okay so um this voiceover deals with chapter 16 and is the first voiceover we'll probably have three for this chapter uh chapter 16 as stated on this slide deals with the topic of conjugation and I start out with the definition of conjugation uh generally speaking I will never ask for a u word description I'll never ask you on an exam to to to give me a definition but you should understand what the terms mean and how to recognize them and how to apply them the different terms so this chapter deals completely with conjugation but in earlier chapters we've talked about things like configurations this is in 351 we've talked about confirmations we've talked about um en animers and diesters and stuff like that and so I never actually ask you to write out a definition but I ask you to demonstrate your understanding of the definition by applying the uh defition in various contexts all right so what is conjugation well it is a term that describes a phenomenon where P orbitals on adjacent carbons and they must there must be three or more adjacent um I just said carbons it could be any atom that has a unhybridized pil on it on three adjacent atoms um they must be continuous right next to each other three or more okay adjacent atoms with unhybridized P orbitals on them and they overlap um in such a way that they form especially stable uh molecular orbitals okay that's what conjugation is all right so molecules that have um three or more atoms that are right next to each other each of which has an unhybridized p orbital on it okay uh will give rise to a unusually stable what we call molecular orbital uh which means very low energy molecular orbital okay so I'm going to give you a little bit of an overview of molecular orbital Theory U from a poor man's perspective um I'm not a big fan of molecular orbital Theory there are some uh chemists that live and die by it and I I think that's wonderful if it works for them fantastic um but in my opinion it's just another model another way of describing things at the end of the today one of my favorite sayings about models um was once stated by a member of our statistics Department who has now long since retired uh his name was Gail Bryce and I'll never forget this never considered this before he said all models are wrong some are useful this was coming from a statistician who they make models all the time about anything I mean you think of it I mean whether you're pulling people for elections or um you're trying to uh predict um the outcome of of I don't know some cells campaign or whatever you you create a model even for the Corona virus you create a model for how it's going to spread and uh statisticians recognize that no model is perfect so all models are wrong some are useful and what by wrong it means that no model completely accurately describes the real situation or reality but there are some models that you know come pretty close and one of those is molecular orbital Theory okay this term here molecular orbital theory is a model that um describes conjugation and how it impacts the uh unusual stability of certain molecules all let's take a look at an example sorry about that and this one right here is a one3 dying system okay that system is said to be conjugated because each of these carbon atoms one two three and four has an unhybridized um p orbital on it and um that's by definition what's required for conjugation okay so this would be expected to be unusually stable so what do we mean by that let's take a look here at the next slide and let me illustrate this by comparing um compounds a and compounds B okay now if you look look at a you will see that we have four carbons one two 3 four adjacent to one another Each of which has an unhybridized pital remember that the pi bond is formed when unhybridized P orbitals overlap to form the second Bond and and that's something that's discussed in chapter one so you may want to go back and review SP2 hybridization and uh What uh this double bond is representing and what it means is there is one single bond that forms by overlap of um a sigma orbital SP2 hybridize on each carbon the second bond is formed by overlap of two P orbitals one on each carbon and they are perpendicular to the plane of the page coming out at us one there one there one there one there so there's four of them so that satisfies requirement for conjugation now here you also have four one two and then three and four but they are not contiguous okay they're not right next to each other you have one two and then there's this sp3 hybridized ch2 so that guy is not conjugated and therefore would not be considered to be very stable okay so let's see what's in the box honestly I don't remember I just reviewed these slides a few seconds ago I don't don't remember what I wanted to show under the blue box but let's take a peek okay we'll find out together okay ah a is low in energy than b well of course it's conjugated and we said that's by definition um what conjugation means they are uh compounds conjugated compounds have P orbitals on three or more adjacent atoms could be carbon or oxygen or whatever uh and they overlap in such a way that makes a molecule very stable very low in energy okay and so since compound a is conjugated it is lower in energy than b now how do we know that okay let me review an experiment that was done many years ago I never read the the original paper but I've read about it in textbooks and so I accept on faith that they know what they're talking about um and this is what they did all right they took this compound is called 13 pentadine and this one is one two three four so 1 14 penine and they treated both with hydrogen and a catalyst and in each case you get the same product okay so I think you'll notice and agree that that product is the same as that one and so you get the same product but different starting materials so if you run these two reactions and they give off a different amount of heat