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NCERT Solutions for Class 12 Chemistry Chapter 9 Amines

NCERT Solutions for Class 12 Chemistry Chapter 9 Amines

Edited By Shivani Poonia | Updated on Jul 14, 2025 11:50 PM IST | #CBSE Class 12th

Ever wondered what happens when ammonia meets alkyl or aryl groups? Let us find out the answer through amines! Amines are a class of organic compounds which is derived by replacing one or more hydrogen atoms of ammonia with alkyl or aryl groups. The chapter will make students learn about different types, structures, properties and reactions of amines. The topics like Hofmann bromamide degradation, Gabriel phthalimide synthesis and how amines react in various situations are well discussed in this chapter.

This Story also Contains
  1. NCERT Solutions for Class 12 Chemistry Chapter 9 Download PDF
  2. NCERT Solutions Of Class 12 Amines (Intext Questions Ex 9.1 to 9.9)
  3. NCERT Solutions for Class 12 Chemistry Chapter 9 Amines (Exercise Questions)
  4. Class 12 Chemistry NCERT Chapter 9: Higher Order Thinking Skills (HOTS) Questions
  5. Approach to Solve Questions of Chapter 9: Amines
  6. Topics of Chapter 9 In NCERT Book Class 12 Chemistry
  7. What Extra Should Students Study Beyond the NCERT for JEE/NEET?
  8. NCERT Solutions for Class 12 Chemistry
  9. NCERT Solutions for Class 12 Subject-wise
  10. NCERT Exemplar subject-wise
  11. NCERT Books and NCERT Syllabus
NCERT Solutions for Class 12 Chemistry Chapter 9 Amines
NCERT Solutions for Class 12 Chemistry Chapter 9 Amines

The chapter is important for anyone looking to do well in organic chemistry. The NCERT solutions for class 12 chemistry chapter 9 offer clear explanations that will help students understand these topics and improve their problem-solving skills. These NCERT solutions will help you prepare for boards as well as for competitive exams. We have also included higher-order thinking skills (HOTS) questions to check your critical thinking and deepen your understanding of the topics.

Also Read

NCERT Solutions for Class 12 Chemistry Chapter 9 Download PDF

Students can download the NCERT Solutions for Chapter 9 in PDF format for free. These solutions are designed to help you understand the fundamental concepts and solve textbook questions with ease.

Download PDF

NCERT Solutions Of Class 12 Amines (Intext Questions Ex 9.1 to 9.9)

The detailed solutions to in-text questions are given below

Page No. 262

Question 9.1 Classify the following amines as primary, secondary or tertiary:

(i) 1596565119000

Answer :

1596565129561

The nitrogen atom is directly connected with only one carbon atom, so it is a primary aromatic amine.

Question 9.1 Classify the following amines as primary, secondary or tertiary:

(ii) 1596565162027

Answer :

1596565172867

In this compound, the nitrogen atom is directly connected to 3 carbon atoms. So, it is a tertiary aromatic amine

Question 9.1 Classify the following amines as primary, secondary or tertiary:

(iii) (C2H5)2CHNH2

Answer :

(C2H5)2CHNH2

1596565186466

Here, the N atom is directly connected to only one carbon atom. So it is a primary aliphatic amine.

Question 9.1 Classify the following amines as primary, secondary or tertiary:

(iv) (C2H5)2NH

Answer :

(C2H5)2NH

1596565203549

In structure, we can see that N is directly connected to 2 carbon atoms. So, it is a secondary amine.

Question 9.2 (i) Write structures of different isomeric amines corresponding to the molecular formula,

C4H11N

Answer :

different isomeric amines corresponding to the molecular formula, C4H11N

(i)CH3CH2CH2CH2NH2(v)CH3CH2CH2NHCH3

(ii)1596565271605 (vi) 1596565310726

(iii)1596565289920 (vii) 1596565320554

(iv)1596565299578 (viii) 1596565330072

Question 9.2 (ii) Write IUPAC names of all the isomers.

Answer :

IUPAC name of all the isomers-

CH3CH2CH2CH2NH2
Butanamine
Butan-2-amine
2-Methylpropanamine
N-Methylpropan-2-amine
1596566740951
2-Methylpropan-2-amine
N,N-Dimethylethanamine
CH3CH2CH2NHCH3
N-Methylpropanamine
1596565424603
N-Ethylethanamine

Question 9.2 (iii) What type of isomerism is exhibited by different pairs of amines?

Answer :

CH3CH2CH2CH2NH2
Butanamine
(Chain isomerism + position isomerism)
1596565439230
Butan-2-amine
(chain isomerism + position isomerism)
1596565448437
2-Methylpropanamine
(chain isomerism)
1596565458211
N-Methylpropan-2-amine
(position isomerism + metamerism)
1596565467597
2-Methylpropan-2-amine
(chain isomerism)
1596565490077
N,N-Dimethylethanamine
CH3CH2CH2CH2NHCH3
N-Methylpropanamine
(Position isomerism)
1596565505529
N-Ethylethanamine
(no isomerism)

Question 9.3 (i) How will you convert

Benzene into aniline

Answer :

1596565516861

Nitration of benzene gives nitrobenzene. And after that, reduce the nitro group by catalytic hydrogenation.

Question 9.3 (ii) How will you convert

Benzene into N, N-dimethylaniline

Answer :

1596565527611

Nitration of benzene gives nitrobenzene, and after catalytic hydrogenation of nitrobenzene, it gives aniline. Aniline reacts with two moles of chloromethane to form N, N-dimethylaniline.

