3-bromo-2-methylpentane in ethanol is heated to produce a major product and a minor product?

Answer 1

I suppose there is KOH in the ethanolic solution.

The major product is (CH3)2C=CHCH2CH3, while the minor product is (CH3)2CHCH=CHCH3.

Answer 2

I'm going to assume you mean in acid and not in base, because you can't form a carbocation in base. Anyways, look at the carbocation you form after the bromine "spontaneously" leaves. It's secondary, yes? Right next to it is a tertiary carbon with a hydrogen. That hydride's going to move over and "attack" the carbocation, thus neutralizing that secondary carbon and producing a positive charge on the tertiary carbon. This is energetically favored because the tertiary carbocation is stabilized more by the added hyperconjugation (more overlap with the adjacent sigma C-H electrons). So that hydride shift occurs, and the oxygen's lone pair from the solvent will come in and attack, and subsequently get deprotonated. Ultimately, what you end up with is called 2-methyl-2-pentanol. But remember, you're going to get some E1 as well. (All of that is assuming it's in an acidic solution, like H2O/H3O+. If it's in base, like NaOH/H2O, you'll get some Sn2 and a lot of E2.)