Write the mechanism of the reaction of $HI$ with methoxymethane.

(N/A) The mechanism of the reaction of $HI$ with methoxymethane involves the following steps:
Step $1$: Protonation of methoxymethane:
$CH_3-O-CH_3 + HI \rightleftharpoons CH_3-O^{+}(H)-CH_3 + I^{-}$
Step $2$: Nucleophilic attack of $I^{-}$ on the protonated ether:
$I^{-} + CH_3-O^{+}(H)-CH_3 \to CH_3-I + CH_3-OH$
Step $3$: When $HI$ is in excess and the reaction is carried out at a high temperature,the methanol formed in the second step reacts with another $HI$ molecule:
$CH_3-OH + HI \rightleftharpoons CH_3-OH_2^{+} + I^{-}$
$I^{-} + CH_3-OH_2^{+} \to CH_3-I + H_2O$