`CH_3-undersetunderset(CH_3)(|)CH-underset(Cl)underset(|)(CH)-CH_3 overset(EtOH//Delta)to` Major elimination product X.
X is :

Options

A. `CH_3-oversetoverset(CH_3)(|)CH-CH=CH_3`
B. `CH_3-oversetoverset(CH_3)(|)C=CH-CH_3`
C. `CH_3-overset(C_2H_5)overset(|)C=CH_2`
D. `CH_3-oversetoverset(CH_3)(|)C=oversetoverset(CH_3)(|)C-CH_3`

Answer

To solve the problem of determining the major elimination product (X) from the given compound when reacted with ethanol (EtOH) under heat, we will follow these steps:

### Step 1: Identify the Structure of the Starting Compound
The starting compound is:
\[ CH_3 - C(CH_3)(Cl) - C(CH_3) - CH_3 \]
This compound has a chlorine atom (Cl) attached to a carbon that is also bonded to two methyl groups (CH3).

### Step 2: Determine the Type of Reaction
Since we are reacting with ethanol (EtOH) in the presence of heat, we expect an elimination reaction to occur. Given that EtOH is a strong base, the reaction will likely follow the E2 mechanism, which is characterized by a concerted elimination of a leaving group (in this case, Cl) and a hydrogen atom from an adjacent carbon.

### Step 3: Identify Possible Elimination Products
In an E2 reaction, the elimination will occur between the carbon bearing the chlorine and a hydrogen atom from a neighboring carbon. The possible elimination products can be determined by considering the removal of Cl and a hydrogen atom from adjacent carbons.

1. **Product 1**: If we remove Cl from the carbon attached to Cl and a hydrogen from the adjacent carbon, we can form:
\[ CH_3 - C(CH_3) = C(CH_3) - CH_3 \]
This results in a double bond between the second and third carbons.

2. **Product 2**: Alternatively, if we remove Cl and a hydrogen from the other adjacent carbon, we can form:
\[ CH_3 - C(CH_3) - C(CH_3) = CH_3 \]
This results in a double bond between the first and second carbons.

### Step 4: Apply the Zaitsev's Rule
According to Zaitsev's rule, the more substituted alkene is generally the major product in elimination reactions.

- **Product 1** has a double bond between the second and third carbons and has three substituents on the double bond.
- **Product 2** has a double bond between the first and second carbons and has only one substituent on the double bond.

### Step 5: Determine the Major Product
Since Product 1 is more substituted (with three alkyl groups attached to the double bond), it will be the more stable and thus the major product formed during the reaction.

### Conclusion
The major elimination product X is:
\[ CH_3 - C(CH_3) = C(CH_3) - CH_3 \]

### Final Answer
X is: **CH3-CH(CH3)=C(CH3)-CH3**

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