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

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    $\begingroup$ This is not cycloalkane. It is a cyclicether with ester grop in it at chiral carbon $\endgroup$ Sep 7, 2020 at 21:32
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    $\begingroup$ Priority is $\ce{-O- \Rightarrow -CO2Et \Rightarrow -CH2- \Rightarrow -CH3}$ $\endgroup$ Sep 7, 2020 at 21:34
  • $\begingroup$ 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. $\endgroup$ Sep 7, 2020 at 21:35
  • $\begingroup$ @MathewMahindaratne Thank you for this correction. $\endgroup$
    – Seminom
    Sep 7, 2020 at 21:35
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    $\begingroup$ 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. $\endgroup$
    – user55119
    Sep 8, 2020 at 0:31

1 Answer 1


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 $S$.


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