4.4: Resonance

Electronegativity and induction effectively describe the distribution of electrons in σ-bonds, but are not sufficient to predict electron distribution in π-bonds. Consider allyl cation, [CH₂CHCH₂]⁺. If you were asked to draw a line-bond structure of this molecule, there are two equally valid possibilities that you could come up with, as shown below (the carbons have been numbered for bookkeeping purposes).

If the structure on the left is accurate, we would expect the bond between C1 and C2 to be around 133 pm long (a standard length for a carbon-carbon double bond) and the bond between C2 and C3 to be around 154 pm (a standard length for a single bond). We’d also expect the positive charge to be isolated on C3. Meanwhile, if the structure on the right is accurate, then we would expect the longer bond to be between C1 and C2, the shorter bond to be between C2 and C3, and the positive charge to be isolated on C1. When we measure the lengths of the C1-C2 and C2-C3 bonds, we find that they have the exact same length of 139 pm. That is, they are too long to be a standard double bond, and yet too short to be a standard single bond. We also find that the positive charge is equally shared by C1 and C3. Let’s look at the molecular orbitals in the π-system, below, to understand what is going on.

Each carbon atom (C1, C2 and C3) has an unhybridized p-orbital. These three p-orbitals combine in three different ways to make three molecular orbitals of increasing energy, as shown. The π-system in allyl cation has two electrons, so only the lowest energy π-molecular orbital is occupied. Notice that this occupied orbital is distributed equally across all three carbon atoms. That is, there are π-electrons equally between C1 and C2 and between C2 and C3 so that the true structure of allyl cation is the average of the two line-bond structures we drew above. When two or more structures are necessary to describe the bonding in a molecule, we call this resonance. The depictions above are two resonance structures of allyl cation. To fully represent the bonding in a molecule with resonance, we draw all reasonable resonance structures with a double-headed arrow between them, as shown below for allyl cation (we will show how to determine how to draw reasonable resonance structures below). We sometimes represent the true structure, or resonance hybrid, with dotted lines across all the atoms that share the π-electrons.

Resonance is able to help us accurately predict the electron distribution in π-bonds. Knowing where the electrons are helps us predict how a molecule will react.

Resonance is possible whenever you can draw multiple valid Lewis dot structures that only differ by their placement of π-electrons. It’s possible to determine resonance structures by guessing and testing with multiple Lewis dot structures; however, a much more methodical way to draw resonance structures and make sure that you don’t miss any is to keep track of the π-electrons using the curved arrow formalism (also called ‘arrow pushing’).

In organic chemistry, π-bonds are between adjacent, parallel p-orbitals (d-orbitals can also participate in π-bonds, but we will not see any examples in this course). In order to ‘push’ a pair of π-electrons along a molecule using a curved arrow, they need to be going from an unhybridized p-orbital (or π-bond) to an adjacent, parallel, unhybridized p-orbital. There are three ways that this can happen: (a) a π-bond can become a new π-bond; (b) a π-bond can become a lone pair; or (c) a lone pair can become a π-bond. We will look at each of these possibilities in turn. It’s important that every structure you draw is a valid Lewis dot structure, so be sure to keep track of all valence electrons and be mindful of the octet rule.

1. π-bond to π-bond
   The first possibility is that we can move the electrons from a π-bond along a molecule to form a new π-bond. This is exactly what we did with allyl cation. It’s important that the atom we move the electrons towards is able to accept electrons without breaking its octet. For example, in the resonance structure of allyl cation on the left, C3 has 6 electrons, so is able to accept the 2 incoming electrons into its unhybridized p-orbital. By contrast, in allyl anion, C3 already has 8 electrons, so we cannot push the π-electrons from C1-C2 to C2-C3, because the resulting structure would have too many electrons (10) on C3.
2. π-bond to lone pair
   The second possibility is that we can take the electrons shared between two atoms as a π-bond, and give them entirely to one of the atoms as a lone pair. When the two atoms have different electronegativities, the better resonance structure is the one that places the negative charge on the more electronegative atom (i.e. the more stable structure). For example, in the structure below, carbon and oxygen share two electrons in a π-bond. We can give both of these electrons either to the carbon atom or the oxygen atom to give a new resonance structure. Since oxygen is more electronegative than carbon, the better resonance structure is the one that places the negative charge on the oxygen atom (structure 2).
3. Lone pair to π-bond
   The final possibility is that we can take two lone pair electrons from one atom and share them with an adjacent atom to form a π-bond. Since π-bonds are between parallel p-orbitals, the lone pair must be held in an unhybridized p-orbital, and the adjacent atom must have a parallel, unhybridized p-orbital capable of accepting electrons. Consider the reverse of the example we used above. We can share one of the lone pairs on oxygen with the positively charged carbon to re-form the original resonance structure.

Resonance and hybridization
You may have noticed something interesting about the hybridization of the oxygen in the resonance structures above. If we look only at structure 1, the oxygen has one bonding group and two lone pairs so is sp² hybridized. If we look only at structure 2, then we would conclude that the oxygen is sp³ hybridized because it has three lone pairs and one bonding group. However, the oxygen can have only one of these hybridizations because there is only one true structure (the weighted average of all contributing resonance structures). In order for structure 1 to contribute, the oxygen must have an unhybridized p-orbital to form the π-bond. Thus, oxygen is sp² hybridized in both of these resonance structures. In structure 2, the oxygen holds one of its lone pairs in the unhybridized p-orbital.

Not all resonance structures are created equal
It is often possible for us to draw a number of resonance structures for a molecule. While resonance structures can all contribute to the overall resonance hybrid, they do not all contribute equally. More stable resonance structures contribute more towards the resonance hybrid in the weighted average than less stable structures. You can assess how ‘good’ or stable a resonance structure is using the same criteria we’ve used previously. As a guideline, the most important factors to consider in order of priority are:

1. Better resonance structures have more atoms with a full octet
2. If a structure has a net charge, better resonance structures place formal negative charges on more electronegative atoms and formal positive charges on less electronegative atoms
3. Better resonance structures have more covalent bonds
4. Better resonance structures have the fewest number of formal charges

Resonance and stability
Like with electronegativity, induction and polarizability, we can use resonance to help us assess the relative stabilities of charged molecules. Charged molecules that have more good resonance structures are more stable than those with fewer or poorer resonance structures. This follows the general principle that charges are more stable when they are more spread out. For example, consider the two molecules below:

The molecule at the top has two resonance structures, each placing the negative charge on a carbon. The molecule on the bottom has three resonance structure, each placing the negative charge on a carbon. Because the resonance structures of each molecule are equally ‘good’ (they all place the negative charge on carbon), the lower molecule is more stable because it has more resonance structures so the negative charge is spread between three carbons rather than two.

Now consider instead the two molecules below.

The molecule at the top and the molecule at the bottom each have two resonance structures. In the top molecule, the negative charge is shared between two oxygen atoms. In the bottom molecule, the negative charge is shared between the oxygen and a carbon. Oxygen is better at stabilizing negative charge than carbon because it is more electronegative, so the top molecule is more stable.

Worked examples:

Interactive: