In case of checking for the optical activity of cyclohexane or its derivative why we check for POS and COS in its cyclic 6 membered planar eclipsed state but not by its real conformation it forms chair, boat,twist boat etc) .

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  • $\begingroup$ Are you asking why we determine if the compound is chiral from the simplified planar hexagonal drawing, rather than its true conformation drawing? $\endgroup$
    – H.Linkhorn
    Jun 3 '20 at 16:23
  • $\begingroup$ Yes I am asking this $\endgroup$
    – user94341
    Jun 3 '20 at 17:22
  • $\begingroup$ Why might you think? Which drawing is easy to show and visualise planes of symmetry? Remember the drawings are saying the same thing - the chair conformation is implied in the drawing of a hexagon. $\endgroup$
    – H.Linkhorn
    Jun 3 '20 at 17:23
  • $\begingroup$ It is not about being visualised easily but it is about the real structures which should be seen for the molecule.The complete planar form doenot exist in any of its conformation.Check out for finding the plane of symmetry of a simple molecule 1,2 cis dimethyl cyclo hexane which has POS according to planar structure,If there is no problem with this then the actual conformer should also have POS ,but where it is $\endgroup$
    – user94341
    Jun 3 '20 at 18:35
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    $\begingroup$ So each of the conformation is chiral because they have no plane of symmetry like you suggested. However the ring is constant flipping between the two chiral conformations, which are enantiomers of each other. So at room temp you moving between each enantiomer so fast you can call the overall compound achiral. But at very low temp when there isn’t the energy to ring flip it is technically a chiral compound as you stuck in one conformation $\endgroup$
    – H.Linkhorn
    Jun 4 '20 at 5:33