NCERT Solutions for Class 12 Chemistry Chemical Kinetics [Old Bok]

NCERT
SOLUTIONS
CLASS - 12th

aglase .co

Class : 12th
Subject : Chemistry
Chapter : 13
Chapter Name : Chemical Kinetics

Q13.1 Classify the following amines as primary, secondary or tertiary.

Answer. Primary: (i) and (iii)
Secondary: (iv)
Tertiary: (ii)

Page : 392 , Block Name : Intext Questions

Q13.2 (i) Write structures of different isomeric amines corresponding to the molecular formula, C H
4 11 N

(ii) Write IUPAC names of all the isomers.
(iii) What type of isomerism is exhibited by different pairs of amines?

Answer.

(iii) The pairs (a) and (b) and (e) and (g) exhibit position isomerism.
The pairs (a) and (c); (a) and (d); (b) and (c); (b) and (d) exhibit chain isomerism.
The pairs (e) and (f) and (f) and (g) exhibit metamerism. All primary amines exhibit functional isomerism with secondary
and tertiary' amines and vice-versa.

Page : 392 , Block Name : Intext Questions

Q13.3 How will you convert
(i) Benzene into aniline
(ii) Benzene into N, N-dimethylaniline
(iii) Cl–(CH2 ) 4 –Cl into hexan-1,6-diamine?

Answer.

Page : 395 , Block Name : Intext Questions

Q13.4 Arrange the following in increasing order of their basic strength:
(i) C H NH , C H NH , NH , C H CH NH and (C H ) 2NH
2 5 2 6 5 2 3 6 5 2 2 2 5 2

(ii) C H NH , (C H ) NH, (C H ) N, C H NH
2 5 2 2 5 2 2 s 3 6 5 2

(iii) CH NH , (CH ) NH, (CH ) N, C H NH , C H CH NH
3 2 3 2 3 3 6 5 2 6 5 2 2

Answer. (i) Considering the inductive effect of alkyl groups,NH C H N and C H N can be arranged in the increasing 3 2 5 2 2 5 2

order of their basic strengths as:
NH3 < C2 H5 NH2 < (C2 H5 ) NH
2

Again, C6 H5 NH2 has proton acceptability less than NH3 . Thus, we have: Due to the —I effect of C H group, the 6 5

C6 H5 NH2 < NH3 < C2 H5 NH2 < (C2 Hs ) NH
2

electron density on the N-atom in C H CH NH is lower than that on the N-atom in C H N . but more than that in
6 5 2 2 2 5 2

NH . Therefore, the given compounds can be arranged in the order of their basic strengths as:
3

C6 H5 NH2 < NH3 < C6 H5 CH2 NH2 < C2 H5 NH2 < (C2 H5 ) NH
2

(ii) Considering the inductive effect and the steric hindrance of the alkyl groups, C H NH , C H N . and their basic 2 5 2 2 5 2

strengths as follows:
C2 H5 NH2 < (C2 H5 ) N < (C2 H5 ) NH
3 2

Again, due to the —R effect of C H group, the electron density on the N atom in C H
6 5 6 5

NH is lower than that on the N atom in C H NH Therefore, the basicity of C H NH is lower than that of C H NH .
2 2 5 2 2 5 2 2 5 2

Hence, the given compounds can be arranged in the increasing order of their basic strengths as follows:
C6 H5 NH2 < C2 H5 NH2 < (C2 Hs ) N < (C2 H5 ) NH
3 2

(iii) Considering the inductive effect and the steric hindrance of alkyl groups, CH CH CH CH can be arranged in the 3 2 3 2

increasing order of their basic strengths as:
(CH3 ) N < CH3 NH2 < (CH3 ) NH
3 2

In C NH N is directly attached to the benzene ring. Thus, the lone pair of electrons on the N—atom is delocalized over
6 2

the benzene ring. In C H CH NH , N is not directly attached to the benzene ring. Thus, its lone pair is not delocalized
6 5 2 2

over the benzene ring. Therefore, the electrons on the N atom are more easily available for protonation in

C6 H5 CH2 NH2 than in C H CH NH i.e., C H CH NH is more basic than C H CH NH
6 5 2 2 6 5 2 2 6 5 2 2

Again, due to the —I effect of C€Hs group, the electron density on the N—atom in C CH NH is lower than that on the N
6 2 2

