Question

\( \mathrm{CH_3-NH_2} \) on reaction with \( \mathrm{(CH_3CO)_2O} \) gives

• A. \( \mathrm{CH_3CONH_2} \)
• B. \( \mathrm{CH_3COONHCH_3} \)
• C. \( \mathrm{CH_3-NH-C(=O)-CH_3} \)
• D. \( \mathrm{CH_3-C(=O)-CH_2-NH_2} \)

Hint:

A primary amine with an acid anhydride undergoes acylation to form an N-substituted amide plus acetic acid. Methylamine gives N-methylacetamide.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Primary and secondary amines react with acid chlorides and acid anhydrides by nucleophilic substitution.
This reaction is called acylation. The hydrogen on nitrogen is replaced by an acyl group.

Step 2: Identify the reagents:
Methylamine, \( \mathrm{CH_3NH_2} \), has a lone pair on nitrogen, so it is a nucleophile.
Acetic anhydride, \( \mathrm{(CH_3CO)_2O} \), is the acylating agent.

Step 3: Write the reaction:
The nitrogen attacks one carbonyl carbon of the anhydride. The acetate group leaves as acetic acid.
\[ \mathrm{CH_3NH_2 + (CH_3CO)_2O \longrightarrow CH_3NHCOCH_3 + CH_3COOH} \]
The product is N-methylethanamide (N-methylacetamide).

Step 4: Why the other options are wrong:
(A) \( \mathrm{CH_3CONH_2} \) would need the methyl group on nitrogen to be lost. It is not lost.
(B) \( \mathrm{CH_3COONHCH_3} \) joins the acyl group to oxygen of an N-O bond. No such bond exists in the reactants.
(D) is an amino ketone. It needs a new C-C bond, but an amine only forms a new C-N bond here.

Final Answer:
The product is \( \mathrm{CH_3-NH-CO-CH_3} \), option (C).
\[ \boxed{\mathrm{CH_3NHCOCH_3}} \]