Question 9.3(iii) How will you convert

Cl(CH2)4Cl into hexan-1,6-diamine

Answer :

On reacting the given reactant with ethanolic sodium cyanide, the CN molecules replace both chlorine atoms. And after that catalytic hydrogenation, we get our desired product. (reduce the CN to CH2NH2 in both sides)

Question 9.4 (i) Arrange the following in increasing order of their basic strength:

C2H5NH2, C6H5NH2, NH3, C6H5CH2NH2 and (C2H5)2NH

Answer :

Considering the inductive effect, solvation effect and steric hindrance of the alkyl group, which decides the basic strength of alkylamines. The order of basic strength in ethyl-substituted amine is-

NH3<C2H5NH2<(C2H5)2NH

and order in the benzene-substituted ring-

C6H5NH2<C6H5CH2NH2

Due to the -R effect of benzene C6H5NH2 has less electron density than ammonia. So, the final order is -

C6H5NH2<NH3<C6H5CH2NH2<C2H5NH2<(C2H5)2NH

Question 9.4 (ii) Arrange the following in increasing order of their basic strength:

C2H5NH2, (C2H5)2NH, (C2H5)3N, C6H5NH2

Answer :

C6H5NH2<C2H5NH2<(C2H5)3N<(C2H5)2NH

On considering the inductive effect, solvation effect and steric hindrance of the alkyl group, the increasing order of basicity in ethyl as a substituted group is shown above.

Question 9.4 (iii) Arrange the following in increasing order of their basic strength:

(iii) CH3NH2, (CH3)2NH, (CH3)3N, C6H5NH2, C6H5CH2NH2.

Answer :

On considering the inductive effect, solvation effect and steric hindrance of the alkyl group, which decides the basic strength of alkyl amines in the aqueous state. The order of basic strength in the case of methyl-substituted amines is -

C6H5NH2<C6H5CH2NH2<(CH3)3N<CH3NH2<(CH3)2NH

Question 9.5 (i) Complete the following acid-base reactions and name the products:

CH3CH2CH2NH2+HCl

Answer :

The above-given reaction is an acid-base reaction. So salt is formed.

CH3CH2CH2NH2+HCl CH3(CH2)2NH3+Cl

1596565551481

Question 9.5 (ii) Complete the following acid-base reactions and name the products:

(C2H5)3N+HCl

Answer :

(C2H5)3N+HCl (C2H5)3N+HCl

It is an acid-base reaction, so the N will accept H from HCl and form a salt.

1596565560887

Question 9.6 Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution.

Answer :

The methyl iodide reacts with aniline to give N, N-dimethylaniline.

1596565570028

With the excess of methyl iodide in the presence of sodium carbonate solution (Na2CO3), N,N-dimethylaniline produces N,N,N-trimethyl anilinium carbonate.
1596565579864

Question 9.7 Write chemical reaction of aniline with benzoyl chloride and write the name of the product obtained.

Answer :

When aniline reacts with benzoyl chloride, HCl is produced as a by-product and N-phenylbenzamide is produced as a major product. The lone pair of the nitrogen atom attacks the acidic carbon of benzoyl chloride.

1596565588671

Question 9.8 Write structures of different isomers corresponding to the molecular formula, C3H9N. Write IUPAC names of the isomers which will liberate nitrogen gas on treatment with nitrous acid.

Answer :

Structures of different isomers corresponding to the molecular formula C3H9N and their IUPAC name-

(i)
1596565598336
Propan-1-amine
(ii)
1596565606603
Propan-2-amine
(iii)
1596565615299
N – Methylethanamine
(iv)
1596565622752
N,N-Dimethylmethanamine

The structures (i) and (ii) will liberate nitrogen gas (N2) on treatment with nitrous acid.

Question 9.9 (i) Convert

3-Methylaniline into 3-nitrotoluene

Answer :

1596565637949

On diazotisation, reaction NH2 will convert to N2Cl . And then reacts with fluoroboric acid, followed by NaNO2/Cu/Δ that converts N2Cl into the nitro group (NO2) and evolution of dinitrogen gas.

Question 9.9 (ii) Convert

Aniline into 1,3,5-tribromobenzene.

Answer :

1596565649772

Aniline on bromination in the presence of water gives 2,4,6 tribromoaniline, which on further reacting with nitrous acid converts NH2 into N2Cl. Now, with the help of H3PO2/water, it removes the N2Cl group from the benzene ring and we get our final product 1,3,5-tribromobenzene.

NCERT Solutions for Class 12 Chemistry Chapter 9 Amines (Exercise Questions)

The exercise questions are solved here with detailed explanations.

Question 9.1 Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

(i) (CH3)2CHNH2

Answer :

The IUPAC name of the compound is Propan-2-amine

Since N (nitrogen) is connected to only one carbon. Hence, it is a primary amine.

1596565661754

Question 9.1 (ii) Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

CH3(CH2)2NH2

Answer :

The IUPAC name is Propan-1-amine

It is a primary amine (nitrogen connects with only one carbon atom)

Here is the structure of the compound CH3(CH2)2NH2 -

1596565672365

Question 9.1 (iii) Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

CH3NHCH(CH3)2

Answer :

CH3NHCH(CH3)2

N-Methylpropan-2-amine

Since N atom is connected with two carbon atoms, it is a two-degree amine.

Question 9.1(iv) Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

(CH3)3CNH2

Answer :

(CH3)3CNH2

2-methyl-propan-2-amine

It is a one-degree amine

Question 9.1 (v) Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

C6H5NHCH3

Answer :

C6H5NHCH3

N-Methylbenzamine or N-methylaniline

Here N atom is connected to two C atoms. So, it is a secondary amine.

