on reaction with $\mathrm{NaIO}_3$ in the present of $\mathrm{KMnO}_4$ gives:The compound on reaction with $\mathrm{NaIO}_3$ in the present of $\mathrm{KMnO}_4$ gives:
on reaction with $\mathrm{NaIO}_3$ in the present of $\mathrm{KMnO}_4$ gives:- $\mathrm{CH}_3 \mathrm{COCH}_3$
- $\mathrm{CH}_3 \mathrm{COCH}_3+\mathrm{CH}_3 \mathrm{COOH}$
- $\mathrm{CH}_3 \mathrm{COCH}_3+\mathrm{CH}_3 \mathrm{CHO}$
- $\mathrm{CH}_3 \mathrm{CHO}+\mathrm{CO}_2$
Asked in: NEET 2003