Video Transcript : Key Arrow Patterns in Resonance Structures

Key Arrow Patterns in Drawing Resonance Structures by Leah4sciBelow is the written transcript of my YouTube tutorial video –  Key Arrow Patterns in Resonance Structures.

If you prefer to watch it, see Video HERE, or catch the entire Resonance Structures in Organic Chemistry series.

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Leah here from and in this video we’ll look at key arrow patterns to help you figure out resonance structures. Say you’re given a molecule like this and ask to show all the contributing resonance structures, how do you even know where to begin? Before you’ll look at patterns to recognize, let’s look at what you should never do when working at resonance.

Rule number 1, only ever move electrons. I know this sounds self-explanatory but I can’t tell you how many times I see students trying to put arrows on positive charges or even hydrogen atoms. Only add remove electrons and more specifically these electrons should be lone pairs or pi electrons. Resonance is the movement of electrons between different atoms, that means you do not change the structure, you don’t move the atoms and you don’t break any sigma bonds. You can move the lone pairs and you can move the double or triple bond pi electrons.

Here’s the Enolene ion we looked at in video 1. And for this ion we can move the lone pair of electron on Oxygen down towards the pi bond and we can also move the pi bond electrons between the two carbon atoms to collapse them onto carbon. So we’re moving a lone pair and we’re moving pi electrons. But when we draw the resulting resonance structure, the skeleton stays the same. The atoms are in the same place, the sigma or single bonds haven’t move, the only thing that’s different is the red lone pair of oxygen is now a pi bond between an atom and Oxygen and the red pi bond between the two carbon atoms is now a lone pair on the oxygen atom. And yes the charge move but not because we had an arrow starting on the negative and pushing that elsewhere but rather because when the electrons shift, the atoms give new formal charges, Oxygen has a new formal charge of zero, carbon has a new formal charge of negative 1 and that’s how the charge is moved.

Rule number 2 is another very very common mistake and that is to never resonate on toward an sp3 carbon atom. Say I have a molecule that looks like this with a negative charge on an oxygen. If I take those electrons on Oxygen and try to resonate them down to carbon, carbon would get an overfull octet. Carbon starts out with a complete octet of 8 and if I bring that pi bond down, I have 10 electrons. And the same thing happened here but as a result of that new pi bond forming, we kick out the other pi bond so we still overfill the octet but we simply shifted the pi bond. In this case because the carbon atom is sp3, it’s single bound to 4 different atoms. Don’t forget there’s an invisible hydrogen atom on carbon. So if Oxygen forms a pi bond we don’t have any other bond to break, that means we can’t resonate those electrons down so there is no resonance form for the starting molecule. Which brings me to rule number 3.

Rule number 3 is just the continuation of rule number 2. But carbon needs its own special reminder. Just keep an eye on your octets. If the atom in question already has a complete octet, then ask yourself, by adding electrons to this atom am I violating the octet? Can I kick anything out or is this allowed to have more than just 8 in the octet? For example in the sulfate iron, sulfur already has a complete octet but I am allowed to bring the electrons from the Oxygen and resonate them towards sulfur to form a pi bond. That’s because sulfur is in period 3, it’s a larger atom and it can hold as many as 12 electrons so we don’t have the issue of violating the octet because it’s an exception to the octet rule. In fact, I can do this one more time to take sulfur from having 10 on its octet to a total of 12. Giving me this 3 valid structures which can continue resonating even more if I’d start out with one of the other negative oxygen atoms.

So if the atom’s an exception to the octet rule and you can have more than 8 electrons, that’s okay. If it’s not an exception and you bring too many, make sure that there’s something else you can kick out potentially in the form of breaking a pi bond. Also keep an eye on having too few electrons in your octet. For example, with this ketone propanone I can take this pi bond and collapse the electron onto oxygen giving me a resonance form as follows. Oxygen had 2 blue lone pairs, still has that but the red pi bond now sits as a lone pair on oxygen giving me a negative charge on oxygen for the formal charge of plus 1 on the carbon atom. Conservation of charge is present, we’re starting with a neutral molecule and here we have the plus and minus to cancel for a net neutral molecule. But carbon does not have a complete octet making it less stable than the starting molecule.

Now we come across structures where the resonance will have less than a complete octet.So keep in mind the electronegativity.The more electronegative the atom, the less stable it’s going to be with that incomplete octet. Carbon will be common with the positive charge but atoms like Oxygen and Hydrogen should only have a positive charge for too many bonds with a complete octet rather than having an incomplete octet. Now that you know what not to do, let’s take a look at the key arrow patterns.

Every book or professor will teach you differently and this is just my method. The first thing I look for are negative electrons. If resonance is the movement of electrons and electrons are negative, they’ll be attracted to positive or partial positive so it makes sense to find the most negative electrons and see where they want to move. For example if I’m given this molecule and ask to show resonance, I have a choice between lone pair and Nitrogen and the lone pair on carbon. But the lone pair on carbon is negative and Nitrogen is neutral so I want to start with a negative electrons. I may want to move them down but the problem is that this carbon here is an sp3 carbon has nothing to kick out so that will violate the octet and is not possible. Instead I’m looking for an sp2 hybridize carbon, one that has other carbocation meaning an incomplete octet or a pi bond so that I can kick out those extra electrons and I wanna move the electrons to form the pi bond between the negative carbon and the carbon double bound to Nitrogen.

