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The reaction of $C_6H_5CH=CHCH_3$ with HBr produces
A
$C_6H_5CH(Br)CH_2CH_3$
B
$C_6H_5CH_2CH(Br)CH_3$
C
$C_6H_5CH_2CH_2CH_2Br$
D
$p\text{-}BrC_6H_4CH=CHCH_3$
Detailed Solution
$H^+$ adds to the terminal carbon of the double bond to give the more stable benzylic carbocation $C_6H_5\overset{+}{C}HCH_2CH_3$.
$Br^-$ then attacks this carbon.
Product: $C_6H_5CH(Br)CH_2CH_3$
$Br^-$ then attacks this carbon.
Product: $C_6H_5CH(Br)CH_2CH_3$