Ozonolysis in the presence of $$O_3/Zn$$-$$H_2O$$ cleaves every $$C=C$$ bond and converts the two vinylic carbons into carbonyl groups (an aldehyde if the carbon was $$\;-CH=$$ and formaldehyde if it was $$\;=CH_2$$).
With the dienes given in List-I, this step therefore furnishes an Ο-dialdehyde (or an Ο-aldehyd-one when a double bond is internally substituted) plus two molecules of $$HCHO$$ that are of no further consequence.
When the dialdehyde is now treated with aqueous $$NaOH$$ the more stable enolate (the one that is more highly substituted) is generated first. That enolate performs an intramolecular aldol addition on the other carbonyl group, producing a Ξ²-hydroxy-carbonyl compound. The size of the ring obtained is decided purely by the number of carbon atoms that originally separate the two carbonyl groups:
Number of carbon atoms between the two $$-CHO$$ groups = $$n$$
Ring size produced in the aldol step = $$(n+1)$$ (because a new C-C bond is formed between the Ξ±-carbon of one carbonyl and the carbonyl carbon of the other).
Case P
The two double bonds of P are situated at C-1 and C-9 of a nonane chain (1,8-nonadiene). After ozonolysis
$$CH_2=CH-(CH_2)_5-CH=CH_2\xrightarrow[Zn/H_2O]{O_3}OHC-(CH_2)_5-CHO+2HCHO$$
Six methylenes are now counted from one carbonyl carbon to the other, so $$n=5$$ and the intramolecular aldol gives a six-membered Ξ²-hydroxy aldehyde. That product is entry (2) of List-II.
Case Q
Q is 1,7-octadiene. Ozonolysis furnishes
$$CH_2=CH-(CH_2)_4-CH=CH_2\longrightarrow OHC-(CH_2)_4-CHO+2HCHO$$
Now $$n=4$$; the aldol cyclisation therefore delivers a five-membered Ξ²-hydroxy aldehyde, which is shown as entryΒ (1) in List-II.
Case R
R is a methyl-substituted analogue of Q (2-methyl-1,7-octadiene). Ozonolysis gives
$$CH_2=CH-CH(CH_3)-(CH_2)_3-CH=CH_2\longrightarrow OHC-CH(CH_3)-(CH_2)_3-CHO+2HCHO$$
Again $$n=4$$, so a five-membered ring is formed. Because the more substituted enolate is generated at the Ξ±-carbon that bears the $$CH_3$$ group, the methyl ends up on the ring carbon bearing the newly formed hydroxyl group. This corresponds to entryΒ (5) in List-II.
Case S
S is 2-methyl-1,8-nonadiene. Its ozonolysis product is
$$CH_2=CH-CH(CH_3)-(CH_2)_4-CH=CH_2\longrightarrow OHC-CH(CH_3)-(CH_2)_4-CHO+2HCHO$$
Now $$n=5$$, hence the aldol condensation builds a six-membered ring. As in caseΒ R the enolate forms preferentially at the $$CH(CH_3)$$ end, so the methyl group becomes attached to the Ξ²-hydroxy carbon of the ring, giving entryΒ (3) of List-II.
Collecting the results:
(P) β (2), (Q) β (1), (R) β (5), (S) β (3)
Therefore the correct option is:
Option CΒ which is:Β (P)β(2), (Q)β(1), (R)β(5), (S)β(3)
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