carboxylic acid and its derivatives_chemistry

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SUBJECT – CHEMISTRY

TOPIC – CARBOXYLIC ACIDS & ITS DERIVATIVE

COURSE CODE – AISM-09/C/CAID

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Contents :- CARBOXYLIC ACIDS & ITS DERIVATIVE

Introduction……………………………………………………………………………...3

Nomenclature ………………………………….……………………………………...5

General methods of preparation……………………………………………...11

Physical properties…………………………………………………………………..24

Chemical reactions…………………………………………………………………..27

Reactions of RCOOH………………………………………………………………..28

Salt formation………………………………………………………………………….31

The mechanism of easterification reaction………………………………35

Carboxylic acid derivatives……………………………………………………….44

Chemical reactions of acid derivatives……………………………………..47

Base promoted hydrolysis of esters: Saponification………………….52

Claisen Condensation……………………………………………………………….55

Amides……………………………………………………………………………………..57

Answers to excersies…………………………………………………………………66

Miscellaneous…………………………………………………………………………..75

Solved examples……………………………………………………………………….83

IIT level questions………………………………………………………………….….87

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INTRODUCTION

Carboxylic acids are characterized by the presence of carboxyl group. The

–COOH group which itself is made up of a carbonyl group (C=O) and a

hydroxyl group ( OH) is called carboxyl group (carb from carbonyl and oxyl

from hydroxyl)

C

O

O H C O

O

H

Carbonyl Hydroxyl Carboxyl

Carboxylic acids may be aliphatic or aromatic

R C

O

O

H

Ar C

O

O

H

Aliphatic carboxylic acid

(Where R = H or any alkyl group)

Aromatic carboxylic acid

(Where Ar is any aryl group)

Comparison of resonating structures of carboxylic group and

carbonyl group

Carbonyl group has two resonance structures (I and II)

C O

(I)

C O

(II)

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However, for a carboxyl group, three resonance structures (A, B and C) can

be written.

C

O

O

H

(A)

C

O

O

H

(B)

C

O

O

H

(C)

In both structures (A) and (C), the C – atom and the two O – atoms have

eight electrons in their respective valence shells while in structure (B), C –

atom has only six electrons. Therefore, structure (B) is less stable than

structure (C), in other words the two important resonance structures of

carboxyl group are structures (A) and (C). In both these structures, carboxyl

carbon is electrically neutral. However in case of aldehydes and ketones,

only one structure i.e. I is electrically neutral. As a result, the carboxyl

carbon of the resonance hybrid is less positive and hence less electrophilic

than the carbonyl carbon of aldehydes and ketones. However, it may be

noted that like carbonyl group, carboxyl group is also polar due to resonance

structures (B) and (C).

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NOMENCLATURE

The aliphatic carboxylic acids are commonly known by their initial names,

which have been derived from the source of the particular acid.

Examples:

HCOOH Formic acid [Latin: Fermica = ant]

CH3COOH Acetic acid [Latin: acetum = Vinegar]

CH3–CH2–COOH Propionic acid [Greek: Proton = First; Pion = Fat]

CH3(CH2)2COOH Butyric acid [Latin: Butyrum = Butter]

CH3(CH2)3COOH Valeric acid

CH3(CH2)14COOH Palmitic acid

CH3(CH2)16COOH Stearic acid

Alternative system of nomenclature is naming the acids as the derivatives of

acetic acids. The only exception being formic acid.

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Example:

CH3 – CH2 – COOH Methyl acetic acid

(CH3)3C – COOH Trimethyl acetic acid

According to the IUPAC system of nomenclature, the suffix of the

monocarboxylic acid is ‘oic acid’, which is added to the name of the alkane

corresponding to the longest carbon chain containing the carboxyl group,

e.g.

HCOOH methanoic acid

CH3 – CH2 – CH2 – COOH butanoic acid

The positions of side-chains (or substituents) are indicated by numbers, the

numbering to be started from the side of the carboxyl group.

CH3 – CH – CHCH2 – COOH 3, 4-dimethylpentanoic acid | | CH3 CH3

1 1

3 2 4 5

NAMING OF ACYL GROUPS, ACID CHLORIDES AND ANHYDRIDES:

The group obtained from a carboxylic acid by the removal of the hydroxyl

portion is known as an acyl group. The name of an acyl group is created by

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changing the - ic acid at the end of the name of the carboxylic acid to –yl,

examples:

O||

C – OH

Benzoyl group

O||

H –C – O – H

O||

H –C –

Formic acid Formyl group

O||

C –

Benzoic acid

Acid chlorides are named systematically as acyl chlorides.

CH3

Cl

O

acetyl chloride

An acid anhydride is named by substituting anhydride for acid in the name of

the acid from which it is derived.

OH

OHO

O

succinic acid

O

O

O

dihydrofuran-2,5-dione

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COOH

COOH

phthalic acid

O

O

O

2-benzofuran-1,3-dione

Naming of Salts and Esters:

The name of the cation (in the case of a salt) or the name of the organic

group attached to the oxygen of the carboxyl group (in the case of an ester)

precedes the name of the acid.

O-Na

+

O

1-phenylethanone

O

O

CH3

CH3

ethyl 4-methylbenzoate

Sodium benzoate

Naming of Amides and Imides:

The names of amides are formed by replacing –oic acid (or –ic acid for

common names) by amide or –carboxylic acid by carboxamide.

CH3

NH2

O

acetamide

NH2

O

cyclohexanecarboxamide

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If the nitrogen atom of the amide has any alkyl groups as substitutents, the

name of the amide is prefixed by the capital letter N; to indicate substitution

on nitrogen, followed by the name(s) of alkyl group(s).

O2N

N

O

CH3

CH3

N-ethyl-N-methyl-4-nitrobenzamide

If the substituent on the nitrogen atom of an amide is a phenyl group, the

ending for the name of the carboxylic acid is changed to anilide

NH

O

CH3

N-phenylacetamide

O

NH

N-phenylbenzamide

Some dicarboxylic acids form cylic amides in which two acyl groups are

bonded to the nitrogen atom. The suffix imide is given to such compounds.

NH

O

O NH

O

O

pyrrolidine-2,5-dione 1N-isoindole-1,3(2H)-dione

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Illustration 1:

Give the IUPAC names and common names of the following

compounds

(i)

COOH

COOH

(ii) HOOC

COOH

(iii) CH3

COOH

OH

iv) HOOC

OH COOH

OH

Solution:

IUPAC Name General name

(i) Ethanedioic acid Oxalic acid

(ii) Butanedioic acid Succinic acid

(iii) 2-Hydroxy propanoic acid Lactic acid

(iv) 2, 3 Dihydroxybutanedioic acid Tartaric acid

Illustration 2:

Which is not a hydroxy acid?

(A) lactic acid (B) tartaric acid

(C) citric acid (D) succinic acid

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Solution:

(D)

GENERAL METHODS OF PREPARATION

1. Oxidation of alcohols, aldehydes and ketones

Refer AEP & AK

Illustration 3:

How will you prepare CH3CH2COOH from CH3 CH=CH2

Solution:

3 4

2 2

BH / THF MnO /H

3 2 3 2 2 3 2H O /OHCH CH CH CH CH CH OH CH CH COOH

Illustration 4:

Bring about the following transformations

CH2 OH

Ag NH OH3 2MnO2

HA B

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Solution:

CHOA : C OH

O

B :

2. Oxidation of alkyl benzenes

Although benzene and alkane are quite unreative towards the usual

oxidizing agents (KMnO4, K2Cr2O7 etc). The benzene ring renders an

aliphatic side chain quite susceptible to oxidation. The side chain is

oxidised down to the ring and only a carboxyl group ( COOH) remains

to indicate the position of the original side chain. Potassium

permanganate is generally used for this purpose, although potassium

dichromate or dilute nitric acid can also be used. (Oxidation of a side

chain is more difficult, however, than oxidation of an alkene and

requires prolonged treatment with hot KMnO4)

CH2CH2CH2CH3

hot KMnO 4

COOH

CO2

n - butyl benzene Benzoic acid

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This reaction is used for two purposes (a) synthesis of carboxylic acids

and (b) identification of alkyl benzenes.

3. Carbonation of Grignard reagents

The Grignard synthesis of a carboxylic acid is carried out by bubbling

gaseous CO2 into the ether solution of the Grignard reagent or by

pouring the Grignard reagent on crushed dry ice (solid CO2). In the

latter method dry ice serves not only as reagent but also as cooling

agent.

The Grignard reagent adds to the carbon – oxygen double bond of CO2

just as in the reaction with aldehydes and keotnes. The product is the

magnesium salt of the carboxylic acid, from which the free acid is

liberated by treatment with mineral acid.

R MgX C

O

O

R COO Mg XH

R COOH Mg2

X

Grignard's reagent

The Grignard’s reagent can be prepared from primary, secondary,

tertiary or aromatic halides. The method is limited only by the

presence of other reactive group in the molecule. The following

synthesis illustrate the application of this method.

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CH3 C

CH3

CH3

OHHCl

CH3 C

CH3

CH3

ClMg

CH3 C

CH3

CH3

MgClH

CH3 C

CH3

CH3

COOH

Trimethylacetic acid

CO2

Illustration 5:

(i)

CH3

CH3 CH3

CH3

COOH

CH3CH3

A BBr2 Mg CO2 H

EtherFe

Write A and B

(ii) Br

CH

C2H5

CH3

Mg CO 2 H

CH

C2H5

CH3

COOH

A B

p-bromo-sec-butyl benzene p-sec-butyl benzoic acid

Ether

Write A and B

Solution:

(i)

CH3

CH3CH3

Br2

CH3

Br

CH3CH3

Bromomesitylene

Mg

CH3

MgBr

CH3

CH3

CO 2 H

CH3 CH3

COOH

CH3Mesitylene

Mesitoic acid

( 2, 4, 6 - trimethyl - benzoic acid)

Ether

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(ii) Br

CH

C2H5

CH3

p-bromo-sec-butyl benzene

Mg

CH

C2H5

CH3

MgBr

CO2

CH

C2H5

CH3

COOMgBr

H

CH

C2H5

CH3

COOH

p-sec-butyl benzoic acid

Ether

Illustration 6:

Bring about the following transformations.

Br

C

O

OH

Solution:

Br Mg/Ether CO2 H O3 C

O

OHMgBr CO2MgBr

Bring about the following transformations

Exercise 1:

(i)

Br OH

O

(ii) CH CH

O

OH

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Exercise 2:

(i) Why we can not get anhydrous formic acid by fractional

distillation of the Fatty acids?

(ii) CH2=CH—CH=CH2

2 3

(i) HBr

(ii) Mg/ether+(iii) CO /H O

A. Identify A?

4. Hydrolysis of nitriles

Aliphatic nitriles are prepared by treatment of alkyl halides with

sodium cyanide in a solvent that will dissolve both reactants. In

dimethyl sulfoxide (DMSO), reaction occurs rapidly and exothermically

at room temperature. The resulting nitrile is then hydrolysed to the

acid by boiling with aqueous alkali or acid.

R C N H2O

H

OH

R COOH NH4

R COO NH3

DMSOR Cl NaCN

Illustration 7:

(i) n - C4H9Br NaCN

n - butyl bromide

NH4

A

BC

Write A, B and C.

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(ii) CH2Cl

Benzyl chloride

NaCN 70% H2SO4, reflux NH4A B

Write A and B.

Solution:

(i) n - C4H9Br NaCNn - C4H9CN

n - butyl bromide n - voleonitrile

(pentanenitrile)

n - C4H9COO NH3

n - C4H9COOH NH4

n - valeric acid

(pentanoic acid)

aq. alc. NaOH, reflux

H

(A)

(B)

(C)

(ii) CH2Cl

Benzyl chloride

NaCN

CH2CN

70% H2SO4, reflux

CH2COOH

NH4

Phenyl acetic acidPhenyl acetonitrile

Exercise 3:

Identify the missing reagents or products

+- i) BH /THFH O/H conc. H SOHCN/OH 32 2 4ii) CH COOH3

(A) (B) (C) (E)CH3

CH3

O

(D)

i) BH3/THF ii) H2O2/OH-

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Exercise 4:

Conc. H2SO4

ACl2/H2O

BCN

CH3O

D E2 - +

2 7Cr O /H

OH

Exercise 5:

A pleasant smelling optically active ester (A) has molecular weight

186. it does not react with Br2 in CCl4. Hydrolysis of (A) gives two

optically active compounds B and C. Compound (C) gives positive

iodoform test and on warming with conc. H2SO4 gives D (saytzeff

product) with no geometrical isomers. (C) on treatment with benzene

sulphonyl chloride gives (E) which on treatment with NaBr gives

optically active F. When Ag2+ salt of (B) is treated with Br2, racemic F

is formed. Give structures of A to F.

