Methyl amine reacts with acetyl chloride and forms 

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

$\mathrm{A}$ is :-

The correct set of the products obtained in the following reactions

$\mathrm{RCN}\stackrel{\text{ reduction }}{⟶}\left(\mathrm{A}\right),\mathrm{RCN}\frac{\left(\mathrm{i}\right){\mathrm{CH}}_3\mathrm{MgBr}}{\left(\mathrm{ii}\right){\mathrm{H}}_2\mathrm{O}}\to \left(\mathrm{B}\right),$

$\mathrm{RNC}\stackrel{\mathrm{hydrolysis}}{\to }\left(\mathrm{C}\right), {\mathrm{RNH}}_2\stackrel{{\mathrm{HNO}}_2}{\to }\left(\mathrm{D}\right)$

the answer is 

The final product $\mathrm{C}$ Obtained in this reaction would be

${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{Br}\stackrel{\mathrm{AgCN}}{\to }\mathrm{A}\stackrel{{\mathrm{H}}_3{\mathrm{O}}^{+}}{\to }\mathrm{HCOOH}+\mathrm{B}; \mathrm{B}\underset{\mathrm{KOH}}{\overset{{\mathrm{CHCl}}_3}{\to }}\mathrm{A}\stackrel{\mathrm{Reduction}}{\to }\mathrm{C}$

A,B,C respectively in the above sequence are:

$\mathrm{B}$ and $\mathrm{C}$ in the above sequence are

The end-product in the reaction sequence would  be :

Ethyl amine $\stackrel{\mathrm{HNO}}{\to }\mathrm{A}\stackrel{{\mathrm{PCl}}_5}{\to }\mathrm{B}\stackrel{{\mathrm{NH}}_3}{\to }\mathrm{C}$

$\mathrm{A}\stackrel{{\mathrm{H}}_2\mathrm{NOH} }{\to }\mathrm{B}\stackrel{\mathrm{reduction}}{\to }\mathrm{C}\stackrel{\mathrm{NOCl}}{\to }{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Cl},$

$\mathrm{A}$ and $\mathrm{C}$ in the above sequence respectively are:-