Question

In the reaction: $A\stackrel{Zn/\Delta }{\to }B\stackrel{H_{3}P{O}_2}{\leftarrow }C$, if $B$ is benzene, then $C$ can be converted into $A$ by using:

• A. $H_{2}O,283K$
• B. ${NaNO}_2+HCl,0-5^{o}C$
• C. $C_2{H}_{5}OH,273K$
• D. ${HBF}_4,0-5^{o}C$

Answer 1

The correct option is A $H_{2}O,283K$

The reactions are:
$C_6{H}_{5}OH\stackrel{Zn/\Delta }{\to }{C}_6{H}_6\stackrel{H_{3}P{O}_2}{\leftarrow }{C}_6{H}_5{N}_2^{+}$
Now,
$C_6{H}_5{N}_2^{+}\stackrel{H_{2}O,283K}{\to }{C}_6{H}_{5}OH$