An organic compound A on reacting with ${\mathrm{NH}}_3$ gives B. on heating B gives C. C in the presence of KOH reacts with ${\mathrm{Br}}_2$ to give ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{NH}}_2.$.A is

$\mathrm{A}\stackrel{{\mathrm{NH}}_3}{⟶}\mathrm{B}\stackrel{\Delta }{⟶}\mathrm{C}\stackrel{\mathrm{KOH}/\mathrm{Br}}{⟶}{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{NH}}_2$

The reaction of C with KOH/Br₂ to give amine is called Hofmann bromamide reaction. This reaction is given by acid amides only in which

group undergoes rearrangement along with the loss of CO₂ molecule. Thus, the compound C must be acid amide with three carbon atoms. Hence, the compound C is 

Since, all the options shows that A is an acid and it forms acid amide on reaction with NH₃. Thus, acid must contain three carbon atoms. Hence, the compound A is ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{COOH}$. The complete series of reaction can be represented as