called the heat of hydrogenation these are exothermic reactions so they'd be expected to give off you know heat they give off a different amount of heat different amount of energy then if they're giving the same product that has the same energy where's that difference in heat of reaction coming from so let's take a look at what the experiment is okay so I need to move something so I can see this a little more clear okay at the top here we've got the 13 penine which is conjugated each carbon has a SP2 hybridized atom on it and an unhybridized p orbital on each atom which makes it the the pi Bond there's one there one there one there one there so there's four that means our criteria must have three or more so that guy is conjugated excuse me the bottom one here is not conjugated um you don't have a a p orbital on that sp3 hybridized carbon okay so this is like two isolated alenes all right so I just said a second ago that they ran this reaction and they found that uh uh they gave the same product not shown right here the energy level of the same products right there okay but they uh gave off different amounts of energy one gave off a lot of energy and one gave off um less okay so which one would give off less and which one give off more okay so I think we call this one a and that one B that's what under the box and I can't remember if I click forward if all the boxes will reveal so this is a up here and that's B so which one of these reaction progress diagrams goes with a and which goes with B okay which of these two goes with A and B which one would give off the has the smallest heat of hydrogenation it's enthalpy okay the enthalpy of of um hydrogenation is the lowest one is this one okay we're measuring the amount of energy get the difference between the starting material and the product the highest one is this one highest most energy given off is this one okay so recall that in terms of stability um when organic chemists talk about stability compounds that are very stable are low in energy and compounds that are not very stable are high in energy okay so we just got through saying that this one is stable or lower in energy so which one should it go with should it go with that fraction par on the right or the one on the left okay it's lower energy than that one it should go with that and in fact that's what was saying okay so let me call the top reaction a bottom reaction B and there we see there's the reaction progress diagrams for for both of those reactions okay um so the fact that a gives off less energy means that a started at a lower energy and the fact that b is sophomore means that it started at a higher energy and and the difference between a and b is that a is conjugated B is not they have the same molecular formula both have two alkenes but something's clearly very different about the alkenes in a versus the alkenes in B that something is a is said to be conjugated okay so let's talk a little bit about um molecular orbital Theory and how that helps us understand why a and systems like a um are more stable than unconjugated systems again unusual stability of a is attributed to conjugation why let's talk about Mo Theory okay now I have a little thing from Zoom blocking my view here so I want to kind of move it up get it out of the way I think I'm going to chck change this pointer to a pen uh and let me draw some things okay so let's change the pointer to a pen let me just review the fact that each one of these carbons is an unhybridized p orbital okay so when you see this thing on the left structure drawn on the left you should think to yourself that each of those Pi bonds results from two P orbitals overlapping so each carbon has a unhybridized p p on like that shown right there and each one of those unhybridized PS contributes one electron each to the piy bond okay so the double bond we showed over there results from these two guys overlapping and the double bond we showed over here results from those two overlapping okay um but what I just talked about with where I said these two overlap and those overlap um that is from kind of a Vesper Theory um perspective I want to talk about the molecular orbital theory that explains why the conjugated system is so low in energy all right Ino Theory what you do as you take a look at um my phone just buzz want make sure that stays off sorry about that um okay sorry about that in ml Theory um what you do is you take what is assumed is that you take four Atomic orbitals and then you combine them mathematically to create what are called molecular orbitals okay so drawn here are Atomic orbitals orbitals that would be on the carbon atom if it were isolated all by itself once you join them together they become a molecule and so you combine them mathematically in all possible ways that you can using basic um you know algebra not not multiplications but addition and subtraction okay so what am I talking about here okay uh we can uh combine all four of those orbitals by adding all four what I'm trying to show here let me get my pen is that the hashed this supposed to represent a colored node not node colored phase excuse me um and so this color here just represents relative to the origin mathematically if we're graphing you know something on a like an algebra you know like on a cartisian coordinate If This Were the origin okay zero anything up above it would be positive anything below it would be negative I'm not talking about charge I'm talking about mathematical sign right so this a plus side of a graph that's a negative side of a graph if that's zero okay so every uh p orbital has a plus phase okay mathematically and a minus phase mathematically I'm not talking charge I'm talking about um just where they lie relative to the to the origin here okay all right so we can add all four okay or we can add two and subtract two okay or we can add one subtract these two and add another we agree that if I were to subtract these