5. Use of alkoxide

H

darkRO Na CO RCOONa RCOOH

3 3CH O Na CO CH COOH

0

8

210 C, pressure

3 3Co COCH OH CO CH COOH

6. Carbonylation of alkenes

3 40

H PO

2 2 2 3 2300 400 CCH CH CO H O CH CH COOH

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3 40

H PO

3 2 2 3300 400 C |CH CH CH CO H O CH CH COOH

CH3

7. Oxidative cleavage of alkenes, alkynes and cyclo alkenes

3 4

2

i O or KMnO

ii H ORC CR' RCOOH R'COOH

3

2

i O

3 3 3ii H OCH C CCH 2CH COOH

4i Alkaline KMnO

ii H /RCH CHR 2RCOOH

4KMnO /Δ/HCH2COOH

CH2COOH

Illustration 8:

Compound A, C5H8O3, when heated with soda lime gives B which

reacts with HCN to give C, C reacts with PCl5 to give D which reacts

with KCN to form E. E, on alkaline hydrolysis gives a salt which is

isolated and heated with soda lime to produce n-butane. A, on careful

oxidation with K2Cr2O7 gives acetic acid and malonic acid. Give

structural formulae of A to E.

Solution:

(C5H8O3) elimSoda

Compound B HCN

Compound (C)

Careful oxidation with K2Cr2O7

Compound E KCN Compound D,

Alk. hydrolysis

CH3COOH + CH2

COOH

COOH Salt

elimsodan-butane

PCl5

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On Oxidation only two –COOH groups can be introduced, i.e., one to

each carbon undergoing C–C fission, but in the resulting products we

have three –COOH groups. Hence, one –COOH group is already there

in compound A, the remaining portion C4H7O–, resembles with Keto

substituted alkyl group

C3H7 O

This indicates that the given organic compound A is Keto substituted

acid. To assign position to Keto group in carbon chain, we know that

keto acids on careful oxidation undergo C–C bond fission at a place

where -C -

O||

is situated, further Keto group is also converted into –COOH

and remains with acid having small number of carbon atoms. From the

above discussion it is clear that acetic acid is formed from:

3CH C arrangement

O||

C3H7CO–has 3 2 2CH C CH CH

O||

structure and compound A is

3 2 2CH C CH CH COOH

O||

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A, with soda-lime undergoes decarboxylation to give

CH3 – CO – CH2– CH3 (B)

B, being Ketone will give addition product with HCN to form

| CN

CH3 – C – CH2 – CH3

OH |

(C)

| CN

CH3 – C – CH2 – CH3

Cl |

(D)

| CN

CH3 – C – CH2 – CH3

CN |

| COONa

CH3 – C – CH2 – CH3 elimSoda CH3 – CH2 – CH2 – CH3

COONa |

(n-butane)

Alkaline hydrolysis

KCN

PCl5

Illustration 9:

How will you affect the following conversions?

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(i) COOH

Cl

CH3

(ii)

COOH

Solution:

(i) CH3

KMnO 4

OH

COOH

Fe/Cl 2

COOH

Cl

(ii)

CH3

OH2O/H

Hg2+

NaOH

I2COOH

Illustration 10:

An organic compound A (C4H8O3), acidic in nature, is oxidised to give B

which on gentle heating produces C, (C3H6O) and CO2. A on heating

yields D (C4H6O2) having an acid neutralization equivalent of 86. What

are A, B and C?

Solution:

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A on heating produces D (C4H6O2). It clearly tells us that it is

dehydration reaction and A must be a - hydroxyl acid.

So A is |

3 2

OH

CH HC CH COOH and D is CH3 CH = CH COOH

CH3C CH2

OH

COOH

H

Oxidation H3C C CH2

O

COOH

(B)

Δ decarboxylation

H3C C

O

CH3 CO2

(Because if a carbonyl group is present to carboxylic group, the

compound undergoes decarboxylation on heating)

Illustration 11:

Formic acid is obtained when

(A) calcium acetate is heated with conc. H2SO4

(B) glycerol is heated with oxalic acid

(C) acetaldehyde is oxidised with K2Cr2O7 and H2SO4

(D) calcium formate is heated with calcium acetate

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Solution:

(B)

Exercise 6:

A neutral organic compound A, C3H6O2 on hydrolysis gives a monobasic

acid B and a neutral compound C. The acid B reduces HgCl2 solution.

The compound C gives iodoform test. Write structures of A, B and C and

give equations to explain the reactions involved.

PHYSICAL PROPERTIES

Some important physical properties of carboxylic acids are given below,

1. Solubility

As the size of the alkyl group increases, the solubility of the acid

decreases and polarity is reduced.

2. Boiling points

Due to intramolecular hydrogen bonding dimerization of acid takes

place and boiling point of carboxylic acid is higher than expected.

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R C

O

O

H

RC

O

OH

3. Melting points

The melting points of aliphatic carboxylic acids do not show a regular

pattern. The first ten members show a alteration effect, i.e. the

melting point of an acid containing even number of carbon atoms is

higher than the next lower and next higher homologues containing odd

number of carbon atoms.

Illustration 12:

On the basis of H-bonding explain that the second ionization constant

K2 for fumaric acid is greater than for maleic acid.

Solution:

We know that H-bonding involving acidic H has an acid weakening

effect and H-bonding in conjugate base has an acid strengthening

effect.

Both dicarboxylic acids have two ionisable hydrogen atoms.

Considering second ionization step.

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Fumarate monoanion (no H-bonding)

C

C

C

H C

O

OH

O

O

C

C

H C

O

O

O

C

O

H

H

Maleate monoanion (H-bonding)

H

Since the second ionisable H of the Maleate ion participates in H-

bonding more energy is needed to remove this H because the H-bond

must be broken. The maleate mono anion is, therefore, the weaker

acid.

Exercise 7:

Explain the following

(i) Carbon-oxygen bond length in formic acid are 1.24 Å and 1.36

Å but in sodium formate both the carbon-oxygen bonds have

same value i.e.1.27 Å.

(ii) Acetic acid in the vapour state shows a relative molecular

weight of 120.

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CHEMICAL REACTIONS

The characteristic chemical behavior of carboxylic acids is, of course,

determined by their functional group, carboxyl, –COOH. This group is made

up of a carbonyl group (C = O) and a hydroxyl group (–OH). As we shall see,

it is the –OH that actually undergoes nearly every reaction. Loss of H+, or

replacement by another group but it does so in a way that is possible only

because of the effect of the C = O.

R C

H

H

C O H

O

(d)

(a)(b)

(c)

Carboxylic acids can also show various types of reactions.

(i) Removal of H+ (due to cleavage of O H bond) by reaction with a

base (at ‘a’ above)

(ii) C O bond breaks at b (by PCl5, PCl3, SOCl2, NH3/ )

(iii) Nucleophilic attack at point (c) in carboxyl carbon (Ester formation)

Reaction in which OH is replaced by NH2, Cl is SN type

R C OH

O

NH2 R C NH2

O

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(iv) Halogenation at –C by P/Br2 at point d (Hell – Volhard Zelinsky

reaction)

(v) Oxidation of - methylene group by SeO2

2

||SeO

2 2

O

RCH COOH RCCOOH H O Se

keto acid

Some reactions are summarized below:

LiAlH 4

R'OH/H

CH2N2

Na

S o d a lim e, Δ

NaOH

RCH 2OH

RCOOR'

RCOOCH 3

RCOONaH2

RH

RCOONa

ROCI

P2O5SOCl 2PCl 3PCl 5

(RCO) 2O

NaHCO 3NH3

3NH , Δ

RCOONH 4 RCONH 2 RCOONa CO 2

R C

O

OH

X

Y

Δ

3N H

2 2 2N CO R NH

Reactions of RCOOH

1. Acidity and salt formation

Acidity of Carboxylic Acids

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The acidity of a carboxylic acid is due to the resonance stabilization of its

anion.

R

O

O+

H

R

O

O H

R

O

O

R

O

O

H

H

(I) (II) (III) (IV)

Because of the resonance, both the carbon oxygen bond in the carboxylate

anion have identical bond length. In the carboxylic acid, these bond lengths

are no longer identical.

The acidity of carboxylic acid depends very much on the substituent

attached to – COOH group. Since acidity is due to the resonance stabilization

of anion, substituent causing stabilization of anion increases acidity whereas

substituent causing destabilization of anion decrease acidity. For example,

electron withdrawing group disperses the negative charge of the anion and

hence makes it more stable causing increase in the acidity of the

corresponding acid, on the other hand, electron-releasing group increases

the negative charge on the anion and hence makes it less stable causing the

decrease in the acidity. In the light of this, the following are the orders of a

few substituted carboxylic acids.

(a) Increase in the number of Halogen atoms on -position increases the

acidity, eg.

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Cl3CCOOH > Cl2CHCOOH > ClCH2COOH > CH3COOH

(b) Increase in the distance of Halogen from COOH decreases the acidity

e.g.

CH3

COOH

Cl

CH3

COOH

Cl

COOH

Cl

This is due to the fact that inductive effect decreases with increasing

distance.

(c) Increase in the electronegativity of halogen increases the acidity.

FCH2COOH > BrCH2COOH > ICH2COOH

Illustration 13:

Which one of the following would be expected to be most highly ionised

in water?

(A) CH2Cl–CH2–CH2COOH (B) CH3CHCl –CH2–COOH

(C) CH3–CH2–CHCl–COOH (D) CH3CH2–CCl2 –COOH

Solution:

(D)

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SALT FORMATION

Carboxylic acids are weak acids and their carboxylate anions are strong

conjugate bases and are slightly alkaline due to the hydrolysis of carboxylate

anion compared to other species, the order of acidity and basicity of

corresponding conjugate bases are as follows:

Acidity RCOOH > HOH > ROH > HC CH > NH3 > RH

Basicity RCOO– < HO– < RO– < HC C– < 2NH < R–

1. The carboxylic acids react with metals to liberate hydrogen and are

soluble in both NaOH and NaHCO3 solutions. For example.

2CH3COOH + 2Na 2CH3COO–Na+ + H2

CH3COOH + NaOH CH3COO–Na+ + H2O

CH3COOH + NaHCO3 CH3COO–Na+ + H2O + CO2

Examples:

3 3 222CH COOH Zn CH COO Zn H

Aceticacid Zinc acetate

3 2 3 2 210 10CH CH COOH NaOH CH CH COO Na H O

lauricacid Sodiumlaurate

COOH

NaHCO 3

Benzoic acid

COO Na

CO2 H2O

Sodium benzonate

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Illustration 14:

Arrange the following illustration in order of increasing acidity

(i) HCOOH, ClCH2COOH, CH3COOH

(ii) CH3COOH, (CH3)2CHCOOH, (CH3)3CCOOH

(iii) ClCH2COOH, Cl2CHCOOH, Cl3CCOOH

(iv) ClCH2COOH, CH3CH2COOH, ClCH2CH2COOH, (CH3)2CHCOOH,

CH3COOH

(v) CH3COOH, Cl2CHCOOH, CH3CH2COOH, Cl3CCOOH, ClCH2COOH

Solution:

(i) CH3COOH HCOOH ClCH2COOH

(ii) (CH3)3CCOOH (CH3)2CHCOOH CH3COOH

(iii) ClCH2COOH < Cl2CHCOOH < Cl3CCOOH

(iv) (CH3)2CHCOOH CH3CH2COOH CH3COOH ClCH2CH2COOH

ClCH2COOH

(v) CH3CH2COOH CH3COOH ClCH2COOH Cl2CHCOOH Cl3CCOOH

Illustration 15:

Acetic acid reacts with chlorine in the presence of catalyst, anhydrous

FeCl3 to give

(A) acetyl chloride (B) methyl chloride

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(C) tri-chloro acetic acid (D) chloral hydrate

Solution:

(C)

2. Conversion into functional derivatives

R C

OH

O

R C

Z

O

(Z = -Cl, -OR', -NH 2)

(a) Conversion into acid chlorides

R C

OH

O

R C

Cl

OSOCl 2

PCl 3

PCl 5

Acid chloride

Examples:

COOH PCl 5

Benzoic acid

COCl

01 0 0 CPOCl 3 HCl

Benzoyl chloride

reflux

17 35 2 17 35 2Thionyl chlorideStearic acid Stearoyl chloride

n C H COOH n SOCl n C H COCl SO HCl

050 C

3 3 3 3 3Acetic acid Acetyl chloride

CH COOH PCl 3CH COCl H PO

Illustration 16:

Benzoyl chloride is prepared from benzoic acid by

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(A) Cl2,h (B) SO2Cl2

(C) SOCl2 (D) Cl2, H2O

Solution:

(C)

(b) Conversion into esters

Conversion into Esters (Esterification):

Carboxylic acid on reacting with alcohols in presence of

dehydrating agent (H2SO4 or dry HCl gas) gives esters. The

reaction is known as esterification.

