INTRODUCTION

Both aldehydes & ketones contain carbonyl group as their functional group.

Structure of carbonyl group

Both aldehydes & ketones have carbonyl group as the functional group. The carbonyl carbon is

sp2 hybridised & it uses sp2 hybrid orbitals to form 3σ bonds, one with oxygen atom & remaining 2

with two other atoms or groups (R or H). All these 3σ bonds lie in same plane at the angle of 120°.

The unhybridized p – orbital of carbonyl carbon form π - bond

with oxygen atom by sidewise overlapping with half filled

p – orbital of oxygen atom.

Since carbon & oxygen have different values of

electronegativity, the bond between carbon & oxygen is polar.

Infact electron density around the oxygen atom is increased

which causes the development of partial positive charge (δ+)

on carbon & partial negative charge (δ−) on oxygen.

Thus the carbonyl carbon is an electrophilic & carbonyl

oxygen is nucleophilic centre.

C O

π bond−

σ bond−

Orbital picture of carbonyl group

Illustration 1. Give the IUPAC name for

(i) CH3CHO

(ii)

CH CHO

Solution: (i) Ethanal

(ii) 2, 2 – diphenyl ethanal

Illustration 2. Write structure of the following compounds:

(i) 3-phenyl 2-propenal

(ii) 1, 5 – pentanedialSolution: (i)

CH CH CHO

(ii) − − − −2 2 2OHC CH CH CH CHO

Exercise 1.Write the formula of the following compound: Cyclopentane carbaldehyde

Exercise 2.Calculate the number of sigma bonds in diphenyl ketone.GENERAL METHODS OF PREPARATION OF ALIPHATIC ALDEHYDES & KETONES

1. From Alcohols

(a) By Direct oxidation: Aldehydes are prepared by oxidation of 1° alcohols. Refer to

Alcohols

(b) By catalytic dehydrogenation

When vapours of 1° or 2° alcohols are passed over copper gauze, they get

dehydrogenated to form aldehydes or ketones.

−→ +Cu/ 573K

3 2 2 3 2 2n propyl alcohol Pr opionaldehyde

CH CH CH OH CH CH CHO H

Dehydrogenation reaction is a better method of preparation because there is no risk of

further oxidation of aldehyde. (c) By using PCC

PCC stands for pyridinium chlorochromate. It is an equimolar mixture of CrO3, HCl and pyridine. It is used to oxidise 1° alcohol to aldehyde and 2° alcohol to ketones without affecting double or triple bond.

Illustration 3. Complete the following reaction by writing down the major product:

(i) OH

→2 2

PCCCH Cl

(ii) 2 2

PCCCH Cl

→OH

Solution: (i) OH

2 2

PCCCH Cl→

O

(ii)2 2

PCCCH Cl

→OH O

H

Exercise 3.Why is it important to distil out the aldehyde as soon as it is formed by the oxidation of primary alcohol?

2. From Acid chlorides

Aldehydes are prepared from acid chlorides by reaction with H2 in the presence of palladium

catalyst supported on BaSO4.

R C Cl

O

H24Pd, BaSO ,S

Boiling xylene→

R C H

O

HCl (Rosenmund’s reduction)

Ketones are obtained by reacting acid chlorides with dialkyl cadmium. + → +2 22RCOCl CdR' 2RCOR' CdCl

Illustration 4. Prepare ethanal by reduction method.

2

Solution:

CH3 C Cl

O

CH3 C H

O

H24Pd/BaSO

xylene→ HCl

Ethanal

Exercise 4.Write the structures of compound A and B in the following reaction:

HOH2C CH2 CH2 C

O

Cl2

PCCCH Cl

←B 2

4

H / PdBaSO

→ A

3. From fatty acids

(a) By heating calcium salt of fatty acid

Aldehydes are obtained by heating calcium salt of fatty acids with calcium formate.

Ca

CH3COO

CH3COO

Ca

OOCH

OOCH

2CH3CHO 2CaCO3

Calcium acetate Calcium formate

Acetaldehyde

Ketones are formed by distilling calcium salt of fatty acids alone.

Ca

CH3COO

CH3COO

675KCH3 C

O

CH3 CaCO3

Similarly mixed ketones, can also be obtained by similar reactions:

Ca

CH3COO

CH3COO

2CH3CC2H5

O

2CaCO3

ΔCa

OOCC2H5

OOCC2H5

(b) By passing vapours of fatty acids over manganese oxide

In this method, formic acid alone gives formaldehyde. Acetic acid gives acetone & the

mixture of two acids gives acetaldehyde. → + +MnO

2 2575K2HCOOH HCHO H O CO

→ − − + +||

MnO3 3 3 2 2575K

O

2CH COOH CH C CH H O CO

+ → + +MnO3 3 2 2575K

CH COOH HCOOH CH CHO H O CO

4. From Alkynes: (Refer to hydrocarbons)

5. By reductive ozonolysis of alkenes: (Refer to hydrocarbons)

Illustration 5. What happens when (give equation only)?

(i) Ethyne is treated with dilute H2SO4 in the presence of HgSO4.

(ii) propan-2-ol is treated with Cu at 573 K.

Solution: (i) ≡ →2 4

4

dil. H SO3HgSO

CH CH CH CHO

(ii)

Cu573K→CH3 CH

OH

CH3 CH3 C CH3

O

Illustration 6.( )

( ) ( )− − → → →32

2

i OBr3 2 3 h Alc. KOH ii H O / Zn mixture

CH CH CH A B Cν

Identify A, B and C.

Solution: CH3CH2CH32Br

hν→ CH3 CH CH3

Br(A)

Alc. KOH

CH3 CH CH2( )

( )3

2

i O

ii H O/ Zn←HCHOCH3CHO

(C) (B)

Exercise 5.

Convert 2 – chloro but-2-ene to 2-butanone.

Exercise 6.

Hydrocarbon ( ) ( ) ( ) ( ) ( )3 4

2

O LiAlH6 12 3 6 3 8Zn H O

A C H B C H O C C H O−→ →

The compound (C) gives yellow precipitate with I2 and NaOH. What is the structure of compound A?

Exercise 7.

(i) What is the hybrid state of carbon 1 & 2 in ethanal?

(ii) What product is obtained in the following reaction?

O3 2

Zn. dustH O→

4

Exercise 8.

4OsO→ (A) H∆+

→ (B) (major). (B) is

(A) O (B) CHO

(C) OH (D) O

PREPARATION OF AROMATIC ALDEHYDES & KETONES

1. By oxidation of alkyl benzene

Aromatic aldehydes are obtained by oxidation of side chain in the aromatic ring.

CH3

[ ]( )

3

3 2

CrO O

CH CO O→

HCOCOCH3

OCOCH3

2H O→

CHO

2CH3COOH

BenzaldehydeCrO2Cl2(Etard reaction)

Illustration 7. How will you prepare benzaldehyde commercially?

Solution: It is prepared form toluene. CH3

2Cl /hheat

ν→

CHCl2

20

H O

100 C→

CHO

2. By Friedel crafts reaction (Refer to hydrocarbons)

Illustration 8. In the following electrophilic substitution reaction.

( )+ →3AlCl3 2

CH CO O

COCH3

A

(i) Identify the compound A.

(ii) Write the structure of the electrophite.

Solution: (i)

CH3 C OH

O

A

(ii)

Electrophile is CH3 C

O

Illustration 9. Complete the following

AlCl3A Zn / Hg

HCl→ B

CH3CH2CCl

O

Solution:

A =

C

O

CH2 CH3

B =

Exercise 9.COCl

N2O

3AlCl X , X is→

(A) O2N CO (B)

N2O

(C) COCH3 (D) C

O

Exercise 10.

O

Cl

H

H

Cl

O3AlCl

heat A; Identify the A.→

(A) O H

ClH

O

(B)

O

(C) O H

OH

(D)

OOH

Cl

6

3. By Gattermann Koch Reaction

In this method aromatic aldehydes are prepared by formylation of aromatic ring with carbon

monoxide.

CO 3AlCl

HCl →

CHO

benzaldehyde

4. From Grignard’s reaction

Both aliphatic & aromatic aldehyde can be obtained by this method.

HCN on treatment with Grignard’s reagent & subsequently followed by hydrolysis yield an

aldehyde.

CH3MgBrDry

Ether →H3CHC NMgBr 2H O+→ Mg

OH

Br

CH3CHO NH3HCN

Similarly

C N

3CH MgI→

C NMgI

CH3

2H O→

C O

CH3

NH3 Mg

I

OH

PHYSICAL PROPERTIES OF ALDEHYDES & KETONES

(i) Physical state

Most of aldehydes (except formaldehyde which is a gas) are liquids at room temperature.

