The reaction of 

with bromine and KOH gives $$RNH_2$$ as the end product. Which one of the following is the intermediate product formed in this reaction?

Explanation: The reaction converts a primary amide $$\mathrm{(R-CONH_2)}$$ into a primary amine $$\mathrm{(R-NH_2)}$$ with one less carbon atom using $$\mathrm{Br_2/KOH}$$.

An N-bromoamide intermediate is first formed: $$\mathrm{R-CONHBr}$$

Rearrangement then occurs with migration of the $$\mathrm{R}$$ group from the carbonyl carbon to nitrogen and simultaneous loss of $$\mathrm{Br^-}$$.

This produces the key intermediate: $$\mathrm{R-N=C=O}$$

The intermediate $$\mathrm{R-N=C=O}$$ is an alkyl isocyanate.

Hydrolysis of the isocyanate finally gives the primary amine.

Thus, the characteristic intermediate in Hoffmann bromamide degradation is: $$\mathrm{R-N=C=O}$$

Correct Option: $$\mathrm{(C)}$$