Consider the following reaction: CH_3-C(CH_3) | vedclass

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Question

(A) The reaction of tert-butyl methyl ether with $HI$ follows the $S_N1$ mechanism because a highly stable tertiary carbocation $(CH_3-C^{+}(CH_3)_2)$

Vedclass · Accepted Answer

$CH_3-C(I)(CH_3)_2 + CH_3-OH$

Vedclass · Answer

(A) The reaction of tert-butyl methyl ether with $HI$ follows the $S_N1$ mechanism because a highly stable tertiary carbocation $(CH_3-C^{+}(CH_3)_2)$ is formed upon protonation and cleavage of the $C-O$ bond.The iodide ion $(I^{-})$ then attacks this carbocation to form tert-butyl iodide $(CH_3-C(I)(CH_3)_2)$,while the other fragment remains as methanol $(CH_3-OH)$.

Consider the following reaction: $CH_3-C(CH_3)_2-O-CH_3 \xrightarrow{HI (1 \text{ mole})} \text{Products}$. The main products of the reaction will be:

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