The reaction of CH_3-CH=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 $(CH_3-CH=CH-)$ attached to the phenol ring. According to Markovniko

Vedclass · Accepted Answer

$CH_3-CHBr-CH_2-C_6H_4-OH$

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(A) The reaction involves the electrophilic addition of $HBr$ to the alkene group $(CH_3-CH=CH-)$ attached to the phenol ring. 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. Therefore,the $Br$ atom will attach to the $CH$ group adjacent to the benzene ring. The product is $CH_3-CHBr-CH_2-C_6H_4-OH$.

The reaction of $CH_3-CH=CH-C_6H_4-OH$ with $HBr$ gives:

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