resonance that's going to be the topic of this last lesson in a chapter on molecular representation and resonance uh in my organic chemistry playlist now this is my brand new organic chemistry playlist i'm releasing lessons weekly throughout the 2020-21 school year so if you don't want to miss one subscribe to the channel click the bell notifications you'll be notified every time i put one of these up this is something you you learn back in gen chem and resonance always implies that you've got delocalized electrons and typically we're talking about delocalized pi electrons now i want to bring this up in a context where you would have seen it in general chemistry and then we'll kind of bring it to more of an organic chemistry focus so back in the day you guys might have made a lewis structure for the nitrate ion so and if we draw a lewis structure here we'll find out there are three equivalent resonance structures now we've got five valence electrons for nitrogen so six each for three of the options three times six is 18 plus five is 23 and the negative sign gives us one extra electron for 24 total valence electrons so nitrogen can make more bonds less electronegative so it goes in the middle so we set up our skeleton and then fill up all the outside atoms and so one two three four five six seven eight nine ten eleven twelve thirteen fourteen fifteen sixteen seventeen eighteen nineteen twenty twenty one two two twenty three twenty four we've used all 24 now we don't have any left over to give that central nitrogen and problem is nitrogen is definitely not happy he's two electrons short of a filled octet so somebody's gonna have to share well in this case we've got three adjacent atoms that are all identical and all equally likely to share and that was our evidence we're gonna have resonance back in a general chemistry class in this case because there's three atoms that can all share so we're going to end up with three resonant structures because they're all identical atoms they're going to be three equivalent resonant structures so i could have it be the oxygen on the left that's going to do the sharing that would leave us with this structure i could have had it been the oxygen down below that did the sharing or i could have had it been the oxygen on the right that did the sharing and got the double bond there and give me a sec just to get my lone pairs in all right we also want to put some formal charges on here in every single one of these the nitrogen has a positive formal charge so five is a normal valence minus one two three four lines is plus one whereas the oxygens that have a double bond are neutral the oxygen's normal number of valence electrons is six minus one two three four five six minus the dotted line the zero but for the options with single bonds it's six minus one two three four five six seven and you get a negative one formal charge for all those oxygens again with a single bond in any one of these structures okay so there's our nitrate ion if you add up those formal charges you'll see that overall this ion has a negative charge so and these would be called three equivalent resonance structures now again we said that resonance always implies delocalized electrons what that really means that electrons are in multiple locations at the same time something that we don't experience in any sense at the macro level but at the atomic level or the molecular level it's totally possible and so these resonant structures we should know are just tools they don't actually exist it's not like these are oscillating back and forth that's not what that double double arrow means so and just a quick word to the wise that double arrow is different than this double arrow you can't use this one for resonance this is your resonance double arrow this double arrow means equilibrium just a quick aside so but these double arrows don't mean these are oscillating back and forth these structures none of these actually exist so they're just a tool that we use to figure out what the real structure molecular ion is going to look like and we call that real structure the resonance hybrid here and in this case i can see that i've got a double bond in this location in this structure but a single bond in the other two so there is a partial bond here and in the same light there's a partial bond here and a partial bond here as well now the nitrogen's got a positive formal charge in all three structures so we're going to give it a positive formal charge so but each oxygen is it is neutral in at least one of the structures but negative and at least one as well and so all the oxygens are going to be partially negative cool and again we call this the resonance hybrid it's a better representation of what the real molecule looks like all right and so in this case you can kind of see that we've got one two three four bonds total so shared in one two three locations and so we'd kind of think of this as equivalent as a four thirds bond a one and one third bond so that's stronger than a single bond shorter than a single bond but weaker than a double bond longer than a double bond that's the kind of way this works cool but that's one extra pi bond shared in all three locations at the same time and so it's an extra third of a bond so that's what we mean by delocalized electrons that pi bond is in multiple locations at the same time now in ocam we like to focus on you know how you get from one resonant structure to the next because we're not going to be just drawing basic lewis structures like this you're typically going to be given one resonant structure and then have to predict all the possible resonant structures said so instead of starting by drawing some simplistic lewis structure we're going to be giving you probably more complex organic molecules giving