In the reaction of \( NaOBr \) with amide, the carbonyl carbon is lost as :
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If you start with propanamide (\( C_{3} \)), you get ethanamine (\( C_{2} \)). The "lost" carbon is always in the carbonate byproduct.
- \( HCO_{3}^{-} \)
- \( CO_{3}^{2-} \)
- \( CO_{2} \)
- \( CO \)
The Correct Option is B
Solution and Explanation
• This question refers to the Hoffmann Bromamide Degradation reaction.
• In this reaction, an amide (\( R-CONH_{2} \)) is treated with bromine (\( Br_{2} \)) and an alkali (like \( NaOH \) or \( KOH \)).
• The combination of \( Br_{2} \) and \( NaOH \) in situ forms sodium hypobromite (\( NaOBr \)).
• The reaction results in the formation of a primary amine containing one carbon atom less than the parent amide.
Step 1: Observe the overall reaction equation
The balanced chemical equation for the Hoffmann Bromamide reaction is: \[ R-CONH_{2} + Br_{2} + 4NaOH \rightarrow R-NH_{2} + Na_{2}CO_{3} + 2NaBr + 2H_{2}O \] Alternatively, using \( NaOBr \): \[ R-CONH_{2} + NaOBr + 2NaOH \rightarrow R-NH_{2} + Na_{2}CO_{3} + NaBr + H_{2}O \]
Step 2: Track the fate of the Carbonyl Carbon
The amide group (\( -CONH_{2} \)) contains a carbonyl carbon (\( C=O \)).
During the rearrangement mechanism (involving an isocyanate intermediate, \( R-N=C=O \)), this carbonyl carbon is expelled from the organic molecule.
Step 3: Identify the inorganic byproduct
The expelled carbon reacts with the excess hydroxide ions (\( OH^{-} \)) present in the alkaline medium to form sodium carbonate (\( Na_{2}CO_{3} \)).
In an aqueous ionic solution, sodium carbonate exists as sodium ions (\( Na^{+} \)) and carbonate ions (\( CO_{3}^{2-} \)).
Step 4: Conclusion
Therefore, the carbonyl carbon is formally lost from the organic chain in the form of the carbonate ion. The final answer is \( CO_{3}^{2-} \).
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