Resonance Structures

residents theory is usedto represent the different ways that the same molecule can distribute its electrons. So what that means is that it turns out that even though the connectivity or how atoms are connected isn't going to change. The electrons between them can move sometimes. And that's what residents theory is all about. So I'm gonna teach us some rules, and you guys are gonna get the hang of it as I go along. All right, So the first thing to know is that atoms will never, ever move. The reason is because remember that I said the connectivity of those atoms, how they're connected to each other doesn't change. The only thing that changes is the kind of electrons that air in between them that are keeping them linked together. Okay, the only thing that moves is the electrons, okay? And when I talk about electrons, what I'm talking about is pi Bonds pi bonds move, and I'm also talking about lone pairs. Okay, So what that means is that literally I'm not moving any atoms. All in moving is double bonds around or triple bonds around. And I'm also moving where lone pairs air at okay and that has to do with the electrons that are moving throughout the molecule. Okay, now, something about resonant structures. We're gonna find out that there's something called contributing structures contributing structures or structures that both contribute to the actual representation of the molecule because they averaged together. And what we're gonna find out is that none of these contributing structures are actually gonna look like the actual molecules. So what that means is the molecule is a blend of all the different possible resident structures that a molecule can have. Okay, so let's go ahead and learn some rules. First of all, on, we're gonna use curved arrows to represent electron movement. Just so you know, these rules are gonna apply to the rest of organic can. We're gonna keep using these rules any time that we're moving electrons, which is pretty much all the time. So what a curved arrow would look like is like this. Okay, so notice that I'm using a full arrow, I'm curving it around. What that means is that two electrons that represents two electrons are moving from one place to another. Okay, um, what we're gonna do is after we've built our resident structures. We're gonna use double sided arrows and brackets toe link related structures together. So that means that once I figure out my resin structures, I link them together using those double sided arrows like I have here and then brackets like I have here. Okay, then finally, we're not. Finally, but arrows are always gonna travel from regions of high density, high electron density toe, low electron density. And like I said, this is a rule that applies for the rest of organic camp. Any time we're moving electrons, we always start from the area of the highest density and moved to the area of lowest density. So what that means is that, for example, a positive charge would be an area of low density. So you because that means that you have electrons missing, right? So what that means is you would never start an arrow from a positive charge. In fact, you would always go towards the positive because that's the area of low density. Okay. And then finally, the net charge of all the structures that we make must be the same. Okay. And the reason for that is that remember that residents structures are different ways to represent the same molecule. And what that means is that all of them should have the same net charge because we're just distributing the electrons different. But we're not adding any electrons or subtracting any electrons. So what that means is they should really all be have the same charge.

So what I want to do now is I want to talk about common forms of residents. These are patterns that I've basically just discovered while teaching organic chemistry. And I want to share these with you guys. Okay, so let's talk about basically three right now. Movement of cat ions and ions and the neutral hetero atoms. Okay, so let's talk about Catalans first. So, Catalans, the way this works is that if you have a cat ion next to a double bonds, let's go ahead and put that next to a double bond. Okay. What that gives us the ability to do is now to switch the place of those electrons. Okay, Because what I have is an area of high density on one side, which is a double bond. Remember that a dull bond not only has a sigma bond, but also as a pie bond. Remember that pie bonds are extra electrons that are shared between two atoms. It's not something that I can actually move. Okay, then I have an area of low density, which is my positive charge. Okay, So if I want to move this around, what do I do? What? Turns out that This is kind of this is one of the easier examples. If you have a positive charge, an adult one next to each other, you can actually kind of swing them open like a door hinge using one arrow. So what that means is I would start from the high density, my dull bond, and I would move towards the positive charge, but I wouldn't make it just towards the positive will take Make it towards that bond. So what I'm gonna get now is that now I get a double bond in the place where the positive used to be. And now my positive moves over here. Okay. What that means is that now my positive is actually distributed from that read from the left side, over here on the red, and then over on the blue side, it's going to the right side as well. Okay, So what I'm trying to say is that any time you have a positive charge next to its old bond, it can be represented by both of these drawings. It's not just going to stay in one place automatically, just by laws of chemistry. It's gonna wind switching places at some point. Okay, The rial molecule is gonna look like a average of both of these or a combination of both of these. That's why I talked about the fact that none of them is a true representation. The best representation is by hybridizing both of these, and I'm going to talk about what? That is in a little bit. Okay, Not so. Get an ions. What if I had a negative charge next? That double bond. Okay, Now I have to ask you guys, what do you think is gonna be the region of the highest electron density? It turns out that the dull bond has a lot. But now that we have a full negative charge, that's gonna have even more electron density, cause a full negative charge means that it just has a lone pair just hanging out. Okay, so if you have a full negative charge, we're actually gonna use two arrows. What I'm gonna do is I'm gonna take these electrons and push them into this bond making a double bond. Okay? But now we have an issue. If I make a double bond there, then let's look at this carbon right here. How many bonds did it already have? Well, it already had a double bond. That's two already had a bond to hydrogen. That's one. And then it already had a bond to carbon. So we had four bonds already. If I make another bond with that negative charge, what is? How many bonds is that carbon