$3 -Phenyl propene$ on reaction with $H B r$ 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 Markovnikov's rule, the negative part of the unsymmetrical reagent adds to less hydrogenated (more substituted) carbon atom of the double bond.

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