14.1: Conjugation vs. Aromaticity

Let’s recall our discussion about conjugation. Remember that as more and more p systems become conjugated, the HOMO-LUMO gap shrinks. The energy required to excite an electron from \(π\) --> \(π^{*}\) decreases, or shifts to longer wavelengths of light. For example, hexatriene has a \(λ\)_max of 330 nm. Now, what if we were to place these same 6 p electrons not in a straight chain, but in a cyclic ring. Well, we get the molecule benzene. Notice that the \(λ\)_max of benzene (255 nm) is considerably different than hexatriene – it takes much more energy to excite an electron from HOMO to LUMO. As an aside, this wavelength of light is often used to detect quantify proteins because they contain phenyl rings in one of their amino acid side chains.

These data tell us that there is something interesting going on. Another clue comes from the bond lengths of benzene and hexatriene. In hexatriene, the double bond is 1.34 Å but the single bond is 1.47 Å. In benzene, all of the C-C bonds are identical in length. In order for this to be the case, the Lewis structure is incomplete, and we draw a resonance form that indicates the bonds are identical. We often abbreviate benzene as a six-membered ring with a circle inside to indicate that the electrons are moving throughout and all bonds are equal. There is also a third resonance form called Dewar benzene, which is only about 6% of the resonance contribution.

Given this, does 1,2-dichlorobenzene have one or two isomers? In other words, are the two structures below the same? Indeed, these molecules are the same.

How about another example – does cyclooctatetraene (COT) have one or two isomers? One might think that since we can draw a circle of \(π\) electrons inside the ring, that there is only one isomer, but that would be incorrect. COT has two isomers, and there is an equilibrium between them (E_a = 12-15 kcal/mol). So, why the difference?

Well, the thing that makes benzene so special is a property called aromaticity. We say that benzene is aromatic (originally because these compounds smelled sweet) and that COT is antiaromatic. You might guess that the difference have something to do with benzene having 6 \(π\) electrons and COT having 8 \(π\) electrons, but that’s not the whole story. We will come back to how one would predict or define aromaticity in a second, but let’s look quickly at the reactive properties of aromatic compounds, for it confirms the structure of benzene we have drawn. During the bromination of 2-butene, a bromonium ion forms, followed by S_N2 attach to yield a trans-dibromide product. Overall, we lose a \(π\) bond (63 kcal/mol) but gain two \(σ\)_C-Br bonds (2 x 67 kcal/mol). Overall, this is exothermic.

So, what happens when we try to brominate benzene? There is no reaction. This tells use that benzene is a worse nucleophile. Recall last time that conjugated systems were both more nucleophilic and more stable. In other words, the result with benzene completely contradicts this principle.

But when we treat benzene with Br₂ in the presence of a Lewis acid, a reaction occurs.

These results suggested to those who originally studied it that there was something “special” about benzene that made it a thermodynamic sink. This turned out to be aromaticity, which is characterized by electrons flowing in a complete circuit in a planar ring. The structure itself was first proposed by August Kekulé who dreamt that a serpent was dancing around a six-membered ring while biting its own tail.

What makes benzene so “special?” Resonance energy, or what I like to call the aromatic stabilization energy.

Aromatic Stabilization Energy

Let’s consider heats of hydrogenation for various alkenes to understand aromaticity better.

a) Single isolated double bond

b) Two isolated double bonds

c) Conjugated double bonds

Overall, the starting material is more stable, so this reaction is exothermic (conjugation adds ~2 kcal/mol of stability)

d) Benzene

Benzene is aromatic, and thus much more stable than cyclohexatriene. One might predict that the ΔH_o by the following formula: 3 x (-28.5 kcal/mol) = -85.5 kcal/mol, and if you factor in the contribution from conjugation, it should be 3 x (2 kcal/mol) = 6kcal/mol higher, so -79.5 kcal/mol. However, there is huge difference between this predicted value and the observed value. It tells us that mere conjugation is not enough to explain the stabilization - there is something greater, which we call the aromatic stabilization energy. How large is it? Well, 85.5 – 50 = 35.5 kcal/mol!!!

Huckel's Rule

So, how do we determine or predict aromaticity? We follow Huckel’s rule – “things that are aromatic have a closed 4n+2 loop of \(π\) electrons that are completely aligned.” Essentially, there are four factors:

1. cyclic – things that are aromatic must be rings

2. fully conjugated – no intervening sp³ hybridized atoms

3. planar – in order to align properly, these molecules must be flat

4. 4n+2 \(π\) electrons in the system, where n is an integer or zero.

Systems that abide by rules 1-3, but have 4n \(π\) electrons are said to be antiaromatic.

Let’s do some examples – are these molecules aromatic or antiaromatic?