Phosgene `(COCl_(2))+1" mol"C_(2)H_(5)OH overset(NH_(3))to X,X` is

Options

A. `NH_(2)-underset(O)underset(||)C-NH_(2)`
B. `NH=C=O`
C. `NH_(2)-underset(O)underset(||)C-Cl`
D. `EtO-underset(O)underset(||)C-NH_(2)`

Answer

To solve the problem, we need to analyze the reaction between phosgene (COCl₂) and ethyl alcohol (C₂H₅OH) in the presence of ammonia (NH₃). Here's the step-by-step solution:

### Step 1: Identify the Reactants
We have:
- Phosgene: COCl₂
- Ethyl alcohol: C₂H₅OH
- Ammonia: NH₃

### Step 2: Deprotonation of Ethyl Alcohol
Ammonia acts as a base and deprotonates ethyl alcohol. This results in the formation of ethoxide ion (C₂H₅O⁻) and ammonium ion (NH₄⁺):
\[ C_2H_5OH + NH_3 \rightarrow C_2H_5O^- + NH_4^+ \]

### Step 3: Nucleophilic Attack on Phosgene
The ethoxide ion (C₂H₅O⁻) acts as a nucleophile and attacks the electrophilic carbon of phosgene (COCl₂). The reaction can be represented as follows:
\[ C_2H_5O^- + COCl_2 \rightarrow C_2H_5O-C(=O)Cl + Cl^- \]

### Step 4: Formation of Intermediate
The product from the previous step is an intermediate compound, which can be represented as:
\[ C_2H_5O-C(=O)Cl \]

### Step 5: Proton Transfer
The ammonium ion (NH₄⁺) can now donate a proton to the oxygen of the intermediate, leading to the formation of a more stable compound:
\[ C_2H_5O-C(=O)Cl + NH_4^+ \rightarrow C_2H_5O-C(=O)H + NH_3 + Cl^- \]

### Step 6: Elimination of Chloride Ion
The intermediate can lose a chloride ion (Cl⁻) to form the final product:
\[ C_2H_5O-C(=O)H \rightarrow C_2H_5O-C(=O) + HCl \]

### Step 7: Final Product
The final product X is an ethyl carbamate (C₂H₅O-C(=O)NH₂), which can be represented as:
\[ X = C_2H_5O-C(=O)NH_2 \]

### Conclusion
Thus, the final product X is ethyl carbamate.

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