 Okay, we're going to try to finish up additions to pi star, well additions to carbonyls today. So, of course, when I think of pi star, I always think of carbonyls, that's sort of the classical functional group for C-C bond formation. When we left off, we were talking about the equilibrium constants for addition of water, reversible addition of water to various different aldehydes and ketones. And so now I want to talk about kinetic rates of addition, not the equilibrium for parameterization but the rates of addition to these various different ketone compounds. It'll give us a chance to look at the effects of rings and ring size. I'm going to give you a set of relative rates for reaction of sodium borohydride. It's kind of a prototypical nucleophile, sort of anionic character to it. And I'm going to assign just a simple acyclic ketone, acetone to a relative rate of 1 so we can compare everything to acetone. So this would be true for any acyclic ketone. What's interesting here is that cyclohexanone, which is a pretty common substrate, reacts about 10 times faster than an acyclic ketone. You know, it looks pretty similar to me. It's not a steric effect. I don't think you could argue that this is less hindered than acetone. And so what's going on here with this 10 times faster? Cyclopentanone is pretty similar. It's a little bit slower. I don't think that that effect is that meaningful. What's important is the comparison between cyclohexanone versus cyclopentanone. That's the kind of effect that you can take advantage of when you're doing a selective reduction of one of two ketones in a molecule. And of course, if you get to a more strange system, this is not happy. This is even a little bit more fast than cyclohexanone. So let me compare these. So, okay, there are some differences here. I don't want to get too caught up in these differences. If you compare this to the difference in reactivity to an aldehyde like Benzaldehyde, that's a pretty common aldehyde. You get a real sense for a difference in rate. This is 12,000 times faster than those ketones. So I'm going to harp a little bit on this tiny difference here. We try to keep it in perspective, right? Aldehydes are way more reactive than any ketone we're talking about here. So how do I explain this difference for cyclohexanone, right? Cyclohexanone is not less hindered, it's more hindered by any sort of measurement you could use. What you have to do to understand the reaction of cyclohexanone is in the transition state as your nucleophile comes in. And it doesn't matter whether it's coming in axial or equatorially. As your nucleophile comes in to cyclohexanone, you're starting to go sp3 hybridized here. And let me help you see that there are eclipsing interactions here in cyclohexanone. If I draw out these substituents at the alpha position, these protons, these bonds right here eclipse with the CO group and that's not good. That destabilizes the cyclohexanone ground state because this carbonyl, because this is sp2 hybridized, this carbonyl is stuck sort of in the same plane as this CH bonds. And as you go in the transition state towards something that's more tetrahedral like, you relieve that eclipsing interaction. Now, if I stood down this right here with my eyeball and I looked down this bond, I would find perfect eclipsing. And if you draw a good chair with cyclohexanes and all the axial and equatorial substituents correctly, you can see that there's perfect, sorry, perfect bond staggering, not perfect eclipsing. You get perfect bond staggering here. So that this CH bond is now not overlapping directly with either of these two bonds. So you relieve this eclipsing interaction. I'll just write here, this is eclipsing and it's bad. And with cyclohexanone, by adding to cyclohexanone, you relieve the eclipsing, you relieve that eclipsing interaction. And not just in the product but in the transition state leading to product. Now, when we come over to a five-membered ring, right, there's multiple different confirmations for a five-membered ring. But you can draw confirmations for a five-membered ring where you can get around that eclipsing. You've got one eclipsing interaction. There's no way to, you'll have at least one eclipsing interaction but over here, this carbonyl can bisect these two CH bonds. So you really only have one eclipsing interaction. I'll just write one eclipsing here. And that's because this CH bond is in the same plane. But these two CH bonds are not in the same plane as the carbonyl. So you just have one bad eclipsing interaction right here between this and between the carbonyl, between this CH bond and that carbonyl bond. When I come over and draw out the products for that reduction, the hydride reduction, now I do have eclipsing interactions. And so you can see why for this, why for cyclopentanone, maybe that's a little bit slower. You didn't relieve. It's about the same as a regular acyclic ketone. You didn't significantly relieve eclipsing interactions by adding to cyclopentanone. So eclipsing destabilizes in a way, hurts cyclohexanone and makes it a little bit more reactive. And these differences between five and six-membered ring ketones, you can take advantage of those when you're doing synthesis, when you're trying to selectively add nucleophiles. Okay, so now let's switch gears from just playing ketones and let's compare them to other types of pi star groups. So I'm going to draw two different types of polar double bonds, polarized double bonds, an imine versus a ketone. Which of those do you expect to be more reactive towards nucleophiles? I would expect the carbonyl, the CO pi bond, to be more reactive. Oxygen is more electronegative than nitrogen. I would expect it to lower the energy of the lumeau, by replacing, if I had a carbon there and I replaced it with oxygen, I would expect all those orbitals to drop more than if I replaced it with a nitrogen atom. So I would expect that to be more reactive. So if I look at reactivity, I would expect nucleophiles to react with this because oxygen is more electronegative than nitrogen. If I compare that, I can take a similar sort of analysis and apply it to the positively charged species. So if I