`R-NH_(2)+underset("excess")(CH_(3)COCl)toA`.
The product (A) Will be

Options

A. `RNHCOCH_(3)`
B. `RN(COCH_(3))_(2)`
C. `RN(COCH_(3))_(3)Cl^(-)`
D. `R-CONH_(2)`

Answer

To solve the question regarding the reaction of R-NH₂ with excess CH₃COCl, we will follow these steps:

### Step 1: Identify the Reactants
The reactants are:
- R-NH₂ (a primary amine)
- CH₃COCl (acetyl chloride)

### Step 2: Understand the Reaction Mechanism
In this reaction, the nitrogen atom in the amine (R-NH₂) has a lone pair of electrons, making it a nucleophile. The carbon atom in acetyl chloride (CH₃COCl) is electrophilic due to the presence of the electronegative chlorine atom, which creates a partial positive charge on the carbon.

### Step 3: Nucleophilic Attack
The nucleophilic nitrogen will attack the electrophilic carbon in acetyl chloride. This forms a tetrahedral intermediate where the nitrogen is bonded to the carbon of the acetyl group and the chlorine atom is displaced.

### Step 4: Formation of the Product
As the reaction proceeds, the chlorine atom leaves (as Cl⁻), resulting in the formation of an amide. The product formed is R-NH-CO-CH₃ (N-acetyl derivative of the amine).

### Step 5: Consider the Excess Reagent
Since acetyl chloride is in excess, it is important to note that the nitrogen atom of the amide (R-NH-CO-CH₃) does not have enough electron density to attack another molecule of acetyl chloride. The electron-withdrawing nature of the carbonyl group (C=O) makes the nitrogen less nucleophilic.

### Final Product
The final product (A) of the reaction is:
**R-NH-CO-CH₃ (N-acetyl amine)**