NCERT Solutions for Class 12 Chemistry Chapter 13 Amines

NCERT Solutions for Class 12 Chemistry Chapter 13 Amines are prepared to help students understand the key topics from this chapter in a simple way. In this chapter, you will learn about amines, which are special compounds made from ammonia. The chapter explains how amines are named, how they are prepared, their types, and how they react with other chemicals. It also covers diazonium salts, which are used to make other organic compounds.

Chemistry Class 12 Amines NCERT solutions make it easier to understand the questions and remember tricky reactions and structures. Amines NCERT solutions are not only useful for board exams but also important for entrance tests like NEET and JEE. So, keep reading to explore NCERT Solutions for Class 12 Chemistry Chapter 13 Amines to understand how to write answers in a correct way.

Check Out: CBSE Class 12 Books

NCERT Solutions for Class 12 Chemistry Chapter 13 Amines

NCERT Solutions for Class 12 Chemistry Chapter 13 Amines are shared here to help you understand each topic in a clear and easy way. These Amines NCERT solutions are made to support your board exam preparation and help you feel confident while answering tricky questions.

Moreover, these CBSE Class 12 Chemistry Chapter 13 NCERT solutions are very useful for quick revision and clearing doubts. Whether you are preparing for school exams or other competitive tests, these Amines NCERT solutions will help you get better at solving questions. Check the detailed Chemistry Class 12 Chapter 13 NCERT solutions for each section here.

Check Out: Class 12th Question Banks

CBSE Class 12 Chemistry Chapter 13 Solutions

Question 1. Write the IUPAC names of the following compounds and classify them into primary, secondary, and tertiary amines.

(i) (CH₃)₂ CHNH₂
(ii) CH₃(CH₂)₂NH₂
(iii) CH₃NHCH(CH₃)₂
(iv) (CH₃)₃CNH₂
(v) C₆H₅NHCH₃
(vi) (CH₃CH₂)₂NCH₃
(vii) m−BrC₆H₄NH₂

Solution:

(i) 1-Methylethanamine (10 amine)
(ii) Propan-1-amine (10 amine)
(iii) N−Methyl-2-methylethanamine (20 amine)
(iv) 2-Methylpropan-2-amine (10 amine)
(v) N−Methylbenzamine or N-methylaniline (20 amine)
(vi) N-Ethyl-N-methylethanamine (30 amine)
(vii) 3-Bromobenzenamine or 3-bromoaniline (10 amine)

Question 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

Solution:

(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.

(ii) Secondary and tertiary amines can be distinguished by allowing them to react with Hinsberg’s reagent (benzenesulphonyl chloride, C6H5SO2Cl). 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 HNO2 (NaNO2 + dil.HCl) at 0-5°C, followed by a reaction with the alkaline solution of 2-naphthol. The dye is usually yellow, red, or orange in colour. Aliphatic amines give a brisk effervescence due to the evolution of N2 gas under similar conditions.

(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 an unstable diazonium salt, which in turn gives alcohol with the evolution of nitrogen gas.

On the other hand, aniline reacts with HNO2 at a low temperature to form a stable diazonium salt. Thus, nitrogen gas is 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 a positive carbylamine test. However, N-methylaniline, being a secondary amine, does not.

Question 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) Althoughthe  amino group is o, 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.

Solution:

(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.

On the other hand, in case of methylamine (due to the +I effect of methyl group), the electron density on the N-atom is increased. As a result, aniline is less basic than methylamine. Thus, pKb of aniline is more than that of methylamine.

(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.

But aniline does not undergo H−bonding with water to a very large extent due to the presence of a large hydrophobic −C6H5 group. Hence, aniline is insoluble in water.

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

Due to the +I effect of −CH3 group, methylamine is more basic than water. Therefore, in water, methylamine produces OH− ions by accepting H+ ions from water.

Ferric chloride (FeCl3) dissociates in water to form Fe3+ and Cl− ions.

Then, OH− ion reacts with Fe3+ ion to form a precipitate of 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).

For this reason, aniline on nitration gives a substantial amount of m-nitroaniline. 

