The reaction of an ether with hydrogen iodide (HI) involves the cleavage of the C-O bond.

1. Protonation: The first step is the protonation of the ether oxygen to form an oxonium ion: $[CH_3-OH^+-CH(CH_3)_2]$.
2. Mechanism Selection: The pathway depends on the nature of the alkyl groups. Neither group is tertiary (methyl is primary, isopropyl is secondary), so the reaction follows an $S_N2$ mechanism.
3. Nucleophilic Attack: In an $S_N2$ reaction, the nucleophile ($I^-$) attacks the less sterically hindered carbon atom . The methyl group ($CH_3-$) is much less hindered than the isopropyl group ($-CH(CH_3)_2$).
4. Product Formation: The iodide ion attacks the methyl carbon, breaking the $CH_3-O$ bond. This results in the formation of methyl iodide ($CH_3I$) and isopropyl alcohol ($(CH_3)_2CHOH$).