When 3,3-Dimethyl-2-butanol is heated with acid `H_(3)PO_(4)`, it is converted into an `80 : 20` mixture of alkene A and B respectively. The structure of A and B are:

Options

A.
B.
C.
D.

Explanation

To solve the problem of identifying the structures of alkenes A and B formed from the dehydration of 3,3-Dimethyl-2-butanol when heated with H3PO4, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure of 3,3-Dimethyl-2-butanol**:
- The compound has a four-carbon chain (butanol) with an -OH group on the second carbon and two methyl groups on the third carbon.
- The structure can be represented as:
```
CH3
|
CH3-C-CH2-CH3
|
OH
```

2. **Protonation of the Alcohol**:
- When 3,3-Dimethyl-2-butanol is heated with H3PO4, the -OH group gets protonated, forming a better leaving group (water).
- The protonation leads to the formation of a carbocation intermediate:
```
CH3
|
CH3-C^+-CH2-CH3
|
H2O
```

3. **Carbocation Rearrangement**:
- The initial carbocation is a secondary carbocation, which can rearrange to form a more stable tertiary carbocation by a methyl shift:
```
CH3
|
CH3-C-CH3
|
CH2^+
```

4. **Elimination to Form Alkenes**:
- From the tertiary carbocation, two possible alkenes can be formed by the elimination of a proton (H+):
- **Alkene A (Major Product)**: Elimination of H from the tertiary carbon leads to:
```
CH3
|
CH3-C=CH2
```
- **Alkene B (Minor Product)**: Elimination of H from the secondary carbon leads to:
```
CH2=C(CH3)-CH3
```

5. **Determine the Ratio of Alkenes**:
- The problem states that the mixture of alkenes A and B is in the ratio of 80:20, indicating that Alkene A is the major product and Alkene B is the minor product.

### Final Structures:
- **Alkene A**: 2,3-Dimethylpropene (Major Product)
- **Alkene B**: 2-Methylpropene (Minor Product)