JEE Main 2025 Apr 7 Shift 2, Chemistry Q16: Stereoisomerism

Complete ozonolysis cleaves every C=C double bond, converting each alkene carbon into a carbonyl carbon.

Cleaving the triene at all three double bonds breaks the chain into two terminal fragments, each becoming CH${}_3$-CHO, and two central fragments, each retaining one of the original stereocentres now flanked by two -CHO groups.

Although each original stereocentre had four different groups while embedded in the long chain, after cleavage the two remaining substituents on that carbon (besides CH${}_3$ and H) both become identical -CHO groups, so the carbon is no longer bonded to four different groups.

Since none of the resulting carbonyl fragments retain a genuine stereocentre, none of the ozonolysis products are optically active.

$\therefore$ The number of optically active products is 0.