gonna have? It's gonna have five. That would be really, really bad. That would be basically impossible. You can't have a carbon with five bonds. That would be super terrible. So this is in a situation where we're gonna use a rule that's called make a Bond break a bond. So if I make a bond on this side, Okay, in order to preserve the octet of the middle Carbon, I must break a bond, Okay? Because if I don't, then I'm going to give this carbon that I'm shading him green. I'm going to give it five bonds, and that just sucks. So if I make this bond, I have to break this bond, okay? And when I break that bond, what winds up happening is that now I get a negative charge over here. Okay? So as you can see with a positive charge, I didn't have to actually break any bonds because I was never breaking. I was never violating any OC tests. But in the movement of anti answer negative charges, I do have to break upon because I am gonna violate an architect. So it's important to note here is that cat ions move with one arrow and then an ions move with two arrows. Not so bad. Right? Okay. So just remember that positive charges they can swing like a door hinge, whereas two arrows, I mean, whereas with the negative charge, I'm going to use makeup on break upon, because the fact that I have to preserve that octet of the middle Adam All right, then let's look at neutral hetero atoms. Okay? There's the last situation. And what this would be is that. Remember that? I said we could move double bonds and we could move lone pairs. So imagine that I have a lone pair here. Well, let's say imagine that I have my two lone pairs there for that oxygen. Because remember that oxygen has a bonding preference of two bonds and two lone pairs. All right, Cool. And then imagine that the nitrogen has one lone pair because remember that the nitrogen has a bonding preference of three bonds and one lone pair. Remember this. This is how it's going to satisfy its octet and how it's also going to satisfy its valence. Alright, so now let me ask you as a question. Is there any way that I can turn these lone pairs one of these lone pairs into a double bond and not breaking octet? Okay. And it turns out, let's look at our options. Basically, the two options or this either I could move one of these green will impairs down here and make a triple bond. Or what I could do is I could move one of these red lone pairs here and make a double bond. Okay, it turns out you guys might be thinking, Well, Johnny, why would I only move in that direction? Why couldn't I move like this? Why wouldn't I move the electrons down, make a double bond there? Well, first of all, the reason is because double bond and electrons are the things that usually switch places, so I would want to go in the direction that's going to go towards the double bond. I wouldn't want to go away from it. So I would not go in destruction, cause that's away from my double bond. But on top of that, check this out. I have ah, hydrogen here, right? If I move these electrons down into this area, I would make a double bond here, okay? And I'm sorry. I actually had more than one hydrogen. I had two. There's two hydrogen, is there okay, because that's a ch two. If I move these electrons in here and make a double bond, I'm gonna break the octet down here, and there's gonna be no fixing that. There's actually no bond that I could break because these were all single bonds. And you can't break single bonds in resonance theory. Okay, So what that means is that I would wind up getting a double bond down here That would violate this octet, and it would suck. That would not be a good resident structure. So instead, I never deal with the other two situations that I was talking about, which is that either the oh jumps down and makes a triple bond or the n lone pair jumps up and makes a double bond. So let's look at the old making a triple bond. If the Almeida triple bond like this. Okay, let's look at this for a second. Don't draw it. Just look at it. How many bonds with this carbon have? It would also have five. So we would break another octet by doing that. Is there any way that we could break upon to make that to make that carbon feel better? No, because it turns out that there's just single bonds on both sides, so there's nothing you could do. We would be stuck. So we're definitely not going to move this lone pair either. So now I have one last choice. The last choice is that I would move these electrons from the end up and make a double bond. OK, if I make a double bond here, how many? How many bonds will that center carbon have still five, So it looks like I'm screwed like any. Either way, I'm always making five bonds, but there's one difference with this one. If I go ahead and go up and make the double bond up towards that carbon, guess what I can do. I can break a bond, so this is a situation where I am making a bond towards a double bond. The good? The reason that a dull bond is helpful is because double bonds I actually can break where a single bonds you're not allowed to break. But double bonds notice that I have these electrons in the stole bond that air free to move. So what I'm gonna do is I'm gonna make up on and then, for the sake of preserving the octet of this carbon right here, I'm gonna break a bond, and that would be right here. So what I would do is I would basically turn two electrons from that bond into a lone pair on the oxygen, and that's gonna preserve the four bonds that I need for that carbon right there, because I'm making one, but I'm also breaking one. Okay, so what that would look like average all the residents structure is I would now have a dove on here. Okay, I would have No, I would have no electrons in the end, because I just use those electrons to make the dole bond. And then instead of having to lone pairs now it have the two lone pairs from before, So let's go ahead and draw those the green ones. But now I'm gonna have one more lone pair. The last loan pair comes from the bond that I broke because basically what I did was I took two electrons from that double bond, and I made them into a lone pair. Now all we have to do is count formal charges, and we're done. So this oxygen it wants toe have six electrons, but it turns out that it has seven. Because, remember, we're kind of sticks and dots, so this would have a negative charge. The end wants toe have five electrons total, but right now just has four bonds, right? It has the double bond. It has the single bond there, and then it has the hydrogen. So that means that the nitrogen wants five, but it only has four. So this sort of a positive charge and that is our resident structure. Okay, so what we have effectively done is we've taken these lone pairs and we were just distributed them around. And now we're showing another way that these electrons can exist in this molecule, but notice that we're never moving single bonds, single bonds are a big no, no, don't break those. And also we're not rearranging the way that atoms are connected.

