`CH_3-undersetunderset(CH_3)(|)CH-oversetoverset(Cl)(|)CH-CH_3 underset(Delta)overset(t-BuO) to` Major Product is :

Options

A. `CH_3-undersetunderset(CH_3)(|)C=CH-CH_3`
B. `CH_3-undersetunderset(CH_3)(|)CH-CH=CH_2`
C. `CH_2=undersetunderset(CH_3)(|)C-CH_2-CH_3`
D. `CH_3-undersetunderset(CH_3)(|)CH-oversetoverset(OH)(|)CH-CH_3`

Answer

To solve the given problem, we need to analyze the reaction of the compound with the reagent provided. The compound in question is:

\[ CH_3-CH(CH_3)-CH(Cl)-CH_3 \]

This compound has a chlorine atom (Cl) attached to a carbon that is also connected to a tertiary carbon (the one with two methyl groups). The reagent used in the reaction is t-butoxide ion (\( t-BuO^- \)), which is a strong base and will promote an elimination reaction.

### Step-by-Step Solution:

1. **Identify the Structure**:
The starting compound can be drawn as:
\[ CH_3-CH(CH_3)-C(Cl)-CH_3 \]
This shows that we have a chlorine atom attached to a secondary carbon.

2. **Determine the Type of Reaction**:
The presence of a strong base like t-butoxide indicates that we will likely undergo an elimination reaction (E2 mechanism). The base will abstract a proton (H) from a carbon adjacent to the carbon bearing the chlorine.

3. **Identify Possible Elimination Sites**:
The adjacent carbons to the carbon with chlorine are:
- One carbon with a hydrogen (from the \( CH \) group)
- The other carbon is a tertiary carbon (with two \( CH_3 \) groups).

4. **Elimination Reaction**:
The elimination reaction will involve the base abstracting a proton from the adjacent carbon, leading to the formation of a double bond. The most stable alkene is formed by removing a hydrogen from the more substituted carbon.

5. **Forming the Alkene**:
By removing a hydrogen from the tertiary carbon (the one with two \( CH_3 \) groups), we can form a double bond with the carbon that has the chlorine. This results in:
\[ CH_3-CH(CH_3)=C(CH_3)-CH_3 \]
This is a more substituted alkene and is therefore more stable.

6. **Final Product**:
The major product of the reaction is:
\[ CH_3-CH(CH_3)=C(CH_3)-CH_3 \]
This product is 2-methyl-2-butene.

### Conclusion:
The major product of the reaction is:
\[ \text{2-methyl-2-butene} \]