Question

Select the best combination of reagents which will bring the above conversion.

• A. $B{r}_2/FeB{r}_3$; $HN{O}_3/H_{2}S{O}_4$; $NaOH$; $Sn/HCl$
• B. $C{H}_{3}COCl/AlC{l}_3$; $B{r}_2/FeB{r}_3$; $NaB{H}_4/C{H}_{3}OH$
• C. $C{H}_{3}Cl/AlC{l}_3$; $KMn{O}_4$; $C{H}_{3}Cl/AlC{l}_3$; $NBS/hv$; $LiAl{H}_4/H_{2}O$; $NaOH$
• D. $C{H}_{3}Cl/AlC{l}_3$; $HN{O}_3/H_{2}S{O}_4$; $NBS/hv/NaOH$

Answer 1

The correct option is A $B{r}_2/FeB{r}_3$; $HN{O}_3/H_{2}S{O}_4$; $NaOH$; $Sn/HCl$

The best combination of reagents which will bring the conversion of
benzene into p amino phenol is $B{r}_2/FeB{r}_3$; $HN{O}_3/H_{2}S{O}_4$;
$NaOH$; $Sn/HCl$
Benzene is brominated with bromine in presence of ferric bromide to obtain bromo benzene. This is followed by nitration with nitric acid in presence of sulfuric acid to obtain p-nitrobromobenzene. The nucleophilic substitution of bromo group with hydroxyl group by using sodium hydroxide results in the formation of p-nitro phenol. the nitro group is then reduced to amino group by using tin and HCl to obtain 4-aminophenol.