Consider the following reaction sequence:C_6H | vedclass

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

Question

(D) The reaction of styrene $(C_6H_5CH=CH_2)$ with $Br_2$ is an electrophilic addition reaction across the double bond.This results in the formation o

Vedclass · Accepted Answer

Styrene dibromide,benzoic acid

Vedclass · Answer

(D) The reaction of styrene $(C_6H_5CH=CH_2)$ with $Br_2$ is an electrophilic addition reaction across the double bond.This results in the formation of styrene dibromide $(C_6H_5CH(Br)CH_2Br)$,which is compound $A$.When styrene dibromide is treated with $KMnO_4$ (a strong oxidizing agent),the side chain is oxidized.Specifically,the alkyl group attached to the benzene ring is oxidized to a carboxylic acid group,resulting in the formation of benzoic acid $(C_6H_5COOH)$,which is compound $B$.Therefore,$A$ is styrene dibromide and $B$ is benzoic acid.

Consider the following reaction sequence:
$C_6H_5CH=CH_2$ $\xrightarrow{Br_2} A$ $\xrightarrow{KMnO_4} B$
Compound $A$ and $B$ respectively are:

Similar Questions

Product $(B)$ of the reaction is

Which of the following is not a mixture of hydrocarbons?

Write chemical equations for the combustion reaction of the following hydrocarbons:
$(i)$ Pentene
$(ii)$ Hexyne
$(iii)$ Toluene

Predict the product $(A)$ of the following reaction:

Complete oxidation of hydrocarbons always produces........... as products.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Consider the following reaction sequence:$C_6H_5CH=CH_2$ $\xrightarrow{Br_2} A$ $\xrightarrow{KMnO_4} B$Compound $A$ and $B$ respectively are: