Tert. butyl chloride on treatment with sodium alkoxide yields

Options

A. An ether
B. An alkene
C. An alcohol
D. An alkane

Explanation

To solve the question of what tertiary butyl chloride yields when treated with sodium alkoxide, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactant is tertiary butyl chloride, which has the structure (CH₃)₃C-Cl.
- The other reactant is sodium alkoxide (R-O-Na), where R can be any alkyl group.

2. **Understand the Reaction Type**:
- The reaction between tertiary butyl chloride and sodium alkoxide resembles the Williamson ether synthesis. However, due to the structure of tertiary butyl chloride, we need to consider the possibility of elimination reactions instead of substitution.

3. **Consider the Structure of Tertiary Butyl Chloride**:
- Tertiary butyl chloride has a tertiary carbon attached to the chlorine atom. This structure is significant because tertiary alkyl halides tend to favor elimination reactions over substitution reactions.

4. **Elimination Reaction**:
- In the presence of sodium alkoxide, the chlorine atom (Cl) can be eliminated along with a hydrogen atom (H) from a neighboring carbon. This results in the formation of a double bond, leading to the formation of an alkene.

5. **Product Formation**:
- The elimination of HCl (where H is from the carbon adjacent to the tertiary carbon) results in the formation of an alkene. The specific alkene formed from the elimination of H and Cl from tertiary butyl chloride is isobutylene (C₄H₈).

6. **Conclusion**:
- Therefore, the product formed when tertiary butyl chloride reacts with sodium alkoxide is an alkene.

### Final Answer:
The product is an alkene (Option B).

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