1596565693795

Question 9.1 (vi) Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

(CH3CH2)2NCH3

Answer:

(CH3CH2)2NCH3

N,N-Diethylmethylamine

1596565707700

N is connected with three carbon atoms so that it is a tertiary amine

Question 9.1 (vii) Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

mBrC6H4NH2

Answer :

mBrC6H4NH2
3-bromoaniline or 3-bromobenzenamine

It is a primary amine

1596565717107

Question 9.2 Give one chemical test to distinguish between the following pairs of compounds.

(i) Methylamine and dimethylamine

Answer :

Carbylamine test

This test is used to distinguish between aliphatic and aromatic primary amines. Heating with chloroform (CHCl3) and potassium hydroxide ( KOH ) gives isocyanides or carbylamines, which have a foul smell. Two-degree and 3-degree amines do not show this reaction.

Here, methylamine is primary and dimethylamine is a secondary amine.

1596565730642

The first one is 1-degree and the second one is 2-degree amine.

Question 9.2 Give one chemical test to distinguish between the following pairs of compounds.

(ii) Secondary and tertiary amines

Answer :

Secondary and tertiary amines can be distinguished by reacting them with Hinsberg's reagent, which is also called benzenesulphonyl chloride. (C6H5SO2Cl) In the case of primary amine, the product is soluble in alkali but not in the secondary amine case. And tertiary amine does not react with this reagent.

In the case of a secondary amine

1596565740393

Tertiary amine + benzenesulphonyl chloride no reaction

Question 9.2 Give one chemical test to distinguish between the following pairs of compounds.

(iii) Ethylamine and aniline

Answer :

Ethylamine and aniline can be differentiated by the azo-dye test.

On reacting aniline with (NaNO2+dil.HCl) followed by 2-naphthol, gives a colored product. But when it is ethylamine, it gives brisk effervescence due to the evolution of N2 gas under the same conditions.

Question 9.2 Give one chemical test to distinguish between the following pairs of compounds.

(iv) Aniline and benzylamine

Answer :

Benzylamine reacts with nitrous acid to form a diazonium salt, which is unstable, and it also gives alcohol with the evolution of nitrogen gas. On the other hand, aniline reacts with nitrous acid to form a stable diazonium salt.

Question 9.2 Give one chemical test to distinguish between the following pairs of compounds.

(v) Aniline and N-methylaniline.

Answer :

Aniline and N-methylaniline can both be distinguished by the carbylamine test. Aniline is the primary aromatic compound so it gives s positive test, but N-methyl aniline is secondary and does not give this test.

Structure of both compounds-

Question 9.3 Account for the following:

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

Answer:

Here is the structure of aniline (left) and methylamine

1596565760265

Due to the resonance in aniline, the electrons of the nitrogen atoms are delocalised over the benzene ring. So, because of that, the electron density at the nitrogen atom decreases and is less available for donation.

1596565770185

On the other hand, in the case of methylamine, the methyl (CH3 ) group is an electron-donating group (+I effect) which increases the electron density at the N atom. Hence pKb value of aniline is higher than that of methylamine.

Question 9.3 Account for the following:

(ii) Ethylamine is soluble in water whereas aniline is not.

Answer:

1596565779599R=C2H5

The extent of intermolecular hydrogen bonding in ethylamine is very high. Hence, it is soluble in water. But aniline does not undergo hydrogen bonding with water to a large extent due to the presence of a bulky hydrophobic group (C6H5). Hence, it is insoluble in water.

Question 9.3 Account for the following:

(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.

Answer :

H3CNH2 OH

Due to the +I effect of the methyl (CH3) group, methylamine is more basic than water. So in water CH3NH2 produces hydroxide ion by accepting H+ ions from water.

CH3NH2+H2OCH3N+H3+OH

Ferric chloride (FeCl3) dissociates in water into- FeCl3Fe3++3Cl

The hydroxide ion (OH) react with Fe3+ and form a precipitate of Fe(OH)3 (reddish brown ppt)

Fe3++3OHFe(OH)3

Question 9.3 Account for the following:

(iv) Although amino group is o– and p– directing in aromatic electrophilic substitution reactions,aniline on nitration gives a substantial amount of m- nitroaniline.

Answer :

Nitration is carried out in a strongly acidic medium, and aniline is protonated to form the anilinium ion, which is meta-directing. Because of this reason, aniline on nitration gives a substantial amount of meta-nitroaniline.

1596565798623 1596565806109

Question 9.3 Account for the following:

(v) Aniline does not undergo Friedel-Crafts reaction.

Answer :

In the Friedel-Crafts reaction, we use AlCl3, which is a Lewis acid, and aniline has basic nature. Thus, there is an acid-base reaction between them and they form a salt.

1596565815066

Due to the positive charge on the N-atom. The electrophilic substitution in the benzene ring is deactivated.

Question 9.3 Account for the following:

(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.

Answer :

In diazonium salt, the structure undergoes resonance due to which the dispersal of positive charge is more and we know that the higher the resonance, the higher is the stability. Therefore, the diazonium salt of aromatic amines is more stable than that of aliphatic amines.

1596565822279

Question 9.3 Account for the following:

(vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines.

Answer :

Gabriel synthesis is used for primary amines. Secondary and tertiary amines are not formed by this method. Therefore, to obtain pure and only 1-degree amine, the Gabriel phthalimide reaction is preferred.

Question 9.4 Arrange the following:

(i) In decreasing order of the pKb values:

C2H5NH2, C6H5NHCH3, (C2H5)2NH and C6H5NH2

Answer :

C2H5NH2 and (C2H5)2NH are aliphatic amines, so they are more basic than the other two compounds. In ethylamine, there is only one electron-donating group (+I effect), but in diethylamine, there are two electron-donating groups, so the electron density is much higher than in ethylamine.