you one structure and say draw the rest and so you're not going to arrive at these resonance structures in the same way we did in general chemistry and so we've got some rules for you for what electrons you're allowed to move to get from one resonant structure to the next and so it turns out you're allowed to move non-bonding electrons first off so non-bonding electrons can move to any adjacent bond but that's it and then pi electrons and pi electrons are allowed to go to two places so if you've got a pi bond it can move to either adjacent atom like this pi bond right here that you see in double or triple bonds it could move to either a json atom or potentially it could go to an adjacent bond as well so you got two options with those pi electrons so but those are the only electrons you can move either non-bonding electrons or pi electrons and we move them in such a way to get to the next resonance structure so well it turns out to go from this structure to this one so i can see that in in this case that i'm going to have a double bond right here but that oxygen on the bottom has three lone pairs now it's only got two lone pairs one of those lone pairs must move well the only place lone pairs are allowed to move is into an adjacent bond so there must be where that pi bond is coming from okay so but i can also see that this pi bond right here is gone in the next structure and so where does it go well i go from two lone pairs in this oxygen to three lone pairs on this oxygen and now we know where it goes and so in this case it implies that these two electrons move to this adjacent atom now again in principle pi electrons have options of moving to an adjacent atom or to an adjacent bond but i can see to get from this structure to this structure if i already have this tron but that's where they must have gone now one thing to note arrows here so we've got a full headed arrow we said or a double-headed arrow so because we'll see in a little bit that you can also have a single headed arrow and double-headed arrows show the movement of two electrons this will be important for all of the rest of your two semesters of ocum in mechanisms we like to show where the electrons are going to make bonds and break bonds and things of this sort and when you show a double-headed arrow you're showing the movement of two electrons at a time whereas when you've got a single-headed area you're going to show the movement of one electron at a time all right so but these two arrows get me to this resonance structure and then to get from this one to this one we do the same kind of thing and say okay where the electrons moving well in this case i can see that we're going to end up with a double bond here and only two lone pairs in the oxygen and so yet again one of these lone pairs must be forming the pi bond we see in the next structure but also i go from double bond to single bonds so these pi electrons need to move and where do they go well we go from two lone pairs on oxygen to three lone pairs on oxygen so they must be forming another lone pair on that oxygen cool so we're drawing resonance structures we like to go sequentially from one to the next and we typically show the arrow pushing of how we get there now here we had already drawn all these by drawing you know kind of the standard way we draw lewis structures but again from an organic chemistry's perspective you're going to have just one of these pictured here and you're going to figure out based on patterns how it is you actually get to the next one based on predicting the arrows and then based on where you've predicted arrows then you'll be predicting the structure so we've kind of done this backwards so but you'll see what i mean very soon so now we're going to get and draw on some resonance structures and you'll find out that we typically learn how to draw resonance structures for three different kinds of compounds and it turns out when we draw resonant structures you're going to have electrons in multiple locations and that lowers the energy of those locations of those electrons because they're seeing multiple nuclei and stuff like that they're attracted to those so it also often spreads charges around and charges get shared on multiple atoms like we saw in nitrate where there was a partial negative charge on three different auctions a negative charge being shared between three different auctions so at least there's actually two negative charges being shared between three different options so that spreads out that charge that also makes things more stable so we often talk about the idea of resonance stabilization it makes things more stable and typically the more resonant structures the more stable so but you also got to pay attention to like where you put negative charges generally you want a negative charge and a more electronegative atom and a positive charge and a lesser election of god things of that sort all right so there's three main structures we can undergo resonance with and staple resonant stabilization with and carbocations are one of them that's carbon with a positive formal charge lone pairs of electrons and here i've got a carb anion which is pretty common to show up but it technically doesn't actually have to be carbon or an anion for that matter so it could be a neutral nitrogen or neutral oxygen or something like that so we'll find out some different situations where we stabilize lone pairs and then radicals where you have a single unpaired electron that is your third example of where we're going to do resonance stabilization now a couple terms i want to introduce you to and the terms allylic and benzylic so if you've got the study guide in front of you they're already on your hand out there so but allylic means you are one bond away from an alkene so one bond away