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Can someone explain to me, why this molecule has an (S)-chirality? I don't get it. I tried to apply the rules, but I get only an (R)-chirality. Is there something special with cyclo-alkanes?

ethyl (2S)-2-methyloxane-2-carboxylate

Seminom
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    This is not cycloalkane. It is a cyclicether with ester grop in it at chiral carbon – Mathew Mahindaratne Sep 07 '20 at 21:32
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    Priority is $\ce{-O- \Rightarrow -CO2Et \Rightarrow -CH2- \Rightarrow -CH3}$ – Mathew Mahindaratne Sep 07 '20 at 21:34
  • There is not much to be explained. This molecule is an S enantiomer according to the rules. There are no special rules for cyclic molecules. That's about the size of it. – Ivan Neretin Sep 07 '20 at 21:35
  • @MathewMahindaratne Thank you for this correction. – Seminom Sep 07 '20 at 21:35
  • @MathewMahindaratne I thought all the time, that whenever two carbons have the same priority, it depends on the priority of their substituents. So CH_3 was higher ranked than CH_2 for me. – Seminom Sep 07 '20 at 21:39
  • Now why would CH3 be ranked higher than CH2, if the latter has some substitute instead of H, which surely has higher priority than H (because everything has higher priority than H)? – Ivan Neretin Sep 07 '20 at 23:02
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    Working outward from the stereocenter, the methylene carbon has its 2 hydrogens and the next carbon in the ring attached to it. This is written as {C,H,H}. The methyl carbon has three hydrogens attached to it. It is designated {H,H,H}. The methylene carbon has priority because a one-to-one comparison has C>H based on atomic number. – user55119 Sep 08 '20 at 00:31
  • You really don't need the bold and dashed bonds as long as you realize that your ring implies that CH2 is in the foreground and the oxygen is in the rear. If not make the CH2 bond bold and the oxygen bond dashed. Once you establish priorities, don't worry about moving the structure but rather look here: https://chemistry.stackexchange.com/questions/99721/twisting-stereoisomers-with-rings-to-determine-r-s/99750#99750 – user55119 Sep 08 '20 at 14:53

1 Answers1

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One subtlety: when you rank the groups and look at the rotational order of ranks 1, 2, 3, you have to consider also whether group 4 is towards you or away from you. A clockwise order for 1, 2, 3 (which the given structure shows) corresponds to $R$ when group 4 is directed away from you. But here group 4 is directed towards you, so the chirality is reversed --- 1,2,3 clockwise now means $S$.

Oscar Lanzi
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