An organic compound 'A'on treatment with $\mathrm{NH}_{3}$ gives ' $\mathrm{B}$ ' which on heating gives '…
An organic compound 'A'on treatment with $\mathrm{NH}_{3}$ gives ' $\mathrm{B}$ ' which on heating gives ' $\mathrm{C}^{\prime},{ }^{\circ} \mathrm{C}$ ' when treated with $\mathrm{Br}_{2}$ in the presence of KOH produces ethylamine. Compound 'A' is:
Since, $\mathrm{C}$ when heated with $\mathrm{Br}_{2}$ in presence of $\mathrm{KOH}$ produces ethylamine, hence it must be propanamide and hence the organic compound (A) will be propanoic acid. The reactions follows.
$\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{COOH} \stackrel{\mathrm{NH}_{3}}{\longrightarrow}$
$\quad$$\quad$$\quad$ (A)