from an alkene so an alkene carbon-carbon double bond we learned in the last lesson and if you're exactly one bond away so this carbon right here i'll highlight with the c and this carbon right here they're both exactly one sigma bond away from our double bond right here and therefore they get the term allylic and the reason they get that term so it's a special term is because if on one of those allylic carbons that we've labeled it has either a positive formal charge a lone pair or a radical it will have resonance so got another term as well and this is benzylic and benzylic is very similar but it means you're one bond away from a benzene ring instead and so one sigma bond from a benzene ring that is the benzylic carbon we'll label him as well and means the same thing though if you have a positive formal charge there a carbocation or a lone pair like a carbanion or a radical at that location exactly there will be resonance with the pi electrons in the benzene ring and we can actually make it a bigger generalization so if you are a carbocation or a lone pair of electrons or a a radical so but notice note that both these could be termed as non-bonding electrons so if you're a carbocation or non-bonding electrons and you're exactly one bond away from pi electrons which includes both allylic and benzylic and others so but if you're exactly one bond away from pi electrons there will be resonance stabilization so that's where these terms get significance in this chapter so because again if you have a carbocation or non-bonding electrons in analytic or benzylic position there will be resonance cool and we're going to use that to kind of predict when we should see resonance so we'll start with carbocations here and let me make some room for drawing all our resonant structures here so but we're going to kind of parallel these but it turns out how we approach drawing resonance structure and the arrows we draw for the movement of electrons are going to be different in these three classes and so that's where we're going to present them kind of uh in the same breath here in the same uh on the same board in parallel with each other so you can see the difference so carbocations are the easiest and so we'll start there and so in this case this positive formal charge is two bonds away from these pi electrons there will be no resonance involving those pi electrons at all cool but this pi this carbocation positive charge one bond away from these pi electrons there's going to be resonance and so this guy's a carbocation because he's only got three bonds so if you look at this he's got two bonds to carbon atoms showing that implies that if there were no formal charge he'd have two hydrogen atoms bonded to him as well but with a positive formal charge he's only got three bonds total so he actually only has one hydrogen and so he's got no filled octet he's really unstable when he doesn't have a field octet here and so our goal is to get him another bond and that is possible when he's exactly one bond away from pi electrons and so and this is 90 of the time if you're going to stabilize a carbocation it's because you're one bond away from pi electrons now with pi electrons we have options we could move these pi electrons to either adjacent atom i could move them over here or i could move them over here and that would put a lone pair either on this atom or this atom in fact if i did the arrow on the left i'm sorry the arrow on the right first it would pull a lone pair and a negative charge on that carbon and leave behind a positive formal charge on that carbon since he'd be missing a bond now too so and then this positive charge would still be there as well that'd be a horrible idea so most of the time you don't want to add additional charges into your resonant structures we'll see one exception for the most part but most of the time that's true so same thing would happen if you move the the pi bond to make a lone pair on the atom on the left here there'd be a negative charge and a lone pair on the that atom and then a positive formal charge left behind on this atom and then we'd still have a positive formal charge in the carbon we've begun with neither one of those would help us our goal is to make him more stable and we make him more stable if we get him his fourth bond and so we won't consider either of those options technically you know pi electrons are allowed to move to an adjacent atom but if it serves a purpose in those cases it wouldn't serve us a good purpose here but if we move these to the adjacent bond it's totally going to serve us a purpose by you know moving to them to form a double bond in this location that carbon is going to pick up his fourth bond and no longer have a positive formal charge so let's draw that next resonance structure here alright so this one hasn't moved but this one has now moved to this location and if you take a look then this carbon which still again had one hydrogen to begin with still has one hydrogen but now has one two three and then the bond of the hydrogen be the fourth bond still is now it has four bonds no longer has a positive formal charge however one of the carbons did lose a bond in the process and will now have a positive formal charge and that is this one right here two carbons away now you might be like why isn't the next one to him well this carbon here has the pi bond on his right and then here has the pi bond on his left and whether it's on his right or on his left he's good he's like i have four bonds no matter how you slice it but it's this carbon right here that used to have this pi bond in this structure and now no longer does and so he only has three bonds now he's now got the positive formal charge he's now the carbocation in that structure cool so that's our first uh resonant structure