
$(1)$ $C{H_3}C{H_2}N{H_2}$ $(2)$ $\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_2}C{H_3}}\\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,}\\
{\,C{H_3}C{H_2}NH\,\,\,\,\,\,\,}
\end{array}$
$(3)$ $\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,C{H_3}}\\
{\,|}\\
{{H_3}C - N - C{H_3}}
\end{array}$ $(4)$ $\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,C{H_3}}\\
{\,|}\\
{Ph - N - H}
\end{array}$
${C_6}{H_5} - N{O_2} + 6[H] \to {C_6}{H_5} - N{H_2} + 2{H_2}O$
The reducing agent used in this reaction is …….

${C_6}{H_5}N{H_2}\mathop {\xrightarrow{{NaN{O_2}/HCl}}}\limits_{0 - 5\,^o C} X\mathop {\xrightarrow{{HN{O_2}}}}\limits_{C{H_2}O} Y + {N_2} + HCl$
$X$ and $Y$ are respectively