Give the reaction of tert-butyl methyl ether with $HI$ and explain its mechanism.

(N/A) The reaction of tert-butyl methyl ether with $HI$ proceeds via an $S_{N}1$ mechanism because the tert-butyl group can form a stable carbocation intermediate.
Step $1$: Protonation of the ether oxygen atom by $H^+$ from $HI$ to form a protonated ether.
$(CH_3)_3C-O-CH_3 + H^+ \rightleftharpoons (CH_3)_3C-O^+(H)-CH_3$
Step $2$: The $C-O$ bond between the tert-butyl group and oxygen breaks to form a stable tert-butyl carbocation $(CH_3)_3C^+$ and methanol $(CH_3OH)$. This is the slow,rate-determining step.
$(CH_3)_3C-O^+(H)-CH_3 \xrightarrow{\text{slow}} (CH_3)_3C^+ + CH_3OH$
Step $3$: The nucleophilic iodide ion $(I^-)$ attacks the tert-butyl carbocation to form tert-butyl iodide.
$(CH_3)_3C^+ + I^- \xrightarrow{\text{fast}} (CH_3)_3C-I$
Overall reaction: $(CH_3)_3C-O-CH_3 + HI \rightarrow (CH_3)_3C-I + CH_3OH$.