right. So there were a few things that you should remember that I told you guys were very important about resident structures. First of all, remember that we use curved arrows. Easy. Okay, remember that we use brackets with little double sided arrows, toe link structures. But also remember that we always start from the area of highest electron density and work our way to the areas of less density. So in that case, that has to be the nitrogen because the nitrogen has a has a full negative charge on it. That means that is the most negative thing. So what I want to do here is I want to try to move those electrons. The only way that I could move them is by becoming a double bond. Okay, So are becoming a pipe on. I'm sorry. So if these electrons move down here and became a pi bon, that would be great. Except I have a problem. If I did that, then this carbon would have 55 electrons on it, okay? Or five bonds. Okay, so five bonds is terrible. That would break the octet rule. It would be 10 electrons, by the way. I'm sorry not 5. 10 electrons would break the octet rule. So if I make that bond, what do I have to dio? Do you guys remember? I have to break a bond. Okay. Why are you drawn at the bottom? You could have drawn it at the top two. But if you make up on, you have to break upon. And this is that pattern that I told you guys that Oops, that was weird that an ions come with two arrows. Okay. And the reason is because anytime you're making that new double bond, you're gonna have Thio break a bond as well. Our new pie bond. Okay, so what that's going to do is it's going to give me a structure that looks like this when I have N with a triple bond carbon and then in oxygen. Now, nitrogen already gave up one of its lone pairs to become a triple bonds. That means it only has one lone pair left. Carbon has the same amount of electrons before. It's just arranged a little differently. And then oxygen has one additional lone pair because the electrons from that double bond became a lone pair. So basically the additional lone pair is this red one. And that red one came from this bond over here breaking. Does that make sense? Cool. So now we have to do formal charges. So I would have It's funny that I put my negative there. I actually would have a negative right here on the, uh Oh. And then would I have any other charges that have to worry about? You know, the carbon is fine and the end is fine. So really, that's it. I just got my resident structure. Okay. Are there any other things that we could do? Not really. Because if I make this negative, let's say that I go back and put this negative back here. Once again, I'm gonna have to break a bond. So what that means is that these two resident structures are going to be basically two different versions of the way this molecule could look. But that's it. That's the only thing that it can do. So I'm gonna put brackets around this, and we're gonna That's gonna be a That's gonna be a rap. Okay, that's gonna be the end of that problem.