that just flip that that's that's what's represented here is these two are added those two are subtracted add subtract those two add add one subtract one add one subtract one those are the four different ways you can combine uh four Atomic P orbitals mathematically okay all right I want to talk about this concept of node now you if you took uh gen cam you've probably heard this before the node is an area in space where there is no electron density okay and so um in this particular molecular orbital um the node um really occurs right there so there's no electron density right there but as far as between the the lobes um these all have the same sign and so there is actually electron density over all four uh of of those um mixed Atomic orbitals but right here is a node because we go from a plus phase right here okay to now our plus phase is down there they call that region a node there's no electron density there there's electron density between uh phases of of like sign okay um but there is no electron density where you change sign so we go from positive uh phase to negative phase so that's a node we go from positive to negative that's a node and we have two positives so those guys are overlapping we're okay there but that's a node okay so that has two nodes this one has three we can have electron density up here but we can't have it here okay we have it here it's okay there but we don't have it there we can have it here you don't have it there you got to have similar phases overlapping in order for there to be electron density um in those combined molecular orbitals okay well it turns out that the more nodes there are the higher an energy the molecular orbital okay um such that the zero node molecular orbital has lower energy than the one node molecular orbital that's lower than two and lower than three okay um it also turns out that with four orbitals two will be bonding and two are anti-bonding so if you put electrons in these orbitals and they can go in there but it's not really to the advantage of the molecule because the electrons go in those orbitals um you don't have as good of overlap and spreading out of the electrons over larger area space which we've talked about time and time again that anytime you can spread out electron density over numerous atoms that stabilizes uh the molecule in this case it stabilizes the the energy of the molecular orbital okay so two bottom ones are lower in energy than the two top ones because they have fewer nodes and when electrons bind in those orbitals um they in the ones below the dotted line in the middle they're said to be bonding so those are bonding molecular orbitals and those are antibonding okay all right so the theory is that when you have four conjugated P orbitals you can join them together like this and make uh two very low energy uh um orbitals that makes it so that two of them are spread out out over all four atoms and the other two two would be kind of in in in that lobe overlapping and two in that okay so that's the idea behind M Mo Theory and so this slide shows resonance Theory versus m Mo Theory and I personally prefer resonance just because it seems a little easier for me a little more intuitive to me uh but basically if we take this one resonance form um we can move those electrons over there like that to form a pi Bond make a pie bond between those two carbons that bumps those two electrons out onto that carbon okay and that gives rise to that resonance form or we can let me erase that okay go back and then forward here um and you might want to look at that line longer you can go back and take a screenshot of it um the other thing we could do is to get to this structure we're going to move those electrons in like that to make a pi bond between those two carbons and that bumps these two out onto the terminal carbon and that gives us that resonance form there's actually three resonance forms we can draw for that um compound and resonance sir would predict that because you can draw more than one resonance form fact the general rule of thumb is if you can draw the more resonance forms you can draw the lower energy the an or the the structure is now there's one exception to that rule that we'll get to later in this in this these slides but generally speaking that true the more resonance forms you can draw the the lower energy something will be all right Amo Theory what we're saying here we got four electrons in unhybridized P orbitals if you take and combine those P orbitals by adding all four or adding two and subtracting two ex me adding two and subtracting two um subtracting one adding those two subtracting uh subtracting adding subtracting adding um you get four different molecular orbitals and by the way there's always as many molecular orbitals as there were Atomic orbitals that go into making the molecular orbitals there were four Atomic orbitals each carbon had one p orbital one there one there one there one there when you combine B them together you make the same number of molecular orbitals so since there are four Atomic ones you make four molecular ones and the two bonding molecular orbitals have different energies the second one is a little bit higher in energy because it has one node okay there is no electron density the electrons do not overlap there here's a node and there's a node so it has two nodes and this one has three nodes and this is a great drawing I think because it shows kind of what's going on in the lowest energy molecular orbital you have all of the positive loes overlapping and the negative lobes overlapping and those two electrons can exist uh in this lobe and in that one now I'm not going to get into wave particle duality this in a case where the electrons can be viewed as waves um and be be better to view them as waves here but I'm going to talk about them like they're particles but if those two electrons are acting as a wave um this is the crest of a wave and there is