Ethylamine when treated with acetyl chloride we get a compound of the formula .

Options

A. `CH_3 - underset(O ) underset( ||) C-underset( CH_3) underset(|) N-NH -C_2H_5`
B. `C_2H_5 -underset(O) underset(||) C-NH -C_2 H_5`
C. `CH_3-underset(O) underset( ||)C- underset(H) underset(|) N-CH_2 -CH_3`
D. `CH_3 -underset(O) underset(||)C- underset(H) underset(|) N-underset(CH_3 ) underset( |) CH -CH_3`

Answer

To solve the problem of what compound is formed when ethylamine reacts with acetyl chloride, we can follow these steps:

### Step 1: Identify the Reactants
- **Ethylamine** has the formula \( \text{C}_2\text{H}_5\text{NH}_2 \) (or \( \text{CH}_3\text{CH}_2\text{NH}_2 \)).
- **Acetyl chloride** has the formula \( \text{CH}_3\text{COCl} \).

### Step 2: Understand the Reaction Mechanism
- Ethylamine contains a nitrogen atom with a lone pair of electrons. This lone pair can act as a nucleophile.
- Acetyl chloride has an electron-deficient carbon atom (the carbon in the carbonyl group), which makes it susceptible to nucleophilic attack.

### Step 3: Nucleophilic Attack
- The lone pair on the nitrogen in ethylamine attacks the carbon atom of the carbonyl group in acetyl chloride.
- This results in the formation of a tetrahedral intermediate.

### Step 4: Formation of the Product
- As the tetrahedral intermediate collapses, the chloride ion (\( \text{Cl}^- \)) is expelled, and a bond is formed between the nitrogen and the carbonyl carbon.
- The product formed is an amide, specifically N-ethylacetamide.

### Step 5: Write the Final Structure
- The structure of the product can be written as:
\[ \text{CH}_3\text{CH}_2\text{NHCOCH}_3 \]
- This compound can also be represented as \( \text{C}_4\text{H}_9\text{NO} \).

### Conclusion
The compound formed when ethylamine reacts with acetyl chloride is N-ethylacetamide, with the formula \( \text{C}_4\text{H}_9\text{NO} \).

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