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During the chirality analysis of various molecules containing rings, we assume for convenience that the rings are completely flat/planar and if the molecule has a plane of symmetry, it is achiral. However, does the fact that say, 6-membered rings actually exist in a chair or boat conformation have any effect on our final answer (it has a plane of symmetry if it is completely planar, but does not if it is in its 3D conformation)?

I'm a little confused because the odd, puckered conformation of various rings don't seem like they would have a plane of symmetry. Take for example, the molecule below. In the 2D drawing, we see 1 plane of symmetry, it goes through $\ce {Br}$ and cuts through the ring in a perpendicular fashion. However, in the 3D drawing, the plane we identified earlier doesn't seem like a plane of symmetry anymore.

enter image description here

Of course, in the 2D picture, the molecules would seem to be equivalent no matter if the bond were wedged or dashed, but why do they appear to be different if they are drawn in 3D?

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    $\begingroup$ Not sure whether your 3D drawings are accurate, but that doesn't matter anyway. In short, if the structures appear to be different in 3D, then they are different, but that doesn't matter either, because they easily change from one conformation to another. $\endgroup$ Apr 30 at 19:01
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    $\begingroup$ Any chiral conformation of bromocyclopentane is matched by an equal population of its enantiomer. Thus, there is a racemate. $\endgroup$
    – user55119
    Apr 30 at 19:45
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    $\begingroup$ No matter how you turn or twist these rings to change from one conformer to an other: the configuration of the stereogenic center remains either (R), or (S) and the atom's connectivity does not change. Some conformations may not be easy accessible (higher energy, thus less likely populated [in parlance of statistical thermodynamics]). But a change of the configuration between (R) and (S) of a stereogenic center however is a change of constitution, only possible by breaking old and establishing new bonds (on/off binary) costing more energy than the wiggles and twists about conformers. $\endgroup$
    – Buttonwood
    Apr 30 at 20:22
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    $\begingroup$ @user55119 How does it work so that the ever fluctuating concentrations have equal concentrations of enantiomers at all times? I may be interpreting your comment incorrectly but wouldn't this mean bromocyclopentane is chiral (even though it's racemic), which isn't true? $\endgroup$ Apr 30 at 21:42
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    $\begingroup$ And, the other problem with your example, is that the rings have a plane of symmetry even in chair and boat-like confirmations. The plane is perpendicular to the ring: we don't have to pretend the ring is flat. $\endgroup$
    – matt_black
    Apr 30 at 23:48

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