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(B) The reaction of $tert$-butyl methyl ether with $HI$ follows an $S_N1$ mechanism because the $tert$-butyl group can form a stable $tert$-butyl carb

Vedclass · Accepted Answer

$tert$-butyl iodide and methanol

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(B) The reaction of $tert$-butyl methyl ether with $HI$ follows an $S_N1$ mechanism because the $tert$-butyl group can form a stable $tert$-butyl carbocation. $1$. The oxygen atom of the ether is protonated by $HI$ to form an oxonium ion: $(CH_3)_3C-O(CH_3) + H^+ \rightarrow (CH_3)_3C-O^+(H)(CH_3)$. $2$. The $C-O$ bond between the $tert$-butyl group and oxygen breaks to form a stable $tert$-butyl carbocation $(CH_3)_3C^+$ and methanol $(CH_3OH)$. $3$. The iodide ion $(I^-)$ then attacks the $tert$-butyl carbocation to form $tert$-butyl iodide $(CH_3)_3C-I$. Thus,the products are $tert$-butyl iodide and methanol.

Which of the following mixtures is obtained when $tert$-butyl methyl ether is heated with $HI$?

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