The major product of the following reaction is:
`CH_(3)-underset(H)underset(|)overset(CH_(3))overset(|)C-underset(Br)underset(|)(CHCH_(3)) overset(CH_(3)OH) to`

Options

A. `CH_(3)-underset(H)underset(|)overset(CH_(3))overset(|)C-underset(OCH_(3))underset(|)(CHCH_(3))`
B. `CH_(3)-overset(CH_(3))overset(|)C=CHCH_(3)`
C. `CH_(3)-overset(CH_(3))overset(|)underset(OCH_(3))underset(|)C -CH_(2)CH_(3)`
D. `CH_(3)-underset(H)underset(|)overset(CH_(3))overset(|)C-CH= CH_(2)`

Explanation

To find the major product of the given reaction, we will follow these steps:

### Step 1: Identify the Reactants
The reactant is a compound with a bromine (Br) substituent and several alkyl groups. The structure can be simplified as follows:
- A carbon chain with a Br atom attached to one of the carbons.
- The presence of a methanol (CH3OH) in the reaction indicates that it will act as a nucleophile.

**Hint:** Look for the functional groups in the reactant and identify the nucleophile in the reaction.

### Step 2: Nucleophilic Attack
In the presence of methanol (CH3OH), the Br atom will leave as a bromide ion (Br⁻) due to the nucleophilic attack by the hydroxyl group (OH) from methanol. The alkyl group (CH3) will stabilize the positive charge that develops during this process.

**Hint:** Remember that nucleophiles attack electrophilic centers, leading to the departure of leaving groups.

### Step 3: Formation of a Carbocation
After the Br⁻ leaves, a carbocation is formed on the carbon that was previously bonded to Br. We need to check if this carbocation is stable. If there are adjacent alkyl groups, they can stabilize the carbocation through hyperconjugation or inductive effects.

**Hint:** Assess the stability of the carbocation formed; tertiary carbocations are more stable than secondary or primary.

### Step 4: Rearrangement (if applicable)
If the carbocation formed is not stable, a rearrangement may occur. This could involve a hydride shift or a methyl shift to form a more stable carbocation.

**Hint:** Look for possible shifts that can lead to a more stable carbocation.

### Step 5: Nucleophilic Attack by OH
Once a stable carbocation is formed, the hydroxyl group (OH) from methanol will attack the positively charged carbon, resulting in the formation of an alcohol.

**Hint:** The nucleophile will attack the most stable carbocation available.

### Step 6: Final Product
The final product will be an alcohol where the OH group replaces the Br atom. If a rearrangement occurred, ensure that the final structure reflects that.

**Hint:** Draw the final structure carefully to ensure all valencies are satisfied.

### Conclusion
The major product of the reaction is an alcohol formed from the original compound after the substitution of Br with OH, considering any rearrangements that may have occurred during the reaction.

### Final Answer
The major product of the reaction is the corresponding alcohol after the bromine atom is replaced by the hydroxyl group from methanol.