the the trough of the wave but if you want to think of this two electrons zipping around kind of like a balloon shaped thing that's okay question becomes how do they get from here down to there and the answer is if you think of them as particles they can't you think them as as waves remember that in a wave you have the direction of waves traveling you have the crest and you go through the note that's a note I just went through ceas to be a wave when it goes through the the node here's a trough here's a wave this isn't very good looking but okay um so there's a node there's a node and you cease to be a wave when you're in the node but when you're a Crest or a trough you're you're a wave okay so here's a Crest there's a trough and you're going through the node okay all righty so that's the mo Theory and mo Theory the idea is we have created a molecular orbital that extends out over all four atoms and if you want to think the electrons acting as particles you can think of them being two electrons kind of zipping around here some of the time down there some of the time they're spread out over all four atoms here they're localized over those two or those two but you don't get any electron density right there so that is more localized okay and therefore um not as low in energy as this in this one you have electron density um on that particular um atom or some over those two or some over that and there's two loes excuse me nodes rather and here you have three nodes you have electron density there or electron density there or there or there but not in between there's no no there's no electron density there or there or there so there three nodes here's there two nodes okay and there's your one node okay that's Theo Theory to explain why conjugated systems are so low in energy all right turns out that Aly carbocations as we discussed in um chapter 15 when we were discussing radicals same thing is true for Radicals as is true for carboca but they are also um very they're very stable so ask this question why is this carbo Canon so stable the answer is because it's conjugated we have a p orbital on that carbon a p orbital on that one and on that one so remember that carbocations are SP2 hybridized that's an important factoid and you could very definitely expect to see that on an MCAT or dat so that guy that carbon okay that went a little bit longer I wanted to is SP2 hybridized I'll write that in here SP2 the carbon's SP2 which means it has an unhybridized peal which not is not explicitly drawn but it is implied on that carbon so there's a p there and one there and one there so it's three p orbitals and that satisfies our requirement for conjugation we can explain the energy loow effect of those three being conjugated by using resonance this is the resonance um Theory and if we just invoke that Pi Bond there moving in between those two carbons okay that will move the plus charge um from the right hand carbon put on the left left and um turns out this is what the resonance hybrid looks like this dot dot dot means that the pi bond isn't stuck here nor is it stuck there it's spread out over all three atoms and that's what makes it so low in energy because it's spread out you can also uh describe this using Mo Theory and just recall and again turn this into a pen that the pi Bond here it's implied that the double bond is made when two unhybridized peals one on each carbon atom overlaps okay that's where the P Bond comes I've drawn each of those independently with one electron in it one there one there and the carbocations we have three p orbitals on adjacent carbons if we do a similar analysis to what we did um with the um the be dining system okay with the system a few minutes ago we talked about system that had four peals on adjacent carbons one at each of the four if we do that with with this one we start out with three Atomic orbitals and we're going to make three molecular orbitals okay so we can add three together oops or we can add one and subtract two or add one subtract one and add one and each time the uh phase of the P orbitals changes again this would be plus phase if this is zero on a cartisian coordinate okay this is plus that's minus here's plus minus minus um here's a node there's a node and there's a node those are areas where there is no electron density okay all right well the more nodes the higher in energy the orbital and we only have two electrons to put in or these two in that Pi Bond so we'll put them in the lowest energy orbital and idea is they're spread out and put in one molecular orbital which is spread out over three atoms and that's a stabilizing Factor makes that carboon fairly low energy okay um shown here is a more General structure for an alyc carbo Caton don't matter you know what's at position R1 R2 R3 R4 R5 there can be different things the key thing here what you're looking for in ailc carbocadon is you have a pi system okay right next to a carbon with the plus charge on that's the alic carbocation and again I said that 2 pi electrons are spread out in one m o which stands for molecular orbital and that means the electrons are spread out over all three atoms and that's that's a good thing this again shows it kind of pictorially what the orbitals might look like um so those two electrons can zip around in the plus lobe or the minus lobe here they can be and um you know one carbon here one there here you have there there and there and you've got a note here note here note there Etc okay let's talk about the relative energies of carbal catons um I mentioned when I was talking about radicals that for all intents and purposes the relative energies of radicals and carbocations are identical there was one exception to that and here it is I'd like you to learn it differently than this even though this apparently is true our author says so I'm not I don't know where she she got her information on that I'm going to accept that that on faith but she