GUJCET 2026 Chemistry Paper

Concept:Oxidative cleavage of an alkenyl side chain on a benzene ring by alkaline $\mathrm{KMnO}_4$ gives the aromatic carboxylate salt, which on acidification gives the carboxylic acid.Explanation:Styrene ($\mathrm{C_6H_5-CH=CH_2}$) is treated with $\mathrm{KMnO_4/KOH}$ in presence of $[\mathrm{O}]$.The $\mathrm{C=C}$ bond of the side chain is completely broken, giving potassium benzoate, so $P = \mathrm{C_6H_5COOK}$.Potassium benzoate is then acidified with $\mathrm{H_3O^+}$.The salt is protonated to give benzoic acid, so $Q = \mathrm{C_6H_5COOH}$.Hence $P$ is potassium benzoate and $Q$ is benzoic acid.Answer:$P$ = Potassium benzoate, $Q$ = Benzoic acid. Option (B) is correct.