The lower ketones are colourless liquids and have pleasant smell.

(ii) Boiling points

Aldehydes & ketones have relatively high boiling points as compared to hydrocarbons of

comparable molecular masses due to polar carbonyl group, which bring stronger

intermolecular dipole – dipole interactions between the opposite ends of C = O dipoles.

Ketones are relatively more polar than their corresponding isomeric aldehydes due to the

presence of two electron repelling alkyl group around the carbonyl carbon.

(iii) Solubility

The lower members of aldehydes & ketones (upto four carbon atoms) are soluble in water. It

is due to their capability of forming hydrogen bonds with water molecules. The solubility of

these compounds in water decreases with the increase in the size of alkyl group because of

the increase in magnitude of non polar part in the molecule.

C O

R

R

OH H O C

R

R

CHEMICAL PROPERTIES

Aldehydes & ketones are highly reactive compounds, they undergo nucleophilic addition

reactions. Their reactivity is due to presence of a polar carbonyl group. The positively charged

carbon atom of carbonyl group is readily attacked by nucleophilic species for initiation of the

reaction. This leads to formation of intermediate anion which further undergoes the attack of

H+ ion or other positively charged species to form the final product. The reaction in general may

be represented as:

C Oδ + δ −

Nu

Planar

C

O

Nu

C

OH

Nu

fast

step-2

slow

step-1

Relative reactivity of aldehydes & ketones

In general ketones are less reactive than aldehydes on a account of following facts:

(i) Electron releasing effect of two alkyl groups, decreases the magnitude of positive charge on

ketones.

(ii) Steric effect caused by two alkyl groups also hinders the approach of the nucleophile to the carbonyl carbon.

C O

H

H

> C O

R

H

> C O

R

R

Reactivity decreases

Illustration 10. Draw (i) resonance structures and (ii) an atomic orbital representation of the

C O

H

H

Solution: (i)

H H

O

CH H

O

8

(ii)

C

H

H

O

π

σσ

σ

Illustration 11. Ethanal is more soluble in water than ethyl chloride. Explain.

Solution: This is due to the ability of ethanal to form hydrogen bonds with water.

Exercise 11.Predict the value of the angle (a) and (b) in the following figure.

C O

H

H

ab

Exercise 12.Explain the following fact: Dipole moment of ethanal = 2.72 D Dipole moment of diethyl ether = 1.18 D

Exercise 13.Give the increasing order of boiling point among following compounds.I : CH3CH2CH2OH II : CH3CH2OCH3 III : CH3CH2CHO IV : CH3CH2CH3

TYPE OF CHEMICAL REACTIONS IN CARBONYL COMPOUNDS

(i) Addition across C = O bond.

(ii) Replacement of carbonyl oxygen by other groups.

(iii) Oxidation

(iv) Reduction

(v) Reaction with alkalies

(vi) Miscellaneous reactions

1. Addition across C = O bond

Sr.

No.

Addition of Substrate Product

1. Hydrogen cyanide C O

C

OH

CN (Cyanohydrin)

2. Sodium bisulphite

(NaHSO3) C OC

OH

SO3 Na

(Bisulphite addition product)

3.Grignard reagent (RMgX)

followed by hydrolysis

HCHO ( )− − 03 2H C CH OH 1 alcohol

Aldehydes

(except

formaldehyde)

2° alcohol

Ketones 3° alcohol

4. Alcohols (ROH)C O

Hemiacetal which finally converts to

acetal

Illustration 12. Write the structure of compound A and B

( ) +

= → →3H OHCN3 2

CH C O A B

Solution:

CH3 C CN

OH

CH3

A CH3 C COOH

OH

CH3

B

Exercise 14.Give the product of the reaction of acetaldehyde with sodium hydrogen sulphite.

Exercise 15.The most reactive compound towards formation of cyanohydrin on treatment with KCN followed by acidification is(A) Benzaldehyde (B) p – Nitrobenzeldehyde(C) Phenyl acetaldehyde (D) p – Hydroxybenzeldehyde

2. Replacement of carbonyl oxygen atom with other groups

10

(a) Reaction with ammonia derivatives

Aldehydes & ketones react with a number of NH3 derivatives such as hydroxyl amine,

hydrazine, semicarbazide etc, in weak acidic medium. In general, if we represent these

derivatives by NH2 G, then their reaction with aldehydes & ketones can be represented

as follows:

C O H2N GH C N G H2O

Ammonia derivatives & their products with carbonyl compounds

G Ammonia Derivative Product obtained

OH NH2OH

Hydroxylamine C NOH

Oxime

NH2 NH2NH2

Hydrazine C NNH2

Hydrazone

NHC6H5 NH2NHC6H5

Phenyl hydrazine C NNHC6H5

Phenyl Hydrazone

NH

O2N

NO2 H2NHN

O2N

NO2

2, 4 – dinitrophenyl hydrazine

C N

NH

O2N

NO2

2,

4 – dinitrophenyl hydrazone

NHCONH2

H2NHNCNH2

O

Semicarbazide

NNHCNH2

O

Semicarbazone

(b) Reaction with ammonia

Like ammonia derivatives, ammonia also reacts with aldehyde (except formaldehyde)

& ketones to form the products, called imines.

C O

CH3

H

NH3 C

CH3

H NH2

OH

Acetaldimine

-H2OC NH

CH3

H

However, formaldehyde reacts with NH3 to form hexamethylene tetramine, (CH2)6N4 also

known as urotropine as shown below:

6CH2O 4NH3-6H2O

Hexamethylene tetramine

N

CH2

NCH2

N

CH2

N

CH2

H2C CH2

Acetone reacts with NH3 to form diacetonamine

2CH3COCH3 NH3

CH3

CH3

NH2

CH2COCH3

H2O

4 - amino - 4 - methyl pentan - 2 - one

(c) Reaction with primary amines

Aldehydes & ketones react with 10 amines to form Schiff;s bases. These compounds are also

called imines. + → = − +2 2Aldehyde AldimineAmine A Schiff ' s base

RCHO H NR RCH N R H O

C O

R

R

C N

R

R

R H2OH2NR

Ketone Ketimine

(d) Reaction with PCl5 or SOCl2 (thionyl chloride)

Aldehydes or ketones with PCl5 or thionyl chloride to form geminal dihalides.

C O C

Cl

Cl

POCl3PCl5

C O C

Cl

Cl

SO2SOCl2Pyridine

Illustration 13. Write the structural formula of the following ammonia derivatives:

12

(i) 2, 4 – dinitrophenyl hydrazine (ii) Semicarbazide (iii) Hydroxyl amine

Solution: (i)

O2N NHNH2

NO2

(ii) H2NNHCONH2 (iii) H2NOH

Exercise 16.

Which NH2 group of semicarbazide ( ) ( )a b

2 2H NCONH NH undergo condensation with the

carbonyl group?

Exercise 17.Complete the reaction:

(i) 2 2NH CONHNH3CH CHO →

(ii) ( ) 2 2NH NH3 2CH CO →

Exercise 18.O

2 2 3NH CH CH A, where A is−+ →

(A) (B)

CH2

(C) N OH (D)

N CH2CH3

3. Oxidation

Aldehydes are easily oxidised to carboxylic acids containing the same number of carbon

atoms, as in parent aldehyde.

C

H

O

R [O] C

OH

O

R

The reason for this easy oxidation is the presence of a hydrogen atom on the carbonyl

carbon, which can be converted into OH group without involving the cleavage of any other

bond. Hence, aldehydes are oxidised not only by strong oxidizing agent but also by weak

oxidizing agents. As a result, aldehydes act as strong reducing agents.

Aldehydes reduce Tollen’s reagent to Ag & appear in the form of silver mirror. This test is

called silver mirror test. It is given by all aldehydes & reducing sugars.

( )( )

3 3 22Silver mirror

RCHO 2 Ag NH 3OH RCOO 2Ag 4NH 2H O+

− ∆ − + + → + ↓ + +

Aldehydes (except benzaldehyde) reduce Fehling’s solution (Cu+2 reduced to Cu+) which

is an alkaline solution of cupric (Cu2+) ion complexed with tartarate ion. + − ∆ − ++ + → + +2

2red pptRCHO 2Cu 3OH RCOO 2Cu 2H O

Aldehydes also reduce Benedict’s solution (Cu2+ complexed with citrate ion) to Cu+.

Aldehydes & ketones with a methyl or methylene group adjacent to the carbonyl group

are oxidised by SeO2

CH3CHO SeO2 CHO

CHO

H2O Se

+ → + +3 3 2 3 2CH COCH SeO CH COCHO H O Se

Ketones are also oxidised by caro’s acid (H2SO5) or peroxybenzoic acid (C6H5CO3H) to

esters.