—atom in (CH ) N . Therefore, (CH ) N is more basic than CH NH . Thus, the given compounds can be arranged in
3 3 3 3 2 2

the increasing order of their basic strengths as follows.
C6 H5 NH2 < C6 H5 CH2 NH2 < (CH3 ) N < CH3 NH2 < (CH3 ) NH
3 2

Page : 395 , Block Name : Intext Questions

Q13.5 Complete the following acid-base reactions and name the products:
(i) CH3 CH2 CH2 NH2 + HCl →

(ii) (C2 H5 ) N + HCl →
3

Answer. (i)
CH3 CH2 CH2 NH2 CH3 CH2 CH2 NH3 Cl
+ HCl ⟶
n − Propylamine n-Propylammoniumchloride

(C2 H5 ) N+ (C2 H3 ) NH3 Cl
3 3
HCl ⟶
(ii) Triethylamine Triemethylammoniumchloride

Page : 395 , Block Name : Intext Questions

Q13.6 Write reactions of the nal alkylation product of aniline with excess of methyl iodide in the presence of sodium
carbonate solution.

Answer. Aniline reacts with methyl iodide to produce N, N-dimethylaniline.

With excess methyl iodide, in the presence of Na CO solution, N, N-dimethylaniline produces N, N, N—
2 3

trimethylanilinium carbonate.

Page : 395 , Block Name : Intext Questions

Q13.7 Write chemical reaction of aniline with benzoyl chloride and write the name of the product obtained.

Answer.

Page : 395 , Block Name : Intext Questions

Q13.8 Write structures of different isomers corresponding to the molecular formula, C H N Write IUPAC names of the
3 9

isomers which will liberate nitrogen gas on treatment with nitrous acid.

Answer. The structures of different isomers corresponding to the molecular formula, C H N are given below:
3 9

(a)
CH3 − CH2 − CH2 − NH2
∘
Propan-1-amine (1 )

(b)

NH 2

CH3 − CH − CH3

0
Propan-2-amine (1 )

CH3 − NH − C2 H5
(c) 0
N -Methylethanamine (2 )

(d)
CH3

CH3 − N − CH3

0
N, N -Dimethylmethanamine (3 )

1' amines, (a) propan-1-amine, and (b) Propan-2-amine will liberate nitrogen gas or

treatment with nitrous acid.

Page : 395 , Block Name : Intext Questions

Q13.9 Convert
(i) 3-Methylaniline into 3-nitrotoluene.
(ii) Aniline into 1,3,5 - tribromobenzene.

Answer.

Page : 407 , Block Name : Intext Questions

Q13.1 Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.
(i) (CH)2CHNH 2

(ii) CH (CH ) NH
3 2 2 2

(iii) CH NHCH(CH )
3 3 2

(iv) (CH ) CNH
3 3

(v) C H N H CH
6 5

(vi) (CH CH ) 2NCH
3 2 3

(vii) m − BrC H NH 6 4 2

Answer. (i)
∘
1 -Methylethanamine (1 amine )

(ii)Propan-l-amine (1 0
amine )

(iii)N − Methyl-2-methylethanamine (2 0
amine )

(iv)2 - Methylpropan-2-amine (1
0
amine )

(v)N -Methylbenzamine or N -methylaniline (2 0
amine )

(vi) N-Ethyl-N-methylethanamin (3 0
amine )

(vii) 3 -Bromobenzenamine or 3 -bromoaniline (1 0
amine )

Page : 408 , Block Name : Exercise

Q13.2 Give one chemical test to distinguish between the following pairs of compounds.
(i) Methylamine and dimethylamine
(ii) Secondary and tertiary amines
(iii) Ethylamine and aniline
(iv) Aniline and benzylamine
(v) Aniline and N-methylaniline

Answer. (i) Methylamine and dimethylamine can be distinguished by the carbylamine test. Carbylamine test: Aliphatic and
aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form foul-smelling isocyanides
or carbylamines. Methylamine (being an aliphatic primary amine) gives a positive carbylamine test, but dimethylamine
does not.
CH3 − NH2 + CHCl3 + 3KOH ⟶ CH3 − NC + 3KCl + 3H2
∘
Methylamine (1 ) Methylisocyanide

(foulsmell)

(ii) Secondary and tertiary amines can be distinguished by allowing them to react with Hinsberg's reagent
(benzenesulphonyl chloride,. Secondary amines react with Hinsberg's reagent to form a product that is insoluble in an
alkali. For example, N, N—diethylamine reacts with Hinsberg's reagent to form N,N—diethylbenzenesulphonamide, which
is insoluble in an alkali. Tertiary amines, however, do not react with Hinsberg's reagent.

