transfer is so stable. In poly(isobutylene), there is no hydrogen on the carbon with the stabi lizing substituents; any hydride transfer would generate a 2° carbocation at the expense of a 3° carbocation at the end of a growing chain-this is an increase in energy and therefore unfavorable.

2° " H Me H Me 1 1 1 1

JVVVV C - C - C -C.JVVVV\ 1 1 1 1

H Me H Me

Me H Me H

+ 1 1 1 1 + C -C -C -C.JVVVV\ no hydride

30 1 1 1 1 transfer Me H Me H

middle of a poly(isobuty lene) chain growing poly(isobutylene) chain

661

Page 666: Solucionario de wade

26-7 . . :0 : :0: I I I

H C-OCH3 H C-OCH3 I _I I I I

I\IVVV' C -C : ....... If---l .. � JVVVV C -C I I I I

H H H H

26-8 0 :0 : II

: 0 : :0 : I II II

H COCH3 H C-OCH3 H C-OCH3 H C-OCH, \ nl

HO: C=C ----• • '----"I \

I _I HO-C-C:

I I I I I

�---l"'� HO -C -C I I

I I ....... !---l ... � HO-C-C

I I I H C N H C N: H C-N: H C=N:

: 0 :

o I I COCH3

\ nl C=C

I \ H C_N

. . -:0:

. .

:0: COOMe I I COOMe I COOMe "

H I H C-OCH3 H I H C-OCH3 H I H C-OCH3 I I _I I I " I I I

HO-C-C-C-C: • ... HO-C-C-C-C ...... !---l ... �HO-C-C-C-C I I I I I I I I I I I I I

H C N H C _ N: H CN H C N: H CN H C = N: ! etc.

COOMe COOMe COOMe COOMe COOMe

CN CN CN CN CN

This polymerization goes so quickly because the anionic intermediate is highly resonance stabilized by the carbonyl and the cyano groups . A stable intermediate suggests a low activation energy which translates to a fast reaction.

662

Page 667: Solucionario de wade

26-9

(a) H H H H I I (I I

rvvVV' C - C -C - C .JVVVV'\ I I I I

CN H CN H middle of a poly(acrylonitrile) chain

H H CN H I I _ I I

..JVVVV C - C - C - C .JVVVV'\ + I I I

CN H H

H H \ nl C=C

I \ H CN

H H CN H I I I I

rvvVV' C - C - C - C .JVVVV'\ I I I I

CN H H

H2C,

""/H C 1-CN

!

H H H H - I I I I :C - C-C-C.JVVVV'\

I I I I CN H CN H

growing poly(acrylonitri le) chain

H H H H I I I I

H-C-C-C-C.JVVVV'\ I I I I

CN H CN H

terminated chain

> CN CN

CN

CN

CN

CN

(b) The chain-branching hydride transfer (from a cationic mechanism) or proton transfer (from an aniontc mechanism) ends a less-highly-substituted end of a chain and generates an intermediate on l more-highly­substituted middle of a chain (a 3° carbon in these mechanisms). This stabilizes a carbocation, but greater substitution destabilizes a carbanion. Branching can and does happen in anionic mechanisms, but it is less like ly than in cationic mechanisms.

26- 1 0 isotactic poly(acrylonitrile)

NC H NC H NC H NC H

syndiotactic polystyrene

663

Page 668: Solucionario de wade

26-1 1

(a) all trans

(b) The trans double bonds in gutta-percha allow for more ordered packing of the chains, that is, a higher degree of crystallinity. (Recall how cis double bonds in fats and oils lower the melting points because the cis orientation disrupts the ordering of the packing of the chains.) The more crystalline a polymer is, the less elastic it is.

26-12 Whether the alkene i s cis or trans is not specified.

H Me H Me H H

I I I I I I � C -- C--C -- C==C -- C �

\

I I I I H Me H H

Y I \. Y

isobutylene isoprene

26-13 The repeating unit in each polymer is boxed.

(a) Nomex®

)

o 0 0 0

IVV'g--o-� g- �--o-� � -g--o-� g - �--o-� �.NV' � H � H � H � H

26-1 4 Kodel® polyester (only one repeating unit shown)

-0 �-o-� .NV' 0 - CH 2 CH2 - 0 - C � !J C.NV'

664

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26-15

o CH,o-"'"

o H 1 1 -0-0

I

-C '/ '\ II I \.

I """"

O II -0-\\ 0 2C-O -C

II CHO

I c-o \\ !J C-O -CH

� _ -rn

2

I I 0

CHOJVV\ H C

11-0-0 C

I

2 -O -C II

H,o-

CH20

� h C-O -CH

,

I CH20

Glycerol is a trifunctional molecule , so not only does it grow in two directions to make a chain, it grows in three directions. All of i ts chains are cross-linked, forming a three-dimensional lattice with very little motion possible. The more cross-l inked the polymer is, the more rigid it is.

26-16 For simplicity in this problem, bisphenol A will be abbreviated as a substituted phenol.

-o-� b�OH HO I � __ --CH3

bisphenol A

mechanism a: .. �OH C I .. · .. >C'CI R�

• • � --

o II

OO/C'OU 10 I �

R R

R

represented as R-o-OH

:?) 0C

1'CI Cl + o .. , H

-HCI -C I-- .---

as above

665

-C l---

� I :R: \.-·CI· H , • • /C, • •

�9+ CI

R

(?¢CI

HCI (g) +

:�D C

O/.'CI

R H 1

:0:

/C'o·u o +1 ......; H I �

R

R

R

Page 670: Solucionario de wade

26-17 Bisphenol A i s made by condensing two molecules of phenol with one molecule of acetone, with loss of a molecule of water. This is an e\ectrophi lic aromatic subsitution (more specifical ly , a Friedel-Crafts alkylation) , and would require an acid catalyst to generate the carbocation. While a Lewis acid could be used, the mechanism below shows a protic acid.

HO

from phenol

plus four resonance forms

from acetone

OH

from phenol

...

_ �H CH3 B O / / ---0-//C� I ...

0 HO '/

C-OH

666

- I CH3 plus three resonance forms

Page 671: Solucionario de wade

26-18

Ph-N = CQ � �R-Et i

..

- }

:0 : :0 : • • I • • ..

- I I • • Ph-N = C-O-Et � Ph-N -C-O -Et +k cf. .. ;i?

H -O- Et f + 1) H-O-Et H • •

:0 : I I

Ph-N-C-O-Et I H

two rapid proton transfers

26-19 Glycerol is a trifunctional alcohol . It uses two of its OH groups in a growing chain . The third OH group cross-links with another chain. The more cross-linked a polymer, the more rigid it is .

26-20 urethanI linkage r ,

Me 0 H H 0 -O--I� II I I I I .1\1\1'0 � /; T� O-C-N� N-C .N\r

Me � \... ) \. Me )

V y bisphenol A toluene diisocyanate

26-21 Please refer to solution 1-20, page 12 of this Solutions Manual.

26-22

(a) H Me H Me H Me I I I I I I

.rvvvv C-C-C-C-C-C .I\I\I' I I I I I I

H Me H Me H Me

(b) Polyisobutylene is an addi tion polymer. No small molecule is lost, so this cannot be a condensation polymer.

(c) Either cationic polymerization or free-radical polymerization would be appropriate . The carhocation or free-radical intermediate would be 3° and therefore relatively stable. Anionic polymerization would be inappropriate as there is no electron-withdrawing group to stabil ize the anion .

26-23

(a) It is a polyurethane.

(b) As with all polyurethanes, it is a condensation polymer.

(c) o I I

.N\r CH2CH2CH2 - N -C -0.1\1\1' I H

HOCH2CH2CH2NH2 + CO2

667

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26-24

(a) It is a polyester.

(b) As with all polyesters, it is a condensation polymer.

(c)

Using the dicarboxylic acid instead of the ester would produce water as the small neutral molecule lost in this condensation.

26-25

(a) Ury lon® is a polyurea.

(b) A polyurea is a condensation polymer.

(c) ° II H20

.JVV'- (CH ) - N - C - N JVV' � 2 9 I I

26-26

H H

(a) Polyethylene glycol , abbreviated PEG, is a polyether.

(b) PEG is usually made from ethylene oxide (first reaction shown). In theory, PEG could also be made by intermolecular dehydration of ethylene glycol (second reaction shown), but the yields are low and the chains are short.

n

° U ethylene oxide

+

� ",OH HO "'" �

ethylene glycol

HO-

(c) B asic catalysts are most l ikely as they open the epoxide to generate a new nuc\eophi le . Acid catalysts are possible but they risk dehydration and ether cleavage.

