4
$\begingroup$

I understand that carbon atoms normally only have up to 4 substituents and the R and S chirality centers are defined for 4 substituents. Is there a similar convention of 'chirality' for an atom with 5 substituents?

The SMILES website (section 3.3.4 General Chiral Specification) shows that general 'chirality' for 5 and 6 substituents exists. But the text there focuses on how to convert them to SMILES strings. Is there any references about the classifications and nomenclatures?

https://www.daylight.com/dayhtml/doc/theory/theory.smiles.html

Sorry I am not a chemistry major and my nomenclature may not be correct.

$\endgroup$
8
  • 1
    $\begingroup$ I don't understand why it's downvoted. They do exist as shown in the link. Conceptually you can easily see such chirality variants could exist $\endgroup$
    – nos
    Apr 29, 2021 at 18:59
  • 2
    $\begingroup$ Chirality is not about bonds at all. $\endgroup$ Apr 29, 2021 at 19:03
  • $\begingroup$ @IvanNeretin How should I phrase it? 4 connected objects (what's the chemical term for that)? $\endgroup$
    – nos
    Apr 29, 2021 at 19:11
  • 1
    $\begingroup$ Your question implies that chirality is caused by certain centers. It is not. $\endgroup$ Apr 29, 2021 at 19:32
  • 2
    $\begingroup$ I think you are not actually asking about chirality, but rather configuration. Notice l-tartaric acid (O[C@H]([C@H](C(O)=O)O)C(O)=O); d-tartaric acid (O[C@@H]([C@@H](C(O)=O)O)C(O)=O); meso-tartaric acid (O[C@@H]([C@H](C(O)=O)O)C(O)=O). The @'s and @@s are not meant to be flags which indicate that the carbon is asymmetric (otherwise there wouldn't be two different flags just to indicate the presence of an asymmetric carbon). They're intended to describe, or fully specify, the configuration of these carbons, i.e. the exact distribution of the substituents in space. $\endgroup$ Apr 29, 2021 at 19:43

1 Answer 1

6
$\begingroup$

Chirality depends not so much on coordination number, but on the symmetry of the substituents. Meaning you have to pay attention to coordination geometry. In the case of four-coordination you can have four different substituents and yet no chirality... if you have square planar instead of tetrahedron coordination.

For five-coordination you lack a mirror plane and thus have a chiral center in the following cases:

  • Square pyramidal geometry: if all four substituents on the base are different, or if just one pair of these basal substituents are identical and the identical pair is on adjacent vertices. The axial substituent does not have any impact because all possible mirror planes pass through it.

  • Triangular bipyramid geometry: if all three equatorial substituents are different and the two axial substituents are different from each other. You may still have chirality if one axial substituent happens to be the same as one equatorial one.

$\endgroup$
3
  • $\begingroup$ Suggested change: «not so much on coordination but on the symmetry of the substituents» to «not so much on coordination number, but on the symmetry of the substituents». Because so far I understood the priorities of substitutents according to the CIP rules to eventually assign (R) or (S) as not constrained to four atoms around one in the centre. If the rules equally account for a lone electron pair (the chiral sulfoxides) as then lowest priority among the other three atoms, why should this not work for 5? $\endgroup$
    – Buttonwood
    Apr 29, 2021 at 19:56
  • $\begingroup$ The group ranking methodology you see for tetrahedral four-coordination needs some patching when we pass th five-coordination, because as we see above, in both cases we have to allow point chirality with identical attached groups. And with six or more it's going to get worse. $\endgroup$ Apr 30, 2021 at 14:09
  • $\begingroup$ I agree that four (or five, six) atoms around one center in common as number is not a sufficient criterion for a chiral pointgroup. Its recognition requires the background a flow chart like this summarizes and some programs assist to see (e.g., Jmol when issuing calculate pointgroup;, show pointgroup; (table), and draw pointgroup; (display in the structure) on its CLI, with adjustable parameter thresholds). $\endgroup$
    – Buttonwood
    Apr 30, 2021 at 14:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.