Which of the following reaction can be used to convert acetamide to methyl amine?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Hoffmann Bromamide reaction
In the Hoffmann bromamide degradation an amide is treated with bromine and aqueous (or ethanolic) sodium hydroxide. The alkyl group moves from the carbonyl carbon to nitrogen, and the product is a primary amine with one carbon less than the amide. Acetamide (two carbons) gives methylamine (one carbon):
\( \displaystyle \ce{CH3CONH2 + Br2 + 4NaOH -> CH3NH2 + Na2CO3 + 2NaBr + 2H2O} \)
Why the other options are incorrect
Option (2) Carbylamine reaction — Incorrect. A test for primary amines; the amine is heated with chloroform and alcoholic KOH to give a foul-smelling isocyanide. It starts from an amine, not an amide:
\( \displaystyle \ce{RNH2 + CHCl3 + 3KOH} \xrightarrow{\Delta} \ce{RNC + 3KCl + 3H2O} \)
Option (3) Gabriel phthalimide reaction — Incorrect. Makes a primary amine from an alkyl halide, with the same number of carbons as the alkyl halide. Phthalimide is converted to its potassium salt, which is alkylated and then hydrolysed:
\( \displaystyle \ce{C6H4(CO)2NH} \xrightarrow{\text{alc. }\ce{KOH}} \ce{C6H4(CO)2N^-K+} \)
\( \displaystyle \ce{C6H4(CO)2N^-K+ + R-X -> C6H4(CO)2N-R + KX} \)
\( \displaystyle \ce{C6H4(CO)2N-R} \xrightarrow{\ce{NaOH(aq)}} \ce{R-NH2 + C6H4(COONa)2} \)
Option (4) Stephen’s reduction — Incorrect. Reduces a nitrile to an aldehyde with \( \ce{SnCl2} \)/HCl followed by hydrolysis; it gives no amine:
\( \displaystyle \ce{RCN} \xrightarrow{\ce{SnCl2 + HCl}} \ce{RCH=NH} \xrightarrow{\ce{H3O+}} \ce{RCHO} \)
NCERT (2026-27 edition):
Hoffmann Bromamide Degradation Reaction – “Hoffmann developed a method for preparation of primary amines by treating an amide with bromine in an aqueous or ethanolic solution of sodium hydroxide. In this degradation reaction, migration of an alkyl or aryl group takes place from carbonyl carbon of the amide to the nitrogen atom. The amine so formed contains one carbon less than that present in the amide.”