[Solution from source] HI protonates the ether oxygen and I⁻ attacks by SN2 at the less hindered carbon, which is the methyl group. The products are CH₃I and isopropyl alcohol, (CH₃)₂CHOH. Attack at the secondary isopropyl carbon, giving CH₃OH and isopropyl iodide, is slower under these conditions. Answer (4).