Chapter 9 Amines Intext Questions NCERT Solutions Class 12 Chemistry- PDF Download

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9.1. Classify the following amine as primary, secondary or tertiary:

(iii) (C₂H₅)₂CHNH₂

(iv) (C₂H₅)₂NH

Solution

(i) Primary

(ii) Tertiary

(iii) Primary

(iv) Secondary

9.2. (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?

Solutions

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

(a) Butanamine (1°)

CH₃-CH₂-CH₂-CH₂-NH₂

(b) Butan-2-amine (1°)

(d) 2-Methylpropan-2-amine (1°)

(e) N-Methylpropanamine (2°)

CH₃-CH₂-CH₂-NH-CH₃

(f) N-Ethylethanamine (2°)

CH₃-CH₂-NH-CH₂-CH₃

(g) N-Methylpropan-2-amine (2°)

(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.

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9.3. How will you convert:

(i) Benzene into aniline

(ii) Benzene into N, N-dimethylaniline

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

Solution

(i) Benzene into aniline

(ii) Benzene into N, N-dimethylaniline

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

9.4. 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, C₂H₅NH_2, and (C₂H₅)₂NH can be arranged in the increasing order of their basic strengths as:

NH₃ < C₂H₅NH₂ < (C₂H₅)₂NH

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

C₆H₅NH₂ <NH₃ < C₂H₅NH₂ < (C₂H₅)₂NH

Due to the −I effect of C₆H₅ group, the electron density on the N-atom in C₆H₅CH₂NH₂ is lower than that on the N-atom in C₂H₅NH_2, but more than that in NH₃. Therefore, the given compounds can be arranged in the order of their basic strengths as:

C₆H₅NH₂ <NH₃ < C₆H₅CH₂NH₂

(ii) Considering the inductive effect and the steric hindrance of the alkyl groups, C₂H₅NH₂, (C₂H₅)₂NH_2,and their basic strengths as follows:

C₂H₅NH₂ < (C₂H₅)₃N < (C₂H₅)₂NH

Again, due to the −R effect of C₆H₅ group, the electron density on the N atom in C₆H₅NH₂ 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:

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

(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:

(CH₃)₃N < CH₃NH₂ < (CH₃)₂NH

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 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

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

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

(i) CH₃CH₂CH₂NH₂ + HCl →

(ii) C₂H₅)₃N + HCl →

Solution

9.6. Write reactions of the final alkylation product of aniline with 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 Na₂CO₃solution, N, N-dimethylaniline produces N, N, N−trimethylanilinium carbonate.

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

Solution

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.

Solution

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

(a) Propan-1-amine (1°)

CH₃-CH₂-CH₂-NH₂

(b) Propan-2-amine (1°)