Nitration of aniline in strong acidic medium also gives $m$-nitroaniline because

Solution & Explanation

\textbf{Answer:} (D) In strong acid, aniline is protonated to the anilinium ion $\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{NH}{}_3{}^{+}$, whose $-\mathrm{NH}{}_3{}^{+}$ group is deactivating and meta directing. \textbf{Solution:} Free $-\mathrm{NH}{}_2$ is a strong activating, ortho/para director through lone-pair donation into the ring. In a strongly acidic nitrating mixture, however, the basic amino group is protonated: $\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{NH}{}_2+\mathrm{H}{}^{+}\to \mathrm{C}{}_6\mathrm{H}{}_5\mathrm{NH}{}_3{}^{+}$. The resulting $-\mathrm{NH}{}_3{}^{+}$ carries a full positive charge, has no lone pair to donate, is electron-withdrawing (deactivating) and therefore meta directing. Hence a significant fraction of nitration occurs at the meta position, yielding $m$-nitroaniline alongside the ortho/para products from any unprotonated aniline. Option (D) gives the correct reason. Options (A) and (C) are false generalisations, and (B) wrongly calls the free amino group meta directing.