Treatment of ammonia with excess of ethyl iodide will yield

Options

A. diethylamine
B. ethylamine
C. triethylamine
D. tetraethylammonium iodide

Answer

To solve the question regarding the treatment of ammonia with excess ethyl iodide, we can break down the process step by step.

### Step-by-Step Solution:

1. **Identify the Reactants**:
- We have ammonia (NH₃) and excess ethyl iodide (C₂H₅I).

2. **Understand the Reaction Mechanism**:
- Ammonia acts as a nucleophile due to the presence of a lone pair of electrons on the nitrogen atom. Ethyl iodide has a carbon atom that is electrophilic because the iodine atom is a good leaving group.

3. **First Reaction**:
- The lone pair on nitrogen attacks the electrophilic carbon in ethyl iodide, leading to the formation of ethylamine (C₂H₅NH₂) and the release of iodide ion (I⁻).
- Reaction:
\[ NH₃ + C₂H₅I \rightarrow C₂H₅NH₂ + HI \]

4. **Second Reaction**:
- Ethylamine (C₂H₅NH₂) can further react with another molecule of ethyl iodide. The nitrogen atom in ethylamine again acts as a nucleophile.
- Reaction:
\[ C₂H₅NH₂ + C₂H₅I \rightarrow C₂H₅NH(C₂H₅) + HI \]

5. **Subsequent Reactions**:
- This process can continue, with each ethylamine molecule reacting with more ethyl iodide, leading to the formation of diethylamine (C₂H₅)₂NH, triethylamine (C₂H₅)₃N, and finally, tetraethylammonium iodide (C₂H₅)₄N⁺I⁻.

6. **Final Product**:
- After multiple reactions, the final product obtained from the treatment of ammonia with excess ethyl iodide is tetraethylammonium iodide (C₂H₅)₄N⁺I⁻.

### Conclusion:
The treatment of ammonia with excess ethyl iodide will yield tetraethylammonium iodide.

---