Between C6H5NHCH3 and C6H5NH2, they are weak bases because of delocalisation of lone pair electrons in the benzene ring. Aniline is less basic than N-methylaniline( C6H5NHCH3, ) due to the absence of an electron-donating group. Thus, the decreasing order of pKb values is-

C6H5NH2>C6H5NHCH3>C2H5NH2>(C2H5)2NH

Question 9.4 Arrange the following:

(ii) In increasing order of basic strength:

C6H5NH2, C6H5N(CH3)2, (C2H5)2NH and CH3NH

Answer :

The increasing order of basic strength-

C6H5NH2<C6H5N(C2H5)2<CH3NH2<(C2H5)2NH

Aliphatic amines are more basic than aromatic amines due to the presence of a high electron density of electron at the nitrogen atom so that it can donate an electron. On the other hand, in aromatic amines, the lone pair of electrons is delocalised in the benzene ring, so the availability of electrons is less.

(C2H5)2NH is more basic because of the +I effect of the two ethyl groups (so electron density is highest in this compound), while methylamine has one ethyl group. In aromatic amines, diethylaniline is more basic due to the presence of a greater number of electron-donating groups.

Question 9.4 Arrange the following:

(iii) In increasing order of basic strength:

(a) Aniline, p-nitroaniline and p-toluidine

Answer :

Structure of the following compounds-

1596565835752

Aniline, p-nitroaniline, p-toluidine

In p-toluidine, the methyl group increases the electron density at the N atom (+I effect of methyl), so it is more basic than aniline. In p-nitroaniline, the nitro group is an electron-withdrawing group, so it decreases the electron density at nitrogen, so it is less basic than aniline.

Question 9.4 Arrange the following:

(iii) In increasing order of basic strength:

(b) C6H5NH2, C6H5NHCH3, C6H5CH2NH2.

Answer :

The increasing order of basic strength- C6H5NH2<C6H5NHCH3<C6H5CH2NH2

C6H5NHCH3 is more basic than aniline due to the presence of one electron-donating group, which increases the electron density at nitrogen atom. But in this compound N atom is directly attached with the benzene ring, so due to -R effect C6H5NHCH3 is less basic than C6H5CH2NH2.

Question 9.4 Arrange the following:

(iv) In decreasing order of basic strength in gas phase:

C2H5NH2, (C2H5)2NH, (C2H5)3N, and NH3

Answer :

Decreasing order of basic strength in the gas phase-

(C2H5)3N>(C2H5)2NH>C2H5NH2>NH3

In the gas phase, there is no solvation effect. So, the basicity directly depends on the number of electron-donating groups. More is the electron electron-donating groups, the higher the +I effect and if the size of the donating group is large +I effect should also be high.

Question 9.4 Arrange the following:

(v) In increasing order of boiling point:

C2H5OH, (CH3)2NH, C2H5NH2

Answer :

The boiling point of the compounds depends on the extent of hydrogen bonding of the compound. The more the extent, the higher the boiling point. C2H5NH2, it contains two H atoms, so it has a higher boiling point than (CH3)2NH. On the other hand, the Oxygen atom is more electronegative than the N atom, so the strength of hydrogen bonding is higher in C2H5OH than in C2H5NH2.

Thus, the increasing order of boiling point is-

(C2H5)2NH<C2H5NH2<C2H5OH

Question 9.4 Arrange the following:

(vi) In increasing order of solubility in water:

C6H5NH2, (C2H5)2NH, C2H5NH2.

Answer :

The more extensive the hydrogen bonding, the greater is the solubility in water. C2H5NH2 contains two H atom while (C2H5)2NH contains only one. So, the C2H5NH2 undergoes extensive hydrogen bonding. Hence its solubility is more than (C2H5)2NH in water. An increase in the molecular mass of amines decreases their solubility in water because of the increase in bulky hydrophobic groups.

Thus, the increasing order of solubility in water is-

C6H5NH2<(C2H5)NH<C2H5NH2

Question 9.5 How will you convert:

(i) Ethanoic acid into methanamine

Answer :

Ethanoic acid on reacting with SOCl2 replaces the OH by Cl and then reacts with an excess of ammonia molecule to produce methanamide (CH3CONH2), which, by the Hoffmann bromamide degradation reaction, gives us desired product (methenamine)

CH3COOH+SOCl2CH3COClNH3 (excess) CH3CONH2Br2+KOHCH3NH2

Question 9.5 How will you convert:

(ii) Hexanenitrile into 1-aminopentane

Answer :

First, hexanenitrile undergoes acid hydrolysis, converting the nitrile group (-CN) into a carboxylic acid (COOH). This carboxylic acid is then treated with thionyl chloride (SOCl2) to form the corresponding acid chloride (COCl). Reacting this with excess ammonia (NH3) replaces the chlorine atom with an amine group, yielding a primary amide (CONH2).
Finally, the Hofmann bromamide degradation is carried out, which removes one carbon atom and produces the desired primary amine as the final product.

Question 9.5 How will you convert:

(iii) Methanol to ethanoic acid

Answer :

When methanol reacts with phosphorus pentachloride, OH- is reduced by Cl- to form chloromethane, which on reacting with ethanolic sodium cyanide, gives CH3CN and followed by acidic hydrolysis, it produces ethanoic acid (CH3COOH)

Question 9.5 How will you convert:

(iv) Ethanamine into methanamine

Answer :

Ethanamine reacts with nitrous acid to form an azo compound, which further reacts with water to form ethanol and this on oxidation gives ethanoic acid. After treating with an excess of ammonia, the ethanoic acid becomes ethanamide, which further reacts with bromine and a strong base (Hoffmann bromamide degradation reaction), to form methanamine.

