When di-isopropyl cadmium is treated with ethanoyl chloride, the product is :

Options

A.
B.
C.
D.

Explanation

To solve the question of what product is formed when di-isopropyl cadmium is treated with ethanoyl chloride, we can follow these steps:

### Step 1: Identify the Reactants
- Di-isopropyl cadmium is represented as (CH3)2CH-Cd. This means it has two isopropyl groups attached to cadmium.
- Ethanoyl chloride (also known as acetyl chloride) is represented as CH3COCl.

### Step 2: Understand the Reaction Mechanism
- Di-isopropyl cadmium is a nucleophile due to the presence of the cadmium atom, which can donate electrons.
- Ethanoyl chloride, on the other hand, is an acyl chloride that can undergo nucleophilic attack due to the electrophilic carbon in the carbonyl group (C=O).

### Step 3: Draw the Structures
- Draw the structure of di-isopropyl cadmium:
```
CH3
|
CH3 - C - Cd
|
CH3
```
- Draw the structure of ethanoyl chloride:
```
O
||
CH3 - C - Cl
```

### Step 4: Perform the Nucleophilic Attack
- The nucleophile (di-isopropyl cadmium) attacks the carbonyl carbon of ethanoyl chloride, leading to the formation of a tetrahedral intermediate.
- The chloride ion (Cl-) is expelled during this process.

### Step 5: Form the Product
- The product formed after the nucleophilic attack is:
```
O
||
CH3 - C - (C(CH3)2)
```
- This product is 2-isopropyl-2-oxopropane (or 2-isopropylacetone) along with cadmium chloride (CdCl2) as a byproduct.

### Step 6: Write the Final Equation
- The overall reaction can be summarized as:
```
2(CH3)2CH-Cd + 2CH3COCl → 2(CH3)2C(=O)CH + CdCl2
```

### Conclusion
The product formed when di-isopropyl cadmium is treated with ethanoyl chloride is 2-isopropyl-2-oxopropane (or 2-isopropylacetone) along with cadmium chloride.