Last updated: October 31st, 2022 |
We’ve spent a lot of time in previous posts “building up” and drawing out the molecular orbitals for various species. In this post we’ll learn an extremely useful shortcut that will help us draw the energy levels of cyclic pi-systems extremely quickly.
The trick is called “Frost Circles”, or, sometimes, the “Polygon method”.
Table of Contents
- The Molecular Orbitals Of Benzene Can Be Superimposed On A Hexagon With the Vertex Pointing Down
- The Molecular Orbitals Of Cyclobutadiene Can Be Superimposed On A Square With the Vertex Pointing Down
- Frost Circles (The Polygon Method): A Useful Mnemonic For Drawing Energy Levels In Cyclic Pi Systems
- Frost Circles For Three-Membered Rings
- Frost Circles For Four-Membered Rings
- Frost Circles For Five-Membered Rings
- Six-Membered Rings
- Seven-Membered Rings
- Eight-Membered Rings
- Summary: Frost Circles
- (Advanced) References and Further Reading
1. The Molecular Orbitals Of Benzene Can Be Superimposed On A Hexagon With the Vertex Pointing Down
First: Recall that we saw the energy levels of the molecular orbitals of benzene look like this:
Useful observation: these energy levels can be superimposed on a hexagon with the vertex pointing down. Like this!
Let’s do another one.
2. The Molecular Orbitals Of Cyclobutadiene Can Be Superimposed On A Square With the Vertex Pointing Down
We’ve seen what the molecular orbitals for cyclobutadiene look like; the two highest energy levels are each singly occupied (and non-bonding, to boot) which helps to explain why cyclobutadiene is so spectacularly unstable.
The energy levels for cyclobutadiene can be superimposed on a diamond, which is just a fancy name for a square with its vertex pointing down.
That’s pretty cool. You might wonder…. does it work for other polygons too?
3. Frost Circles (The Polygon Method): A Useful Mnemonic For Drawing Energy Levels In Cyclic Pi Systems
Indeed it does. Back in 1953, Frost published an article describing this method for drawing out the energy levels in cyclic systems, with a simplified version as follows [Note 1]
“A circle… is inscribed with a polygon with one vertex pointing down; the vertices represent energy levels with the appropriate energies”.
Vertices below the halfway mark of the circle are considered bonding orbitals, and vertices above the halfway mark are considered antibonding orbitals. If vertices are exactly in the middle (as they are for 4- and 8- membered rings) they represent non-bonding orbitals.
This idea is presented in the diagram below for 3, 4, 5, 6, 7, and 8-membered rings:
This is an extremely useful mnemonic! This saves us a lot of work in building energy levels from the ground up.
To draw the molecular orbitals of a cyclic pi system, all we have to do is draw the appropriate polygon, vertex-down, and then fill it up with electrons.
Let’s see how we can apply these “Frost Circles” for 3, 4, 5, 6, 7, and 8 membered rings. (Although we’re going to skip drawing the actual circles, and just focus on the positions of the orbitals).
4. Frost Circles For Three-Membered Rings
There are two important configurations of energy levels for 3-membered cyclic pi systems, depending on the number of pi electrons.
With two pi electrons, we’d expect an aromatic molecule. One example is the cyclopropenium cation (below left), which is indeed aromatic.
With 4 pi electrons, an antiaromatic molecule is expected. Oxirene (below right) which has never been isolated, is in this category.
5. Frost Circles For Four-Membered Rings
Cyclobutadiene was covered above, and in the last post. Note that the molecular orbital diagram predicts that if you rip off two of the pi-electrons, the resulting cyclobutene di-cation should be aromatic, (Substituted cyclobutene dications have indeed been synthesized and found to be aromatic, by the group of the estimable, late, George Olah) [Note 2]
6. Frost Circles For Five-Membered Rings
Cyclic 5-membered pi systems with 6 pi electrons are predicted to be aromatic. Examples are abundant. such as the cyclopentadienyl anion (below left), furan (below right), pyrrole, thiophene, imidazole, and many others. And yes, arsoles are aromatic too, but you probably didn’t need me to tell you that.
Although not drawn here, removing 2 pi electrons would result in an antiaromatic system. The cyclopentadienyl cation is a classic example.
