Chemistry - What does the circle inscribed in a benzene ring represent?

Solution 1:

The circle, if anything, represents the inability of our everyday physical intuition to cope with the quantum phenomena. See, you would often encounter those two pictures with "double bonds this way" and "double bonds that way", intended to give a vague impression that the molecule switches quickly between the two, but that's not true.

Kekulé benzene

It does not switch; it just stays there, in some sense "halfway between" these structures. There is no movement of double bonds, and in fact there are no double bonds. (Also, there is no movement of electrons either. They just sit there on their molecular orbitals, which by itself is an approximation, though a pretty decent one). All $\ce{C-C}$ bonds in benzene are the same; they are neither double nor ordinary. Read the Wikipedia page on aromaticity, or just look at this picture from it:

"Real" benzene

That's what the circle stands for.

Solution 2:

This is a typical example of the downfall of the Lewis structures. In fact, $\ce{C6H6}$ can be represented by more than one Lewis structure:

Dewar and Kekule structures of benzene

Do any single one of them actually represent benzene? No! If it did, then the $\ce{C-C}$ bond lengths would differ from each other. In reality, it is observed that all the bond lengths are equal.

There are two approaches to solving this problem: valence bond theory (VBT) and molecular orbital theory (MOT).


An electron is described by the wavefunction $\psi$ where $$\psi(x)\equiv \langle x|\psi\rangle= \textrm{probability amplitude of finding the electron at} \, x;$$

where the state $|\psi\rangle$ is defined as $$|\psi\rangle = \int_{\textrm{all}\, x} |x\rangle\langle x|\psi\rangle \, \mathrm dx.$$

The probability of finding the electron in the region $x \pm \mathrm dx$ is given by $$\textrm{prob}(x,\mathrm dx)= |\psi(x)|^2\;\mathrm dx$$

Also, from the First Principles of Quantum Mechanics:

When an event can occur in several alternative ways, the probability amplitude for the event is the sum of the probability amplitudes for each way considered separately.$^\dagger$

Valence bond theory

Consider hydrochloric acid. How can it be represented by Lewis structure?

$$\ce{H-Cl}\,\,\,\textrm{or}\,\,\, \ce{H^+Cl^-}$$

The former is described by the wavefunction $$\psi_{\ce{H-Cl}}(1,2)= \psi_{\ce{H}}(1)\psi_{\ce{Cl}}(2) + \psi_{\ce{H}}(2)\psi_{\ce{Cl}}(1)$$ and the latter by $$\psi_{\ce{H^+Cl^-}}(1,2)= \psi_{\ce {Cl}}(1)\psi_{\ce {Cl}}(2)$$

However, we know that $\ce{HCl}$ is neither 100% ionic nor 100% covalent. To properly describe the wavefunction of $\ce{HCl}$, quantum superposition is applied viz.

$$\psi_{\ce{HCl}}= \psi_{\ce{H^+ Cl^-}}C_1 + \psi_{\ce{H-Cl}}C_2$$

where $C_1$ is the probability-amplitude to find the electron-pair at $|\psi_{\ce{H^+ Cl^-}}\rangle$ provided it was prepared in $|\psi_{\ce{HCl}}\rangle$ that is, $C_1 =\langle \psi_{\ce{H^+ Cl^-}}|\psi_{\ce{HCl}}\rangle;$ also $C_2 = \langle \psi_{\ce{H-Cl}}|\psi_{\ce{HCl}}\rangle$ and $C_1^2 + C_2^2= 1.$

A better description of the wavefunction for the molecule is a superposition of the covalent & ionic descriptions, & we write (with a slightly simplified notation) $$\psi= \psi_{\ce{H-Cl}} + \lambda\psi_{\ce{H^+ Cl^-}}$$ with $\lambda$ some numerical coefficient. In general, we write $$\psi = \psi_\text{covalent} + \lambda \psi_\text{ionic}$$ [...]According to the general rules of quantum mechanics, in which probabilities are related to squares of wavefunctions, we interpret the square of $\lambda$ as the relative proportion of the ionic contribution. If $\lambda^2$ is very small, the covalent description is dominant. If $\lambda^2$ is very large, the ionic description is dominant.$^{\ddagger}$

Resonance is nothing but quantum 'superposition of the wavefunctions representing different electron distributions in the same nuclear framework.'

In the same way, we can describe benzene as the following (un-normalised) wavefunction :

$$\psi = c_1\psi_{\text{Kek}1} + c_2\psi_{\text{Kek}2} + c_3\psi_{\textrm{Dew}1} + c_4\psi_{\textrm{Dew}2} + c_5\psi_{\textrm{Dew}3}$$

[It has been estimated that each Kekule canonical form has a relative contribution of about $0.408$ while the Dewar structures each has a relative contribution of $0.177\;.$ Thus the weight of two Kekule structures each is $0.39$ ie. they contribute to about $39\%$ to the resonance while each Dewar structure has relative weight is $0.073$ ie. they contribute to $7.3\%$ to the resonance hybrid. And thus many times, Dewar structures aren't taken into consideration in creating the resonance hybrid of benzene.]

Remember, the wavefunction on the left is real and all in the right are fictitious.

Resonance thus explains the delocalisation of electrons in unhybridised $p$ orbitals that gundergo $\pi$ bonding throughout the whole molecule. Thus, the circle represents the delocalisation of $\pi$ bonds over the whole molecule.

Molecular orbital theory

Molecular orbitals are one-electron wavefunctions which are formed due to linear combination of atomic or hybrid orbitals. Unlike in VBT where bonding occurs due to the coupling between atomic orbitals of the participating atoms and thus remain localised between those two atoms, in MOT, all atomic orbitals apart from those two atoms participate in the combination and thus the MOs are spread over all the nuclei of the molecules.

Delocalisation is thus inherent in the approach of MOT. The bonding influence of a single electron pair is spread over the entire molecule and not just the adjacent participating atoms of the molecule.

Now, let's come to the case of benzene.

All the unhybridised p orbitals of $\ce{C}$ atoms in $\ce{C6H6}$ combine to form MO. Since there are six such orbitals, six MOs would be formed.

The (un-normalised) wavefunction thus can be written as:

$$\psi= c_1\psi_1 + c_2\psi_2 + c_3\psi_3 + c_4\psi_4 + c_5\psi_5 + c_6\psi_6$$

Three bonding MOs and three antibonding MOs are formed; the former being lower in energy:

MOs of benzene

There are two pairs of degenerate orbitals; one is lower in energy while the other is in higher level. The electrons occupy the MOs of lower energy level.

Each MO is spread either all around, or partially around, the $\ce{C6}$ ring which corresponds to delocalised $\pi$ bonding. Thus, each electron pair helps in binding together several of the $\ce C$ atoms.


Due to delocalisation, all $\ce{C-C}$ bonds get extra stability which is due to the lowering of energy levels of the electrons. While VB theory explains with imaginary structures, MO theory explains it with the inherent delocalisation in its definition.

But the crux of the point is, the bonded electron pairs never remain confined or localised between two participating atoms; it spreads over the whole molecule; it delocalises.

The circle in that structure you mentioned represents this delocalisation of $\pi$ bond over the entire ring.


$^\dagger$ The Feynman Lectures of Physics Vol. III by Richard P. Feynman, Matthew Sands, Robert B. Leighton.

$^\ddagger$ Elements of Physical Chemistry by Peter Atkins and Julio de Paula.