The reaction shown above is known as
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Hoffmann bromamide degradation reaction
The reaction shown, an amide with \( \ce{Br2} \) and NaOH giving a primary amine and sodium carbonate, is the Hoffmann bromamide degradation. The amine has one carbon atom less than the amide; for example, acetamide gives methylamine:
\( \displaystyle \ce{CH3CONH2 + Br2 + 4NaOH -> CH3NH2 + Na2CO3 + 2NaBr + 2H2O} \)
Why the other options are incorrect
Option (1) Gabriel phthalimide synthesis — Incorrect. It makes primary amines from alkyl halides: phthalimide is turned into its potassium salt, alkylated, and hydrolysed. The amine keeps all the carbons of the alkyl group.
\( \displaystyle \ce{C6H4(CO)2NK + R-X -> C6H4(CO)2NR + KX} \)
\( \displaystyle \ce{C6H4(CO)2NR + 2NaOH -> R-NH2 + C6H4(COONa)2} \)
Option (2) Gattermann reaction — Incorrect. It replaces the diazonium group of an arenediazonium salt by Cl or Br, using the halogen acid and copper powder.
\( \displaystyle \ce{ArN2+X-} \xrightarrow{\ce{Cu/HBr}} \ce{ArBr + N2 + CuX} \)
Option (3) Carbylamine reaction — Incorrect. It is a test for primary amines: on heating with chloroform and alcoholic KOH they give foul-smelling isocyanides.
\( \displaystyle \ce{R-NH2 + CHCl3 + 3KOH} \xrightarrow{\Delta} \ce{R-NC + 3KCl + 3H2O} \)
NCERT (2026-27 edition):
Hoffmann Bromamide Degradation Reaction
