Question

2-bromo-2-methylbutane undergoes an e1 elimination reaction in the presence of ethanol. In the next reaction, only one of the possible products is represented. Although the product shown is not the major product of the reaction, notice that there is more than one way it can be produced. Draw the missing intermediate and complete the mechanism (using electron-flow arrows) in the second box. (You can automatically show hydrogen atoms by clicking the H�� button, but you only need to explicitly show hydrogen atoms that participate in the mechanism.)

Answer 1

The question is about illustrating the E1 elimination reaction of 2-bromo-2-methylbutane with ethanol. The mechanism includes formation of a carbocation intermediate followed by deprotonation to produce the alkene.

Step-by-step explanation:

The question involves the E1 elimination reaction of 2-bromo-2-methylbutane in the presence of ethanol, where the major product is not shown but needs to be explained through the mechanism. In the E1 elimination process, the first step is the formation of a carbocation intermediate following the departure of the bromide ion (Br-) as a good leaving group. This intermediate is then deprotonated by a base (in this case, ethanol acting as the base) to form the alkene product. Since the student has asked for the mechanism, it will include showing the loss of Br- to form the carbocation, followed by the removal of a proton (H+) by ethanol leading to the formation of the double bond.

1. Departure of the bromide ion from 2-bromomethylbutane to form a tertiary carbocation.
2. Base (ethanol) removes a hydrogen atom from an adjacent carbon, causing the electrons from the C-H bond to form a double bond and resulting in the formation of water (H2O).