we can draw on then you've got to ask yourself a question in the case of carbocation is is he a lilic or benzylic to any additional pi bonds so now he's definitely going to be a lick from the one we just came from he's going to be one bond away from the pi electrons we just moved so that's always the case and moving them back just gets you back to the structure you came from but is he one bond away from any additional pi electrons and in this case he's exactly one sigma bond away from those pi electrons and so we should anticipate more resonance and so here we can also move these over to that bonding position and again we could consider moving them to either adjacent atom it's just not going to help us stabilize this carbocation in this case so we'll not consider those but if we move it to the adjacent bond we will indeed get a new structure here that gets this carbon four bonds and we'll make it not a carbocation so it will lead to stabilization here all right so we've got that there and now we've got these here and now indeed this carbon right here picked up its fourth bond is no longer carbocation but two carbons away that's a pattern here we now have a carbon that is missing a bond and is now our carbocation and he's not one bond away from any individual you know pi electrons he's always going to again be one bond away from the pi electrons we just moved to get to this structure but he's not one bond away from any additional structures which means we are now out of resonance structures and there we go cool so these are our corresponding resonance structures so and we can draw analogous structures if you're stabilizing a lone pair notice i made a pretty much identical structure except we have a carbocation here versus a lone pair of electrons here so and pretty much you're going to get analogous structures here so let's just go down the list and make the corresponding structures so i'm going to have the pi bonds in all the same locations without the carbocation and instead of a carbo you know a cation here we're going to have the lone pair and the anion there and same thing on the last one and it actually works exactly the same way with the radical so we're going to get the exact same analogous structures so but instead of a cation instead of a lone pair we'll just have a single electron there cool now the reason we went through and drew the structures first on this one is i wanted to show you that it is just as easy to predict the resonant structures when stabilizing lone pairs or radicals as it is with the carbocation but the arrow pushing of the electrons is more challenging and so if you know what to expect for what the structure should look like it'll make the arrow pushing a little easier and so the truth is if you learn how to push arrows it'll make predicting structures easier but if you can predict the structures it'll make learning the arrow pushing easier and they'll feed into each other all right so we started with the easy one with carbocations the question is how do we get now from this structure to this one when stabilizing a lone pair of electrons well a lot of students look at this and they just look at where things need to go and they're like oh i see this yeah we just need to move this bond to right there and then we can move this lone pair over to here and life is good beautiful except that totally violates our rules for how we're allowed to move electrons so if we call lone pairs the only place they're allowed to go is to an adjacent bond they are not allowed to jump even to the next atom over let alone two atoms away over to this one so this would definitely not get us our movement of electrons we want it looks like it should get you the right structure it's just totally a violation of what electrons can do in drawing resonance structures and so what's actually going on since this lone pair is not present on that atom in the next structure these actually have to move but the only place they can go to an adjacent bond either here or here well it turns out they can't go here because you can never move them towards an sp3 hybridized atom and the reason you can't is it would give it a fifth bond and it would violate the octet rule so that's the third rule in moving electrons you can't move towards an sp3 and it just means you're going to violate the octet rule if you do so but we could move them towards an sp2 over here which is what we're going to have to do we're going to form a pi bond there we can't move them all the way to the next atom but i can move them towards that atom by forming a pi bond here and there indeed is that pi bond in that next structure now the problem is if that's the only thing i do we're going to have a double bond here and a double bond here and this carbon would still have a hydrogen and would have five bonds and violate the octet rule so it can't be the only thing we do and so in this case i see this next structure that there is no pi bond in this location so and i need to get a lone pair well there's an easy solution then we're going to move this to that atom and put the lone pair on that carbon and so when you're stabilizing a lone pair typically most of the time your resonant structures to get from one to the next is not going to involve just a single arrow a movement of two electrons but two arrows of movement of four electrons and typically you get this pattern you're gonna have a lone pair become a pi bond and a pi bond becomes a lone pair so the lone pair becomes the pi bond and the pi bond becomes the lone pair that's almost always how it works all right so we go to the next one and we're going to have the same kind of pattern so our lone pair is going to become this pi bond and the pi bond is going to become this lone pair and there's our movement of electrons