In the given reaction `CH_3-underset(CH_3)underset(|)overset(CH_3)overset(|)C-CH_2-Broverset(Delta)rarr[X]`
[X] will be

Options

A. `CH_3-underset(CH_3)underset(|)(CH)-CH=CH_2`
B. `CH_3-underset(CH_3)underset(|)C=CH-CH_3`
C. `CH_3-CH=CH-CH_3`
D. `CH_3-CH_2-CH=CH_2`

Answer

To solve the question, we need to analyze the given reaction step by step. The reaction involves a compound with a bromine atom (Br) that is likely undergoing elimination to form an alkene.

### Step-by-Step Solution:

1. **Identify the Structure**: The starting compound is CH₃-CH(CH₃)-C(CH₃)-CH₂-Br. This structure indicates that we have a bromine atom attached to a carbon that is also connected to other carbon atoms.

**Hint**: Visualize the structure of the compound to understand the positioning of the bromine and the surrounding groups.

2. **Heat the Compound**: The reaction is heated, which suggests that an elimination reaction (dehydrohalogenation) will occur. The bromine atom will leave, and a hydrogen atom will be removed from an adjacent carbon, leading to the formation of a double bond.

**Hint**: Remember that heating often promotes elimination reactions, leading to the formation of alkenes.

3. **Formation of Carbocation**: When the bromine leaves, a carbocation is formed. Since the bromine is on a primary carbon, this carbocation is initially a primary carbocation.

**Hint**: Identify the type of carbocation formed after the elimination of Br.

4. **Carbocation Rearrangement**: Primary carbocations are less stable than secondary or tertiary carbocations. Therefore, the primary carbocation will undergo rearrangement to form a more stable tertiary carbocation through a methyl shift.

**Hint**: Consider the stability of carbocations and how they can rearrange to form more stable structures.

5. **Methyl Shift**: The methyl group from the adjacent carbon will shift to the carbon with the positive charge, converting the primary carbocation into a tertiary carbocation.

**Hint**: Visualize the movement of the methyl group and how it affects the structure of the carbocation.

6. **Formation of Alkene**: The newly formed tertiary carbocation can now lose a proton (H⁺) to form an alkene. There are two possible positions from which the proton can be removed, leading to two different alkene products.

**Hint**: Identify the two possible sites for proton removal and the resulting alkene structures.

7. **Identify the Major Product**: The two possible alkenes formed will have different degrees of substitution. The more substituted alkene is generally more stable due to hyperconjugation and the inductive effect.

**Hint**: Compare the stability of the two alkenes formed to determine which one is the major product.

### Conclusion:
The major product [X] will be the more stable alkene formed after the rearrangement and elimination reaction.

### Final Answer:
The product [X] will be the more substituted alkene resulting from the elimination reaction.