The reaction of R-CO-NH₂ with bromine and KOH gives RNH₂ as the end product. Which one of the following is the intermediate product formed in this reaction?

R-N=C=O

The correct answer is option 3. R-N=C=O.

The reaction of an amide (R-CO-NH₂) with bromine (\(Br_2\)) and potassium hydroxide (\(KOH\)) is known as the Hofmann degradation reaction. This reaction typically involves the conversion of an amide to a primary amine with the loss of one carbon atom.

The steps involved in this reaction are:

Formation of an Intermediate: The amide reacts with bromine and a strong base like potassium hydroxide to form an intermediate known as N-haloamide.

Formation of an Isocyanate Intermediate: The N-haloamide decomposes to form an isocyanate intermediate.

Hydrolysis of Isocyanate: The isocyanate intermediate is then hydrolyzed to form the primary amine and carbon dioxide.

Reaction Mechanism: Let us take an example to understand the mechanism:

Intermediates in the Reaction:

R-CO-NH-Br: This is not typically a stable intermediate in the Hofmann degradation reaction; instead, it's the N-haloamide.

R-NH-Br: This is not an intermediate in the Hofmann degradation reaction.

R-N=C=O: This is the isocyanate intermediate, which is formed after the N-haloamide decomposes and is then hydrolyzed to give the primary amine.

R-CO-NBr₂: This is the N-haloamide intermediate, which reacts to form the isocyanate.

The intermediate product formed in this reaction is R-N=C=O (isocyanate), which is an important intermediate in the Hofmann degradation process.

Therefore, the correct answer is 3. R-N=C=O.