The reaction of `R-underset (O)underset("||")C-NH_(2)` with bromine and KOH gives `RNH_(2)` as the end product .Which one of the following is the intermediate product formed in this reaction ?

Options

A. `R-underset (O)underset("||")C-NH-Br`
B. `R-NH-Br`
C. `R-N=C=O`
D. `R-underset (O) underset("||")C -NBr_(2)`

Explanation

To solve the problem, we need to analyze the reaction of an amide with bromine and KOH, and identify the intermediate product formed during the reaction that ultimately yields RNH2 (a primary amine) as the end product.

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactant is an amide represented as R-CO-NH2.

2. **Reaction with KOH**:
When the amide reacts with KOH, the hydroxide ion (OH-) acts as a base and abstracts a proton (H+) from the amide nitrogen (NH2). This forms a negatively charged nitrogen species.

\[ R-CO-NH_2 + OH^- \rightarrow R-CO-NH + H_2O \]

3. **Formation of Brominated Intermediate**:
The next step involves the reaction of the brominated species. The nitrogen atom can now react with bromine (Br2). The nitrogen atom will bond with a bromine atom, forming an N-bromoamide.

\[ R-CO-NH + Br_2 \rightarrow R-CO-NHBr \]

4. **Formation of Isocyanate**:
The N-bromoamide can undergo further reaction. The base (OH-) can again abstract a proton from the nitrogen, leading to the formation of isocyanate (R-N=C=O).

\[ R-CO-NHBr + OH^- \rightarrow R-N=C=O + HBr + H_2O \]

5. **Hydrolysis to Form RNH2**:
Finally, the isocyanate can be hydrolyzed in the presence of water to yield the primary amine (RNH2).

\[ R-N=C=O + H_2O \rightarrow RNH_2 + CO_2 \]

6. **Identify the Intermediate**:
The intermediate product formed during the reaction is isocyanate (R-N=C=O).

### Conclusion:
The intermediate product formed in the reaction of the amide with bromine and KOH is **isocyanate (R-N=C=O)**.