A given nitrogen-containing aromatic compound $A$ reacts with $\mathrm{Sn}/\mathrm{HCl}$, followed by $\mathrm{HNO}{}_2$, to give an unstable compound $B$. $B$, on treatment with phenol, forms a beautiful coloured compound $C$ with molecular formula $\mathrm{C}{}_{12}\mathrm{H}{}_{10}\mathrm{N}{}_2\mathrm{O}$. The structure of compound $A$ is

Solution & Explanation

\textbf{Answer:} (B) Nitrobenzene is reduced by $\mathrm{Sn}/\mathrm{HCl}$ to aniline, diazotised to an unstable diazonium salt, which couples with phenol to give the azo dye $\mathrm{C}{}_{12}\mathrm{H}{}_{10}\mathrm{N}{}_2\mathrm{O}$. \textbf{Solution:} The sequence is reduction $\to$ diazotisation $\to$ azo coupling. $\mathrm{Sn}/\mathrm{HCl}$ reduces a nitro group to a primary amine: $\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{NO}{}_2\stackrel{\mathrm{Sn}/\mathrm{HCl}}{\to }\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{NH}{}_2$. Treatment with nitrous acid ($\mathrm{HNO}{}_2$, from $\mathrm{NaNO}{}_2+\mathrm{HCl}$ at 273-278 K) diazotises aniline to benzenediazonium chloride $B$: $\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{NH}{}_2\stackrel{\mathrm{HNO}{}_2}{\to }\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{N}{}_2{}^{+}\mathrm{Cl}{}^{-}$. This aromatic diazonium salt is stabilised by resonance with the ring but is still unstable on isolation. As an electrophile it couples at the para position of activated phenol to give $p$-hydroxyazobenzene (an orange azo dye): $\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{N}{}_2{}^{+}+\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{OH}\to \mathrm{C}{}_6\mathrm{H}{}_5-\mathrm{N}=\mathrm{N}-\mathrm{C}{}_6\mathrm{H}{}_4-\mathrm{OH}$. This product is $\mathrm{C}{}_{12}\mathrm{H}{}_{10}\mathrm{N}{}_2\mathrm{O}$, matching $C$. Working backward, $A$ must be the nitro compound that yields aniline on reduction, i.e. nitrobenzene (B). Aniline (A) would not need $\mathrm{Sn}/\mathrm{HCl}$; benzonitrile and benzamide on reduction/hydrolysis do not directly fit the stated route giving the azo dye.