Explain the reaction of methoxyethane with concentrated $HI$ and describe its mechanism.

(N/A) Reaction of methoxyethane with concentrated $HI$: When methoxyethane is heated with concentrated $HI$,the $C-O$ bond breaks to form methyl iodide and ethanol. At higher temperatures,ethanol is further converted into ethyl iodide.
$(b)$ Mechanism: This reaction proceeds via an $S_{N}2$ mechanism in the following steps:
Step-$1$: Protonation of the ether occurs due to the presence of acidic $HI$,forming a protonated ether.
Step-$2$: The nucleophilic attack by $I^{-}$ results in the formation of methyl iodide $(CH_{3}I)$ and ethanol $(C_{2}H_{5}OH)$.
$(i)$ The nucleophile $I^{-}$ attacks the smaller alkyl group $(CH_{3})$ of the protonated ether from the opposite side,forming a transition state $(T)$ where the $C-I$ bond is partially formed and the $C-O$ bond is partially broken.
$(ii)$ In the transition state,the bond breaks. Since the rate depends on two species,it is a second-order or bimolecular reaction.
$(iii)$ Since the nucleophile $I^{-}$ displaces the $OCH_{2}CH_{3}$ (ethoxy) group from the ether,the reaction follows a nucleophilic substitution mechanism.