(iii) Ethylamine and aniline can be distinguished using the azo-dye test. A dye is obtained when aromatic amines react
with (HNO ) at 0-50C, followed by a reaction with the alkaline solution of 2-naphthol. The dye is usually yellow, red, or
2

orange in colour. Aliphatic amines give a brisk effervescence due (to the evolution of N gas) under similar conditions.
2

(iv) Aniline and benzylamine can be distinguished by their reactions with the help of nitrous acid, which is prepared in situ
from a mineral acid and sodium nitrite. Benzylamine reacts with nitrous acid to form unstable diazonium salt, which in
turn gives alcohol with the evolution of nitrogen gas.

On the other hand, aniline reacts with (HNO ) at a low temperature to form stable diazonium salt. Thus, nitrogen gas is
2

not evolved.

(v) Aniline and N-methylaniline can be distinguished using the Carbylamine test. Primary amines, on heating with
chloroform and ethanolic potassium hydroxide, form foul- smelling isocyanides or carbylamines. Aniline, being an
aromatic primary amine, gives positive carbylamine test. However, N-methylaniline, being a secondary amine does not.

Page : 408 , Block Name : Exercise

Q13.3 Account for the following:
(i) pKb of aniline is more than that of methylamine.
(ii) Ethylamine is soluble in water whereas aniline is not.
(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
(iv) Although amino group is o– and p– directing in aromatic electrophilic substitution reactions, aniline on nitration
gives a substantial amount of m-nitroaniline.
(v) Aniline does not undergo Friedel-Crafts reaction.
(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
(vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines

Answer. (i) pKb of aniline is more than that of methylamine:

Aniline undergoes resonance and as a result, the electrons on the N-atom are delocalized over the benzene ring.
Therefore, the electrons on the N-atom are less available to donate.

(ii) Ethylamine is soluble in water whereas aniline is not: Ethylamine when added to water forms intermolecular H—bonds
with water. Hence, it is soluble in water.

(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide:
CH3 ⟶ NH2 H − OH

Methylamine

Due to the +1 effect of CH group, methylamine is more basic than water. Therefore, in water, methylamine produces OH-
3

ions by accepting ions from water.
−
CH3 − NH2 + H − OH ⟶ CH3 − NH3 + OH

Ferric chloride dissociates in water to form Fe3+ and Cl- ions.
3+ −
FeCl3 ⟶ Fe + 3Cl

Then, OH- ion reacts with Fe3+ ion to form a precipitate of hydrated ferric oxide.
3+ −
2Fe + 6OH ⟶ Fe2 O3 ⋅ 3H2 O

Hydrated .
ferric oxide

(iv) Although amino group is o,p— directing in aromatic electrophilic substitution reactions, aniline on nitration gives a
substantial amount of m- nitroaniline:
Nitration is carried out in an acidic medium. In an acidic medium, aniline is protonated to give anilinium ion (which is
meta-directing).

(v) Aniline does not undergo FriedeI-Crafts reaction:
A Friedel-Crafts reaction is carried out in the presence of AlCl . But AlCl is acidic in nature, while aniline is a strong
3 3

base. Thus, aniline reacts with AlCl to form a salt (as shown in the following equation).
3

Due to the positive charge on the N-atom, electrophilic substitution in the benzene ring is deactivated. Hence, aniline
does not undergo the Friedel-Crafts reaction.

(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines:
The diazonium ion undergoes resonance as shown below:

This resonance accounts for the stability of the diazonium ion. Hence, diazonium salts of aromatic amines are more stable
than those of aliphatic amines.
(vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines:
Gabriel phthalimide synthesis results in the formation of 10 amine only. 20 or 30 amines
are not formed in this synthesis. Thus, a pure 10 amine can be obtained. Therefore, Gabriel phthalimide synthesis is
preferred for synthesizing primary amines.