All right, So remember that I said that we can move electrons as long as we're not breaking octet. It's okay. We can't break out tats. We can't make more than eight electrons. Whatever. So, in this case, I really only have one set of electrons that has my attention. Will always want to start with the most negative thing. And that would be my lone pair because my lone parents just these free electrons. So if I were to move these electrons and make them into a double bond, would that be okay? What are you breaking any octet? It's actually I would be if I just left it like that. Okay, because remember this carbon here already has. Ah, hydrogen. Okay, so if I made that double bond, I would now have five bonds in that carbon. That would suck. So can you guys see anything that I could do to fix that? What I could do was break a bond so I could break this double bond and put those two electrons. Remember that there's two electrons in that double bond. We could in the additional pi bon. We could take those two electrons and make them into a lone pair. So what that means is that, um Let's just go ahead and draw this as double sided arrow. Since we're gonna draw a new resident structure, What I would get is something like this where I have an n h two here. But now I have a double bond, and now I have a lone pair here. Okay, But remember that with bond line structures, usually we don't include a lot of lone pairs. We instead want to use formal charges. So let's compute the formal charges here. Okay? By the way, that h is still there. I just didn't draw because ages could be implied. Okay, So what would be the formal charge of this carbon right here now? Well, it wants four electrons, And how many does it have? Five has five valence electrons, so this is gonna have a negative charge. Okay, so I'm just gonna erase the lone parent. I'm just gonna replace it with the negative, because I think that's a little easier to look at. Okay, Now, let's look at any at the at the nitrogen. Does that one have a formal charge? Well, nitrogen wants five electrons, and it has four, so kind of like they swapped the nitrogen has a positive. All right, so there we have it. That is a resident structure. Is there anything else that could happen? I know that. So you guys were wondering OK, but couldn't I do something else? Couldn't my like, let's say, make this negative. Do a double bond there. Couldn't I do that? And the answer is No, you couldn't. That would be terrible. Please don't do that. The reason is because think about it. There's already two. Hydrogen is here. Okay. If I went ahead and tried to make a double bond here, first of all, that carbon would now have five bonds. Secondly, there's nothing else that I can break to make that work. You can never break single bonds with resonant structures. So what that means is that I would have to either break off one of the h is or I would have to cut off this carbon carbon bonds, which would suck so that negative charges stuck. It can't go there, you say. Oh, what if it goes down? How it if it goes? How about if I put it down here? Same exact thing. Once again, I got to h is. And by making a double bond, I will be forced to break off a hydrogen or break off a carbon. It would suck so that negative charge is stuck there. The only other thing that I could do is it could go back in the direction it came from. So if I made a double bond there, then that would be fine. Because then I could break this bond and make it alone. Pair there, see how this works. So you basically keep going with that charge until you get stuck until there's nothing else you can dio. So those are my resident structures for this compound. Cool.

All right. So remember that positive charges. I said they swing like a door hinge. So imagine that you're just opening up this door and you could just do that. Okay, So what I would get is in my first resonance structure, By the way, this thing resident structure that I'm showing you is gonna be super important for or go to. Okay, but right now, we're not gonna concentrate on it too much. We're just going Thio do this. So now I have a double bond here, and I have a positive charge here. The reason is because remember that the double bond and the positive switch places when you do this resonance structure. Okay, So now what I ask myself is okay. Is that positive charge stuck? Is there nothing else that it could do? Actually, no, it's not stuck, because now it's next to another door hinge. So what I could do now is swing this one up like that, and now I would have another resident structure. This resonance structure is now gonna have a dull bon. Their adult bon, their adult bon there. And a positive church there. So now is that one stuck? Nope. still, if not stuck because it could do swing another door open. And that's gonna be this one. Okay, now, some of you guys. So you smart guys out there might be saying, Johnny, isn't that the same thing that I did over there? You know, where I'm basically moving the dull bond up or whatever, and it's similar, but actually, with resident structures, we want to draw every single movement that can happen even if all of them look similar to you. Okay, so even if it looks like we're doing the same exact thing on both sides, you would still draw them because you want to indicate the motion of these electrons all over the molecules. So now what I'm gonna do is draw that. I'm gonna draw double sided arrow. And if this was actually a test, I probably wouldn't do this because it could be a little bit confusing. But I'm gonna continue the resident structure down here. Okay, so then what I would have is double bond double bind. It's old bond positive charge. And then that's it. Those of your four resident structures, if you want, you could then show how you get back the other one, and you could show that that is in residence. That's fine. Okay, if you wanted to do that, that's fine. Now we just have to set this off in brackets, so I'm just gonna do bracket bracket. Okay, So the resident structures of the important part the fact that I have double sided arrows reported brackets are important, Then the way that I laid this out probably could have been better. It would have been also have Could have would have put all four in a in a vert in a horizontal row. But I couldn't fit all of them. I made my arrows too big. But don't worry about it too much. If anything, you could do something like this. If you're ever like running out of space, you could just do some point. This double sided arrow, double sided arrow that takes care of it. Okay, your professor will know exactly what you're doing. Cool. Alright. So hopefully that helped residents make a little bit more sense to you. And let me know if you have any questions