668

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26-26 continued

(d) Mechanism of ethylene oxide polymerization (showing hydroxide as the base) :

/"""-.... ..,0 .... /"""-.... 0 0-HO' � � '0 :

f))" etc. /"""-.... .., 0 .... /""'0.... /"""-.... ..,0 .... /"""-.... 0 0-

HO' � � '0 ' � 'v/ ' 0 :

26-27 (a) Polychloroprene (Neoprene®) is an addition polymer. (b) Polychloroprene comes from the diene, chloroprene, just as natural rubber comes from isoprene:

26-28

H Cl \ I C -C chloroprene

II \\ H2C CH2

(a) H H H H 1 1 1 1

oJV\r C -0 -C - 0 -C -O-C - OoJV\r Delrin® (polyformaldehyde) 1 1 1 1 H H H H

(b) All of these intermediates are resonance-stabilized.

H /

H+ O=C---'---/ \ H

H

• 1 +

H-O-C",---, • • I , H /

O=C H

\ H

H H 1 1 +

H-O-C-O-C� 1 0 0 1

H H H '\0 / O=C 0 0 \

H H H H

etc. I I 1+ 4 H-O-C-O-C-O-C

I I I H H H trimer

(c) Delrin® is an addition polymer; instead of adding across the double bond of an alkene, addition occurs across the double bond of a carbonyl group.

669

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26-29

(a) cis

trans

(b) Each structure has a f ully conjugated chain. It is reasonable to expect e lectrons t o be able to be transferred through the 1t system, just as resonance effects can work over long distances through conjuga ted systems. (c) It is not surprising that the conductivity is directional. Electrons must flow along the 'It system of the chain, so if the chains were aligned, conductivity would be greater in the direction parallel to the polymer chains. (It is possible, though less likely, t hat electrons could pass from the 'It system of one chain to t he 'It system of another, that is, perpendicular to the direction of the chain; we would expect reduced conductivity in that direction.)

26-30

(a) A Nylon is a polyamide. Amides can be hydro lyzed in aqueous acid, cleaving the polymer chain in t he process. 0 0 0 0

I I I I H30+ + II I I + .JVV' N -C JV'V\. C -N .JVV' ---I'-� .JVV' NH 3 + HO -C .JVV' C -OH + H 3N .JVV'

I I H H

(b) A po lyester can be saponified in aqueous base, cleaving the polymer chain in the process . o 0 0 0 I I I I NaOH _ I I I I

.JVV'O -C'VVV'C -O.JVV' .. .JVV' OH + 0 -C.JVV'C -0 + HO.JV\r

26-31 0 0 0 I I I I I I (a) CCH3 CCH3 CCH3 I I I

H 0 H 0 H 0 H OH H OH H OH I I I I I I H2O I I I I I I

.JV\rC -C -C -C -C -C.JV\r .- .JVV'C -C-C -C -C -C.JV\r I I I I I I H+ or I I I I I I

H H H H H H HO- H H H H H H poly (vinyl acetate) poly (vinyl alcoho l)

(b) A polyester is a condensation polymer in which monomer units are linked through ester groups as part of the polymer chain . Poly(vinyl acetate) is rea lly a substituted polyethylene, an addition polymer, wi th only carbons in the chain ; the ester groups are in the side chains, not in t he polymer backbone .

( c) Hydrolysis of t he esters in poly (vinyJ acetate) does not affect the chain because the ester groups do not occur in the chain as they do i n Dacron® .

(d) Viny l a lcohol cannot be polymerized because it is unstable, tautomerizing to aceta ldehyde. OH I

H2C==CH --

o I I

CH3-CH

670

Page 675: Solucionario de wade

26-32 OCOCH3 (a) �

cellulose acetate

o I

COCH3

(b) Cellulose has three OH groups per gl ucose monomer, which form hydrogen bonds with other polar groups . Transforming t hese OH groups into acetates makes the polymer much less polar and therefore more soluble in organic solvents . (c) The acetone dissolved t he cellulose acetate in the fibers . As the acetone evaporated, t he cellulo se acetate remained but no longer had t he fibrous, woven structure of clot h . It recrystallized as white fluff . (d) Any article of clothing made from synthetic fibers is susceptible to the ravages of organic solvents . Solvent splas hes leave dimples or blotc hes on Corfam shoes . (Yet, Corfam shoes co uld still provide protection for t he toenail polish!) 26-33 OH

mechanism

..

l-w

OH OH OH

plus four resonance forms

Bakelite is highly cross -linked through t he ortho and para positions of phenol; each phenol can form a chain at two ring positions, then form a branch at the third position.

H

H°-o- i-o-°H H

further coupling at ortho positions leads to cross-linked Bake lite

671

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26-34 ·0·

i

·0· .O • • ;} . . . . . . II I I

....... C, . ..... H ....... C,.: .... H ....... C� .... H H2N � N . .. H2N N .. .. H2N N \.. I .NH3 '--__ --_ H.J � :0: II

....... c, .. -H2N 7,) H-C-H

C6H

I :0 :

....... C, H2N 'N: I H-C-H I OH

� H ' c / H l cg

:0: II H3N: ....... C, ...... H ....... C, ...... H

.. H2N N:...J .. + H2N N

-HO-

I H-C-H I OH

:0 : I I

H3N-H I U" H-?-H � :o: ..

from above :0:

I I ....... C, ....... C, • • -.... H

.. H2N eN: + H2N N gA-" �

H""'" 'H �

:0: :0: a I I II I I

....... C,.. .......C, •• � ....... C, this leads to H2N C N + H2N N N NH2 cross-linking .............. g�

I w � H""'" 'H H

/""" 0 plus another resonance form o II II

....... C,................. .......C, H2N N N NH2 HN

) A I

C:--, H2N""'" "0

672

Page 677: Solucionario de wade

2

6-3�O�

O�O�

O�O�

O-y�

o 0 0 '-----T---" '-----T---" '-----T---" '----r-' '-----T---" '-----T---"

glycolic lactic glycolic lactic glycolic lactic acid acid acid acid acid acid

26-36 OH

� OH OH

OH O�

cellulose = OH cotton OH

polypropylene

As we have seen repeatedly through this presentation of organic chemist ry, physical and chemical behavior depend on structure . The structure of cotton, i.e. cellulose, has multiple oxygen atoms that form hydrogen bonds with water. When cotton gets wet, it holds onto the water tightly, as you have seen if you have put cotton clothes in a clothes dryer-it takes a long time to dry. Polypropylene is a hydrocarbon with no hydrogen bonding groups; the fiber feels dry because it cannot hold the water the way cotton can. Athletic garments are increasingly using polypropylene because they allow evaporation and cooling during periods of exertion ; cotton is just the opposite .

26-37 (a) This addition polymer is called polyvinylidene chloride, trade name Saran®. It could be made by any of the three mechanism types : radical, cationic, or anionic .

(b) When the substituent is on every fourth carbon, and one double bond in the chain in every 4-carbon unit, the polymer mus t come from addition across a diene, probably under ca tionic conditions.

(c) This polyester is a condensation polymer of two monomersO a diol and a derivative of phthalic acid, either the anhydride, an ester, the acid chloride, or the acid itself. Heating the monomers wil l make the polymer; no catalyst is required if done at high temperature.

HO---<>--OH

monomer 673

Cl >=== Cl

H3CO

h 0

one of these is the other monomer

monomer

monomer

X X 0

X = OH orCl or OR

0

Page 678: Solucionario de wade

26-37 continued (d) This polyamide (Nylon) is a condensation polymer made from two monomers, a diamine and a derivative of succinic acid, either the anhydride, an ester, the acid chloride, or the acid itself. Heating the monomers will make the polymer; no catalyst is required if done at high temperature.

H2N�NH2

monomer

o�o

one of these is the other monomer

26-3 8 o (a) COOH }-OCH2CH20H

X=OH or Cl or OR

===< + HO .... /"".... ............, 'OH =====\ hydroxyethyl methacrylate

CH3 CH3 j polymerize using radical or anionic conditions

polymer

(b) This polymer has a few properties that make it useful as the material in soft, extended-wear contact lenses. First, carboxylic acids usually are crystalline solids with high melting points, but esters and alcohols are low melting, often liquids, so the polymer with this ester is softer than the carboxylic acid or even the methyl ester. (The methyl ester, polymethyl methacrylate or Plexiglas, was the first material used in the original hard contact lenses.) Second, the ability of the free OH to form hydrogen bonds with water makes the contact lens more fluid and less irritating to the cornea. Third, a hidden advantage but very important for ocular health: the fluidity of the contact lens also permits oxygen to go through the lens. Because the cornea does not have a large blood flow, it needs to absorb oxygen from the air to maintain its health, and this enhanced gas permeability permits the contact lens to be worn for days at a time without compromising the health of the cornea. Thanks, polymers!

Note to the student: BON VOYAGE! I hope you have enjoyed your travels through organic chemistry.

Jan Simek

674

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Appendix I-Summary of IUPAC Nomenclature ofOq:anic Compounds Introduction The purpose of the IUPAC system of nomenclature is to establish an international standard of naming compounds to facilitate communication. The goal of the system is to give each structure a unique and unambiguous name, and to correlate each name with a unique and unambiguous structure.