claims that a primary alyc um uh is uh is less stable or higher an energy than a tertiary carbocadon so in terms of relative energies we're looking at how energetic these species are this is saying that a methyl cation is higher in energy than a cation that has an R Group such as a methyl or an eth or whatever alkal group here and that's true that makes sense that this would be high energy because it has fewer hyper conjugative interactions and if you're still hyper conjugation seems kind of weird till you go back and look at um chapter 15 where we talk about hyper conjugation so if the methyl has zero hyper conjugative stabilizations this guy has one this one has um two and this one has three and so um uh you but I did say for our purposes um let's move that one over here and say that it is lower in energy than the trary carbocadon okay that is not true but I think it's going to be easier for you to remember um and moving forward you will never be tested by me or nor an MC thatb will want you to know that an Al primary alic is about the same energy as a secondary so I would to me makes more sense where this guy has resonance destabilization um that it would be less in energy than tertiary um so I'm a little bit surprised that she says that this is higher in energy than a tertiary so again if you go back and look at the energy levels I gave you in chapter 15 I showed it as a methyl radical being higher in energy than an ethyl than um a secondary then a tertiary and then um and then I had this being lower in energy the primary then secondary alic is lower and low EST is a tertiary alic okay so for our purpose as I State we will consider that alyc carbocations whether or not they're primary secondary or tertiary alyc okay are more stable than a non-alc tertiary carbocation okay all right let's talk about some common examples of residents now our author gives I think us four types and here's type one is called the uh P alic system and I need to move my zoom control so I can see what we're talking about here the three alic Pi system uh she calls it type one I don't care that you call it type one no other author uses the exact you know um classification they talk about the the principle but what I'm trying to say is you won't go to text every textbook and say that look up the word type one and have them show it this way she's decided to call this type one almost nobody else I know of calls it type one but they do call it a pi alil uh three atom pyal system and in that case you have some atom X found to an atom y via double bond and that's bound via single bond to some atom Z and on atom Z is I'm putting an Aster the asteris can be um one electron or a plus charge or a negative charge and a pair of electrons would would be along with the negative charge okay that's called the pial system uh okay let's get the next one type two the conjugated double bond we just talked about that a second ago and this is the resonance uh model here where the piie bonds are not stuck where they're drawn but they delocalize out over that four atom system there's type three cat I that have a positive charge they're adjacent to lone pair we'll see some examples all these in a minute and type four double bonds with one atom more electronegative than the other that's something like a carbonal found in alahh or ketones or Esters or carboxilic acids again I don't care that you learn the types you know I wouldn't memorize type one type two type three type four don't memorize the name of the types but recognize the concept so if you see a type one know that it's that it's stabilized by residents and how it's stabilized you don't need to know that it's called type one you see one of these know it's resent stabilized but you don't need to know it's called type two Etc get what I'm saying okay if you see a type three know that it's resident stabilized you see a type four know it's resident stabilized but don't don't worry about what oh man was that four is that one is that two drive you nuts so don't worry about remembering the type names all right let's take a look at some examples of the three atom Pi Al system and she calls this type one and somehow I managed to get the X Y and Z in here don't how that happened maybe because I have an animation flying in let's see let's take a look didn't turn out quite how I wanted it to um oh well we I'll go back and fix it but here are five examples of uh three atom I'll call it a pi I usually call it Pi Al system okay we have X bound to Y bound to Z and these letters just floating around because I'm going to assign them to all five of these here in a minute and hopefully you can see the X bound to Y be a double bond bound to bound to Z and again the Z has something on it either a single electron a plus charge or a negative charge uh uh with a pair of electrons these kind of go together okay so here's your plus charge on Z here's your negative charge on Z another negative charge on z a radical on Z and negative charge on Z so I think my next slide should asside okay I didn't practice the animation sorry so that that's not quite as smooth I thought it was going to all show up as as as one slide but um I'm putting on the X Y and Z here for you okay so we have a double bond between X and Y and a plus charge on Z double bond between X and Y and negative charge on z uh double bond between X and Y and a negative charge on Z double long between X and Y and an electron on Z X and Y there is double bond and another Bond but all you're looking for is just one double bond at least and the Z on the next atom okay so these are all unusually stable because they have that General format okay sorry for the crummy animations okay we've already talked about resonance Deo delocalization of a uh alic carbocation and we mentioned that in an earlier slide that there's two different resonants for forms and what that means is that the F bond isn't localized there nor is it localized