+ →2 5H SORCOR' O RCOOR'

O

6 5 3C H CO H →OO

O

C6H5CCH3

O

6 5 3C H CO H→ H5C6 O C CH3

O

It is called Bayer villiger oxidation.

It is exactly oxygen insertion between carbonyl carbon & the larger of two groups

attached to it.

Haloform Reaction

CH3CHO 3I2NaOH I3CCHO 3HI

Hydrolysis

CHI3 HCOONa

iodoform (yellow ppt)

Due to the formation of yellow ppt. of iodoform in this reaction, it is known as iodoform test & used

in for characterizing compound containing CH3CO or a group like CH3CH2OH which can be

easily oxidised to CH3CO group by halogens.

Illustration 14. Give a chemical test to distinguish between each of the following pair of organic compounds. (i) propanal an propanol (ii) propanone and propanal

Solution: (i) Propanal is an aldehyde and gives a silver mirror with Tollen’s reagent while propanol is an alcohol which do not respond silver mirror test positively.

(ii) (a) Propanone give yellow ppt. of iodoform on reaction with I2/NaOH while propanal does not react.

(b) Propanal gives silver mirror with Tollen’s reagent while propanone does not.

14

Illustration 15. What is Fehling’s solution?

Solution: Fehling solution is a mixture of alkaline copper sulphate (Fehling A) and sodium potassium tartrate (Fehling B).

Exercise 19:How does the oxidation state of copper changes when Fehling solution is added to acetaldehyde?

Exercise 20: Which of the following compound gives yellow precipitate with iodine and sodium hydroxide? (i) 3-methyl-4-phenyl but–3–en–2–one (ii) 1- phenyl ethanone (iii) Butanal (iv) Pentan–3–one

4. Reduction

Carbonyl compounds can be reduced to 1° or 2° alcohol, by LiAlH4, NaBH4 or direct reduction

with H2/Ni.

(a) 4LiAlH3 3 2CH CHO CH CH OH→

= →4LiAlH3 3 2 2 2CH CH CHCHO CH CH CH CH OH

with LiAlH4 CHO group is reduced to CH2OH (1° alcohol) and C = C bond is also

reduced when it is in conjugation with carbonyl groups. O

4LiAlH

H+→

OH

C O Group is reduced to CHOH (2° alcohol)

LiAlH4 also reduces ester & acid chloride to alcohols.

(b) NaBH4 has similar function. But this reagent does not affect (C = C) double bond.

OH2H /Ni← O 4NaBH

Ethanol→ OH

NaBH4 does not reduce ester & acid chloride CH2OH

OH

COOC2H5

4NaBH

Ethanol←

CHO

O

COOC2H5

4LiAlH →

CH2OH

OH

CH2OH

C2H5OH

(c) Amalgamated zinc, Zn(Hg) & conc. HCl (Clemmensen reduction) & hydrazine

(NH2 –NH2) followed by reaction with strong base like KOH in alkaline glycol

(Wolf Kishner reduction) reduces carbonyl group to alkyl group.

C NNH2−←2 2NH NH

Wolf kishner C O( ) +→2Zn Hg Conc.HCl /H O

Clemmensen reduction CH2

Glycol, KOH

CH2

(d) Reduction to pinacol

2CH3COCH3Mg Hg, Benzene−→ CH3 C

CH3

C

CH3

O O

Mg

CH32H O→ CH3 C

CH3

C

CH3

OH OH

CH3

(pinacol)

Illustration 16. Find A and B:

CHO

O

COOC2H5

→2H / NiWolf KischnerorClemmensen

−←A B

Solution: CH3

COOC2H5

CH2OH

COOC2H5

OH

A = B =

Illustration 17. Explain Clemmensen’s reduction.

Solution: In Clemmensen’s reduction, we reduce the carbonyl group to CH2 using zinc amalgam in concentrated HCl.

C O Zn /HgHCl→ CH2

Zn /HgHCl→CH3 C

H

O CH3 CH2

HEthaneEthanal

16

Illustration 18. Name the major product of the following reactions:

(i)Zn / Hg

3 2 HClCH CH CH CH CHO= − →

(ii)2 4N H

KOH / glycol→

CH3

CH3

O

Solution: (i) ( )= − − −3 2 3CH CH CH CH CH Pent 2 ene

(ii)CH3

CH3

1, 1 - dimethylcyclohexane

Exercise 21:Convert propanone into propane.

Exercise 22:Write the structures of A and B.

OH 2 4N HCu573K KOH / glycol

A B→ →

Exercise 23:The reduction:

O

O

O O

OH

O

can be brought about by(A) LiAlH4 (B) NaBH4

(C) Clemmensen’s reduction (D) Wolff−Kishner reduction

Exercise 24: The appropriate reagent for the transformation:

HO

CH3

O

HO

CH3

(A) Zn – Hg / HCl (B) (i) NH2 – NH2 (ii)OH − (C) Both (A) & (B) (D) None of these

Exercise 25:

Under Wolff Kishner reduction conditions, the conversion which may be brought about is

(A) H

OH

OH (B)

OH

(C)

O OH

(D)

O

5. Reaction with Alkalies

(A) Aldol Condensation

Two molecules of an aldehydes or a ketone having atleast one α - hydrogen atom, condense in presence of a dilute alkali to give a β - hydroxyaldehyde or β - hydroxy ketone.

CH3 C

O

H

HCH2CHOdil. NaOH

CH3 C

OH

H

CH2CHO

( 3- hydroxy butanal)Ethanal

The products of aldol condensation when heated with dilute acids undergo dehydration to form α, β - unsaturated aldehydes or ketones.

H3CHC CH2CHO

OH

2

H , heatH O

+

−→ H3CHC CHCHO

But-2-enal(crotonaldehyde)

In general all aldehydes & ketones which contain α - hydrogen can undergo this reaction.

Those which do not contain α - hydrogen like HCHO, C6H5CHO etc, do not undergo this

reaction.

Mechanism

Mechanism involves formation of carbanions (i) a nucleophile form first molecules which is

condensed with second molecule.

OH H CH2 C

O

H H2O C

O

HH2C

H2C C H

O

H2C C H

O

CH

O

H2C CH2H3CHC C

O

H

O

CH

O

CH3

18

CHCH3 CH2 C

O

H

O

H2O CHCH3 CH2 C

O

H

OH

OH

Aldol

Aldol product on dehydration give α, β - unsaturated ketones.

CHCH3 CH2 C

O

H

OH

CHCH3 CH CHOΔ

Illustration 19. Convert ethanal into 2 – butenal.

Solution:

dil. NaOH3CH CHO→ CH3 CH

OH

CH2CHO Heat

H+→ HCH3C CHCHO

Exercise 26:base

3HCHO CH CHO Major product will be+ →

(A)

CH3

OH

OH

(B)

CHOOH

(C)CHOOH

(D)

None

Exercise 27:

1. OH

2.X (ultimate ).

∆

−

→

O

CH2

CH3

O

+ The product X is:

(A)

O

(B)

CH3

O

(C)

O

(D)

O

(B) Cannizzaro’s reaction

Aldehydes that have no α-hydrogen atom (or acidic hydrogen) undergo cannizzaro reaction (CR) in which disproportionation reaction takes place one being reduced to alcohol & other being oxidised to salt of the corresponding carboxylic acid. The reaction lakes place with 50% aqueous or ethanolic alkali solution. 2HCHO NaOH CH3OH HCOONa

50%

ReductionOxidation

( ) ( ) ( )+ → +3 3 2 33 3 350%2 CH CCHO NaOH CH CCH OH CH CCOONa

When an aldehyde (showing CR) is treated with HCHO & 50% base, then HCHO undergo

oxidation (rather than any other aldehyde). This reaction is called crossed CR.

CHO HCHO NaOH CH2OH HCOONa

CR involving different aldehydes or same aldehydes is proton (H+), hydride (H−) transfer

reaction.

Mechanism

2NaOH HCH2OH2HCH

O

HCONa

O

Step I

H C

H

O H C

H

O

OH

OH

Step II

H C

H

O

OH

H C O

H

slowH C O

OH

H C O

H

H

HCOO HCH2OH

When the reaction is carried out in D2O instead of in H2O, it is found that there is no new

C – D bond formation. This indicate that the hydrogen must come from aldehyde & not from

the solvent.

Illustration 20. Identify aldehydes which can give cannizaro reaction (CR):

(a) CCl3CHO (b) (CH3)2CHCHO

(c) (CH3)3CCHCl2 (d) C6H5CHO

Solution: Aldehydes which do not have H at α - C give CR.

