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NEET CHEMISTRYMedium

The reaction of C6H5CH=CHCH3\text{C}_6\text{H}_5\text{CH=CHCH}_3 with HBr\text{HBr} produces:

A

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B

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C

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D

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Step-by-Step Solution

The reaction proceeds via an electrophilic addition mechanism. The proton (H+\text{H}^+) from HBr\text{HBr} attacks the alkene double bond to form the most stable carbocation intermediate. Between the two possible carbocations, C6H5C+HCH2CH3\text{C}_6\text{H}_5\text{C}^+\text{HCH}_2\text{CH}_3 (a benzylic secondary carbocation) is significantly more stable than C6H5CH2C+HCH3\text{C}_6\text{H}_5\text{CH}_2\text{C}^+\text{HCH}_3 (a standard secondary carbocation) because of strong resonance stabilization from the adjacent phenyl ring. The bromide ion (Br\text{Br}^-) then attacks this stable benzylic carbocation to form 1-bromo-1-phenylpropane (C6H5CH(Br)CH2CH3\text{C}_6\text{H}_5\text{CH(Br)CH}_2\text{CH}_3) as the major product. Since the exact structural options are missing from the input text, the precise correct option cannot be visually verified.

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