The intermediate during the addition of HCl to propene in the presence of peroxide is

\cdot C H_{3} C H C H_{2} C l

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+ C H_{3} C H C H_{3}

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C H_{3} C H_{2} \cdot C H_{2}

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C H_{3} C H_{2} + C H_{2}

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Solution

The correct option is B $+ C H_{3} C H C H_{3}$

Anti markonvnikov's addition prevails here, due to peroxide. Thus, the product will be propane.

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