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

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(B) The reaction of $CH_3-CH=CH-C_6H_4-OH$ with $HBr$ follows Markovnikov's rule.In this reaction,the proton $(H^+)$ adds to the carbon atom of the do

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$CH_3-CH_2-CHBr-C_6H_4-OH$

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(B) The reaction of $CH_3-CH=CH-C_6H_4-OH$ with $HBr$ follows Markovnikov's rule.In this reaction,the proton $(H^+)$ adds to the carbon atom of the double bond that has more hydrogen atoms,and the bromide ion $(Br^-)$ adds to the more substituted carbon atom.The carbocation formed at the benzylic position is more stable due to resonance with the benzene ring.Therefore,the product is $CH_3-CH_2-CHBr-C_6H_4-OH$.

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

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