JEE Main Chemistry question

*p*-Toluamide (CH₃–C₆H₄–CONH₂) is treated with (1) NaOH/Br₂ and then (2) benzoyl chloride (C₆H₅COCl). The product is

(A)*p*-CH₃–C₆H₄–C(=O)–O–C(=O)–C₆H₅ (an anhydride)
(B)C₆H₅–NH–CO–C₆H₄–CH₃ (wrong amine partner)
(C)*p*-CH₃–C₆H₄–NH–CO–C₆H₅Correct
(D)*p*-CH₃–C₆H₄–CO–NH–CO–C₆H₅ (an imide)

Answer: (C)

Solution

NaOH/Br₂ effects a Hofmann bromamide degradation on the primary amide, converting it to the one-carbon-shorter primary amine, *p*-toluidine (CH₃–C₆H₄–NH₂). This

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Question ID VK-140291