`2DCDO overset(OH^(-))rarr [X] and [Y]` are

Options

A. `DCOO^(-), D_(2)CHOH`
B. `HCOO^(-), CH_(3)OH`
C. `HCOO^(-), CD_(3)OH`
D. `DCOO^(-), CD_(3)OH`

Answer

To solve the question regarding the reaction of 2DCDO with OH⁻, we can break down the process step by step.

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactant is 2DCDO, which is a carbonyl compound (specifically a ketone or aldehyde). The nucleophile involved in the reaction is hydroxide ion (OH⁻).

2. **Nucleophilic Attack**:
The hydroxide ion (OH⁻) acts as a nucleophile and attacks the carbonyl carbon of 2DCDO. This carbon is electrophilic due to the partial positive charge created by the oxygen's electronegativity.

3. **Formation of Tetrahedral Intermediate**:
Upon the nucleophilic attack, a tetrahedral intermediate is formed. The carbon now has five bonds (four from the tetrahedral structure and one from the carbonyl oxygen). This intermediate is unstable.

4. **Rearrangement and Proton Transfer**:
The tetrahedral intermediate will rearrange. One of the bonds must break to accommodate the new bond formed by the hydroxide. The negative charge on the oxygen can facilitate a proton transfer, leading to the formation of an alcohol.

5. **Final Products**:
The final products of the reaction will be:
- **[X]**: A compound with a hydroxyl group (alcohol).
- **[Y]**: A compound resulting from the rearrangement of the original molecule.

6. **Conclusion**:
The products can be summarized as:
- **[X]**: DCOH (an alcohol)
- **[Y]**: CD3OH (another alcohol)

### Final Answer:
The products [X] and [Y] are DCOH and CD3OH respectively.

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