For the reaction: CH_3-CH(CH_3)-O-CH_3 + HI r | vedclass

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Question

(A) The reaction of an unsymmetrical ether with $HI$ follows an $S_N1$ mechanism when one of the alkyl groups is tertiary or secondary,as it can form

Vedclass · Accepted Answer

$CH_3-OH \& (CH_3)_2CH-I$

Vedclass · Answer

(A) The reaction of an unsymmetrical ether with $HI$ follows an $S_N1$ mechanism when one of the alkyl groups is tertiary or secondary,as it can form a stable carbocation. In the given ether,$CH_3-CH(CH_3)-O-CH_3$,the isopropyl group $(CH_3)_2CH-$ is secondary. Upon protonation of the oxygen atom by $HI$,the bond between the oxygen and the secondary carbon breaks to form a stable secondary carbocation,$(CH_3)_2CH^+$. The iodide ion $(I^-)$ then attacks this carbocation to form isopropyl iodide,$(CH_3)_2CH-I$. The remaining part,$CH_3-O-H$,forms methanol,$CH_3-OH$. Thus,the products are $CH_3-OH$ and $(CH_3)_2CH-I$.

For the reaction: $CH_3-CH(CH_3)-O-CH_3 + HI \rightarrow X + Y$,identify $X$ and $Y$.

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