JEE Advanced 2014 · Paper 1 · Chemistry Q23

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ChemistryMultiple correct+3Question 23 of 60

In the reaction shown below, the major product(s) formed is/are

3-(Aminomethyl)benzamide treated with acetic anhydride in CH₂Cl₂ gives product(s).

N-(3-carbamoylbenzyl)acetamide (the CH₂NH₂ acetylated, the CONH₂ unchanged) + CH₃COOH.

3-(Aminomethyl)-N-acetylbenzamide (the amide N acetylated, CH₂NH₂ free) + CH₃COOH.

Both nitrogens acetylated (CH₂NHCOCH₃ and CONHCOCH₃) + H₂O.

The ammonium acetate salt ⁺NH₃CH₂- with the amide N acetylated (CONHCOCH₃), counter-ion CH₃COO⁻.

Answer and worked solution

Official answer: A

Nucleophilicity. The benzylic $\ce{CH2NH2}$ is a basic, nucleophilic amine. The amide nitrogen of $\ce{CONH2}$ has its lone pair delocalised into the carbonyl and is barely nucleophilic.

Reaction. Acetic anhydride therefore acylates only the amine, giving $\ce{CH3CONHCH2-C6H4-CONH2}$ (N-(3-carbamoylbenzyl)acetamide) and acetic acid as the by-product.

(A)

Topics:Amines Acid derivatives

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JEE Advanced 2014 · Paper 1, Chemistry question 23