(i) a & d do not H at α - C hence give CR.

(ii) b has H at α - C but due to steric hindrance it gives CR.

(iii) c with OH− reactant is first converted to aldehyde which does not have H at

α - C hence give CR.

20

(CH3)3CCHCl2OH−

→(CH3)3CCH

OH

OH

2H O−→ (CH3)3CCHO

Exercise 28:

Identify A & B in the following reactions:

O dil . NaOH→ A ( )Sn Hg conc.HCl→ B

Exercise 29:Write the structure of A and B in the following reaction.

CHO dil . NaOH3CH CHO A− → Heat B→

Exercise 30:

→-

2OD /D OPhCHO

Find product of the reaction.

Exercise 31:

( conc.)OH X Y−+ → +D

O

D

( 2 moles)

where X and Y are:

(A)

D

O-

O

andOD

D

H

D (B)

H

O-

O

andOH

D

D

D

(C)

H

O-

O

andOH

D

H

D (D)

D

O-

O

andOH

D

D

D

Exercise 32:In the Cannizzaro reaction given below:

Ph CHO Ph CH2OH PhCOO the slowest step isOH−

→(A) The attack of OH − at the carbonyl group (B) The transfer of hydride to the carbonyl group.(C) The abstraction of proton from carboxylic acid (D) The deprotonation of Ph – COOH.

(C) Perkin reaction

In this reaction aromatic aldehyde is heated with an acid anhydride & its corresponding

sodium salt to form condensation products which on hydrolysis gives α, β - unsaturated acids.

Acetic anhydride & sodium acetate are commonly used in this reaction.

(CH3CO)2OCH3COONa

Cinnamic acidCH3COOHPhCHO Ph CH CH COOH

6. Miscellaneous reactions

(i) Formation of phorone

Three moles of acetone condense in the presence of dry HCl to form phorone.

C O

CH3

CH3

dry HCl gas-H2O

C CH

CH3

CH3

C

O

CH CCH3

CH3

Phorone

3

(ii) Formation of mesitylene

Three moles of acetone on refluxing with conc. Sulphuric acid produces mesitylene as one of

the products.

C O

CH3

CH3

3 2 4

2

conc.H SO

3H O− →

CH3

CH3 CH3

Mesitylene

(iii) Reaction with alc. KCN

On heating with ethanolic solution of KCN, two molecules of aromatic aldehyde undergo

condensation to form benzoin. It is called benzoin condensation.

2 CH O

benzaldehyde

alc.KCNΔ

→ C CH

O OH

benzoin

(iv) Reaction with chloroform

Ketones condenses with chloroform in presence of alkali to form chloretone.

22

C O

CH3

CH3

C

CH3

CCl3

OH

CHCl3KOH

CH3

Illustration 21. Convert acetone into mesityl oxide.

Solution:

( )2dil. Ba OH→CH3 C

CH3

O CH3 C

CH3

CH2COCH3

OH

Heat→ CH3 C

CH3

CHCOCH3

Exercise 33:What is trioxane? Give its structure.

ANSWER TO EXERCISES

Exercise 1:

CHO

Exercise 2:

The structure of diphenyl ketone is

C

O

HH

H

H

H

H H

H

H

H

Number of sigma bonds = 25

Exercise 3:

We remove aldehyde as soon as it is formed thus preventing its further oxidation to

carboxylic acid.

Exercise 4:

A → HOCH2 CH2 CH2 CHO

B OHC CH2 CH2 C

O

Cl

Exercise 5:

2NaNH→ CH3 C C CH3

H2SO4 & HgSO4

CH3 C

Cl

CH CH3

CH3 C CH2 CH3

O

Exercise 6:

The hydrocarbon (A) is 2, 3 – dimethylbut-2-ene. CH3 C C CH3

CH3 CH3

24

Exercise 7:

(i) sp2 & sp3 respectively.

(ii)O

O

2HCHO

Exercise 8: (A)

Exercise 9: (A)

Exercise 10: (C)

Exercise 11:The angle (a) is 120° while angle (b) is 116.5°, slightly less than 120°.

Exercise 12:The large value of the dipole moment of ethanal is due to larger contribution of the

dipolar ion structure (II).

H H

O

CH H

O

(I) (II)

Exercise 13: The correct order of increasing boiling point is IV < II < III < I

Exercise 14:

CH3 C O

H

3NaHSO→

CH3

H

ONa

So3H

H ion+

→

CH3

H

OH

SO3Na

Crystalline product

Exercise 15:(B)

Exercise 16:

The NH2 group labelled (b) undergo condensation with carbonyl group. The NH2 group

(a) is involved in resonance with CO group and is deactivated.

NH2 C NHNH2

O

H2N C NHNH2

O

Exercise 17:

(i) → =2 2NH CONHNH3 3 2CH CHO CH CH NNHCONH

(ii) ( ) 2 2NH NH3 2

CH CO→ CH3 C

CH3

NNH2

Exercise 18: (D)

Exercise 19: The oxidation state of copper change from +2 to +1.

Exercise20:

(i) 3-methyl-4-phenyl but – 3 – en – 2 – one

(ii) 1- phenyl ethanone

Exercise 21:

→Zn/HgHCl

CH3C

O

CH3 CH3 CH2 CH3

Exercise 22:

OA

cyclohexanone

B

Cyclohexane

Exercise 23:(B)

Exercise 24: (B)

Exercise 25:(D)

Exercise 26:(B)

Exercise 27:(A)

Exercise 28:

Cyclohexanone undergoes aldol condensation in presence of dil. NaOH.

O OH−

→H

O

OH

O

(A)

OH

(B)

Sn(Hg) conc.HCl

26

Exercise 29:

CH

OH

CH2 CHOA

OB

Exercise 30:

Ph H

OD

HPh O

O

Exercise 31:(D)

Exercise 32:(B)

Exercise 33:

Trioxane is a cyclic trimer of methanal. Its structure is

CH2

O

O

CH2

H2C

O

MISCELLANEOUS EXERCISES

Exercise 1: Give the industrial preparation of ethanal.

Exercise 2: Write an equation for making aldehydes by the oxo process.

Exercise 3: What happens when ethanal reacts with excess of methanol in the presence of

trace amount of dry HCl?

Exercise 4: Convert acetic acid into acetaoxime.

Exercise 5: Give some physical properties of acetone.

Exercise 6: Which aldehyde smells like bitter almonds? Give its one use.

Exercise 7: Give one difference between the paraldehyde and metaldehydes.

Exercise 8: What type of aldehyde undergo cannizzaro reaction?

Exercise 9: What happens when ethanal is distilled with dilute H2SO4 at 273 K?

Exercise 10: What is formalin? Give its one use.

28

ANSWER TO MISCELLANEOUS EXERCISES

Exercise 1: Wacker’s process

= + →2

2

Pd Cl2 2 2 3CuCl

1CH CH O CH CHO

2

Exercise 2: Alkene on reaction with (CO + H2) in the presence of dicobalt octacarbonyl as a

catalyst giving an aldehyde with one carbon more. ( )= + + →2 8

Co CO

2 2 2 2RCH CH CO H RCH CH CHO

Exercise 3:CH3 CH O

H OCH3

H OCH3

HCl→ HCH3COCH3

OCH3

1, 1 - dimethoxyethane

Exercise 4: ( )→ →2 2Ca OH H NOH

3 3 3dry distill HClCH COOH CH COCH

CH3 C

CH3

NOH

Exercise 5: (i) It is a colourless, pleasant-smelling liquid.

(ii) It is miscible in water.

Exercise 6: Benzaldehyde smells like bitter almonds. It is used in perfumes.

Exercise 7: Paraldehyde is a cyclic trimer of CH3CHO and is a pleasant smelling liquid.

Metaldehyde is a cyclic tetramer of CH3CHO and is a white solid.

Exercise 8: Aldehyde without α-hydrogen atom undergo cannizzaro reaction.

Example, HCHO,

O

Exercise 9: Ethanal form metaldehyde (cyclic tetramer of ethanal).

Exercise 10: 40% aqueous solution of formaldehyde is called formalin. It is used as a

disinfectant.

SOLVED PROBLEMS

Subjective:

Board Type Questions

Prob 1. Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reaction: Benzaldehyde, p – tolualdehyde, p – nitrobenzaldehyde, acetophenone.