R – C

O

OH

R OH R –C

O

OR

+ H2O

Acid

H+

+

This reaction is reversible and the same catalyst, hydrogen ion,

that catalyzes the forward reaction, esterification, necessarily

catalyzes the reverse reaction hydrolysis.

The equilibrium is particularly unfavourable when phenols

(ArOH) are used instead of alcohol; yet if water is removed

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during the reaction, phenolic esters [RCOOAr] are obtained in

high yield.

The presence of bulky group near the site of reaction, whether in

alcohol or in the acid, slows down esterification (as well as its

reverse, hydrolysis).

Reactivity CH3OH > 1° > 2° > 3°

In esterification HCOOH > CH3COOH > RCH2COOH > R2CHCOOH

> R3CCOOH

The Mechanism of the Esterification Reaction:

The step in the mechanism for the formation of an ester from an acid and an

alcohol are the reverse of the steps for the acid-catalyzed hydrolysis of an

ester, the reaction can go in either direction depending on the conditions

used. A carboxylic acid does not react with an alcohol unless a strong acid is

used as a catalyst, protonation makes the carbonyl group more electrophilic

and enables it to react with the alcohol, which is a weak nucleophile.

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O || C

CH3 O – H

+ H+

Protonation of the carbonyl group

O – H || C

CH3 O – H

+

+

O – H | CH3 – C – O – H

B

O

CH3 H

O – H | CH3 – C – O – H

O

CH3

O – H || C

CH3 O – H

O

CH3 H

+

H+

-BH

Nucleophilic attack at the carbonyl group

Deprotonation of the intermediate

Tetrahedral intermediate being protonated at another site

O || C

CH3 O – CH3

Deprotonation of the carbonyl group

O – H || C

CH3 O – CH3

+

O – H | CH3 – C – O – H

O

CH3

H

+

O

H H

B H – B+

+

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Examples:

COOH CH3OH

Benzoic acid

COOCH 3 H2O

Methyl benzoate

Methanol

H

CH2OH

Benzyl alcohol

HCH3COOH

Acetic acid

CH3 C

O

O CH2

2 52 C H OHSOCl

3 3 3 2 53 3 3Trimethyl acetic acid Ethyl trimethyl acetate

CH CCOOH CH CCOCl CH CCOOC H

Illustration 17:

Assign a structure to each compound indicated by a letter in the

following equations.

OH C

O

(CH2)9 C

O

OH 3

2 4

CH OH (excess)

H SO , ΔA

Solution:

OCH 3(CH2)9O

CH3

O O

A =

Exercise 8:

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Assign a structure to each compound indicated by a letter in the

following equations.

OH

OH

O

3 2 2CH CH -OH SOCl

HCl(g), Δ ΔA B

(c) Conversion into amides

R C

OH

O

R C

Cl

O

SOCl 2

An acid chloride

NH3 R C

NH2

O

An amide

Example:

32 NHSOCl

6 5 2 6 5 2 6 5 2 2Phenyl acetic acid Phenyl acetyl chloride Phenyl acetatamide

C H CH COOH C H CH COCl C H CH CONH

Illustration 18:

Discuss the reason for that a characteristic reaction of aldehydes and

Ketones is one of nucleophilic addition while Acyl compounds yield

Nucleophilic substitution product.

Solution:

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R

R

C = O

Nu–

Nu

R

C – O

R

Nu

R

C – O – H

R

H – Nu

Aldehyde or Ketone

Nucleophilic addition

An acyl compound

R

L

C = O

Nu – H

Nu+

R

C – O

L

H

Nu

R

C – O

HL+

Nu

R

C = O + HL

Another acyl compound

Nucleophilic addition

(L = –OH, –Cl, –NH2, OR etc.) Elimination

Nucleophilic substitution

The initial step in both reactions involves nucleophilic addition at the

carbonyl carbon atom. It is after the initial nucleophilic attack has

taken place that the two reactions differ. The tetrahedral intermediate

formed from an aldehyde or ketone usually accepts a proton to form a

stable addition product. By contrast, the intermediate formed from an

acyl compound usually eliminates a leaving group: this elimination

leads to regeneration of the carbon oxygen double bond and to a

substitution product. The overall process in the case of acyl

substitution occurs, therefore, by a nucleophilic addition–elimination

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mechanism. Acyl compounds react as they do so because they all have

good leaving groups attached to the carbonyl carbon atom.

3. Reduction

4

0

LiAlH

21 Alcohol

R COOH R CH OH

Examples:

4(CH3)3CCOOH 3LiAlH 4Ether [(CH3)3CCH 2O]4AlLi 2LiAlO 2 4H2

(CH3)3CCH 2OH

Neopentyl alcohol

( 2, 2 - dimethyl - 1 - propanol)

COOH

CH3

LiAlH 4

CH2OH

CH3

m - toluic acid m - methyl benzyl alcohol

4. Substitution in alkyl or aryl group

(a) Halogenation of Aliphatic Acids (Hell-Volhard-Zelinsky

Reaction)

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In the presence of phosphorus, aliphatic carboxylic acids react

smoothly with chlorine or bromine to yield a compound in which -

hydrogen has been replaced by halogen.

2 2 2Cl /,P Cl ,/ P Cl ,/ P

3 2 2 3CH COOH Cl CH COOH Cl CH COOH Cl CCOOH

The function of the phosphorus is ultimately to convert a little of the acid

into acid halide so that it is the acid halide, not the acid itself, that

undergoes this reaction.

P + X2 PX3

R – CH2 – COOH + PX3 RCH2 – COX

R COX2X R

X

COX

2H OHX R

X

CO2H

The halogen of these halogenated acid undergoes nucleophilic displacement

and elimination same as it does in the simple alkyl halides. Halogenation is

therefore the first step in the conversion of a carboxylic acid into many

important substituted carboxylic acid.

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R

Br

COOH

3NH R

NH2

COOH

An halogenated acid

An hydroxy acid

R

Cl

COOH

HNaOH R

OH

COOH

An amino acid

Examples:

CH3COOH

Acetic acid

Cl 2/PClCH 2COOH

Chloroacetic acid

Cl 2/PCl 2CHCOOH

Dichloroacetic acid

Cl 2/P

Cl 3CCOOHTrichloroacetic acid

CH3CHCH 2COOH

CH3

Isovaleric acid

CH3CH

CH3

CHCOOH

Br

α Bromoisovaleric acid

Br2/P

(b) Ring substitution in aromatic acids:

–COOH deactivates and directs incoming electrophilic to meta

position.

Example:

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COOH

HNO 3/H2SO4/Heat

NO2

COOH

benzoic acid m - nitrobenzoic acid

Illustration 19:

Hydrolysis of a compound A (C7H3Cl5) gives an acid B of the formula

C7H4Cl2O2. Decarboxylation of acid yields a neutral substance (C), the

nitration of which forms only one mono derivative (D). Identify A, B, C

and D.

Solution:

A =

Cl

Cl

CCl 3

B =

Cl

Cl

COOH

C =

Cl

Cl

D =

Cl

Cl

NO2

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Illustration 20.

Which of the following will not undergo HVZ reaction?

(A) 2, 2-dimethyl propanoic acid

(B) propanoic acid

(C) acetic acid

(D) 2-methyl propanoic acid

Solution:

(A)

CARBOXYLIC ACID DERIVATIVES

There are four carboxylic acid derivatives. These are generally represented

as||O

R C Z , where Z is halogen (usually Cl), OCOR , OR or NH2 (or NHR or

NR2 ).

(a) When Z is halogen (usually Cl), the derivatives are called as acid

chlorides.

R C Cl

O

(b) When Z is OR , the derivatives are called as esters.

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(c) When Z is||O

O C R' , the derivatives are called carboxylic anhydrides.

(d) Where Z is NH2, the derivatives are called amides. When Z is NHR

or NR2 they are called N – substituted amides.

Synthesis of acid derivatives

Carboxylic acid derivatives are exclusively prepared from carboxylic acids.

The preparation methods of carboxylic acid derivatives are already discussed

under the chemical reactions of carboxylic acids.

Illustration 21:

(a) 21. Excess CO

2. HPhLi A

(b) 32 eq. CH Li

3 3 HCH COOH B 2

2

Ag O

3 Br ,3CH COOH C

Solution:

(a)

Ph O C O Ph C O

O

Ph C OH

O

H

(A)

(b) ||

3 3 33 3

O

CH CCOOH CH C C OCH B

2

2

Ag O

3 3 2Br ,3 3CH CCOOH CH CBr CO

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Illustration 22:

Suggest a likely mechanism for each of the following reactions.

(a) OH R C

O

CX3

H2ORCOO CHX 3

(b)

C

O

F

NH2

NH3

F

C

O

NH2

Solution:

(a)

OH R C CX3

O

R C OH

O

CX3

(b) C

O

F

NH2

NH3

C NH2

O

C

O

FNH2

F

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Exercise 9:

(i)

O

O

O

AlCl3A(C10H10O3)

Zn(Hg)

HClB(C10H12O2)

SOCl2

CAlCl3D(C10H10O)

LiAlH4E(C10H12O)

H2SO4F(C10H10)

Galc. KOH

H(C10H8)

NBS

(ii) When heated with acid (e.g. conc. H2SO4), O – benzoyl benzoic

acid yields a product of the formula C14H8O2. What is the

structure of this product?

CHEMICAL REACTIONS OF ACID DERIVATIVES

(i) Acyl chloride

We have already seen that acyl chlorides are the most reactive of all

acid derivatives. As a result, acyl chlorides are often selected as the

starting material for the preparation any other acid derivative. Let us

see how this is done.

R C Cl

O

R C O Na

O

R C O

O

C

O

R'

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R C Cl

O

R OH R C O

O

R'

R C Cl

O

R C NH2

O

NH3

R C Cl

O

R C NHR'

O

R'NH 2

R C Cl

O

R C NR'R"

O

R'R"NH

CH3 C Cl

O

H2NC6H5 C6H5NHC

O

CH3HCl

Acetanilide

OH

COOH

CH3COCl

O

COOH

C CH3

O

Acetyl salicylic acid (aspirin)

Reaction of acetyl chloride with olefins

Acetyl chlorides add on to the double bond of an olefin in the presence of a

catalyst (AlCl3 or ZnCl2) to form a chloro ketone which on heating, eliminates

a molecule of hydrogen chloride to form an unsaturated ketone.

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CH3 CH CH2 CH3COClZnCl 2 CH3 CH CH2

Cl

C

O

CH3

Δ

CH3 CH CH C

O

CH3HCl

Conversion into acids: Hydrolysis.

R C

O

Cl H2O R C

O

OH HCl

An acid

Illustration 23:

Acetyl chloride reacts with water more readily than methyl chloride.

Explain.

Solution:

Alkyl halides are much less reactive than acyl halides in nucleophilic

substitution because nucleophilic attack on the tetrahedral carbon of

RX involves a hindered transition state. Also, to permit the attachment

of the nucleophile a bond must be partly broken. In CH3COCl, the

nucleophile, attack on > C = O involves a relatively unhindered

transition of acyl halides occurs in two steps. The first step is similar to

addition to carbonyl compound and the second involves the loss of

chlorine in this case.

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+H2O

O

C

CH3 Cl

O-

C

CH3 Cl

+OH2

O–

C

CH3 Cl

+OH2

Transition state Intermediate –

O

C

CH3 OH2

Cl

O

C

CH3 OH

–H+

+Cl–

+

Transition state Product

–

Illustration 24:

Hydrogenation of benzoyl chloride in the presence of Pd and BaSO4

gives

(A) benzyl alcohol (B) benzaldehyde

(C) benzoic acid (D) phenol

Solution:

(B)

(ii) Carboxylic acid anhydrides

Carboxylic acid anhydrides can be used to prepare esters and amides.

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R C

O

O C

O

RR'OH

R C

O

OR' R C

O

OH

NH3 excess

R C

O

NH2 R C

O

O NH 4

Anhydrides can be hydrolysed to get back acids.

