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(D) The reaction of ethyl bromide $(CH_3-CH_2-Br)$ with alcoholic $KOH$ is a dehydrohalogenation reaction (elimination reaction). In this reaction,the

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Alcoholic $KOH$

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(D) The reaction of ethyl bromide $(CH_3-CH_2-Br)$ with alcoholic $KOH$ is a dehydrohalogenation reaction (elimination reaction). In this reaction,the base $(OH^-)$ abstracts a proton from the $\beta$-carbon,leading to the elimination of $HBr$ and the formation of ethene $(CH_2=CH_2)$. The reaction is: $CH_3-CH_2-Br + KOH 	ext{ (alc)} 	o CH_2=CH_2 + KBr + H_2O$.

Ethyl bromide gives ethylene when reacted with

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