
$C{H_3}CN\xrightarrow{{Na + {C_2}{H_5}OH}}X\xrightarrow{{HN{O_2}}}Y\mathop {\xrightarrow{{{K_2}C{r_2}{O_7}}}}\limits_{{H_2}S{O_4}} Z$
$\begin{array}{*{20}{c}}
{\,\,\,\,\,C{H_3}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} O{\mkern 1mu} } \\
{\,\,\,|{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,\,\,{\mkern 1mu} {\mkern 1mu} ||} \\
{C{H_3} - CH - C - OH}
\end{array}$ $+ C{H_3} - N{H_2} \to 'A'\xrightarrow[\Delta ]{}'B'\xrightarrow{\begin{subarray}{l}
{\text{LiAl}}{{\text{H}}_4} \\
{\text{(excess)}}
\end{subarray} }'C'$
The final product $‘C’$ will be
$(A)$ $\begin{matrix}
O\,\,\,\,\,\,\,\,\,\, \\
||\,\,\,\,\,\,\,\,\,\,\,\, \\
R-C-NHBr \\
\end{matrix}$ $(B)$ $R-NH-Br$
$(C)$ $R-N=C=O$ $(D)$ $\begin{matrix}
\,\,\,\,O\,\,\,\,\,\,\,\,\,\, \\
\,\,\,\,\,\,||\,\,\,\,\,\,\,\,\,\,\,\, \\
R-C-NB{{r}_{2}} \\
\end{matrix}$