The product of the reaction CH_3-C(CH_3)_2-O- | vedclass

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(C) The reaction of an ether with $HI$ follows the $S_N1$ mechanism when one of the alkyl groups is tertiary,as the tertiary carbocation is highly sta

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$CH_3-OH + (CH_3)_3C-I$

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(C) The reaction of an ether with $HI$ follows the $S_N1$ mechanism when one of the alkyl groups is tertiary,as the tertiary carbocation is highly stable. $1$. Protonation of the ether oxygen occurs: $(CH_3)_3C-O(H^+)-CH_3$. $2$. The 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 $I^-$ attacks the tertiary carbocation to form tert-butyl iodide $(CH_3)_3C-I$. Thus,the products are $CH_3OH$ and $(CH_3)_3C-I$.

The product of the reaction $CH_3-C(CH_3)_2-O-CH_3 \xrightarrow{cold \ HI}$ is:

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