1. The reaction of an amide with bromine and $KOH$ giving $RNH_2$ is the Hofmann bromamide degradation (also called Hofmann rearrangement).

2. General reaction: $RCONH_2 \xrightarrow{Br_2/KOH} RNH_2$ This converts an amide into a primary amine with loss of one carbon atom.

3. Mechanism outline:

   1. The amide first reacts with bromine in base to form N-bromoamide: $RCONH_2 \rightarrow RCONHBr$
   2. On treatment with base, rearrangement occurs to give an isocyanate intermediate: $RCONHBr \rightarrow R-N=C=O$
   3. The isocyanate then hydrolyzes to form the amine: $R-N=C=O + H_2O \rightarrow RNH_2 + CO_2$

4. Therefore, the key intermediate product formed in this reaction is: $R-N=C=O$ which is an isocyanate.

5. Checking options:

• Option B: $R-NH-Br$ is not the correct rearranged intermediate; the actual initial brominated species is $RCONHBr$, not simply $R-NH-Br$.
• Option C: $R-N=C=O$ ✔ correct, the isocyanate intermediate.

Hence, the correct answer is Option C.

6. Comparison with stored correct answer: Stored correct answer = C Derived answer = C They match.