`I overset(COcl_2) larr "Aniline"overset(C_2 H_5 MgI)to II`. Products I and II are :

Options

A. Diphenylurea , ehtane
B. Diphenylurea ,` C_6H_5MgI`
C. Dipphenlamine , ethane
D. Diphenylamine , benzene

Answer

To solve the question regarding the reaction of aniline with COCl₂ and C₂H₅MgI, we will break it down into two parts: the reaction with COCl₂ and the reaction with C₂H₅MgI.

### Step 1: Reaction of Aniline with COCl₂
1. **Identify Aniline Structure**: Aniline (C₆H₅NH₂) contains an amino group (-NH₂) attached to a benzene ring.
2. **Reaction with COCl₂**: When aniline reacts with phosgene (COCl₂), the lone pair of electrons on the nitrogen atom of the amino group attacks the carbon atom of the carbonyl group (C=O) in COCl₂.
3. **Formation of Intermediate**: This leads to the formation of an intermediate where the nitrogen is bonded to the carbonyl carbon (C=O) and one of the chlorine atoms is displaced.
4. **Formation of Urea Derivative**: The reaction continues as the nitrogen attacks another COCl₂ molecule, leading to the formation of diphenylurea (C₆H₅NHCOCl) after the loss of two chlorine atoms.
5. **Final Product**: The final product from this reaction is **Diphenylurea**.

### Step 2: Reaction of Aniline with C₂H₅MgI (Ethylmagnesium Iodide)
1. **Identify Reagent**: C₂H₅MgI is an organomagnesium compound (Grignard reagent).
2. **Nucleophilic Attack**: The nitrogen atom in aniline is nucleophilic and can react with the ethyl group from C₂H₅MgI.
3. **Formation of Ethyl Aniline**: The ethyl group will add to the nitrogen of aniline, forming ethyl aniline (C₆H₅NH(C₂H₅)).
4. **Final Product**: The final product from this reaction is **Ethyl Aniline**.

### Summary of Products:
- **Product I (from COCl₂)**: Diphenylurea
- **Product II (from C₂H₅MgI)**: Ethyl Aniline