1. Let compound $A$ be an alkene.

   The statement says that on reaction with reagents $X$ and $Y$, it gives the same major product but different by-products $a$ and $b$.

2. We are told:

   • Oxidation of by-product $a$ gives a substance produced by ants.
   • The substance produced by ants is formic acid, $\mathrm{HCOOH}$.

   Therefore, $a$ must be formaldehyde, $\mathrm{HCHO}$, because: $\mathrm{HCHO} \xrightarrow{[O]} \mathrm{HCOOH}$

3. Now think of reactions of alkenes that can produce the same carbonyl compound(s) but with different small by-products.

   Two important oxidative cleavage reactions are:

   (i) Ozonolysis with reductive workup $\mathrm{O_3/Zn,H_2O}$ This cleaves an alkene into aldehydes/ketones.

   (ii) Hot acidic $\mathrm{KMnO_4}$ $\mathrm{KMnO_4/H^+}$ This also cleaves the double bond, but any aldehyde formed gets further oxidized to carboxylic acid.

4. To get the same major product from both reactions, consider a terminal alkene of type $\mathrm{RCH=CH_2}$

   • On ozonolysis: $\mathrm{RCH=CH_2 \xrightarrow[Zn/H_2O]{O_3} RCHO + HCHO}$ Here by-product $a = \mathrm{HCHO}$.

   • On oxidative cleavage with hot acidic $KMnO_4$: $\mathrm{RCH=CH_2 \xrightarrow{KMnO_4/H^+} RCOOH + CO_2}$ Here the terminal carbon is fully oxidized to $\mathrm{CO_2}$.

   Thus the small by-products are different: one gives $\mathrm{HCHO}$ and the other gives $\mathrm{CO_2}$.

5. The clue about $a$ confirms that $a$ is $\mathrm{HCHO}$, so the reagent producing $a$ must be: $X = \mathrm{O_3, Zn/H_2O}$

   Then the other reagent must be: $Y = \mathrm{KMnO_4/H^+}$

6. Check the options:

   • A: wrong order and dilute $KMnO_4$ at $273\,K$ gives dihydroxylation, not cleavage.
   • B: same issue.
   • C: order reversed.
   • D: $X=\mathrm{O_3, H_2O/Zn}, \quad Y=\mathrm{KMnO_4/H^+}$

   This matches perfectly.

Therefore, the correct option is D.