Hofmann bromamide degradation method is used for preparation of
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Primary amines
The reaction is:
\( \displaystyle \ce{R-CONH2 + Br2 + 4NaOH -> R-NH2 + Na2CO3 + 2NaBr + 2H2O} \)
For example, ethanamide gives methanamine:
\( \displaystyle \ce{CH3CONH2 + Br2 + 4NaOH -> CH3NH2 + Na2CO3 + 2NaBr + 2H2O} \)
The nitrogen of the amide carries two hydrogens and receives only the one group R, so the product is always R–NH2, a primary amine.
Why the other options are incorrect
Option (2) Secondary amines — Incorrect. A secondary amine needs two carbon groups on nitrogen; in this reaction only the single group R moves to the nitrogen.
Option (3) Tertiary amines — Incorrect. Three carbon groups on nitrogen cannot arise from a \( \ce{-CONH2} \) group, which supplies one group R.
NCERT (2026-27 edition):
Hoffmann Bromamide Degradation Reaction
