In this post, we will see to an even greater extent than before how (simple) graph theory can be applied to the theory of alkanes and other hydrocarbons. We will discuss bipartite graphs, and see their application to chemistry.

Say there is some group of boys and girls that go to some dance. We can draw a graph responding to this group, pairing two people of opposite genders if they dance with one another. The resulting graph has an interesting property. We say it is bipartite. This means we can divide the vertices into two groups, the red vertices (girls) and the blue vertices (boys), such that no two vertices of the same color (gender) have an edge between them.

Now, we saw in the last post that any hydrocarbon with only single bonds can be represented as a graph all of whose degrees are at most 4. It turns out in chemistry there is a particular kind of hydrocarbon called an alternant hydrocarbon. The quickest definition is that it is a hydrocarbon whose carbon skeleton is a bipartite graph. We can see then that a class of chemical compounds can be described mathematically. All acyclic hydrocarbons are alternant hydrocarbons, because all trees are bipartite (seehttps://docs.google.com/document/d/1m2gRfmlF1MAVmbwc7NDGosU7bMThYuwaCqAfPi6EXq0/edit). The ones with an even number of atoms (vertices) even alternant and those with an odd number are old alternant. It turns out that these alternant carbons have several neat properties. These links: (http://www.bama.ua.edu/~blacksto/CH435_CH531/handouts/alternant%20pi%20systems.pdf

) and http://chemistry.umeche.maine.edu/CHY556/Alternant.html

describe many of them, and of course to be bipartite is to be “starrable.” The pi electrons and molecular orbitals (those not involved in the single bonds, but rather in additional bonds) are evenly distributed, and the electron densities are the same everywhere. This even distribution means there are no partial positive charges anywhere. Therefore, there is no charge difference anywhere; it is not merely that just different dipoles cancel out.

Whoa, wait a second. Aren’t there never any charge differences in a hydrocarbon, though? Aren’t the C-H bonds nonpolar? Well, any acyclic hydrocarbon, and quite a few cyclic ones such as benzene, are alternating and thus nonpolar, so hydrocarbons people deal with daily tend to be nonpolar. But it is possible for a molecule to have only C-H bonds and for it to be polar! So why is this? And why is it alternant hydrocarbons not have a dipole moment?

Consider Azulene.

There are two cycles of carbon with an odd length- the cycle on the left with 7 vertices and the one on the right with 5 vertices. So it is not bipartite. And in fact it is polar. Whoa! What’s going on there? It turns out, if you look at all possible resonance structures, some of the carbon atoms get double bonds more than other ones- it’s not like the alternant hydrocarbon benzene.

To fully explain this, we need to talk about dual graphs. For any graph, we can construct its dual graph whose edges correspond to vertices of the old graph, and whose vertices correspond to edges of the new graph. We connected two vertices in the new graph if the edges they represented in the old graph shared a vertex. Note that a cycle on N vertices in a graph will correspond exactly to a cycle on N vertices in its dual graph (the roles of the vertices and edges switch). So a graph is bipartite if and only if its dual graph is bipartite. Now, in an alternant hydrocarbon (say benzene), we can therefore color the bonds (edges of the graph, vertical of the dual graph) red and blue so that no two are adjacent. Then in one resonance (say in benzene), all of the red bonds will be double and the blue single, whereas in the other all the reds will be single and the blues double. So you have this nice symmetry. With azulene, however, there is no such symmetry. So there ends up being a net dipole.

Who knew that hydrocarbons could be polar? The fact that they are not is dependent on this notion of bipartiteness or alternateness, which we so often take for granted but not every hydrocarbon has.

Cool, huh? Next time, we will talk about multigraphs.