All right, guys, we just talked about resonance structures and how one single molecule could have several different contributing structures. That's what we called each structure that has a slightly different, um, distribution of electrons. We call that a contributing structure. Well, it turns out now we want to talk about is hybrids, how they blend together. And also which one would be the major structure in terms of which one represent the way that the molecule looks the most. So let's go ahead and begin. So basically, the resonance hybrid is going to be a mathematical culmination of all the contributing structures. Okay, and what it does is it indicates where the resonating electrons within a molecule are most likely oops, most likely to reside. So what we do for this is we literally combine the two different resonance structures in tow one drawing or 234 etcetera, and we combine them all into one drawing. And then we try to analyze, which would be the the resident structure that would contribute the most of that hybrid. So here's a molecule that we're going to deal with a lot in or go to. It's called Isocyanate, and I don't really care that you guys know that much about it. But what's interesting is let's look at the contributing structures here. So for one of these, I have to double bonds. And then what I've done here is I've done I've used the negative charge rule to make a bond break a bond. So what I'm doing here is I'm taking these electrons here making a triple bond. But then if I made that triple bond, that carbon would violate a talk Tet right. It would have five bonds so that I'm gonna break this bond and make a negative charge over there. So at the end, what I'm going to get is two different structures, one that has a negative charge in the end, one that has a negative charge in the okay, What the residents hybrid is it's a blend of both of these. So what I would do is I would just draw the parts of the bond that are not changing. So, for example, notice that here I always have it. Least two bonds between the carbon and the nitrogen in this structure. Action of three bonds. But in this one, I have to so I would draw those two. Okay. On the oxygen side, I always have a least one bond between the carbon and the oxygen. But in this, in this case, I have to. Okay, then what I would do is I would draw partial bond from the nitrogen to the carbon and from the carbon to the oxygen. What that indicates is that this bond is being created and destroyed at the same time. Okay, But it also indicates Is that basically I'm in between both okay. And then finally, I put partial charges in all the places that have a negative charge. Why? Because the hybrid, Like I said, it's not in equilibrium. This is not like, okay, This is not like we've talked about in came to We have a reaction that favors the right or favors the left, and it goes back and forth. No, what? This is It's a mathematical concepts where I say, Okay, this gets, let's say, 40% of the molecule, this is 60% and the actual molecule looks like a blend of both of them. Does that kind of makes sense? So in this case, I've drawn my hybrid notice that basically everything that's changing is shown on this hybrid. I'm showing that the bonds are being broken and destroyed, broken and create at the same time. On I'm also showing that the negative charges moving from one place to another, okay?

now, in terms of major contributors, that's for us. That's when we determine. Okay, Which of these is the one that looks the most, like the hybrid? Is it number one, or is it number two? Okay. And to figure that part out, we have to use just a few rules. Okay? So often it turns out that one of the residents structures will be more stable. Okay, So it turns out, let's say you have more than one resident structure. One of them is the most stable. So that's gonna be the one that we use. And that means that it's going to contribute to the hybrid more than the others will. Okay, and major contributors will often have the following characteristics. Okay, so the first thing is that neutral structures are almost always going to be more stable than charged ones. Okay, so if I have a choice between let's say, have a residence structure that's neutral and a resin structure that has charges on it, I'm gonna pick the neutral one to be my major contributor and to be the one that looks most like the resident like the residents hybrid. Why? Because that's the one that's over almost stable. Okay, so even if the other one is possible, it may exist to some extent, but the one that's really gonna exist in excess or not exist. But the one that's going to contribute in excess is gonna be the neutral. Does that make sense? Cool. So a good example for that would be where I showed you guys the neutral, hetero atom example on the other page, where there was one that had basically a neutral structure and then one that had a positive and a negative. The major contributor would be the one that was just fully neutral, the one that had a positive and the negative would be a minor contributor because that one already has charges. It's not a stable. Okay, so let's keep looking at this. Another rule is that, if possible, every atom should feel it's octet. So what that means is that we're gonna look towards resin structures that are not satisfying The octet. Let's say ones that have too few electrons, those air usually gonna be minor contributors. If I have a choice between a resident structure that fulfills all of the talk pets and one that doesn't I'm always gonna pill. Pick the one that does full, full of talk tests. And then finally, the electron negativity trends are going to determine the best placement of charges. Now, I know it's been a really long time since you talked about Elektra negativity. In fact, for a lot of you guys, you haven't heard about it since Gen Com. Okay, so I just want to remind you guys that this is the Elektra Elektra negativity scale. What? It basically says that is that as you go to the right and as you go up, your election negativity gets higher. What that means is that Florian is the atom that is most comfortable having a negative charge or having electrons on it. So if I were to pick that the negative charges on a flooring or the negative charges on a carbon, which one is gonna be more stable? What do you guys think? The flooring, right, Because that's electro negative. That means that it likes toe, have electrons or negative charges on it, whereas carbon is not as to the right as flooring. So carbon is gonna be a lot less comfortable having that negative charge. Does that make sense?

Alright, guys. So it turns out that there were no neutral structures, so I couldn't use the neutral rule. All of these molecules fulfilled their octet, so I couldn't use the octet rule. But we have differences in Electra negativity. It turns out that the O being with a negative charge is gonna be more stable. Okay, so that would be my major contributor. Why? Because it turns out that it was more Electra negative. I'm just gonna use e n for Elektra. Negative. The nitrogen. What that means is that oxygen is more comfortable having that lone pair on it than nitrogen is. So this would be less Electra Negative. Okay, because of that, this is going to be the minor contributor. Okay, So when I go ahead and draw my resonance hybrid, we can draw it the same exact way. But we have to acknowledge that lets say that I'm drawing it like this and c o partial bond. Partial bond. One of the ways that we could draw this is we could draw the partial negative on the O bigger. So we draw bigger, partial negative on the O and a smaller partial negative on the end Why is that? Because, remember, we just said that even though both of these could exist, the negative on the, uh oh is going to be the most stable. So that means that most of the time it's gonna look more like this. One more. Sorry, that kind of got blurry, more like this one and less like the other one. So that means that my hybrid would be a bigger share of the major contributor. Does that kind of makes sense?