I. Fundamental Principle IUPAC nomenclature is based on naming a molecule's longest chain of carbons connected by single bonds, whether in a continuous chain or in a ring. All deviations, either multiple bonds or atoms other than carbon and hydrogen, are indicated by prefixes or suffixes according to a specific set of priorities. II. Alkanes and Cycloalkanes (also called "aliphatic" compounds)

Alkanes are the family of saturated hydrocarbons, that is, molecules containing carbon and hydrogen connected by single bonds only. These molecules can be in continuous chains (called linear or acyclic), or in rings (called cyclic or alicyclic). The names of alkanes and cycloalkanes are the root names of organic compounds. Beginning with the five-carbon alkane, the number of carbons in the chain is indicated by the Greek or Latin prefix. Rings are designated by the prefix "cyclo". (In the geometrical symbols for rings, each apex represents a carbon with the number of hydrogens required to fill its valence.)

CI C2 C3 C4 Cs C6 C7 Cs C9 CIO Cll

CH4 CH3CH3 CH3CH2CH3 CH3[CH2hCH3 CH3(CH2hCH3 CH3(CH2]4CH3 CH3(CH2lsCH3 CH3[CH2]6CH3 CH3[CH2hCH3 CH3[CH2lsCH3 CH3[CH2]9CH3

D cyclopropane

0 cyclohexane

methane ethane propane butane pentane hexane heptane octane nonane decane undecane

H , , C

H

> H / \ H 'C-C'" I \ H H

C12 C13 CI4 C20 C21 C22 C23 C30 C31 C40 Cso

D

CH3(CH2] IOCH3 CH3(CH2] II CH3 CH3(CH2] 12CH3 CH3(CH2] ISCH3 CH3[CH2]19CH3 CH3(CH2hoCH3 CH3[CH2hICH3 CH3[CH2hsCH3 CH3(CH2h9CH3 CH3(CH2hsCH3 CH3[CH2]4SCH3

dodecane tridecane tetradecane icosane henicosane docosane tricosane triacontane hentriacontane tetracontane pentacontane

0 cyclobutane cyclopentane

0 0 cycloheptane cyclooctane

The IUPAC system of nomenclature is undergoing many changes, most notably in the placement of position numbers. The new system places the position number close to the functional group designation; however, you should be able to use and recognize names in either the old or the new style. Ask your instructor which system to use.

675

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Appendix 1, Summary of IUP AC Nomenc lature, continued

III. Nomenclature of Molecules Containina: Substituents and Functional Groups

A . Priorities of Substi tuen ts and Functional Groups LISTED HERE FROM HIGHEST TO LOWEST PRIORITY, except that the s ubsti tuents within Group C have equivalent priority .

Group A-Functional Groups Named By Prefix Or Suffix

Functional Group Structure Prefi x Suffi x 0 II

Carboxylic Acid R - C-OH carboxy -0 \I

Aldehyde R -C-H oxo-0 (formyl) \I

Ketone R -C-R oxo-

Alcohol R-O -H hydroxy-

/ Amine R-N amino-

"

Group B-Functional Groups Named By Suffix Only

Functional Group

Alkene

Alkyne

Struct ure

\ / C=C

/ \ -C:::C-

Group C--Substituent Groups Named By Prefix Only

Substituent Struct ure

Alkyl (see next page) R- alkyl-

Alkoxy R -0 - alkoxy-

-oic acid (-carboxylic acid)

-al (carbaldehyde)

-one

-01

-amine

-ene

-yne

(alkoxy groups take the name of the alkyl group, like methyl or ethyl, drop the "yl", and add "oxy"; CH30 is me thoxy; CH3CH20 is ethoxy) Halogen F­

CI­Br -1-

fluoro­chloro­bromo­iodo-

Miscellaneous s ubsti tuents and their prefixes

-N02 ni tro

-CH=CH2 vinyl

-CH2CH=CH2 allyl

676

< > phenyl

Page 681: Solucionario de wade

Appendix 1, Summary of IUPAC Nomenclat ure, continued Common alkyl groups-rep lace "ane" ending of alkane name with "yl". A lternate names for complex substit uents are given in brackets. CH3

I CH3

I -CH3

methyl -CH -CH -CH2CH3

ethyl

-CH2CH2CH3 propy l (n-propyl)

-CH2CH2CH2CH3 butyl (n-butyl)

\ CH3 isopropyl [ 1 -methylethyl]

isobutyl [2 -methylpropyl]

\ CH2CH3 sec-butyl [ I -methylpropyl]

CH3 I -C-CH I 3 CH3

t-butyl or ter t-butyl [ 1,I -dimethylethyl]

B . Naming Substit uted Alkanes and Cycloalkanes-Group C Substit uents Only

Organic compounds containing substituents from Group C are named following this sequence of steps, as indicated on the examples below :

·Step l. Find the longest continuous carbon chain . Determine the root name for this parent chain. In cyclic compounds, the ring is us ually considered the parent chain, unless it is attached to a longer chain of carbons ; indicate a ring with the prefix "cyclo" before the r oot name. (When there are two longest chains of equal length, use the chain with the greater n umber of s ubstituents .)

·Step 2. Number the chain in the direction such that the position number of the first substituent is the smaller number. If the first substituents have the same number, then number so that the second substit uent has the smaller number, etc.

·Step 3. Determine the name and position number of each s ubstit uent. (A substituent on a nitrogen is designated with an "N" instead of a n umber ; see Section 111.0. 1 . below .)

·Step 4. Indicate the n umber of identical groups by the prefixes di, tri, tetra, e tc. ·Step 5. Place the position numbers and names of the substituent groups, in alphabetical order,

before the root name. In alphabetizing, ignore prefixes like sec -, tert -, di, tri, etc ., b ut include iso and cyclo. Always include a position number for each substit uent, regardless of redundancies .

CH3 CH2CH2CH3 CH2CH2CH3 I 2 3 1 1 I CH3 -CH -CH 4C -sCH -I CH2CH3 l-. __ 1 __ 1 __ 1 __ -"

1 1 CH3 -TH TFH -3T_H�_ 2T _H _CH_3-, CI B r CH3 CH3 F CH3

3 -bromo-2-chloro- 5-ethyl-4,4-dimethyloctane 3 -fluor o-4-isopropyl-2-methylheptane H3Ca-lCHCH2CH3 6 2 5 3 N02 4

I-sec-butyl-3-nitrocyclohexane (numbering determined by t he alphabetical order of s ubstit uents, "b" comes before "n")

677

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Appendix 1 , Summary of IUPAC Nomenclature , continued C . Naming Molecules Contain ing Functional Groups from Group B-Suffix Only

1 . Alkenes-Follow the same steps as for alkanes , except : a. Number the chain of carbons that includes the C= C so that the C=C has the lower pos it ion number, since it has a higher priority than any substituents; b. Change "ane" to "ene" and assign a position number to the first carbon of the C=C ; place the position number just before the name of functional group(s); c . Designate geometrical isomers with a cis, trans or E,Z pre fix .

F \ H 2 I

4 CH -C - C =CH2 / 3

1 H F CH3

4,4-di fluoro-3 -methylb ut - l -ene

F \ 2 3 4 I C ==C - C = CH2 I I H

F CH3 1 , I-di fluoro-2 -methyl­buta- l ,3 -diene

2 3 5-methylcyclopenta -1 ,3-diene

Special case : When the chain cannot include an alkene , a s ubstituent name is used. See Numbering must be on EITHER Section V.A.2.a. 0,2

6 C = CH2 3 H 3 . I I h 1 a ring OR a chain, but not both. -vmy cyc 0 ex- -ene

5 4 2. Alkynes-Follow the same steps as for alkanes , except : a . Number the chain of carbons that includes the C-C so that the alkyne has the lower pos ition number ; b . Change "ane" t o "yne" and assign a position number to the first carbon of the C=C ; place the position number just before the name of functional group (s). Note : The Group B f unctional groups (alkene and alkyne) are considered to have equal priori ty : in a molecule with both an ene and an yne , whichever is closer to the end of the chain determines the di rection of numbering. In the case where each would have the same position n umber, the alkene takes the lower number. In the name , "ene" comes before "yne" because of alphabetization .

F \ H 2 I CH - C - C == CH

4; 3 / -F CH3

4,4-difluoro-3 -methylbut - l-yne

HC = C - C = C - CH -H H

3

pent -3-en- l -yne ( "yne" closer to end of chain)

H2 HC C -C -C =CH2

H pent- l-en -4-yne

( "ene " and "yne " have equal priority un less they have the same posit ion n umber, when "ene" takes the lower number)

(Notes : 1 . An "e" is dropped if the lette r following it is a vowel: "pent-3 -en- l-yne" , not "pent -3-ene- l-yne" . 2 . An "a" is added if inclusion of di , tri, etc. , would put two consonants togethe r: "buta- l ,3 -diene" , not "but - l ,3 -diene".)