7. Six-Membered Rings
Already covered above – but note that the rules can be applied not only for benzene, but also “heterocycles” (i.e. aromatic rings with at least one non-carbon ring atom) such as pyridine, pyrimidine, and even the “pyrylium cation”.
8. Seven-Membered Rings
Cyclic 7-membered pi systems with 6 pi electrons are predicted to be aromatic.
For a ring entirely comprised of carbon atoms, this corresponds to the cycloheptatrienyl cation (sometimes known as the “tropylium ion”).
We’re used to thinking of carbocations as being unstable intermediates with a short lifetime. But the aromatic tropylium ion is such a stable salt you can actually put it in a bottle. Aldrich sells it.
The Frost circle method gives us the energy levels for the tropylium ion:
9. Eight-Membered Rings
With 8 pi electrons, cyclooctatetraene is predicted to be antiaromatic, and its molecular orbitals are predicted to look like this:
If you read the earlier post on antiaromaticity, however, you’ll recall that cyclooctatetraene can “escape” from antiaromaticity by adopting a tub-like shape. The molecular orbital levels we’d predict from the polygon method thus do not exactly correspond to the actual energy levels of cyclooctatetraene itself – it’s not a diradical, for instance.
That’s not quite the end of the story, however. Just like anti aromatic cyclobutadiene can be made aromatic through the removal of two electrons, the stability of (theoretically antiaromatic) cyclooctatetraene can likewise be adjusted by the removal or addition of two pi-electrons to give 6 or 10 pi-electrons, respectively.
Chemical relatives (“derivatives”) of the cyclooctatetraene dication (6 pi electrons) and the cyclooctatetraene dianion (10 pi electrons) have been synthesized and found to have aromatic character.
Therefore we would predict that these molecules are planar, unlike cyclooctatetraene itself. [In fact, this has been confirmed for the cyclooctatetraene dianion by X-ray crystallography – reference]
10. Summary: Frost Circles
While there’s no end of interesting things to learn about aromaticity (homoaromaticity, anyone?) this post concludes our treatment of it for the time being.
In the next series of articles we’ll finally get in to discussing the reactions of aromatic compounds, beginning with the most important reaction type of all: electrophilic aromatic substitution.
Many thanks to Matt Knowe for assistance with this article.
Note 1. The more complex version is:
A circle centered at α with radius 2β is inscribed with a polygon with one vertex pointing down; the vertices represent energy levels with the appropriate energies
What are these terms “α” and “2β”, you may ask? The answer is beyond the scope of what we’ll discuss here, but the essence is that they are important terms in Hückel Molecular Orbital Method, a way of simplifying the Schrödinger Wave Equation for pi systems by treating the pi electrons and sigma electrons separately. It’s a powerful and useful method for calculating the energies of pi systems and understanding their reactions.
Note 2. The stability arising from aromaticity in the cyclobutadiene dication should be reduced considerably, however, by the Coulombic repulsion arising from the di-cationic nature of the molecule (like charges repel, after all). Similarly, the stability arising from aromaticity of the cyclooctatetraene dianion is lessened somewhat by the repulsion of its two negative charges.
(Advanced) References and Further Reading
Synthesis and Isolation of cyclooctatetraene dianion:
- THE CYCLOÖCTATETRAENYL DIANIONThomas J. KatzJournal of the American Chemical Society 1960 82 (14), 3784-3785
In this paper Prof. Thomas Katz of Columbia University shows how the cyclooctatetraene dianion can be prepared by reducing cyclooctatetraene with two equivalents of potassium metal.
- Structure of the 10.pi. electron cyclooctatetraene dianion in potassium diglyme 1,3,5,7-tetramethylcyclooctatetraene dianion, [K((CH3OCH2CH2)2O)]2[C8H4(CH3)4]
Stephen Z. Goldberg, Kenneth N. Raymond, C. A. Harmon, and David H. TempletonJournal of the American Chemical Society 1974 96 (5), 1348-1351
This 1974 work has the first published crystal structure of the cyclooctatetraene dianion and show that it is indeed planar and has structural features consistent with aromaticity.
