IIT-JEE 2010 · Paper 2 · Chemistry Q3

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ChemistrySingle correct+5 / -2Question 3 of 57

In the reaction (4-methylbenzamide) $\xrightarrow[\text{(2) } \ce{C6H5COCl}]{\text{(1) } \ce{NaOH/Br2}}$ T, the structure of the Product T is

4-Methylbenzamide (H₃C-C₆H₄-CONH₂) treated with (1) NaOH/Br₂ and then (2) benzoyl chloride (C₆H₅COCl) to give T.

4-Methylbenzoic benzoic anhydride: H₃C-C₆H₄-CO-O-CO-C₆H₅.

N-Phenyl-4-methylbenzamide: C₆H₅-NH-CO-C₆H₄-CH₃ (para).

N-(4-Methylphenyl)benzamide: H₃C-C₆H₄-NH-CO-C₆H₅ (para).

N-Benzoyl-4-methylbenzamide (an imide): H₃C-C₆H₄-CO-NH-CO-C₆H₅.

Answer and worked solution

Official answer: C

Step 1, Hofmann bromamide reaction. $\ce{NaOH/Br2}$ converts the amide into an amine with one carbon fewer:
$\ce{CH3-C6H4-CONH2 -> CH3-C6H4-NH2}\ (\text{4-methylaniline}).$

Step 2, benzoylation. Benzoyl chloride acylates the amine: $\ce{CH3-C6H4-NH-CO-C6H5}$ , N-(4-methylphenyl)benzamide.

(C)

Topics:Amines Acid derivatives

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IIT-JEE 2010 · Paper 2, Chemistry question 3