Question 9.5 How will you convert:

(v) Ethanoic acid into propanoic acid

Answer :

Acetic acid, on reduction with lithium aluminium hydride followed by acid hydrolysis, gives ethanol, which on reacting with PCl5 gives ethyl chloride. Ethyl chloride further reacts with ethanolic sodium cyanide to form propanenitrile, which on acidic hydrolysis gives the desired product.

Question 9.5 How will you convert:

(vi) Methanamine into ethanamine

Answer :

Methanamine on diazotisation gives methane diazonium salt, which on further hydrolysis forms methanol. Methanol on reacting with PCl5, followed by ethanolic sodium cyanide (NaCN), produces ethane nitrile. This on reduction with H2/Pd gives ethanamine.

Question 9.5 How will you convert:

(vii) Nitromethane into dimethylamine

Answer :

Conversion of Nitromethane into dimethylamine is shown below:

CH3NO2HSn/HClCH3NH2CHCl3/KOHCH3NC

Reduction of Isocyanide

CH3NCHNa/C2H5OHCH3NHCH3

Question 9.5 How will you convert:

(viii) Propanoic acid into ethanoic acid?

Answer :

Propanoic acid on reacting with an excess of ammonia, gives propanamide, which further reacts with bromine and potassium hydroxide (Hoffmann bromamide degradation reaction) to give ethanamine. On diazotisation, ethanamine is converted into an azo salt, which on treatment with water forms ethanol. Ethanol on oxidation, provides ethanoic acid.

Question 9.6 Describe a method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved.

Answer :

Primary, secondary, and tertiary amines are distinguished by Hinsberg's reagent test. In this test, the amines are allowed to react with benzene sulphonyl chloride (Hinsberg's reagent). All three types of amines give different products.

1.Primary amine reacts with benzene sulphonyl chloride (C6H5SO2Cl) gives N-ethylbenzene sulphonamide which is soluble in alkali.

1596566276020

2. When secondary amines react, it gives N,N-diethylbenzene sulphonamide, which is insoluble in alkali.

1596566321487

3. Tertiary amine does not react with Hinsberg's reagent.

Question 9.7 Write short notes on the following:

(i) Carbylamine reaction

Answer :

Carbylamine reaction-

Aliphatic and aromatic primary amines on reacting with chloroform and ethanolic KOH form isocyanides or carbylamine, which is a foul-smelling substance. Secondary and tertiary amines do not give this reaction. This reaction is known as the carbylamine reaction or isocyanide test. It is used to distinguish primary amines.

RNH2+CHCl3+KOHheatΔRNC+3KCl+3H2O

Question 9.7 Write short notes on the following:

(ii) Diazotisation

Answer :

Aromatic primary amines react with nitrous acid (NaNO2+2HCl) at low temperature to form diazonium salts. This conversion of aromatic primary amines into diazonium salts is known as diazotisation.

C6H5NH2+NaNO2+2HCl273278 KC6H5 N2+Cl+NaCl+2H2O

Question 9.7 Write short notes on the following:

(iii) Hofmann’s bromamide reaction

Answer :

Hofmann’s bromamide reaction-
When an amide is treated with bromine in aqueous solution or an ethanolic solution of sodium hydroxide, it produces primary amines with one less carbon atom than the parent compound. This reaction is known as the Hoffmann bromide degradation reaction.
1596566544782

R = alkyl group like CH3,C2H5....etc

Question 9.7 Write short notes on the following:

(iv) Coupling reaction

Answer :

The reaction of joining diazonium salt to the aromatic ring (like phenol) through a N=N bond is known as a coupling reaction. Benzene diazonium chloride reacts with phenol in which the phenol molecule at its para position is coupled with the diazonium salt (C6H5N2+Cl) to give p-hydroxyazobenzene (orange in colour).

Reactions-

1596566555567

Diazonium salt reacts with aniline to give p-aminoazobenzene (yellow colour)

Question 9.7 Write short notes on the following:

(v) Ammonolysis

Answer :

Ammonolysis-
When an alkyl or benzyl halide is going to react with the solution of ammonia(an ethanolic solution), it undergoes a nucleophilic substitution reaction in which the halogen atom is replaced by an amino (NH2) group. This process of cleavage of the carbon-halogen (C-X) bond by the ammonia molecule is known as ammonolysis.

1596566372072

When this substituted ammonium salt is treated with a strong base, it gives a primary amine
RN+H3X+NaOHRNH2+H2O+NaX

  • Here, primary amine behaves as a nucleophile and can further react with an alkyl halide to form secondary and tertiary amines and finally quaternary ammonium salt.

1596566366031

Question 9.7 Write short notes on the following:

(vi) Acetylation

Answer :

Acetylation-
The introduction of an acetyl group ( CH3CO )in a molecule is known as acetylation.
1596565937421

Example-

1596566385638

Aliphatic and aromatic primary or secondary amines undergo acetylation reaction when treated with acid chlorides, anhydrides, or esters by nucleophilic substitution. Here Hydrogen atom of NH2 or NH is replaced by CH3CO the group.
1596566392759

Question 9.7 Write short notes on the following:

(vii) Gabriel phthalimide synthesis.

Answer :

Gabriel phthalimide synthesis.-
It is mainly used for aliphatic primary amine preparation. Phthalimide reacts with ethanolic potassium hydroxide to produce the potassium salt of phthalimide. Upon heating with an alkyl halide and subsequent alkaline hydrolysis, this salt yields the corresponding primary amine. Aromatic amines cannot be produced by this method due to the inability of aryl halides to undergo nucleophilic substitution reaction with the anion produced by phthalimide.