R C O C R

OO

H2O2R C OH

O

Illustration 25:

MeO CHO 3

3

CH COONa

H O(X)

COOH

The compound (X) is

(A) CH3COOH (B) BrCH2 – COOH

(C) (CH3CO)2O (D) CHO – COOH

Solution:

(C)

Exercise 10:

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O

O

O

3 2CH -CH -OH

TsOH ΔA

(iii) Esters

Ester hydrolysis

Acid catalysed esterification is an essentially reversible reaction. If you

follow the backward course of reactions of esterification it gives you

the mechanism for ester hydrolysis.

R C OR'

O

H2OR C OH

O

HR'OH

Base promoted hydrolysis of esters: Saponification

Esters undergo base promoted hydrolysis also. This reaction is known as

saponification, because it is the way most of the soaps are manufactured.

Refluxing an ester with aqueous NaOH produces an alcohol and the sodium

salt of the acid.

R C OR'

O

H2ONaOH R C O Na

O

R'OH

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This reaction is essentially irreversible because carboxylate ion is inert

towards nucleophilic substitution.

Mechanism

R C OR'

O

OH

R C OR'

O

OH

R C OH

O

R'O

R C O

O

ROH

If an ester is hydrolysed in a known amount of base (taken in excess), the

amount of base used up can be measured and used to calculate the

saponification equivalent; the equivalent weight of the ester, which is

similar to the neutralization equivalent of an acid.

Transesterification

Esters can also be prepared by transesterification (an alcohol displacing

another from an ester)

C

OR'R

O

R" OHHA

C

OR"R

O

heatR' OH

The mechanism of this reactions is quite similar to esterification.

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H2C CHCOCH 3

O

CH3CH2CH2CH2OH H2C CHCOCH 2CH2CH2CH3

O

Transesterification is an equilibrium reaction. To shift the equilibrium to

right, it is necessary to use a large excess of the alcohol whose ester we

wish to make or else to remove one of the products from the reaction

mixture.

Reduction of esters

(i) Catalytic hydrogenation

20

||H , CuO, Hg/H

3 2 2 3 3 2 2 2 3150 C high pressure

O

CH CH CH C O CH CH CH CH CH OH CH OH

(ii) Chemical reduction

4

2 5

||LiAlH /H

3 2 2 3 3 2 2 2 3or Na / C H OH

O

CH CH CH C O CH CH CH CH CH OH CH OH

Reaction of esters with Grignard reagents

The reaction of carboxylic esters with Grignard’s reagent is a good method

for the preparation of 3 alcohols.

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R C

O

OR' R"MgXR C

O

R" R"MgXR C

OMgX

R"

R"

R C

OH

R"

R"

H

Initially ketones are formed. However, as we know, ketones themselves

readily react with Grignard reagent to yield teriary alcohols.

Claisen Condensation

When ethyl acetate reacts with sodium ethoxide, it undergoes a

condensation reaction. After acidification, the product is a - keto ester,

ethyl aceto acetate (commonly known as – aceto acetic ester)

2CH3C

O

OC2H5

NaOC 2H5 CH3 C CH

O

C

O

OC2H5

Na

C2H5OHHCl

CH3C

O

CH2C

O

OC2H5

Condensation of this type is known as Claisen Condensation. For esters, it is

the exact counterpart of the Aldol Condensation. Like the Aldol

Condensation, the Claisen Condensation involves nucleophilic attack by a

carbanion on an electron – deficient carbonyl compound. In the Aldol

Condensation, nuclephilic attack leads to addition (the typical reaction of

aldehydes and ketones). In the Claisen Condensation, nucleophilic attack

leads to substitution (the typical reaction of acyl compounds)

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Mechanism

Step-1 R CH

H

C

O

OC2H5 OC2H5 C

O

OC2H5RCH C

O

OC2H5RCH

CH2C

OC2H5

O

R HC C

R

O

OC2H5 RCH 2 C

O

OC2H5

CH

R

C

O

OC2H5

RCH 2 C

O

CH C

O

OC2H5

R

C2H5O

Step- 2

Step-3 RCH 2C C

H

COC 2H5

R

OO

C2H5O C CRCH 2C

O

R

O

OC2H5 C2H5OH

This step is highly favourable and draws the overall equilibrium toward the

product formation

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Step-4 C CRCH 2C

O

R

O

OC2H53H O

rapidRCH 2C

O

CH

R

C

O

OC2H5RCH 2C C

OH

R

COC 2H5

O

Keto form Enol form

When planning a claisen condensation with an ester it is important to use

alkoxide ion that has the same alkyl group as the alkoxyl group of the ester.

This is to avoid the possibility of transesterification.

An intramolecular claisen condensation is called Dieckmann condensation.

In general, the Dieckmann condensation is useful only for the preparation of

five and six membered rings.

Amides

Preparation:

In the laboratory amides are prepared by the reaction of ammonia with acid

chlorides or, acid anhydrides.

Basic Character of Amides:

Amides are very feebly basic and form unstable salts with strong inorganic

acids. e.g. RCONH2HCl. The structure of these salts may be I or II

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R – C

NH2

O – H

Cl–

+

R – C

NH3

O

Cl–

+

or

I II

Acidic Character of Amides:

Amides are also feebly acidic e.g. they dissolve mercuric oxide to form

covalent mercury compound in which the mercury is probably linked to the

nitrogen.

2RCONH2 + HgO (RCONH)2Hg + H2O

Exercise 11:

PhHC CH CH CH3

O

C O

Optically active

PhHC CH CH CH3

OH

(Optically active)

OH

Exercise 12:

An ester A (C4H8O2) on treatment with excess methyl magnesium

chloride followed by acidification gives an alcohol B as the only

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organic product. Alcohol B on oxidation with NaOCl followed by

acidification, gives acetic acid. Deduce the structures of A & B. Show

the reactions involved.

(a) Hydrolysis of amides

Amides undergo hydrolysis when they are heated with aqueous acid or

aqueous base.

Acidic hydrolysis

R C

O

NH22

Δ

H OH3O R C

O

OH NH4

Basic hydrolysis

R C

O

NH22

Δ

H OOH R C

O

O NH3

(b) Reduction of amides

Amides are reduced by Na/C2H5OH or by LiAlH4 to a primary amine.

R C

O

NH22 5Na / C H OH

4H R CH2 NH2 H2O

(c) Reaction with P2O5

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When heated with P2O5 amides are dehydrated to cyanides.

R C

O

NH22 5

2

P O

H OR C N

Amides may also be converted to cyanides by PCl5.

R C

O

NH25PCl

R C NRCCl 2NH22HCl

(d) Hofmann rearrangement

Amides with no substitutents on the nitrogen react with solutions of

Br2 or Cl2 in NaOH to yield amines through a reaction known as

Hofmann rearrangement.

R C

O

NH2Br2 4NaOH 2NaBr Na2CO3RNH 2

2H2O

Mechanism

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R C

O

N

H

H

OHR C

O

N

H

H

R C N H

O

Br Br

R C N Br

O H

Br

OH

R C N Br

O

-BrR N C O

OH

R N C O

OH

R N C

OH

O

R N C

O

O

H

Carbamate ion

H OH

2 2

3

RNH CO OH

HCO

Illustration 26:

Acetamide reacts with NaOBr in alkaline medium to form

(A) NH3 (B) CH3NH2

(C) CH3CN (D) C2H5NH2

Solution:

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(B)

Illustration 27:

A(C8H8O)NH2OH/HCl

B and C

H H

D E

F G

Alc. KOHCH3 C Cl

O

alc. KOH

(C7H6O2) (C6H7N)

Keeping in mind the fact that B, C, D and E are all isomers of

molecular formula (C8H9NO), identify A to G.

Solution:

(A)

C

CH3

O

(B) C N

Ph

CH3

OH

(C) C N

Ph

CH3 OH (D) Ph C

O

NH CH3

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(E) CH3 C

O

NH Ph (F) Ph COOH

(G) Ph NH2

Illustration 28:

Formic acid and acetic acid may be distinguished by reaction with

(A) sodium

(B) dilute acidic permanganate

(C) 2, 4 dinitrophenylhydrazine

(D) sodium ethoxide

Solution:

(B)

Exercise 13:

(a) CH2 CH CH CH2

CHCOOH

CHCOOH

A

(b) 3NH

2 Δ2CH COOH B

Exercise 14:

Identify A, B, C and D?

(i) R—COOH 3ND /Δ A 2

KOH

Br /ΔB

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(ii) R—COOH 3NH /Δ C 2

KOD

Br /Δ D

Exercise 15:

(i)

Ph C OH

OSOCl2

CH2N2

A

B

H2/Pd(BaSO4)

H

1 eq. DIBALH

C

Find out A, B and C

(ii)

Ph C

O

OHConvert to PhNH2

(iii) (a)

CH3C OH

O

3

+

i 2 eq. CH Li

ii H A

(b)

C

O

C

O

O

1 eq. CH3CH2OH

1. LiAlH4

2. H

B

C

What are A, B and C?

Exercise 16:

An organic compound (A), C8H14O forms an oxime and gives positive

haloform reaction. On ozonolysis it gives acetone and a compound

(B), C5H8O2 (B) forms a dioxime and on subjecting to haloform

reaction gives an acid (C), C4H6O4. On treatment with excess of

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ammonia and strong heating (C) gives a neutral compound (D),

C4H5O2N (D) on distillation with Zn dust forms pyrrole. Suggest

structures for (A), (B), (C) and (D). Give the IUPAC names of

compounds (A) to (D).

Exercise 17:

An acidic compound (A), C4H8O3 loses its optical activity on strong

heating yielding (B), C4H6O2 which reacts readily with KMnO4. (B)

forms a compound C with SOCl2, which on reaction with (CH3)2NH

gives (D). The compound (A) on oxidation with dilute chromic acid

gives a compound (E) which on gentle heating readily gives (F),

C3H6O. Give structures of (A) to (F) with proper reasoning.

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ANSWERS TO EXERCISES

Exercise 1:

(i) Br MgBr

H O3

O

OMgBrMg/Ether

OHKMnO /H4

OH

O

(ii) CH CH

O

OHi COHBr Mg/Ether 2

2 2ii H O3

CH CHBr CH CH MgBr

Exercise 2:

(i) Boiling points of formic acid (100.3°C) and water (100°C) are

almost equal.

(ii) CH2

CH2

HBrCH3 Br

Mg/ ether

CH3Mg

Br2

3

CO

H O CH3COOH

Exercise 3:

A = CH3

CH3

OH

CN

B = CH3

CH3

OH

OH

O

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C = CH3

CH2

OH

O

D =

CH3

OH

O

OH

E =

CH3

CH3 OH

O

Exercise 4:

A = B =

OH

Cl

C =

OH

CN

D =

OH

COOH

E =

O

COOH

Exercise 5:

(A)

(H3C)2HC CH

CH3

C

O

O CH

CH3

CH(CH 3)2

(B) CH3 CH

CH3

CH COOH

CH3

(C) CH3 CH

OH

CH CH3

CH3

(D) H3CHC C(CH3)2

(E) CH3 CH

OSO 2C6H5

CH(CH 3)2

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(F) CH3 CH

Br

CH CH3

CH3

Exercise 6:

Since neutral compound A on hydrolysis gives an acid B and neutral

compound C, it must be an ester.

H

A B CEster Acid Alcohol

Further since the acid B reduces mercuric chloride, it must be formic

acid (HCOOH) which is the only reducing carboxylic acid. So A must be

HCOOC2H5.

H

2 5 2 5BA C

HCOOC H HCOOH C H OH

2 2 2 2HCOOH 2HgCl Hg Cl 2HCl CO

2

NaOH

2 5 3IC

C H OH HCOOH CHI

Exercise 7:

(i) In formate ion resonance gives rise to identical bond lengths

H—C—O– H—C = O

O O–

H—C O

O

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Whereas no such resonance is noticed in formic acid (H — COOH) and

thus C—O bonds are different in HCOOH

(ii) Acetic acid undergoes intermolecular H–bonding to form dimeric

state (CH3COOH)2 and thus results in double molecular weight.