D. Naming Molecules Containing Functional Groups from Group A-Pref ix or Suffix

In naming molecules containing one or more of the f unctional grou ps in Group A, the group of highest pr iority is indicated by suffix ; the others are indicated by p refix , with priority equivalent to any other substituents . The table in Section lIlA. defines the priorities ; they a re discussed on the following pages in order of increasing pr iority .

678

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Appendix 1 , Summary of IUP AC Nomenclature, continued Now that the functional groups and substituents from Groups A , B , and C have been described, a modified set of steps for naming organic compounds can be app li ed t o all simple structures :

-Step 1 . Find the hi ghest priority functional group. Detennine and name the longest continuous carbon cha in that includes this group.

-Step 2 . Number the c hain so that the highest priority functional grou p is assigned the lower number . (The number " I " is oft en omitted when there is no confusion about where the group must be . Aldehydes and carboxylic acids must be at the first carbon of a chain, s o a " I " is rarely used with those functional groups.)

-Step 3 . If the carbon chain includes multiple bonds (Group B) , rep lace "ane" with "ene" for an alkene or "yne" for an alkyne. Designate the position of the multiple bond with the number of the first carbon of the multiple bond .

-Step 4. If the molecule includes Group A functional groups, replace the last "e" with the suffix of the highest priority functional group , and include its position number just before the name of the highest priority functional gr oup .

-Step 5 . Indicat e all Group C substituents, and Group A functional groups of lower priority, with a prefix . Place the prefixes , with appropriate position numbers, in alphabetical order before the root name. 1. Amines : prefix : amino-; suffix : -amine-substituents on nitrogen denoted by "N'

propan - l -amine

CHPyY NH2 CH3CH, ,�/CH2CHJ

V H2C =C -CHCH3 3-methoxycyclohexan - l -amine (" 1 " is optional in this case)

H N ,N-diethylbut-3 -en -2 -amine

2 . Alcohols : prefix : hydroxy- ; su ffix : -0 1 c(0H

ethanol

OH I

H3C-C -C =CH2 H H

but -3 -en-2-01

3. Ketones: prefix : oxo- ; suffix : -one (pronounced "own ")

o I I

CH3-CH -C -CH3 I

o

NH2 2 -aminocyc lobutan - l-01 ( " 1 " is optional in this case)

H3C, /CH3 o N I I I

CH3 -C - CH2 - C = CH2 OH 6 4- (N ,N-dimethylamino )pent A-co-2-one

3-h ydrox ybutan -2 -one cyclohex-3-en- l -one (" 1" is optional in this case)

4. Aldehydes : prefix : oxo-, or formyl - (O=CH-); suffix : -al (abbreviation : --CHO)

An aldehyde can only be on carbon I , so the " 1" is generally omitted from the name. o I I

HCH

methanal ; formaldehyde

o "

CH3 -CH

ethana l; acetaldehyde

OH 0 I "

H2C-�=�-CH

4-hydrox ybut -2-enal

679

o 0 I I "

CH3CCH2CH2CH

4-oxopentana l

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Appendix 1, Summary of IUP AC Nomenclature , continued Specia l case : When the chain cannot inc lude the carbon of the aldehyde , the suffix "carba ldehyde " is used: 3 2 a

4 I CH 0-11

cyc lohexanecarba ldehyde 5 6

5 . Carboxylic Acids : prefix : carboxy- ; suffix : -oic acid (abbreviation : -COOH) A carboxy l ic aci d can on ly be on carbon 1 , so the "1" is genera l ly omitted from the name . (Note : Chemists traditiona l ly use , and IUPAC accepts , the names "formic acid" and "acetic ac id" in p lace of "methanoic acid" and "ethanoic acid" .)

a I I

HC - OH methanoic ac id; fo rmic acid

a I I

CH3C - OH ethanoic acid; acetic acid

a O� CH - CH -gOH _ 2 I

NH2

a a CH I I I I I 3

HC - C - C - COOH I

CH3 2 -amino-3 -pheny lpropanoic aci d 2 ,2 -dimethy l-3 ,4-

dioxobutanoic acid

Specia l case : When the chain numbering cannot inc lude the carbon of the carboxy lic acid, the suffix "carboxy lic ac id" is used: CHO K,

O�COOH

2-formy l-4-oxocyclohexanecarboxylic ac id ( "formy l" is used to indicate an aldehyde a s a substituent when its carbon cannot h e in the chain numbering)

5 6

E . Naming Carboxylic Acid Derivatives The six common groups derived from carboxylic acids are , in decreasing priority after carboxylic aci ds : s al ts , anhydrides, esters , acyl ha lides , amides , and nitriles .

1. Sa lts of Carboxy lic Acids Sa lts are named with cation first , fol lowed by the anion name of the carboxy lic acid, where " ic acid" is rep laced by "ate" :

becomes acetate becomes butanoate

acet ic acid butanoic acid

cyc lohexanecarboxy lic acid becomes cyc lohexanecarboxy late

NH2 I

CH3 - CHCOO- Li +

l ithium 2 -aminopropanoate CICH2COO- Na+

sodium ch loroacetate

2 . Anhydrides : a

"oic acid" is replaced b y "oic anhydride "

I I R - C - OH alkanoic acid

> a a I I I I

R - C - O - C - R a lkanoic anhydride

680

CH30

UCOO­

ammonium 2-methox y­cyclobutanecarbox y late

a a �O� benzoic anhydride

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Appendix 1 , Summary of IUPAC Nomenclature, continued 3 . Esters Esters are named as "organic salts" that is, t he alkyl name comes first, f ollowed by the name of t he carboxylate anion . (common abbreviation : -COOR)

a CH3 a CH3 I I I : I ( carbJ,xylate \ �

a I I

R - C - O � R

I I H3C - C - O + CH2CH3

H C - C - C - O � CHCH 3 I :

3

CH3 "a lkanoate" . "alky l" "alkyl alkanoate"

ethyl acetate isopropyl 2,2 -dimethylpropanoa te a I I

a I I .

H2C = C - C - 0 7" C = CH2

HOVC - �H3 < }-CH2COO +-0 H . H

viny l prop -2 -enoate methyl 3 -hydroxycyc lo­pentanecarbox y late cyclohexyl 2 -pheny lacet at c

4. Acyl Halides : "oic acid" is replaced by "oyl halide" a a a a �Cl I I I I �CI

R - C - OH > R - C - Cl butanoyl I � benzoyl alkanoic acid alkanoyl ch loride chloride ch loride

5. Amides : "oic ac id" is replaced by "amide" a 0

a a � NH2 I I I I (i.:NH2 R - C - OH > R - C - NH2 I � benzamide alka no ic acid a lkanamide butanamide Amides are notable for their role in biochemistry, i .e . , the special amide bond between two amino acids is called a peptide bond.

6. Nitriles : "oic acid" is replaced by "enitri le" a I I

R - C - OH alkanoic acid alkanenitrile

IV. Nomenclature of Aromatic Compounds

� C = N

butanenit rile V:�z:ni l'i IC

(common spelling di ffers from IUPAC)

"Aromatic" compounds are those derived from benzene and similar ring systems . As with aliphatic nomenclature described above , the process is : determining the root name of t he parent ring ; determin ing pr iority, name, and pos iti on number of subst ituents ; a nd assembling the name in alphabetical order . Functional group priorities are the same in aliphatic and aromatic nomenclature. See p . 676 f or the list of priorities . A . Common Parent Ring Systems

benzene

(a) 70)8 1 2 (t})

I '-':: '-':: 6 � � 3

5 4 naphthalene

681

8 9 1 7 � 2

6� 3 5 1 0 4

a nthracene

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Appendix 1 , Summary of IUP AC Nomenclature , continued B . Monosubstitute d Benzenes 1 . Most substituents keep the ir designation , followed by the word "benzene " :

6

6 8

CH3

chlorobenzene nitrobenzene ethylbenzene

2 . Some common substituents change the root name of the r ing . IUPAC accepts these a s root names , li sted here in decreasing p riority (same as Group A, p . 676):

COOH

6

H CHO OH

6 6

3

6 6 6 6 benzoic benzene - benzaldehyde phenol acid su lfonic aci d

C. Disubstituted Benzenes 1. Designation of sub stitution--only three possibilities :

common: IUPAC:

X X �Y � U �y ortho- (0-) meta- (m-) 1 ,2 - 1 ,3 -

aniline anisole toluene

X

¢ y

para- (p-) 1 ,4-

2 . Naming disubstituted benzenes-Priorities from Group A, p . 676 , dete rmine root name and substituents

Br { }- Br COOH

Ql H0'Q

6 I � � OCH3 � NH2 CHO CH] p-dibromobenzene m-aminobenzoic acid o-methoxybenzaldehyde m-methylphenol 1 ,4-dibromobenzene 3 -aminobenzoic acid 2 -methoxybenzaldehyde 3 -methylphenol

