Question

The final product (IV) obtained in the reaction sequence :
Toluene $\underset{1.KMn{O}_4/{OH}^{-}}{\overset{{2.H}^{+}}{\to }}I\stackrel{SOC{l}_2}{⟶}II\stackrel{N{H}_3}{⟶III}\stackrel{{OBr}^{-}}{⟶}IV$
is :

• A. $C_6{H}_5{CONH}_2$
• B. $P-{{CH}_{3}C}_6{H}_4{NO}_2$
• C. $C_6{H}_5{CH}_2{NH}_2$
• D. $C_6{H}_5{NH}_2$

Answer 1

Answer: (D)

$C_6{H}_{5}C{H}_3\underset{2H^{+}}{\overset{KMn{O}_4/O{H}^{-}}{\to }}{C}_6{H}_{5}COOH\stackrel{SOC{l}_2}{\to }{C}_6{H}_{5}COCl\stackrel{N{H}_3}{\to }{C}_6{H}_{5}CON{H}_2\stackrel{OB{r}^{-}}{\to }{C}_6{H}_{5}N{H}_2$

First, Toluene undergoes oxidation to form benzoic acid which on reaction with $SOC{l}_2$ forms benzoyl chloride.

This benzoyl chloride reacts with $N{H}_3$ to form the corresponding amide.

This amide on reaction with (hoffmann bromamide) $OB{r}^{-}$ forms Aniline as a product.-