The major product in the reaction is
`CH_(3)-underset(CH_(3))underset(|)CH-CH_(2)Br underset(CH_(3)OH)overset(CH_(3)O^(-))to`

Options

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

Explanation

To determine the major product of the reaction between the given alkyl halide and methoxide ion in methanol, we can follow these steps:

### Step 1: Identify the Reactants
The reactants are:
- An alkyl halide: `CH3-CH(CH3)-CH2Br`
- A nucleophile/base: Methoxide ion (`CH3O^(-)`)
- A solvent: Methanol (`CH3OH`)

### Step 2: Analyze the Reaction Conditions
The methoxide ion is a strong nucleophile and a strong base. In the presence of a bulky alkyl halide, we need to consider whether a substitution (SN2) or elimination (E2) reaction will occur.

### Step 3: Consider Steric Hindrance
The alkyl halide has two methyl groups adjacent to the carbon that is bonded to the bromine atom. This creates significant steric hindrance, making it difficult for the methoxide ion to perform a backside attack necessary for an SN2 reaction.

### Step 4: Determine the Reaction Type
Due to the steric hindrance from the two adjacent methyl groups, the SN2 pathway is not favored. Instead, the methoxide ion will act as a base and abstract a proton from the β-carbon (the carbon adjacent to the one bonded to Br). This leads to an elimination reaction (E2).

### Step 5: Perform the Elimination Reaction
1. The methoxide ion abstracts a proton from the β-carbon.
2. The bromine atom (leaving group) departs, resulting in the formation of a double bond between the α and β carbons.

### Step 6: Write the Major Product
The major product of the elimination reaction is:
- An alkene: `CH3-CH=CH-CH3` (butene)

### Final Product
The major product of the reaction is **butene**.

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