1596565944101

Question 9.8 Accomplish the following conversions:

(i) Nitrobenzene to benzoic acid

Answer :

First, reduce nitrobenzene into aniline with the help of H2/Pd. Now convert aniline to diazonium salt, followed by Sandmeyer's reaction (CuCN/KCN) to obtain benzonitrile, and after acid hydrolysis, this benzonitrile becomes benzoic acid.

1596565952459

Question 9.8 Accomplish the following conversions:

(ii) Benzene to m-bromophenol

Answer :

First, benzene is nitrated to form nitrobenzene, which is a meta-directing group and thus directs incoming substituents to the meta position. Next, bromination of nitrobenzene is carried out, introducing a bromine atom at the meta position relative to the nitro group.

The nitro group is then reduced to an amine using Sn/HCl that forms m-bromoaniline. This is followed by diazotization (reaction with NaNO2 and HCl at 05C) to form the diazonium salt, which is finally subjected to hydrolysis with warm water that gives m-bromophenol as the final product.
1596565960886

Question 9.8 Accomplish the following conversions:

(iii) Benzoic acid to aniline

Answer :

On reacting benzoic acid with SOCl2, we get benzene sulphonyl chloride and when it reacts with ammonia the chlorine is replaced by NH2. Then the Hoffmann bromide degradation reaction is carried out to remove -CO group.

Question 9.8 Accomplish the following conversions:

(iv) Aniline to 2,4,6-tribromofluorobenzene

Answer :

First, we will do the bromination of aniline which gives 2,4,6-tribromoaniline. After that, we will perform diazotisation reaction which converts NH2 group into N2Cl and then we will react it with fluoroboric acid( HBF4+Δ) to give us the desired product.

1596565973725

Question 9.8 Accomplish the following conversions:

(v) Benzyl chloride to 2-phenylethanamine

Answer :

On reacting benzyl chloride with the ethanolic sodium cyanide (NaCN), it replaces the Chlorine with the CN molecule. And then reduction with the help of H2/Ni

1596565980143

Question 9.8 Accomplish the following conversions:

(vi) Chlorobenzene to pchloroaniline

Answer :

On nitration of chlorobenzene, it gives p-nitrochlorobenzene, which is on further reaction with H2/Pd, reduces NO2 to NH2

Question 9.8 Accomplish the following conversions:

(vii) Aniline to pbromoaniline

Answer :

Aniline on reacting with the acetic anhydride in the presence of pyridine, gives N-phenylethanamide, which on further reaction with Br2/CH3COOH, directs the bromine to the para-position. After hydrolysis, we can recover the original NH2 group at the same position

1596565991838

Question 9.8 Accomplish the following conversions:

(viii) Benzamide to toluene

Answer :

Use the Hoffmann bromamide degradation reaction so that it gives aniline and then do diazotisation reaction so that NH2 group is replaced by N2Cl and then reduce it with the help of (H2O/H3PO2) so that it reduced into the benzene ring. And then alkylation of benzene will give toluene as a final product.

1596566000675

Question 9.8 Accomplish the following conversions:

(ix) Aniline to benzyl alcohol

Answer :

Aniline on treating with nitrous acid followed by sandmeyers reaction gives benzonitrile ( C6H5CN), which on acidic hydrolysis form benzoic acid (C6H5COOH). The benzoic acid is then reduced by LiAlH4 (lithium aluminium hydride) to get the desired product.

1596564302098

Question 9.9 Give the structures of A, B and C in the following reactions:

(i) CH3CH2INaCNAPartialhydrolysisOHBNaOH+Br2C

Answer :

In the first reaction, the iodine is replaced by CN, so the structure of (A) is CH3CH2CN.

On partial hydrolysis, the CN converted in to CONH2, so the structure of B is CH3CH2CONH2 (propanamide). The third reaction is the Hoffmann bromide degradation reaction. So, the -CO group will be removed and it will become a primary aliphatic amine. Thus the structure of C is CH3CH2NH2 (ethanamine).

Question 9.9 Give the structures of A, B and C in the following reactions:

(ii) C6H5N2ClCuCNAH2O/H+BΔNH3C

Answer :

The first part is the Sandmeyers reaction so CN replaces the N2Cl of the reactant and the product (A) is cyanobenzene (C6H5CN). On acidic hydrolysis of cyanobenzene give product (B), benzoic acid (C6H5COOH) and benzoic acid reacts with ammonia (heat) to produce

(C) benzamide (C6H5CONH2).

Question 9.9 Give the structures of A, B and C in the following reactions:

(iii) CH3CH2BrKCNALiAlH4B0CHNO2C

Answer :

The structure of A,B and C is

CH3CH2CN,CH3CH2CH2NH2 and CH3CH2CH2OH, respectively.

In the first reaction, the bromine is replaced by a cyanide molecule (CN) and after a reduction by LiAlH4, the CN group becomes CH2NH2.

Question 9.9 Give the structures of A, B and C in the following reactions:

(iv) C6H5NO2Fe/HClA273KNaNO2+HClBH2O/H+ΔC

Answer :

A- C6H5NH2

B- C6H5N2+Cl

C- C6H5OH

The first nitrobenzene reduced to aniline and then it reacts with (NaNO2/HCl) to form benzene diazonium salt and is hydrolysed to form phenol( C6H5OH).

Question 9.9 Give the structures of A, B and C in the following reactions:

(v) CH3COOHΔNH3ANaOBrBNaNO2/HClC

Answer :

A- ethanamide (CH3CONH2)

B- methenamine (CH3NH2)

C- methanol (CH3OH)

Acetic acid reacts with ammonia, giving ethanamide and the second reaction is the Hoffmann bromamide degradation reaction, so it removes the -CO group and forms methanamine. The reaction with NaNO2/HCl gives methanol.

