In the following reaction,the compound C is:{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Step $1$: Toluene $({C_7}{H_8})$ reacts with $3Cl_2$ in the presence of $\Delta$ (heat) to undergo side-chain chlorination,forming benzotrichlorid

Vedclass · Accepted Answer

$m-$bromotoluene

Vedclass · Answer

(C) Step $1$: Toluene $({C_7}{H_8})$ reacts with $3Cl_2$ in the presence of $\Delta$ (heat) to undergo side-chain chlorination,forming benzotrichloride $({C_6}{H_5}CCl_3)$ as compound $A$.Step $2$: Benzotrichloride $({C_6}{H_5}CCl_3)$ contains a $-CCl_3$ group,which is a strongly deactivating and meta-directing group. Reaction with $Br_2/Fe$ leads to electrophilic aromatic substitution at the meta position,forming $m-$bromobenzotrichloride as compound $B$.Step $3$: Reduction of the $-CCl_3$ group using $Zn/HCl$ converts it back into a methyl $(-CH_3)$ group. Thus,$m-$bromobenzotrichloride is reduced to $m-$bromotoluene,which is compound $C$.

In the following reaction,the compound $C$ is:
${C_7}{H_8}$ $\xrightarrow[\Delta]{3Cl_2} A$ $\xrightarrow{Br_2/Fe} B$ $\xrightarrow{Zn/HCl} C$

Similar Questions

What is the product $X$ in the following Wurtz-Fittig reaction?

Consider the following compounds:
$a$: Chlorobenzene
$b$: Nitrobenzene
$c$: Anisole
Arrange these compounds in the increasing order of reactivity towards electrophilic aromatic substitution (nitration).

Oxidation of toluene with dilute $HNO_3$ gives:

When $C_6H_5Cl$ is heated with $NaOH$ at $300 \, ^\circ C$ under pressure,phenol is obtained. However,the yield is low because a side reaction produces ...............

Which of the following is the correct order of electrophilic substitution for the given compounds?
$(i)$ Toluene
(ii) Isopropylbenzene
(iii) tert-Butylbenzene
(iv) Ethylbenzene

Vedclass Test Series

Exam Paper Generator

Online Exam Module

In the following reaction,the compound $C$ is:${C_7}{H_8}$ $\xrightarrow[\Delta]{3Cl_2} A$ $\xrightarrow{Br_2/Fe} B$ $\xrightarrow{Zn/HCl} C$