D. Polysubstituted Benzenes-must use numbers to indicate subs ti tuent positi on

r(YCI

HNVCl I CH3

3 ,4-d ichloro-N-methy lan i line

CH3 02NAN02

Y N02

2 ,4,6 -trinit rotoluene (TNT)

682

�OCH2CH3

YOH

NH2 ethy I 4-amino-3 -hydroxybenzoate

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Appendix 1, Summary of IUPAC Nomenclat ure, conti nued

E. Aromatic Ketones A special group of aromatic compounds are ketones where the c arbonyl is att ached t o at lea st one benzene ring. Such compounds are named as "phenones ", the prefix depending on the si ze and nature of the group on the other side of the carbonyl. These are the common examples :

o < }-C-CH3 o < }-C - CH2CH3 acetophenone propiophenone

benzophenone

V. Nomenclature of Bicyclic Compounds

"Bicyclic" compounds are those that contai n two ri ngs. There are f our possible arrangements of two ri ngs that depend on how many atoms are shared by the two rings . The first arrangement in whic h the ri ngs do not share any atoms does not use any special nomenclat ure, but the other t ypes require a

method to designate how the ri ngs are put t ogether. Once the ring s ystem is named, then functional groups and substit uents follow the standard rules described above.

Type 1 . Two rings with no common atoms These follow the standard r ules of choosing one parent ring system and describing the other ring as a s ubstituent . 5 4

6y,D o

ketone is the highest priorit y functional group, phenyl is substi tuen t c::==�> 3-pheny Icyclohexan - l-one (" 1" could be omit ted here)

benzene is the parent ring system as it is larger than c yclopentane and it has three substituents

c::=�> l-cyclopentyl-2,3-dinitr obenzene

Type 2. Two rings wi th one common atom-spiro ring system The ring s ystem in s piro compounds is indicated by the wor d "spiro" (instead of "cyelo"), followed by brackets i ndicating how many atoms are contained in each path around the rings , ending with the alkane name describing how many carbons are in the ring systems including the spiro carbon. (If any atoms are not carbons, see section VI.) Numbering follows the smaller path first, passmg through the sprio carbon a nd around the second ring.

S 1

:00, 5 3

spiro[3.4 ]octane

683

4 6 7 spiro[4.5 ]decane

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Appendix 1, Summary of IUPAC Nomenclatu re , continued Substituents and func tional groups are indicated in the usual ways . Spiro ring systems are a lways numbered smaller before larger, and numbered in such a way as to give the highest priority functional g roup the lower position number. I 10 9 o 7 r\A 2

6 l0V 5 3

8 '---+-CH3

CH3 spiro [3.4 ]oct -5-ene 7,7 -d imethylspiro [4.5 ]decan -2 -one

Type 3 . Two rings with t wo common a tom-fused ring sys tem Two rings tha t sha re two common atoms are called fused rings. Th is ring system and the next type called bridged rings share the same des igna tion of ring system. Each of the two common a toms i s called a b ridgehead atom, and the re a re three paths between the two bridgehead atoms . In contrast with naming the spi ro rings, the longer path is coun ted first, then the shorter, then the shortest. in fused rings, the shortest pa th is always a zero, meaning ze ro atoms be tween the two bridgehead atoms . Numbering s tarts at a b ridgehead, continues around the largest ring, through the other bridgehead and around the shorte r ring . (In these structures, bridgeheads are marked with a dark ci rcle for cla rity .) 2

9COL O 1 2

3

4 bicyclo [2 . i .O]pen tane (path of 2 atoms and a

path of I atom)

8 4 6 7 5

b icyclo [4.4.0]decane (path of 4 atoms in

each d irection)

9 4 6 bicyclo [5.3 .0]decane (path of 5 atoms and

path of 3 atoms)

Substituents and func tional groups a re indica ted in the usual ways . Fused rings systems are always numbered larger b efore smaller, and numbered in such a way as to give the h ighest priority functional group the lower posi tion number. Type 4. Two rings with more than two common atom-bridged ring system Two rings that share more than two common a toms are called bridged rings . B ridged rings "hare the same designation of rin g system as Type 3 in which the re are three paths be tween the tw'.) bridgehead atoms. The longer path is counted f irst, then the medium, then the shortest . Numbering starts at a b ridgehead, continues around the la rgest ring, through the other bridgehead and around the medium path, ending with the shortes t path numbered from the original b ridgehead a tom . (In these s tructures, b ridgeheads a re marked with a dark circle for clarity .) �6

4 �8

3 4 5 3

2 1 5 6 1

b icyclo [2. 1 . 1 ]hexane (paths of 2 atoms, I atom, and 1 atom)

2 bicyc 10[2 . 2 . 2 ]oc tane (three paths of 2 atoms)

684

8

7 3

bicyclo [3 . 2 . 1 ]octane (paths of 3 atoms, 2 atoms,

and 1 atom)

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Appendix 1 , Summary of IUPAC Nomenclature, continued

3

6

� B, 2 �5

Br 5 ,5-dibromo ­bicyclo [2 . 1 . 1 ]hexane

3 6

o bicyclo [2.2.2 ]oct -5-en -2-one

VI. Replacement Nomenclature of Heteroatoms

3 8 ,8 -dimethyl ­bicyclo [3 . 2 . 1 ]octan - l -ol

The teon "heteroatom" a pplies to any atom other than carbon or hydrogen . It is common for heteroatoms to appear in location s that are inconvenient to name following ba sic rule s , so a simple sy stem called "replacement nomenclature" ha s been devi sed . T he fundamental principle i s t o name a compound a s i f it contained only carbon s in t he skeleton , plus any functional groups or sub stituent s , and then indicate which carbon s are "replaced" by heteroatoms . The pre fixe s u sed to indic ate t hese sub stition s are li sted here in decreasing priority and listed in this order in the name:

Element Prefix Example o oxa S thia N aza P pho spha

H H H2

OH

nonan - l -01

Si sila B bora /" N...........".. B '-.,./ Si � S '-.,./OH

2-thia -8-aza -4-sila -6 -boranonan - l -ol

In t he above example , note that the (imaginary) compound no longer ha s nine carbon s, even t hough the name still includes "nonan" . The heteroatom s have replaced carbon s , but the compound i s named a s i f it still had tho se carbon s. Where the replacement sy stem i s particularly useful i s in polycyclic compounds . Shown below are three example s of commercially available and synthetically u seful reagent s that u se thi s system .

parent hydrocarbon

S 7�84 3

6� 2

bicyclo [2.2.2 ]octane

4� �O � 6 8

b icyclo[5 .4.0]undec -7 -ene

reagent

i(53

6 I 2 1 ,4-DiAzaBiCyclo [2 . 2 . 2 ]Octane (upper case added to explain abbreviation)

4 r�:) �O �N 6 8

1 ,8-DiazaBic yclo [5.4.0]Undec -7-ene (upper case added to explain abbreviation)

685

abbreviation

DABCO

DBU

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Appendix I , Summary o f IUPAC Nomenclature , cont inued

9

3

H 9 B ....

3�5 6 9-BBN

7

bicyclo [3 . I. l ]nonane

2 I 8 7

9-BoraBicyclo [3 . 1. 1 ]Nonane (upper case added to explain abbreviation)

VII. Designation of Stereochemistry; Cahn-Ingold-Prelog system

Is this alkene c is or trans ? F I >=< Cl Br

How can we dist ingu ish this st ructure from its mirror image ?

F

" , I I \\'1"CI Br

Compounds that exhibit stereo isomer ism , whether geometric isomers around double bonds , substituent groups on rings , or mo lecules with asymmetric tetrahedral atoms (wh ich are almost always carbons) , require a system to des ignate re lative and absolute orientation of the groups. The terms cis/trans , DIL in carbohydrates and amino acids , and dlf for opt ically act ive compounds, are limited and cannot be used generally , although each st ill is used in appropr iate s ituat ions . For example , cis/trans st ill is used to indicate relative positions of substituents around a ring .

A system developed by chemists Cahn , Ingold , and Prelog , uses a series of steps to determine group pr ior ities , and a def inition of position based on the relative arrangement of the groups. In alkenes , the system is relatively simple :

high high

>===< low low this arrangement is defined as Z = zusammen , together , from both high priority groups on the same side o f the C=C

h igh low

>===< low h igh this arrangement is def ined as E = entgegen , opposite , from the two h igh priority groups on opposite s ides of the C=C

As with alkenes , the orientation around an asymmetic carbon can be only one of two choices . In three d imensions , clockwise and counte rclockwise are the only two d irect ions that are defin ite , and even that description requires a f ixed reference point . To des ignate configurat ion , the lowest (fourth) prior ity group is always placed farthest away from the v iewer ( indicated by a dashed line) , and the group priorit ies w ill follow 1 to 2 to 3 in e ither a clockwise or a counterclockwise direction.

