What are A and B respectively in the followin | vedclass

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(C) The reaction of an ether with $HI$ proceeds via an $S_N1$ mechanism when one of the alkyl groups is tertiary,as it can form a stable carbocation.$

Vedclass · Accepted Answer

$A = CH_3OH, B = (CH_3)_3C-I$

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(C) The reaction of an ether with $HI$ proceeds via an $S_N1$ mechanism when one of the alkyl groups is tertiary,as it can form a stable carbocation.$1$. The oxygen atom of the ether is protonated by $HI$ to form an oxonium ion: $(CH_3)_3C-O^+(H)-CH_3$.$2$. The $C-O$ bond between the tertiary carbon and oxygen breaks to form a stable tertiary carbocation $(CH_3)_3C^+$ and methanol $CH_3OH$.$3$. The iodide ion $I^-$ then attacks the tertiary carbocation to form tert-butyl iodide $(CH_3)_3C-I$.Thus,$A$ is $CH_3OH$ and $B$ is $(CH_3)_3C-I$.

What are $A$ and $B$ respectively in the following reaction?
$(CH_3)_3C-OCH_3 + HI \rightarrow A + B$

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What are $A$ and $B$ respectively in the following reaction?$(CH_3)_3C-OCH_3 + HI \rightarrow A + B$