Exercise 8:

OH

O

O

CH3(b) A =

Cl

O

O

CH3B =

Exercise 9:

(i)

A =

C CH2 CH2 COH

OO

B =

CH2 CH2 CH2 COH

O

C =

CH2 CH2 CH2 C

O

Cl

D =

O

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E =

OH

F =

G =

Br

H =

(ii)

C

O

C O

OH

C

O

C

O

(Friedel - crafts alkylation)

Exercise 10:

O

O

CH3

O

OH(c) A =

Exercise 11:

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CH CH CH CH3

O

CPh O

Ph

OH

CH CH CH CH3

O

CPh O

Ph

O

C OHPh

O

CH CHPh CH

O

CH3C OPh

O

CH CHPh CH

OH

CH3

Exercise 12:

A = H C

O

O HC

CH3

CH3

B = CH3 CH

OH

CH3

Exercise 13:

(a) COOH

COOH(A)

(Diel's - Alder Reaction)

(b)

H2C

H2C

C

C

NH

O

O

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Exercise 14:

(i) R—COOH 3ND ,2RCOND

(A) 2

KOH

Br2RND

(B)

(ii) R—COOH 3NH ,2RCONH

(C)

2

KOD

Br2R NH

(D)

Exercise 15:

(i)

Ph C

O

ClA = Ph C

O

OCH 3B =

Ph C

O

HC =

(ii)

Ph C

O

OHSOCl 2

Ph C

O

ClNH3

ExcessPh C

O

NH2

OH Br2

Ph NH2

(Hoffmann rearrangement)

(iii) (a)

CH3C OH

O

CH3LiH3C C

O

OLi CH4

CH3LiH3C C

OLi

OLi

CH3

H

H3C C

OH

OH

CH3

H3C C

O

CH3

(A)

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(b)

C

C

O

O

O CH2 CH3

OH

CH2OH

CH2OH

(B) (C)

Exercise 16:

A is (CH3)2 C = CH–CH2–CH2–COCH3

6–methyl–5–hepten–2–one

B is CH3–CO–CH2–CH2–CHO

4–oxopentanal

C is HO2 CCH2 CH2 CO2 H

1, 4–butandioic acid

D is succinimide

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CH2 – CH2

O = C C = O

N | H

Exercise 17:

|

3 2 3

|| ||

3 3 3 2

|| || ||

3 2 3 3

OH

A. CH C H CH COOH B. CH CH CH COOH

O O

C. CH CH CH C Cl D. CH CH CH C N CH

O O O

E. CH C CH C OH F. CH C CH

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MISCELLANEOUS EXERCISES

Exercise 1:

Suggest a suitable oxidising agent for the conversion.

3 3 3 22 2CH C = CHCOCH CH C = CHCO H

Exercise 2:

Arrange the following compounds in order of increasing acid strength:

3 2 3 2 3 2 3 2 22

i CH CH CH Br COOH,CH CH Br CH COOH, CH CHCO H, CH CH CH COOH

(ii) Benzoic acid, 4-nitrobenzoic acid, 3, 4-dinitrobenzoic acid, 4-

methoxybenzoic acid.

Exercise 3:

Why does benzoic acid not undergo Friedel-Crafts reaction?

Exercise 4:

Why carboxylic acids do not form oximes?

Exercise 5:

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Arrange the following in decreasing order of the boiling points:

CH3CH2CH2CH2OH, CH3CH2OCH2CH3.CH3CH2CH2COOH.

Exercise 6:

Why is an amide more acidic than amines?

Exercise 7:

Identify (A) and (B) in the following sequence of reactions.

(i) 3

4

CH COOH excess distil

3HgSOCH CH A B CH CHO

(ii) 34 CH COOHAIPO

3 1075KCH COOH A B

Exercise 8:

Why excess amine is necessary in the reaction of an acyl chloride with

an amine?

Exercise 9:

Write the structure of the products

(a) acetic acid + benzyl alcohol

(b) propionic acid + isobutanol

Exercise 10:

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Complete the following:

H5C6

O

OH

heat

5PCl

CH3

O

OK

Electrolysis

2H O

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ANSWERS TO MISCELLANEOUS EXERCISES

Exercise 1:

Alkaline KMnO4, acidified K2Cr2O7 or HNO3 can not be used since all of

these will cleave the molecule at the site of double bond giving a

mixture of ketone/acids. The most suitable reagent for this oxidation is

NaOI (I2/NaOH) since methyl ketones on treatment with NaOH

undergo Iodoform reaction to give Iodoform alongwith the Na salt of a

carboxylic acid having one carbon atom less than the starting methyl

ketone.

CH3 C

CH3

CH C

O

CH3 3NaOI CH3 C

CH3

CHCOONa 3CHI 2NaOH

2H /H O

CH3 C

CH3

CHCOOH

4 Methylpent 3 en 2 one

(Mesityl oxide)

3 Methylbut 2 en 1 oic acid

( , Dimethylacrylic acid)

Exercise 2:

(i) We know that +I, effect decreases while I effect increases the acid

strength of carboxylic acids. Since +I effect of isopropyl group is more

than that of propyl group, therefore, (CH3)2CHCOOH is a weaker acid

than CH3CH2CH2COOH.

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Further since I effect decreases with distance, therefore,

CH3CH2CH(Br)COOH is a stronger acid than CH3CH(Br)CH2COOH. Thus,

the overall acid strength increases in the order:

CH3 CH

CH3

COOH < CH2 COOHH2C

< CH3 CH CH2

Br

COOH < CH3 CH2 CH COOH

Br

CH3

(ii) Since electron donating groups decrease the acid strength,

therefore, 4–methoxybenzoic acid is a weaker acid than benzoic

acid.

Further since electron–withdrawing groups increase the acid

strength, therefore, both 4–nitrobenzoic acid and 3,

4–dinitrobenzoic acids are stronger acids than benzoic acid.

Further due to presence of an addition NO2 at m – position w.r.t

COOH group, 3, 4–dinitrobenzoic acid is a little stronger acid

than 4–nitrobenzoic acid. Thus the overall acid strength

increases in the order:

4 methoxybenzoic acid benzoic acid 4 nitrobenzoic acid

3,4 dinitrobenzoic acid

Exercise 3:

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Due to deactivation of the benzene ring by electrons withdrawing

effect of the COOH group.

Exercise 4:

Due to resonance between lone pairs of electrons on the O-atom of the

OH group and C = O, the carboxyl carbon is less electrophilic than

carbonyl carbon in aldehyde and ketones. Therefore, nucleophilic

addition of NH2OH to the C = O group of carboxylic acid does not occur

and hence carboxylic acids do not form oximes.

Exercise 5:

CH3CH2CH2COOH > CH3CH2CH2CH2OH > CH3CH2OCH2CH3

Hint: Due to hydrogen bond

Exercise 6:

R NH2

O

R NH2

O

Firstly due to delocalization of lone pair of electrons of N atom over the

C = O group, amino group gets a positive charge which makes N–H

bond weak. Secondly the anion left after removal of a proton is

stabilized by resonance.

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R NH

O

R NH

O

In amines no such stabilization is possible.

Exercise 7:

(i) 3

4

CH COOH excess distil

3 3 3 3HgSO 2 2Acetylene AcetaldehydeEthylidene diacetate Acetic anhydride

CH CH CH CH OOCCH CH CO O CH CHO

(ii) 34

2

CH COOHAIPO , 1075K

3 2 3H O 2KetoneAcetic acid Acetic anhydride

CH COOH CH C O CH CO O

Exercise 8:

R Cl

O

2R NH

R NHR'

O

HCl

The products of this reaction are an amide and HCl. HCl formed in this

reaction protonates unreacted amine and since protonated amines are

not nucleophiles, the reaction with acyl chloride stops.

2 3

No nucleophilic site

R NH HCl RNH Cl

R Cl

O

3RNH No reaction

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Hence, reaction must be carried out with twice as much as amine as

acyl chloride.

Exercise 9:

(a) CH3—COOCH2—Ph

(b) CH3—CH2COOCH2—CH(CH3)2

Hint: Esterification

Exercise 10:

(a) COOH

5PCl

COCl

3HCl POCl

(b) electrolysis

3 2 3CH COOK H O CH COO K

At anode:

CH3

CH3

22CO 2e32CH COO

At cathode:

2 22e 2H O 2OH H

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SOLVED PROBLEMS

Subjective:

Board Type Questions

Problem 1:

Write chemical reactions to effect the following transformation.

(a) Butan-1-ol to butanoic acid

(b) Benzyl alcohol to phenylacetic acid

(c) Bromobenzene to benzoic acid

(d) p-methylacetophenone to benzene-1, 4-dicarboxylic acid

Solution:

(a) 4 2 4Aq. KMnO Dil. H SO

3 2 2 2 3 2 2 3 2 2CH CH CH CH OH CH CH CH COOK CH CH CH COOH

Butan-1-ol Pot. Butanoate Butanoic acid

(b) 3

2

PBr KCN Hydrolysis

6 5 2 6 5 2 6 5 2 6 5 2or HBr H /H O

Benzyl alcohol Benzyl bromide Phenylacetic acid

C H CH OH C H CH Br C H CH CN C H CH COOH

(c)

2

3H OMg Dry ice

6 5 6 5 6 5 6 5dry ether

Bromo benzene Phenyl mag. bromide Benzoic acid

i.e. CO (s)

C H Br C H MgBr C H COOMgBr C H COOH

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(d) CH3 COCH3 4KMnO /KOH, KOOC COOK 2 4Dil. H SO HOOC COOH

P-methylacetophenone Dipotassium benzene- 1, 4-

dicarboxylate

Benzene-1, 4-dicarboxylic acid

Problem 2:

Arrange the following compounds in increasing order of their boiling

points:

acetic acid, methyl formate, acetamide, propan-1-ol

Solution:

(i) Out of all these compounds methyl formate does not undergo H-

bonding therefore, its boiling point is the lowest.

(ii) Amongst the remaining three compounds intramolecular H-

bonding is most extensive in acetamide, followed by acetic acid

and least in propan-1-ol. Therefore, their boiling points decrease

in the same order i.e. acetamide>acetic acid>propan-1-ol. Thus,

the overall increasing order of their boiling points is methyl

formate<propan-1-ol<acetic acid<acetamide.

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Problem 3:

CH3 C

O

OHSOCl2 A

H2/Pd

BaSO 4B

HCNC

H3OD E

Δ

Solution:

A = CH3 C

O

Cl B = CH3 C

O

H C = CH3 C

OH

CN

H

D = CH3 C

OH

COOH

H

E =

O

CH

C

C

CH

O

O

CH3

O

CH3

Problem 4:

A dicarboxylic acid (A), C4H6O4, gave a compound (B), C6H10O4 upon

treatment with excess of methanol and a trace of H2SO4. Subsequent

treatment of (B) with LiAlH4 followed by usual work up gave C,

C4H10O2. Heating of A yielded D, C4H4O3. Assign structures to A, B, C

and D.

Solution:

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A =H2C

H2C

COOH

COOH

B =H2C

H2C

C

C

OCH 3

OCH 3

O

O

C =CH2CH2OH

CH2CH2OH

D =H2C

H2C

C

C

O

O

O

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IIT Level Questions

Problem 5:

Two organic compounds A and B have vapour densities 15 and 30

respectively. (A) reduces Fehling solution, but does not react with

Na2CO3. (B) does not reduce Fehling solution but gives effervescences

with Na2CO3. B when treated with a concentrated solution of base

followed by prolonged heating gives a compound C(C3H6O). Identify A,

B and C.

Solution:

(A) H C

O

H

(B) CH3 C

O

OH

(C) CH3 C CH3

O

Problem 6:

An organic acid A, C3H4O3 is catalytically reduced in presence of

ammonia to give B, C3H7NO2. B reacts with acetyl chloride,

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hydrochloric acid and alcohols. It can also react with nitrous acid to

give another compound C, C3H6O3, along with the evolution of

nitrogen. What are A, B and C. Give reasons?

Solution:

Compound A is acid having one –COOH group only, the remaining part

C2H3O can be

O || CH3 – C – only. Hence, structural formula of A is

CH3 COOH

O

on catalytic reduction keto group is converted into secondary alcohol

which with ammonia will give amino acid, i.e.,

O || CH3 – C – COOH reduction CH3 – CH – COOH 2HNH

CH3 – CH – COOH –H2O

OH |

NH2

|

with nitrous acid, B, react to give

NH2

| CH3 – CH – COOH + HNO2 CH3 – CH – COOH + N2 + H2O

OH |

(B) (C)

Problem 7:

Compound A (C6H12O2) on reduction yields two compounds B and C.

The compound B on oxidation gave D, which on treatment with

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aqueous alkali and subsequent heating furnished E. The latter on

catalytic hydrogenation gave C. The compound D was oxidised further

to give F which was found to be a monobasic acid (mw = 60). Deduce

the structures of A to E.

Solution:

ReductionA B C

OHD E

H2/Pt[O]

O

monobasic acidmol. wt. 60

F

From the molecular weight data F comes out to be CH3COOH. If we

work back B is CH3CH2OH and D is CH3CHO.