there it spread out over the entire system and that gives each of the two car terminal carbons kind of a about a half plus charge or you can also draw like that to show the plus charge right there that's the resonance hybrid and here's a couple other examples we'll go into those one more detail okay let's go to the next slide and we're going to take a look at the pial system now what I didn't show in the earlier slide and this is important is that the double bond starts out between X and Y and the other resonance form it moves and takes occupies a position between Y and Z and X picks up the the asteris and the asterisk is a plus charge a negative charge or radical okay that has the effect of spreading out that Pi Bond so it's not stuck between X and Y but it's also uh attached to Y and Z so it spreads it out over all three atoms and that's why um it is so effective at lowering the energy of these species okay here's some examples um we have an a little cadine which we already saw before if we swing those the electrons between carbons one and two out and form a bond between two and three we get that cation if we move the negative charge in like this and swing the negative charge at out or the pi Bond on to one we get a negative charge on one in the right hand structure whereas it was on three on on the left hand structure so here is x y z here we have uh x y z but now the X has the asteris on it and the double bond is between Y and Z okay so take a look at what's in the box here and compare it to what's going on here all of these structures these two structures have that format where you have uh either a charge a plus charge or or a negative charge in this case on the X and your double bond between Y and Z okay whereas before you had the plus and minus charge on Z and the double bond between X and Y okay so just uh by way of review I just wanted to review that um the use of fish Oaks we've already talked about this quite a bit so maybe I didn't need to have this slide in um but if a bond breaks in this way it's called homolytic where each react each atom that's bound together when a bond breaks each one picks up one electron they call it a A homolysis or a homolytic reaction um or if you join them two together each one contributes one electron you have to use the fish hook Arrow okay that means moving of one electron they call it a homogenic reaction or a homogenesis um you can break bonds heterolytically where both electrons go into one atom in this case both went on b and a didn't get any you call that heterolytic because the different atoms get a different number of electrons or you can make a bond where one atom contributes two electrons and one doesn't contribute it just picks them up so they're they're contributing a different number of electrons B is not contributing any a is giving two and so they call a heterogenic reaction okay all right so I did that to kind of set us up for looking at this example here so here's our X double bond y z if we break that Bond homolytically and then join those two together homogenic now we have our x with the radical on it and our Y and Z atom with the double bond okay this was X that's y That's Z here we have X Y and Z the double bonds between Y and Z here bonds between X and Y in this bottom one okay and we can do a similar thing I think we already talked about these two in the earlier slide okay um there's some additional examples we have X bound to Y with a double bond and a z z has a negative charge on it if we move the electron in and bump those out we now have a double bond between Y and Z and negative charge on X so here's your X atom there's your Y and there's your Z double bond like that and over here you have your X why the double bonds there and your Z had the lone pair on it okay similarly we have X double bond y z z has a negative charge if we move those electrons in like that that pushes those two out okay um and that gives us a double bond between uh z and y and there's X let me label that for a second I think that one's may be a little more confused than the others okay so uh we have x y and z now do that kind of backwards I don't have the X on the left and and then y to the right and the Z to the right of that I just flip that so I have X and I'm going to the left y and Z bottom line is we want a double bond between two atoms and a single bond to the third and either a lone pair or cation or radical on the third atom okay so now over here we have a negative charge on the X Y still involved in the double bond but now it's double b bound to Z okay and that has the effect of spreading out that negative charge such that it's not stuck on Z nor is it stuck on X it's kind of spread out over the entire system and that lowers um that that conjugated system um uh in energy quite a bit we already saw this one okay where we had this carbocation here um this is uh you can move those electrons in like that to form a double bond move those over there to form a double bond and so um it's a little more extended this is how can I put this I'm trying to Ill r that there are some examples that are a little more this is almost like having uh an X I'm making this up on the Fly just notice XY uh z w well I don't know bottom line is You' got an additional Pi bond out here and that can also undergo the same kind of delocalization although it doesn't quite fit that that three atom model like we took and conjugated onto the XY maybe what I ought to do is do it like this okay so let me go back and let's call this your X and your Y and your Z and call this maybe like um X Prime and Y Prime okay and that also can undergo resonance to where X Prime now picks up the plus charge and Y Prime has a double bond but the Double B Bond now goes to that X right there and um the Y atom now has a double bond with the Z okay all right so it's not identical but it's similar and it's also uh resident stabilized okay that's that um here's an example