JEE Main · 2025 · 28 Jan 2025 · Shift 1 · Haloalkanes and Haloarenes

Chlorobenzene first forms phenylmagnesium chloride with Mg in dry ether; carbonation with $C{O}_2$ followed by acidic workup gives benzoic acid, and heating benzoic acid with $N{H}_3$ converts it to benzamide (A). Benzamide undergoes Hofmann bromamide degradation with $B{r}_2/NaOH$: the carbonyl carbon is lost as carbonate/$C{O}_2$ via isocyanate and carbamate intermediates, and the ring nitrogen ends up directly attached to the ring as aniline (B), one carbon fewer than the amide. The benzene ring is conserved throughout, so moles of chlorobenzene equal moles of aniline formed: $\frac{11.25\times 10^{-3}}{112.5}=\frac{x\times 10^{-1}\times 10^{-3}}{93}$, giving $x\times 10^{-1}=93\times 0.1=9.3$, so $x=93$.