Chemistry - Carboxylic acid and their Derivatives Question with Solution | TestHub

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.

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