Page : 408 , Block Name : Exercise

Q13.4 Arrange the following:
(i) In decreasing order of the pKb values:
C2 H5 NH2 , C6 H5 NHCH3 , (C2 H5 ) NH and C6 H5 NH2
2

(ii) In increasing order of basic strength:
C6 H5 NH2 , C6 H5 N(CH3 ) , (C2 H5 ) NH and CH3 NH2
2 2

(iii) In increasing order of basic strength:
(a) Aniline, p-nitroaniline and p-toluidine
(b)C H NH , C H NHCH , C H CH NH
6 5 2 6 5 3 6 5 2 2

(iv) In decreasing order of basic strength in gas phase:
C2 H5 NH2 , (C2 H5 ) NH, (C2 H5 ) N and NH3
2 3

(v) In increasing order of boiling point:
C2 H5 OH, (CH3 ) NH, C2 H5 NH2
2

(vi) In increasing order of solubility in water:
C6 H5 NH2 , (C2 H5 ) NH, C2 H5 NH2
2

Answer.
(i) In C2 H5 NH2 , only one − C2 H5 group is present while in (C2 H5 ) NH, two − C2 H5 groups
2

are present. Thus, the + I effect is more in (C2 H5 ) NH than in C2 H5 NH2 . Therefore, the
2

electron density over the N-atom is more in (C2 H5 ) NH than in C2 H5 NH2 . Hence
2

(C2 H5 ) NH is more basic than C2 H5 NH2 .
2

Also, both C6 H5 NHCH3 and C6 H5 NH2 are less basic than (C2 H5 ) NH and C2 H5 NH2 due to
2

the delocalization of the lone pair in the former two. Further, among C6 H5 NHCH3 and

C6 H5 NH2 , the former will be more basic due to the + T effect of − CH3 group. Hence, the

order of increasing basicity of the given compounds is as follows:

C6 H5 NH2 < C6 H5 NHCH3 < C2 H5 NH2 < (C2 H5 ) NH
2

(ii) C6 H5 N(CH3 ) is more basic than C6 H5 NH2 due to the presence of the + I effect of two
2

−CH3 groups in C6 H5 N(CH3 ) . Further, CH3 NH2 contains one − CH3 group while (C2 H5 ) NH
2 2

contains two − C2 H s groups. Thus, (C2 H5 ) NH is more basic than C2 H5 NH2 .
2

Now, C6 H5 N(CH3 ) is less basic than CH3 NH2 because of the − R effect of − C6 H5 group
2

Hence, the increasing order of the basic strengths of the given compounds is as follows:

C6 H5 NH2 < C6 H5 N(CH3 ) < CH3 NH2 < (C2 H5 ) NH
2 2

(iii) (a)

In p-toluidine, the presence of electron-donating − CH3 group increases the electron

density on the N-atom.

Thus, p-toluidine is more basic than aniline.

On the other hand, the presence of electron-withdrawing

−NO2 group decreases the electron density over the N -atom in p -nitroaniline. Thus, p−

nitroaniline is less basic than aniline.

Hence, the increasing order of the basic strengths of the given compounds is as follows.

p-Nitroaniline < Aniline < p-Toluidine

C6 H5 NHCH3 is more basic than C6 H5 NH2 due to the presence of electron-donating
(b) Again, in C H NHCH —
6 5 3
−CH3 group in C6 H5 NCH3 .

group is directly attached to the N-atom. However, it is not
so in C H NHCH . Thus, in C H NHCH the —R effect of group decreases the electron density over the N-atom.
6 5 3 6 5 3

Therefore, C H NHCH is more basic than C H NHCH .
6 5 3 6 5 3

Hence, the increasing order of the basic strengths of the given compounds is as follows:
C6 H5 NH2 < C6 H5 NHCH3 < C6 H5 CH2 NH2

(iv) In the gas phase, there is no solvation effect. As a result, the basic strength mainly depends upon the +1 effect. The
higher the +1 effect, the stronger is the base. Also, the greater the number of alkyl groups, the higher is the +1 effect.
Therefore, the given compounds can be arranged in the decreasing order of their basic strengths in the gas phase as
follows:
(C H ) N > (C H ) NH > C H NH > NH .
2 5 2 s 2 5 2 3
3 2

(v) The boiling points of compounds depend on the extent of H-bonding present in that compound. The more extensive
the H-bonding in the compound, the higher is the boiling point.
(CH ) NH contains only one H—atom whereas (CH ) NH contains two H-atoms. Then, (CH ) NH undergoes more
3 3 3
2 2 2

extensive H-bonding than (CH ) NH . Hence, the boiling point of (CH ) NH is higher than that of (CH ) NH.
3 2 3 2 3 2

Further, O is more electronegative than N. Thus, C H OH forms stronger H—bonds than C H NH As a result, the 2 5 2 5 2

boiling point of C H OH is higher than that of C H NH and (CH ) NH Now, the given compounds can be arranged in
2 5 2 5 2 3 2

the increasing order of their boiling points as follows
(CH3 ) NH < C2 H5 NH2 < C2 H5 OH
2

