Some 2-methyl-2-butene may be produced in the reaction as a by-product. Give a mechanism for its production.

We can't draw pictures on the board; even if we could you didn't give the reaction.

This sounds like an elimination reaction problem, in which case the answer is usually a discussion of E1 v E2 so see

http://en.wikipedia.org/wiki/Elimination_reaction

The production of 2-methyl-2-butene as a by-product can occur through a mechanism known as the E1 elimination reaction. In this mechanism, a proton is abstracted from a carbon adjacent to a carbocation, resulting in the formation of a double bond.

Here is the step-by-step mechanism for the production of 2-methyl-2-butene:

1. The reaction begins with the protonation of a molecule, typically an alcohol or alkene, to generate a carbocation. In this case, let's consider the example of ethanol reacting with sulfuric acid (H2SO4) as an acid catalyst:

CH3CH2OH + H2SO4 ⟶ CH3CH2OH2+ + HSO4-

2. The strong acid, H2SO4, can act as a catalyst, facilitating the departure of a water molecule (H2O) from the protonated molecule. This process forms the oxonium ion intermediate:

CH3CH2OH2+ ⟶ CH3CH2OH2+ ⟶ CH3CH2OH + H2O+

3. The oxonium ion intermediate undergoes deprotonation, resulting in the formation of a carbocation:

CH3CH2OH ⟶ CH3CH2O+ + H2O

4. The carbocation formed can then undergo elimination of a proton from a carbon atom to generate a double bond. In this case, the neighboring carbon, which bears a methyl group, is favored due to increased stability:

CH3CH2O+ ⟶ CH2=CH2+ + CH3

5. Finally, another base, such as water (H2O), can remove the proton from the neighboring carbon atom to complete the elimination process and form 2-methyl-2-butene:

CH2=CH2+ + H2O ⟶ CH3C(CH3)=CH2

Overall, the production of 2-methyl-2-butene as a by-product occurs through the E1 elimination mechanism, involving the protonation of a molecule, formation of a carbocation, and subsequent elimination of a proton to form the double bond.

To determine the mechanism for the production of 2-methyl-2-butene as a by-product, we need to understand the applicable reaction and starting materials involved.

Assuming we have a reactant that can give rise to 2-methyl-2-butene, one possible mechanism for its production is through the process of elimination (specifically, an E2 mechanism). This mechanism involves the removal of a leaving group and the simultaneous formation of a double bond. Here's a step-by-step breakdown of the E2 mechanism:

1. Starting with a molecule of alkyl halide, such as 2-bromo-2-methylbutane, the reaction begins with a base, typically a strong base like hydroxide ion (OH⁻), abstracting a proton from a nearby carbon atom (β-carbon). This results in the formation of a negatively charged, highly reactive carbon, called a carbanion.

2. The carbanion attacks the carbon atom adjacent to the leaving group (bromine atom in this case), leading to the formation of a new bond between the β-carbon and the carbon of the leaving group.

3. As the bond between the β-carbon and the leaving group begins to break, the leaving group (bromine ion) departs, resulting in the loss of a proton, and the formation of an alkene.

4. Simultaneously, a double bond is formed between the β-carbon and the alpha-carbon (the carbon to which the β-carbon is attached). This forms the structure of 2-methyl-2-butene.

It's important to note that this is just one possible mechanism for the production of 2-methyl-2-butene. The specific reaction conditions, solvent, and other factors can influence the mechanism and product formation. Additionally, multiple competing reactions can occur simultaneously, so this explanation provides a general overview.