`H_(3)C-CH_(2)-underset(O)underset(||)(C)-CH_(3)underset(D_(2)O)overset(NaBH_(4))to` Product ?

Options

A. `H_(3)C-underset(H)underset(|)overset(OD)overset(|)(C)-CH_(2)-CH_(3)`
B. `H_(3)C-underset(D)underset(|)overset(OH)underset(|)(C)-CH_(2)-CH_(3)`
C. `H_(3)C-underset(D)underset(|)overset(OD)overset(|)(C)-CH_(2)-CH_(3)`
D. none

Answer

To solve the problem, we need to analyze the reaction step by step. The starting material is a ketone, and we will be reducing it using sodium borohydride (NaBH4) in the presence of deuterium oxide (D2O). Let's break down the process:

### Step 1: Identify the Reactants
The starting material is:
\[ \text{CH}_3\text{CH}_2\text{C}(\text{O})\text{CH}_3 \]
This is a ketone (specifically, butan-2-one).

### Step 2: Reduction of the Ketone
Sodium borohydride (NaBH4) is a reducing agent that will reduce the ketone to an alcohol. The mechanism involves the hydride ion (H-) from NaBH4 attacking the carbonyl carbon (C=O) of the ketone.

The reaction can be represented as:
\[ \text{CH}_3\text{CH}_2\text{C}(\text{O})\text{CH}_3 + \text{NaBH}_4 \rightarrow \text{CH}_3\text{CH}_2\text{C}(\text{OH})\text{CH}_3 \]

### Step 3: Formation of the Alcohol
After the hydride attack, the carbonyl group is converted into a hydroxyl group (OH), resulting in:
\[ \text{CH}_3\text{CH}_2\text{C}(\text{OH})\text{CH}_3 \]
This is butan-2-ol.

### Step 4: Hydrolysis with D2O
Now, we perform hydrolysis using deuterium oxide (D2O) instead of regular water. During this step, the hydroxyl group (OH) will be replaced by a deuterated hydroxyl group (OD).

The reaction can be represented as:
\[ \text{CH}_3\text{CH}_2\text{C}(\text{OH})\text{CH}_3 + \text{D}_2\text{O} \rightarrow \text{CH}_3\text{CH}_2\text{C}(\text{OD})\text{CH}_3 \]

### Step 5: Final Product
The final product after the reaction with D2O will be:
\[ \text{CH}_3\text{CH}_2\text{C}(\text{OD})\text{CH}_3 \]

### Conclusion
The product of the reaction is:
\[ \text{CH}_3\text{CH}_2\text{C}(\text{OD})\text{CH}_3 \]

### Matching with Options
From the options given, we can conclude that the correct product corresponds to:
- **Option 1: CH3ODCH2CH3**