The enol form acetone after treatment with `D_(2)O` gives:

Options

A. `CH_(3)-overset(OD)overset(|)C=CH_(2)`
B. `CD_(3)-overset(O)overset(||)C-CD_(3)`
C. `CH_(2)=overset(OH)overset(|)C-CH_(2)D`
D. `CD_(2)=overset(OD)overset(|)C-CD_(3)`

Explanation

To solve the problem of what the enol form of acetone yields after treatment with D₂O (heavy water), we will follow these steps:

### Step 1: Identify the enol form of acetone
Acetone (C₃H₆O) has the following structure:
```
O
||
CH₃-C-CH₃
```
The enol form of acetone is formed by the migration of one of the alpha hydrogens to the carbonyl oxygen, resulting in:
```
OH
|
CH₂=C-CH₃
```
This can be represented as CH₂=COH-CH₃.

### Step 2: React the enol form with D₂O
When the enol form of acetone is treated with D₂O, the hydroxyl group (OH) will interact with D₂O. The deuterium (D) from D₂O will replace the hydrogen (H) in the enol form:
```
OD
|
CH₂=C-CH₃
```
This gives us the intermediate:
```
OD
|
CH₂=C(=O)H-CH₃
```

### Step 3: Elimination of H⁺
The next step involves the elimination of H⁺ from the intermediate. The remaining OD⁻ will remove the H⁺ from the enol form, leading to:
```
OD
|
CH₂=C(=O)-CH₃
```

### Step 4: Convert back to keto form
Now, we convert this back to the keto form:
```
O
||
CH₃-C-CH₂D
```
This structure indicates that one of the alpha hydrogens has been replaced by deuterium.

### Step 5: Repeating the process
This process can be repeated. Each time the keto form is converted back to the enol form, the alpha hydrogens will be replaced by deuterium. After several iterations, all alpha hydrogens will be replaced by deuterium.

### Final Structure
After all alpha hydrogens are replaced by deuterium, the final product will be:
```
O
||
CD₃-C-CH₃
```
This can also be represented as:
```
CD₂
|
CD₃-C(=O)-CD₃
```
Thus, the final product after treatment with D₂O is:
```
C(=O)CD₃CD₃
```

### Conclusion
The final product of the enol form of acetone after treatment with D₂O is:
**C(=O)CD₃CD₃**