In forming that pi bond, that carbon that’s double bound to Nitrogen will now have too many electrons but I can fix them by taking the pi bond and showing in arrow moving towards the Nitrogen atom kicking out those electrons and collapsing them onto Nitrogen. Show the double headed arrows and first draw your skeleton. Everything that hasn’t change then look for what changed. The red electrons are now sitting as a pi bond between the former negative and the former double bound to Nitrogen carbon. The green electrons are now sitting on the Nitrogen atom. Next we want to do a formal charge. It’s hard to do this on carbon in skeleton structure so remember the trick.

If carbon starts out with a lone pair, and then forms a pi bond, the lone pair has one more electron directly attached to carbon compared to the carbon with the pi bond so the charge has to go up by 1. If we start with the negative carbon we go up to zero giving it no formal charge or formal charge of zero. And the carbon that has a pi bond and still has a pi bond even if it’s in a different direction will not change its charge because the total number of electrons haven’t changed. Nitrogen is much easier to see, we should have 5 directly attached, I count 6, 5 minus 6 is negative 1 giving me formal charge of minus 1 and don’t forget your brackets. You quick conservation of charge starting with negative 1, we end with negative 1, charge is conserved and we are good to go.

If your molecule doesn’t have negative electrons, the next thing to look for are lone electron pairs. But here’s the key, you’re not just looking for lone electron pairs, you’re looking for the electrons to be sitting on an atom that does not yet have a pi bond. Now this thing is tricky because you’ll see lone electron pairs on atoms that are already pi bound and can’t move. So make sure that if the atoms is neutral and has a lone electron pair, it shouldn’t already have a pi bond. What do I mean by that? These lone electrons can resonate, it’s on a negative Nitrogen but they can resonate back towards that carbon to reform the starting structure. The lone electron here that remains a Nitrogen cannot resonate further. And the reason for that is the carbon atom has nothing left to kick out and that means you’ll be violating the octet.

So make sure that if the atom has a lone electron pair, if it’s already has a pi bond, ask yourself if I form another pi bond are there any electrons to kick out? Let’s take a look at this enol. For this molecule we have two lone pairs on Oxygen so we can take one of those lone pairs and collapse it down towards that double bound sp2 carbon atom. Carbon would have too many electrons on its octet but we can take the pi bond and collapse it on a lone on a carbon atom that will not have a pi bond to Oxygen. We draw the double headed arrows, re-draw the skeleton and let’s see what changed. We have a green lone pair that hasn’t moved, the purple lone pair that now formed the pi bond to carbon and the red lone pair sitting as a lone pair on carbon.

A formal charge on Oxygen shows a plus one, and a formal charge on carbon shows negative one. Notice that the electrons collapse onto carbon rather than moving to form a pi bond to the end and that’s because the terminal carbon is sp3 and you cannot resonate onto sp3 carbon atoms. Don’t forget your brackets and you will quick net charge analysis. We’re starting with the neutral molecule, our final structure has a plus one and minus one which cancel out to give us a neutral molecule and we’re good to go. Contrast this with a lone pair of electrons on the ketone Oxygen. In this case, if you try to resonate the electrons down, carbon is sp2 but the sp2 is, is sp2 double bound to Oxygen there’s nothing to kick out. And if you try to show a resonance here, carbon has too many electrons in its octet. And that is not allowed.

If your molecule doesn’t have negative lone electrons or partially negative lone electrons, the next thing you want to look for are pi bonds especially in a conjugated system. A conjugated system is when you have alternating pi bond such as 1,3-butadiene. In this case we have no lone pairs, negative, or neutral. So the next best negative electrons to look at are the pi electrons. I can take these pi electrons and resonate them towards the third carbon atom so that will give me an overfull octet. So as a result I have to take those electrons and collapse them as a lone pair on the end of the molecule. That gives me a resonance structure like this and then for formal charge, if carbon used to have pi bond, lost that electron and got nothing in return, it has a formal charge of +1. If carbon used to have a pi bond and now has a lone pair, it gets a charge of -1. Don’t forget the brackets and the conservation of charge we start with neutral plus and minus cancel out, once again it’s neutral.

For this molecule you can break the resonance of even more to show it step by step. We can start by showing the red electrons collapse onto carbon as a lone pair giving me an intermediate resonance structure that still has a pi bond on the end but now has a lone pair on the second carbon with a negative charge and a positive charge on the first carbon. We can then show this lone electrons collapsing down towards the second pi bond causing the second pi bond to collapse as a lone pair giving me the third structure that we’ve already seen with positive and negative on other end. Both sets of resonance are correct, it’s simply a question of what your professor is looking for.

Rule number 4 isn’t a rule, it’s just something I want you to keep in mind. Till now it started with negative or neutral atoms but you’ll sometimes get a starting molecule that has a carbocation or a positive charge. Since electrons are moving and electrons are negative, they’re going to be attracted to that positive charge. So for example, if we start with a conjugated system but this time we have two pi bonds and a positive charge, the pi bond closest to the positive charge has negative electrons that are attracted to that positive charge so you show that as your first resonance where the electrons go towards the positive. As a result we now have a pi bond on the end, the pi bond on the left hasn’t moved and the carbocation in the middle of the molecule. Once again, electrons are attracted to positive. So we’ll show the pi bond on the left now attracted to those positive electrons.

Show the double headed arrow and see what changed. The green pi bond hasn’t move, the blue pi bond now sets one carbon over, and carbon on the left has a positive charge. Put brackets around it, check out the net charge is the same for every structure, we have plus one, plus one, and plus one, we are good to go. Using all of this information, I challenge you to try this question which came out of my resonance practice quiz. You can find this entire video series, practice quiz, solutions and resonance, study guide by visiting my website,

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