Question 9.9 Give the structures of A, B and C in the following reactions:

(vi) C6H5NO2Fe/HClA273KHNO2BC6H5OHC

Answer :

Initially, product A is the reduction of nitrobenzene means it is an aniline. When A react with HNO2 it gives benzene diazonium chloride (B). And when B reacts with another aromatic ring (phenol), it's a coupling reaction, so it gives azodye (p-Hydroxyazobenzene)

A= Aniline

1596566051588

B = benzene diazonium chloride

1596566042855

C= p-Hydroxyazobenzene

1596566034767

Question 9.10 An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Br2 and KOH forms a compound ‘C’ of molecular formula C6H7N. . Write the structures and IUPAC names of compounds A, B and C.

Answer :

The molecular formula of C = C6H7N

AaqNH3ΔB

B+Br2+KOHC(C6H7N)

It is a Hoffmann bromamide degradation reaction so the product should be C6H5NH2 (aniline).

So B should be benzamide (C6H5CONH2) and A should be benzoic acid ( C6H5COOH )

Therefore, the above reaction proceeds as

Question 9.11 Complete the following reactions:

(i) C6H5NH2+CHCl3+alc.KOH

Answer :

The reaction given above is carbylamine reaction in which primary aromatic amines react with chloroform and alcoholic potassium hydroxide to give isocyanide or carbylamine as a product, which is a foul-smelling substance.

C6H5NH2+CHCl3+alc.KOH C6H5NC+KCl+H2O

Question 9.11 Complete the following reactions:

(ii) C6H5N2Cl+H3PO2+H2O

Answer :

When benzene diazonium chloride reacts with H3PO2 it reduced into benzene and H3PO2 oxidised and HCl produced as a by-product in this reaction.

C6H5N2Cl+H3PO2+H2O C6H6+H3PO3+HCl

Question 9.11 Complete the following reactions:

(iii) C6H5NH2+H2SO4(conc.)

Answer :

Aniline is a base and H2SO4 is acid. So, basically, it is an acid-base reaction. Aniline accepts H+ ions from sulphuric acid and forms a product anilinium hydrogen sulphate.

C6H5NH2+H2SO4(conc.) C6H5NH3+HSO4

Question 9.11 Complete the following reactions:

(iv) C6H5N2Cl+C2H5OH

Answer :

Ethanol reduces the benzene diazonium chloride to benzene and oxidises into ethanal ( CH3CHO), hydrochloric acid is produced as a by-product, and the evolution of nitrogen gas also occurs.

C6H5N2Cl+C2H5OH C6H6+CH3CHO+N2+HCl

Question 9.11 Complete the following reactions:

(v) C6H5NH2+Br2(aq)

Answer :

Aniline on bromination in the presence of water at room temperature, gives a white precipitate of 2,4,6-tribromoaniline

1596564428916

Question 9.11 Complete the following reactions:

(vi) C6H5NH2+(CH3CO)2O

Answer :

Aniline reacts with acetic anhydride by a nucleophilic substitution reaction. Here, the H atom from NH2 group is replaced by an acyl group. The product formed is N-phenylethanamide. The reaction is carried out in the presence of a base stronger than the amine, like pyridine

C6H5NH2+(CH3CO)2O C6H5NHCOCH3+CH3COOH

1596564455558

Question 9.11 Complete the following reactions:

(vii) C6H5N2Cl(ii)NaNO2/Cu,Δ)(i)HBF4)

Answer :

In this reaction, the chloride ion is replaced by BF4 ion. When diazonium fluoroborate is heated with aqueous sodium nitrite in the presence of copper, the diazonium group is replaced by the –NO2 group.

1596566636152

Question 9.12 Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?

Answer :
Gabriel phthalimide synthesis is mainly used for aliphatic primary amine preparation. Phthalimide reacts with ethanolic potassium hydroxide and produces the potassium salt of phthalimide, which on heating with alkyl halide, followed by alkaline hydrolysis, gives the corresponding primary amine.

1596566079767

Aromatic amines cannot be produced by this method due to the inability of aryl halides to undergo nucleophilic substitution reaction with the anion produced by phthalimide.

Hence, aromatic amines cannot be produced by this method.

1596566087216

No reaction {if R = Aromatic compound}

Question 9.13 Write the reactions of

(i) aromatic primary amines with nitrous acid.

Answer :

Aromatic primary amines react with nitrous acid( NaNO2+HCl) at 273-278 K to form stable diazonium salts. Aniline reacts with nitrous acid to form benzene diazonium chloride.

1596566104222

Question 9.13 Write the reactions of

(ii) aliphatic primary amines with nitrous acid.

Answer :

When aliphatic primary amines react with nitrous acid to form unstable diazonium salts, which on further reaction produce alcohol and hydrochloric acid (HCl) with the evolution of dinitrogen gas (N2).


1596566113004

Question 9.14 Give a plausible explanation for each of the following:

(i) Why are amines less acidic than alcohols of comparable molecular masses?

Answer :

Protonation of amines gives-

RNH2RNH+H+ (amide ion)

Protonation of alcohol gives -

ROHRO+H+ (alkoxide ion)

The negative charge is easily accommodated by the O atom because it is more electronegative than N atom. So, the amide ion is less stable than the alkoxide ion. Hence, amines are less acidic than alcohols.

Question 9.14 Give a plausible explanation for each of the following:

(ii) Why do primary amines have a higher boiling point than tertiary amines?

Answer :

The extent of intermolecular hydrogen bonding in primary amines is higher than that of secondary amines. Secondary amine has only one hydrogen, and tertiary amines have no hydrogen atom for intermolecular hydrogen bonding, whereas primary amine has two hydrogen atoms available for intermolecular hydrogen bonding. And we know that the higher the extent of hydrogen bonding, the higher is the boiling point.

