Question

Identify the total number of atoms in product $Q$ in the given sequence of reaction
$C_6{H}_{5}N{H}_2\stackrel{NaN{O}_2+HCl}{\to }X\stackrel{C{u}_2(CN{)}_2}{\to }Y\stackrel{H_{2}O/H^{+}}{\to }Z$

$\stackrel{NaOH/CaO,\Delta }{\to }P\stackrel{C{l}_2,\text{ hvExcess }}{\to }Q$

Answer 1

Answer: 18

$C_6{H}_{5}N{H}_2\stackrel{NaN{O}_2/HCl}{\to }{C}_6{H}_5{N}_2^{+}C{l}^{-}\stackrel{C{u}_2(CN{)}_2}{\to }{C}_6{H}_5$
$CN\stackrel{H_{2}O/H^{+}}{\to }{C}_6{H}_{5}COOH$
$C_6{H}_{5}COOH\stackrel{NaOH/CaO,\Delta }{\to }{C}_6{H}_6\stackrel{C{l}_2,hv\text{ (Excess) }}{\to }{C}_6{H}_{6}C{l}_6$
It is diazotisation reaction to produce $X$, then sandmayer's reaction to give $Y$, followed by hydrolysis to produce benzoic acid $(Z)$. Benzoic acid on decarboxylation gives benzene, which on excess photochlorination gives benzene hexachloride.