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Discussion

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

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

  • A.3-(Acetamidomethyl)benzamide with the CONH2 group intact, plus CH3COOH
  • B.3-(Aminomethyl)-N-acetylbenzamide with a free CH2NH2 group, plus CH3COOH
  • C.Benzene ring with an acetamidomethyl group and an N-acetylcarboxamide group, plus H2O
  • D.N-Acetylbenzamide with a CH2NH3+ group and CH3COO- as the acetate salt

Answer: A

The major product is (A): the CH2NH2 group is acetylated, giving CH2NHCOCH3, with acetic acid as by-product.

Acetic anhydride acylates the most nucleophilic nitrogen. The benzylic amine nitrogen has a free lone pair and is strongly nucleophilic. The amide nitrogen of CONH2 is not: its lone pair is delocalised into the C=O group, so it does not react under these mild conditions. One equivalent of acetic anhydride therefore gives the N-acetylated amine and CH3COOH.

(B) acetylates the wrong nitrogen. (C) acetylates both and also gives water, which is not a by-product of acylation by an anhydride. (D) again acetylates the amide nitrogen while leaving the more basic amine as a salt.

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