Dodecahedrane has 12 faces, right?

So there was this guy called Euler, and he had a formula that goes something like F = E - V + 2. Well, actually it is χ = V - E + F, where χ is the Euler characteristic, and this is equal to 2 for polyhedra. Anyway, the point is that dodecahedrane has 12 faces (cycles).

For the SSSRFinder, however, it has only 11; which is annoying. Moreover the ring equivalence class method only distinguishes based on the underlying simple graph - in other words it ignores bond order. In some applications this might be exactly what is needed, but I'm glad that my method gives a more detailed result:

So, apart from being a ridiculously detailed image, the above shows the face (ring, cycle) equivalence classes for dodecahedrane with a particular double bond network. Clearly any face could be 'glued' to another along one of the edges, following the vertex classes. All possible combinations of faces are shown in the 'face quotient graph' at the bottom right.

Comments

Egon Willighagen said…

12 faces = 11 bonds to cut to remove all cycles... I think the SSSR is build around the latter concept. alpha-pinene has three unique cycles, but only two bonds to cut, and two a SSSR of size two?

20 December 2010 at 19:44

gilleain said…

I really should read the papers that the code references. What I suspect is happening is that the SSSRFinder is properly implemented, but that the algorithm doesn't work the way I expect it to. Oh, and I checked and yes alpha-pinene does have 2 rings in its SSSR.

20 December 2010 at 20:22

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

Some Stuff

Search This Blog