Question 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:

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 <C₂H₅NH₂ C₆H₅CH₂NH₂<C₂H₅NH₂<(C₂H₅)₂NH

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

C₂H₅NH₂ <(C₂ H₅)₃N <(C₂ H₅)₂NH

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:

C₂H₅NH₂ <C₂ H₅NH₂ <(C₂ H₅)₃N <(C₂ H₅)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.