compare an oxonium ion, so that's oxygen with three bonds, that could be a protonated carbonyl versus a protonated emine. I expect the same trend to hold. Oxygen is more electronegative than nitrogen. I expect it to be easier to add to a protonated oxygen than to a protonated emine, than to an aminium ion. But what you can't possibly know is the comparison between these two. What happens if you compare just a simple ketone versus in a aminium ion? You couldn't know that, right? One of them is charged, but it's nitrogen. Nitrogen is less electronegative, but what about when I put a formal positive charge on there? And so I'll tell you straight up, this is an important relationship if you didn't know, is that aminium ions are more reactive than ketones. They are more reactive than ketones. You couldn't know that. You couldn't know whether it was more important that it's nitrogen, right? Nitrogen over here is less electronegative, or whether that positive charge makes more of a difference. So remember this always, that aminium ions are more reactive than ketones. Here's how you take advantage of that fact. Let me give you an example. I'll show you something called a Bortre reduction. A Bortre reduction is based on this reagent. There's other reagents that you can use for this. If you take sodium cyanoborohydride and you replace one of the hydrudo groups with an electron withdrawing group, it makes it very choosy. So here somebody went and took borane and added cyanide. So now with that electron withdrawing cyanobrope on there, this reagent becomes very choosy. I don't know how to spell choosy. I'll put an E in there. You get the idea it's choosy. Now, that cyanobroup is electron withdrawing. So if you take an amine and you mix that with a ketone under these Bortre reductive amination conditions and you use this kind of choosy reagent, cyanoborohydride, I mean it's commonly drawn like this even though the carbon's attached to boron. And these would be typical conditions, the polyloquous methanol, pH 5, that's acidic conditions. You run this under acidic conditions. It works best if you run it under acidic conditions. If you don't, if you allow the conditions to become basic as the reaction proceeds, which is what happens, you'll get burned. It doesn't work very well. But as long as you control the pH and keep it slightly acidic, you'll get good, you'll get good formation of the amine. This is called a reductive amination reaction. So what's going on here? What's going on is that you're forming aminium ions. And let me draw out the equilibrium more appropriately. You don't have to generate a lot of aminium ion. You just have to generate a little bit of aminium ion. Even the tiniest amount of aminium ion. So the byproduct in this equilibrium is water. This is a five step mechanism. I'm not going to draw all five steps for aminion, for aminium ion formation. I expect you to know that we talked about that in discussion section, I think in our first discussion section. And that's one of the fundamental mechanisms that you need to know. So who cares? It's five steps. They are all fast. Whenever under acidic conditions, not too acidic, not in hydrochloric acid. But whenever you mix amines and ketones, you will have an equilibrium amount of aminium ions. And what did I just say about ketones versus aminium ions? What I said is that aminium ions react faster than ketones. So when you put in a choosy reagent like sodium borohydride, you don't see reduction to make an alcohol. This just sits there. And the cyanoborohydride will just collide and collide and collide and never react. But when cyanoborohydride collides with this, it very quickly reduces this. And so I'll draw that hydride reduction. That's fast. So reduction of the ketone by cyanoborohydride is slow. Reduction of aminium ions is fast. Any reaction, any space, any time you have something that can generate a carbocation that will very quickly be reduced by cyanoborohydride. Cyanoborohydride only reduces things that have positive charge, like carbocation or aminium ions. If you dump in too much acid here, let's suppose you squirt in hydrochloric acid, you'll start to protonate the ketone and then your ketone will reduce. So you have to carefully control the pH for those reactions. Okay, aminium ions react faster. Keep that in mind. We'll use that in just a little bit. Okay, we're going to take a little excursion here and stop talking about ketones and aldides for a moment. And we'll start talking about carboxylic acid derivatives. Let's suppose I take a carboxylic acid derivative and it's reacting too slowly. And so I put a really good leaving group there, like bromide. Maybe the ester was slow or the acid chloride was slow. And so then I can rely on my nucleophile coming in and displacing this, this carbon bromine to make some sort of carboxylic acid derivative. And I hope you're not drawing this on your paper because it's killing me to draw this. This is wrong at every single level of detail. And from the beginning of class, I've tried to convince you that this is wrong at every level of detail. And so let me just remind you, it is always easier to add to pi star, right? If I draw this carbon bromine bond breaking at the same time that the nucleophile is adding, that's an SN2 reaction. I've just drawn an SN2 reaction if I draw it like this, right? And if you do this, then you're claiming it's going through SN2. Whether you drew it on the same side or the opposite side, you just drew an SN2 reaction. So it's never faster in this kind of comparable case to add to sigma star. It's always faster to add to pi star. So let me just remind you, every single time that there is some leaving group on a carbonyl, yes, there is a sigma star orbital that's on the back side of this carbon X bond. So let me label it sigma star carbon X. Yeah, there is some sort of sigma star orbital back here that a nucleophile could add into, but no nucleophile would ever do that. All nucleophiles will prefer to add to pi star. So let me just write out pi star for the CO pi bond. That's where you add. It is always faster. No matter how good you make the leaving group, the leaving group simply makes it even