The major product of the reaction between `CH_(3) CH_(2) O Na` and `(CH_(3))_(3) C Cl` in ethanol is

Options

A. `CH_(3) CH_(2)OC (CH_(3))_(3)`
B. `CH_(2) = C(CH_(3))_(2)`
C. `CH_(3) CH_(2) C(CH_(3))_(3)`
D. `CH_(3) CH = CH CH_(3)`

Explanation

To solve the question regarding the reaction between `CH₃CH₂ONa` (sodium ethoxide) and `(CH₃)₃CCl` (tert-butyl chloride) in ethanol, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- We have sodium ethoxide (`CH₃CH₂ONa`), which is a strong base.
- We also have tert-butyl chloride (`(CH₃)₃CCl`), which is a tertiary alkyl halide.

2. **Determine the Type of Reaction**:
- Since sodium ethoxide is a strong base, it can promote elimination reactions. The presence of a tertiary alkyl halide suggests that an E2 elimination reaction is likely to occur.

3. **E2 Mechanism**:
- In an E2 reaction, the base (sodium ethoxide) abstracts a β-hydrogen from the alkyl halide while the leaving group (Cl) departs simultaneously.
- For tert-butyl chloride, the β-hydrogens are located on the methyl groups adjacent to the carbon bearing the chlorine.

4. **Formation of the Alkene**:
- The elimination of HCl will lead to the formation of a double bond between the carbon that was attached to the Cl and the carbon from which the β-hydrogen was removed.
- The product of this reaction will be 2-methylpropene (also known as isobutylene).

5. **Final Product**:
- The major product of the reaction is 2-methylpropene (C₄H₈).

### Summary:
The major product of the reaction between `CH₃CH₂ONa` and `(CH₃)₃CCl` in ethanol is **2-methylpropene (C₄H₈)**.