The reaction of CH_3CH=CH-C_6H_4-OH with HBr | vedclass

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Question

(A) The reaction involves the electrophilic addition of $HBr$ to the alkene group $CH_3CH=CH-$. According to Markovnikov's rule,the $H^+$ ion attaches

Vedclass · Accepted Answer

$CH_3CHBr-CH_2-C_6H_4-OH$

Vedclass · Answer

(A) The reaction involves the electrophilic addition of $HBr$ to the alkene group $CH_3CH=CH-$. According to Markovnikov's rule,the $H^+$ ion attaches to the carbon atom with more hydrogen atoms,and the $Br^-$ ion attaches to the carbon atom with fewer hydrogen atoms. However,the presence of the $-OH$ group on the benzene ring makes the ring highly activated towards electrophilic substitution. In the presence of $HBr$,the alkene double bond undergoes addition,and the $-OH$ group is a strong ortho/para directing group. Given the structure $CH_3CH=CH-C_6H_4-OH$ (where the $-OH$ is at the para position relative to the propenyl group),the addition of $HBr$ to the double bond follows Markovnikov's rule to form $CH_3CHBr-CH_2-C_6H_4-OH$.

The reaction of $CH_3CH=CH-C_6H_4-OH$ with $HBr$ gives:

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