122 Ch 4 Aromatic

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Aromatic Compounds

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Discovery of Benzene

Isolated in 1825 by Michael Faraday who

determined C:H ratio to be 1:1.

Synthesized in 1834 by Eilhard Mitscherlich

who determined molecular formula to be

C6H6.

Other related compounds with low C:H ratios

had a pleasant smell, so they were classified

as aromatic.

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Kekul Structure

Proposed in 1866 by Friedrich Kekul, shortly

after multiple bonds were suggested.

Failed to explain existence of only one isomer

of 1,2-dichlorobenzene.

CC

CC

C

C

H

H

H

H

H

H

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Resonance Structure

Each sp2hybridized C in the ring has an

unhybridizedporbital perpendicular to the

ring which overlaps around the ring.

=>

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Unusual Reactions

Alkene + KMnO4diol (addition)

Benzene + KMnO4 no reaction.

Alkene + Br2

/CCl4

dibromide (addition)

Benzene + Br2/CCl4no reaction.

With FeCl3catalyst, Br2reacts with benzene to

form bromobenzene + HBr

(substitution!). Double bonds remain.

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Unusual Stability

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Annulenes

All cyclic conjugatedhydrocarbons wereproposed to bearomatic.

However,cyclobutadiene is soreactive that it

dimerizes before it canbe isolated.

And cyclooctatetraeneadds Br2readily.

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What Does It Take to Be Aromatic?

Alternating double and single bonds

A magic number of pi electrons

Resonance structures must be able to movethe pi electrons in a circular manar.

Non bonding electrons can also participate in

the resonance structures.

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Hckels Rule

Hckels Rule is used to generate the magic

number of pi electrons.

If the compound has a continuous ring of

overlappingporbitals and has 4N+ 2 pi

electrons, it is aromatic.

If the compound has a continuous ring of

overlappingporbitals and has 4N electrons, it

is antiaromatic.

Nis any integer, starting at zero

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Generation of Magic Pi Electrons

Value of N 4N+ 2

0 2

1 62 10

3 14

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[N]Annulenes

[4]Annulene is antiaromatic (4Ne-s)

[8]Annulene would be antiaromatic, but its

not planar, so its nonaromatic.

[10]Annulene is aromatic except for the

isomers that are not planar.

Larger 4Nannulenes are not antiaromatic

because they are flexible enough to become

nonplanar.

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Tropylium Ion

The cycloheptatrienyl cation has 6pelectronsand an emptyporbital.

Aromatic: more stable than open chain ion.

H OH

H+, H2O

H

+

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Which of the Following are Aromatic?

a. b. c. d.

e. f. g. h.

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Disubstituted Benzenes The prefixes ortho-, meta-, and para- are

commonly used for the 1,2-, 1,3-, and 1,4-

positions, respectively.

=>

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3 or More Substituents

Use the smallest possible numbers, but

the carbon with a functional group is #1.

NO2

NO2

O2N

1,3,5-trinitrobenzen

NO2

NO2

O2N

OH

2,4,6-trinitrophenol

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Common Names of Benzene

Derivatives

OH OCH3NH2CH3

phenol toluene aniline anisole

C

H

CH2 C

O

CH3C

O

HC

O

OH

styrene acetophenone benzaldehyde benzoic acid

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Phenyl and Benzyl

Phenyl indicates the benzene ring

attachment. The benzyl group has

an additional carbon.Br

phenyl bromide

CH2Br

benzyl bromide

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Fused Ring Hydrocarbons Naphthalene

Anthracene

Phenanthrene

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Larger Polynuclear

Aromatic Hydrocarbons

Formed in combustion (tobacco smoke).

Many are carcinogenic.

Epoxides form, combine with DNA base.

pyrene =>

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Physical Properties

Melting points: More symmetrical thancorresponding alkane, pack better intocrystals, so higher melting points.

Boiling points: Dependent on dipole moment,so ortho> meta>para, for disubstitutedbenzenes.

Density: More dense than nonaromatics, lessdense than water.

Solubility: Generally insoluble in water.

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Chapter 17 21

Electrophilic Aromatic Substitution

Electrophile substitutes for a hydrogen onthe benzene ring.

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Chapter 17 22

Mechanism

Step 1:Attack on the electrophile forms the sigma complex.

Step 2:Loss of a proton gives the substitution product.

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Chapter 17 23

Bromination of Benzene

Requires a stronger electrophile than Br2.

Use a strong Lewis acid catalyst, FeBr3.

Br

HBr+

Br Br FeBr3 Br Br FeBr3+ -

Br Br FeBr 3

HH

H

H

H

H

H

H

H

H

H H

Br+ + FeBr4

_+ -

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Chapter 17 24

Comparison with Alkenes

Cyclohexene addsBr2, H = -121 kJ

Addition to benzene is endothermic, not

normally seen.

Substitution of Br for H retains aromaticity, H

= -45 kJ.

Formation of sigma complex is rate-limiting.

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Chapter 17 25

Energy Diagram for Bromination

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Chapter 17 26

Chlorination and Iodination

Chlorination is similar to bromination. Use

AlCl3as the Lewis acid catalyst. Iodination requires an acidic oxidizing

agent, like nitric acid, which oxidizes the

iodine to an iodonium ion.

H+

HNO3 I21/2 I+

NO2 H2O+ ++ +

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Chapter 17 27

Nitration of Benzene

Use sulfuric acid with nitric acid to form the

nitronium ion electrophile.

H O N

O

O

H O S O H

O

O

+ HSO4

_H O N

OH

O+

H O N

OH

O+

H2O + N

O

O

+

NO2+then forms a sigma complex with

benzene, loses H+to form nitrobenzene.

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Chapter 9 28

Sulfonation

Sulfur trioxide, SO3, in fuming sulfuric

acid is the electrophile.

S

O

O O

S

O

O O

S

O

O O

S

O

O O

+ + +

_

_ _

S

O

O O

H

S

O

O

OH

+

_

S

HOO

O

benzenesulfonic a

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Chapter 17 29

Desulfonation

All steps are reversible, so sulfonic

acid group can be removed by heating in

dilute sulfuric acid.

This process is used to place deuterium in

place of hydrogen on benzene ring.

Benzene-d6

D

D

D

D

D

D

D2SO4/D2O

large excess

H

H

H

H

H

H

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Chapter 17 30

Nitration of Toluene

Toluene reacts 25 times faster than benzene.

The methyl group is an activating group.

The product mix contains mostly ortho and para

substituted molecules.

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Chapter 17 31

Sigma Complex

Intermediate is

more stable if

nitrationoccurs at the

ortho

orpara

position.

=>

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Chapter 17 32

Energy Diagram

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Chapter 17 33

Activating, O-, P-Directing Substituents

Alkyl groups stabilize the sigma complex byinduction, donating electron density throughthe sigma bond.

Substituents with a lone pair of electrons

stabilize the sigma complex by resonance.

OCH3

H

NO2

+

OCH3

H

NO2

+

b l

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Chapter 17 34

Substitution on Anisole

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Chapter 17 35

The Amino Group

Aniline, like anisole, reacts with bromine water

(without a catalyst) to yield the tribromide.

Sodium bicarbonate is added to neutralize

the HBr thats also formed.

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Chapter 17 36

Summary of Activators

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Chapter 17 37

Deactivating Meta-Directing Substituents

Electrophilic substitution reactions fornitrobenzene are 100,000 times slowerthan for benzene.

The product mix contains mostly the meta

isomer, only small amounts of the orthoand para isomers.

Meta-directors deactivate all positions onthe ring, but the meta position is lessdeactivated.

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Chapter 17 38

Ortho Substitutionon Nitrobenzene

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Chapter 17 39

Para Substitution on Nitrobenzene

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Chapter 17 40

Meta Substitution on Nitrobenzene

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Chapter 17 41

Energy Diagram

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Chapter 17 42

Structure of Meta-Directing Deactivators

The atom attached to the aromatic ring will

have a partial positive charge.

Electron density is withdrawn inductively

along the sigma bond, so the ring is less

electron-rich than benzene.

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Chapter 17 43

Summary of Deactivators

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Chapter 17 44

More Deactivators

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Chapter 17 45

Halobenzenes

Halogens are deactivating toward

electrophilic substitution, but are ortho,

para-directing!

Since halogens are very electronegative, they

withdraw electron density from the ringinductively along the sigma bond.

