Benzene is treated with CH3Cl and anhydrous AlCl3 to give A. A is heated with alkaline KMnO4 and then acidified to give B. B is converted to its amide C by heating with NH3 (via the ammonium salt). C is treated with Br2/KOH to give D. D is

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Benzene is treated with $CH X_{3} Cl$ and anhydrous $AlCl X_{3}$ to give A. A is heated with alkaline $KMnO X_{4}$ and then acidified to give B. B is converted to its amide C by heating with $NH X_{3}$ (via the ammonium salt). C is treated with $Br X_{2} / KOH$ to give D. D is

Abenzylamine, $C X_{6} H X_{5} CH X_{2} NH X_{2}$
Baniline, $C X_{6} H X_{5} NH X_{2}$
Cbenzamide, $C X_{6} H X_{5} CONH X_{2}$
Dbenzoic acid, $C X_{6} H X_{5} COOH$

Answer: B

A is toluene, B is benzoic acid and C is benzamide. The Hofmann bromamide reaction removes the carbonyl carbon as carbonate, giving an amine with one carbon less:

C X_{6} H X_{5} CONH X_{2} C X_{6} H X_{5} NH X_{2}

, aniline.

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Benzene is treated with CHX3​Cl and anhydrous AlClX3​ to give A. A is heated with alkaline KMnOX4​ and then acidified to give B. B is converted to its amide C by heating with NHX3​ (via the ammonium salt). C is treated with BrX2​/KOH to give D. D is

Answer: B

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Benzene is treated with $CH X_{3} Cl$ and anhydrous $AlCl X_{3}$ to give A. A is heated with alkaline $KMnO X_{4}$ and then acidified to give B. B is converted to its amide C by heating with $NH X_{3}$ (via the ammonium salt). C is treated with $Br X_{2} / KOH$ to give D. D is