(v) Aniline does not undergo a Friedel-Crafts reaction: A Friedel-Crafts reaction is carried out in the presence of AlCl3. But AlCl3 is acidic in nature, while aniline is a strong base. Thus, aniline reacts with AlCl3 to form a salt (as shown in the following equation).

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 1° amine only. 2° or 3° amines are not formed in this synthesis. Thus, a pure 1° amine can be obtained. Therefore, Gabriel phthalimide synthesis is preferred for synthesizing primary amines.

Read More: NCERT Solutions for Class 12 Chemistry Chapter 2

Question 4. Arrange the following:

(i) In decreasing order of the pKb values: C₂H₅NH₂, C₆H₅NHCH₃, (C₂H₅)₂NH, and C₆H₅NH₂
(ii) In increasing order of basic strength: C₆H₅NH₂, C₆H₅N(CH₃)₂, (C₂H₅)₂NH and CH₃NH₂
(iii) In increasing order of basic strength:
  (a) Aniline, p-nitroaniline, and p-toluidine
  (b) C₆H₅NH₂, C₆H₅NHCH₃, C₆H₅CH₂NH₂
(iv) In decreasing order of basic strength in gas phase: C₂H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃,N, and NH₃
(v) In increasing order of boiling point: C₂H₅OH, (CH₃)₂NH, C₂H₅NH₂
(vi) In increasing order of solubility in water: C₆H₅NH₂, (C₂H₅)₂NH, C₂H₅NH₂

Solution:

(i) In C2H5NH2, only one −C2H5 group is present while in (C2H5)2NH, two −C2H5 groups are present. Thus, the +I effect is more in (C2H5)2NH than in C2H5NH2. 

• Therefore, the electron density over the N-atom is more in (C2H5)2NH than in C2H5NH2. Hence, (C2H5)2NH is more basic than C2H5NH2. Also, both C6H5NHCH3 and C6H5NH2 are less basic than (C2H5)2NH and C2H5NH2 due to the delocalization of the lone pair in the former two. Further, among C6H5NHCH3 and C6H5NH2, 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: C6H5NH2 < C6H5NHCH3 < C2H5NH2 < (C2H5)2NH We know that the higher the basic strength, the lower is the pKb values. C6H5NH2 > C6H5NHCH3 > C2H5NH2 > (C2H5)2NH

(ii) C6H5N(CH3)2 is more basic than C6H5NH2 due to the presence of the +I effect of two −CH3 groups in C6H5N(CH3)2. Further, CH3NH2 contains one −CH3 group while (C2H5)2NH contains two −C2H5 groups. 

• Thus, (C2H5)2 NH is more basic than C2H5NH2. Now, C6H5N(CH3)2 is less basic than CH3NH2 because of the−R effect of −C6H5 group. 

• Hence, the increasing order of the basic strengths of the given compounds is as follows: C6H5NH2 < C6H5N(CH3)2 < CH3NH2 < (C2H5)2NH

(iii) (a)

In p-toluidine, the presence of an 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 an 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 

(b) C6H5NHCH3 is more basic than C6H5NH2 due to the presence of an electron-donating −CH3 group in C6H5NHCH3. Again, in C6H5NHCH3, −C6H5 group is directly attached to the N-atom. 

• However, it is not so in C6H5CH2NH2. Thus, in C6H5NHCH3, the −R effect of −C6H5 group decreases the electron density over the N-atom. 

• Therefore, C6H5CH2NH2 is more basic than C6H5NHCH3. Hence, the increasing order of the basic strengths of the given compounds is as follows: C6H5NH2 < C6H5NHCH3 < C6H5CH2NH2. 

(iv) In the gas phase, there is no solvation effect. As a result, the basic strength mainly depends upon the +I effect. The higher the +I effect, the stronger is the base. Also, the greater the number of alkyl groups, the higher is the +I effect. Therefore, the given compounds can be arranged in the decreasing order of their basic strengths in the gas phase as follows: (C2H5)3N > (C2H5)2NH > C2H5NH2 > NH3 

(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. (CH3)2NH contains only one H−atom, whereas C2H5NH2 contains two H-atoms. 

