The reaction of n-propyl bromide with ethanol | vedclass

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(B) The reaction of $n$-propyl bromide $(CH_3CH_2CH_2Br)$ with ethanolic $KOH$ 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 ethanolic $KOH$ is a dehydrohalogenation reaction (an elimination reaction).In this reaction,the base $(OH^-)$ abstracts a proton from the $\beta$-carbon,and the bromide ion is eliminated,resulting in the formation of an alkene.$CH_3CH_2CH_2Br + KOH (	ext{ethanolic}) ightarrow CH_3CH=CH_2 + KBr + H_2O$.The product formed is $CH_3CH=CH_2$,which is propene.

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

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