2H /PtOH

3 3 3 2 2 2D E C

CH CHO CH HC CH CHO CH CH CH CH OH

Aldol condensation

From here we can deduce that reduction of A with LiAlH4 gives two

alcohols. So A must be an ester. Hence (A) comes out to be

CH3CH2CH2COOC2H5.

Problem 8:

An organic compound (A) on treatment with acetic acid in the

presence of sulfuric acid produces an ester (B). (A) on mild oxidation

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gives (C). (C) with concentrated KOH followed by acidification with dil.

HCl regenerates (A) and produces (D). (D) with phosphorous

pentachloride followed by reaction with ammonia gives (E). (E) on

dehydration produces hydrocyanic acid. Identify compounds A, B, C, D

and E.

Solution:

E-H2O

HCN So (E) must be H C

O

NH2

H C

O

OH

(D)

PCl 5H C

O

ClNH3

H C

O

NH2

(E)

(A) reacting with acetic acid is giving an ester (B). So (A) must be an

alcohol. (A) alcohol on mild oxidation gives (C). Hence (C) is an

aldehyde. (C) with conc. KOH followed by acidification is yielding (A)

alcohol and | |O

H C OH (D). So this reaction is cannizzaro reaction. So A

must be CH3OH.

||

3 3 3 2

O

CH OH CH C O CH HCHO HCOOH HCONH

A (B) C D E

Problem 9:

An aromatic compound A on treatment with CHCl3 and KOH gives B &

C, both of which, in turn give the same compound D when distilled

with Zn dust.Oxidation of D yields E of formula C7H6O2. The sodium

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salt of E on heating with soda lime gives F which can also be obtained

by distilling A with Zn dust. Identify A, B, C, D, E and F.

Solution:

Molecular formula of (E) is C7H6O2 and reaction of its sodium salt with

soda lime (decarboxylation) to form (F) indicates that (E) and (F)

should be C6H5COOH and C6H6 respectively. Since (F) is also obtained

from (A) by reaction with Zn dust, it indicates that (A) should be

phenol. Nature of (A) as phenol is confirmed by the fact that it

explains all the given reactions.

OH

(A)

CHCl 3

OH

CHO

OH

CHOB & C

Zn dust

CHO

(D)

[O]

COOH

(E)

NaOH

Soda lime

(F)

KOH

Zin

c dust

dis

tilla

tion

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Problem 10:

Five isomeric para – disubstituted aromatic compounds A to E with

molecular formula C8H8O2 were given for identification. Based on the

following observations, give structures of the compounds?

(i) both A and B form a silver mirror with Tollen’s reagents; B gives

a positive test with FeCl3 solution also.

(ii) C gives positive iodoform test.

(iii) D is readily extracted in NaHCO3 solution.

(iv) E on acid hydrolysis gives 1, 4 dihydroxy benzene.

Solution:

(A) and (B) gives tollen’s test. Hence must contain aldehyde group.

But (B) shows phenolic test also. Hence (A) & (B) are

OCH 3

CHO

CH2OH

CHO

or

(A)

OH

CH2CHO

(B)

(C) shows Iodoform test show it must contain ||

3

O

C CH group.

Therefore (C) is

COCH 3

OH

(C)

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(D) shows presence of carboxylic group. Hence (D) and (E) are

COOH

CH3

OH

O CH CH2

(D) (E)

Problem 11:

An organic compound A, C8H4O3 in dry benzene in the presence of

anhydrous AlCl3 gives compound B. The compound B on treatment

with PCl5, followed by reaction with H2/Pd (BaSO4) gives compound C,

which on reaction with hydrazine gives cyclised compound D

(C14H10N2). Identify A, B, C and D.

Solution:

Nature of reagents in the conversion of A to B indicates that the

reaction must be Friedal – crafts reaction.

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C

C

O

O

O

C

C C6H5

OH

O

O(A)

C6H6

AlCl 3

PCl 5

C

C C6H5

Cl

O

O

H2/Pd BaSO 4

CHO

C C6H5

O

(C)

N

N

C6H5

(D)

(B)

NH2NH2

Problem 12:

An organic compound A, C6H10O on reaction with CH3MgBr followed by

acid treatment gives compound B. The compound B on ozonolysis

gives compound C, which in presence of a base gives 1- acetyl

cyclopentene. The compound B on reaction with HBr gives compound

D. Write the structures of A, B, C and D.

Solution:

As B undergoes ozonolysis to form C, if must have a double bond. C in

the presence of base gives , unsaturated ketone, so it is aldol

condensation.

O

(A)

(i) CH 3MgBr

(ii) H

CH3

(B)

O3CH3

H

O

O

(C)

OH

COCH 3

OH

-H2O

COCH 3

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HBr

CH3 CH3 Br

(B) (D)

Problem 13:

A phenolic compound (A), C7H6O2 on mild oxidation gives a highly

volatile oil (B). A forms (C) on reaction with dimethyl sulphate in

alkali. Oxidation of (C) with hot KMnO4 gives (D), the silver salt of

which reacts with bromine water followed by heating gives (E)

containing about 72% bromine. Give structures A to E.

Sol.

A can be

OH

CHO (ortho, meta or para)

O H

C H

O

But it must be ortho because on oxidation, it gives a highly volatile oil

indicating that there is intramolecular hydrogen bonding.

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OH

CHO

(A)

OH

CHO

(A)

OH (CH3)2SO4

OCH 3

CHO

(C)

Oxidation

OH

COOH

Oxidation

OCH 3

COOH

(D)

Ag2O

Br2/H2O

OCH 3

BrBr

Br

(E)

(B)

240 100% of bromine in E 69.56%

345

Problem 14:

An aromatic compound (A) gave a mixture of two isomeric products B

and C on reaction with NH2OH. C rearranged to D (C8H9NO) on heating

with H2SO4. D on hydrolysis produced E and F. A was oxidised with per

benzoic acid to G. Hydrolysis of G gave H and E. An anhydride of E and

its sodium salt on condensation with PHCHO produced cinnamic acid. H

on reaction with phthalic anhydride in H2SO4 gave phenolphthalein.

Suggest structures for A to H.

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Sol.

C

CH3H5C6

O

NH2 OH C N

H5C6

CH3

OH

(A)(B)

per benzoic acid

C

OH3C

O

C6H5

H

CH3 C

O

OH

(G)

(E)

C6H5OH

(H)

C N

H5C6

CH3 OH

(C)

2 4H SO , Δ

CH3CONHC 6H5

(D)

H2O

C6H5CO 3H

CH3CONHC 6H5

OH

H2OCH3COOH C6H5NH2

(D) (E) (F)

(CH3CO) 2O C6H5CHO

CH3COONa H5C6

COOH

Anhydride of (E) (Perkin Reaction)

Formation of cinnamic acid proves that the anhydride is acetic

anhydride. Hence E is acetic acid.

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Objective:

Problem 1:

End products of the following sequence of reactions is

O

C CH3

O

2+

1. I +NaOH

2. H , Δ

(A) CHI3 ,

O

COOH

(B) CHI3 ,

O

CHO

(C) CHI3 ,

O

(D) CHI3 ,

COOH

COOH

Solution:

Intermediate is

O

COOH which loses CO2 on heating.

(C)

Problem 2:

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End product of the reaction || ||

3 2 2 2

O O

H C C CH CH CH COH 4

2

1. NaBH

2. H O, H is

(A)

O

CH3

O

(B

)

O

O

CH3

(C) H3CHC

OH

CH2CH2CH2CH2OH

(D

) CH3C

O

CH2CH2CH2CH2OH

Solution:

NaBH4 reduces ketonic group but not acidic group

CH3 CH

OH

CH2 CH2 CH2 COOH

is the intermediate which forms cyclic ester.

O

CH3

O

(A)

Problem 3:

R CH2 CH2OH can be converted into R CH2CH2COOH. The

correct sequence of reagents is

(A) 3 2PBr , KCN, H /H O (B) 3 2PBr , KCN, H

(C) KCN, H (D) HCN, H

Solution:

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3

2

PBr KCN H

2 2 2 2 2 2 2 2H OR CH CH OH R CH CH Br R CH CH CN R CH CH COOH

(A)

Problem 4:

The ease of alkaline hydrolysis is more for

(A) COOCH 3

NO2

(B) COOCH 3

Cl

(C) COOCH 3

(D) COOCH 3

OCH 3

Solution:

There is more electron deficiency on carbonyl carbon in

COOCH 3

NO2

(A)

Problem 5:

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Which of the following undergoes HVZ reaction?

(A) HCOOH (B) CCl3COOH

(C) C6H5COOH (D) none

Solution:

None of these contains - hydrogen.

(D)

Problem 6:

The product C in the reaction 5 2 4PCl H /Pd, BaSO dil. NaOH

3CH COOH A B C

(A) CH3CH = CHCHO (B) CH3CH = CHCOOH

(C) CH3CH2CHO (D) CH3CH2CH2COOH

Solution:

5 2 4

||PCl H /Pd, BaSO OH

3 3 3 3

O

CH COOH CH COCl CH C H CH CH CHCHO

Problem 7:

The end product of the reaction is

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Br2A

H2, Pt KCN

H3OCB

C is

(A) propanoic acid (B) adipic acid

(C) malonic acid (D) succinic acid

Solution:

Br

Br

Br

Br

COOH

COOH

Adipic acid(A) (B) (C)

Problem 8:

Find out D in the following sequence

OH

conc.

H2SO4

A NBS B KCN CH3O D

(A)

CH3

COOH

(B)

COOH

CH3

(C)

COOH

COOH

(D)

CH3

COOH

Solution:

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OH

conc.

H2SO4

NBS

Br

KCN

CNCOOH

Problem 9:

The order of the acidic strength of following acids is

(i) CH3CH2COOH (ii) CH2 = CHCOOH (iii) CH CCOOH

(A) i > ii > iii (B) iii > ii > i

(C) ii > i > iii (D) iii > i > ii

Solution:

sp hybridized carbon is most electronegative and it stabilizes the

carboxylate ion. to the maximum extent.

Problem 10:

CH3

COOH

OH

CrO3 A Δ

B

A and B are

(A) CH3COCH2COOH, CH3COCH3

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(B) CH3CH(OH)CH3, CH3COCH3

(C) CH3COCH3, CH3COOH

(D) CH3CH = CHCOOH, CH3CH = CH2

Solution:

COOH

OH

CrO3 COOH

O

Δ

O

(A) (B)

(A)

Problem 11:

*

3

3

i Me MgXNaHCO

3 2 ii H OCH CH COOH A a gas B

B is *

14C is C

(A) *

3CH COOH (B) *

3CH COOH

(C) *

3 2 3CH CH COCH (D) *

3 3CH COCH

Solution:

*

3

*NaHCO

3 2 3 2 2 2CH CH COOH CH CH COONa H O CO

3

* *H O

2 3CO MeMgX CH COOH

(A)

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Problem 12:

OC2H5

CH2CO2C2H5

O

C2H5OA

A is

(A) O

OC2H5

(B) O

CO2C2H5

(C)

O

O

(D) O

O

Solution:

OC2H5

CH2CO2C2H5

O

C2H5OOC2H5

CH

O

C O

O

C2H5

O

C

O

O C2H5

Problem 13:

The carboxylic acid which has maximum solubility in water is

(A) phthalic acid (B) succinic acid

(C) malonic acid (D) salicylic acid

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Solution:

Salicylic acid has intramolecular hydrogen bonding. So it is least

soluble. Out of the first three acids malonic acid has smallest non –

polar alkyl part. So it is most soluble.

(C)

Problem 14:

Br

CH3

H

O

O

2

3

i Mg/ether

ii CO

iii H O

The final product is

(A) COOH

CH3

H

O

O

(B) MgBr

CH3

CHO

(C) COOH

CH3

CHO

(D) None of these

Solution:

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Br

CH3

H

O

O

Mg/ ether

MgBr

CH3

H

O

O

2CO

COO

CH3

H

O

O

H3O

COOH

CH3

CHO

Problem 15:

In the reaction 3 2 5NH P O

3 2CH CH COOH A B C,

Identify C.

(A) CH3CH2CN (B) CH3CH2CH2NH2

(C) CH3CH2CH2OH (D) (CH3CH2CO)2O

Solution:

CH3CH2COOHNH3 H2CH3C C

O

ONH 4Δ CH2 C

O

NH2CH3

P2O5

CH2 CNCH3

Fill in the blanks

Problem 16:

Benzoic acid has a………….pKa value than phenol.

Solution:

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Lower

Problem 17:

The acid derivatives are hydrolysed…………..readily in either alkaline or

acidic medium than in neutral medium.

