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

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

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Solution

The correct option is B
Addition of HBr to alkene:

Benzyl carbocation is a stable intermediate, hence, formed product is option (d).

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Similar questions

C_{6} H_{5} C H = {C H C H}_{3}

H B r

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

K M n O_{4} / H O^{-} / H e a t

C_{6} H_{5} - C H O + B r - C H_{2} - C O O C_{2} H_{5} (i) Z n - -------------- \to (i i) H O H / N H_{4} C l (i i i) Δ

C_{6} H_{5} - C H = C H - C O O C_{2} H_{5}

C_{6} H_{5} \oplus C H_{2}

C_{6} H_{5} C H_{2} \oplus C H_{2}

C_{6} H_{5} \oplus C H C H_{3}

C_{6} H_{5} \oplus C (C H_{3})_{2}

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