Evaluating Resonance Forms (2): Applying Electronegativity

by James

in Organic Chemistry 1, ResCC, Where Electrons Are

The last time I talked about evaluating resonance structures the molecules were, to be honest – pretty simple. Evaluating the resonance structures of ethene, the allyl carbocation, and the allyl carbanion are a pretty far cry from some of the more complicated structures you’ll see in a typical course. So today I thought I’d start to talk about the (much more common) situation when you have a π bond between two dissimilar atoms. How do you evaluate the resonance forms in these cases?

We’re going to have to go back to using our old friendly measuring stick, electronegativity for this task. Here’s the bottom line lesson for the resonance structures we’ll evaluate today.

Hopefully this makes some sense! Charged resonance forms are less stable than neutral resonance forms. So if we absolutely must form a charged resonance form, it makes sense to put the negative charge on the atom best able to stabilize it. How do we know which atoms stabilize negative charge the best? Well, a pKa table will give you a really good idea. But beyond that, if you look at the five key factors that influence acidity, one of the most important factors is the electronegativity of an atom. After all, electronegativity is ultimately a measure of to what extent an atom is able to stabilize negative charge.

So hopefully it should come as no surprise as we walk through these three examples that the second-best resonance forms are the ones where negative charge ends up on the more electronegative atom.

Let’s start by looking at a simple carbonyl compound, acetone (2-propanone).

If we look at the possible curved-arrow “moves” for drawing the resonance forms of this molecule, there’s two possibilities. In the first possibility, we draw an arrow from the π bond to the oxygen atom, putting a negative charge on the oxygen and leaving behind a positive charge on the carbon. In the second, we’re making the carbon negative and leaving behind a positive charge on the oxygen. Not only is there a positive charge on the oxygen, it has less than a full octet. This is an extremely unstable situation. So hopefully it’s clear that the resonance form where there is a negative charge on oxygen is the second-best resonance form next to the neutral one, and the resonance form where there is a negative charge on carbon is insignificant. Experiment bears this out. Calculations of the charge density on acetone reveal that the carbon is electropositive and the oxygen is electronegative, as per what we’d expect from electronegativity differences. So the molecule can be thought of as a hybrid of the best and second-best resonance forms.
Likewise, the resonance forms for the imine below similar behavior. As expected, calculations of electron density for this imine show that there is considerable positive charge density on the carbon and a high density of negative charge on the nitrogen. As you’d expect, the resonance form where there’s negative charge on carbon gets very little weight.

Finally we come to the acetate ion, which we discussed previously. Again, the second-best resonance form is that where there’s a positive charge on the carbon (and the worst is the one where it bears a negative charge).

With these examples in mind, can you apply the rule to determine the “second-best” resonance forms for each of these molecules?

Click here for answers

Next time I’ll go into a few more details on evaluating resonance forms based on lessons we learn from acidity and basicity.

Next Post: Evaluating Resonance Forms (3) – Where to Put Negative Charges 



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{ 5 comments… read them below or add one }

Pujan Vakharia

Can I get the answer to second best resonating forms? Willing to match my answers


James Ashenhurst

Yep – just put them up. Thanks for commenting!



Love this website, help me a lot for the DAT exam! Introduced it to lots of friends.


Kumar Neelabh

Please tell me where am I wrong… After reading the next article in this series, this came to my mind :
In the 2nd problem of the second best resonating forms (S=O), shouldn’t the negative charge be on S (least basic) ?

In fact there’s even an example in the next article in the “Factor-2” section, which shows that -ve charge would be more stable on S atom.



Good point. If the negative charge was on S however, the oxygen would have less than a full octet, which is very unstable. I should have mentioned in the section on positive charge that sulfur can have less than a full octet.


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