

$\begin{array}{*{20}{c}}
{\,\,\,\,\,C{H_3}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} O{\mkern 1mu} } \\
{\,\,\,|{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,\,\,{\mkern 1mu} {\mkern 1mu} ||} \\
{C{H_3} - CH - C - OH}
\end{array}$ $+ C{H_3} - N{H_2} \to 'A'\xrightarrow[\Delta ]{}'B'\xrightarrow{\begin{subarray}{l}
{\text{LiAl}}{{\text{H}}_4} \\
{\text{(excess)}}
\end{subarray} }'C'$
The final product $‘C’$ will be
${C_6}{H_5} - N{H_2}\xrightarrow[{HCl}]{{NaN{O_2}}}A\xrightarrow{{CuCN}}B\xrightarrow{{{H_2}/Ni}}C\xrightarrow{{HN{O_2}}}D$
product $(D)$ would be


