Sodium salt of acetic acid reacts with ethyl iodide to give

Answer

To solve the problem 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 in this reaction are:
- Sodium salt of acetic acid (sodium acetate), which has the formula CH₃COONa.
- Ethyl iodide (iodoethane), which has the formula CH₃CH₂I.

### Step 2: Understand the Reaction Mechanism
The reaction involves a nucleophilic substitution mechanism. The acetate ion (CH₃COO⁻) acts as a nucleophile and attacks the electrophilic carbon in ethyl iodide (CH₃CH₂I). This results in the displacement of the iodide ion (I⁻).

### Step 3: Write the Reaction
The reaction can be represented as follows:
\[ \text{CH}_3\text{COONa} + \text{CH}_3\text{CH}_2\text{I} \rightarrow \text{CH}_3\text{COOCH}_2\text{CH}_3 + \text{NaI} \]

### Step 4: Identify the Product
The product formed from this reaction is ethyl acetate (CH₃COOCH₂CH₃) and sodium iodide (NaI). Ethyl acetate is an ester formed from the reaction of acetic acid and ethanol.

### Step 5: Conclusion
Thus, when sodium salt of acetic acid reacts with ethyl iodide, the product formed is ethyl acetate.

### Final Answer
The correct answer is ethyl acetate (CH₃COOCH₂CH₃).

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