1 to 2 to 3 is clockwise, defined as the R = rectus configuration

1 )< 3 2

686

1 )< 2 3

1 to 2 to 3 is counterc Iock WIse,

def ined as the S =

sinister conf iguration

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Appendix 1 , Summar y of IUPAC Nomenc lature, continued

The on ly step remaining is t o determine the prior it y of groups , for which t here is a c arefu lly def ined set of rules . Rule 1. Consi der the first atom of the group , the point of attachment. Atoms with higher atomic number receive higher pr iorit y. Heavier isotopes have higher priorit y t han lighter isotopes .

I > Br > C I > F

Rule 2. If the f irst atoms of two or more groups are the same , go out t o the next atoms t o brea k the tie . One high priorit y atom takes priority over any number of lower-priority atoms .

H I

-C -OH I

H

CH3 I

> -C -H I CH3

> H I

-C -H I CH3

> H I

-C - H I

H

Rule 3. Treat mult iple bonds as if they were all s ingle bonds ; one wil l be to the rea l atom , the others will be t o imaginar y atoms. This is the hardest rule to put int o practice . This example of an alkyne shows stepwise how to accomplish t his. In the pictures , imaginary at oms (ones that di d not start out in the structure , we

had to invent them) are ind icated by italics.

--C C - H all at oms are real

>

C started with3 bonds to carbon, C C I ( so have to replace them I I -C -C -H I > --C -C - H I I I

C C C halfway there-had to add two fake carbons

fina l-had to add four fake carbons

o I I

....... C ..... R H

becomes 0, /O - C r:::===�> ....... C .....

R H

becomes R -C :: N >

N C I I

R -C - N I I N C added a fake 0 to

the real C , and a fake C to the real 0

Examples applying the Cahn -Ingold-Prelog sytem

compari ng '- __ J comparing i low F I high}

F and Cl t==\- I and Br

high CI Br low

E

687

ad ded two fake Ns to the C , and two Cs to the N

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Appendix 1 , Summary of IUPAC Nomenclature , continued

4 F

" , I I " '1"CI

Br2

3

fl ip � Br up so'\) , F goes back

Often, the hard part of applying the RlS system to asymmetric carbons is orienting the molecule so that the fourth priority group is farthest away. Sometimes it is easier to put the fourth priority group toward you, then take the opposite of what the direction of 1 to 2 to 3 says.

2 Br

I " F 4

� I CI 1 3

R

I 1

= R ,\\\\1"�',/1 Br" " CI

2 F � 3 up

� Putting the F (4th priority) pointing up toward you, the groups 1 to 2 to 3 appear counterclockwise, but we have to take the opposite, so instead uf S, it is R.

688

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Appendix 2: Summary of Acidity and Basicity Imagine that you are at a family reunion where you can observe t he competition for i ce cream cones among your nieces and nephews . Pretty s oon , you formulate a generalization : t he older kids can hold onto t heir ice cream cones more strongly t han the younger ones. Another way of saying it is t hat t he older ones are less l ikely to give up their cones . You could even represent this information in a table showing a series of equilibria between the child with ice cream , and the free ice cream plus the hungry child . The difference s in strength could also be quantitated : the larger t he hunger factor , pKH' the less likely the child will gi ve up the ice cream .

pKH

1 2

1 0

8

6

4

Approximate pKH Values of Children

1 2-year-old with ice cream --- + --- Ice cream

1 O-year-old with ice cream ------ Ice cream +

8 -year-old with ice cream ---

ice cream + ---

6-year-old with ice cream --- + --- Ice cream

4-year-old w ith ice cream --- + --- Ice cream

hungry 1 2-year-old

hungry l O-year-old

hungry 8-year-old

hungry 6-year-old

hungry 4-year-old

What good is this table ? It allows anyone to make predictions about what will happen when kids wit h ice cream are mixed with hungry kids .

hungry 1 O-year-old } ? 1 1 O-year-Old with ice cream + --- +

4-year-old with ice cream ---

hungry 4-year-old

If the hungry lO-year -old was left unattended in a room with the 4-year-old with i ce cream, whi ch side of this equilibr ium would be fa vored w hen you came back in a few minutes: "reactants" or "products " ? More likely than not , there would be an ice cream transfer in your absence ; "products" would be fa vored . You could ha ve predicted this from t he table, and you could generalize : the hungry lO- year-old will be strong enough to rip the ice cream away from any kid lower on the table than the lO-year-old himlherself .

Let 's predic t the results of another equilibrium :

hungry 6-year-old } ? i 6-year-old with ice cream + --- +

l 2-year-old with ice cream ---

hungry 1 2-year-old

Is the hungry 6-year-old stron g enough to pu ll the ice cream away from the 1 2 -year-old ? Not in most families . The table shows that the only chance for the hungry 6-year -old is t o find a 4-year-old with ice cream . The 1 2-year -old with ice cream is pretty safe as long as the hunger tab le doesn 't go to higher ages.

If you understand this analogy and can make predictions of ice cream transfer using the table, then you can understand how to predict the direction of equilibrium in acid-base reactions. Turn the page .

689

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Appendix 2 continued, Summary of Acidity and B asicity

To set the stage ....

A few generalizations : A) This Appendix deals with pro tic acids and bases, called Bronsted-Lowry acids . Similar statements can be made about Lewis acids but they are not the focus of this discussion. B) Values of pKa are measures of equil ibrium constants , described further below. Values between 0 and 15 .7 are measured by titration in water solution and are known accurately , to within 0. 1 pK unit and sometimes better. Values outside of thi s range cannot be measured in water because of the leveling effect of water, and there is no universal ly accepted method for measuring these pKa values . This lack of a single standard of measurement means that the values below 0 and above 15 .7 should be considered relative, not absolute . If your instructor says the pKa of methane is 46 and this book says it is 50, those should be considered the same value within experimental variation . C) Acidi ty is a thermodynamic property, and the acid equi l ibrium constant , Ka, is a measure of the relative concentrations of species in the protonated and unprotonated form. As most organic acids are weak acids, meaning they are present mostly in the protonated form at equi librium, the Ka < l. Since the pKa = -log Ka, the pKa values wil l be greater than 0, with the larger pKa representing a weaker acid. If this is not clear, review text section 1 - 13 . D) Acids do not spontaneously spit out a proton ! Despite our way of writing ionization equil ibria as shown on the next page, acids do not give up a proton unless a base comes by to take the proton away . The reactions as drawn in the table should be considered half-reactions, just as the reactions in the electromotive series were half-reactions for balancing oxidation-reduction reactions in general chemistry.

I. Predicting equilibrium position

Look at the table on p. 6 9 1 and notice how it looks just l ike the "chi ldren-with-ice-cream" table. We can use this table to make predictions about equi librium position in acid-base reactions just as we did for the children with ice cream. 1) A base will deprotonate any acid stronger than its conjugate acid. This is the most important principle of predicting acid-base reactions . On the table, this means that any base, hydroxide for example, can react with any acid more acidic than the conjugate acid of itself, water in our example. So hydroxide is a strong enough base to pull the proton from any of these : bicarbonate ion, a phenol , carbonic acid, a carboxylic acid, or a sulfonic acid. We can also predict that hydroxide is NOT a strong enough base to react with any acid above water on the table; for example , a mixture of hydroxide with an alkyne wil l favor the reactants at equilibrium, with only a small amount of products.

reactants favored

HO- + RC _ C - H pKa 25

weaker base

weaker acid

.. ___ H20 + RC - C :

pKa 15.7

stronger acid

stronger base

2) Another way of predicting the position of an equilibrium is to assign "stronger" and "weaker" to the acid and base on each side of the equation, using the table to determine which is stronger and which i s weaker. Equilibrium will always favor the weaker acid and base. This method wil l always give the same answer as the principle in # 1 above.

To lead into the next section, look again at the table on p. 69 1 and notice two things: a) with only a couple of exceptions, a l l the acidic protons are on either oxygen or carbon; and b) generalizations can be made about the acidity of functional groups. Learning to correlate acidity with functional group is important in predicting reactivity of the functional group.

690

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Appendix 2 continued, Summary of Acidity and Basic ity

A��roximate �Ka Values of Oq�anic Com�oun ds pKa

I I weaker acid alkane "" 50 R - C - H H+ + R - C : stronger base

I I / /

a lkene "" 45 ==C H

H+ + == �.

