In the reaction: CH_3-CH(CH_3)-CH_2-O-CH_2-CH | vedclass

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(C) The reaction of an ether with $HI$ follows the $S_N2$ mechanism when the alkyl groups are primary or secondary. The iodide ion $(I^-)$ attacks the

Vedclass · Accepted Answer

$CH_3-CH(CH_3)-CH_2OH + CH_3CH_2I$

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(C) The reaction of an ether with $HI$ follows the $S_N2$ mechanism when the alkyl groups are primary or secondary. The iodide ion $(I^-)$ attacks the less sterically hindered alkyl group. In isobutyl ethyl ether $(CH_3-CH(CH_3)-CH_2-O-CH_2-CH_3)$,the ethyl group is less sterically hindered than the isobutyl group. Therefore,the $I^-$ ion attacks the ethyl group,resulting in the formation of isobutyl alcohol $(CH_3-CH(CH_3)-CH_2OH)$ and ethyl iodide $(CH_3CH_2I)$.

In the reaction: $CH_3-CH(CH_3)-CH_2-O-CH_2-CH_3 + HI \xrightarrow{\text{Heated}}$ Which of the following compounds will be formed?

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