Sol. Acetophenone is ketone, all other are aldehydes, ∴ it is less reactive. +M group increases electron density and so makes the compound unfit for nucleophilic addition & vice versa is true for M group. Order is: Acetophenone < p – tolualdehyde < benzaldehyde < p – nitrobenzaldehyde

Prob 2. Identify A, B & C in the following

CH3

→NBS A NaCN→ B →

MgBr

H3OC

Sol. CH3

NBSallylic substitution

→

CH2Br

N

NaCNS

→

CH2CN

MgBr

H3OH2C C

O

(A) (B)

(C)

Prob 3. Complete the following:

(i)

CH3

O

CH3NaOCl→

(ii)

C CH3O2N

O

NaOI→

30

(iii)

2I / NaOH→

CH3 CH CH2 C

O

CH3

Sol. (i)CH CH

CH3

COONa CHCl3

(ii)COONaO2N CHI3

(iii)CH3 CH CH2 COONa CHI3

Prob 4. What is the final product of?

O OH , ∆−

→

Sol.O OH H2O O

O O

O

O

H2O

OH

O

Δ

O

Prob 5. Arrange the following in increasing extent of hydration: O O

Br

CHO CHO

Br

Sol. The extent of hydration increases as groups tendency to form H – bonding increases. In general hydration of an aldehyde is greater than that of a ketone.

O O

Br

CHO CHO

Br

< < <

Prob 6. Identify A, B & C in the following O

APt, 1 equ. H2

B

C

Raney Ni, H2

NaBH4, CH3OH

Sol. With Pt, only double bond is affectedO

A is

With Raney Ni, −

||O

Cgroups as well as double bond is reduced.

OH

B is

NaBH4, CH3OH does not reduce double bond.

OH

C is

Prob 7. Compound A, having the empirical formula C7H8 is chlorinated in sunlight to give a product which is hydrolysed to produce B. B after oxidation reacts with acetic anhydride in the Perkin reaction to produce an acid C, which has an equivalent weight of 148. Give the name & structure of A, B & C.

32

Sol. Since B on oxidation undergoes Perkin reaction, ∴ the oxidised compound is benzaldehyde & A is toluene.

CH3

(A)

2Cl , hν →

CH2Cl

2H O / OH−

→

CH2OH

(B)

CHO

Oxidation

Perkinreaction

←

HC CHCOOH

(C)

Prob 8. Which of the following ketone is more acidic? Give a reason. O O

Sol. O

Base

O

This ketone is more acidic because the resulting enolate ion obeys Huckel’s rule & is thus more stable.

Prob 9. Explain why acetophenone but not benzophenone forms an adduct with NaHSO3.

Sol. Because of steric hinderance of two phenyl groups.

Prob 10. Which of the following compound gives positive iodoform test? (a) 2 – butanol (b) 1 – pentanol (c) Acetone (d) 3 – pentanone

Sol. Compounds containing ||

3

O

C CH− − &

− 3IOH

C HCH will give a positive iodoform test.

∴ 2 butanol & acetone will give positive iodoform test. Prob 11. Suggest a chemical test that will distinguished the following pairs of compounds

(a) Acetic acid & propanal (b) Acetaldehyde & acetone

Sol. (a) Propanal gives Tollens test. (b) Acetaldehyde gives Tollen’s test.

Prob 12. Complete the following reaction by giving the principal organic product. Br

Cl

2Mg / ether→ 6 5C H CHO

2 moles→ HOH / H +

→A B

Sol.

Br

Cl

2Mg / ether →

MgBr

MgCl

CHO

(i)

(i) HOH/H

HOHC

CHOH-C6H5

C6H5

(A) (B)

Prob 13. Starting with cyclohexanone suggest reasonable synthesis for the following: D

D

O

DD

Sol. D

D

O

DD

D2O/OHO

Prob 14. With in each pair which compound will react faster in carbonyl addition reactions:

(a) O

or

O

(b)

CHO or C

O

CH3

Sol. (a) Cyclopropanone

(b) Benzaldehyde

34

Prob 15. Complete the following reaction: CHO

OCH3

HCHOKOH

P

Sol. CH2OH

OCH3

HCOO KProducts =

Objective:

Prob 1. The correct order of reactivity towards nucleophilic addition reaction will be

(A) ( )3 3 3 2 52CH CHO CH CO CH COC H> >

(B) ( )3 3 3 2 52CH CO CH CHO CH COC H> >

(C) ( )3 2 5 3 3 2CH COC H CH CHO CH CO< >

(D) ( )3 3 2 5 3 2CH CHO CH COC H CH CO< >

Sol. Aldehydes are more reactive than ketones due to lesser number of +I group & again the reactivity of carbonyl group depends on the major nature of alkyl group attached to it. So correct order is

( )> >3 3 3 2 52CH CHO CH CO CH COC H

∴ (A)

Prob 2. Which of the following will not give aldol condensation? (A) RCH2CHO (B) RCOCH2R (C) C6H5CHO (D) C6H5CH2CHO

Sol. Aldol condensation is given by those carbonyl compounds, which possess at least one α - H. Q (C) is not having α hydrogen hence it will not give aldol condensation ∴ (C)

Prob 3. Aldehydes & ketones can be reduced to corresponding hydrocarbons by (A) refluxing with strong acids (B) passing the vapours under heated PbO2 (C) Refluxing with zinc amalgam (D) Refluxing with strong alkali

Sol. ∴ (C)

Prob 4. Which of the following compound gives a positive iodoform test? (A) pentanal (B) 1 – phenyl ethanol (C) 2 – phenyl ethanol (D) 3 – pentanol

Sol. Haloform test is used to identify COCH3 group or CH3 CH(OH) group which can

be oxidised to − −||

3

O

CH C. ∴ 1 – phenyl ethanol will give this test.

∴ (B)

Prob 5. Which of the following compound will give Cannizzaro’s reaction?

(A) CH3CH2CH2CHO (B)3

|

3 2

CH

CH CH CH CHO− −(C) (CH3)2CHCH2CHO (D) (CH3)3C. CHO

Sol. The compounds which do not contain α - hydrogen atom undergo cannizzaro’s reaction. ∴ (D)

36

Prob 6. Compound A(C6H10O) form a phenyl hydrazone and gives a negative Tollen’s reagent test & iodoform test. On reduction with Zn/Hg, HCl compound A gives n – pentane. A is (A) 1° alcohol (B) aldehyde (C) 2° alcohol (D) ketone

Sol. As A forms phenyl hydrazone it must be a carbonyl compound again, it is giving negative test with Tollen’s reagent test, it is a ketone. ∴ (D)

Prob 7. The end product of the reaction is

NBS → X Mg / ether→ Y ( )( )

3

2

i CH CN

ii H O→ Z

(A)

COCH3

(B)

CO COOH

(C)

COCH2OH

(D) none

Sol.

NBS→ Mgether→

Br Mg Br

(i) CH3CN (ii) H2O

CO CH3

∴ (A)

Prob 8. Arrange the following compounds in order of their decreasing activity towards Tollen’s reagent: HCHO CH3CHO CH3COCH3 C6H5CHO I II III IV (A) I > II > IV > III (B) I > II > III > IV (C) II > I > III > IV (D) none

Sol. Aldehydes are more reactive than ketones ∴ (A)

Prob 9. The most reactive compound towards formation of cyanohydrin on treatment with KCN followed by acidification is (A) benzaldehyde (B) p – nitrobenzaldehyde (C) phenylacetaldehyde (D) p – hydroxybenzaldehyde

Sol. Presence of electron withdrawing group in para position increases nucleophilic addition.∴ (B)

Prob 10. Which of the following chemical system has most acidic hydrogen? (A) 3 – hexanone (B) 2, 4 – hexanedione (C) 2, 5 – hexanedione (D) 2, 3 – hexanedione

Sol. 2, 4 – hexanedione has CH2 group surrounded on both sides of electronegative (>C=O) group, hence it is acidic in nature, the carbanions

CH3 C

O

CH C

O

CH2CH3, once formed stabilizes due to resonance.

∴ (B)

Prob 11. Predict the product ‘B’ in the sequence of reaction:

( )+≡ → →2 42

3% H SO NaOH

HgCH CH A B

(A)CH3COONa (B

)CH3COOH

(C)

CH3CHO (D)

CH3 CH

OH

CH2CHO

Sol. ( )+≡ → →2 42

H SO NaOH3 3 2Aldol condensationHg

CH CH CH CHO CH CH OH CH CHO

∴ (D)

Prob 12. Acetophenone can be prepared by (I) Oxidation of 1 – phenylethanol (II) Reduction of benzaldehyde with methyl magnesium bromide (III) Friedel – crafts reaction of benzene with acetylchloride(IV) Distillation of calcium benzoate Which of above is correct? (A) I & III (B) II & IV(C) III & IV (D) I & II

Sol. → ← +3AlCl[O]6 5 3 6 5 3 6 6 3C H CHOHCH C H COCH C H CH COCl

∴ (A)

Prob 13. An organic compound ‘A’ has the molecular formula C3H6O. It undergoes iodoform test. When saturated with HCl it gives ‘B’ of molecular formula C9H14O, ‘A’ & ‘B’ respectively are (A) propanal & mesitylene (B) propanone & mesityl oxide (C) propanone & 2, 6 – dimethyl – 2, 5 – heptadien – 4 – one (D) propane & meistylene oxide

Sol. Since the compound A has the molecular formula C3H6O & undergoes iodoform test, it should be CH3COCH3. Further reaction of B with HCl gives C9H14O, this should be trimer of acetone, i.e. It should be phorone.

