Here we see an reaction with rearrangement. The bromine, an excellent leaving group, leaves the carbon chain and a carbo-cation (positively charged carbon) is formed on that carbon. A positive charge is more stable on a more substituted carbon, and so the positive charge rearranges itself onto the branched carbon. Essentially, the positive charge and a hydrogen on the branched carbon switched positions. The methanol was then free to attack the branched carbon to form the major product shown.

This reaction adds and (eliminate II). The reaction is Markovnikov (Eliminate I). A hydride shift occurs putting the carbocation on the more substituted carbon before addition of (eliminate III and V).

After carbocation is formed, a rearrangement reaction stabilizes positive charge by putting it on a tertiary carbon. This is done by a methyl shift. Recall that tertiary carbocations are the most stable due to the inductive effect of alkyl groups on the electron-deficient carbocation.

The first step of this reaction will be protonation of the hydroxyl oxygen to create a good leaving group. When the leaving group leaves, what's left is a secondary carbocation that is vicinal to (next to) a quaternary carbon. A methyl shift is thermodynamically favored in this case, as the rearrangement will leave a tertiary carbocation. Following the rearrangement the nucleophile (bromide) will attack the tertiary carbocation, forming a sigma bond with the carbon. The answer is thus .