2-Methyl propene can be prepared by the reaction of

The correct answer is option 1.

When t-butyl bromide reacts with ethanol, a small amount of elimination products is obtained via the \(E_1\) mechanism.

Mechanism:

Step 1: The bromide dissociates and forms a tertiary (3°) carbocation. This is a slow bond-breaking step, and it is also the rate-determining step for the whole reaction. Since only the bromide substrate was involved in the rate-determining step, the reaction rate law is first order. This is the reaction rate only depends on the concentration of \((CH_3)_3Br\) and has nothing to do with the concentration of the base.

Step 2: The hydrogen on β-carbon (β-carbon is the one beside the positively charged carbon) is acidic because of the adjacent positive charge. The base, \(CH_3ONa\), reacts with the β-H by removing it, and the C-H bond electron pair moves in to form the C-C π bond.