The correct increasing order of basic strength for the following compounds is (I) Aniline, $\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{NH}{}_2$ (II) $p$-Nitroaniline, $\mathrm{NH}{}_2-\mathrm{C}{}_6\mathrm{H}{}_4-\mathrm{NO}{}_2$ ($-\mathrm{NO}{}_2$ at the para position) (III) $p$-Toluidine, $\mathrm{NH}{}_2-\mathrm{C}{}_6\mathrm{H}{}_4-\mathrm{CH}{}_3$ ($-\mathrm{CH}{}_3$ at the para position)

Solution & Explanation

\textbf{Answer:} (D) Electron-withdrawing $-\mathrm{NO}{}_2$ lowers basicity below aniline, while electron-donating $-\mathrm{CH}{}_3$ raises it above aniline. \textbf{Solution:} Basicity of an aniline depends on the availability of the nitrogen lone pair, which is affected by the para substituent. In (II) $p$-nitroaniline, the $-\mathrm{NO}{}_2$ group is strongly electron-withdrawing ($-I$ and $-R$); it delocalises the $\mathrm{N}$ lone pair into the ring, greatly reducing electron density on nitrogen $\to$ \textbf{weakest base}. In (I) aniline, the lone pair is delocalised into the ring (resonance) but there is no extra deactivating group, so it is intermediate. In (III) $p$-toluidine, the $-\mathrm{CH}{}_3$ group is electron-donating ($+I$, hyperconjugation); it increases electron density on nitrogen $\to$ \textbf{strongest base}. Therefore increasing basic strength: II $<$ I $<$ III, which is option (D).