List - I List - II (P) (1) Odd number of double bond equivalent (Q) (2) Only sp 2 & sp 3 carbons are present (R) (3) Even number of double bond of equivalent (S) CH 3 − CH = CH − C ≡ CH (4) Contains

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List - I List - II (P) (1) Odd number of double bond equivalent (Q) (2) Only sp 2 & sp 3 carbons are present (R) (3) Even number of double bond of equivalent (S) CH 3 ​ − CH = CH − C ≡ CH (4) Contains 2 ∘ H only (5) All carbon is sp 2 hybridised carbon

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P → 2 : Toluene contains only sp² and sp³ hybridised carbons.

Q → 3 : The cyclic compound has even number of double bond equivalents (DBE = 2).

R → 3,4 : The structure has even DBE and contains only secondary hydrogens.

S → 1 : CH₃–CH=CH–C≡CH has odd number of double bond equivalents.

List - I	List - II
(P)	(1) Odd number of double bond equivalent
(Q)	(2) Only sp² & sp³ carbons are present
(R)	(3) Even number of double bond of equivalent
(S) $CH_{3} - CH = CH - C \equiv CH$	(4) Contains $2^{\circ} H$ only
	(5) All carbon is $sp^{2}$ hybridised carbon

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