38

∴ (C)Prob 14. What will be the product when 2-Butenal reacts with sodiumborohydride?

(A) Butanal (B) Butanol(C) But-2-ene-1-ol (D) None

Sol. ∴ (C)

Prob 15. Which will give a yellow precipitate with iodine & alkali? (A) 2 – hydroxyl propane (B) benzophenone (C) o – methyl toluene (D) acetamide

Sol. ∴ (A)

ASSIGNMENT PROBLEMS

Subjective:

Level – O

1. Suggest a reason for large difference in the boiling points of butanol and butanal, although they have same solubility in water.

2. Arrange the following in order of their increasing reactivity towards HCN?

3 3 3 2 5 3CH CHO, CH COCH , HCHO, C H COCH

3. To what oxidation state does ethanal reduce Cu (II).

4. Write the I.U.P.A.C. name of

CH3 CH

OH

CH2 C

O

CH3

5. Identify A, B and C:

CH3C OH

O

2 2 2Ca(OH) NH . NHA B C∆→ → →

6. Complete the following reaction:CHO

NaOH+ →

7. Find the product:

CH3 CH CH2

OH

CO CH3∆

8. Give the equation associated with Fehling’s Test given by CH3CHO.

9. How will you convert acetone into 2 – methyl – 2 – propanol?

10. Mention an industrial product manufactured from methanal.

11. Account for the following: (i) In the reactions of aldehyde/ketones with ammonia derivative, the pH has to be carefully

controlled. (ii) There are two –NH2 groups in semicarbazide. However, only one is involved in the

formation of semicarbazones. (iii) Cyclohexanone forms cyanohydrin in good yiled but 2, 2, 6 – trimethyl cyclohexanone

does not.

40

12. Effect the following conversions:(a) Acetone to propyne(b) Acetaldehyde to acetaldehyde semicarbazone.

13. Give one chemical test to distinguish following pair of compounds in solution:Acetone and acetaldehydes.

14. Give one chemical test to distinguish :Acetaldehyde and benzaldehyde.

15. Write the equations and conditions to show how the following conversions are carried out:(a) Benzaldehyde to acetophenone(b) Acetaldehyde to 2 – Butenal

Level – I

1. Arrange the following in order of decreasing acidity: CH2CH2CHO CH2CH2CHO

Cl

CH3CHCHO

NO2

(I) (II) (III)

2. The following ketone is unusually basic. Explain. O

3. Name the alkyne which can give the following carbonyl compounds on acid catalysed hydration.

(i)

(CH3)3CCCH3

O (ii)

C

O

CH3

4. Which alkene would give the following products on ozonolysis?

OHCO

O

HCHO

5. Complete the following reaction O

HCN

6. Explain why only α-hydrogen atom in aldehydes & ketones is involved in the aldol condensation.

42

7. Which of the following compound gives a positive iodoform test? 2 pentanone & 3 – hexanol

8. Arrange the following compounds in order of increasing reactivity towards the addition of HCN. Acetone, acetaldehyde, methyl t – butyl ketone & di – t – butyl ketone.

9. Show the product of following reaction NH2

NH2

C

CO

O

H

H

2H O

NaHSO4→

10. Identify products when PhCOCHCl2 undergoes Cannizzaro reaction:

11. Give the aldol condensation product of: (a) CH3CH = CHCHO

(b) Mixture of CH CHCHO & CH3CH CHCHO

12. What happens when cyclopentanone is treated with ethyl magnesium bromide & the product is hydrolysed?

13. What starting compound would you use in an aldol cyclisation to prepare each of?

(a) O

CH3

(b)

O

14. What reagents would you use to carry out following reactions? Ethyl bromide → 1-butyne → 2 – butanone

15. Arrange the following in increasing extent of hydration:

CH3C

O

CH3 , CH3C

O

CH2Cl , CH3 C

H

O , ClCH2C

O

H , HC

O

H

Level – II

1. Identify the product of

(i)

CHO C

O

CH3Δ, OH−

→

(ii)O OH ,Δ−

→

2. Complete the following reaction:

CH CH CH3MgBr4CH−→ X 2

3

CO

H O+→ Y 4 2 4HgSO H SO+→ZKMnO4

M

3. Write the final product of

O O

H3O

4. Complete the following reaction:

CH3C

O

CH2C

O

CH2CH3( )

( )3

3

i CH MgBr

ii H O (excess)+→

5. Identify A, B & C in the following O

O

OH−

→A Δ→B 4LiAlH→ C

6. Complete the following reaction: + →HCl

3 2 2 3 3 2 gasCH CH COCH CH CH CH ONO X

7. Show the mechanism of following reaction: O

HCl

O

H

H

HCl

44

8. Supply the structure from A to E:

PhCH CHCCH3

O

CH3CH2MgBr

H3O

H2

Pt

NaBH4

H

LiCu(CH3)2

H

O3

Zn/H

A

B

C

D

E

9. Effect the following conversion:

(a) O O

(b) O O

CH2CH2C

O

CH(CH3)2

10. How will you bring about the following conversions in not more than two steps? (i) Propanone to propene (ii) Propanal to butanone (iii) Benzaldehyde to benzophenone (iv) Ethanol to 3-hydroxybutanal (v) Benzaldehyde to 3 – phenyl propan – 1- ol(vi) Benzene to m-Nitroacetophenone

11. Illustrate the mechanism of reaction of carbonyl group with an ammonia derivative, H2NZ.

12. Give simple chemical test to distinguish between the following pairs of compounds: (i) Propanal and propanone (ii) Ethanal and propanal (iii) Propanal and Diethyl ether (iv) Propanone and propanol

13. Explain haloform reaction.

14. An organic compound with the molecular formula C9H10O form 2, 4-DNP derivative, reduces Tollen’s reagent and undergo Cannizzaro reaction. On vigorous oxidation, it gives benzene-1, 2- dicarboxylic acid. Identify the compound and write the reactions involved.

15. Compound A on treatment with 5PCl gives compound B which on reduction with 2H /Pd in

presence of 4BaSO gives compound C. C gives Tollen’s test, Fehling’s test and iodoform test. When C is treated with dil. NaOH, D is obtained, which on heating gives crotonaldehyde. Identify A, B, C, D and complete sequence of reaction.

Objective:

Level-I

1. Polarization of electrons in acrolein may be written as:

(A)δ− δ+

= − =2CH CH CH O (B)δ− δ+

= − =2CH CH CH O

(C)δ+ δ−

= − =2CH CH CH O (D)δ+ δ−

= − =2CH CH CH O

2. Which reagent will perform the following reduction?− = − → − = −3 3 2CH CH CH CHO CH CH CH CH O H

(A) LiAlH4 (B) NaBH4

(C) H2/Ni (D) Mg/Hg + H2O

3. The ketone that does not form a cyanohydrin is

(A) C6H5COCH3 (B) CH3COCH3

(C) C6H5COC6H5 (D) CH3COC6H5

4. The decreasing order of solubility of methanal (I), propanaldehye (II), benzaldehyde (III)

& acetophenone (IV) is

(A) I, II, III, IV (B) IV, III, II, I

(C) IV, I, II, III (D) II, I, III, IV

5. Which of the following compounds will undergo self aldol condensation in presence of

dil. alkali?

(A) C6H5CHO (B) CH ≡ C CHO

(C) CH3CH2CHO (D) CH2 = CHCHO

6. Which of the following does not undergo benzoin condensation?

(A) CHO (B) CHO

OCH3

(C) CHO

CH3

(D) C6H5CH2CHO

46

7. In the reaction

C

H

CH3CH2

O( )

( )2 4

4

i HCN / H SO

ii LiAlH →X

,

What is X?

