Ethyl bromide reacts with alcoholic KOH to gi | vedclass

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(B) The reaction of ethyl bromide $(CH_3CH_2Br)$ with alcoholic $KOH$ is a dehydrohalogenation reaction.In this reaction,the alcoholic $KOH$ acts as a

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Ethene

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(B) The reaction of ethyl bromide $(CH_3CH_2Br)$ with alcoholic $KOH$ is a dehydrohalogenation reaction.In this reaction,the alcoholic $KOH$ acts as a strong base and abstracts a proton from the $\beta$-carbon,leading to the elimination of a molecule of $HBr$.This results in the formation of an alkene.The chemical equation is: $CH_3CH_2Br + KOH (	ext{alc.}) ightarrow CH_2=CH_2 + KBr + H_2O$.Thus,the product formed is ethene.

Ethyl bromide reacts with alcoholic $KOH$ to give.....

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