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(A) The reaction of $C_6H_5CH = CHCH_3$ with $HBr$ follows Markovnikov's rule. In this reaction,the electrophile $H^+$ attacks the double bond to form

Vedclass · Accepted Answer

$C_6H_5CH(Br)CH_2CH_3$

Vedclass · Answer

(A) The reaction of $C_6H_5CH = CHCH_3$ with $HBr$ follows Markovnikov's rule. In this reaction,the electrophile $H^+$ attacks the double bond to form the most stable carbocation. The carbocation formed is $C_6H_5CH^+ - CH_2CH_3$,which is a benzylic carbocation stabilized by resonance with the phenyl ring. Subsequently,the nucleophile $Br^-$ attacks this carbocation to form the product $C_6H_5CH(Br)CH_2CH_3$. Thus,the correct option is $A$.

What is the product formed by the reaction of $C_6H_5CH = CHCH_3$ with $HBr$?

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