Which product is formed on reaction between ethanamine and ethanoic anhydride?
A. \(CH_3CH_2NHCOCH_3\)
B. \(CH_3NHCOC_2H_5\)
C. \(CH_3COOH\)
D. \(C_2H_5COOH\)
E. \(CH_3CH_2NHCOC_2H_5\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A and C only
Amines react with acid anhydrides by acylation: one H of –\( \ce{NH2} \) is replaced by the acyl group. Ethanoic anhydride hands over one acetyl group, and the other half leaves as ethanoic acid:
\( \displaystyle \ce{CH3CH2NH2 + (CH3CO)2O -> CH3CH2NHCOCH3 + CH3COOH} \)
The products are N-ethylethanamide and ethanoic acid.
(A) \( \ce{CH3CH2NHCOCH3} \) — Correct. The ethyl group of the amine carries the acetyl group of the anhydride.
(B) \( \ce{CH3NHCOC2H5} \) — Incorrect. This is N-methylpropanamide; it would need methylamine and a propanoyl reagent.
(C) \( \ce{CH3COOH} \) — Correct. The second acetyl half of the anhydride leaves as ethanoic acid.
(D) \( \ce{C2H5COOH} \) — Incorrect. Propanoic acid would come from propanoic anhydride, not ethanoic anhydride.
(E) \( \ce{CH3CH2NHCOC2H5} \) — Incorrect. N-ethylpropanamide carries a propanoyl group, which ethanoic anhydride cannot supply.
NCERT (2026-27 edition):
Acylation – “Aliphatic and aromatic primary and secondary amines react with acid chlorides, anhydrides and esters by nucleophilic substitution reaction. This reaction is known as acylation. You can consider this reaction as the replacement of hydrogen atom of –NH2 or >N–H group by the acyl group. The products obtained by acylation reaction are known as amides.”