The given reaction involves the hydrolysis of an alkyl halide \( X \), which follows second order kinetics. The alkyl halide is \( C_6H_5Br \), which undergoes hydrolysis to form \( Y \). The reaction with \( C_6H_5Cl \) in Na/dry ether produces an ester, specifically methyl ester \( C_6H_5COOCH_3 \), which is \( Y \). Oxidation of \( Y \) with \( KMnO_4 \) yields benzoic acid (\( C_6H_5COOH \)), which is \( Z \).