When CH_2 = CH - O - CH_2 - CH_3 reacts with | vedclass

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(D) The reaction of ethyl vinyl ether $(CH_2 = CH - O - CH_2 - CH_3)$ with $HI$ proceeds via the protonation of the oxygen atom. $CH_2 = CH - O - CH_2

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iodoethane

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(D) The reaction of ethyl vinyl ether $(CH_2 = CH - O - CH_2 - CH_3)$ with $HI$ proceeds via the protonation of the oxygen atom. $CH_2 = CH - O - CH_2 - CH_3 + H^+ \rightarrow CH_2 = CH - O^+(H) - CH_2 - CH_3$ Then,the iodide ion $(I^-)$ attacks the ethyl group $(CH_2CH_3)$ because the vinyl group $(CH_2=CH-)$ is not susceptible to $S_N2$ attack due to the partial double bond character of the $C-O$ bond. $I^- + CH_2 = CH - O^+(H) - CH_2 - CH_3 \rightarrow CH_2 = CH - OH + CH_3CH_2I$ The product $CH_2 = CH - OH$ (vinyl alcohol) is unstable and tautomerizes to form ethanal $(CH_3CHO)$. The other product formed is iodoethane $(CH_3CH_2I)$. Therefore,among the given options,iodoethane is one of the products.

When $CH_2 = CH - O - CH_2 - CH_3$ reacts with $1 \ mol$ of $HI$,one of the products formed is

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