JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Compounds Containing Nitrogen

The amide A (${\mathrm{C}}_8{\mathrm{H}}_9\mathrm{NO}$) reacting with bromine in alkali undergoes the Hofmann bromamide degradation, which shortens the chain by one carbon to a primary aromatic amine B that dissolves in dilute acid. Diazotisation of B at $0$-$5^{\circ }\mathrm{C}$ gives the diazonium salt C, and CuCN/NaCN (Sandmeyer) replaces the diazonium group by a nitrile to give D; hydrolysis of this nitrile gives the acid E, which must also be the hydrolysis product of A. For the two routes to meet, the ring methyl that survives in E must sit where oxidation by acidified permanganate turns it into a second carboxyl and yields an acid F with only two kinds of hydrogen, namely terephthalic acid, whose four ring hydrogens are equivalent and whose two carboxyl hydrogens are equivalent. This places the methyl para to the amide, so A is 4-methylbenzamide ($p$-toluamide).