n-propyl bromide on treating with alcoholic K | vedclass

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Question

(B) Alcoholic $KOH$ acts as a dehydrohalogenating agent, which removes a molecule of $HBr$ from the alkyl halide to form an alkene.When $n-$propyl bro

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propene

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(B) Alcoholic $KOH$ acts as a dehydrohalogenating agent, which removes a molecule of $HBr$ from the alkyl halide to form an alkene.When $n-$propyl bromide $(CH_{3}CH_{2}CH_{2}Br)$ reacts with alcoholic $KOH$, it undergoes dehydrohalogenation to produce propene $(CH_{3}CH=CH_{2})$.The reaction is: $CH_{3}CH_{2}CH_{2}Br + 	ext{alc. } KOH \longrightarrow CH_{3}CH=CH_{2} + KBr + H_{2}O$.

$n-$propyl bromide on treating with alcoholic $KOH$ produces

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