The reaction of C_6H_5CH=CHCH_3 with HBr prod | vedclass

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Question

(C) The reaction follows Markovnikov's rule,where the electrophile $H^+$ adds to the carbon atom with more hydrogen atoms,and the nucleophile $Br^-$ a

Vedclass · Accepted Answer

$C_6H_5CH(Br)CH_2CH_3$

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(C) The reaction follows Markovnikov's rule,where the electrophile $H^+$ adds to the carbon atom with more hydrogen atoms,and the nucleophile $Br^-$ adds to the more substituted carbon atom.In the reaction of $C_6H_5CH=CHCH_3$ with $HBr$,the proton $H^+$ adds to the $CH$ group (adjacent to the methyl group) to form a more stable benzylic carbocation,$C_6H_5CH^+CH_2CH_3$.Subsequently,the bromide ion $Br^-$ attacks this carbocation to form the final product,$C_6H_5CH(Br)CH_2CH_3$.

The reaction of $C_6H_5CH=CHCH_3$ with $HBr$ produces

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