Answer
Iodine leaves first to form carbocation. Then, methyl shift happens to form a 7-membered ring. H2O then attacks the new carbocation, deprotonation follows, and product is formed.
Work Step by Step
See image.
You can help us out by revising, improving and updating this answer.
Update this answerAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.