Solution:

More

Problem 18:

O-nitrobenzoic acid is…………acid than p-nitrobenzoic acid.

Solution:

Stronger

Problem 19:

Malonic acid on heating with phosphorous pentoxide gives…………………..

Solution:

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Carbon suboxide (C3O2)

Problem 20:

Cinnamic acid can be prepared from benzaldehyde

using…………….reaction and shows…………..isomerism.

Solution:

Perkin, geometrical

True and False

Problem 21:

Methanoic acid like ethanoic acid can be halogenated in the presence

of red phosphorous and Br2.

Solution:

False:

Due to the absence of - hydrogen, methanoic acid can not be

halogenated by using Br2/red P.

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Problem 22:

As the number and size of the substituents around the COOH group

increases the rate of esterification slows down.

Solution:

True

Problem 23:

The boiling point of propionic acid is less than that of n-butanol, an

alcohol of comparable molar mass.

Solution:

False:

Due to extensive intermolecular hydrogen bonding, and the cyclic

dimer formation, propanoic acid has higher b.p. than n-butanol.

Problem 24:

The conjugate base of chloro acetic acid is more resonance stabilized

than the conjugate base of acetic acid.

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Solution:

True:

Due to the presence of chlorine atom.

Problem 25:

The correct order of acid strength of the given acid is

2 4 3 2 2 5H SO > CH COOH > H O > HC ≡ CH > C H OH

Solution:

False:

The correct order is2 4 3 2 2 2 5H SO CH CO H H O C H OH HC CH .

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ASSIGNMENT PROBLEMS

Subjective:

Level-O

1. Describe the following:

(a) Saponification (b) Decarboxylation

2. How is acetic acid prepared from?

(i) Acetylene (ii) Ethyl alcohol

3. How is Grignard reagent employed to prepare a carboxylic acid?

4. Explain the following about acetic acid:

(i) Its boiling point is higher than that of n – propanol.

(ii) It is weaker acid than formic acid.

(iii) It is stronger acid than phenol.

5. Acid anhydrides have higher boiling points than the corresponding

carboxylic acids.

6. Give IUAPC name for each of the following compounds.

(a) H3CHC CHCOOH (b) COOH

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(c) CHCOOH

CH3

(d) Cl2CHCOOH

(e)

HOC

O

(CH2)3COOH

7. HOCH2CH2CH2CH2COOH can be prepared from HOCH2CH2CH2CH2Br in

two ways. One is through Grignard synthesis and the other through

nitrile synthesis. Which method would you prefer?

8. Why the bond length of C = O in carboxylic acids is bit longer than in

aldehydes?

9. Acetic acid can be halogenated in presence of phosphorus and chlorine

but formic acid cannot be halogenated in the same way.

10. During the preparation of esters from a carboxylic acid and an alcohol

in the presence of an acid catalyst, the water or the ester should be

removed as fast as it is formed. Explain.

11. Identify the products in the following reactions

(a) 3 2

3

i) CH CH MgBr /ether

3 ii) H OHCOOCH (b) dil. NaOH

6 5 3 2C H CHO CH CH CHO

12. Explain why benzamide is less easily hydrolysed than methyl benzoate.

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13. There are two –NH2 groups in semicarbazide. However only one is

involved in the formation of semicarbazones. Explain.

14. Carbon-oxygen bond length in formic acid are 1.24 Å and 1.36 Å but in

sodium formate both the carbon-oxygen bonds have same value

i.e.1.27 Å.

15. Acid and acid derivatives although contain >C=O group, do not

undergo the usual properties of carbonyl group explain.

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Level – I

1. Convert the following:

(a) Acetaldehyde Butane 1, 3 diol

(b) Acetaldehyde But 2 enoic acid

2. Discuss the action of heat on - hydroxy carboxylic acids.

3. Compare the Ka1 and Ka2 values of maleic and fumaric acids.

4. R NH2 are soluble in aqueous acids like HCl, H2SO4 etc but ||

2

O

R C NH

are almost insoluble in those acids. Explain.

5. What happens when:

(i) Dry chlorine is passed through acetic acid in presence of

sunlight.

(ii) Formic acid is reacted with ammonical silver nitrate solution.

6. Complete the following equations by writing the missing A, B, C.

3 4H O KMnO

3 2 3 2 2 2CH CH MgBr A CH CH CH CH OMgBr B C

7. What is A and B?

3CrO warm

4 8 3C H O A BCH3 CH3

O

2CO

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8. (a) ||

3 2

O

CH CH CH CH C OH A

(b) 2 3 2|| ||HO C CH C CH C OH B

O O

9. Cite 2 reasons why malonic acid does not form a cyclic anhydride.

10. 6 62 4 2

4 3

O C HH SO SOCl

HgSO anhy AlClCH CH A B C D

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Level – II

1. A hydrocarbon A, (C8H10), on ozonlysis gives compound B (C4H6O2)

only. The compound B can also be obtained from the alkyl bromide, C

(C3H5Br) upon treatment with magnesium in dry ether, followed by

CO2 and acidification. Identify A, B and C and also gives equations for

the reactions.

2. Predict the product of the reaction given below, given that it is formed

under high – dilute conditions. EtO cat

2 2 24EtO C CH CO Et

3. Compound (A), C5H8O2, liberates CO2 on reaction with sodium

bicarbonate. (A) exists in two forms neither of which is optically active.

It yields compound (B), C5H10O2 on hydrogenation. Compound (B) can

be separated into two enantiomorphs. Write the structural formulae of

A and B.

4. An organic compound (A) reacts with PCl5 to form a compound (B). A

also reacts with NaOH to form a compound (C). Both B and C react

together to form a compound D (C6H10O3). What are A, B, C and D?

5.

a Δ

CH3

CH3

O

CH3

H

`

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b 2 6B H

HCCH3

O

CH2OH

c 2

2

1 Ag O

2. BrO2N Br

6. The active ingredient of an insect repellent is N, N–diethyl–m–

toluamide. Outline a synthesis of this compound starting with benzene.

7. A carboxylic acid (A) of unknown structure was found to contain only

C, H and O. 154.5 mg of this acid required 11.9 mL of 0.22 N NaOH to

reach equivalence point.

Gentle heating of A led to evolution of CO2 and formation of new

carboxylic acid (B) with nutralisation equivalent of 74. Suggest

structures for A & B.

8. Compound (A), C5H11NO is not soluble in cold dil. alkaline or acidic

solutions. When (A) is refluxed with NaOH solution, a gas (B) is

evolved and a salt (C) is formed. Acetyl chloride reacts with (B) to give

(D), C4H9NO. (B) reacts with HNO2/HCl to give a yellow oil (E). Give

structures of A to E.

9. Compound (A) (C6H12O2) on reduction with LiAlH4 yielded two

compounds B and C. The compound B on oxidation gave (D) which on

treatment with aqueous alkali and subsequent heating furnished (E).

The latter on catalytic hydrogenation gave C. The compound D was

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oxidised further to give F which was found to be monobasic acid with

molecular weight 60. Deduce the structures of A, B, C, D, E and F.

10. Suggest a mechanism for each of the following tranforamations.

O

CO 2Et

CO 2Et t - BuO

CO 2Et

O

CO 2Et

t - BuOH

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Objective:

Level – I

1. Which of the following will not give cyclic anhydride on heating?

(A)

H2C

COOH

COOH

(B) H2C

H2C

COOH

COOH

(C)

H2C

CH2COOH

CH2COOH

(D) COOH

COOH

2. In the reaction

2Cl / red P alc. KOH

3 2CH CH COOH A B

the compound B is

(A) CH3CH2OH (B) CH3CH2COCl

(C) CH2 = CH COOH (D) CH3CHClCOOH

3. Give the structure of the product of the following reaction

O

CH3

O

CH3NH2

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(A)

O

CH3

N CH3

(B)

O

CH3

NHCH 3OH

(C)

CH3 NH C

O

CH2 CH2 CH

OH

CH3

(D) None

4. Malonic and succinic acids are distinguished by

(A) heating (B) NaHCO3

(C) both A and B (D) none

5. Consider the following acids

1. MeCH2COOH

2. Me2CHCOOH

3. Me3CCOOH

4. Et3CCOOH

Correct order of rate of esterification.

(A) 1 > 3 > 2 > 4 (B) 1 > 2 > 3 > 4

(C) 4 > 3 > 2 > 1 (D) 3 > 2 > 1 > 4

6. Which of the following compound on heating with caustic potash gives

ammonia?

(A) CH3CONH2 (B) CH3CH2NH2

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(C) (CH3)2C = NOH (D) ||

2 3 2

O

CH C N CH

7. Identify the final product in the following sequence of reactions

CH3CH2MgBrH2C CH2

O

XH

YKMnO 4

Z

(A) CH3CH2CH2COOH (B) CH3CH(CH3)COOH

(C) CH3CH2COOH (D) CH3CH2CH2CH2OH

8. The typical reaction of carbonyl group is suppressed the most in

(A) CH3 CHO (B) CH3CONH2

(C) CH3COCH3 (D) CH3COOCH3

9. Which of the following carboxylic acids undergoes decarboxylation

easily?

(A) C6H5COCH2COOH (B) C6H5COCOOH

(C) C6H5 HC

OH

COOH

(D) CH

NH2

COOHH5C6

10. Benzoic acid gives benzene on being heated with X and phenol gives

benzene on being heated with Y. Therefore X and Y respectively are

(A) sodalime and copper (B) Zn dust and NaOH

(C) Zn dust sodalime (D) sodalime and Zn dust

11. 2 5 2

||P O Br

2 OH

O

X R C NH Y

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X and Y respectively are

(A) R CN, RCH2NH2 (B) RNH2, RCN

(C) RCN, RNH2 (D) RCN, RCH2NH2

12. 2 5 3Al OC H

32CH CHO

Product and the name of the reaction respectively are

(A) CH3COOCH2CH3, Tisckenko

(B) CH3COOH, CH3CH2OH, cannizzaro

(C) 2CH3CH2OH, perkin

(D) none

13. Which of the following reactions sounds correct?

1.

COOH

COOH Br2

BrHOOC

Br COOH

2.

COOH

COOH alk.KMnO 4

COOHOH

OH COOH

Cold

(meso)

3. Br2

COOHBr

Br COOH

HOOC

COOH

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4. alk.KMnO 4

COOHOH

OH COOH

HOOC

COOH

(meso)

(A) 1 and 2 (B) 2 & 3

(C) 3 & 4 (D) 1 & 4

14.

CH3ONa

C

O

OH

(A)

C O CH3

O

(B)

C O Na

O

CH3OH

(C)

C

O

CH3

(D)

C

O

OH

15. The reaction of ||

2

O

R C NH with a mixture of Br2 and KOH gives R NH2

as a product. The intermediate involved in this reaction is

(A) ||O

R C NHBr (B) R N C O

(C) R NHBr (D) ||

2

O

R C NBr

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16. Among the carboxylic acids:

CH3

O O

OH

OHOO

H

OH

O

and

COOH

NO2

(a) (b) (c) (d)

the, one or more than one undergoing decarboxylation readily is (are)

(A) Only a (B) Both a and b

(C) a, b and d (D) a, b and c

17. In the following reaction,

+

i) DIBALH

ii) Hhydroxy carboxylic acid

A is

(A) lactone (B) Lactide

(C) a hemiacetal (D) , - unsaturated acid

18. Which of the following can form an anhydride on heating?

(A) H

OH

O

(B)

O

OH O

OH

(C)

O

OH OH

O

(D)

OH

O

O

OH

19. i) ND , 3

ii) KOH, Br , H O2 2Product is

OH

O

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(A)

NH2

(B)

ND 2

(C)

NH2O

(D)

ND2O

20. Acetamide is treated separately with the following reagents. Which one

of these would give methylamine?

(A) PCl5 (B) Sodalime

(C) NaOH + Br2 (D) None of these

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Level-II

1. ||

2 2

O

Ph C OH CH N Product

(A) PhCOOCH2N2 (B) PhCOOCH3

(C) PhCH3 (D) none of these

2.

H C

O

O

H

A

B

H C

O

OC

D

(C O) bond lengths designated by A, B, C, D are in order

(A) A = C < B = D (B) A < B < C = D

(C) A < C = D < B (D) all are equal

3. A (monobasic acid) 3NH /Δ

BKBrO

ΔC 2HNO

(H3C)3C OH

or

(H3C)3C Br

or

H3C C CH3

CH2

A is

(A) 3 3CH C COOH (B)

CH3 CH CH2 COOH

CH3

(C) CH3 CH2 CH2 COOH (D)

CH3 C CH COOH

CH3

CH3

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4. O

O

O

O

on saponification forms

(A) OH

OH

O

H

H

O

(B) 2CH3OH, 2HCOOH

(C) OH

OH

O

OH

OH

O

(D) OH

OH

H C

O

OH

5.

