The reaction of n-propyl bromide with alcohol | vedclass

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(B) The reaction of $n$-propyl bromide $(CH_3CH_2CH_2Br)$ with alcoholic $KOH$ is a dehydrohalogenation reaction (an elimination reaction).In this rea

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Propene

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(B) The reaction of $n$-propyl bromide $(CH_3CH_2CH_2Br)$ with alcoholic $KOH$ is a dehydrohalogenation reaction (an elimination reaction).In this reaction,a hydrogen atom is removed from the $\beta$-carbon and the bromine atom is removed from the $\alpha$-carbon,resulting in the formation of a double bond.The reaction is: $CH_3CH_2CH_2Br + KOH (	ext{alc.}) ightarrow CH_3CH=CH_2 + KBr + H_2O$.The product formed is propene.

The reaction of $n$-propyl bromide with alcoholic potassium hydroxide $(KOH)$ produces ...........

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