AIPMT – 2007

Hint: In this reaction, I^– will attack the less substituted carbon.

Explanation:

Since mixed ether is given in this reaction, note that iodide will bond with the smaller alkyl group to minimise the steric repulsion as iodide ion is bulky itself.

The reaction is as follows:

${\mathrm{CH}}_3-\stackrel{}{}\stackrel{|}{\mathrm{CH}}{\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{O}-{\mathrm{CH}}_2-{\mathrm{CH}}_3+\mathrm{HI}\stackrel{∆}{\to }{\mathrm{CH}}_3-\stackrel{}{}\stackrel{|}{\mathrm{CH}}{\mathrm{CH}}_3-{\mathrm{CH}}_2\mathrm{OH}+{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{I}$