The ether shown is treated with one equivalent of $\mathrm{HI}$. Iodide attacks the less hindered carbon of the protonated ether by ${\mathrm{S}}_{\mathrm{N}}2$. Which structure is the alkyl iodide formed?

The ether is ethyl isopropyl ether. After protonation, iodide attacks the less hindered ethyl carbon rather than the secondary isopropyl carbon. The products are iodoethane and propan-2-ol; 2-iodopropane would need attack at the more hindered carbon.