`Me_3 C-MgCl` on reaction with `D_2 O` produces :

Options

A. `Me_3 CD`
B. `Me_3 COD`
C. `(CD)_3 CD`
D. `(CD)_3 COD`.

Explanation

To solve the question regarding the reaction of the Grignard reagent \( \text{Me}_3\text{C-MgCl} \) with \( \text{D}_2\text{O} \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Grignard Reagent**:
The compound \( \text{Me}_3\text{C-MgCl} \) is a Grignard reagent where \( \text{Me}_3\text{C} \) (tert-butyl group) is the alkyl group, and \( \text{MgCl} \) is the magnesium halide.

2. **Understand the Reaction with D2O**:
Grignard reagents react with water or deuterated water (D2O) to form hydrocarbons and magnesium hydroxides. In this case, \( \text{D}_2\text{O} \) will provide deuterium (D) instead of hydrogen (H).

3. **Reaction Mechanism**:
The Grignard reagent can be represented as \( \text{R-MgX} \), where R is the alkyl group (in this case, \( \text{Me}_3\text{C} \)), and X is the halide (Cl in this case). The alkyl group carries a partial negative charge, while magnesium carries a partial positive charge.

4. **Formation of Products**:
When \( \text{Me}_3\text{C-MgCl} \) reacts with \( \text{D}_2\text{O} \), the alkyl group \( \text{Me}_3\text{C}^- \) will react with the deuterium oxide. The reaction can be summarized as:
\[ \text{Me}_3\text{C-MgCl} + \text{D}_2\text{O} \rightarrow \text{Me}_3\text{C-D} + \text{MgCl(OD)} \]
Here, \( \text{Me}_3\text{C-D} \) is the product where one of the hydrogen atoms in the tert-butyl group is replaced by deuterium.

5. **Final Products**:
The final products of the reaction are:
- \( \text{Me}_3\text{C-D} \) (tert-butyl deuteride)
- \( \text{MgCl(OD)} \) (magnesium deuteroxide chloride)

### Conclusion:
Thus, the reaction of \( \text{Me}_3\text{C-MgCl} \) with \( \text{D}_2\text{O} \) produces \( \text{Me}_3\text{C-D} \) and \( \text{MgCl(OD)} \).

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