Written by tutor Amelia P.
Part I: Resonance
The delocalization of electrons can be complicated, confusing and confounding! How do those
little dots move about the molecule on my paper and what does all of it mean? On paper, we only
estimate the location of electrons. In reality they occupy areas of space around atoms called
orbitals. Further complicating things, electrons are often shared between atoms. In order to
clarify the process of electron delocalization and the many effects that it can have we will start
with a molecule that has received a lot of media buzz in recent years, ozone. Ozone is O3, three
molecules of oxygen bonded together. Let’s take a look at the Lewis Structure of ozone.
At this point we notice that the green oxygen (in the center) has one
double bond and one single bond. Each bond represents a shared pair of electrons. The pink
oxygen is sharing 2 electrons with the green oxygen while the green oxygen shares 4 electrons
with the blue. The lone pairs associated with each oxygen atom are there to give each oxygen
atom a full octet. Count the electrons in the lone pairs and the electrons shared by each oxygen
atom. You will see that each atom has 8 electrons. If you’re particularly observant you might be
asking- isn’t there another way for each atom to have 8 electrons? Why, yes there is! What if we
had drawn the structure this way:
This is also an accurate depiction of ozone! The relationship
between these two images is that they are resonance structures. All that changed from one to
the other is the location of the electrons! Let’s take a closer look at the movement of the
To keep track of the movement of electrons notice in the second structure that the pi-bond
formed between the pink oxygen and the green oxygen is pink to indicate that it was formed
from the electrons that were previously on the pink oxygen. Also note that the “new’ lone pair on
the blue oxygen is drawn in black to indicate that it came from the pi-bond between the green
and blue oxygen molecules. So, how does ozone exist in nature? In fact, ozone exists “in
between” its two resonance forms. We can consider each bond to be a “1&1/2 bond.” It is
neither a single bond nor a double bond, instead, it has characteristics of both.
Note the use of curved arrows in the illustration of the resonance structure. Curved
arrows always indicate the movement of electrons (not atoms). Electrons always move away
from more electronegative atoms and towards more positive atoms. By being aware of and
following the rules of curved arrows drawing resonance structures can be very straightforward.
For practice, draw the resonance structures of NO3 (Hint: this compound has 3 resonance
Part II: An Example of The effects of Electron Delocalization
The specific location of electrons can have far reaching effects. For example, certain
reactions require a specific electron configuration around certain atoms. In a benzyne reaction,
for example, both resonance structures of the substituted benzene must be considered to
determine the products. I will spare you the gory details (ie- the mechanism) of this reaction but
we will see the effect that electron delocalization has on the reaction, regardless. Observe the
In the first illustration we see that our starting material is metafluoro-toluene. Both structures
are valid representations of metafluoro-toluene. We will now observe the product(s) of each
resonance structure upon treatment with NaH and H3O+.
Note that there are three unique products from this reaction. The second and fourth product
illustrated are the same. Without knowledge of electron delocalization and the know-how to
react BOTH resonance structures of metafluoro-toluene one would incorrectly resolve that
there are 2 products of this reaction. Hopefully this example helped to illustrate the far-reaching
implications of electron delocalization.
Part III: Aromaticity
Aromatic compounds are extremely important in chemistry. Aromatic compounds satisfy
a specific set of criteria. They are conjugated (consist of double bonds alternated with single
bonds), planar (all atoms are sp2 hybridized), have a continuous pi system, and satisfy the
Huckel 4n+2 rule (4n+2= # π electrons where n=positive integer). When this specific set of
criteria is met the compound is extremely stable. Aromatic compounds may contain anions,
cations and any heteroatom.
The ability of electrons to delocalize throughout an aromatic molecule is what makes the
molecule exceptionally stable. Specifically the second criterion, continuous pi system, relies on
the concept that electrons can delocalize/resonate throughout the molecule. Review the
following compounds and determine if they are aromatic:
All three compounds are aromatic as they satisfy the criteria outlined above. As an example, the
first one is analyzed below.
A resonance structure illustrates another form of a molecule as it exists in nature.
It is important to consider electron delocalization when writing a reaction mechanism.
Electrons do not move from or change after a bond is formed.
An aromatic compound must meet the following criteria:
I. Continuous pi system
III: Huckel Rule
IV: Tetrahedral Carbon molecules
VI: Contain only carbon and hydrogen
All of the above
I, II, III, IV, V
I, II, III, V
I, II, III, IV
Which of the following is a valid resonance structure for carbon dioxide?