The products formed during the following reac | vedclass

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(A) The reaction involved is: $(CH_{3})_{3}C-O-CH_{3} + HI \rightarrow (CH_{3})_{3}C-I + CH_{3}OH$ This is because the reaction occurs by the $S_{N}1$

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$CH_{3}OH + (CH_{3})_{3}C-I$

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(A) The reaction involved is: $(CH_{3})_{3}C-O-CH_{3} + HI \rightarrow (CH_{3})_{3}C-I + CH_{3}OH$ This is because the reaction occurs by the $S_{N}1$ mechanism. The formation of products is governed by the stability of the carbocation formed from the cleavage of the $C-O$ bond in the protonated ether (oxonium ion). Since the tert-butyl carbocation $[(CH_{3})_{3}C^{+}]$ is more stable than the methyl carbocation $[CH_{3}^{+}]$,the cleavage of the $C-O$ bond gives the more stable carbocation $[(CH_{3})_{3}C^{+}]$ and methanol as products. Then,the iodide ion $(I^{-})$ attacks the tert-butyl carbocation to form tert-butyl iodide.

The products formed during the following reaction are:
$(CH_{3})_{3}C-O-CH_{3} + HI \rightarrow ?$

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The products formed during the following reaction are: $(CH_{3})_{3}C-O-CH_{3} + HI \rightarrow ?$