Amines Part 3

Concept:The compound $x$ is a primary aromatic amine because it is basic, gives carbylamine test, and forms an alkali-soluble sulfonamide.Explanation:Molecular formula $\mathrm{C}_6\mathrm{H}_7\mathrm{N}$ matches aniline ($\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_2$).It is sparingly soluble in water but dissolves in dilute mineral acid due to protonation.Carbylamine reaction with $\mathrm{CHCl}_3$ and alc. $\mathrm{KOH}$ gives phenyl isocyanide ($\mathrm{C}_6\mathrm{H}_5\mathrm{NC}$) with unpleasant smell, confirming a primary amine.With benzenesulphonyl chloride, aniline forms $z$, $N$-phenylbenzenesulfonamide ($\mathrm{C}_6\mathrm{H}_5\mathrm{SO}_2\mathrm{NH}\mathrm{C}_6\mathrm{H}_5$), which is soluble in alkali due to acidic $\mathrm{NH}$ proton.To count different $\mathrm{H}$ atoms in $z$, note two monosubstituted benzene rings: one attached to $\mathrm{SO}_2$ and one attached to $\mathrm{NH}$.For the ring attached to $\mathrm{SO}_2$: ortho (2 $\mathrm{H}$, equivalent), meta (2 $\mathrm{H}$, equivalent), para (1 $\mathrm{H}$).For the ring attached to $\mathrm{NH}$: ortho (2 $\mathrm{H}$, equivalent), meta (2 $\mathrm{H}$, equivalent), para (1 $\mathrm{H}$).Plus the $\mathrm{NH}$ hydrogen (1 $\mathrm{H}$).Since the two rings are not symmetric, all these positions are distinct in chemical environment.Total distinct $\mathrm{H}$ atom types = 3 (from first ring) + 3 (from second ring) + 1 ($\mathrm{NH}$) = 7.Answer:D. 7