Skip to main content

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 according to the degree of each vertex.

A different interesting case is a structure with the excellent name of twistane.


This structure is also 3-colored by signatures (if that's the right terminology...) but is not bipartite with respect to degree. Or, to put it more simply, there are carbons with two carbon neigbours connected together. It seems quite clear from looking at the colored structure on the right that atoms with the same color are similar in the context of the structure.

Comments

Anonymous said…
Nice post and this enter helped me alot in my college assignement. Thank you as your information.
Anonymous said…
Hey, I am checking this blog using the phone and this appears to be kind of odd. Thought you'd wish to know. This is a great write-up nevertheless, did not mess that up.

- David

Popular posts from this blog

How many isomers of C4H11N are there?

One of the most popular queries that lands people at this blog is about the isomers of C4H11N - which I suspect may be some kind of organic chemistry question on student homework. In any case, this post will describe how to find all members of a small space like this by hand rather than using software.

Firstly, lets connect all the hydrogens to the heavy atoms (C and N, in this case). For example:


Now eleven hydrogens can be distributed among these five heavy atoms in various ways. In fact this is the problem of partitioning a number into a list of other numbers which I've talked about before. These partitions and (possible) fragment lists are shown here:


One thing to notice is that all partitions have to have 5 parts - even if one of those parts is 0. That's not strictly a partition anymore, but never mind. The other important point is that some of the partitions lead to multiple fragment lists - [3, 3, 2, 2, 1] could have a CH+NH2 or an NH+CH2.

The final step is to connect u…

Havel-Hakimi Algorithm for Generating Graphs from Degree Sequences

A degree sequence is an ordered list of degrees for the vertices of a graph. For example, here are some graphs and their degree sequences:



Clearly, each graph has only one degree sequence, but the reverse is not true - one degree sequence can correspond to many graphs. Finally, an ordered sequence of numbers (d1 >= d2 >= ... >= dn > 0) may not be the degree sequence of a graph - in other words, it is not graphical.

The Havel-Hakimi (HH) theorem gives us a way to test a degree sequence to see if it is graphical or not. As a side-effect, a graph is produced that realises the sequence. Note that it only produces one graph, not all of them. It proceeds by attaching the first vertex of highest degree to the next set of high-degree vertices. If there are none left to attach to, it has either used up all the sequence to produce a graph, or the sequence was not graphical.



The image above shows the HH algorithm at work on the sequence [3, 3, 2, 2, 1, 1]. Unfortunately, this produce…

Generating Trees

Tree generation is a well known (and solved!) problem in computer science. On the other hand, it's pretty important for various problems - in my case, making tree-like fusanes. I'll describe here the slightly tortuous route I took to make trees.

Firstly, there is a famous theorem due to Cayley that the number of (labelled) trees on n vertices is nn - 2 which can be proved by using Prüfer sequences. That's all very well, you might well say - but what does all this mean?

Well, it's not all that important, since there is a fundamental problem with this approach : the difference between a labelled tree and an unlabelled tree. There are many more labeled trees than unlabeled :


There is only one unlabeled tree on 3 vertices, but 3 labeled ones
this is easy to check using the two OEIS sequences for this : A000272 (labeled) and A000055 (unlabeled). For n ranging from 3 to 8 we have [3, 16, 125, 1296, 16807, 262144] labeled trees and [1, 2, 3, 6, 11, 23] unlabeled ones. Only 23 …