What is the product of the reaction of CH_3CH | vedclass

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(B) The reaction of $CH_3CH=CH-C_6H_4-OH$ with $HBr$ is an electrophilic addition reaction across the double bond.According to Markovnikov's rule,the

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$CH_3CH_2CHBr-C_6H_4-OH$

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(B) The reaction of $CH_3CH=CH-C_6H_4-OH$ with $HBr$ is an electrophilic addition reaction across the double bond.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 to form a more stable carbocation intermediate.The carbocation formed at the benzylic position (adjacent to the phenyl ring) is stabilized by resonance with the benzene ring.Therefore,the $Br^-$ attacks the carbon adjacent to the benzene ring,resulting in the product $CH_3CH_2CHBr-C_6H_4-OH$.

What is the product of the reaction of $CH_3CH=CH-C_6H_4-OH$ with $HBr$?

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