The major product of the following reaction is :
`CH_(3)underset(overset(|)(Br))(C)HCH_(2)underset(overset(|)(Br))(C)HCH_(2)CH_(3) underset("heat")overset(KOH, CH_(3)OH)(to)`

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 solve the problem, we need to analyze the reaction of the given dibromoalkane with KOH in methanol under heat. The process involves elimination reactions, specifically an E2 mechanism, where bromine atoms and hydrogen atoms are eliminated to form alkenes. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the Structure of the Reactant
The reactant is a dibromoalkane with the following structure:
```
Br
|
CH3-CH-CH2-CH-CH2-CH3
|
Br
```
This compound has six carbon atoms with bromine substituents on the second and fourth carbons.

### Step 2: Determine the Elimination Mechanism
Since KOH is a strong base and methanol is an alcohol solvent, the reaction will favor an elimination mechanism (E2). In this mechanism, the base will remove a hydrogen atom (H+) from a carbon adjacent to the carbon bearing the bromine (beta hydrogen), while the bromine (Br-) will leave simultaneously.

### Step 3: Identify Beta Hydrogens
For the given structure, we can identify the beta hydrogens:
- Beta 1: H on the carbon adjacent to the second carbon (C2)
- Beta 2: H on the carbon adjacent to the fourth carbon (C4)
- Beta 3: H on the carbon adjacent to the third carbon (C3)

### Step 4: Predict the Possible Products
1. **Elimination of Beta 1 and Beta 2:**
- Removing H from Beta 1 and Br from C2 leads to:
```
CH2=CH-CH=CH-CH2-CH3
```
- This is a conjugated diene.

2. **Elimination of Beta 1 and Beta 3:**
- Removing H from Beta 1 and Br from C4 leads to:
```
CH2=CH-CH2-CH=CH-CH3
```
- This is also a conjugated diene.

3. **Elimination of Beta 2 and Beta 3:**
- Removing H from Beta 2 and Br from C2 leads to:
```
CH3-CH=CH-CH=CH-CH3
```
- This is a conjugated diene.

4. **Elimination of Beta 2 and Beta 2:**
- Removing H from both Beta 2 positions leads to:
```
CH3-CH=CH-CH=CH-CH3
```
- This is a conjugated diene as well.

### Step 5: Determine the Major Product
Among the products formed, we need to analyze their stability:
- The product with the most substituted double bonds will be the most stable.
- The product formed by removing Beta 2 and Beta 3 has two double bonds that are conjugated and is more substituted.

Thus, the most stable product is:
```
CH3-CH=CH-CH=CH-CH3
```

### Conclusion
The major product of the reaction is:
```
CH3-CH=CH-CH=CH-CH3
```