Question

Arrange the following in increasing order of their basic strength:

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

Answer 1

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., 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:

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