The reaction of t-butyl bromide with sodium methoxide mainly produces:A. isobutaneB. isobutyleneC. t-butyl methyl etherD. sodium tert-butoxide

Let's break down the reaction of t-butyl bromide with sodium methoxide to determine the main product.

### Reaction Components:
1. **T-butyl bromide**: This is a tertiary alkyl halide, specifically 2-bromo-2-methylpropane.
2. **Sodium methoxide**: This is a strong base and a nucleophile. Its formula is NaOCH₃.

### Possible Reactions:
When t-butyl bromide reacts with sodium methoxide, there are a couple of pathways it can follow:

1. **Nucleophilic Substitution Reaction (S₁ Reaction):**
- In a nucleophilic substitution reaction, the methoxide ion (CH₃O⁻) would act as a nucleophile and attack the t-butyl bromide.
- However, because t-butyl bromide is a tertiary alkyl halide, it is less likely to undergo an S₁ reaction. Instead, it is more prone to undergo an elimination reaction.

2. **Elimination Reaction (E₂ Reaction):**
- In an E₂ elimination reaction, the methoxide ion (a strong base) would abstract a hydrogen atom from the t-butyl bromide, resulting in the formation of isobutylene (2-methylpropene) and the expulsion of the bromide ion.
- The reaction can be written as:
\[
(CH_3)_3CBr + CH_3O^- \rightarrow (CH_3)_2C=CH_2 + NaBr + CH_3OH
\]
- Here, isobutylene (2-methylpropene) is formed as the main product.

3. **Ether Formation (if considering other possibilities):**
- Though not the main pathway in this case, t-butyl bromide could react with sodium methoxide to form t-butyl methyl ether (an SN2 reaction), but this is less favored with t-butyl bromide because it is a tertiary substrate. The formation of ethers typically occurs more readily with primary or secondary substrates.

### Conclusion:
For the reaction of t-butyl bromide with sodium methoxide, the primary product is **isobutylene** (2-methylpropene) due to the E₂ elimination reaction.

**Correct Answer: B. isobutylene**