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

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$K M n O_{4} / H^{+}$ and dil. $K M n O_{4},$ 273 K

$K M n O_{4}$ (dilute), 273 K and $K M n O_{4} / H^{+}$

$O_{3}, H_{2} O / Z n$ and $K M n O_{4} / H^{+}$

$K M n O_{4} / H^{+}$ and $O_{3}, H_{2} O / Z n$

answer is D.

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Detailed Solution

H₂C = C – CH₂ – C(CH₃)₂ – CH₃
| |
CH₃ CH₃
Compound 'A'

With X:

a + O=C–CH₂–C(CH₃)₂–CH₃
|
CH₃

With Y:

b + O=C–CH₂–C(CH₃)₂–CH₃
|
CH₃

Clue: Oxidation of ‘a’ gives the substance produced by ants (formic acid, HCOOH).

Analysis:

The byproduct ‘a’, when oxidized, gives formic acid, so ‘a’ is formaldehyde (HCHO).
Ozonolysis splits terminal alkenes to give formaldehyde and a ketone; hot KMnO₄ gives acids.
Major product after both reactions is the same, but only ozonolysis gives HCHO as a byproduct.

‘X’ and ‘Y’ respectively are: Ozonolysis and hot KMnO₄.