When mixed ethersare used, the formation of alkyl iodide depends on the nature of alkyl groups. Methyl iodide is formed when one group is methyl and the other a primary or secondary alkyl group. Here reaction follows $\mathrm{S}_{\mathrm{N}} 2$ mechanism and because of the steric effect of the larger group, $\mathrm{I}^{-}$attacks the smaller (Me) group.
$\mathrm{CH}_3 \mathrm{OC}_2 \mathrm{H}_5+\mathrm{HI} \longrightarrow \mathrm{CH}_3 \mathrm{I}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$
When the substrate is a methyl $t$-alkyl ether, the products are $t-R \mathrm{I}$ and $\mathrm{MeOH}$. Here reaction follows $\mathrm{S}_{\mathrm{N}} 1$ mechanism and formation of products is controlled by the stability of carbocation. Since carbocation stability order is $3^{\circ}>2^{\circ}>1^{\circ}>\stackrel{+}{\mathrm{C}} \mathrm{H}_3$, therefore alkyl halide is always derived from tert-alkyl group.