Visualising Ring Equivalence Classes in Jmol

As promised (in the previous post) I've now made Jmol scripts to show the atom/ring equivalence classes. I still think that the ring ones are more clear, but I suppose it depends on what aspect of the symmetry of the structure is needed. As an example:

Shown here is a C70 structure, with coloured circular plates at the centre of each face. It should be clear that there is an axis of symmetry running through the middle, from one blue plate to the other. Around the blue is a ring of green, and 5 rings in between.

The slight difficulty in all this was working out the ring equivalence classes. There is an existing CDK method to do this - in the SSSR ring finder - but it seems to give too many classes. The way I did it was to first find atom equivalence classes (or 'orbits') using signatures. Then each ring is a circular list of the orbit indices : which I'm going to call a 'ring code'. See this image for illustration:

These two rings (A and B) have the same ring code, written as the smallest concatenated string formed from their orbit indices. In other words, the signatures of each atom in the ring is converted to a number based on that signatures index in a list of all the signatures for all the atoms. Obviously other atom-equivalence class methods could be used to find the initial orbits; the rest of the procedure would be the same.

I do wonder if it would be quicker to just find the orbits of the dual of the embedding. However, that involves making that embedding first, so probably not...

Comments

Anonymous said…

Sharing some common points with chiral compound. Chiral synthesis study may benefit from the post.

5 January 2016 at 07:22

Adamantane, Diamantane, Twistane

After cubane, the thought occurred to look at other regular hydrocarbons. If only there was some sort of classification of chemicals that I could use look up similar structures. Oh wate, there is . Anyway, adamantane is not as regular as cubane, but it is highly symmetrical, looking like three cyclohexanes fused together. The vertices fall into two different types when colored by signature: The carbons with three carbon neighbours (degree-3, in the simple graph) have signature (a) and the degree-2 carbons have signature (b). Atoms of one type are only connected to atoms of another - the graph is bipartite . Adamantane connects together to form diamondoids (or, rather, this class have adamantane as a repeating subunit). One such is diamantane , which is no longer bipartite when colored by signature: It has three classes of vertex in the simple graph (a and b), as the set with degree-3 has been split in two. The tree for signature (c) is not shown. The graph is still bipartite accordin...

1,2-dichlorocyclopropane and a spiran

As I am reading a book called "Symmetry in Chemistry" (H. H. Jaffé and M. Orchin) I thought I would try out a couple of examples that they use. One is 1,2-dichlorocylopropane : which is, apparently, dissymmetric because it has a symmetry element (a C2 axis) but is optically active. Incidentally, wedges can look horrible in small structures - this is why: The box around the hydrogen is shaded in grey, to show the effect of overlap. A possible fix might be to shorten the wedge, but sadly this would require working out the bounds of the text when calculating the wedge, which has to be done at render time. Oh well. Another interesting example is this 'spiran', which I can't find on ChEBI or ChemSpider: Image again courtesy of JChempaint . I guess the problem marker (the red line) on the N suggests that it is not a real compound? In any case, some simple code to determine potential chiral centres (using signatures) finds 2 in the cyclopropane structure, and 4 in the ...

General Graph Layout : Putting the Parts Together

An essential tool for graph generation is surely the ability to draw graphs. There are, of course, many methods for doing so along with many implementations of them. This post describes one more (or perhaps an existing method - I haven't checked). Firstly, lets divide a graph up into two parts; a) the blocks, also known as ' biconnected components ', and b) trees connecting those blocks. This is illustrated in the following set of examples on 6 vertices: Trees are circled in green, and blocks in red; the vertices in the overlap between two circles are articulation points. Since all trees are planar, a graph need only have planar blocks to be planar overall. The layout then just needs to do a tree layout on the tree bits and some other layout on the embedding of the blocks. One slight wrinkle is shown by the last example in the image above. There are three parts - two blocks and a tree - just like the one to its left, but sharing a single articulation point. I had...

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