When diethyl cadmium `[(C_(2)H_(5))_(2)Cd]` is treated with acetyl chloride `[CH_(3)COCl]`, the main product likely to be:

Options

A. Acetone
B. Methyl ethyl ketone
C. Diethyl ketone
D. Acetaldehyde

Explanation

To solve the question regarding the reaction of diethyl cadmium with acetyl chloride, we can follow these steps:

### Step 1: Identify the Reactants
We have diethyl cadmium, which is represented as \[(C_2H_5)_2Cd\], and acetyl chloride, represented as \[CH_3COCl\].

### Step 2: Understand the Reaction Mechanism
Diethyl cadmium acts as a nucleophile due to the presence of the alkyl groups. In this case, the ethyl groups (\[C_2H_5\]) will attack the carbonyl carbon of acetyl chloride. The carbonyl carbon in acetyl chloride is electrophilic because of the presence of the electronegative chlorine atom.

### Step 3: Nucleophilic Attack
When the nucleophile (ethyl group from diethyl cadmium) attacks the carbonyl carbon of acetyl chloride, a tetrahedral intermediate is formed. This intermediate has the structure \[CH_3C(OH)(C_2H_5)Cl\].

### Step 4: Elimination of Chlorine
In the next step, the chlorine atom (which is a good leaving group) will leave, resulting in the formation of a ketone. The oxygen will carry a negative charge, which will stabilize the structure.

### Step 5: Final Product Formation
After the elimination of chlorine, the resulting product will be \[CH_3C(=O)C_2H_5\], which is methyl ethyl ketone (also known as 3-pentanone).

### Conclusion
The main product formed when diethyl cadmium is treated with acetyl chloride is methyl ethyl ketone.

### Final Answer
The correct answer is **methyl ethyl ketone**.

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