• Then, C2H5NH2 undergoes more extensive H-bonding than (CH3)2NH. Hence, the boiling point of C2H5NH2 is higher than that of (CH3)2NH. Further, O is more electronegative than N. 

• Thus, C2H5OH forms stronger H−bonds than C2H5NH2. As a result, the boiling point of C2H5OH is higher than that of C2H5NH2 and (CH3)2NH. 

• Now, the given compounds can be arranged in the increasing order of their boiling points as follows: (CH3)2NH < C2H5NH2 < C2H5OH 

(vi) The more extensive the H−bonding, the higher the solubility. C2H5NH2 contains two H-atoms, whereas (C2H5)2NH contains only one H-atom. Thus, C2H5NH2 undergoes more extensive H−bonding than (C2H5)2NH. Hence, the solubility in water of C2H5NH2 is more than that of (C2H5)2NH. 

• Further, the solubility of amines decreases with an 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 C6H5NH2 is greater than that of C2H5NH2 and (C2H5)2NH. Hence, the increasing order of their solubility in water is as follows: C6H5NH2 < (C2H5)2NH < C2H5NH2.

Question 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

Solution:

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

Solution:

Primary, secondary, and tertiary amines can be identified and distinguished by Hinsberg’s test. In this test, the amines are allowed to react with Hinsberg’s reagent, benzenesulphonyl chloride (C₆H₅SO₂Cl). The three types of amines react differently with Hinsberg’s reagent. Therefore, they can be easily identified 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 a 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.

Question 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.

Solution:

(i) Carbylamine reaction: The Carbylamine reaction is used as a test for the identification 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 NaNO2 and a mineral acid such as HCl) at low 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 NaNO2 and HCl at 273−278 K, aniline produces benzenediazonium chloride, with NaCl and H2O as by-products.

(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 the 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 −N=N−bond is known as a coupling reaction. Arenediazonium salts, such as benzene diazonium salts, react with phenol or aromatic amines to form coloured azo compounds.

It can be observed that the para-positions of phenol and aniline are coupled with the diazonium salt. This reaction proceeds through electrophilic substitution. 

(v) Ammonolysis: When an alkyl or benzyl halide is allowed to react with an ethanolic solution of ammonia, 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 bond is known as ammonolysis.

When this substituted ammonium salt is treated with a strong base such as sodium hydroxide, an amine is obtained.

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 −NH2 or > 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 HCl 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.

When amines react with benzoyl chloride, the reaction is also known as benzoylation. For example,

 

(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 the potassium salt of phthalimide. This salt is further heated with an alkyl halide, followed by alkaline hydrolysis to yield the corresponding primary amine.

Read More: NCERT Solutions for Class 12 Chemistry Chapter 4

Question 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-tribromofluorobenzene
(v) Benzyl chloride to 2-phenylethanamine
(vi) Chlorobenzene to p-chloroaniline

Solution:

Question 9. Give the structures of A, B, and C in the following reactions:

Solution:

Question 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.

Solution:

It is given that compound ‘C’, having the molecular formula C₆H₇N, is formed by heating compound ‘B’ with Br₂ 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 C₆H₇N is aniline (C₆H₅NH₂).

Therefore, compound ‘B’ (from which ’C’ is formed) must be benzamide, (C6H5CONH2).

Further, benzamide is formed by heating compound ‘A’ with aqueous ammonia. Therefore, compound ‘A’ must be benzoic acid.

The given reactions can be explained with the help of the following equations:

Question 11. Complete the following reactions:

Solution:

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

Solution:

Gabriel phthalimide synthesis is used for the preparation of aliphatic primary amines. It involves nucleophilic substitution (SN2) of alkyl halides by the anion formed by the phthalimide.

But aryl halides do not undergo nucleophilic substitution with the anion formed by the phthalimide.

Hence, aromatic primary amines cannot be prepared by this process.