so for stabilizing lone pairs so typically again most of the times can involve moving two sets of arrows at a time so four electrons moving from one to the next now radicals are just plain unique so they're different than either one of these and they're the most confusing of the three and oftentimes uh professors will even leave this out at this stage so maybe half the time or so so half of you guys this will be relevant half it won't and if it's not i'm sorry but it will be more relevant later on in the semester if even if it's not right now so the way this works here is this electron right here the only place he's allowed to go being a non-bonding electron is to the adjacent bond but problem is so first off i can only move that one electron it's only moving one electron it's only a half-headed arrow okay well then where's the other half of this pi bond come from and where's that radical come from well it comes from the fact that this pi bond here is actually going to split in half and move in opposite directions so the other half of the pi bond comes from one of these you know electrons from the old pi bond moving to where the new one is and then the other electron moves to the other side to form the radical on that carbon as well and so when two electrons in a bond move in opposite directions like this we call this homolytic cleavage not the most important word now but it'll be relevant later and we'll have a whole chapter on radicals typically at the end of first semester so but that's how the move in electrons works and it's very unique definitely distinct between these three so but a lot of students will learn resonance they won't realize that you know arrow pushing we call it arrow pushing in carbocations versus lone pairs versus radicals is different and they don't realize there's three distinct cases they gotta know and it's just a big jumble in their heads and that's why i'm taking the time to kind of spell it out in all three parallel paths so let's go to this last one here we'll use the radical to make half of a pi bond and then we'll split the pi bond up to make the other half of the new pi bond and to make the radical in the last structure as well cool and this is kind of your normal arrow pushing now with resonant structures we also got something else to identify that's pretty common in organic molecules now in the nitrate ion we just showed we had three equivalent resonance structures they were all equal they all contributed equally to what the resonance hybrid would look like but that doesn't always happen in typical organic molecules and so with the way it works when you've got multiple resonance structures if any of those structures are more stable than the others then the resonance hybrid is going to look more like that structure than the rest so if we look at this we've got to know some trends on stability then so for carbocations there's some trends on stability and the trends on stability go from tertiary to secondary to primary so let's write that up here so we'll do it in a different color to highlight it so tertiary is more stable than secondary is more stable than primaries more stable than methyl if we cared and so if we look at this we identity identify these carbocations so this one's secondary it's a carbon bonded directly to two other carbons this one is also secondary it's bonded to two other carbons but this one over here is primary and so if we draw that overall resonance hybrid here so we can see that we've got delocalization going on all through there and we're going to have a partial positive charge here here and here for that resonance hybrid all right now it turns out all three of these carbons though are not going to be sharing that partial positive equally because they all three of these structures don't contribute equally to this resonance hybrid and the way this works since these two carbocations being secondary are more stable than this one over here that's primary we would refer to these two structures as being major resonance contributors and that means that the resonance hybrid is going to look more like them and then this guy would be a minor resonance contributor which means that the resonance hybrid looks less like him and so what we can conclude is that there's more partial positive on these two carbons than there is on this carbon here so most of the time it's going to work out that you usually have just one major and then one or more miners but this one actually had two major resonance contributors now for the carbanions carbanions actually follow an opposite trend instability so a carbocation here is electron deficient doesn't have enough electrons and it turns out when it's bonded to carbon atoms so they have a way of donating electrons towards it which make it less positive and more stable and so being bonded to more carbons when you're electron deficient makes you more stable but in the carbanions case he's not electron deficient he has too many electrons as evidence by having a negative formal charge and having too many electrons if there's carbons next to him that are giving him even more electron density that's actually going to make him less stable and so the trend is flip-flopped here and so for carbanions we're going to have methyl and i'll symbolize that by me here instead of ch3 means the same thing but it's more stable than primary it's more stable than secondary is more stable than tertiary and once again this is a secondary carbon in this structure secondary in this structure primary in this structure and so now all of a sudden for carbanions it's the primary that's going to be more stable than the secondary and that's going to make this the major resonance contributor and both of these minor cool what that means is in the overall resonance hybrid where once