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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?

timeinbaku
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    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. – Ivan Neretin Apr 30 '21 at 19:01
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    Any chiral conformation of bromocyclopentane is matched by an equal population of its enantiomer. Thus, there is a racemate. – user55119 Apr 30 '21 at 19:45
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    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. – Buttonwood Apr 30 '21 at 20:22
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    @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? – timeinbaku Apr 30 '21 at 21:42
  • You can draw bromocyclopentane with 4 ring carbons in a plane with C5-Br as the "flap of the envelope". With the flap up or down, there are two achiral conformations having a plane of symmetry. Any other conformation, and there many of them, is chiral as a pair of enantiomeric conformations. All of these conformations are rapidly interconverting. It is the chiral conformations that prevail and, as racemates, account for the majority of the lack of optical rotation. The achiral conformations, having a plane of symmetry, also do not contribute to optical rotation. – user55119 Apr 30 '21 at 22:20
  • ...continued. Your 2nd structure is chiral. But there is an equal amount of its isoenergetic mirror image. The same is true of the 3rd structure. Go back to meso-tartaric acid in the eclipsed conformation having a plane of symmetry. Rotate CW about the C2-C3 bond by 60 deg. giving a chiral conformation of a meso cmpd. It is statistically probable that rotation occurs in the CCW direction, forming the mirror image. Now you have a racemate of conformations. This may help: https://chemistry.stackexchange.com/questions/48995/is-2r-3s-butane-2-3-diol-chiral/87415#87415 – user55119 Apr 30 '21 at 22:29
  • Wonky ring structures could, in principle, create chirality, but only if you could force the ring to get stuck in a wonky conformation. This might be possible with very bulky groups that could force one side of the ring to bend differently to the other. But how to do that with symmetrical bulky substituents (if they are not symmetrical then the molecule might already be chiral ignoring the ring conformation). – matt_black Apr 30 '21 at 23:47
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    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. – matt_black Apr 30 '21 at 23:48

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