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

Solution: 

(i) Aromatic amines react with nitrous acid (prepared in situ from NaNO2 and a mineral acid such as HCl) at 273 − 278 K to form stable aromatic diazonium salts i.e., NaCl and H2O.

(ii) Aliphatic primary amines react with nitrous acid (prepared in situ from NaNO2 and a mineral acid such as HCl) to form unstable aliphatic diazonium salts, which further produce alcohol and HCl with the evolution of N2 gas.

Question 14. Give a 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 a higher boiling point than tertiary amines?
(iii) Why are aliphatic amines stronger bases than aromatic amines? 

Solution:

(i) Amines undergo protonation to give an amide ion.

Similarly, alcohol loses a proton to give an alkoxide ion.

In an amide ion, the negative charge is on the N-atom, whereas in an 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.

As a result, extra energy is required to separate the molecules of primary amines. Hence, primary amines have higher boiling points than tertiary amines. 

(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. 

Question 15. Classify the following amines as primary, secondary, or tertiary: 

(i)

(ii)

(iii) (C2H5)2CHNH2 (iv) (C2H5)2NH 

Solution:

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

Read More: NCERT Solutions for Class 12 Chemistry Chapter 10

Question 16. 

(i) Write structures of different isomeric amines corresponding to the molecular formula, C4H11N.
(ii) Write IUPAC names of all the isomers.
(iii) What type of isomerism is exhibited by different pairs of amines? 

Solution:

(i), (ii) The structures and their IUPAC names of different isomeric amines corresponding to the molecular formula, C4H11N, are given below: 

(a) CH3-CH2-CH2-CH2-NH2 Butanamine (10) 

(b)Butan-2-amine (10) 

(c)2-Methylpropanamine (10) 

(d)2-Methylpropan-2-amine (10) 

(e) CH3-CH2-CH2-NH-CH3 N-Methylpropanamine (20) 

(f) CH3-CH2-NH-CH2-CH3 N-Ethylethanamine (20) 

(g)N-Methylpropan-2-amine (20) 

(h)

N,N-Dimethylethanamine (3°) (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.

Question 17. How will you convert? 

(i) Benzene into aniline
(ii) Benzene into N, N-dimethylaniline
(iii) Cl−(CH2)4−Cl into hexan-1, 6-diamine?

Solution:

(i)

(ii)(iii)

Question 18. Arrange the following in increasing order of their basic strength: 

(i) C2H5NH2, C6H5NH2, NH3, C6H5CH2NH2, and (C2H5)2NH
(ii) C2H5NH2, (C2H5)2NH, (C2H5)3N, C6H5NH2
(iii) CH3NH2, (CH3)2NH, (CH3)3N, C6H5NH2, C6H5CH2NH2. 

Solution:

(i) Considering the inductive effect of alkyl groups, NH3, C2H5NH2, and (C2H5)2NH can be arranged in the increasing order of their basic strengths as:

Again, C6H5NH2 has a proton acceptability less than NH3. Thus, we have:

Due to the −I effect of the C6H5 group, the electron density on the N-atom in C6H5CH2NH2 is lower than that on the N-atom in C2H5NH2, but more than that in NH3. Therefore, the given compounds can be arranged in the order of their basic strengths as:

(ii) Considering the inductive effect and the steric hindrance of the alkyl groups, C2H5NH2, (C2H5)2NH2, and their basic strengths as follows:

Again, due to the −R effect of the C6H5 group, the electron density on the N atom in C6H5 NH2 is lower than that on the N atom in C2H5NH2. Therefore, the basicity of C6H5NH2 is lower than that of C2H5NH2. Hence, the given compounds can be arranged in the increasing order of their basic strengths as follows:

(iii) Considering the inductive effect and the steric hindrance of alkyl groups, CH3NH2, (CH3)2NH, and (CH3)3N can be arranged in the increasing order of their basic strengths as:

In C6H5NH2, N is directly attached to the benzene ring. Thus, the lone pair of electrons on the N−atom is delocalized over the benzene ring. In C6H5CH2NH2, N is not directly attached to the benzene ring. Thus, its lone pair is not delocalized over the benzene ring. 

