Treatment of ammonia with excess of ethyl chloride will yield

Options

A. Triethyl amine
B. Diethyl amine
C. Ethyl amine
D. Tetraethl ammonium chlorides

Answer

To solve the question regarding the treatment of ammonia with excess ethyl chloride, we will follow these steps:

### Step 1: Understand the Reactants
We have ammonia (NH₃) and ethyl chloride (C₂H₅Cl). Ammonia is a nucleophile, meaning it has a lone pair of electrons that can attack electrophiles, such as the carbon atom in ethyl chloride.

**Hint:** Identify the nature of the reactants. Ammonia is a nucleophile, while ethyl chloride is an electrophile.

### Step 2: First Reaction
When ammonia reacts with ethyl chloride, a substitution reaction occurs. The nitrogen in ammonia attacks the carbon in ethyl chloride, leading to the formation of ethylamine (C₂H₅NH₂) and the release of hydrochloric acid (HCl).

**Reaction:**
\[ \text{NH}_3 + \text{C}_2\text{H}_5\text{Cl} \rightarrow \text{C}_2\text{H}_5\text{NH}_2 + \text{HCl} \]

**Hint:** Recognize that the first product formed is ethylamine, and HCl is released as a byproduct.

### Step 3: Second Reaction
Since ethyl chloride is in excess, the ethylamine can react again with another molecule of ethyl chloride. The nitrogen in ethylamine can again act as a nucleophile, leading to the formation of diethylamine (C₂H₅)₂NH and another molecule of HCl.

**Reaction:**
\[ \text{C}_2\text{H}_5\text{NH}_2 + \text{C}_2\text{H}_5\text{Cl} \rightarrow (\text{C}_2\text{H}_5)_2\text{NH} + \text{HCl} \]

**Hint:** Each step involves the nucleophilic attack of the nitrogen on the ethyl chloride, leading to the formation of a new amine.

### Step 4: Third Reaction
Continuing this process, diethylamine can react with another molecule of ethyl chloride to form triethylamine (C₂H₅)₃N, with the release of HCl.

**Reaction:**
\[ (\text{C}_2\text{H}_5)_2\text{NH} + \text{C}_2\text{H}_5\text{Cl} \rightarrow (\text{C}_2\text{H}_5)_3\text{N} + \text{HCl} \]

**Hint:** Keep track of the number of ethyl groups attached to the nitrogen as you progress through the reactions.

### Step 5: Final Reaction
Finally, since ethyl chloride is still in excess, triethylamine can react with another molecule of ethyl chloride. The nitrogen now has four bonds (three ethyl groups and one lone pair), resulting in a positively charged quaternary ammonium ion, specifically tetraethylammonium chloride (C₂H₅)₄N⁺Cl⁻.

**Reaction:**
\[ (\text{C}_2\text{H}_5)_3\text{N} + \text{C}_2\text{H}_5\text{Cl} \rightarrow (\text{C}_2\text{H}_5)_4\text{N}^+ + \text{Cl}^- \]

**Hint:** Recognize that the final product is a quaternary ammonium salt due to the nitrogen forming four bonds.

### Conclusion
The treatment of ammonia with excess ethyl chloride yields tetraethylammonium chloride as the final product.

**Final Answer:** Tetraethylammonium chloride (Option D).