`(CH_(3))_(3)CMgBr` 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)`OD

Answer

To solve the question, we need to analyze the reaction of the compound `(CH₃)₃CMgBr` (which is tert-butylmagnesium bromide) with `D₂O` (heavy water). Here’s a step-by-step breakdown of the reaction:

### Step 1: Identify the structure of the compound
The compound `(CH₃)₃CMgBr` is tert-butylmagnesium bromide. It consists of a tert-butyl group `(CH₃)₃C-` bonded to a magnesium bromide. The carbon in the tert-butyl group carries a negative charge, making it a nucleophile.

**Hint:** Recognize that Grignard reagents like `(CH₃)₃CMgBr` have a carbon atom that behaves as a nucleophile due to the presence of the magnesium atom.

### Step 2: Understand the reaction with D₂O
When `(CH₃)₃CMgBr` reacts with `D₂O`, the nucleophilic carbon `(CH₃)₃C-` will attack the deuterium in `D₂O`. The oxygen in `D₂O` is more electronegative and will hold a partial negative charge, making the deuterium a good electrophile.

**Hint:** Remember that nucleophiles attack electrophiles, and in this case, the nucleophile is the carbon with a negative charge.

### Step 3: Write the reaction
The nucleophilic attack will lead to the formation of a new bond between the carbon and deuterium. The reaction can be represented as:
\[ (CH₃)₃C^- + D₂O \rightarrow (CH₃)₃CD + OD^- \]
Here, `(CH₃)₃CD` is the product formed, and `OD^-` is the byproduct.

**Hint:** Be careful to account for the byproducts in the reaction. The deuterium replaces the hydrogen in the alkyl group.

### Step 4: Identify the product
The product of the reaction is `(CH₃)₃CD`, which is deuterated tert-butyl. This indicates that one of the hydrogen atoms in tert-butyl has been replaced by deuterium.

**Hint:** Look for the final product and ensure that the substitution of hydrogen by deuterium is correctly represented.

### Final Answer:
The product of the reaction of `(CH₃)₃CMgBr` with `D₂O` is `(CH₃)₃CD`.