How many of the following amines react with ethanoyl chloride (CH3COCl) to form an amide?
Ethanamine, N-Methylethanamine, N,N-Dimethylaniline, Aniline, N,N-Diethylethanamine, N-Methylaniline, Propan-2-amine
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 5
Acylation replaces a hydrogen atom of the –NH2 or >N–H group by an acyl group, and the product is an amide. So both primary and secondary amines, aliphatic and aromatic, are acylated. A tertiary amine has no hydrogen on nitrogen and gives no amide.
\( \displaystyle \ce{R-NH2 + CH3COCl} \xrightarrow{\text{Base}} \ce{R-NHCOCH3 + HCl} \)
\( \displaystyle \ce{R2NH + CH3COCl} \xrightarrow{\text{Base}} \ce{R2NCOCH3 + HCl} \)
Form an amide:
Ethanamine (primary amine)
N-Methylethanamine (secondary amine)
Aniline (primary amine; gives acetanilide)
N-Methylaniline (secondary amine)
Propan-2-amine (primary amine; one alkyl group on nitrogen)
No amide:
N,N-Dimethylaniline — tertiary amine; no hydrogen on nitrogen.
N,N-Diethylethanamine — tertiary amine; no hydrogen on nitrogen.
Form an amide: 5 amines.
NCERT (2026-27 edition):
9.6 Chemical Reactions – “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.”