A compound ‘A’ on reaction with ‘X’ and ‘Y’ produces the same major product but different by product 'a' and 'b′. Oxidation of 'a' gives a substance produced by ants.
'X' and 'Y' respectively are

Solution

Chemistry ma’am · 4 steps

Look at the CH₂= end: the reagent that leaves it as HCHO is X, because HCHO oxidises to formic acid.

1. Read compound A. A is CH₂=C(CH₃)-CH₂-C(CH₃)₃. Both reactions break the C=C bond and give the same ketone, CH₃COCH₂C(CH₃)₃, from the substituted carbon. So both X and Y must cleave the double bond. They differ only in what the terminal CH₂ becomes.

   Reagent	Effect on the double bond	Ketone formed?	Fate of the =CH₂ carbon
   O₃, then Zn/H₂O	reductive cleavage	yes	stops at HCHO
   KMnO₄/H⁺	oxidative cleavage	yes	HCHO → HCOOH → CO₂ + H₂O
   dil. KMnO₄, 273 K	adds two OH groups, no cleavage	no, gives a diol	stays as CH₂OH in the diol

2. Identify a. Ants produce formic acid, HCOOH. The only by-product that oxidises to it is formaldehyde, HCHO. So a is HCHO, and X is ozonolysis with Zn/H₂O.
3. Identify b. Hot acidic KMnO₄ gives the same ketone, but it oxidises the CH₂ end all the way to CO₂ and water. So b is CO₂, and Y is KMnO₄/H⁺.
4. So. X is O₃, H₂O/Zn and Y is KMnO₄/H⁺, which is option (d).

The idea to keep. A terminal =CH₂ gives HCHO on reductive ozonolysis but CO₂ with acidic KMnO₄. Cold dilute KMnO₄ never cleaves the double bond. It only makes a diol.

Answer from the NTA final answer key. Solution by PrepWiser.