3-Phenylpropene on reaction with HBr gives (as a major product)

A
$C_{6} H_{5} {CH}_{2} CH (Br) {CH}_{3}$
B
$C_{6} H_{5} CH (Br) {CH}_{2} {CH}_{3}$
C
$C_{6} H_{5} {CH}_{2} {CH}_{2} {CH}_{2} Br$
D
$C_{6} H_{5} CH (Br) CH = {CH}_{2}$

Solution:

According to Markownikoffs rule. the negative part of the unsymmetrical reagent adds to less hydrogenated (more substituted) carbon atom of the double bond.

$C_{6} H_{5} CH = CH - {CH}_{3} + HBr \to C_{6} H_{5} \overset{}{CH (B r) {CH}_{2}} {CH}_{3}$

3-Phenylpropene on reaction with $HBr$ gives (as a major product)

Solution:

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