of of the um atom with a plus charge on it right next to uh a negative charge which is in resonance with a double bond between the two the most common examples I've shown here we've got a carbonal and this we you you see that so that that group right there the c bond o that's called a carbonal you see it in ketones you see it in amage you see it in carboxilic acids you see it in Esters and I'm just focusing right now on uh ketones or aldhy and bottom line is if you swing that pie Bond up onto the O there's already two lone pairs I have not drawn them in in explicitly uh explicitly but there are two lone pairs on that oxygen you break that bot and bump it up on onto o now you've got three lone Pairs and that leaves behind a plus charge on the carbon and uh that's res stabilized you can draw two resonance forms the negative Char the electrons are spread out over both atoms and that lowers the energy of of this considerably this is called an imine and we'll learn more about those in I want to say chapter 23 or 24 somewhere along that ballpark but later and it's very similar to a a carbonal and we have a double bond to the nitrogen if we break that Bond move lone pair on the end we now have negative charge on that guy and a plus charge on the carbon okay so these fit that General scheme right okay aromatic systems are also resonant stabilized and we're going to learn a whole lot more about these in uh chapter 18 so that's coming um but essentially what we've done is we chase a pie Bond over here okay so now we get a pie bond between between two and three or we get and we get those two moved to form a pie bond between four and five and that bumps those two over here to get a pie bond between one and six that's what I'm showing over here okay so here we have the pie bond between two and three four and five and one and six whereas on leftand structure the double Bonds were between one and two three and four and six and five and this curved Arrow shows how those electrons move so let me go back and I'm going to repeat what I just did I'm going to circle just so you see that those two are different okay in this one we have a pie bomb between one and two three and four and five and six right over here now we have the pie bom between two and three four and five and one and six so technically these are different and um they are resonance forms and this and here shows how I move the electrons and inter vered those two uh aromatic systems it turns out are unusually stable and we'll talk more about that in chapter 18 and it's way cool I think super cool um what the reactivity what the ramifications are of that stability so Benzene is so low in energy that generally speaking it's not very reactive you can get it to react but you have to really beat on it in special conditions you use special Catalyst and stuff so it's way cool that Benzene actually is and any aromatic hydrocarbon is is very low in energy because of this resident stabilization all right I just wanted to conclude um this section of the slides with the discussion of um the impact of resonance on reactivity okay so here's an interesting factoid and we'll learn more about this in a later chapter so you may as well kind of pay attention to it now but you won't be tested on it till later and that is um because of the fact of the product of a deprotonation of a carbon right next to a carbonal okay this is this is actually called the alpha carbon I don't know if that'll show up in red that's the alpha carbon if you pull a proton off the alpha carbon it generates a carbanion which is resonance delocalized that's an XY system okay there's the X there's the Y and there is the Z okay and if we move that pair in like this and bump that pair out onto the O that gives another resonance form with a negative charge on the X a double bond between the Y and the Z and so that makes that anine unusually stable and it lowers the pka amazingly okay so if you compare the pka of an alkan pka of 50 and if you had me for 351 I you'll remember that I had you memorize the pka because it helps you kind of uh compare different physical properties which are kind of cool alkanes have pka of 50 that's a measured empirical fact and but you put a ch right next to a carbonel so if you just forget about the carbonel and just look at that that CH okay if you're if you ignore the carbonel that CH is identical to that but you can't ignore the carbony it's right next to that carbon such that once you deprotonate it you make a conjugated system okay there SP2 hybridize um p orbital here p orbital there p orbital there this is a p alal system you can resonance delocalize a negative charge and that lowers the energy of this consequently the pka of this proton is 18 okay 18 to 20 that means that that CH is 10 to the 30 I say one more time it is 10 to the 30 times more acidic that is amazing 10 to the 30 I can't even imagine how ginormous that is it's like almost Infinity that's almost infinitely more acidic than an alcane 10 to 30 times man I'd settle like if you gave me 10 to the 122 I'd be delighted you know or 10 to the nine 10 to the nine is a billion right and 10 to the 30 um is a billion billions 10 to the 18 right and um what's a billion trillion that's uh nine and 12 so it's that 21 so trillion trillion is 10 24 uh I can't remember what comes after a trillion is a quadrillion is that what it is 10 15 anyway 10 30 is ginormous so just by putting that carbonal right next to that CH you ginormously okay increase the acidity of that PKA that that CH to where it's 10 the 30 times more acidic than that CH because that annion is not localized it's delocalized and you can draw res two resonance forms for it this one is stuck on that carbon you cannot draw another resonance form and that makes it so it is nowhere near as low in energy as that one above it is so pretty amazing this is a similar slide drawn several days later uh maybe in later that summer or