`R-NH_(2)+CH_(3)COCl overset("(excess)")rarr A.`
The product (A) will be -

Options

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

Explanation

To solve the reaction of R-NH₂ with CH₃COCl in excess, we can break down the process step by step:

### Step 1: Identify the Reactants
We have two reactants:
- R-NH₂ (a primary amine)
- CH₃COCl (acetyl chloride)

### Step 2: Understand the Mechanism
The nitrogen atom in the amine (R-NH₂) has a lone pair of electrons. Acetyl chloride (CH₃COCl) has a carbonyl group (C=O) which is electron-deficient due to the electronegativity of oxygen. This makes the carbon atom in the carbonyl group susceptible to nucleophilic attack.

### Step 3: Nucleophilic Attack
The lone pair on the nitrogen atom of R-NH₂ attacks the carbon atom of the carbonyl group in CH₃COCl. This leads to the formation of a tetrahedral intermediate.

### Step 4: Formation of the Intermediate
The tetrahedral intermediate can be represented as:
- R-NH₂⁺ (the nitrogen now has a positive charge)
- CH₃C(OH)(Cl) (the carbonyl carbon is now bonded to an -OH and -Cl)

### Step 5: Elimination of the Leaving Group
Chlorine (Cl) is a good leaving group. It leaves, resulting in the formation of an imine-like structure:
- R-NH-C(=O)-CH₃ (the product formed after the elimination of Cl⁻)

### Step 6: Proton Transfer
In the presence of excess acetyl chloride, a proton (H⁺) can be transferred, leading to the formation of:
- R-NH-C(=O)-CH₃ + HCl

### Step 7: Final Product
The final product (A) is:
- R-NH-C(=O)-CH₃ (an N-acyl derivative of the amine)

### Conclusion
The product (A) formed from the reaction of R-NH₂ with CH₃COCl in excess is:
- **R-NH-C(=O)-CH₃**