Page No 396: - Chapter 13 Amines Intext Solutions class 12 ncert solutions Chemistry - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Question 13.4:

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.

Answer:

(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:

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

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:

(ii) Considering the inductive effect and the steric hindrance of the alkyl groups, C₂H₅NH₂, (C₂ H₅)₂NH_2, and their basic strengths as follows:

Again, due to the −R effect of 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., C₆H₅CH_2­NH₂ 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.

Question 13.5:

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

(i) CH₃CH₂CH₂NH₂ + HCl 

(ii) (C₂H₅)₃N + HCl 

Answer:

(i) 

(ii) 

C2H53N Triethylamine+ HCl → C2H53N+HCl-Triethylammoniumchloride

Question 13.6:

Write reactions of the final 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 Na2CO3 solution, N, N-dimethylaniline produces N, N, N−trimethylanilinium carbonate.

Question 13.7:

Write chemical reaction of aniline with benzoyl chloride and write the name of

the product obtained.

Answer:

Question 13.8:

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:

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.