In the cleavage of mixed ethers with two different alkyl groups, the alcohol and alkyl iodide that form depend on the nature of alkyl group. When primary or secondary alkyl groups are present, it is the lower alkyl group that forms alkyl iodide therefore

In case of unsymmetric ethers when treated with HI then I^– attaches to smaller and less complex alkyl group as it follows S_N2 mechanism

CH₃ $$-$$ O $$-$$ CH(CH₃)₂ + HI $$ \to $$ CH₃I + (CH₃)₂CHOH