The stereochemical relationship of $\mathrm{H}_{\mathrm{a}}$ and $\mathrm{H}_{\mathrm{b}}$ in the following is
(a)Enantiotopic
(b)Homotopic
(c)Diastereotopic
(d)Constitutionally heterotopic
Answer
Answer: C ✓ checked by 4AB · confidence high
Explanation
Ring 1 of the biaryl carries two different ortho groups (OMe and Me). Ring 2 carries two identical ortho ethyl groups. Rotation about the aryl–aryl bond is hindered, so the rings sit perpendicular. The plane of ring 1 contains the biaryl axis and swaps the two ethyl groups, so the molecule is achiral (Cs). That mirror does not exchange $\mathrm{H_a}$ and $\mathrm{H_b}$ within the same CH₂: one of them points towards the OMe face and the other towards the Me face. Replacing either one by D creates a CHD stereocentre and also makes the axis chiral, and the two products are diastereomers. Hence $\mathrm{H_a}$ and $\mathrm{H_b}$ are diastereotopic (c).