. . amine 35-40 - N - H H+ + -N :

I I "" 35 H -H H+ + H :

alkyne "" 25 RC ::C -H H+ + RC ::C :

0 I 0 I " "

ketone and ester 20-25 -C -C -H H+ + -C -C : I I

R R I I

"" 1 8 R -C -O -H H+ + R -C - O : I I R R

H H I I

alcohol "" 1 7 R -C -O -H .. H+ + R -C - O : I I

R R lcanno! be H H I I

meas ured in "" 1 6 R -C - O -H .. H+ + R -C - O : water solution I I

H H - - - � - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -t measured in 1 5 .7 H2O H+ + HO-water solution

1 0.3 HC03- H+ + col-phenol ",, 1 0 Ar - O -H .. H+ + Ar - O :

6.4 H2C03 .. H+ + HC03-

0 0 I I I I

carboxylic acid 4-5 R -C - O -H H+ + R -C - O :

s ulfonic aci d < 0 RS02 - O -H H+ + RSO -0 : 2 • • stronger acid weaker base

691

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Appendix 2 continued, Summary of Acidity and Basicity

II. Correlation of Acidity with Functional Group

A. Ox�gen Acids

0 0 " I I

R - S - O - H > R - C - O - H I I 0

pKa < O pKa 4-5 sulfonic acid carboxylic acid

> ( }O-H pKa 10 .0 phenol

> HO - H pKa 1 5 .7 water

> R

I R - C - O - H

I

R

pKa 16 - 1 9 alcohol

Sulfonic acids are the strongest of the oxygen acids but are not common in organic chemistry . Carboxylic acids, however, are everywhere and are considered the strongest of the common organic oxygen acids. (Note that "strong" and "weak" are relative terms: acetic acid, pKa 4 .74, was a " weak" acid in general chemistry in comparison to sulfuric and hydrochloric acids , but acetic acid is a "strong" acid in organic chemistry relative to the other oxygen acids . ) Phenols having OR groups on benzene or other aromatic rings are sti l l stronger acids than water.

Why are phenols, carboxylic acids , and sulfonic acids stronger acids than water? Because their anions are stabi lized by resonance. (Refer to text sections 1 - 13 , 10-6 , and 20-4, especially Figure 20- 1 .) Let's look at that statement in more detai l .

Remember that acidity i s a thermodynamic property ; that i s , acidity equi l ibrium depends on the difference in energy between the reactants and products . The more the anion is stabilized by resonance, the lower in energy it is , and the less positive the 1\G, as shown on the reaction energy diagram:

...

Energy

reaction -.

, ,

,

, , ,

, - alcohol

water

phenol

-- - - . carboxylic acid

sulfonic acid

oxygen anions stabil ized by resonance

Why are alcohols weaker acids than water? There are two effects that contribute, both of which are consistent with the trend that 1 ° alcohols (pKa 16) are slightly stronger than 2 ° alcohols (pKa 17) which are sl ightly stronger than 3° alcohols (pKa 1 8) . Alkyl groups are mildly electron-donating in their inductive effect (more about thi s later) and destabilize the anion , as shown in the energy diagram above. Second, the more crowded the anion is, the less it can be stabilized by hydrogen bonding with the solvent.

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Appendix 2 continued, Summary of Acidity and Basicity

B. Carbon Acids

When we think of "acids", we do not usually think of protons on carbon, yet carbon acids and the carbanions that come from them are of tremendous importance in organic chemistry.

"Unstabilized" carbon acids are those that do not have any substituent to stabilize the anion. Alkanes, with only sp3 carbons are the weakest acids with pKa around 50. The vinyl carbon in a carbon-carbon double bond is sp

2 hybridized with the electrons of the anion slightly closer to the positive nucleus,

leading to some stabilization of the anion. This type of stabilization is particularly important in alkynes with sp hybridized carbons.

I I / / R -C-H R -C: =C

= c: . RC::C-H .. RC::C: I I ,

H

sp3

hybridized sp2

hybridized sp hybridized

pKa 50 pKa 45 pKa 25

C. Carbon Acids Alpha to Carbonyl (This topic is described in detail in text section 22-2B.)

Look at these huge differences in pKa when the acidic group is next to a carbonyl. o II

O - H e=:> RCH2O - H pKa 16- 18 R -C-O-H pKa 4-5

0 II H

N-H e=:> RCH2NH - H pKa 35-40 R-C- N-H pKa 16

0 H2 II

C- H e=:> RCH2CH2 - H pKa 50 R-C-C -H pKa 20

Hydrogens alpha to carbonyl are unusually acidic because of resonance stabilization of the anionic conjugate base.

D. Acidities of Acyl Functional Groups

In addition to the significant variation in the acidity of alpha hydrogens depending on which atom the H is bonded to, what is on the other side of the carbonyl also has a dramatic influence. In this case, the stabilization is more important on the starting material, not on the conjugate base. See the energy diagram on p. 694.

aldehyde ketone ester amide

H O H O H O H O I II

H2C -C-H pKa 17

no stabilization of starting material

I II

H2C-C-CH3 pKa 20

mild stabilization of starting material by weak electron donation from CH3

693

I II

H C-C - OCH 2 • • 3

pKa 25

H 0 I I +

H2C-C=�CH3 significant resonance stabilization of starting material

I I I ••

H2C - C - N(CH3}z

pKa 30 t H 0 I I +

H2C -C = N(CH3}z s trongest resonance stabilization of starting material

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Appendix 2 continued, Summary of Acidity and Basicity

Energy

, ... , , , " , ...... \

" . " '" \ , ,

,' '�

, ,I' , ,I

'

least , ,.-

(tabilization ,' ,,/ aldehyde I , "/,,

ketone

ester

amide __ greatest stabilization of starting material, largest LlG

reaction --

E. Carbon Acids Between Two Carbonyls (This topic is described in detail in text section 22-15.) While a hydrogen alpha to one carbonyl moves into the pKa 20-25 range for ketones and esters respectively, a hydrogen between two carbonyls (cyano and nitro are similar to carbonyl electronically) is more acidic than water. The increased resonance stabilization of the conjugate base is largely responsible, but there are subtle variations depending on the type of functional group as noted at the bottom of p. 693. Look at the enormous influence of the nitro group.

0 0

EtO¥OEt

H

pKa 13.5

0 II +

N -H2C' '0

I H

pKa 10.2

N:--... /N �

C C�

y H

pKa 1 1.2

0 0 N -�"+ H3CO '0

H

pKa 5.8

0 0

EtO¥CH3

H

pKa 10.2

0 0 +" "+ -O,N,N,O-

H

pKa 3.6

III. Correlation of Basicity with Functional Group

o 0

H3CVCH3

H

pKa 9,0

The bulk of this Appendix is on acidity because many more functional groups are acidic than are basic. Basically (oooh, sorry), only one functional group is basic: amines. There is variation among aliphatic, aromatic, and heteroaromatic amines; these are covered thoroughly in text sections 19-5 and 19-6. One point in the text, just before Table 19-3, deserves emphasis: for any conjugate acid-base pair:

pKa + pKb = 14

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Appendix 2 continued, Summary of Acidity and Basicity

This simple algebraic relationship is very useful:

Sample problem. Is triethylamine (pKb 3.24) a strong enough base to de proton ate phenol (pKa 1O.0)? We need to calculate either the pKa of the conjugate acid of triethylamine or the pKb of the conjugate base of phenol to see which is stronger and weaker. Then we can say with certainty which side of the equilibrium will be favored.

( } OH ( }o- H +1

+ Et-N-Et + Et - N -Et 1 1

Et Et pKa 10.0 pKb 3. 24 pKb = 4.0 pKa = 10.76

(14 - 10.0) (14 - 3.24)

stronger stronger weaker weaker acid base base acid

Aha! Products are favored at equilibrium, so the correct answer to the question is "Yes, triethylamine is a strong enough base to deprotonate phenol."

Try this for fun: How weak must a base be before it does NOT deprotonate phenol? What algebraic rule can you formulate to predict whether any combination of acid and base will favor products or reactants?

IV. Substituent Effects on Acidity

So far, we have focused on acidities of different functional groups. Let's tum to more minor, more subtle, structural changes to see what effect substituents will have on the acidity of a group. Primarily, we imply electronic effects as opposed to steric effects, but this Appendix will conclude with a discussion of how steric and electronic effects can work together.

A. Classification of Substituents-Induction and Resonance

Substituent groups can exert an electronic effect on an acidic functional group in two different ways: through sigma bonds, where this is called an inductive effect, or through p orbitals and pi bonds which is called a resonance effect. Groups can also be electron-donating or electron-withdrawing by either of the mechanisms, so there are four possible categories for groups. Note that a group can appear in more than one category, even in conflicting groups!

a) Electron-donating by induction: only alkyl groups (abbreviated R) have electrons to share by induction;

b) Electron-withdrawing by induction: every group that has a more electronegative atom than carbon is in this category; some examples: F, Cl, Br, I, OH, OR, NH2, NHR, NR2, N02, C=O, CN, S03H, CX3 where X is halogen;

c) Electron-donating by resonance: groups that have electron pairs to share: F, Cl, Br, I, OH, OR, NH2, NHR,NR2;

d) Electron-withdrawing by resonance: N02, C=O, CN, S03H.

