Propene reacts HCl to form ‘P’ (major product). Propene reacts with HCl in presence of Benzoyl peroxide to form ‘Q’ (major product). P and Q are

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$({CH}_{3})_{2} CHCl, {CH}_{3} {CH}_{2} {CH}_{2} Cl$

$({CH}_{3})_{2} CHCl, ({CH}_{3})_{2} CHCl$

${CH}_{3} {CH}_{2} {CH}_{2} Cl, {CH}_{3} {CH}_{2} {CH}_{2} Cl$

${CH}_{3} {CH}_{2} {CH}_{2} Cl, ({CH}_{3})_{2} CHCl$

answer is A.

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Detailed Solution

HCl cannot be broken to free radicals by a peroxide.

$∴$ HCl even in presence of peroxide gives Markonikoff’s product (Electrophilic addition product).

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