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

C_{6} H_{5} C H_{2} C H (B r) C H_{3}

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C_{6} H_{5} C H (B r) C H_{2} C H_{3}

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C_{6} H_{5} C H_{2} C H_{2} C H_{2} B r

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C_{6} H_{5} C H (B r) C H = C H_{2}

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Solution

The correct option is B $C_{6} H_{5} C H (B r) C H_{2} C H_{3}$
According to Markownikoff's rule, the negative part of the unsymmetrical reagent adds to less hydrogenated (more substituted) carbon atom of the double bond.
$B r | C H C H = C H - C H + H B r \to C_{6} H_{5} C H C H_{2} C H_{3}$

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