JEE Main 2022 · 29 July · evening · Single correct · +4 / −1
Given below are two statements.
Statement I : The compound
is optically active.
Statement II :
is mirror image of above compound A.
In the light of the above statement, choose the most appropriate answer from the options given below.
- (a) Both Statement I and Statement II are correct.
- (b) Both Statement I and Statement II are incorrect.
- (c) Statement I is correct but Statement II is incorrect.
- (d) Statement I is incorrect but Statement II is correct.
Answer and solution
Answer (c) Statement I is correct but Statement II is incorrect.
Solution
Chemistry ma’am · 4 steps
Two propenyl arms that differ only as cis and trans still make C4 chiral; then compare A and B carefully.
- Is C4 chiral. C4 carries H, NO₂, a Z-propenyl group and an E-propenyl group. The two propenyl groups have the same atoms but different geometry, so they are different groups. Four different groups make C4 a stereocentre, and A has no plane of symmetry, so A is optically active.
- Rank the groups. By CIP rules, NO₂ ranks first. The propenyl groups tie on atoms, and in that case Z ranks above E. So the order is NO₂ (1) > Z-propenyl (2) > E-propenyl (3) > H (4), with H pointing to the back in both drawings.
Compound Where 1, 2, 3 sit Turn 1 → 2 → 3 Configuration A NO₂ down (wedge), Z right, E upper left anticlockwise S B NO₂ upper left (wedge), Z lower left, E right anticlockwise S - Compare A and B. Both are S, so B is only A rotated in the plane of the page. It is the same molecule, not its mirror image. The true mirror image would be the R form.
- So. Statement I is correct and Statement II is incorrect, which is option (c).
The idea to keep. Before calling two drawings mirror images, assign R or S to each. Same label means same molecule, however differently it is drawn. And cis and trans versions of one group count as different groups.
Answer from the NTA final answer key. Solution by PrepWiser.