Question

$\text{Chlorobenzene}\underset{\Delta }{\overset{\text{conc}{H}_{2}S{O}_4,\text{conc}HN{O}_3}{\to }}A\underset{ii.H^{+}}{\overset{i.KOH,\Delta }{\to }}\left(B\right)$

$\left(B\right)\underset{\text{(1 equiv.)}}{\overset{B{r}_2/H_{2}O}{\to }}\left(C\right)\stackrel{Sn/HCl}{\to }\left(D\right)\underset{\text{(1 equiv.)}}{\overset{A{c}_{2}O}{\to }}\left(E\right)$

E will be:

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Product A
Nitration of chlorobenzene gives p-chloronitrobenzene as the the major product.

Product B
Nucleophilic aromatic subtitution of $Cl$ by $OH$ group gives p-chlorophenol.

Product C
Bromination occurs ortho to the $OH$ group since it is a activating group.

Product D
Reduction of nitro group with $Sn/HCl$

Product E
N is more nucleophilic than $O$ and hence acetylation occurs at the amino group.