4 10P OA

3

MeMgBr

H OBC NH2

O

2

2 3

i Ca OH

ii I , H O C

(A)

C CH3

O

(B) COOH

(C)

C

O

(D)

6. 2 5 2 5 3C H O / C H OH i) H O

ii)(A) (B)

O

COOC 2H5

COOC 2H5

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The compound B is

(A)

O

COOH

(B) O

COOH

COOH

(C) O

COOEt

COOEt

(D) O

COOEt

7. Which of the following reaction/s will yield carboxylic acid?

1. RMgX + CO2

2. PhOH + CHCl3 + NaOH

3. PhCN + H3O+

4. PhOH + NaOH + CO2 H

(A) 3 and 4 (B) 2 and 4

(C) 2 only (D) 4 only

8. An ester (A) with molecular formula C9H10O2 was treated with excess

of CH3MgBr and the complex so formed was treated with H2SO4 to

give an olefin (B). Ozonolysis of (B) gave a ketone with molecular

formula C4H8O which shows positive iodoform test. The structure of

(A) is

(A) 6 5 2 5C H COOC H (B)

||

2 5 6 5

O

C H C O C H

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(C) C4H9COOC4H9 (D) ||

5 11 3 7

O

C H C O C H

9. An ester (A) with molecular formula C11H14O2 was treated with LiAlH4

to give two compounds C9H12O (B) and C2H6O (C). B on heating with

an acid forms C9H10(D). Compound (D) on oxidation with KMnO4 forms

phthalic acid. Compound A is

(A) CH2COOC 3H7

(B)

H2C

CH3

C

O

O C2H5

(C)

H2C C

O

O C2H5

CH3

(D)

H2C C

O

O C2H5

CH3

10. O

CH3CO2OHP. P is

(A)

O

O

(B)

O

O

O

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(C)

O

O

(D) O

OHOH

11. Identify the product ‘A’ in the following reaction.

H2C

COOH

COOH

ΔCH3COOH A

(A) CO2 (B) CH3CHO

(C) CH3OH (D) none of theses

12. Which carboxylic acid of equivalent mass 52 g losses CO2 when heated

to give an acid of equivalent mass of 60 g?

(A) CH3COOH

(B) CH2(COOH)2

(C) CH2(COOH) CH2(COOH)

(D) HOOC(CH2)3COOH

13. CH3CH2COOHNaN3, conc. H2SO4

ΔX by reaction R 1

Y by reaction R 2

Br2/P

Choose the correct alternate

X Y R1 R2

(A) CH3CH2NH2 3|

CH C HCOOH

Br

Curtius HVZ

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(B) CH3CH2CONH2 CH3CH2OBr HVZ Hoffmann

(C) CH3CH2NH2 CH3CH2OBr Curtius HVZ

(D) none of these

14.

O

O

O

NaBH 4X LiAlH4Y X & Y respectively are

(A)

O

OH

OH

and

(B)

O

O

OH

andOH

OH

(C) O

O

both are

(D) OH

OHboth are

15.

Prolonged

O

COOH

COOH

COOH

Heating

(A) O

COOH

COOH

(B) COOH

COOH

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(C)

C

C

O

O

O

(D)

C

C

O

O

O

O

16. When acetic acid reacts with ketene, product formed

(A) ethyl acetate (B) aceto - acetic ester

(C) acetic anhydride (D) no reaction

17. R—CH2—CH2OH can be converted in R—CH2CH2COOH. The correct

sequence of reagents is

(A) PBr3, KCN, H+ (B) PBr3, KCN, H2

(C) KCN, H+ (D) HCN, PBr3, H+

18. The pKa of acetylsalicylic acid (aspirin) is 3.5. The pH of gastric juice in

human stomach is about 2-3 and pH in the small intestine is about 8.

Aspirin will be.

(A) Unionized in the small intestine and in the stomach

(B) Completely ionized in the stomach and almost unionized in the

small intestine.

(C) Ionized in the stomach and almost unionized in the small

intestine

(D) Ionised in the small intestine and almost unionised in the

stomach

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19. On subjecting mesityl oxide to the iodoform reaction, one of the

products is the sodium salt of an organic acid. Which acid is obtained?

(A) (CH3)2C=CH—CH2COOH (B) (CH3)2CH—COOH

(C) (CH3)2C=CH—COOH (D) (CH3)2C=CH—CO—COOH

20. The ease of alkaline hydrolysis is more for

(A)

COOCH 3

NO 2

(B) COOCH 3

Cl

(C) COOCH 3

(D) COOCH 3

OCH 3

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ANSWERS TO ASSIGNMENT PROBLEMS

Subjective:

Level – O

1. See the text

2. (i) 4

o2 4

1% HgSO

dil. H SO , 80 CCH CH

CH3 H

O

4

[O]

KMnO

CH3 OH

O

(ii)

2 2 7 2 4

[O]

3 2 K Cr O /H SO ,CH CH OH

CH3 H

O

4

[O]

KMnO

CH3 OH

O

3.

2

(i) Dry ether

3 (ii) H /H OCH MgBr O C O

CH3 OH

O

4. (i) In acetic acid O–H bond is more polar so H-bonding will be

stronger and boiling point will be high.

(ii) Due to the presence of +I effect of –CH3 group acetic acid is

weaker than formic acid.

(iii) It is stronger acid than phenol due to the presence of

symmetrical resonating structure whereas in case of phenoxide

ion negative charge comes on less electronegative carbon atom

and contributions of resonating structures are less.

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5. Due to the high molecular mass of anhydride van der Waals’ force

increases hence boiling point increases.

6. (a) But-2-enoic acid

(b) 2-Cyclohexenecarboxylic acid

(c) 2-Cyclopropylpropanoic acid

(d) 2, 2 – Dichloroethanoic acid

(e) Pentane-1, 5-dioic acid

7. A nitrile synthesis. Preparation of Grignard reagent from

HOCH2CH2CH2CH2Br is not possible because of the presence of the

acidic hydroxyl group.

8. Due to resonance.

9. This is HVZ reaction. It occurs only in those carboxylic acids which

have -hydrogen atoms. Acetic acid posseses three -hydrogen atoms

but formic acid does not have even a single -hydrogen atom. Thus,

formic acid does not undergo this reaction.

10. Because water form can again hydrolyse the ester into acid and

alcohol.

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11. (a) H

O

OCH3

3 2

3

(i) CH CH MgBr / ether

(ii) H O

H5C2 C2H5

OH

(b) H5C6 H

O

CH3

H

O

dil. NaOH

6 5 2C H CH CH CH CHO

12. In Benzamide carbonyl group is in resonance with lone pair of N so

positive charge decreases therefore nucleophilic attack inhibit.

13. Lone pair on nitrogen atom adjacent to the carbonyl group will

participate in the resonance with the carbonyl group so this lone pair is

not available for the attack at carbonyl group.

14. In formate ion resonance gives rise to identical bond lengths

H—C—O– H—C = O

O O–

H—C O

O

Whereas no such resonance is noticed in formic acid (H — COOH) and

thus C—O bonds are different in HCOOH.

15. It is because of the possibility of resonance which compensates the

electron deficiency of carbonyl carbon to some extent for example.

R C

O

NH2 R C

O

NH2

or

R C

O

OH R C

O

OH

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Level – I

1. (a) CH3

O

H2dil. NaOH

(aldol) CH3

OH

O

H

4LiAlH

CH3 OH

OH

(b) CH3

O

H2dil. NaOH

(aldol) CH3

OH

O

H CH3 H

O

Tollen's reagent

CH3 OH

O

2. - Hydroxy acids are not capable of forming lactones because of

presence of strain in three – membered ring lactone. But they undergo

intramolecular self-esterification to yield a cyclic product containing

two carbonyl groups and is called a lactide.

CHOH

COOH

CH3 COOH

CHCH 3HO22H O

O

O

CH3

O

O

CH3

3. Ka1 of maleic acid is more than that of fumaric acid. This is explained

by the fact that conjugate base of maleic acid is stabilized by hydrogen

bonding

C C

CC

OO

O OH

H H

But Ka2 of maleic acid is lesser than that of fumaric acid because it is

difficult to remove hydrogen bonded H in the conjugate base of maleic

acid.

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4. This can be explained by the fact that amines are more basic than

amides. The less basicity of amides are due to cross conjugation.

5. (i) Cl

3 2sunlightCH COOH CH ClCOOH

(ii) 3 2 3 4 3 2HCOOH Ag(NH ) NO Ag 2NH NO CO

6.

3 2CH CH MgBr

O

H2C CH23H O

3 2 2 2 3 2 2 2CH CH CH CH OMgBr CH CH CH CH OH(A) (B)

4KMnO

3 2 2CH CH CH COOH(C)

7.

4 8 3(C H O )

CH3

OH

O

OH

(A)

CH3

O O

OH

(B)

warm

CH3

O CH3

2CO3CrO

8. (a) CH3 CH = CH CH3 (A) (decarboxylation)

(b)

O

CH3

CH3

O

O

(B)

(cyclic anhydride formation)

9. 1. The cyclic anhydride of malonic acid would have a strained four –

membered ring.

2. The heating causes decarboxylation instead.

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HOC

O

CH2C

O

OHΔ

HO CH3C

O

CO 2

10.

A = CH3 C

O

H B = CH3 C

O

OH C = CH3 C

O

Cl

C

O

CH3

D =

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Level – II

1. A = COOHB =

BrC =

2. High dilute conditions favour intramolecular condensation over inter –

molecular process.

EtO

2 2 2 2 2 2 2 24EtO C CH CO Et EtO CCHCH CH CH CO Et

CO2Et

OEt

O

CO2Et

O

3.

A = C C

CH3

H COOH

CH3

(or) C C

CH3

H CH3

COOH

B = (or)H5C2 C COOH

CH3

H

HOOC C C2H5

CH3

H

4. ||

3 2

O

A. CH CH COH

||

3 2

O

B. CH CH C Cl

||

3 2

O

C. CH CH C O Na

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3 2 2 3|| ||

D. CH CH C O C CH CH

O O

5. (a) CH3

CH3

O

CH3

COOH

(b)

CCH3

O

COOH

(c) O2N COOH

6.

CH3Cl

AlCl 3 anhy

CH3

KMnO 4

OH, heat

COOH

CH3Cl

Anhy AlCl 3

COOH

CH3

SOCl 2

C

CH3

O

Cl

(C2H5)2NH

CH3

C N(C2H5)2

O

7. The neutralization equivalent of A can be determined as follows

= 154.5

59 gm11.9 0.22

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The fact that A on gentle heating is giving another carboxylic acid

suggests that A is a dicarboxylic acid and that too it must be malonic

acid or its derivative.

OH C

O

C C

R'

R O

OH Δ H C C

R'

R O

OH CO2

Acid (B) has equivalent weight 74. So its molecular weight is 74. It can

be deduced that it must be propanoic acid. So

OH C

O

CH C

CH3 O

OH Δ CH3 CH2 COOH CO2

(A) (B)

8. ||

2 5 3 2

O

A. C H CN CH 3 2B. CH NH

2 5C. C H COONa 3 32

D. CH NCOCH

3 2E. CH NNO

9. ||

3 2 2 2 3

O

A. CH C O CH CH CH CH 3 2B. CH CH OH

3 2 2 2C. CH CH CH CH OH

3D. CH CHO

||

3

O

E. CH CH CH C H 3F. CH COOH

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10.

O

CO 2Et

O2C 1. t - BuO

O

CO2Et

O2C

CO 2Et

O2C O

CO 2Et

O

O2C

CO 2Et

O

O2C

Et Et

EtEt Et

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Objective:

LEVEL - I

1. A 2. C 3. C

4. C 5. B 6. A

7. A 8. B 9. A

10. D 11. D 12. A

13. C 14. A 15. B

16. C 17. C 18. D

19. D 20. C

LEVEL – II

1. B 2. B 3. A

4. C 5. B 6. A

7. A 8. B 9. D

10. B 11. B 12. B

13. A 14. B 15. D

16. C 17. A 18. D

19. C 20. A

carboxylic acid and its derivatives_chemistry

Documents

oxygen bond

mg ether

chloro acetic

ch 2cl nacn

aliphatic

acid strength

acetyl chloride

butyl benzoic