Given below are two statements: Statement I: Aniline does not undergo Friedel-Crafts alkylation reaction. Statement II: Aniline cannot be prepared through Gabriel synthesis. In the light of the above statements, choose the correct answer from the options given below:

Solution & Explanation

\textbf{Answer:} (D) Both statements are true. \textbf{Solution:} Statement I (TRUE): In Friedel-Crafts alkylation the Lewis acid catalyst (e.g. $\mathrm{AlCl}{}_3$) reacts with the basic nitrogen lone pair of aniline $\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{NH}{}_2$ to form a salt ($\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{N}{}^{+}\mathrm{H}{}_2\cdot \mathrm{AlCl}{}_3{}^{-}$ type complex). The resulting positively charged nitrogen strongly deactivates the ring (acts as a meta-directing $-I$ group) and the catalyst is consumed, so aniline does not undergo Friedel-Crafts alkylation. Statement II (TRUE): Gabriel phthalimide synthesis proceeds by an $\mathrm{S}{}_{\mathrm{N}}2$ attack of potassium phthalimide on an alkyl halide. Aryl halides do not undergo nucleophilic substitution under these conditions, so aromatic primary amines like aniline cannot be prepared by Gabriel synthesis. Both statements are correct, so the answer is (D).