(A)CH3CH2CHCH2NH2

OH

(B)CH3CH2CH2CHCH2NH2

OH

(C)CH3 CHCH2CH2CH3

OH

(D)

C NH2

CH3

OH

CH3CH2

8. Identify the final product Z in the following sequence of reaction:

[ ] +

= + → → →3 2 4H O H SO2Me C O HCN X Y Z

(A) (CH3)2C(OH)COOH (B) CH2 = C(CH3)COOH

(C) HO CH2CH(CH3)COOH (D) CH3CH = CHCOOH

9. A dihalo alkane on hydrolysis produces a ketone with formula C3H6O. The dihalo alkane is

(A) 2, 2 – dichloropropane (B) 1, 1 – dichloropropane

(C) 1, 2 – dichloropropane (D) 1, 3 – dichloropropane

10. ON2 COCl 3LiAl(O Bu i) A, A can be− −→

(A) CHO (B) H2N COCl

(C) ON2 CHO (D)ON2 C CH3

O

11. Identify ‘X’ in the sequence +→ →2 2 7 2

2 4

K Cr O I NaOH3 6 3H SO water

X C H O CHI

(A) CH3 CH2 CH2OH (B)CH3 CH CH3

OH

(C) CH3 O CH2CH3 (D) CH3 CH2CHO

12.CH3

O

CH3

HCl32CH OH A, where A is+ →

(A)

CH3

CH3

OCH3H3CO

(B)

An acetal

(C) Propanaldehyde dimethyl acetal (D) All the above

13. Aldol condensation will not be observed in

(A) Chloral (B) Phenyl acetaldehyde

(C) Hexanal (D) None of these

14. Greatest amount of hydration is in

(A) O (B) O

Br

(C)

C H

O

Br

(D) All are equal

15. Identify Z in the series

= → → →2 3

2

Na COHBr hydrolysis2 2 I excess

CH CH X Y Z

(A) C2H5I (B) C2H5OH

(C) CHI3 (D) CH3CHO

16. The product (s) obtained via Oxymercuration (HgSO4 +H2SO4) of butyne - 1 would be

(A) CH3 CH2 COCH3 (B) CH3 CH2 CH2 CHO

(C) CH3 CH2 CHO CH2O (D) CH3 CH2 COOH HCOOH

17. Which one of the following reactions can not be used for the reduction of

C

R

R

O CH2

R

R

48

(A) Clemmensen reduction (B) Wolf – Kishner reduction

(C) Wurtz reaction (D) HI and red phosphorous at 2000C.18. An organic compound C3H6O does not give a precipitate with 2, 4 – dinitro phenyl hydrazine

and does not react with metallic sodium. It could be(A) CH3CH2CHO (B) CH3COCH3

(C) CH2 = CH – CH2OH (D) CH2 = CH – OCH3

19. A ketone reacted with Grignard reagent followed by hydrolysis gave a product which on dehydration gave an alkene. The alkene on ozonolysis gave diethyl ketone and acetaldehyde. The ketone is(A) Dimethyl ketone (B) Ethyl methyl ketone (C) Diethyl ketone (D) Ethyl isopropyl ketone

20. Iodoform test is applicable for

(A)

H CH3

O (B)

CH3 CH3

O

(C)

CH3

O (D) All the above

Level – II

1. Acetaldehyde reacts with

(A) electrophiles only (B) nucleophiles only

(C) free radicals only (D) both electrophiles & nucleophiles

2. Aldehydes & ketones give addition (not condensation) reactions with

(A) hydrazine (B) phenyl hydrazine

(C) semicarbazine (D) hydrogen cyanide

3. ( ) ( )+

→ →3H ONaCN3 HCl2

CH CO A B

In the above sequential reactions A & B are

(A) (CH3)2C(OH)CN, (CH3)2C(OH)COOH (B) (CH3)2C(OH)CN, (CH3)2C(OH)2

(C) (CH3)2C(OH)CN, (CH3)2CHCOOH (D) (CH3)2C(OH)CN, (CH3)2C=O

4. In which of the following reactions, aldehydes & ketones are distinguished

(A) reactions with phenyl hydrazine

(B) reactions with hydroxylamine

(C) reactions with semicarbazide

(D) reactions with silver nitrate mixed with ammonia

5. The most probable structural formula for the compound whose empirical formula is C3H6O

& which can react with Benedict reagent is

(A)CH3CH CH2

O(B) CH3CH2CHO

(C) CH3OCH = CH2 (D) CH2 = CHCH2OH

6. Base catalysed aldol condensation occurs with:

(A) propionaldehyde (B) 2 – methyl propionaldehyde

(C) both (D) none

7. Aromatic aldehydeso undergoes disproportionation in presence of sodium or potassium

hydroxide to give corresponding alcohols & acid. The reaction is known as

(A) Wurtz reaction (B) cannizzaro reaction

(C) friedel crafts reactions (D) claisen reaction

50

8. CHO OHC

OHCCHO

( )( )

0

2

i NaOH/100 C

ii H /H O+→ Major prodcut is

(A)

COOH

COOHHOOC

HOOC

(B)

COOH

HOOCOH

OH

(C)

O

O

O

O

(D)

CH2OH

HOH2C

HOH2C

CH2OH

9. Compound ‘A’ (molecular formula C3H8O) is treated with acidified K2Cr2O7 to form a product

(B) (molecular formula C3H6O). ‘B’ forms a shining silver mirror on warming with ammoniacal

AgNO3. ‘B’ when treated with an aq. Solution to H2NCONHNH2, HCl & Sodium acetate gives

a product ‘C’. Identify the product ‘C’.

(A) CH3CH2CH = NNHCONH2 (B)CH3 C NNHCONH2

CH3

(C)CH3 C NCONHNH2

CH3

(D) CH3CH2CH = NCONHNH2

10. In a cannizzaro reaction, the intermediate that will be best hydride donar is

(A)

OH

H

O

(B)

O

H

O

(C)

O

H

O

OMe

(D)

O

H

O

O2N

11. Mixture of CH3CH2OH & CH3CHO can be separated by using

(A) NaHSO3 (B) NH2OH

(C) HCN (D) NH2 NH2

12.

( )( )( )

i KOBr,Δ

ii HiiiΔ

+→

O

COCH3

CH3

Y, Y is

(A)O

COOH

CH3

(B)O

CH3

(C)COOH

OH

(D)COOH

OH

CH3

13.

2H O→

O

Cl2 X, 3CHCl→

CHO

Br2 Y

X and Y are

(A)O

Cl CHO

Br,

(B)O

Cl

CHO

Br

,

(C)OH

ClCHO

Br,

(D) None is correct

52

14. Following is hydrated maximum at the position: O

O

O

1

2

3

(A) 1 (B) 2

(C) 3 (D) equal

15.CH3 C

O

CH2 CH2 CH2 C

O

CH3OH , Δ−

→ A

A is

(A)

CH3 C C CH2

O

CH2CCH3

(B)

CH3 C C CH

O

CH2CHCH3

(C)

CH3 C CH CH

O

CHCHCH3

(D) None is correct

16. 3

2 2

CrO , pyridineHBr alc. KOH HBr aq. KOH2 Peroxide CH Cl

RCH CH A B C D E= → → → → →

In above conversion E can be

(A) Propanol (B) Butanal

(C) Pentanal (D) Both (B) and (C)

17. Identify the final product (Z) in the following sequence of reactions.

→4KMnO .dil (X) →4HIO (Y) →−OH (Z)

(A)OH

OH

(B) OH

O

(C)

O

(D)

O

18. The conversion; CH3–CH=CH–CHO → CH3–CH=CHCOOH can be effected by

(A) alk. KMnO4 (B) K2Cr2O7/H+

(C) ammoniacal AgNO3 (D) SeO2

19. The abstraction of proton will be fastest, in which carbon in the following compound,

CH3 CH3

O

xy

z p

(A) x (B) y(C) z (D) p

20. Which of the following will be most acidic?

(A)

CH3

O

(B)CH3

O

(C)CH3 CH3

O

(D) CH3

O

CH3

CH3

54

ANSWERS TO ASSIGNMENT PROBLEMS

Subjective:

Level – O

1. Butanol undergoes hydrogen bonding but butanal does not.

2. < < <2 5 3 3 3 3C H COCH CH COCH CH CHO HCHO

3. +1 oxidation state.

4. 4 – Hydroxypentan – 2 – one

5. = 3 2A (CH COO) Ca

B = CH3 C

O

CH3

= =3 2 2C (CH ) C NNH

6. CH2OH COONa

7. CH3 CH CH CO CH3

8. + + → + + +3 4 3 2 2 4 2CH CHO 2CuSO 4NaOH CH COOH Cu O Na SO 2H O

Red ppt.

9.

