Intermediate formed during reaction of `RCONH_(2)` with `Br_(2)` and `KOH` are .

Options

A. `RCONHBr` and `RNCO`
B. `RNHCObr` and `RNCO`
C. `RNHBr` and `RCONHbr`
D. `RCONBr_(2)`

Explanation

To solve the question regarding the intermediate formed during the reaction of `RCONH2` with `Br2` and `KOH`, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**: The reaction involves the compound `RCONH2` (a primary amide) reacting with bromine (`Br2`) in the presence of potassium hydroxide (`KOH`). This is known as the Hofmann degradation reaction.

2. **Initial Reaction**: When `RCONH2` is treated with `Br2` and `KOH`, the first step involves the formation of an N-bromoamide. The reaction can be represented as:
\[ RCONH2 + Br2 \rightarrow RCO-NHBr \]
Here, `RCO-NHBr` is the intermediate formed.

3. **Formation of Nitrene**: The N-bromoamide (`RCO-NHBr`) can undergo further reactions. In the presence of KOH, the nitrogen atom can lose a bromine atom, leading to the formation of a nitrene intermediate:
\[ RCO-NHBr \rightarrow RCO-NH + Br^- \]

4. **Rearrangement**: The nitrene can rearrange to form an isocyanate:
\[ RCO-NH \rightarrow R-N=C=O \]

5. **Hydrolysis**: The isocyanate can then undergo hydrolysis, leading to the formation of a primary amine and carbon dioxide:
\[ R-N=C=O + H2O \rightarrow R-NH2 + CO2 \]

6. **Final Products**: The final products of the reaction are primary amine (`R-NH2`) and carbon dioxide (`CO2`), but the intermediate we are interested in is `RCO-NHBr`.

### Summary of Intermediates:
- The primary intermediate formed during the reaction is `RCO-NHBr`.
- Other intermediates that can be formed during the reaction include `R-N=C=O` (isocyanate) before hydrolysis.

### Final Answer:
The intermediate formed during the reaction of `RCONH2` with `Br2` and `KOH` is `RCO-NHBr`.