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\( \boldsymbol{A} \stackrel{\text {sodalime}}{\longrightarrow} \boldsymbol{C}_{\boldsymbol{6}} \boldsymbol{H}_{\boldsymbol{6}} \stackrel{\boldsymbol{C l}_{\boldsymbol{\lambda}}, \boldsymbol{h} \boldsymbol{v}}{\boldsymbol{B}} \) In this reaction \( A \) and \( B \) are respectively:

1. Pheno
2. Chlorobenzene
3. Sodium benzoate, Внс
4. \( C_{2} H_{2} \& \) ВНС

Correct answer: Sodium benzoate, Внс

Solution

Sodium benzoate undergoes decarboxylation with soda lime to form benzene. Benzene then reacts with chlorine in the presence of light (hν) to give benzene hexachloride (BHC), not chlorobenzene.

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