But halogens have lone pairs of electrons that

can stabilize the sigma complex byresonance.

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Chapter 17 46

Sigma Complex for Bromobenzene

Ortho and para attacks produce a bromonium ion

and other resonance structures.

No bromonium ion

possible with meta attack.

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Chapter 17 47

Energy Diagram

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Chapter 17 48

Summary of Directing Effects

l i l S b i

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Chapter 17 49

Multiple Substituents

The most strongly activating substituent will

determine the position of the next

substitution. May have mixtures.

OCH3

O2N

SO3

H2SO4

OCH3

O2N

SO3H

OCH3

O2N

SO3H

+

F i d l C f Alk l i

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Chapter 17 50

Friedel-Crafts Alkylation

Synthesis of alkyl benzenes from alkyl

halides and a Lewis acid, usually AlCl3.

Reactions of alkyl halide with Lewis acid

produces a carbocation which is the

electrophile. Other sources of carbocations:

alkenes + HF, or alcohols + BF3.

l fC b i i

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Chapter 17 51

Examples ofCarbocation Formation

CH3 CH CH3

Cl

+ AlCl3

CH3C

H3C H

Cl AlCl3+ _

H2C CH CH3

HF

H3C CH CH3

F+

_

H3C CH CH3

OHBF3

H3C CH CH3

OH BF3+

H3C CH CH3+

+ HOBF3

_

F i f Alk l B

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Chapter 17 52

Formation of Alkyl Benzene

C

CH3

CH3

H+

H

H

CH(CH3)2+

H

H

CH(CH3)2

B

F

F

F

OHCH

CH3

CH3

+HF

BF

OHF

+

-

Li it ti f F i d l C ft

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Chapter 17 53

Limitations of Friedel-Crafts

Reaction fails if benzene has a substituent that

is more deactivating than halogen.

Carbocations rearrange. Reaction of benzene

with n-propyl chloride and AlCl3produces

isopropylbenzene.

The alkylbenzene product is more reactive

than benzene, so polyalkylation occurs.

F i d l C ft A l ti

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Chapter 17 54

Friedel-CraftsAcylation

Acyl chloride is used in place of alkyl

chloride.

The acylium ion intermediate is resonance

stabilized and does not rearrange like a

carbocation. The product is a phenyl ketone that is less

reactive than benzene.

M h i f A l ti

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Chapter 17 55

Mechanism of Acylation

C

O

R

+

H

C

H

O

R

+

Cl AlCl3

_C

O

R +

HCl

AlCl3

C t l ti H d ti

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Chapter 17 56

Catalytic Hydrogenation

Elevated heat and pressure are required.

Possible catalysts: Pt, Pd, Ni, Ru, Rh.

Reduction cannot be stopped at an

intermediate stage.

CH3

CH3Ru, 100C

1000 psi3H2,

CH3

CH3

Sid Ch i O id ti

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Chapter 17 57

Side-Chain Oxidation

Alkylbenzenes are oxidized to benzoic acid by

hot KMnO4or Na2Cr2O7/H2SO4.

CH(CH3)2

CH CH2

KMnO4, OH-

H2O, heat

COO

COO_

_

Sid Ch i H l ti

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Chapter 17 58

Side-Chain Halogenation

Benzylic position is the most reactive.

Chlorination is not as selective as

bromination, results in mixtures.

Br2reacts only at the benzylic position.

CHCH2CH3

Br

hBr2,

CH2CH2CH3

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AROMATIC REVIEW

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Give the shape of benzene.

a. Tetrahedral

b. Bent

c. Trigonal pyramidal

d. Planar

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Answer

a. Tetrahedral

b. Bent

c. Trigonal pyramidal

d. Planar

All six carbons and six hydrogens are inthe same plane.

Give the hybridization of each carbon

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Give the hybridization of each carbon

in benzene.

a. sp

b. sp2

c. sp3

d. sp4

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Answer

a. sp

b. sp2

c. sp3

d. sp4

Each carbon in benzene is sp2hybridized.

Give the bond angle of the atoms in

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Give the bond angle of the atoms in

benzene.

a. 45

b. 60

c. 90

d. 109.5

e. 120

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Answer

a. 45

b. 60

c. 90

d. 109.5

e. 120

The carbons are 120oapart in benzene.

