${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{Br}\stackrel{\mathrm{AgCN}}{\to }\mathrm{A}\stackrel{{\mathrm{H}}_3{\mathrm{O}}^{+}}{\to }\mathrm{HCOOH}+\mathrm{B}\stackrel{{\mathrm{CHCl}}_3/\mathrm{KOH}}{\to }\mathrm{A}\stackrel{\mathrm{reduction}}{\to }\mathrm{C}$ (on reduction)

$\mathrm{A},\mathrm{B},\mathrm{C}$ respectively in the above sequence are

A is carbyl aminoethane, $\mathrm{B}$ is ethanamine, and $\mathrm{C}$ is ethyl methyl amine.

 Ethyl bromide in reaction with silver cyanide gives carbyl amino ethane, which upon further acidic hydrolysis gives $\text{HCOOH}$ and ethanamine. Ethanamine in addition to $\mathrm{KOH}$ and chloroform gives carbyl amino ethane which on further reduction gives ethyl methyl amine.