Therefore, the electrons on the N atom are more easily available for protonation in C6H5CH2NH2 than in C6H5NH2, i.e., C6H5CH2NH2 is more basic than C6H5NH2. 

Again, due to the −I effect of the C6H5 group, the electron density on the N−atom in C6H5CH2NH2 is lower than that on the N−atom in (CH3)3N. Therefore, (CH3)3N is more basic than C6H5CH2NH2. Thus, the given compounds can be arranged in the increasing order of their basic strengths as follows.

Read More: NCERT Solutions for Class 12 Chemistry Chapter 11

Question 19. Complete the following acid-base reactions and name the products: 

(i) CH3CH2CH2NH2 + HCl

(ii) (C2H5)3N + HCl

Solution:

(i)

(ii) (C 2 H 5 ) 3 N Triethylamine+ HCl → (C 2 H 5 ) 3 N+HCl−Triethylammoniumchloride

Question 20. Write reactions of the final alkylation product of aniline with an excess of methyl iodide in the presence of sodium carbonate solution. 

Solution: 

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

.

With excess methyl iodide, in the presence of Na2CO3 solution, N, N-dimethylaniline produces N, N, N−trimethylanilinium carbonate.

Question 21. Write the chemical reaction of aniline with benzoyl chloride, and write the name of the product obtained. 

Solution:

 

Question 22. Write structures of different isomers corresponding to the molecular formula, C3H9N. Write IUPAC names of the isomers that will liberate nitrogen gas on treatment with nitrous acid. 

Solution:

The structures of different isomers corresponding to the molecular formula, C3H9N, are given below: 

(a)Propan-1-amine (10)

(b)Propan-2-amine (10)

(c)

 

(d)

N,N-Dimethylmethanamine (30) 10amines, (a) propan-1-amine, and (b) Propan-2-amine will liberate nitrogen gas on treatment with nitrous acid.

Read More: NCERT Solutions for Class 12 Chemistry Chapter 12

Question 23. Convert 

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

Solution:

(i)

(ii)

Class 12 Chemistry Chapter 13 Amines Summary

Now that you've gone through the NCERT Solutions for Class 12 Chemistry Chapter 13 Amines, let's also understand the summary of this chapter. Amines Chemistry Chapter of Class 12 helps you understand how amines behave, how to make them, and how they are used in real life. From making medicines to dyes, amines are used everywhere.

What are Amines?

Amines are organic compounds made from ammonia (NH₃). When we replace one or more hydrogen atoms in ammonia with alkyl or aryl groups, we get amines. These compounds are very important in making medicines, dyes, and many other chemicals.

Types of Amines

Amines are divided into three main types based on how many groups are attached to nitrogen:

• Primary amine (1°): One group is attached (R-NH₂)

• Secondary amine (2°): Two groups are attached (R₂NH or R-NHR′)

• Tertiary amine (3°): Three groups are attached (R₃N or R₂NR′)

• If all the groups are the same, it is called a simple amine.

• If they are different, it's called a mixed amine.

Structure and Basic Nature of Amines

• Nitrogen in amines has a lone pair of electrons. This lone pair makes amines basic in nature.

• Because of this, amines can accept protons (H⁺) and act as Lewis bases.

Nomenclature (Naming Amines)

• In this chapter, you will learn how to name different types of amines.

• Common names and IUPAC names are both used.

Preparation of Amines

There are many methods to prepare amines, such as:

• From alkyl halides using ammonia

• Reduction of nitro compounds

• Gabriel phthalimide synthesis (good for making primary amines)

• Hoffmann bromamide reaction (gives one less carbon atom than the starting material)

Physical Properties of Amines

• Boiling point: Amines have higher boiling points than alkanes due to hydrogen bonding.

• Solubility: Smaller amines dissolve in water. Bigger ones are less soluble.

Chemical Reactions of Amines

• Alkylation: Adding alkyl groups to amines

• Acylation: Adding acyl groups (like CH₃CO–)

• Reaction with nitrous acid: Primary, secondary, and tertiary amines react differently.

