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Evaluating Resonance Forms (1) – The Rule of Least Charges

So far I’ve talked about resonance, and introduced the curved arrow formalism to show the movement of electrons. Importantly, we’ve talked about how the “true” picture of a molecule is a hybrid of its resonance forms (and not an equilibrium between forms).

Here’s a recap of the different “moves” we can perform on a molecule to get different resonance forms. Every resonance form we can draw for a molecule can be made through a combination of these three moves. 


We can apply these “moves” so long as we don’t break the octet rule. The thing is that if we start looking at even simple molecules, we can soon get lost in a maze of potential resonance forms if we apply the arrow-pushing rules willy-nilly. So let’s cut to the main problem: how do we evaluate the stability of different resonance forms? In other words, which are lower energy (more significant) and which are higher energy (less significant).

Here’s the punch line for today:

Let’s look at three examples. Really simple ones, but they’ll do.

The simplest molecule with a π bond is ethene. If we draw a resonance form for it, we can move the  π bond to the lone pair of one of the end carbons (doesn’t matter which one) to give a carbocation and a lone pair. This is a “legal” resonance form, since we’re not breaking the octet rule.

The question is, how significant is the resonance form with the two charges? The answer from experiment is: not very much. If you measure the electronic and structural properties of ethene, it’s clear that the carbon-carbon bond length is shorter than that of a normal carbon-carbon single bond, implying a strong interaction between the two atoms.  Furthermore, if the resonance form on the right were significant, we’d expect ethene to have a relatively high boiling point (since it’s charged) and be quite soluble in polar solvents like water. In fact, ethene boils at a very low temperature (–88 C) and is practically insoluble in polar solvents.  So it’s safe to conclude that the resonance form on the right is quite insignificant.

(One thing to note – the NET charge of each resonance form is the same here (i.e. they both have a net charge of zero) since on the resonance form on the right, the opposite charges cancel).

Let’s now look at the resonance forms of the allyl carbocation. The net charge on the allyl carbocation is +1. When we apply the relevant arrow-pushing “moves” to it, we get something like this, with 4 resonance forms – A, A’ , B, and C. Resonance forms A and A’ both have one charge. Resonance forms B and C have three charges. [Although again, note that B and C each have a net charge of +1 when you cancel the opposite charges].

So what does experiment tell us? Our best experimental evidence for the structure of the allyl carbocation tells us that the positive charge is distributed equally between the two end carbons, with a bond length intermediate between that of a C–C single bond (1.50 Å) and a C–C π bond (1.40 Å).  That is to say, the resonance hybrid of fhe allyl carbocation is a 1:1 mixture of resonance forms A and A’. Resonance forms B and C don’t contribute to the resonance hybrid to any significant extent.

Finally, let’s look at the allyl anion. We can use a similar application of the arrow pushing rules to get the different resonance forms, although here’s a twist. Note that we can’t move the end lone pair to form a new π bond without breaking the octet rule! So in this case we actually have to draw two arrows to make it legal. This gets us to the equivalent resonance form, A’.

On the other hand the other resonance forms (B and C) can be obtained through a fairly straightforward process (move a π bond to form a lone pair on either carbon.) Again, however, these resonance forms don’t contribute much to the overall resonance hybrid. Experiment tells us that the electron density on the allyl anion is on the ends, and the bond lengths are again intermediate between a C–C single bond and a C–C π bond. The best interpretation of this data is that the allyl carbanion is a 1:1 mixture of the resonance forms A and A’.

So again, the bottom line is that the most stable (and significant) resonance forms will be those with the fewest charges. 

You’ve probably noticed however that these examples are pretty simple – we’re just dealing with carbon atoms. What happens when we deal with unsymmetrical resonance forms? We’ll need to introduce a new principle for that.

Next Post: Evaluating Resonance Forms (2): Applying Electronegativity

Edit: removed “we’d expect to see a dipole for ethene” from the paragraph beginning with “The question is?..”. Thanks Jess for the tip.


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