(vi) The more extensive the H—bonding, the higher is the solubility. C H NH contains two H-atoms whereas C H NH 2 5 2 2 5 2

contains only one H-atom. Thus, C H NH undergoes more extensive H—bonding than (C H ) N H . Hence, the
2 5 2 2 5 2

solubility in water of C H NH is more than that of (C H ) N H
2 5 2 2 5 2

Further, the solubility of amines decreases with increase in the molecular mass. This is
because the molecular mass of amines increases with an increase in the size of the
hydrophobic part. The molecular mass of ( \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\) is greater than that of (
\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\) and (C H ) N H . Hence, the increasing order of their solubility in 2 5 2

water is as follows:
C6 H5 NH2 < (C2 H5 ) NH < C2 H5 NH2
2

Page : 409 , Block Name : Exercise

Q13.5 How will you convert:
(i) Ethanoic acid into methanamine
(ii) Hexanenitrile into 1-aminopentane

(iii) Methanol to ethanoic acid
(iv) Ethanamine into methanamine
(v) Ethanoic acid into propanoic acid
(vi) Methanamine into ethanamine
(vii) Nitromethane into dimethylamine
(viii) Propanoic acid into ethanoic acid?

Answer.

Page : 409 , Block Name : Exercise

Q13.6 Describe a method for the identi cation of primary, secondary and tertiary amines. Also write chemical equations of
the reactions involved.

Answer. Primary, secondary and tertiary amines can be identi ed and distinguished by Hinsberg's test. In this test, the
amines are allowed to react with Hinsberg's reagent,
benzenesulphonyl chloride (C H SO Cl). The three types of amines react differently with Hinsberg's reagent. Therefore,
6 5 2

they can be easily identi ed using Hinsberg's reagent. Primary amines react with benzenesulphonyl chloride to form N-
alkylbenzenesulphonyl amide which is soluble in alkali.

Due to the presence of a strong electron-withdrawing sulphonyl group in the sulphonamide, the H—atom attached to
nitrogen can be easily released as proton. So, it is acidic and dissolves in alkali.
Secondary amines react with Hinsberg's reagent to give a sulphonamide which is insoluble in alkali.

There is no H—atom attached to the N-atom in the sulphonamide. Therefore, it is not acidic and insoluble in alkali. On the
other hand, tertiary amines do not react with Hinsberg's reagent at all.

Page : 409 , Block Name : Exercise

Q13.7 Write short notes on the following:
(i) Carbylamine reaction
(ii) Diazotisation
(iii) Hofmann’s bromamide reaction
(iv) Coupling reaction
(v) Ammonolysis
(vi) Acetylation
(vii) Gabriel phthalimide synthesis.

Answer.(i) Carbylamine reaction is used as a test for the identi cation of primary amines. When aliphatic and aromatic
primary amines are heated with chloroform and ethanolic potassium hydroxide, carbylamines (or isocyanides) are formed.
These carbylamines have very unpleasant odours. Secondary and tertiary amines do not respond to this test.

(ii) Diazotisation
Aromatic primary amines react with nitrous acid (prepared in situ from NaNO and a mineral acid such as HCI) at low
2

temperatures (273-278 K) to form diazonium salts. This conversion of aromatic primary amines into diazonium salts is
known as diazotization. For example, on treatment with NaNO and HCI at 273—278 K, aniline produces
2

benzenediazonium chloride, with NaCl and H O as by-products.
2

(iii) Hoffmann bromamide reaction
When an amide is treated with bromine in an aqueous or ethanolic solution of sodium hydroxide, a primary amine with
one carbon atom less than the original amide is produced. This degradation reaction is known as Hoffmann bromamide
reaction. This reaction involves the migration of an alkyl or aryl group from the carbonyl carbon atom of the amide to the
nitrogen atom.

(iv) Coupling reaction
The reaction of joining two aromatic rings through the is known as coupling reaction. Arenediazonium salts such as
benzene diazonium salts react with phenol or aromatic amines to form coloured azo compounds.

(v) Ammonolysis
When an alkyl or benzyl halide is allowed to react with an ethanolic solution of ammonia, it undergoes nucleophilic
substitution reaction in which the halogen atom is replaced by an amino NH group. This process of cleavage of the
+

2

carbon-halogen bond is known as ammonolysis.

