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(A) In the reaction of an unsymmetrical ether with $HI$,if the alkyl groups are primary or secondary,the reaction follows the $S_N2$ mechanism.The iod

Vedclass · Accepted Answer

$CH_3I + (CH_3)_2CHOH$

Vedclass · Answer

(A) In the reaction of an unsymmetrical ether with $HI$,if the alkyl groups are primary or secondary,the reaction follows the $S_N2$ mechanism.The iodide ion $(I^{-})$ attacks the smaller (less sterically hindered) alkyl group.In $CH_3-O-CH(CH_3)_2$,the methyl group $(-CH_3)$ is smaller than the isopropyl group $(-CH(CH_3)_2)$.Therefore,the $I^{-}$ ion attacks the methyl group,leading to the formation of $CH_3I$ and $(CH_3)_2CHOH$ as the major products.

The major organic product in the reaction is: $CH_3-O-CH(CH_3)_2 + HI \to$ products

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