Below is the written transcript of my YouTube tutorial video – Major and Minor Resonance Structures.
Leah here from leah4sci.com and in this video we’ll look at how to identify major and minor resonance contributing structures. This is video 3 in the resonance series, you can all the videos along with the practice quiz and study guide by visiting my website leah4sci.com/Resonance.
In video 2, we looked at different molecules and the resonance forms to learn what key patterns identify and where to move your arrows. But once you come up with more than one structure where they don’t look alike,how can you tell which is the major contributing structure and which is the minor contributing structure? And the answer is not to memorize rules. Instead the answer’s understand stability of each contributing structure. In Organic Chemistry, and in life it’s important to recognize this key key principle. If a molecule is happy, meaning it’s stable, it’s going to be unreactive. If it’s comfortable where it is, it has no desire to change the situation but if it’s unhappy, it’s unstable, and it wants to change its situation making it very reactive. So as we’re looking at the resonance contributing structures, ask yourself, which one is more stable and which one is less stable and more importantly, why is it more stable and why is it less stable?
The first rule to look out for is minimize charge. Charge is a burden. A positive charge and a negative charge is a burden to atoms and if you can avoid having charge, then the molecule will be stable or happier. When minimizing charge, keep in mind the concept to avoid separation of charge. A positive and negative charge next to each other will attack. The negative will attack the positive in an effort to get rid of charge going back to the principle of charge as a burden, let’s get rid of it. We look at the resonance for propanone where we decided that the pi bond can be broken to collapse the electrons onto Oxygen giving me a resonance structure that looks like this. Oxygen with an extra lone pair and a negative charge, carbon with an incomplete octet and a positive charge. The net charge on both sides is a zero so it is a viable structure. But in terms of which one is more stable, the one on the left has absolutely no charge. If charge is a burden, it’s not carrying that burden. The one on the right, not only does it have a positive and a negative charge, but it also has a separation of charge two opposite charges next to each other. In fact, this structure’s considered so unstable that the negative charge will attempt to attack the positive charge giving me the neutral molecule as the ideal resonance form and therefore the major structure.
On your exams, we’ll call the neutral one major and the charge one minor. Now what does this mean in terms of the resonance hybrid? If we said that in reality it doesn’t exist as both. What you want to think of is that yes, the resonance hybrid is going to be some intermediate of the two. But that intermediate will look much more like a pi bond and much less like the separation of charge. So rather than having something exactly in the middle, you’ll tend to see more of the pi bond pair. To compare it’s like negative at the top and a very slight positive all the carbon. Now keep this in mind because in reactions, when you hear a carbonyl carbon is partially positive, this is what we’re talking about, this is where it gets its partial positive charge.
Let’s look at the molecule Ozone. In my Lewis structure video, I showed you how to setup the Lewis structure where we came up with a positive and negative charge. A negative charge on the Oxygen with three lone pairs, and a positive charge on the Oxygen with three bonds and one lone pair. We can show resonance for a lone pair on a negative Oxygen reaches to form a pi bond between itself in a positive Oxygen causing a pi bond between a positive and neutral Oxygen to collapse towards the right. Start by re-drawing the skeleton showing everything that didn’t change then let’s add in the changes. Between electrons formal pi bond, I will show these as red electrons forming the lone pair on the right oxygen atom. We still have the positive on this central Oxygen and a negative on the Oxygen to the right.
But here’s another structure that we could show. Instead of showing the negative Oxygen forming the pi bond for the positive Oxygen we can show the electron from the pi bond simply collapsing on to the Oxygen atom so that there are only single bonds in this molecule. The only thing that change is that we get a pink lone pair to the Oxygen on the right but look what happens when we work out the formal charge. Negative one for Oxygen on the right or left because they have extra lone pairs. The Oxygen on the center should have 6 electrons attached to it but I only count 4. A quick formal charge tells me 6 minus 4 is positive 2 and we still have net charge of zero, we’ll look at the separation of charge on this molecule. This is going to be a very very very minor structure because it’s so unstable. These two are the same so we don’t have to rank them. But now these two are also the same because they each have a negative Oxygen and a positive Oxygen on the center. So for this structure, we really have one major which can be shown right or left or you can call them two majors if you want to differentiate between the Oxygen negative on the right and negative on the left. And then a very minor structure because we have way too many charges going on.
The next rule to keep in mind is electronegativity. An electronegative atom is one that likes and wants electrons. So it makes sense that the more electronegative the atom, the more likely it’ll be to hold the negative charge and therefore the happier it’ll double the negative charge. If you have an atom that is very low in electronegativity, it’ll prefer to hold a positive charge meaning it’ll be more stable holding that positive charge. Let’s take another look at the ketone. We showed the resonance where the pi bond will collapse onto the Oxygen atom giving it an extra lone pair. A quick formal charge gives me a negative on Oxygen and a positive on Carbon. But there is another way that you can show this. You can take the pi bond and collapse it down onto carbon instead of up onto Oxygen. The resulting structure only has 2 lone pairs on Oxygen, an extra lone pair on carbon with a negative one on the carbon atom and a positive one on the Oxygen. In terms of conservation of atoms and charge, they work! The atoms are all there, we didn’t any sigma bonds and the net charge is zero on each structure.
