[Kate]

Hi.

My question is: why is this structure of azulene not considered aromatic?



It's planar, with sp² carbons and obeys Huckel's rule with 10 pi electrons.

[Corribus]

Short answer is that Huckel's rule only applies perfectly to monocyclic ring systems. When you start making other carbon-carbon connections, this changes the secular determinant and the resulting energy solutions. There isn't a convenient generic solution as a function of n anymore. You'll have to solve the problem on a case-by-case basis.

[Kate]

My teacher didn't go into Huckel's method to explain this. He just drew the other resonance structure and added the number of electrons on each ring: 6 + 6. And so, each ring obeys Huckel's rule of 4n+2 pi electrons.

What do you mean by "You'll have to solve the problem on a case-by-case basis"? You're talking about constructing the matrix for each different molecule in this situation and going from there to determine the energy levels of the MO's?

[Corribus]

Yes. For regular cyclic annulene with n carbon atoms (or linear molecule with n carbon atoms) you can arrive at an analytical expression for the energy levels as a function of n. I don't remember what that expression is off the top of my head but most p-chem books have it in there somewhere. From that expression you can derive Huckel's rule for aromaticity.

As soon as you introduce other branches or connectivities*, those general solutions no longer apply, so Huckel's rule is basically out the window. You'd have to set up a unique matrix in that case and solve for the energy levels. Straightforward but the algebra can get crazy for anything over 5 or 6 carbon atoms. Symmetry treatments help but even then you'd probably want to use a good math package to solve for the roots of your determinant.

*This is particularly the case for polycyclic examples in which at least one ring has an odd-number of carbons, due to the fact that they are not alternants. (An alternant is one in which carbons can be divided into two groups A and B, in which there is always an A next to only B's, and vice-versa.)

[Kate]

Yeah, I was going to do the matrix for benzene actually but then realized that I don't know how to calculate the determinant of a 6x6 matrix. I'm sure there are some math softwares where you can put your matrix in and they give you the determinant, but I never used any of that before so I'll have to check it later.

[Corribus]

If you know how to work with character tables and generate symmetry orbitals you can simplify the problem significantly into smaller dimensional matrices. McQuarrie and Simon walks you through how to do this for benzene, if you have access to that textbook.

One of the interesting things you can show using Huckel theory for azulene is that it has a pretty significant dipole moment, which you might not expect from a hydrocarbon. This contributes to some of the molecule's interesting spectroscopic properties compared to more symmetric molecules like naphthalene. This is all resulting from the manner of connectivity. I still find it quite impressive that such a simplistic theoretical treatment has the power to predict and explain the physical properties of such complex organic molecules.

[Kate]

Well, I know how to work with character tables and irreducible representations but I never really understood the logic behind it as it's something that was never really explained in class. And since this was only covered for 2 or 3 weeks in my physical chemistry 2 course, I always thought I wouldn't need it. Lately though, I've realized that symmetry is really important in organic chemistry because of chirality, so it's definitely something I have to look up to see how it's done. I have McQuarrie's textbook, so I'll check it there. Thanks.

[Corribus]

Well, I know how to work with character tables and irreducible representations but I never really understood the logic behind it as it's something that was never really explained in class.

Not surprising. Symmetry can be one of the most useful concepts in physical chemistry and spectroscopy but this is contrasted by how poorly it is usually taught. Most physical chemistry professors I've interacted with do a fine job of teaching how it works, but going the extra step to explain why it works is often neglected. The truth is that if you can understand the concepts behind the math of group theory and start to recognize patterns and their chemical significance, you can use it to make solving a lot of complicated problems very simple and explain a lot of physical phenomena. The Huckel treatment of azulene or napthalene is virtually impossible to do by hand, but with some simple symmetry operations you can solve it in less than an hour.

I have McQuarrie's textbook, so I'll check it there. Thanks.

If you need any further help on the matter, you know where to come.

[Kate]

Not surprising. Symmetry can be one of the most useful concepts in physical chemistry and spectroscopy but this is contrasted by how poorly it is usually taught. Most physical chemistry professors I've interacted with do a fine job of teaching how it works, but going the extra step to explain why it works is often neglected. The truth is that if you can understand the concepts behind the math of group theory and start to recognize patterns and their chemical significance, you can use it to make solving a lot of complicated problems very simple and explain a lot of physical phenomena. The Huckel treatment of azulene or napthalene is virtually impossible to do by hand, but with some simple symmetry operations you can solve it in less than an hour.

Well, I often suspected my professor didn't know why he was constructing those tables either. But most topics in chemistry that really need a mathematical approach were often neglected by chem teachers and what we got was a superficial/qualitative understanding. And my math classes were always given by mathematicians themselves, with no concern or clue for how students might apply it to solve problems in chemistry.

I had one algebra class in undergrad and we didn't even cover group theory.

If you need any further help on the matter, you know where to come.

Absolutely. Thanks.