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Appendix 2 continued, Summary of Acidity and Basicity

B. Generalizations on Electronic Effects on Acidity (refer to text section 20-4B)

Electric charge is the key to understanding substituent effects. An acid is always more positive than its conjugate base; in other words, the conjugate base is always more negative than the acid. Electron­donating and electron-withdrawing groups will have opposite effects on the acid-base conjugate pair.

Electron-donating groups stabilize the more positive acid form and destabilize the more negative conjugate base. From the diagram, it is apparent that electron-donating groups widen the energy gap between reactants and products, making L\G more positive, favoring reactants more than products. In essence, this weakens the acid strength.

Energy acid

reaction -

conjugate base

- "", � ... - electron-donating substituent

no substituent effect

electron-withdrawing substituent

Electron-withdrawing groups destabilize the more positive acid form and stabilize the more negative conjugate base, narrowing the energy gap between reactants and products, making L\G less positive. Products are increased in concentration at equilibrium which we define as a stronger acid.

Electron-withdrawing groups increase acid strength; electron-donating groups decrease acid strength.

C. Inductive Effects on Acidity

Text section 20-4B gives a thorough explanation of the inductive effect of electron-withdrawing groups on simple carboxylic acids, from which three generalizations arise:

A) Acidity increases with stronger electron-withdrawing groups. (See problem 20-33.) B) Acidity increases with greater number of electron-withdrawing groups. C) Acidity increases with closer proximity of the electron-withdrawing group to the acidic group.

We don't usually look to aromatic systems for examples of inductive effects, because the pi system of electrons is ripe for resonance effects. However, in analyzing the resonance forms of phenoxide on the next page, it becomes apparent that the negative charge is never distributed on the meta carbons. Meta substituents cannot exert any resonance stabilization or destabilization; at the meta position, substituents can exert only an inductive effect. The series of phenols demonstrates this phenomenon, consistent with aliphatic carboxylic acids.

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Appendix 2 continued, Summary of Acidity and Basicity

:0: :0: :0: :0: the de localized

6 �-6 (] 6- negative charge does not increase electron

---- ---- density at meta position

'meta C

OH OH OH OH OH

0 CH3 O H 0 � OCH3 0 CI 0 N02 pKa 10.09 pKa 10.00 pKa 9.65 pKa 9.02 pKa 8.39 donating withdrawing withdrawing withdrawing

D. Resonance Effects on Acidity-benzoic acids and phenols (review the solution to problem 20-45)

Resonance effects can be expressed with placement of substituents at ortho or para positions, but ortho has the complication of steric effects, so just para substitution is shown here.

Electron-withdrawing substituents increase the acidity of benzoic acids and phenols:

OH �/hO

I -...;::: I -...;:::

� �

H'

para

pKa 10.00 pKa 8.05

OH OH

¢ ¢ CN N02

pKa 7.95 pKa 7.15

COOH COOH COOH

¢ ¢ ¢ H CN N02

pKa 4.20 pKa 3.55 pKa 3.42

Electron-donating by resonance but electron-withdrawing by induction:

There is a group of substituents that donate by resonance but withdraw by induction: alkoxy groups and halogens are the most notable examples, and the acidity data provide an insight into which effect is stronger.

COOH COOH COOH

¢ 0 � OCH3 ¢

H OCH3

pKa4.20 pKa 4. 09 pKa 4.47

See another example on the next page.

meta-Methoxybenzoic acid is stronger than benzoic acid, consistent with electron-withdrawing by induction which is expressed at the meta position. But the para isomer is weaker than benzoic acid; electron donation by resonance has not only compensated for the inductive effect (which is still operative at the para position) but has decreased the acidity even further. Thus, the donating effect by resonance must be stronger than the withdrawing effect by induction for the methoxy group.

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Appendix 2 continued, Summary of Acidity and Basicity

OH OH OH meta-Fluorophenol is stronger than phenol, consistent

¢ 61

¢ with electron-withdrawing by induction which is "'" "'" "'" expressed at the meta position. The para isomer is still

I I 1 stronger than phenol; electron donation by resonance has h h F h not compensated for the inductive effect (which is still

operative at the para position). Thus, the donating effect H F by resonance must be weaker than the withdrawing

effect by induction for the fluoro group. pKa 10.00 pKa 9.28 pKa 9.81

Studying substituent effects on acidity is the standard method of determining whether a group is donating or withdrawing by induction and resonance.

E. Proximity Effects of Substituents

pKa 10.0 pKa 8.05 OH OH

6 A B

pKa 9.19

OH

9ro

pKa 9. 90 OH 0

cr D

Three effects influence the pKa values of these substituted phenols. In C, the acetyl group at the meta position is electron-withdrawing by induction only. In B, the acetyl group at the para position exerts both resonance and inductive effects, both of which are electron-withdrawing, making the acid stronger. In theory, substituents at the ortho position should be like para, exerting both resonance and inductive effects; in fact, the inductive effect should be stronger because of closer proximity to the acidic group. So we would predict D to be a stronger acid than B, yet it is not. What other effect is operating?

Structure E shows that because of the proximity of the acetyl group to the OH, intramolecular hydrogen bonding is possible. Hydrogen bonding stabilizes the starting material, lowering the energy of the starting material and making �G more positive. Intuitively, it should be apparent that a hydrogen held between two oxygens will be more difficult to remove by a base. Also, after the proton has left as shown in structure F, the negative charge on the phenolic oxygen is close to the partial negative charge on the oxygen of the carbonyl, destabilizing product F, raising its energy, also making AG more positive. The proximity of the acetyl group influences both sides of the equation to make the acid weaker.

intramolecular � hydrogen bond cr

I"'"

h E

.. F

8-o 0

cr Here are two more examples where intramolecular hydrogen-bonding influences acidity.

OH

( )- COOH

Hooe

� eOOH

Ho-o-eOOH

pK\ 4.6; pK2 9.3 pK\ 2.75; pK2 13.4 pK\ 3.02; pK2 4.38

698

Hooe eOOH \::=./

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Appendix 2 continued, Summary of Acidity and Basicity

F. Steric Inhibition of Resonance

Another type of proximity effect arises when the placement of a substituent interferes with the orbital overlap required for resonance stabilization. This can be seen clearly in the acidity of substituted benzoic acids and in the basicity of substituted ani lines.

Let's analyze this series of carboxylic acids.

COOH COOH COOH COOH

OH COOH I 6

¢ 0

H H3C

CHJ pKa 3.75 , " , " � h CH3

pKa 4.20 CH3 pKa 4.27 pKa 3. 46 pKa 3.21

pKa 4. 34

Formic acid, pKa 3. 75, serves as the reference carboxylic acid. Benzoic acid is weaker because the phenyl group is electron-donating by resonance, stabilizing the protonated form. Methyl substituents are known to be electron-donating by induction, strengthening the electron-donating effect, making the meta- and para­substituted acids weaker than benzoic acid.

Then come the anomalies. Alkyl groups are electron-donating by induction and should weaken the acids, but the ortho-t-butyl and the 2,6-dimethylbenzoic acids are not only stronger than benzoic acid, they are stronger than formic acid! Something has happened to tum the phenyl group into an electron-withdrawing group.

Phenyl is electron-donating by resonance but electron-withdrawing by induction, so what has happened is that the ortho substituents have forced the COOH out of the plane of the benzene ring so that there is no resonance overlap between the benzene ring and the COOH orbitals. The COOH "feels" the benzene ring as simply an inductive substituent. Resonance has been "inhibited" because of the steric effect of the substituent. •

""CH3 \\\\\\ 0

<----.l�<O-H

• •

• CH3

COOH group is not parallel with the plane of the benzene ring­no resonance interaction.

this three-dimensional view down the C-C bond between the COOH and the benzene ring shows that COOH is twisted out of the benzene plane

The same phenomenon is observed in substituted ani lines. Anilines are usually much weaker bases than aliphatic amines because of resonance overlap of the nitrogen's lone pair of electrons with the pi system of benzene. When that resonance is disrupted, the aniline becomes closer in basicity to an aliphatic amine. (See problem 19-49(c). )

More examples on the next page.

699

0-'\ N.�H3 - CH3

pKb 8.94

CH3

�N:�H3

�- CH3 CH3

pKb =:: 6-7 (estimated)

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Appendix 2 continued, Summary of Acidity and Basicity

Examples of steric inhibition of resonance:

o amide-not basic because of resonance sharing of N lone pair with carbonyl

pKb 8. 9

3° aromatic amine

o strong base similar to aliphatic amine; geometry of bridged ring prevents overlap of N lone pair with carbonyl

pKb 3.4 3° aliphatic amine

pKb - 2.3 (yes, negative!)

700

�ND � pKb 6.2

3° amine, and the N is bonded to a benzene, but the bridged ring system prevents overlap of N lone pair with benzene

Not only is steric inhibition of resonance important in this example, but so is intramolecular hydrogen­bonding in the protonated form. Draw a picture.