The major product formed in the following reation is
`CH_(3)-underset(H)underset(|)overset(CH_(3))overset(|)C-CH_(2)-Br underset(CH_(3)OH)overset(CH_(3)ONa)rarr` .

Options

A. `CH_(3)CH=CH-CH=CHCH_(3)`
B. `CH_(2)=CHCH=CHCH_(2)CH_(3)`
C. `CH_(3)CH=C=CHCH_(2)CH_(3)`
D. `CH_(2)=CHCH_(2)CH=CHCH_(3)`

Answer

To determine the major product formed in the reaction of the given alkyl halide with sodium methoxide (CH3ONa) and methanol (CH3OH), we will follow these steps:

### Step 1: Identify the Reactants
The reactant is an alkyl halide, specifically 2-bromo-2-methylbutane (CH3)2C(CH2Br)CH3. It is reacting with sodium methoxide (CH3ONa) and methanol (CH3OH).

### Step 2: Determine the Type of Reaction
Since we have a strong base (CH3ONa), the reaction can proceed via either an E2 elimination reaction or an SN2 substitution reaction. However, given that we are dealing with a tertiary alkyl halide, the E2 mechanism is favored.

### Step 3: Identify the Alpha and Beta Carbons
In E2 elimination, the alpha carbon is the one bonded to the leaving group (Br), and the beta carbons are the adjacent carbons. In our case:
- Alpha carbon: The carbon attached to Br.
- Beta carbons: The adjacent carbons.

### Step 4: Elimination Process
In the E2 mechanism, the bromine (Br) will leave as a leaving group, and a hydrogen (H) will be abstracted from one of the beta carbons by the strong base (methoxide ion, CH3O-). This results in the formation of a double bond between the alpha and beta carbon.

### Step 5: Form the Product
When Br leaves and H is abstracted, the double bond forms between the alpha carbon and one of the beta carbons. The major product formed will be 2-methylpropene (CH2=C(CH3)CH3).

### Step 6: Write the Final Product
The final product of the reaction is 2-methylpropene (CH2=C(CH3)CH3), along with the by-products sodium bromide (NaBr) and methanol (CH3OH).

### Final Answer
The major product formed in the reaction is 2-methylpropene (CH2=C(CH3)CH3).

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