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Question

CH3−CH(Br)−CH3alc.KOH−−−−−−→AHBr−−−−−→peroxideBNaI−−−−→acetoneC

The compound C is:

A
CH3CH2CH2I
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B
CH3CH(I)CH3
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C
CH3CHICH2I
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D
CH3CH=CHI
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Solution

The correct option is A CH3CH2CH2I
This process is used to form alkyl iodide as a product.

Alcohol KOH is dehydrohalogenating agent and peroxide is used to form anti-markovnikov product.

CH3CHBrCH3alc.KOH−−−−CH3CH=CH2HBr−−−−peroxideCH3CH2CH2BrNal−−−acetoneCH3CH2CH2I

Option A is correct.

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