Alright guys. So my resident structures were as follows. I always start from the thing that's most negative and that's my negative charge and I can actually go in two different directions here. I could either go in this direction or I could go in this direction. Now the reason that I know that I could go in both those directions is because my negative doesn't get stuck because if I make that bond I could break a bond. So if I go towards the blue direction, I know that I would be able to break this bond in order to keep the octet okay in order not to violate the October that carbon. If I were to go in the red direction then it could break that double bond in order Thio not violate the octet of this carbon Does that make sense? So I have two different directions that we could go. Obviously this notation is horrendous. I should that you should never draw two different resident structures on the same compound. So let's just go with the blue one first. Okay, so the blue one would look like this. What I would get now is a dull one still there. But now I have a dull bon here. The O H. Stays the same. And now I have an extra lone pair on that O, or what I could just put is an O negative, because the negative charge has now transferred toe. Okay, so that's one resident structure. Is there anywhere else that that negative could go? Well, the only thing I could do is it could go back here. And then what that would do is that would send these electrons back here. Now, think about it. You might be thinking Well, couldn't go towards the Ohh. Well, no. Think about it. Um, if the sole bonne went there, the only other option that I would have besides breaking the stole bond is to just kick off the O. H altogether in order to preserve the octet of that carbon in order to make sure that it has four bonds. Still, But that's crazy. Like I said, you can't break single bonds. So my only option here is really to go backwards. The exact way that I came. All right, so that shows you that's one set. But I also told you is that there's another possibility. What if I went in the other direction? Well, then that would lead to a structure that looks like this. So what I would have is that now I have a double bond here, because remember I said that I'm going this way, and then this would break so I would get a negative charge there, and then I would still have this double bond here, so I haven't Oh, in an Ohh. Okay, so that one's a little ugly. Let me try to clean it up a little bit. There, There, There. Okay, so now I have to ask you guys Okay. Is there anywhere else that that negative could go? Well, that negative could only go back where it came from, and then that would just cause the first resident structure that we had. All right, so those are three major residence structures. We basically made the negative charge go as far as it could until it got stuck. And then that's it. Okay, so now it's our job to figure out what the major contributor is gonna be. So which one is the major contributor here? Which one looks like it's going to be the most stable. I remember there were two rules. One was preserving octet. It's and the other one had to do with election negativity. All right, so in this case, do we have any octet? It's that we're breaking. No, All of them have octet. It's okay. But I do have differences in election negativity. In the first one, I had a negative charge on a carbon in the second one. I had a negative charge on an oxygen. Which one is more? Electra? Negative. The oxygen. Okay, So what that means is that this is gonna be my major contributor. Why? Because it is the one that has the negative charge on the most stable, Adam, the one that's most likely to be okay. Having a negative charge on it. Okay, so now we just have to do one more thing. And that is to draw my hybrid. So my resonance hybrid is gonna have all the single bonds exactly the same. So if you have a single bond draw at the same but then everywhere the that the negative charges moving, you have to draw a partial bond. That means that bonds, air braking and being made at the same time. So then I would have partial bond there, partial bond there, partial bond there and partial bond there. Why? Because noticed that the negative charge had double bonds moving throughout all of those atoms. And where is the negative charge of any one time? It could be in the middle or could be on the O or could be on the end. Okay. And in all reality, it's gonna be a mathematical combination of all three of those. So we're gonna do is we're gonna put partial negatives on each of the Adams that it could be on. So I'll be those three and just, you know, another way to know Tate that that is sometimes used is instead of using partial negatives, it would just be to simply use a negative charge and just draw it right in the middle. And then that would show that the negative is being distributed throughout all of those Adams. Alright. Does that make sense? So far,

all right, we can see that this example is something called in a mini, um, Cat ion, which I'll explain more later. But for right now, that doesn't really mean anything in terms of resident structures. But what's the first thing we always wanna look at when you look at a resident structure and it's where to start the arrow from. So right now, what do I have going for me? Well, I've got a positive charge, and I've got two double bonds. So if I had to start my arrow from somewhere, where do you think we would start from one of the double bonds? Right, Because double bonds have electrons. We just wanna start from high density toe low density. So I want to start from one of the double bonds and then go to where? Go to the positive charge, because the positive charge is the thing that's missing electrons. Okay, On top of that, there is one other pattern that we talked about that might be helpful here. Remember that positive charges tend to move with how maney arrows. What do you remember? One. So most likely you're gonna using one. Let's double check. But most like you're gonna be using