When this substituted ammonium salt is treated with a strong base such as sodium hydroxide, amine is obtained.
↔
¯
¯¯¯
R− NH3 X + NaOH ⟶ R − NH2 + H2 O + NaX

Amine

Though primary amine is produced as the major product, this process produces a mixture of primary, secondary and
tertiary amines, and also a quaternary ammonium salt as shown.

(vi) Acetylation
Acetylation (or ethanoylation) is the process of introducing an acetyl group into a molecule.

Aliphatic and aromatic primary and secondary amines undergo acetylation reaction by nucleophilic substitution when

treated with acid chlorides, anhydrides or esters. This reaction involves the replacement of the hydrogen atom of NH or +

2

> NH group by the acetyl group, which in turn leads to the production of amides. To shift the equilibrium to the right hand
side, the HCI formed during the reaction is removed as soon as it is formed. This reaction is carried out in the presence of a
base (such as pyridine) which is Stronger than the amine.

(vii) Gabriel phthalimide synthesis
Gabriel phthalimide synthesis is a very useful method for the preparation of aliphatic primary amines. It involves the
treatment of phthalimide with ethanolic potassium hydroxide to form potassium salt of phthalimide. This salt is further
heated with alkyl halide, followed by alkaline hydrolysis to yield the corresponding primary amine.

Page : 409 , Block Name : Exercise

Q13.8 Accomplish the following conversions:
(i) Nitrobenzene to benzoic acid
(ii) Benzene to m-bromophenol
(iii) Benzoic acid to aniline
(iv) Aniline to 2,4,6-tribromo uorobenzene
(v) Benzyl chloride to 2-phenylethanamine
(vi) Chlorobenzene to p-chloroaniline
(vii) Aniline to p-bromoaniline
(viii) Benzamide to toluene
(ix) Aniline to benzyl alcohol.

Answer.

Page : 409 , Block Name : Exercise

Q13.9

Answer.

Page : 409 , Block Name : Exercise

Q13.10 An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating
with Br and KOH forms a compound ‘C’ of molecular formula C H N. Write the structures and IUPAC names of
2 6 7

compounds A, B and C.

Answer. It is given that compound 'C' having the molecular formula, C H N is formed by heating compound 'B' with Bra
6 7

and KOH. This is a Hoffmann bromamide degradation reaction. Therefore, compound 'B' is an amide and compound 'C' is
an amine. The only amine having the molecular formula, C H N is aniline, C H NH .
6 7 6 5 2

Page : 410 , Block Name : Exercise

Q13.11

Answer.

Page : 410 , Block Name : Exercise

Q13.12 Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?

Answer. Gabriel phthalimide synthesis is used for the preparation of aliphatic primary amines. It involves nucleophilic
substitution of alkyl halides by the anion formed by the phthalimide.

Page : 410 , Block Name : Exercise

Q13.13 Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.

Answer. (i) Aromatic amines react with nitrous acid (prepared in situ from NaNO and a mineral acid such as HCI ) at 273
2

— 278 K to form stable aromatic diazonium salts i.e., NaCl and H O.
2

(ii) Aliphatic primary amines react with nitrous acid (prepared in situ from NaNO and a mineral acid such as HCI) to
2

form unstable aliphatic diazonium salts, which further produce alcohol and HCI with the evolution of Nz gas.

Page : 410 , Block Name : Exercise

Q13.14 Give plausible explanation for each of the following:
(i) Why are amines less acidic than alcohols of comparable molecular masses?
(ii) Why do primary amines have higher boiling point than tertiary amines?

(iii) Why are aliphatic amines stronger bases than aromatic amines?

Answer.

In an amide ion, the negative charge is on the N-atom whereas in alkoxide ion, the negative charge is on the O-atom.
Since O is more electronegative than N, O can accommodate the negative charge more easily than N. As a result, the amide
ion is less stable than the alkoxide ion. Hence, amines are less acidic than alcohols of comparable molecular masses.

(ii) In a molecule of tertiary amine, there are no H—atoms whereas in primary amines, two hydrogen atoms are present.
Due to the presence of H—atoms, primary amines undergo extensive intermolecular H—bonding.

(iii) Due to the —R effect of the benzene ring, the electrons on the N- atom are less available in case of aromatic amines.
Therefore, the electrons on the N-atom in aromatic amines cannot be donated easily. This explains why aliphatic amines
are stronger bases than aromatic amines.

Page : 410 , Block Name : Exercise