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Classify .

a. Aromatic

b. Antiaromaticc. Nonaromatic

d. Acyclic

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Answer

a. Aromatic

b. Antiaromatic

c. Nonaromatic

d. Acyclic

The compound gives a whole number forNin Hckels rule (4N+ 2 = 6, N= 1).

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Classify .

a. Aromatic

b. Antiaromaticc. Nonaromatic

d. Acyclic

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Answer

a. Aromatic

b. Antiaromatic

c. Nonaromatic

d. Acyclic

The compound is cyclic and hascontinuous delocalized electrons, butdoes not give a whole number forHckels rule (4N+ 2 = 8, N= 3/2).

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Classify .

a. Aromatic

b. Antiaromatic

c. Nonaromatic

d. Acyclic

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Answer

a. Anthracene

b. Naphthalene

c. Phenanthrene

d. Benzene

Naphthalene contains two benzenerings fused together.

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Name

a. Anthracene

b. Naphthalenec. Phenanthrene

d. Benzene

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Answer

a. Anthracene

b. Naphthalene

c. Phenanthrene

d. Benzene

Anthracene contains three benzene ringsfused together.

Identify how carbon diamond and

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Identify how carbon, diamond, and

graphite are related.

a. They are enantiomers of carbon.

b. They are diastereomers of carbon.

c. They are allotropes of carbon.

d. They have the same properties.

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Answer

a. They are enantiomers of carbon.

b. They are diastereomers of carbon.

c. They are allotropes of carbon.

d. They have the same properties.

NH2Cl

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Name

a. 4-Bromo-3-

chloroanilineb. 4-Bromo-3-

chlorophenol

c. 4-Bromo-3-chloroanisole

d. 1-Bromo-2-chloro-4-

anilinee. 1-Bromo-2-chloro-4-

phenol

Br

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Answer

a. 4-Bromo-3-

chloroaniline

b. 4-Bromo-3-

chlorophenol

c. 4-Bromo-3-

chloroanisole

d. 1-Bromo-2-chloro-4-

aniline

e. 1-Bromo-2-chloro-4-

phenol

Aniline is the parent compound. The NH2isat position one.

CH3

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Name

a. p-Methylphenol

b. m-Methylphenol

c. o-Methylphenol

d. 4-Methylphenol

e. 3-Methylphenol

OH

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Answer

a. p-Methylphenol

b. m-Methylphenol

c. o-Methylphenol

d. 4-Methylphenol

e. 3-Methylphenol

The groups are on adjacent carbons,

which is ortho.

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Name

a. 3-Amino-5-

benzaldehydeb. 5-Amino-3-

benzaldehyde

c. 3-Amino-benzaldehyde

d. 5-Nitro-3-

benzaldehydee. 3-Nitro-

benzaldehyde

NO2C

O

H

A

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Answer

a. 3-Amino-5-benzaldehyde

b. 5-Amino-3-

benzaldehyde

c. 3-Amino-

benzaldehyde

d. 5-Nitro-3-benzaldehyde

e. 3-Nitro-

benzaldehyde

Benzaldehyde is the parent compound.

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Name

a. o-Amino-

benzaldehydeb. m-Amino-

benzaldehyde

c. p-Amino-benzaldehyde

d. o-Nitro-

benzaldehydee. m-Nitro-

benzaldehyde

NO2C

O

H

A

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Answer

a. o-Amino-benzaldehyde

b. m-Amino-

benzaldehyde

c. p-Amino-

benzaldehyde

d. o-Nitro-benzaldehyde

e. m-Nitro-

benzaldehyde

Benzaldehyde is the parent compound.

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A

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Answer

a. 1-Phenyl-3-pentyne

b. 5-Phenyl-2-pentyne

c. 4-Phenyl-2-pentyne

d. 1-Phenyl-2-butyne

e. 1-Phenyl-3-butyne

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Identify the slow step in

electrophilic aromatic substitution.a. Formation of a stronger nucleophile.

b. Formation of the benzenonium ion.

c. Deprotonation to regain aromaticity.

d. Formation of a carbanion.

A

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Answer

a. Formation of a stronger nucleophile.

b. Formation of the benzenonium ion.

c. Deprotonation to regain aromaticity.

d. Formation of a carbanion.