one arrow and we're gonna moving from negative to positive. So where would we start? What we've got? Let's say Delavan A until one B. And even though I could start from either of these, I think B is the easiest one to visualize because it's the closest to the positive charge. So how could we move the electrons from double bond be towards that positive and well, we learn that there's two things that double bonds conduce. One is that they can donate electrons directly to an atom that there adjacent to. So what could happen is that the double bond becomes a lone pair on the end. Okay, so we'll explore that. We'll see. But we also learned that double bonds can move, swing like a door hinge toe, other neighboring carbons or another other neighboring atoms. So we kind of wanna evaluate both of these possibilities. Is it possible to move it over as it like? Open it like a door? Or would it be? Just add it to the nitrogen. Well, in order to figure out if you could move it like a door, you need to look at the atom that you would be attaching it to. And what we see is that, for example, this carbon here we learned how to calculate how many hydrogen has How many does it have? It has three, one to three. Okay, So if I were to swing this double bond over, like a door hinge, would I run into any problems? Yes, guys, because now you have a double bond on that carbon. You'd be breaking the octet, right? So, actually, even though I kind of I'm thinking I want to swing it open, that's not possible there. Okay, but maybe you're saying. But, Johnny, there's another carbon at the top. How about that one? I have a carbon here. This one is how maney ages to write one too, couldn't I maybe try to swing it open up to here? Well, if I did that, check it out. I'd be breaking the octet again, because once again, now this carbon has four bonds with double bond here, it would have five. So both of those motions aren't possible. Okay, So that means what can I do with my double bond? Well, what I could do is I could take the electrons and I could donate them directly to the end, making a lone pair. So what that means is that for this resonance structure, what it would look like is like this and draw the ring just like before. But now what changed? Well, this double bond stayed exactly the same. There's still a methyl group there. Or just a carbon a ch three, right? That's what we call it for now. But now, instead of having a double bond now, I'm going to get a loan pair on this end. That lone pair came from the electrons being donated to the end. Okay, Now notice that guys remember, I always like to count hydrogen when I'm doing these Russian structures, at least at the beginning, because you're still getting your feet wet. You're still trying to understand these, so we can't be too careful with the way we calculate these. Notice that this carbon here on Lee has one age. Why? Because it's got three bonds to carve a three bonds so it can only have one each. Well, now it still only has one age. So what kind of charge should that carbon now have well going based on our rules of formal charges. We know that Carbon wants four bonds. How maney does it actually have as three? This carbon that I'm looking right here on Leah's three. So four minus my sticks in my dots, which is equal to three equals positive. One slip means I should have a positive charge here. Okay, So what that means is that my first resonance structure? I'll just erase this each now looks like this. Okay. I took my electrons from the double bond and made a lone pair on the end on a positive charge on the carbon. If you guys want to verify the charge of the nitrogen, you'll find that it's neutral cause nitrogen with a lone pair and three bonds is always neutral. So that's one resident structure. Great job, guys. So now I'm just gonna move this over so we have more space. You're gonna grab this and move it over here. So now, guys, what is the next step? Do we have any other resident structures possible? Well, what I like to say is, let's take that positive and keep moving it all the way down until it can't move anymore. So is there anything else that it could possibly move with. And what I see is that I haven't used this double bond yet. So is there a way that that double bond could perhaps react with or resonate to the positive? And what I could try to do is swing it like a door hinge and see if that's gonna help me. So what if I were to swing it like a door hinge? What? I break the octet. Well, right now remember this hydrogen? I mean, this carbon has one h. So if I draw that, what I'm going to get is this. I'm gonna get in. But now meh, Thel or ch three My bad ch three. And then what I have is an h here. So it has three bonds. But now I just added a double bond here. So did I violate the octet of that carbon? It's perfect. Now it has four bond. So actually, in this case, I actually can move the double bond down and notice it's because it's next to a carbon with a positive charge, which we said when you have that specific situation, you can swing your door open like a door hinge. So that just shows that you could do that. Now, what should be the charge on this Adam here. What should be the charge on that one? Well, we could just use the same method. How many hydrogen is? Did it originally have One. How many does it have now? Still one. So what were the charge? Will? It only has three bonds, so it should be a positive. So this is another resident structure. Okay, so I've drawn three resonance structures. I've drawn the original. Now the positive at the bottom and the positive now resonated to the left side. Now, no disguise that. There's these two rules that air like thanks. The most important rules of resident structures. Which is one you can't move atoms. Have I moved any atoms so far? Have I move? Any moved any hydrogen? Xeni carbons. Any nitrogen? No. So that's good. That means I'm probably on the right track. To are all the net charges of my structure is the same net charges. Meaning they all add up to the same number of