Discussion
The topicity relationship of Ha and Hb in X, Y and Z are, respectively,
![X: Fischer projection of tartaric acid with Ha and OH on the upper carbon and HO and Hb on the lower carbon, the hydrogens on opposite sides; Y: [2.2]paracyclophane with Ha on the upper ring and Hb on the lower ring; Z: ClCH=C=CHaHb](/uploads/question-images/gate-cy-2023/q16.png)
Answer: B
The relationships are Homotopic, Enantiotopic and Enantiotopic.
- X is a chiral tartaric acid: in the Fischer projection the two hydrogens lie on opposite sides (and so do the two OH groups). This molecule has a C2 axis that exchanges the two CH(OH) carbons, so Ha and Hb are interchanged by a rotation: homotopic.
- Y is [2.2]paracyclophane. Ha on the upper ring and Hb on the lower ring sit at the same position relative to the bridges, so the mirror plane between the two rings maps one onto the other, but no rotation does: enantiotopic.
- Z, ClCH=C=CHaHb, is an allene whose CH2 plane is perpendicular to the plane of the CHCl end. That CHCl plane is a mirror plane bisecting the H-C-H angle and exchanging Ha and Hb, and no axis does so: enantiotopic.
Calling X diastereotopic would fit meso-tartaric acid, not the chiral isomer drawn. Calling Y homotopic or Z homotopic overlooks that only a reflection relates the two hydrogens.
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