$3 - P h e n y l$ propene on reaction with $H B r$ gives (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 attacking reagent adds to less hydrogenated carbon atom of the double bond<\span>

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3 -Phenyl propene

H B r

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The reaction of $C_{6} H_{5} C H = C H C H_{3}$ with HBr produces :

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