In 10 amine

1596566122228

(In 20 amine),

1596566129657

(In 30 amines)

1596566136983

Question 9.14 Give plausible explanation for each of the following:

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

Answer :

Aliphatic amines are stronger bases than aromatic amines due to resonance. In aromatic amines, the lone pair of electrons at N atom is delocalised over the benzene ring, which makes it less available for electron donation. On the other hand, in aliphatic amines (RNH2), the electron donor group is attached which increases the electron density at the N atom. Thus, the aliphatic amines are more basic than aromatic amines.

1596566484941

Class 12 Chemistry NCERT Chapter 9: Higher Order Thinking Skills (HOTS) Questions

Practice the questions below to improve your problem-solving skills.

Question: When a concentrated solution of sulphanilic acid and 1-naphthylamine is treated with nitrous acid (273 K) and acidified with acetic acid, the mass (g) of 0.1 mole of product formed is : (Given molar mass in g mol–1 H : 1, C : 12, N : 14,O : 16, S : 32)

1) 343

2) 330

3) 33

4) 66

Answer:

Molar mass of the product formed

=(16×12)+(14×3)+(16×3)+32+13=327 g

Mass of 0.1 mol product

32.7 g33 g
Hence, the correct answer is option (3).

Question: The major product (A) formed in the following reaction sequence is

1)

2)

3)

4)

Answer:

Hence, the correct answer is option (1).

Question:

The ratio of the number of oxygen atoms to bromine atoms in the product Q is ______ ×101.

Answer:

No. of oxygens/ No.of bromine atoms = 3/2 = 1.5

= 15 x 10-1.

Hence, the answer is (15).

Approach to Solve Questions of Chapter 9: Amines

To effectively solve questions of Amines, one should follow a structured approach. Here are some points that can help you with problem-solving strategies

1) Understand the key concepts

Firstly, the understanding of basic concepts is necessary to solve the problems related to amines like

  • Classification of primary, secondary and tertiary amines.
  • Classification of Aliphatic and aromatic amines.
  • IUPAC and common names of compounds
  • Distinguishing amines from other functional groups
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2). Memorise preparation methods

Preparation methods are quite important from the exam point of view. Some of them includes,

  • Reduction of nitriles, amides, and nitro compounds.
  • Hoffmann Bromamide degradation
  • Reduction of Nitrobenzene
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3). Properties and trends

Then, carefully study the physical properties of compounds like

  • Boiling point, melting point and solubility trends.
  • Physical properties and the effect of Hydrogen bonding on them.

4). Systematically memorise the chemical reaction and tests

Students can make use of short notes or charts to remember the concepts

  • Basicity of Amines
  • Reaction involving amines

Also, the test to distinguish the amines is too crucial, such as the Carbylamine test, Hinsberg's test and Nitrous acid reaction.

5). Practice more

Solve the problems involving molecular weight, percentage yield and reaction stoichiometry. Also, solve the in-text and exercise problems provided in the NCERT textbook

Topics of Chapter 9 In NCERT Book Class 12 Chemistry

The topics given in the chapter are

9.1 Structure of Amine

9.2 Classification

9.3 Nomenclature

9.4 Preparation Of Amines

9.5 Physical Properties

9.6 Chemical Properties

9.7 Method of Preparation of Diazoniun Salts

9.8 Physical Properties

9.9 Chemical Reactions

9.10 Importance of Diazonium Salts in the Synthesis of Aromatic Compounds

What Extra Should Students Study Beyond the NCERT for JEE/NEET?

The links below will help you classify the topics on the basis of exams and prepare accordingly.

NCERT Solutions for Class 12 Chemistry

The NCERT solutions for the other chapters have also been designed. Click on the links below-

NCERT Solutions for Class 12 Subject-wise

Excel your preparation by solving the NCERT solutions for other subjects as well.

NCERT Exemplar subject-wise

Follow the links below to get your hands on the NCERT exemplar for other subjects for better learning.

NCERT Books and NCERT Syllabus

Get the syllabus and books from the links provided in the table below

Frequently Asked Questions (FAQs)

1. What are amines and why are they important in organic chemistry?

Amines are organic compounds derived from ammonia (NH₃) by replacing one or more hydrogen atoms with alkyl or aryl groups. They are important because they form the basis of many biologically active compounds, including amino acids, proteins, vitamins and neurotransmitters. They are also used in the synthesis of dyes, polymers and pharmaceuticals.

2. What is a quaternary ammonium salt?

A quaternary ammonium salt is formed when the nitrogen atom in an amine is bonded to four alkyl or aryl groups and carries a positive charge. It has the general formula [NR₄]⁺ X⁻, where X⁻ is a counterion (e.g., Cl⁻, Br⁻). They are salts and therefore are often water soluble.

3. What is a major drawback of the ammonolysis of alkyl halides? How can this be overcome?

 Ammonolysis yields a mixture of primary, secondary and tertiary amines and also quaternary ammonium salts. This is because the amine formed in the first step can further react with the alkyl halide. This issue is addressed to some extent by using a large excess of ammonia which favors the formation of the primary amine.

4. What is the advantage of the Gabriel phthalimide synthesis?

The Gabriel synthesis produces pure primary amines without contamination by secondary or tertiary amines.

5. What is carbylamine test?

The carbylamine test is a chemical test used to detect primary amines (both aliphatic and aromatic). When heated with chloroform (CHCl3) and alcoholic KOH , primary amines produce a foul-smelling isocyanide (carbylamine). Secondary and tertiary amines do not give this test.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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