Reaction between `(C_(2)H_(5))_(2)Cd and CH_(3)COCl` leads to the formation of

Options

A. Diethyl ketone`
B. Ethyl methyl ketone
C. Dimethyl ketone
D. Acetaldehyde

Answer

To solve the problem regarding the reaction between `(C2H5)2Cd` (diethylcadmium) and `CH3COCl` (acetyl chloride), we will follow these steps:

### Step 1: Identify the Reactants
The reactants in the reaction are:
- Diethylcadmium `(C2H5)2Cd`
- Acetyl chloride `CH3COCl`

### Step 2: Understand the Nature of the Reactants
Diethylcadmium is a nucleophile because cadmium (Cd) has a partial positive charge, and the ethyl groups `(C2H5)` carry a negative charge. Acetyl chloride contains a carbonyl group (C=O) which is electron-deficient due to the electronegativity of oxygen.

### Step 3: Nucleophilic Attack
The nucleophile (the ethyl group from diethylcadmium) will attack the carbonyl carbon of the acetyl chloride. The carbonyl carbon is electrophilic due to the partial positive charge created by the electronegative oxygen.

### Step 4: Formation of a Tetrahedral Intermediate
When the nucleophile attacks the carbonyl carbon, it forms a tetrahedral intermediate. This intermediate will have a negative charge on the oxygen atom.

### Step 5: Elimination of Chloride Ion
The tetrahedral intermediate will then collapse, leading to the elimination of the chloride ion (Cl⁻) from the acetyl chloride. This results in the formation of a ketone.

### Step 6: Identify the Product
The product formed from this reaction is `ethyl methyl ketone`, also known as `butan-2-one`. The structure can be represented as:
- Ethyl group (C2H5) on one side
- Methyl group (CH3) on the other side

### Final Product
Thus, the final product of the reaction between `(C2H5)2Cd` and `CH3COCl` is **ethyl methyl ketone** (or **butan-2-one**).

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