1, 1-dimethyl ethylene oxide when treated with water in presence of acid catalyst produces –

Options

A. ` CH_ 3 - overset( CH_3 ) overset( | ) underset( OH) underset( | ) C - CH _ 2 - OH `
B. `CH _ 3 - overset( CH_3) overset(| ) underset( OH ) underset( | ) (CHCH _ 2 OH ) `
C. `CH_3 - overset( CH_3 ) overset(| ) underset(OH ) underset(| ) C - CH _ 3 `
D. `CH_3 - underset( O) underset("||")C - CH _2 CH _ 3 `

Explanation

To solve the problem of what 1,1-dimethyl ethylene oxide produces when treated with water in the presence of an acid catalyst, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure of 1,1-Dimethyl Ethylene Oxide**:
- 1,1-Dimethyl ethylene oxide is a cyclic ether with the formula C5H10O. Its structure consists of a three-membered ring with two methyl groups attached to the first carbon atom.
- The structure can be represented as:
```
CH3
|
CH2 - O
|
CH2
```

2. **Protonation of the Ether Oxygen**:
- In the presence of an acid (H+), the oxygen atom in the epoxide ring will be protonated. This increases the electrophilicity of the carbon atoms in the ring.
- The oxygen donates its lone pair to the H+, forming a positively charged oxonium ion:
```
CH3
|
CH2 - O^+H
|
CH2
```

3. **Nucleophilic Attack by Water**:
- Water (H2O) acts as a nucleophile and attacks one of the carbon atoms adjacent to the positively charged oxygen.
- The nucleophilic attack opens the epoxide ring, leading to the formation of a carbocation intermediate:
```
CH3
|
CH2 - OH
|
CH2^+
```

4. **Deprotonation**:
- The positively charged oxygen can lose a proton (H+) to form the final product. The remaining hydroxyl group (OH) will be attached to the carbon atom.
- The final structure will be:
```
CH3
|
CH2 - OH
|
CH2
```

5. **Final Product**:
- The final product of this reaction is 1,1-dimethyl-2-hydroxyethane, which can be represented as:
```
CH3
|
CH2 - OH
|
CH3
```