Reaction of `t-`butyl bromide with sodium methoxide produces

Options

A. `t-`butyl methyl ether
B. isobutylene
C. isobutane
D. Sodium `t-`butoxide

Explanation

To solve the question regarding the reaction of t-butyl bromide with sodium methoxide, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactants are t-butyl bromide (C4H9Br) and sodium methoxide (CH3ONa).
- The structure of t-butyl bromide is (CH3)3CBr, which has a tertiary carbon attached to a bromine atom.

2. **Determine the Type of Reaction**:
- Sodium methoxide is a strong base and a good nucleophile. When it reacts with t-butyl bromide, we need to consider the mechanism.
- Tertiary halides typically undergo elimination reactions rather than substitution due to steric hindrance.

3. **Mechanism of Reaction**:
- The reaction will proceed via an E1 mechanism (unimolecular elimination).
- In the E1 mechanism, the first step involves the formation of a carbocation after the bromine leaves, resulting in a t-butyl cation (C4H9+).

4. **Elimination Step**:
- The methoxide ion (CH3O-) can then abstract a proton from a β-carbon (the carbon adjacent to the carbocation).
- This results in the formation of an alkene through dehydrohalogenation.

5. **Product Formation**:
- The major product formed from this reaction is isobutylene (or 2-methylpropene), which has the structure CH2=C(CH3)2.
- The elimination leads to the formation of a double bond between the two carbons.

6. **Final Product**:
- The final product of the reaction between t-butyl bromide and sodium methoxide is isobutylene (2-methylpropene).

### Conclusion:
The product of the reaction of t-butyl bromide with sodium methoxide is isobutylene (2-methylpropene).

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