The major organic product in the reaction,${\text{CH}}_3-\text{O}-\text{CH}({\text{CH}}_3{)}_2+\text{HI}\to \text{Product}$ is :

Cleavage of an unsymmetrical ether by hydroiodic acid (HI) typically follows an ${\text{S}}_{\text{N}}2$ mechanism unless a tertiary alkyl group is present. Since the ether contains a primary (methyl) and a secondary (isopropyl) group, the iodide nucleophile attacks the less sterically hindered methyl carbon, producing methyl iodide (${\text{CH}}_3\text{I}$) and isopropyl alcohol ($({\text{CH}}_3{)}_2\text{CHOH}$).