again we have delocalization of pi electrons all through those regions a localized pi bond right there and now instead of partial positives we have partial negatives so but now since this is the major resonant contributor we see that we're gonna have more negative on that primary carbon on the far right more partial negative charge there than on the two secondary carbons cool so finally we'll look at the radical as well and for radicals so these don't have a formal charge but they are electron deficient in the fact that they don't have a filled octet and being electron deficient just like with carbocations they want to be bonded to as many carbons as possible and so they follow the same trend tertiary is more stable than secondary is more stable than primaries more stable than methyl and so in this case again we've got a secondary radical here secondary radical here primary radical there and so once again the secondaries are more stable so these would now be the major resonance contributors and this guy the minor cool not usually as important something to to kind of diagram out for radicals but definitely for uh cations and lone pair stabilization at this stage and once again half you guys probably won't even see resonance with radicals just yet but it will show up before the end of the semester i promise all right so we just want to work a couple more examples there's a couple other things we need to show and one involving carbocations and then one involving stabilizing lone pairs so for this one here we've got a carbocation so we definitely want to stabilize that and get him his fourth bond if possible and the first question we should ask ourselves is is he one bond away from pi electrons and indeed he is exactly one sigma bond away from these pi electrons and we should totally expect resonance so we'll move these over to the adjacent bond like we do to stabilize carbocations typically and our next structure cool so we got him his fourth bond and just like the trend follows two atoms over is now the atom missing a bond and he has a positive charge and so no then we normally ask ourselves well is he one bond away from any additional pi electrons and no the only pi electrons in that molecule are these guys at least from the outside of the look of it but for carbocations there is one other way to stabilize them and get another resonant structure and again ninety percent of the time nine times out of ten will be about being one bond away from pi electrons but the second case is does any of your adjacent does any do any of your adjacent atoms have lone pairs because lone pairs can be moved to an adjacent bond which means i could use this lone pair to form a double bond with from oxygen which would give him a fourth bond or i could use this lone pair to form a pi bond there form a double bond with nitrogen which would also get him his fourth bond and that's your other scenario so again nine times out of ten resonance comes down to being one bond away from pi electrons the other ten percent is are you a carbocation next to an atom with lone pairs cool and in this case we got two options so we're just going to pick one of them i'm going to pick the nitrogen first and you might think my reasoning is that you know oxygen is more electronegative and so we hold on to electrons more so maybe nitrogen we should consider first we'll do both essentially but we will find out that nitrogen will be better at sharing than oxygen being less electronegative all right so next structure we'll draw then it's gonna have a double bond of the nitrogen cool and that's gonna leave the positive formal charge now on the nitrogen it might be easier to see if i actually draw in those hydrogens so normal number of valence for nitrogen is five minus one two three four lines is plus one cool now we're there but we also could have done the sharing from the oxygen so let's do this in a different color we also could have said well what if oxygen did the sharing and maybe we might be inclined to like branch it off right here so and if we look the structure we'd get would have the double bond with the oxygen so the last resonant structure we might have arrived at had we branched it off there differently would have a double bond now to the oxygen and that would leave a positive formal charge on the oxygen oxygen's normal valence is uh number of valence electrons is 6 minus one two three lines and two dots six minus five is plus one all right and that is the last resonant structure however there's a temptation to try and get there from here and a lot of students will try and do this they'll put another double arrow right here and just branch it off in the middle two different directions and they'll draw this last resonance structure over here turns out you're not allowed to do that for proper resonance you have to have a linear continuous chain of resonance structures and so to get to this last resonant structure what you're going to have to do because since we can't branch off it violates the rules so is we're going to have to undo the one we just made so from here i need to get it back to here by undoing this arrow and so we're going to take those electrons and move them back onto the nitrogen that way we can move the electrons in and have the oxygen being the one forming the double bond cool and so in this case we actually had to move two sets of arrows but that's only because we had to satisfy the rule of having a linear progression of resonance structures here so but there are all our resonance structures so we saw that something new here that you can get resonance for a carbocation if he's next to an atom with lone pairs but we'll see one other thing new here in the last example we had all carbocations in our resonance structures but now we also have a