The compound (structure missing) on reaction with $\text{NaIO}_4$ in the presence of $\text{KMnO}_4$ gives:

The combination of sodium periodate ($\text{NaIO}_4$) and a catalytic amount of potassium permanganate ($\text{KMnO}_4$) is known as the Lemieux-von Rudloff reagent. It acts as a strong oxidizing agent used for the oxidative cleavage of carbon-carbon double bonds. The reaction yields products similar to oxidative ozonolysis or oxidation by hot acidic $\text{KMnO}_4$ .

In this process, an alkene carbon bonded to two alkyl groups is oxidized to a ketone, while an alkene carbon bonded to one alkyl group and one hydrogen atom is oxidized to a carboxylic acid (the intermediate aldehyde is further oxidized by $\text{KMnO}_4$). To obtain acetone ($\text{CH}_3\text{COCH}_3$) and acetic acid ($\text{CH}_3\text{COOH}$), the original missing compound must have been 2-methylbut-2-ene ($\text{CH}_3\text{-C(CH}_3\text{)=CH-CH}_3$).

Note: The specific chemical structure of the reactant was missing from the input data. Option 2 is marked correct based on the provided probable answer.