# Why six C atoms are usually seen in cyclic compounds?

When it gets to Carbon-based molecules, one very possible structure when there are more than six C atoms is the hexagon; though not mostly perfect, it emphasizes that six Carbon atoms tend to bond with each other.* Why?
I assume this is because of the "stability" that is attributed to the hexagon. Explainers have discussed mainly about this issue.
Neither me nor my teacher were convinced by the description; (Actually, all of this is because of my teacher!) since he asked that for example, why not octagon? No response.

I wondered that if I asked this in here what ideas I am going to face.... Sorry, but I'm accustomed to doing lots of "whys".
*For example; look at the pic below:

In the image above, the number of hexagonal rings is twice the number of pentagonal rings and eight times more than octagonal rings. This macro-molecule is the micro of what's happening in the nature. I don't have citations; but it takes a minute or two to see how much the number of hexagonal rings is surpassing the number of other rings. Perhaps it was so obvious, nobody mentioned it.
Update: According to @ron, and with a little payment of thought, I reached the fact that these rings aren't necessarily geometric; the important thing is that six C atoms, instead of five or eight or whatever else number of C atoms are the majority.
Image credit: Pseudopotential model for Dirac electrons in Graphene...

• Could you be more clear about what you are asking, perhaps by giving some examples. The shape of a ring system depends on how many atoms are in it but I am not sure if this is what you are asking.
– bon
Dec 24, 2014 at 18:11
• You see how most of organic compounds have a hexagonal ring if there are more than 6 Carbon atoms; (if the structure is not linear) For example, an organic compound with 8 C atoms is more likely to have a hexagonal ring with two other C atoms attached rather than an octagonal structure. There are compounds with other rings, but they are the minority. My Q is about the reason C atoms mostly preferred to be hexagonal. I am adding a pic for demonstrating what I mean. Dec 24, 2014 at 18:27
• @MARamezani Those 6-membered rings in your drawing are not geometric (e.g. perfect) hexagons.
– ron
Dec 24, 2014 at 18:42
• @ron, do you think it's better to modify the Q like this: Why is it usual that six C atoms are mainly bonded to each other to form organic compounds? Dec 24, 2014 at 18:45
• @MARamezani If you have 6 carbon atoms, you could make 6 methanes, 3 ethanes, 1 butane and 1 ethane, 1 benzene, etc. 6 carbon atoms can be arranged in a variety of ways, one way is not more or less "usual" than another.
– ron
Dec 24, 2014 at 18:51

It is all about minimizing the energy of a molecule.

In the case of carbon, the only molecule that adopts a perfect hexagonal geometry in its ground state is benzene (and its derivatives that possess a 6-fold rotational axis). In this case the hexagonal geometry is adopted because all of the carbons are $$\ce{sp^2}$$ hybridized. The ideal geometry (lowest energy) for an $$\ce{sp^2}$$ hybridized carbon involves three 120 degree bond angles around the carbon atom. This can be accomplished in the case of benzene by placing 6 carbon atoms in a plane and connecting them as shown to produce the benzene structure.

Other carbon-based molecules may project as hexagons, but they are not hexagons when viewed in 3 dimensions. For example, cyclohexane is not flat like benzene, but rather typically exists in what is called a "chair conformation" in its ground state where the C-C-C angle is around 109.5 degrees.

why not octagon?

Some other regular polygon shapes can be found in molecules. For example, the cyclopentadiene anion exists as a regular pentagon, and the cyclooctatetraene dianion exists as a regular octagon. Both of these molecules contain only $$\ce{sp^2}$$ hybridized carbons.

As mentioned above, these atoms prefer to exist in a geometry with 3 equivalent 120 degree bond angles. However, the internal angle in a regular pentagon and a regular octagon are 108 and 135 degrees respectively. Since these angles deviate from the 120 degree ideal, these molecules will be strained, unlike benzene.

What other ring shapes are preferred, low-energy minima?

We've already discussed the $$\ce{sp^2}$$ case where the preferred 120 degree angle is found in benzene which exists as a hexagonal structure. $$\ce{sp}$$ hybridized carbon is most stable when the X-C($$\ce{sp}$$)-X bond angle is 180 degrees. Clearly small or medium size rings comprised solely of $$\ce{sp}$$ hybridized carbon cannot exist as stable entities. In the case of $$\ce{sp^3}$$ hybridized carbon, the closer we get to the ideal tetrahedral angle (an angle whose cosine = -1/3, ca. 109.47 degrees) the lower the energy of the molecule. We've already mentioned cyclohexane and while it's C-C-C angle is close to the ideal tetrahedral angle, it is not exactly the tetrahedral angle (because each of its carbon atoms does not exist in a perfect tetrahedral environment). If we fuse a few more cyclohexane rings onto cyclohexane, we can produce the molecule adamantane with 4 cyclohexane faces all in the chair geometry.

There is less strain energy in adamantane (per carbon) than in cyclohexane. The geometry at each carbon atom is closer to the ideal tetrahedral geometry, but still not quite there. How can we further elaborate adamantane into a ring system where all carbons have the ideal tetrahedral angle? Does adamantane look vaguely familiar? By continuing to add more and more cyclohexane rings we build the following ring structure...

... Diamond, where all of the $$\ce{sp^3}$$ carbon atoms are in the ideal tetrahedral environment, an extremely stable arrangement.

EDIT: response to OPs comment

So, the lesser energy state must be the reason that after six number of atoms, five is the common one, right?

Kinda. Here's a table showing estimated ring strains for saturated hydrocarbon rings of various sizes.

[Image source Also, see this link for a discussion about why the ring strain for cycloheptadecane (-3.4 Kcal/mol) is suspect and probably should be higher.]

5- and 7-membered saturated hydrocarbon rings have similar strain energies, yet 5-membered rings are more common. We must consider the free energy of the process, enthalpy and entropy changes. In addition to strain (an enthalpic factor) we must also consider entropy. In ring forming reactions it is entropically more difficult for a 7-membered chain to bring both ends together for closure to a ring, then it would be for a smaller 5-membered chain.