- Free Cyclooctatetraene Dianion: Planarity, Aromaticity, and Theoretical ChallengesAlexander Yu. Sokolov, D. Brandon Magers, Judy I. Wu, Wesley D. Allen, Paul v. R. Schleyer, and Henry F. Schaefer, IIIJournal of Chemical Theory and Computation 2013 9 (10), 4436-4443
So is cyclooctatetraene actually aromatic? This paper suggests that while the cyclooctatetraene dianion has considerable resonance stabilization (25 kcal/mol, vs. about 33 kcal/mol for benzene) this stabilization is swamped by Coulombic repulsion of the anions, and as noted by commenter Frosty the Circle, : “Like many multiply charged anions, COT2− exists in isolation only… as a short-lived resonance state lying above neutral COT” “….Charge-compensating complexation of COT2−with two sodium cations results in a thermodynamically stable Na2COT compound”
13 thoughts on “Frost Circles”
How to make molecular orbital energy level of non-cyclic organic compounds?? Is there any perfect method for all compounds?? Please describe
If we’re strictly discussing pi systems, I suggest looking here: https://www.masterorganicchemistry.com/2017/02/16/molecular-orbitals-of-the-allyl-cation-allyl-radical-and-allyl-anion/
Is cyclobutadienyl anion aromatic or not
Cyclobutadienyl di-anion, yes, because it would have six pi electrons. Cyclobutadienyl anion, no, because it can only be formed from deprotonation of C-H and the negative charge (and ensuing lone pair) would be at right angles to the pi system.
Your diagram of the cyclooctatetraene dianion shows the added electrons in a non-bonding orbital. With 10 electrons it would obey the numerology of Huckel’s rule but the diagram argues against it with the added electrons in the non-bonding orbitals.
The compound has been made as a disodium salt but the most recent theory is that it exists because it is stabilized by interactions with counterions: J. Chem. Theory Comput. 2013, 9, 4436−4443.
“Like many multiply charged anions, COT2− exists in isolation only as a short-lived resonance state lying above neutral COT.”
“….Charge-compensating complexation of COT2−with two sodium cations results in a thermodynamically stable Na2COT compound”
Very interesting! Also from the paper: “Comparison with all-trans octatetraene indicates that COT2− has a substantial aromatic stabilization energy (25 kcal mol−1) approaching that of benzene (33 kcal mol−1), but this favorable influence is swamped by Coulomb repulsion. “
TYPO: Note 2 line 1 “The stability arising from antiaromaticity in the cyclobutadiene dication ” should refer to aromaticity.
Overall this is an excellent series of articles.
Fixed it, thank you!
So to conclude: fulfilling all of Huckel’s Rules means a molecule is aromatic or antiaromatic. Drawing a Frost Circle can then determine whether a molecule is aromatic or whether it is antiaromatic.
Is this correct?
The Frost circle is mostly useful for visualizing what the energy levels will look like.
Once you know that the energy levels look like, then you fill them up with the number of available pi electrons.
When you do this with cyclobutadiene you end up with one doubly-occupied orbital and two singly-occupied orbitals (unstable). But when you do this with benzene you end up with three doubly-filled orbitals (stable!)
Just realised my mistake! It’s the number of pi electrons that tell you if it is aromatic or antiaromatic.
4n+2 is aromatic
4n is antiaromatic.
Like Dipita Karmakar, I also do not understand why electrons in non-bonding orbitals are destabilising, making the compound anti-aromatic, rather than just non-aromatic. I would think that these electrons contribute to neither stabilisation or destabilisation.
I also do not understand how I would know from the Frost diagram, why, if I take a 4n+2 system, and remove 2 electrons, the result would be the anti-aromatic system expected from a 4n compound. Because all filled orbitals are stabilising? I’m not sure how to see the aromatic stabilisation or lack thereof in this context.
The key issue is not so much that they are in non-bonding orbitals, it is more that these non-bonding orbitals are singly-occupied (due to Hund’s rule).
This means that these compounds are very similar in behavior to free radicals – except because there are two free radicals, they are what are known as “diradicals”.
So imagine the instability of free radicals, times two.
(The barrier to recombination of free radicals is basically zero, so these types of compounds very readily combine with each other even if they manage to form – they also react rapidly with oxygen)