Answer:
Step-by-step explanation:
In this reaction, the ether is cleaved by SN2 , as I⁻ is a strong nucleophile
So we have the following steps
- Protonation of oxygen in ether, improving the ability of both sides to be a leaving group as alchohols
- I⁻ performs nucleophilic attack on 1st carbon of butyl side, releasing Isopropanol as leaving group and leaving Iodobutane behind
Now, even though butanol can become a leaving group in this reaction, this is unfavourable, as isopropyl group has large steric hindrance which prevents nucleophilic attack on isopropyl side of the ether.