Reaction of 2,3-dimethyl-1-butene with HBr leads to an alkyl bromide, C6H13Br. On treatment of this alkyl bromide with KOH in methanol, elimination of HBr occurs and a hydrocarbon that is isomeric with the starting alkene is formed. What is the structure of this hydrocarbon, and how do you think it is formed from the alkyl bromide

See explanation and image attached

Explanation:

The image attached shows the entire scheme of reactions mentioned in the question.

The first reaction is an addition reaction which yields a tertiary alkyl halide as shown in accordance with Markovnikov rule.

The second reaction is a dehydrohalogenation in which the base abstracts a proton from the alkyl halide followed by loss of a bromide ion to yield the corresponding alkene.

This alkene is an isomer of the starting material.