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(B) The reaction proceeds via the formation of a carbocation intermediate. When $H^+$ adds to the double bond,it forms a carbocation at the benzylic p

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$CH_3CH_2CH(Br)-C_6H_4-OH$

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(B) The reaction proceeds via the formation of a carbocation intermediate. When $H^+$ adds to the double bond,it forms a carbocation at the benzylic position,which is stabilized by resonance due to the electron-donating $-OH$ group at the para position. The carbocation $I$ $(CH_3-CH_2-CH^+-C_6H_4-OH)$ is more stable than other possible carbocations. Subsequently,the $Br^-$ ion attacks this stable carbocation to form the final product: $CH_3-CH_2-CH(Br)-C_6H_4-OH$.

What will be the product of the reaction with $HBr$?

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