Hydrolysis of an alkyl halide \( X \) (\(C_6H_5Br\)) follows second order kinetics. Reaction of \( X \) with \( C_6H_5Cl \) in the presence of Na/dry ether gave \( Y \). Oxidation of \( Y \) in the presence of \( KMnO_4 / OH^- \) gave \( Z \). What are \( Y \) and \( Z \) respectively?
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The hydrolysis of alkyl halides and the oxidation of esters can follow predictable reaction mechanisms and lead to simple aromatic compounds such as alcohols, acids, and esters.
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 \( Z \).