JEE Main 2013 — Carboxylic Acid Derivatives Question with Solution

Ammonia reacts with carboxylic acid to form ammonium carboxylate (B). Which on heating form amide (C). Amide on reaction with bromine and KOH undergo Hoffmann Bromamide reaction to form amine with one carbon less.

$$\begin{gathered} \begin{array}{c}{\mathrm{CH}}_3-{\mathrm{CH}}_2-\stackrel{\underset{\left|\right|}{\mathrm{O}}}{\mathrm{C}}-\mathrm{OH}+{\mathrm{NH}}_3\\ \left(\mathrm{A}\right)\end{array}\stackrel{}{\to }\begin{array}{c}{\mathrm{CH}}_3{\mathrm{CH}}_2-\stackrel{\underset{\left|\right|}{\mathrm{O}}}{\mathrm{C}}-{\mathrm{O}}^{\ominus }\\ \left(\mathrm{B}\right)\end{array}\stackrel{\oplus }{\mathrm{N}}{\mathrm{H}}_4 \\ \begin{array}{c}{\mathrm{CH}}_3{\mathrm{CH}}_2-\stackrel{\underset{\left|\right|}{\mathrm{O}}}{\mathrm{C}}-{\mathrm{O}}^{\ominus }\\ \left(\mathrm{B}\right)\end{array}\stackrel{\oplus }{\mathrm{N}}{\mathrm{H}}_4\stackrel{\mathrm{\Delta }}{\to }\begin{array}{c}{\mathrm{CH}}_3{\mathrm{CH}}_2-\stackrel{\underset{\left|\right|}{\mathrm{O}}}{\mathrm{C}}-{\mathrm{NH}}_2\\ \left(\mathrm{C}\right)\end{array} \\ \begin{array}{c}{\mathrm{CH}}_3{\mathrm{CH}}_2-\stackrel{\underset{\left|\right|}{\mathrm{O}}}{\mathrm{C}}-{\mathrm{NH}}_2\\ \left(\mathrm{C}\right)\end{array}\underset{(\mathrm{Hoffmann} \mathrm{bromamide} \mathrm{degradation})}{\overset{{\mathrm{Br}}_2/\mathrm{KOH}}{\to }}{\mathrm{CH}}_3{\mathrm{CH}}_2-{\mathrm{NH}}_2 \end{gathered}$$