1. Identify the reaction

   The reaction of an amide with hypobromite is the Hofmann bromamide degradation (or Hofmann rearrangement).

   General reaction: $RCONH_2 \xrightarrow[\text{base}]{Br_2/NaOH} RNH_2$

   In this reaction, the amide loses one carbon atom and gives a primary amine with one less carbon.

2. What happens to the carbonyl carbon?

   During the rearrangement, the carbonyl carbon is removed from the amide skeleton. The intermediate formed ultimately gives an isocyanate, which on hydrolysis produces: $R-N=C=O + H_2O \rightarrow RNH_2 + CO_2$

   Thus, the carbonyl carbon is initially liberated as carbon dioxide.

3. Role of hypobromite medium

   Since the reaction is carried out in basic medium (hypobromite is generated in alkali), the $CO_2$ formed does not remain free for long. It reacts with hydroxide: $CO_2 + 2OH^- \rightarrow CO_3^{2-} + H_2O$

   Therefore, in the reaction mixture, the carbonyl carbon is finally present as carbonate ion.

4. Check the options

   • A: $CO_2$ — formed initially, but in alkaline medium it is converted further.
   • B: $CO$ — incorrect.
   • C: $HCO_3^-$ — not the final species in strongly basic medium.
   • D: $CO_3^{2-}$ — correct final form in alkaline hypobromite medium.

5. Conclusion

   Hence, the carbonyl carbon is lost as: $\boxed{CO_3^{2-}}$

So the correct option is D.