something but took another picture of it uh just to see which one I likeed best I don't know what do you think the first one the the chalkboards not as clean that was cleaner um yeah anyway all right here's some other impacts or effects of resonance stabilization if you deprotonate a carboxilic acid you make that anion okay it's conjugate base of carboxilic acid the negative charge this one's not stuck there you can delocalize it in like that to make a d double bond between that carbon and that o and that pushes electrons out onto that o okay and there's another resonance form which I won't draw here but it has a single Bond of the o and a negative charge and a double bond to that O Okay so that negative charge is spread out over all three atoms and look what it does the pka so that negative charge compared to this one which is this is the conjugate base of an alcohol and once we pull off a proton those electrons move back onto the O we're talking about cleavage of this the C the O Bond those electrons move back onto the O we have an anion which is localized the negative charge is stuck on that o it's not moving at all okay stuck right there so it's all stuck on one pinpoint it can't delocalize and look what it does is the pka okay the pka of this guy is 16 all right so uh these powers of 10 remember that PKA is a logarithmic log base 10 function and so what we're saying here is that um this guy is 10 to the 11 times more acidic than that one 10 the 11 okay that's a 100 billion that's a 100 billion times more acidic than this one which means that that anion is a 100 billion times lower in energy than than that one totally amazing now and that washes out of the Resonance delocalization of that negative charge um turns out that the pka of a phol is 10 okay and that's a million times more acidic okay than a than a regular alcohol and the reason there is that this negative charge can also be delocalized into the ring okay now it turns out out that for re reasons we'll talk about in chapter 18 um even though you can draw more resonance forms for that annion than you can for this one several of them are not very happy they're fairly high in energy compared uh to this one and so you don't get quite the same degree of resonance stabilization even though you can draw about five uh different resonance forms of this depending on how you count them because some of them look like repeats if you flip them all over looks like the same couple repeats but bottom line is you can draw more resonance forms for this one than you can for that one this is one exception to the general rule the general rule of thumb is if you can draw more resonance forms that anion will be lower in energy that's a general rule but there are exceptions to all rules and this is one of them and that is in this particular case even though you can draw more resonance forms and you can for that one and this one has several that are fairly high in energy and so that does it makes it so it's not quite as acidic as that one is okay and in fact this one is 10 5times less acidic than that one five five power of 10 all right so that's 100,000 it's 100,000 times less acidic than that one but it's still a million times this is still a million times more acidic than a straight up alcohol because that negative charge never leaves the O it's stuck on that o doesn't move at all this one here can delocalize into the ring but this delocalization is not as good as the delocalization up here we'll come back and talk about why this one is not quite as good delocalizing as that one you should I think memorize these PKA if you've forgotten them make some flash cards and practice them and be aware of why the carboxilic acid is more acidic than an alcohol and why e phol is more acidic than an alcohol and that washes out of of uh resonance stabilization okay so here this slide shows the five different resonance forms we can draw for that an of the phenol you'd think that that would make it more stable but it turns out that resin forms b c d b c and d um are um less happy or they're higher in energy than than these two in Aggregate and so bottom line is uh if anol is not as acidic as a carboxilic acid and I think that's pretty near the end of the slides yeah it just says again that this anion is completely localized so it's not resonance stabilized and oh we got some type two are those all type one we still got type two to go I so excited I thought we were just about done oh we are just about done I'm talk about some type twos uh we've already kind of I'll let you look at these there's not much to say about them um yeah just be aware of the different resonance forms and be able to recognize resonance forms classic kind exam question is for me to give you um several structures and ask you which ones are resonance forms so be able to recognize resonance forms also be able to predict the properties that result for them for example be aware that a carboxilic acid is more acidic than an alcohol um pka of carboxilic acid is five no why it's that low and uh no that it the PK of the carboxilic acid is lower than for a phenol and know kind of why that is um May able to recognize don't don't be able to I'm not asking you to say okay well clearly that's that's type two as I said before don't worry about memorizing the type but you should be able to recognize that for example that that is a resonance form of that and that this one is a resonance form of that one okay all right when we come back next next time for part two of chapter 16 we're going to talk about the structure of benzene and uh whether or not it's prism or Durer Benzene or uh one of these two and uh we'll talk about the properties of benzene and how how it was that people were able to deduce um after some study uh that it's not prismane or dur Benzene but uh maybe one more like uh one of these two okay all right so that's it for now we'll see you in the next uh set of recordings