• Hinsberg test: Used to identify primary, secondary, and tertiary amines using p-toluenesulphonyl chloride

Diazonium Salts (From Aromatic Amines)

• Formed by treating aromatic amines (like aniline) with nitrous acid at low temperature

• These salts are very useful in making azo dyes and other aromatic compounds.

• They easily get converted into other useful products like phenols, halides, and cyanides.

Basic strength of amines depends on the groups attached to nitrogen.

• Alkyl amines are usually stronger bases than ammonia.

• Aromatic amines (like aniline) are weaker bases due to the effect of the benzene ring.

• The electronegativity difference between atoms affects how reactive an amine is.

• Nucleophilic substitution: Amines can replace leaving groups in reactions with acid chlorides, esters, etc.

Along with CBSE Class 12 Chemistry Chapter 13 NCERT solutions, it is also important to study these key topics carefully, as most of the questions asked in board or other entrance exams come from these areas.

Check Out: Class 12th Sample Papers

Why Are NCERT Solutions for Class 12 Chemistry Chapter 13 Important?

NCERT Solutions for Class 12 Chemistry Chapter 13 can be useful study material if you want to understand the topic of amines in a simple and clear way. These solutions explain all the important concepts in an easy to understand manner, which helps those who find this topic difficult to learn.

Chemistry Class 12 Amines NCERT solutions given above cover all textbook questions with correct and step-by-step answers. These Amines Class 12 NCERT solutions help you during your revisions to boost your preparations for exams and also clear any doubts related to tricky topics like reactions, basicity, and structure of amines.

Here are some reasons why these CBSE Class 12 Chemistry Chapter 13 NCERT solutions are important for you:

• Easy to Understand: The Amines NCERT solutions class 12 simplifies hard topics like classification of amines, preparation methods, and chemical reactions.

• Based on Latest Syllabus: These Chemistry Class 12 Chapter 13 NCERT solutions follow the latest CBSE syllabus, so you are always learning what is actually needed for exams.

• Helps in Exam Practice: Many board exam questions come directly from NCERT. Solving the textbook questions using Amines NCERT solutions gives you good practice and improves answer-writing skills.

• Quick Revision: You don’t have to search for answers. All important questions from the chapter are explained in one place, which saves your time during revision.

• Useful for NEET and JEE: Amines are an important topic in entrance exams too. These Amines NCERT solutions class 12 help build a strong base for competitive exams.

Therefore, using these Amines Class 12 NCERT solutions along with additional Class 12 Books can greatly help you study better for boards and score well.

Also Check, Class 12th Sample Papers

NCERT Solutions for Class 12 Chemistry Chapter 13 Amines FAQs

Q.1. Why should I study NCERT Solutions for Class 12 Chemistry Chapter 13 Amines?

Ans. These Amines Class 12 NCERT solutions help you understand the chapter in an easy way. All textbook questions are solved step by step, which makes revision faster and preparation stronger for board exams and entrance tests.

Q.2. What are amines in Class 12 Chemistry?

Ans. Amines are organic compounds that come from ammonia. When one or more hydrogen atoms in ammonia are replaced by alkyl or aryl groups, we get amines. They are important in medicines, dyes, and chemical industries.

Q.3. Are Chemistry Class 12 Amines NCERT solutions hard to understand?

Ans. The Amines Class 12 NCERT solutions are easy to understand. If you read the important topics from the textbook with these solutions, you will be able to understand the concepts and their solutions better.

Q.4. How do these Amines NCERT solutions class 12 help in boards?

Ans. CBSE Class 12 Chemistry Chapter 13 NCERT solutions cover all important questions from the textbook. They also help you learn how to frame your answer in a neat and clear way, which is very important if you want to score well in boards.

Q5. Are Amines NCERT solutions class 12 useful for NEET or JEE?

Ans. Amines chapter is a part of the NEET and JEE syllabus. Therefore, through these Chemistry Class 12 Chapter 13 NCERT solutions, you can understand the basics clearly, which is important for solving higher-level questions in entrance exams.