Aromaticity Introduction

Aromaticity Introduction

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[edit] Introduction

Consider the following six carbon member rings,



Drawn at this level, the two are very similar. One has single bonds, the other has double bonds, both look like a hexagon -however, if one looks at their structure in 3D this is not the case at all.



In cyclohexane, every carbon is in an sp3 hybridization, while in benzene they are sp2 as a result of the double bonds (which are left in their p orbitals to pi bond). This difference drastically changes not only the shape of the molecule, but also the reactivity. Cyclohexane, which orients each carbon in a tetrahedral fashion, is capable of many reactions, while benzene lies flat on a plane and because of its unique pi bond structure, is very unreactive in most situations.

[edit] Delocalization of Pi Electrons


One of the most fundamental properties of benzene (along with all other aromatic molecules) is the delocalization of its pi electron bonds. While the drawing of benzene above shows three double bonds (and thus three pairs of p orbitals interacting) there is no reason for the p orbitals to be paired and thus, every p orbital mixes together in a “delocalized” fashion, sharing the double-bond character between all of them. It is this unique formation of six electrons sharing a complex of multiple p orbitals which gives benzene its stable properties.


[edit] Effects of Aromaticity

The properties of benzene are in fact not unique to benzene at all. In general, aromatic compounds will have a planar organization of all the carbon atoms that share the delocalized pi bond system. These molecules as a result, will resist most reactions due to their natural stability and resonance stabilized organization.

[edit] The Huckel Rule

Molecular Orbital theory predicts the stability of aromatic systems and while an advanced knowledge in understanding bonding and antibonding systems is not necessary for the MCAT, a general understanding is. The theory can be summarized through the “Huckel Rule”. In summary, aromatics will have 4n+2 electrons in their delocalized pi bond system (where n can be any whole number). For instance, benzene has 6 electrons in its pi bonds, this works with the Huckel rule when n=1. When a compound has a number of pi electrons that does not obey the Huckel rule it is generally unstable because unwanted antibonding states are formed.

Consider the two systems below,

The four carbon system has two double bonds, and at first glance those pi bonds look like they could easily exist in a resonance system delocalized around the whole molecule. We could image that each carbon is sp2 hybridized and the extra p orbital from each carbon is working together to form a delocalized system. If we apply the Huckel rule however, no number for n can satisfy the value of 4. Thus one should predict that it is NOT aromatic, and in fact experimental measurements have shown it does not behave aromatically.

Looking now at the five carbon system we can easily see two problems with it. Firstly, if we try to make alternate resonance structures for the two pi bond electrons it looks as though they aren't completely conjugated around all the carbons. Secondly, if we apply the Huckel rule, we have the same problem as the 4 carbon system. Thus, this molecule is also NOT aromatic, but could we fix it some how?

Consider what would happen if a hydronium ion were removed from the carbon which is not participating in the pi system. If H+ were to leave, then the remaining carbon would be negatively charged with a lone-pair containing the two electrons which were originally in a sigma bond with the hydrogen. If these electrons were promoted to a p orbital then they would be inline with the other p orbitals of the pi bonding system and we would have a total of six electrons present instead of four. Six, just like benzene, works well with the Huckel rule, thus we can predict that this negatively charged ion of the compound is aromatic. In fact it is aromatic, and as a result, this molecule easily loses hydrogen in reactions to become the more stable aromatic ion.





1. Which of the following compounds is aromatic?

a
b
c
d
→ Following the Huckel rule, (a) has six electrons but does not involve ALL the carbons in the ring, thus it is not aromatic. (b) has eight electrons, which fails the Huckel rule. (c) has six electrons and is deficient of an electron, thus it has an empty orbital, an empty p orbital specifically, which allows it to participate in the pi system. (d) is both unconjugated and has triple bonds which will not double how many electrons are in a pi system as all the p orbitals must be in the same plane (triple bonds have p orbitals in two planes).

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[edit] Heteroatomic Atoms

In the previous examples we were focused on homoatomic systems (all carbons), but there is no reason other types of atoms can’t participate. The general rule for any atom in a possible aromatic ring is, if electrons can be promoted to a p orbital to obey the Huckel rule, they will, otherwise they will just exist in a regular lone pair or sigma bond (sp2 orbital).



Consider the sulfur containing heteroatomic ring thiophene above. If all the carbons are sp2 hybridized then we have four electrons remaining for pi bonding. This does not obey the Huckel rule, but we are not finished with the molecule yet. Sulfur has two lone pairs and two bonds (which would normally make four sp3 hybridized orbitals). If sulfur were to promote one of its lone pairs to a p orbital, it could become hybridized like all the other carbons (three sp2 and one p) and as a result there would be a total of four + two electrons from sulfur in the pi bonding network. This works with the Huckel rule and in fact this is exactly what happens.


(Notice that only ONE of the two lone pairs on sulfur was moved to a p orbital while the other remained in a hybridized sp2 orbital sticking out to the side on the same plane as all the other sigma bonds in the compound.)




1. The energy required to break a double bond is roughly 120 kJ/mole. Benzene has the equivalent of three double bonds yet the energy calculated to convert it to cyclohexane is 208 kJ/mol. What is the most likely reason for this discrepancy?

Benzene’s double bonds often combine to form triple bonds of greater stability.
Benzene’s carbons are closer together making its double bonds stronger.
Benzene’s delocalized pi bonds resist breaking due to their stability.
Benzene’s cyclic structure prevents it from being readily broken.
→ One would expect that bezene, with the equivalent of three double bonds, would take 3x120kj/mol, which is 360. This significant difference is due to the pi system which stabilizes the whole molecule.


2. How many pi electrons are involved in the above aromatic compound Naphthalene?

8
10
12
14
→ At first glance, naphthalene looks as though it were two fused benzene rings. Based on the double bonds we can determine that the pi system must wrap around both and through the center, thus involving all the carbon atoms. Thus, if each carbon contributes one electron (like in benzene) we get 10.


3. In the diagram of imidazole above, a total of ____ electrons from the two nitrogens are in sp2 orbitals while a total of ____ electrons from the two nitrogens are in p orbitals.

4; 6
2; 8
3; 5
7; 3
→ Each nitrogen must give 5 electrons and will want to be in an octet. The upper nitrogen has two bonds and a lone pair in sp2 orbitals, each 120 degrees apart. The p orbital contains the last electron and takes part in the pi system. For the bottom nitrogen, the three bonds use up three electrons in sp2 orbitals and the final two electrons are placed in the p orbital to take part of the pi bonding system.


4. Which of the above compounds has a delocalized pi bond system?

a
b
c
d
→ While the answer is not an aromatic ring, the sample principles apply for all delocalized pi systems, conjugated double bonds that allow multiple p orbitals to interact.

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