Account for the following with equations of the reactions involved:a) Gabriel phthalimide synthesis is preferred for preparation of primary amines.b) Explain why pKb of methanamine is 3.38 while that of aniline is 9.38.

Gabrielphthalimidesynthesis results in the formation of 1° amine only. 2° or 3° amines are not formed in this synthesis. Thus, a pure 1° amine can be obtained. Therefore, Gabrielphthalimidesynthesis is preferred forsynthesizingprimary amines.


Aniline undergoes resonance and as a result, the electrons on theN-atomaredelocalizedover the benzene ring. Therefore, the electrons on theN-atomare less available to donate.

On the other hand, in case ofmethylamine(due to the +I effect of methyl group), the electron density on theN-atomis increased. As a result, aniline is less basic thanmethylamine. Thus,pKbof aniline is more than that of methylamine.