nitrogen cation and an oxygen cation and the question now we have to say is major versus minor who's the most stable resonance contributor all right so for a positive charge it's opposite of negative charge negative charge you typically want to put it on the most electronegative atom period for a positive charge you typically want to put it on the least electronegative atom but not period so all things being equal you want to put it on the least electronegative atom and so in this case you're like well carbon carbon nitrogen oxygen the least electronegative is carbon done okay that's gonna be one of these and he's secondary and we'll figure out what he is and you know however you have to say all things equal and all things are not equal here so turns out that when you have a positive formal charge on carbon you only have three bonds on each of these carbons and no filled octet whereas when you've got a positive formal charge of nitrogen you've got a filled octet when you've got a positive formal charge in auction you've got a filled octet and so having a filled octet much more stable than not having a filled octet and so in this case actually these two resonant structures are better than when the positive charge is on the less electronegative atom which seems counterintuitive you could also look at this one other way if you notice the number of pi bonds in this structure is one and there's one two three lone pairs and in this structure you got one pi bond and three lone pairs but in this structure you've got two pi bonds and only two lone pairs in this structure same thing you've got two pi bonds and only one and then i should draw the other end and only two lone pairs and so electrons are more stable lower energy when they're bonding than where they're non-bonding and since these two structures have more electrons that are now pi bonds and less that are non-bonding lone pairs you could look at them as being more stable that way so whether you look at it from the octet rule perspective or from the number of bonding electrons perspective these are more stable resonance structures than these two even though the positive charge are on more electronegative atoms like nitrogen and oxygen now between these two all things are equal they have the same number of pi bonds same number of lone pairs and everybody's got a filled octet and so now between these two you would say i'd rather have the positive charge on the less electronegative atom and so picking nitrogen would be your choice because nitrogen's less electronegative oxygen so this guy's going to be our major resonance contributor this guy would be minor and these guys even more minor but we don't really usually kind of rank them we just say one major usually and bunch of minors so unlike that first example where we actually had two majors in one minor but most times it's usually one major and a bunch of minors cool and that just means the the the resonance hybrid even though the positive charge is going to be shared on two carbons a nitrogen and an oxygen more of that partial positive charge on the residence hybrid will be on the nitrogen then on the carbons or the oxygen all right let's look at one more example with lone pairs and in this case we've got this lovely carbana and we want to stabilize that and in this case we have to ask ourselves the question is it one bond away from pi electrons in this case yep it's totally one bond away from pi electrons and so we should expect resonance and we're going to form a pi bond so lone pair becomes a pi bond pi bond becomes a lone pair now a lot of students ask me and be like chad hey hey hey why didn't you just move these over to the adjacent bond over here and then move these up here well we could but we'd miss some of our resonance structures and we'd miss the fact that some of that electron density is going to be on this carbon right here if we just skip that resonance structure we'd miss that and so that's why the pattern goes lone pair becomes a pi bond pi bond comes a lone pair is you want to consider moving it if possible to the adjacent atom before you move it further to the adjacent bond or you might miss some of your resonance structures so let's get that resonant structure up here now we'll ask ourselves the question is this one bond away from any additional pi electrons well yeah we're one sigma bond away from these pi electrons and so we can then pi lone pair becomes a pi bond pi bond becomes a lone pair and i forgot a couple of my lone pairs let's just get those on and draw one more resonance structure cool and that'll be our last resonant structure we're not one bond away from any other pi electrons at this point except the ones we just moved to get there so there's our resonant structures we have a negative charge on a carbon here a carbon here and an oxygen here and here we don't have to worry about like that whole all things being equal and stuff with a negative charge put it on the most electronegative atom done and so the one on oxygen here negative formal charge in oxygen is definitely our major resonance contributor whereas these other two with a negative charge on carbon will be minor contributors and again that means that the resonance hybrid is going to have more partial negative charge on the oxygen than on either one of these carbon atoms so last thing in this lesson we've got to talk about a connection between resonance and hybridization we can see a slight contradiction that arises when we start trying to figure out what's the hybridization of a particular atom so and if you judge that based on a single resonant structure you might not get the right answer so one thing to remind so in this case we've got a couple of different resonance structures and just recall that these resonance structures don't