What will the major product of the following :
`C_2H_5ONa + CH_3-underset(CH_3)underset(|)overset(CH_3)overset(|) C - Cl to ?`

Options

A. `CH_3 underset(C_2H_5)underset(|)overset(CH_3)overset(|)C-OCH_3`
B. `CH_3 - underset(CH_3)underset(|) C = CH_3 `
C. `CH_2 = CH_2`
D. `CH_3 - underset(CH_3)underset(|) overset(CH_3)overset(|)C - OC_2H_5`

Explanation

To determine the major product of the reaction between sodium ethoxide (C2H5ONa) and the given tertiary halide (CH3)2C(Cl)CH2CH3, we will follow these steps:

### Step 1: Identify the Reaction Type
The reaction involves a tertiary halide and a strong base (sodium ethoxide). In this case, we need to consider the potential pathways: substitution (SN1 or SN2) and elimination (E1 or E2).

**Hint:** Tertiary halides typically favor elimination reactions over substitution due to steric hindrance.

### Step 2: Analyze the Tertiary Halide
The given halide is a tertiary halide, which means that the carbon bonded to the halogen (Cl) is attached to three other carbon atoms. This structure makes it less favorable for SN2 reactions because of steric hindrance.

**Hint:** Tertiary halides do not undergo SN2 reactions due to steric hindrance.

### Step 3: Determine the Elimination Pathway
Since SN2 is not viable, we consider elimination reactions. Given that sodium ethoxide is a strong base, it will promote E2 elimination. In the absence of a solvent, E1 elimination is also not favored.

**Hint:** Strong bases promote E2 eliminations, especially in the absence of solvents.

### Step 4: Identify the β-Hydrogen
In E2 elimination, we need to remove a β-hydrogen that is anti to the leaving group (Cl). The β-hydrogens are the hydrogens on the carbon adjacent to the carbon with the halogen.

**Hint:** Look for β-hydrogens that can be eliminated in an anti-periplanar fashion with respect to the leaving group.

### Step 5: Perform the Elimination
In this case, we can remove a β-hydrogen from the CH2 group adjacent to the carbon with the Cl. This will lead to the formation of a double bond between the carbon that had the Cl and the carbon from which the hydrogen was removed.

**Hint:** The elimination will result in the formation of a double bond between the two carbons involved.

### Step 6: Write the Major Product
The major product of this reaction will be an alkene. The product will have the structure of a double bond formed between the carbon that had the Cl and the adjacent carbon from which the hydrogen was removed.

The final product will be:
- **Product:** CH2=C(CH3)CH2CH3 (2-methylpropene)

**Hint:** The major product is often the most stable alkene formed from the elimination reaction.

### Conclusion
The major product of the reaction between sodium ethoxide and the given tertiary halide is 2-methylpropene (CH2=C(CH3)CH2CH3), formed through an E2 elimination mechanism.