JEE Main 2025 · 7 April · evening · Single correct · +4 / −1
The number of optically active products obtained from the complete ozonolysis of the given compound is :
- (a) 4
- (b) 0
- (c) 2
- (d) 1
Answer and solution
Answer (b) 0
Solution
Chemistry ma’am · 4 steps
Cut every C=C and then look again at each old stereocentre: it now carries two identical CHO groups.
- Read the chain. The compound is CH₃-CH=CH-CH(CH₃)-CH=CH-CH(CH₃)-CH=CH-CH₃. It has three C=C bonds, and the two CH(CH₃) carbons are its stereocentres.
- Cut the double bonds. Complete ozonolysis breaks every C=C and turns each double-bond carbon into a C=O. Each stereocentre sits between two double bonds, so it ends up flanked by two aldehyde groups.
Fragment Product Four different groups? Optically active? C1–C2 and C9–C10 CH₃CHO (2 molecules) no stereocentre no C3–C4–C5 and C6–C7–C8 OHC-CH(CH₃)-CHO (2 molecules) no: H, CH₃, CHO, CHO no - Chirality is lost. In 2-methylpropanedial the central carbon has two identical -CHO groups, so it is not a chiral centre. The molecule has a mirror plane and is optically inactive.
- So. The products are only ethanal and 2-methylpropanedial, and neither is chiral. The number of optically active products is 0.
The idea to keep. After ozonolysis, re-check every old stereocentre in the product. If two of its groups become identical (like two CHO groups), the chirality is gone.
Answer from the NTA final answer key. Solution by PrepWiser.