Identify compound X in the following reaction: $\mathrm{C}{}_6\mathrm{H}{}_5-\mathrm{CH}{}_3\underset{\text{h}\nu }{\overset{\text{Cl2}}{\to }}\mathrm{X}\underset{\text{373 K}}{\overset{\text{H2O}}{\to }}\mathrm{C}{}_6\mathrm{H}{}_5-\mathrm{CHO}$

Solution & Explanation

\textbf{Answer:} (A) Benzal chloride $\mathrm{C}{}_6\mathrm{H}{}_5-\mathrm{CHCl}{}_2$ \textbf{Solution:} Free-radical side-chain chlorination of toluene with $\mathrm{Cl}{}_2$ under light introduces two chlorines on the methyl carbon to give benzal chloride $\mathrm{C}{}_6\mathrm{H}{}_5-\mathrm{CHCl}{}_2$ (X). Hydrolysis of this gem-dichloride with water at 373 K gives a gem-diol that loses water to form benzaldehyde. (Benzyl chloride would give benzyl alcohol and benzotrichloride would give benzoic acid, so X must be benzal chloride.)