Natural rubber on complete ozonolysis (O₃/Zn-H₂O) gives compound X as the major product. X gives positive iodoform and Tollen’s tests. X on heating with aqueous NaOH gives Y as the major product. Y is

The correct option is A.

Step 1: Ozonolysis of Natural Rubber to find Compound X
Natural rubber is cis-polyisoprene, which is a polymer of isoprene (2-methylbuta-1,3-diene). Its structure consists of repeating units:

$-[{\text{CH}}_2-\text{C}({\text{CH}}_3)=\text{CH}-{\text{CH}}_2]{-}_n$

Upon complete reductive ozonolysis using ${\text{O}}_3/\text{Zn}-{\text{H}}_2\text{O}$, the double bond in each unit is cleaved to form 4-oxopentanal (levulinic aldehyde) as the major product X:

$\text{X}={\text{CH}}_3-\text{C}(=\text{O})-{\text{CH}}_2-{\text{CH}}_2-\text{CHO}$

Step 2: Verification of Tests for Compound X
1. Iodoform Test: Compound X possesses a methyl ketone group (${\text{CH}}_3-\text{C}=\text{O}$), so it reacts with ${\text{I}}_2/\text{NaOH}$ to give a positive iodoform test (yellow precipitate of ${\text{CHI}}_3$).
2. Tollen's Test: Compound X possesses a terminal aldehyde group ($-\text{CHO}$), so it reduces Tollen's reagent to give a positive silver mirror test.

Step 3: Intramolecular Aldol Condensation of Compound X to form Compound Y
When 4-oxopentanal (Compound X) is treated with aqueous $\text{NaOH}$ and heated, it undergoes an intramolecular aldol condensation:
1. Hydroxide ion (${\text{OH}}^{-}$) deprotonates the methyl group of the ketone to generate an enolate carbanion:

${[}^{-}{\text{CH}}_2-\text{CO}-{\text{CH}}_2-{\text{CH}}_2-\text{CHO}]$

2. The carbanion attacks the more reactive aldehyde carbonyl carbon, resulting in ring closure to form a thermodynamically favored 5-membered ring intermediate.
3. Subsequent dehydration (elimination of ${\text{H}}_2\text{O}$) yields a stable α,β-unsaturated cyclic ketone, cyclopent-2-en-1-one, as product Y.

Thus, compound Y corresponds to option (A):