JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Aldehydes Ketones and Carboxylic Acids

In the Hofmann bromamide degradation a primary amide is treated with bromine and alkali and loses one carbon as carbonate, turning $\mathrm{R}-\mathrm{CONH}{}_2$ into the amine $\mathrm{R}-\mathrm{NH}{}_2$. Propanamide therefore gives ethanamine: $\mathrm{CH}{}_3\mathrm{CH}{}_2\mathrm{CONH}{}_2+\mathrm{Br}{}_2+4\mathrm{KOH}\to \mathrm{CH}{}_3\mathrm{CH}{}_2\mathrm{NH}{}_2+2\mathrm{KBr}+\mathrm{K}{}_2\mathrm{CO}{}_3+2\mathrm{H}{}_2\mathrm{O}$. The other routes do not hold: Gabriel\textquoteright s phthalimide synthesis does not work with an aryl halide, so it cannot give aniline; reduction of benzyl isocyanide gives a secondary amine ($\mathrm{PhCH}{}_2\mathrm{NHCH}{}_3$), not benzylamine; and nitration of the carbanilate directs to the ring bearing the $-\mathrm{NH}-$ group rather than to the far ring. The correct reaction is the Hofmann degradation of propanamide to ethanamine.