In the reaction, the products formed are ${\left({\mathrm{CH}}_3\right)}_2\mathrm{CH}\cdot {\mathrm{CH}}_2\cdot \mathrm{O}\cdot {\mathrm{CH}}_2{\mathrm{CH}}_3+\mathrm{HI} \stackrel{\mathrm{heated}}{\to }$

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