The Hoffmann bromamide degradation reaction is used for conversion of
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Amide to amine
In the Hofmann bromamide degradation, an amide is treated with bromine in an aqueous or ethanolic solution of sodium hydroxide. It gives a primary amine with one carbon atom less:
\( \displaystyle \ce{R-CONH2 + Br2 + 4NaOH -> R-NH2 + Na2CO3 + 2NaBr + 2H2O} \)
For example, ethanamide gives methanamine:
\( \displaystyle \ce{CH3CONH2 + Br2 + 4NaOH -> CH3NH2 + Na2CO3 + 2NaBr + 2H2O} \)
So the reaction converts an amide to an amine.
Why the other options are incorrect
Option (1) Nitro to amine — Incorrect. A nitro compound is converted to an amine by reduction, with hydrogen over a metal catalyst or with Sn or Fe and HCl:
\( \displaystyle \ce{C6H5NO2 + 3H2} \xrightarrow{\ce{Pd}} \ce{C6H5NH2 + 2H2O} \)
Option (3) Cyanide to amine — Incorrect. A nitrile is reduced to an amine with \( \ce{LiAlH4} \) or \( \ce{H2} \)/Ni; this keeps all the carbon atoms, so the amine has one carbon more than the Hofmann product from the corresponding amide:
\( \displaystyle \ce{R-C#N + 2H2} \xrightarrow{\ce{Ni}} \ce{R-CH2-NH2} \)
Option (4) Amine to diazonium salt — Incorrect. This is diazotisation, done with sodium nitrite and hydrochloric acid at 273–278 K:
\( \displaystyle \ce{C6H5NH2 + NaNO2 + 2HCl} \xrightarrow{273-278\ \text{K}} \ce{C6H5N2+Cl- + NaCl + 2H2O} \)
NCERT (2026-27 edition):
Hoffmann Bromamide Degradation Reaction
