`(CH_(3))_(3) CMgCl` on reaction with `D_(2)O` produces

Options

A. `(CH_(3))_(3)CD`
B. `(CH_(3))_(3)COD`
C. `(CD_(3))_(3)CD`
D. `(CD_(3))_(3)CD`

Answer

To solve the problem of what `(CH₃)₃C MgCl` produces when it reacts with `D₂O`, we can follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are `(CH₃)₃C MgCl` (tert-butyl magnesium chloride) and `D₂O` (deuterated water).

### Step 2: Understand the Mechanism
In this reaction, the carbon-magnesium bond in `(CH₃)₃C MgCl` is nucleophilic. The carbon atom bonded to magnesium has a strong tendency to donate its electron pair to a positively charged region, which in this case is the deuterium atom in `D₂O`.

### Step 3: Reaction Process
1. The carbon atom in `(CH₃)₃C MgCl` donates its electron pair to a deuterium atom from `D₂O`.
2. This results in the formation of a new bond between the carbon and deuterium, creating `(CH₃)₃C-D`.
3. The remaining part of `D₂O` (which is `OD`) will combine with `MgCl` to form `MgClOH`.

### Step 4: Write the Reaction
The overall reaction can be represented as:
\[ (CH₃)₃C MgCl + D₂O \rightarrow (CH₃)₃C-D + MgClOH \]

### Step 5: Identify the Products
From the reaction, we can see that the primary product is `(CH₃)₃C-D`, which is tert-butyl deuteride.

### Final Answer
The product of the reaction `(CH₃)₃C MgCl` with `D₂O` is `(CH₃)₃C-D`.

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