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, we need to determine the structures of the alkenes A and B that are formed when 3,3-Dimethyl-2-butanol is heated with phosphoric acid (H₃PO₄). The reaction leads to an 80:20 mixture of these alkenes.

### Step-by-Step Solution:

1. **Identify the Structure of 3,3-Dimethyl-2-butanol**:
- The compound has 4 carbon atoms (butanol) with two methyl groups on the third carbon.
- The structure can be drawn as:
```
CH3
|
CH3-C-CH2-CH3
|
OH
```

2. **Protonation of the Alcohol**:
- When heated with phosphoric acid, the hydroxyl (OH) group of the alcohol is protonated, forming a better leaving group (water).
- The reaction can be represented as:
```
CH3-C(OH)-CH2-CH3 + H⁺ → CH3-C(OH₂⁺)-CH2-CH3
```

3. **Formation of Carbocation**:
- The protonated alcohol loses a water molecule, generating a carbocation. The most stable carbocation will form:
```
CH3-C⁺-CH2-CH3
```

4. **Rearrangement to More Stable Carbocation**:
- The carbocation can undergo a rearrangement (methyl shift) to form a more stable tertiary carbocation:
```
CH3-CH-C⁺-CH3
```

5. **Elimination to Form Alkenes**:
- From the carbocation, elimination of a proton (H⁺) will lead to the formation of alkenes.
- The two possible alkenes formed are:
- Alkene A (more stable):
```
CH3-CH=CH-CH3 (2-butene)
```
- Alkene B (less stable):
```
CH2=C(CH3)-CH2-CH3 (1-butene)
```

6. **Ratio of Alkenes**:
- The more stable alkene (A) is formed in a greater amount (80%) compared to the less stable alkene (B), which is formed in a smaller amount (20%).

### Final Structures:
- Alkene A (80%): 2-butene (CH3-CH=CH-CH3)
- Alkene B (20%): 1-butene (CH2=C(CH3)-CH2-CH3)