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(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.

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.