If heavy water is taken as solvent of normal water while performing cannizzaro reaction, the products of the reaction are

Options

A. `RCOO^(-)+RCH_(2)OH`
B. `RCOO^(-)+RCH_(2)OD`
C. `RCOOD+RCD_(2)OD`
D. `RCOO^(-)+RCD_(2)OD`

Explanation

To solve the problem regarding the Cannizzaro reaction in the presence of heavy water (D2O), we will follow these steps:

### Step-by-Step Solution:

1. **Understanding the Cannizzaro Reaction**:
The Cannizzaro reaction occurs with aldehydes that do not have an alpha hydrogen. In this reaction, one molecule of the aldehyde is oxidized to a carboxylic acid, while another molecule is reduced to an alcohol.

2. **Identifying the Solvent**:
In this case, instead of normal water (H2O), we are using heavy water (D2O). This means that the hydroxide ion (OH-) from water will be replaced by deuteroxide ion (OD-).

3. **Reaction Mechanism**:
- The aldehyde (RCHO) reacts with the deuteroxide ion (OD-).
- The base (OD-) attacks the carbonyl carbon of the aldehyde, forming a tetrahedral intermediate.
- This intermediate collapses, leading to the formation of a carboxylate ion (RC(=O)O-) and a deuterated alcohol (RCH2OD).

4. **Final Products**:
- The products of the reaction will be:
- A carboxylate ion: RC(=O)O- (where the O is negatively charged)
- A deuterated alcohol: RCH2OD

5. **Conclusion**:
Therefore, the products of the Cannizzaro reaction when heavy water is used as a solvent are:
- RC(=O)O- (carboxylate ion)
- RCH2OD (deuterated alcohol)

### Final Answer:
The products of the reaction are:
- RC(=O)O- (carboxylate ion)
- RCH2OD (deuterated alcohol)