Sodium salt of acetic acid reacts with ethyl iodide to give

Options

A. `CH_3-overset(O)overset(||)C-OCH_3`
B. `CH_3-CH_2-O-overset(O)overset(||)C-CH_3`
C. `CH_3-CH_2-overset(O)overset(||)C-O-OCH_3`
D.

Answer

To solve the question of what product is formed when sodium salt of acetic acid reacts with ethyl iodide, we can follow these steps:

### Step 1: Identify the Reactants
The reactants are:
- Sodium salt of acetic acid (sodium acetate, CH₃COONa)
- Ethyl iodide (C₂H₅I)

### Step 2: Understand the Structure of Sodium Acetate
Sodium acetate can be represented as:
- CH₃COO⁻ Na⁺
This indicates that the acetate ion (CH₃COO⁻) has a negative charge.

### Step 3: Recognize the Nucleophilic Nature of Acetate Ion
The acetate ion (CH₃COO⁻) acts as a nucleophile due to its negative charge. It can attack electrophilic centers.

### Step 4: Identify the Electrophilic Center in Ethyl Iodide
In ethyl iodide (C₂H₅I), the carbon atom bonded to iodine is electrophilic because iodine is more electronegative and pulls electron density away from the carbon, making it partially positive.

### Step 5: Nucleophilic Attack
The acetate ion (CH₃COO⁻) will attack the electrophilic carbon in ethyl iodide (C₂H₅I). The reaction can be represented as:
- CH₃COO⁻ + C₂H₅I → CH₃COC₂H₅ + I⁻

### Step 6: Formation of the Product
The product formed from this reaction is ethyl acetate (CH₃COC₂H₅) and iodide ion (I⁻) as a leaving group.

### Conclusion
Thus, the final product of the reaction between sodium salt of acetic acid and ethyl iodide is **ethyl acetate (CH₃COC₂H₅)**.

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