CH3 C

O

CH3 3 3CH MgBr H O+

→ → CH3 C

OH

CH3

CH3

10. Bakelite.

11. (i) The reaction

C O C

OH

NHZ2H N Z+ − → 2H O−→ C N Z is pH sensitive.

At lower pH (solution is highly acidic), the protonation of N takes place thereby decreasing the concentration of free base available.

NH2

Z

3H H N Z+ ++ −��������

The H3N+ Z can not attack now at the carbonyl carbon. At higher pH, (solution is basic), the base will remove H+ ion from OH in the intermediate formed thereby slowing the dehydration step.

(ii) In semicarbazide, the NH2 group closer to the − =

|C O

group is resonance stabilized

and is deactivated (lone pair of N is involved in resonance) compared to the other end NH2 group).

H2NHN C

O

NH2 HNNH2 C

O

NH2

deactivated

(iii) 2, 2, 6 – trimethyl cyclohexanone does not form cyanohydrin as the carbonyl carbon is highly crowded and the electron density is very high due to the presence of three e−

donating methyl groups.

O HCN→

OH

CN

O

CH3CH3

CH3

HCN→ no reaction

12. (a) CH3 C

O

CH3 2SOCl→ CH3 C CH3

Cl Cl

2NaNH→ CH3 C CHPropyne

2, 3 - dichloro propane

(b)

CH3 C O

H

H2N. NH C

O

NH2H+

→ CH3 C N.NHCNH2

H

OSemicarbazide

13. Acetaldehydes on warming with Fehling’s solution gives red precipitate of copper oxide, where as acetone does not.

+ + → + +3 2 2 2CH CHO 2Cu(OH) NaOH RCOONa Cu O 3H O

(Red ppt.)

14. These two compounds can be distinguished by Fehling’s test. Acetaldehyde gives red coloured ppt. with Fehling’s solution but benzaldehyde does not.

15. (a) CHO

4Alk. KMnO→

COOH

Benzoic acid

2SOCl→

COCl

3 2Cd(CH )→

COCH3

(b)

dil. NaOH32CH CHO → CH3CH CH2CHO

OH

Aldol3CH CH CHCHO∆→ =

2 - Butenal

56

Level – I

1. III > II > I

2. The salt of ketone is resonance stabilized.

O OH OH

etcH

+

3. (i) (H3C)3C C CH

(ii)C CH

4.

5. CNOH

6. Because an α-hydrogen atom is acidic & the resulting enolate anion is stabilized by

resonance.

7. 2 – pentanone

8. di – t – butyl ketone < methyl – t – butyl ketone < acetone < acetaldehyde.

9.N

N

10. COCHCl2

KOH

COCH(OH)2 COCHO

OH

C

O

C

O

H

OH

-HC

O

C

O

O

CH

OH

COO

H

11. (a) CH3CH CHCHO H CH2CH CHCHO

Δ 2H O−

CH3CH CHCH CH

CH

CHOHC

(b)

CH CHCHO & H2CH CH CHCHO

(CH=CH)3CHO

12. OCH2CH3OH

+→3 2

3

CH CH MgBr

H O

13. (i) 3 2 2 3|| ||CH CCH CH CCH

O O

(ii) 3 2 2 2||CH CCH CH CH CHO

O

58

14. CH3CH2BrHC CNa

CH3CH2C CH

HgSO4/H2SO4

CH3CH2CCH3

O

15.

CH3C

O

CH3 CH3C

O

CH2Cl CH3 C

H

O ClCH2C

O

H HC

O

H< < < <

Level – II

1. (i)

CH CHC

O (ii)

O

2.

CH C MgBr 2

3

CO

H O+→X = CH C COH

O

Y

22 4Hg , H SO+

HC CHCOOHOHCH2 COOHCHOH2C

COOH

COOH

KMnO4

3.

O O

H3O

OH OH OH OH

-H2O

OH O H OH

4.

CH3C

O

CH2 C

OH

CH3

CH3 CH2( )

( )3

3

i CH MgBr

ii H O +→CH2C

O

CH2CH3 CH2C

O

CH2CH3

cyclisation

O OHCH3

CH3

CH3

5.

O

OH

CH3

(A)(B) (C)

O

CH3

OH

CH3

6. + → +3 2 3 2CH CH ONO HCl CH CH Cl HONO

CH3COCH2CH3 3 3O N OH CH COCCH+ = − →

NOH(X) oxime

60

7.

H

O

-Cl

H ClH

O H

H

O H

H

O

H

Cl

Cl

8.

A = PhCH CHC

OH

CH3

CH2CH3

D = PhCHCH2

CH3

CCH3

O

C

O

CH3PhCH2CH2B =

E = PhCHO HC C

O O

CH3

C = PhCH CHC

OH

H

CH3

9. (a) O

( )( )

( ) 2 2

i LDAii PhSeBr

iii H O→

O

2 2H C CHCH CH= =→

O

H2 Pt

O

(b) O

( )( ) ( )

2

2 3 2| |O

3

i R NH, H

ii H C CHCCH CH

(iii) H O

+

+

=→

O

CH2CH2C

O

CH(CH3)2

10. (i)

CH3 C

O

CH34LiAlH→ CH3 CHCH3

OH

2 4conc. H SOHeat→ CH3 CH CH2

(ii)

3CH MgBr

H+→CH3CH2CHO CH3CH2CH

OH

CH3Cu573k→CH3CH2C

O

CH3

(iii) CHO

MgBrH+

→CH

OH

PCC→ C

O

(iv)

→ →

OH

Cu dil.NaOH3 2 3 3 2573K

|CH CH OH CH CHO CH C HCH CHO

(v) CHO

dil.NaOH3 HeatCH CHO+ → CH CH CHO →2

0

H /Ni

60 C CH2CH2CH2OH

(vi)

3

2 4

HNOH SO→

NO2

( )3 2

3

CH CO O

AlCl→

NO2

COCH3

11. The reaction of carbonyl group with H2NZ is an acid catalysed reaction.

C O C OH 2H N Z−→ C

OH

N

H

Z

H

-H

C

OH

NH

Z

2H H O+ −←C N Z

(i)Test Propanal (CH3CH2CHO)

CH3COCH3

(Propanaone)

(a) Tollen’s reagent Silver mirror −ve test

(b) Iodoform test (I2/NaOH)

−ve Yellow ppt.

(ii)

Test

Ethanal

CH3 C

O

H

Propanal

CH3CH2CHO

Iodoform test (I2/NaOH) Yellow ppt. −ve test

(iii)Test

Propanal

CH3CH2CHO

Diethyl Ether

CH3CH2OCH2CH3

(a) Tollen’s test Silver mirror −ve test

(b) Fehling’s test Reddish brown ppt. −ve

(iv)Test

Propanone

CH3COCH3

Propanol

CH3CH2CH2OH

62

Iodoform test Yellow ppt. −ve test

13. Methyl ketones CH3 C

O

R on oxidation with X2/NaOH (X = Cl, Br, I) gives sodium salt of

carboxylic acid with one carbon atom less. The CH3− group bonded to carbonyl group is

converted to haloform, CHX3. With iodine in NaOH, a yellow precipitate of Iodoform CHI3, is

obtained. (Iodoform test).

+ → + +2 2X 2NaOH NaOX NaX H O

CH3 C

O

R 3NaOX→ CH3 C

O

R 3NaOH

CX3 C

O

R NaOX3CHX RCOONa→ +

14. The compound is a carbonyl compound (+ve, 2, 4 – DNP test)

The carbonyl group is an aldehyde (+) ve Tollen’s reagent.

The aldehyde group is directly bonded to benzene, no α - H (Cannizzaro reaction).

The other side chain is at the ortho position as it gives phthalic acid on oxidation. Hence the

compound is CHO

C2H5

The reactions are

CHO

C2H5

2,4 DNP−

( )3 2Ag NH

+

( )O

conc. NaOH

COONa

C2H5

COOH

COOH

COOH

C2H5

CH

C2H5

NNH

O2N

NO2

CH2OH

C2H5

15. CH3C OH

O

CH3 C

O

Cl

(A)

5PCl→

(B)

2

4

H /PdBaSO

→ CH3 C

O

H

(C)

NaOH

CH3 CH CH2CHO

OH

(D)

∆←CH3 CH CH CHO

Crotonaldehyde

64

Objective:

Level - I

1. D 2. B 3. C

4. A 5. C 6. D

7. A 8. B 9. A

10. C 11. B 12. D

13. A 14. C 15. C

16. A 17. C 18. D

19. C 20. D

Level - II

1. B 2. D 3. A

4. D 5. B 6. C

7. B 8. B 9. A

10. D 11. A 12. B

13. A 14. B 15. A

16. D 17. B 18. C

19. A 20. A