



(1) $C{H_3}CON{H_2}$
(2) $C{H_3}C{H_2}N{H_2}$
(3) $ Ph-C{H_2}CON{H_2}$
In the above reaction, $3.9\, g$ of benzene on nitration gives $4.92\, g$ of nitrobenzene. The percentage yield of nitrobenzene in the above reaction is............. $\%$. (Round off to the Nearest Integer).
(Given atomic mass: $C : 12.0\, u , H : 1.0\, u$$O : 16.0\, u , N : 14.0\, u )$
$C{H_3}CN\xrightarrow{{Na + {C_2}{H_5}OH}}X\xrightarrow{{HN{O_2}}}Y\mathop {\xrightarrow{{{K_2}C{r_2}{O_7}}}}\limits_{{H_2}S{O_4}} Z$