charges. Yes. The original mini, um cat ion was plus one. My second structure is plus one. My third structures plus one Awesome. Have I? And then the third rule, which I consider like the third important rule is have I always gone from negative to positive? Yes, every single time I was going from a double bond to something positive. So I fulfilled my three rules of resident structure. I'm on the right track now. What you might think is, well, now that we have the positive there, is there anywhere else that we can put it and guys, the answer is no, because notice that over here on this carbon, there's nothing to react with it. There's nothing. And I keep saying the word react. But that's the wrong word. What I mean is resonate with it. There's nothing to resonate with it. I don't have charges. I don't have double bonds. I just have to ages. And those two ages can't resonate with positive charge because that would mean that I'm moving atoms and I can't move atoms. So that means that this thing is done. Okay, there's no other residents structures. So these are the three. I'm gonna call it a day. Okay, so that is the end of the first part, which is to find all the resident structures. We found them, which is three. Okay, so now we have to move on to the second part, which is to predict which one is the major contributor and which ones are the minor contributors or whatever. Okay, well, what did we learn? What we learned is that first of all, um, the more election negative something is, the less it wants to have a positive charge. Okay, so one thing that we learned is that you've got your periodic table, right, And nitrogen is here, and carbon is here. Okay, So of those two, I'm sorry. Like that's that they're actually next to each other, but whatever. I'm just I always draw these very like, ugly looking, periodic tables. Case you have carbon e of nitrogen. Remember that electro negativity goes in this direction. Okay, Now, if you haven't covered this topic yet, don't worry too much. This is something just from Gen. Kem that it's really not hard to remember. It just means that flooring is your most electro negative and you go away and you know it gets less election negative. The more you go away from that. So which one is the more negative C or n en is the more negative. Which means, see, is the more positive? Okay, so you would think that the best answer is gonna be that C wants to have the positive charge because it's less Electra. Negative. But remember, that was just the first rule. Remember, the second rule for major contributors was try to fill all octet. It's can't remember that not having a full octet is bad. You do not want to have an unfilled octet because that's gonna be very unstable. So in this case, the carbons with the positive charges. What's wrong with them? Well, this carbon here, for example, it's a carbon was sick with three bonds, it's got three bonds like this. That means that it only has six electrons since I was three bonds its six electrons a full of tech for carbon. No, carbon wants to have eight. How about this one? Over here, this carbon it has again three bonds like this that the ones Ah, hydrogen positive. This one also has six electrons. So is that gonna be good for an octet? No, that's terrible. What about the first one? The first one is nitrogen nitrogen When it has a positive charge, it has a double bond, and it has to bonds like this, and it has a positive How many octet electrons does the nitrogen have? Well, guys, nitrogen. Even though it has a positive charge, it actually has eight octet electrons. So off the three structures that I'm choosing from which one is gonna be the most stable, is it gonna be one of the carbons that has the six electrons? Or is it going to be the nitrogen with the eight electrons and guys? It turns out that it's gonna be the nitrogen. Okay, so the major contributor is actually going to be the A mini, um, cat iron, just like we drew it. And the minor contributors are gonna be these guys. Okay, Now, it turns out something that I like to do. A clutch is. I'd like to introduce topics ahead of times that when you see them, you'll know more about them. So this thing called in a mini, um, Cat ion is something that you're going to see later on in further chapters of organic chemistry. And what we're gonna find is that let me if you guys don't mind. I'm just gonna start erasing some stuff. The A mini, um cat ion. When you draw medium Catalans, you always draw them with the positive charge on the end. Why? Because that's the most stable that it could be. If you draw the positive charge in the carpet, that's not a stable. Okay, so I'm actually showing you why The a Medium Catalan is always drawn in that way because that's the major contributor versus the minor contributors. Okay, guys, one more thing we have to do, let's draw our residents hybrid and be done with this problem. So our residents hybrid guys is just, ah positive charge everywhere that the positive is resonating too. So that would be all along these bonds here, so you could just put a full positive there. And that just means that along, basically, this entire area, you always there's a possibility of getting a positive charge. And that would be a resonance hybrid. Okay, awesome guys. So I hope that residents structures are making a little bit more sense to you. By the way, if you're ever wondering, Johnny, isn't there another resident structure that you didn't cover? Draw it yourself and count out your hydrogen and make sure that it actually is possible because nine out of 10 times if I didn't draw it, it's because it's not possible. It's because when you draw that double bond there, you're gonna find that it breaks in octet for something. So draw it yourself on. What you're gonna find is that if you're systematic and methodical about it, you can actually get all the resident structures just like I did. And you can avoid making mistakes with the wrong ones because you made sure you counted all your bonds. Alright, awesome guys. So let's move on to the next page.