Benzene attacking the electrophile to formthe benzenonium ion is the slow step.

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a. Hexachlorobenzene

b. Hexachlorocyclohexanec. 5,6-Dichloro-1,3-cyclohexadiene

d. Chlorobenzene

Cl2

AlCl3

A

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Answer

a. Hexachlorobenzene

b. Hexachlorocyclohexane

c. 5,6-Dichloro-1,3-cyclohexadiene

d. Chlorobenzene

One chlorine atom substitutes on thebenzene.

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a. Nitrobenzene

b. Anilinec. Chlorobenzene

d. Benzenesulfonic acid

1. HNO3, H2SO4

2. Zn, aq. HCl

Answer

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Answer

a. Nitrobenzene

b. Aniline

c. Chlorobenzene


A nitro group is added, which is thenreduced to an amino group.

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a. Nitrobenzene

b. Anilinec. Chlorobenzene


SO3

Answer

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Answer

a. Nitrobenzene

b. Aniline

c. Chlorobenzene


A sulfonic acid group is added to thebenzene.

CH CH

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a. 2- and 4-nitroethylbenzene

b. 3-Nitroethylbenzenec. 2- and 4-ethylbenzenesulfonic acid

d. 3-Ethylbenzenesulfonic acid

CH2CH3

HNO3

H2SO4

Answer

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Answer

a. 2- and 4-nitroethylbenzene

b. 3-Nitroethylbenzene

c. 2- and 4-ethylbenzenesulfonic acid

d. 3-Ethylbenzenesulfonic acid

The ethyl group is an ortho and paradirector.

Classify a bromide

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Classify a bromide.

a. Meta, activating group

b. Meta, deactivating group

c. Ortho and para, deactivating groupd. Ortho and para, activating group

Answer

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Answer



c. Ortho and para, deactivating group

d. Ortho and para, activating group

The electrons can delocalize into thebromide, making another benzenoniumion intermediate.

Classify a nitro group

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Classify a nitro group.



c. Ortho and para, deactivating groupd. Ortho and para, activating group

Answer

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Answer



c. Ortho and para, deactivating group

d. Ortho and para, activating group

OCH3

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a. 3-Propylanisole

b. 2- and 4-propylanisolec. 3-Isopropylanisole

d. 2- and 4-isopropylanisole

AlCl3

CH3CH2CH2Cl

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Give the intermediate in a Friedel

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Crafts acylation.

a. Carbocation

b. Carbanion

c. Radicald. Acylium ion

Answer

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Answer

a. Carbocation

b. Carbanion

c. Radical

d. Acylium ion

An RCO+is an intermediate in aFriedelCrafts acylation.

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a. Chlorobenzene

b. Benzoic acid

c. Benzaldehyde

d. Benzene

CO, HCl

AlCl3, CuCl

Answer

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Answer

a. Chlorobenzene

b. Benzoic acid

c. Benzaldehyded. Benzene

The GattermanKoch formylation formsbenzaldehyde.

CH2CH2CH3

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a. 2-Bromo-

propylbenzeneb. 3-Bromo-

propylbenzene

c. 4-Bromo-propylbenzene

d. -Bromo-

propylbenzenee. -Bromo-

propylbenzene

Br2

light

Answer

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Answer

a. 2-Bromo-propylbenzene

b. 3-Bromo-

propylbenzenec. 4-Bromo-

propylbenzene

d. -Bromo-propylbenzene

e. -Bromo-

propylbenzene

Bromine substitutes on the benzylic

carbon.

OH

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a. 2-Isopropylphenol

b. 3-Isopropylphenolc. 4-Isopropylphenol

d. 2-Isopropylphenol and 4-isopropylphenol

e. 2-Isopropylphenol and 3-isopropylphenol

(CH3)2CHOH

HF

Answer

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Answer

a. 2-Isopropylphenol

b. 3-Isopropylphenol

c. 4-Isopropylphenold. 2-Isopropylphenol and 4-isopropylphenol

e. 2-Isopropylphenol and 3-isopropylphenol

Phenols are highly reactive substrates

for electrophilic aromatic substitution.

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End of Chapter 9

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122 Ch 4 Aromatic