AP EAPCET 23 May 2025 Shift 2 Paper

Statement IIn the nitration of aniline, more amount of $m$-nitroaniline is formed than expected.Normally, $-\mathrm{NH}_2$ is an activating, ortho/para-directing group.Hence, we expect mainly $o$ - and $p$-nitroaniline.However, nitration is carried out in strongly acidic medium ( $\mathrm{H}_2\mathrm{SO}_4/\mathrm{HNO}_3$ ), and under these conditions, aniline gets protonated to form the anilinium ion $(\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_3^{+})$.The $-\mathrm{NH}_3^{+}$group is an electron-withdrawing, deactivating, meta-directing group.So the nitration mixture now contains protonated aniline, which gives meta nitro product.As a result, the actual nitration product mixture contains a higher proportion of $m$-nitroaniline than we would expect based on the activating nature of $-\mathrm{NH}_2$ alone.Hence, Statement I is correct.Statement IIIn the presence of a strongly acidic medium, aniline is protonated to form anilinium ion, which is meta-directing.As discussed above, this is true - in a strongly acidic medium, aniline is converted to the anilinium ion $(\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_3^{+})$.The anilinium ion is meta-directing because the positively charged nitrogen withdraws electrons from the ring, deactivating it, especially at ortho and para positions.