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Nilanjan Bhowmick AIR 3, CSIR NET (Earth Science)
Smitkriti
If we check the symmetry operation ( Plane of symmetry, centre of symmetry, Improper axis of rotation) both B and D have none, therefore should be optically active. But now we have to see which would be the most stable one out of both. In B as the methyl group is on equatorial position while it's on axial position in D. So, B is more stable than D. Hence, more chance of existing in that form.