The reaction of CH_3-CH=CH-C_6H_4-OH with HBr | vedclass

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Question

(B) The reaction involves the electrophilic addition of $HBr$ to the alkene group $(CH_3-CH=CH-)$ attached to the benzene ring.According to Markovniko

Vedclass · Accepted Answer

$CH_3-CH_2-CHBr-C_6H_4-OH$

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(B) The reaction involves the electrophilic addition of $HBr$ to the alkene group $(CH_3-CH=CH-)$ attached to the benzene ring.According to Markovnikov's rule,the proton $(H^+)$ adds to the carbon atom with more hydrogen atoms,and the bromide ion $(Br^-)$ adds to the more substituted carbon atom.In the molecule $CH_3-CH=CH-C_6H_4-OH$,the double bond is between $C_2$ and $C_3$ of the propenyl chain.The carbocation formed at the benzylic position $(CH_3-CH_2-CH^+-C_6H_4-OH)$ is more stable due to resonance with the benzene ring.Therefore,the $Br^-$ attacks the benzylic carbon,resulting in the product $CH_3-CH_2-CHBr-C_6H_4-OH$.

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

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