exist the real molecule is an average of the two and again we call it the resonance hybrid so we've got a partial negative charge here and a partial negative charge here and so if i ask you hey what's the hybridization of that carbon atom right there you should not think of what is it in this exact resonance structure because this structure doesn't exist you should think about what would it look like in the context of the actual molecule which the resonance hybrid depicts and so if you look at this it might be helpful if i draw in some hydrogen atoms so we've got a couple on that carbon and we've also got a couple on that atom same thing here got one here and here and here and here and so if you look at this carbon on the left here and i said what's his hybridization you'd look and you'd say well he's bonded with three atoms and no lone pairs that's three electron domains total and so he's going to be sp2 problem is that we look at that same carbon atom in the next resonant structure and here you say oh wait a minute he's bonded to one two three atoms and has a lone pair that's four total domains and so maybe he's sp3 and so the question is what is he well again the key is that neither one of these resonance structures even exists so what actually exists is an average of the two at the same time so we're like well then is the sp two and a half chad well no no such thing as sp2 and a half or at least nothing we're going to consider in this class uh so the question then becomes uh which is a better a more accurate reflection of the real structure well if we take a look at what's going on here you guys might recall with uh carbon's four uh valence electrons so he promotes one from the s into the p so he's got four unpaired electrons and then we've got an option if he's going to be sp3 hybridized then he's going to mix the s's and p's all four of them to make four brand new sp3 hybrids or if he decides to be sp2 hybridized he'll only mix one of the s with only two of the p's making only three hybrid orbitals and that third 2p orbital will just be left unhybridized and the key is this if you have a left over unhybridized p orbital you've got it so that you can make a pi bond and so if you're sp3 hybridized you don't have an unhybridized p orbital and you can't have any pi bonding and so unhybridized p orbitals are the only thing that can be used for pi bonding in this class so if we look at that and say okay so looking back again at that carbon is he involved in pi bonding well in this structure i'd say yes and in this structure i'd say no but the key is again we shouldn't be evaluating these structures we should be looking at the resonance hybrid and in this case he does have a partial pi bond it's not a full pi bond it's a partial pi bond but if i said is he involved in pi bonding you'd have to say yes you could not say no and if he's involved in pi bonding then he cannot be sp3 sp3s are never involved in pi bonding and so as a result he must be sp2 and so this is the connection between residence and hybrid hybridization if you look at just one resonance structure like if i looked at this one i'd get it wrong and if i looked at this one i might have just by default got it right so but the truth is i should be looking at the resonance hybrid realizing ah can't be sp3 has to be the sp2 usually it works out is if you have in one structure it's sp2 and one it's sp3 it's just going to be sp2 that's usually the way it ends up working out let's look at a couple of examples all right so the examples we're going to look at are two different nitrogen atoms on the same molecule we want to give both their hybridizations and and if we didn't know anything about resonance and its connection to potential connection to hybridization we might just look at this and be like nitrogen's bonded to one two three atoms has a lone pair sp3 so okay and look at this one same thing so monitor three atoms lone pair four total domains sp3 problem is then we'll say okay well now that we do know about hybridization if either of these is involved in resonance then it's going to have delocalized pi bonding going on and having partial pi bonding it can't be sp3 it would have to be sp2 instead and so the question is are either one of these lone pairs involved in resonance and recall that lone pairs are involved in resonance if they're exactly one bond one sigma bond away from pi electrons so this guy here is he one bond away from pi electrons well he is one sigma two sigma bonds away from pi electrons no resonance so he really is sp3 no no nothing new there but this lone pair right here on the nitrogen is exactly one sigma bond away from these pi electrons there's going to be resonance here we could draw another resonance structure where we'd have a double bond right here which means there's really a partial double bond right here and this nitrogen cannot be sp3 he must be sp2 instead and so that's kind of the approach we take in in looking at resonance hybridization if you if you have a situation where there's no resonance that's possible well then it's exactly how you'd see it in the structure you have but if resonance is possible you might have to consider that other resonance structure in this case i never drew it out didn't really have to i just realized this guy couldn't have been sp3 but we could have moved lone pair into a pi bond pi bond into a lone pair and having a double bond there i could envision that that nitrogen would have been sp2 in that structure so in this case again because of resonance it just couldn't be sp3 had to be sp2 instead cool if you got something out of this lesson please consider giving me a like and a share and if you're looking for practice problems or the study guides that go with us check out my premium course on chadsprep.com