But look at the difference in electronegativity. Remember the Orgo Basics series I had you memorize the following atoms and their placement on the periodic table. The Hydrogen, Carbon, Nitrogen, Oxygen, and Flourine, and phosphorous, Sulfur, Chlorine, then Bromine, then Iodine. I had you remember the trends where electronagativity goes up on towards the right and size is down on towards the left. Well this is one of those situations where knowing that stable will help you. If electronegativity goes up and towards the right and the two atoms we’re comparing are a carbon and Oxygen. Oxygen is to the right of carbon making it more electronegative. Carbon is to the left of Oxygen making it less electronegative. The goal is to have the negative charge on the more electronegative atom so the positive charge should go on the less negative atom. That means Oxygen would rather have the negative charge and carbon would rather have the positive charge. So for this molecule, the neutral structure is the major contributing structure because no charges are better than distributed charges.
The second or minor structure would be a negative Oxygen and positive carbon. And the very very minor, you probably don’t even have to show this on your exam is a negative carbon and a positive Oxygen. Not only does this structure have a separation of charge in the wrong direction, we’re also violating the next rule which is to pay attention to your octets. The octet rule tells you an atom wants to have 8 electrons in its valence shell through lone pairs or bonds. If you setup a resonance structure where you’re violating that octet by putting in too many electrons if it can’t handle it or taking away electrons so that does not have a complete octet, it would be stable.
Atoms in period three and lower are larger and can have more than 8 in their octet.The common atoms that want to have 8 and only 8 are carbon, nitrogen, Oxygen, and Fluorine. Remember that within these atoms, electronegativity goes towards the right and the more electronegative the atom, the less likely it is to hold the positive charge. So when you see this atoms, carbon being the least electronegative of all of them is okay with the positive charge but you do not want to have a positive charge due to an incomplete octet on Oxygen, Nitrogen, or Fluorine. If you have resonance that puts an extra bond on Oxygen or Nitrogen to give it a positive charge but still have a complete octet, that’s okay.
So let’s go back to this example and notice that carbon has a complete octet. It has 2, 4, 6, electrons in bonds to carbon and Oxygen. Two electrons as a lone pair for a total of 8. Oxygen only has 6 in its octet instead of 8. We have two in the bond, two more for each of the lone pairs for a total of 6. An Oxygen being more electronegative than carbon will be very very unhappy with an incomplete octet and a resulting positive charge. We definitely want to keep this in mind when you’re trying to decide which electrons to use in resonance. For this molecule, we have three Nitrogen atoms that each have a lone electron pair and the question is which one do we use to start a resonance? First notice that if we cut it down the middle we have symmetry so the Nitrogen on right and left are the same. The question then is do we use the upper Nitrogen which will show with purple electrons or the lower Nitrogens with the red electrons?
If we try to use the purple electrons then we’ll violate the octet for carbon as follows. Everything stay the same but we now have a triple bond between Nitrogen and Carbon. Carbon now has a total of 10 electrons in its octet. Octet means 8, 10 are not allowed. But Nitrogen still has a complete because it has 4 bonds and 8 electrons. Another clue to help you recognize that this is incorrect is a quick formal charge on Oxygen. We have a total of 4 electrons, should have a total of 5 giving us a charge of plus one. Carbon should have a total of 4 directly attached. Given that we have five bond there are five directly attached, 4 minus 5 is negative one and everything about this should scream something’s wrong.
So let’s try it again and this time we’ll use the red electrons. If these electrons move towards the sp2 hybridized carbon it would again violate the octet, but this time carbon can compensate by taking the green pi electrons and collapsing them as a lone pair onto Nitrogen. The resulting structure has a pi bond between the left Nitrogen and carbon and then an extra green lone pair on the upper Nitrogen atom. A quick formal charge tells you that the Nitrogen on the left has a positive charge. Carbon has 4 bonds still has 4 bonds, the octet has obeyed and it’s good to go. Nitrogen on top now has a negative charge due to having another lone pair on the top. Now check it out, Nitrogen is more electronegative than carbon so if it’s a choice between putting a positive or negative on Nitrogen, we prefer to have the negative instead of positive.
For even more practice make sure you check out my resonance practice quiz on my website along with this entire video series and study guide by visiting leah4sci.com/resonance. And be sure to join me in the next video where we look at Radical Resonance which is the movement of a single electron.
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