faster to add to the carbonyl pi star. So you always get a two-step substitution process through addition followed by elimination. This is the way carboxylic acid derivatives always substitute. So you go through a tetrahedral intermediate. That's the tetrahedral intermediate. I'm just going to abbreviate that TIM, tetrahedral intermediate. And then in a second step, you push out the leaving group. Usually the first step is the slow step. Let me draw the lone pairs on the opposite side so I can see that now they're ready to push, to donate into sigma star and break that carbon X bond. It's in the second step that you can donate into sigma star, and that's the lone pairs right nearby that are doing that. So two steps, addition, elimination. And I expect you to know that this already, that this is the way carboxylic acid substitutes. So it's not elimination to make a C-C pi bond, it's elimination to make a carbonyl bond. Okay, so if you're not up on that, go practice some carboxylic acid substitution reactions like fissurist erifications, ester hydrolysis, acid catalyzed hydrolysis of amides. They all involve some step that looks like that. Okay, let me just remind you of a problem that often arises when you try to make ketones. So let's suppose I wanted to convert this ester into a ketone by replacing this methoxide leaving group. Esters are very common functional groups. And if I wanted to convert this into a ketone, and I guess my example here, I have phenylithium. I'm going to get a poor yield for this reaction. That is 100 percent sure. And it's not that phenylithium is not good enough. Nucleophile, it is a great nucleophile. And it's not that methoxide is not a good enough leaving group, it's plenty good to leave when there's an anionic alkoxide. The problem is this. The problem is that the initial addition, let me draw the bond, the nucleophilic bond there, there's no problem with that, that's fast even at low temperature until you'll generate this tetrahedral intermediate, there's the leaving group. And then this is going to push that out. Here's that lone pair on the oxygen is very nucleophilic and this can now be pushed out to give you a ketone. And unless I do otherwise, that lithium still needs to be on that ketone. Don't have it suddenly not there, right? It's still there. So I'm not showing how the lithium got on there. I made a mistake here, that should be phenyl. So I've just added my new phenyl group. The problem with this is not getting the phenylithium to add, the problem is keeping the phenylithium from adding a second time. This is now more reactive than the ester I started with. Esters are less reactive because that donation into the carbonyl group makes esters unreactive. And once you form a ketone, it instantly reacts with any leftover phenylithium. And the more ketone you generate, the faster this reacts. So what you end up with is double addition. And if you didn't add enough reagent, you'll have leftover starting ester present as well. So it's very hard to stop this double addition. Don't draw out some synthesis where you simply add nucleophiles to esters because you can't stop it when addition. But there's a trick you can use that's commonly used in synthesis. So if you have a carboxylic acid and you want to convert it to a ketone without double addition, what you do is you convert that carboxylic acid not into an ester but into something called a wine-rebs-amid. And it's very specifically this. It's the N-methyl N-methoxy-amid. So this is what you add alkyl-lithium to. So for example, if I had now phenylithium to this, it generates a tetrahedral intermediate. Here's my phenyl group. And again, it's the same initial step here. I add to pi star. And now I've got this species in here that has this methoxy group poised five atoms away from the lithium. This low-emparant oxygen is poised to donate into this and form a stable chelate. And that chelate now, let me just make this a generic R group here, and that chelate is stable. It doesn't fall apart to regenerate a super-duper reactive ketone. That just sits there until you're ready to work it up. So I'll go ahead and draw the bonding here. But that's stable. This tetrahedral intermediate that you get is stable. It's all right, stable tetrahedral intermediate. And then upon workup, you're throwing some sort of aqueous solution. The whole contraption decomposes and falls apart to generate the ketone. So this is workup here, workup. And so if you want to stop at the ketone and not get over addition, you use wine-rebs amide because it creates this stable tetrahedral intermediate. Okay, so let's go ahead and talk about making carboxylic acid. So this is a way to make carboxylic acid derivatives kind of in essence less reactive towards substitution because you slow down the elimination step. And so what do you do if you want to accelerate the addition step to a carboxylic acid derivative? And I want to ponder a fundamental question because one of the most reactive types of carboxylic acid derivatives you should know about are acid chlorides. So you should know that you can take a carboxylic acid and treat it with thionyl chloride, SOCl2, and make an acid chloride. If you don't remember that, I'll draw it out for you. That's money. We call that a money reaction. It works every time. And it's super reactive. I want to compare some acid halides so we can talk about the relative rates of reactivity. What you find as you look in organic synthesis is you don't use these with equal frequency. You use acid chlorides. When was the last time you saw somebody make an acid bromide? Or when was the last time you saw somebody make an acid fluoride? You can find cases, but it's very rare. So let me go ahead and talk about the order of reactivity. Acid bromides are more reactive. In fact, they are too reactive. When you make acid bromides and you try to store them, the slightest bit of water and it's gone. So that's very inconvenient. Acid fluorides are way too unreactive. Why would you go through the trouble of making an acid halide if it's not going to be very reactive? So I want to talk about this ordering of reactivity here. So it turns out that acid chlorides have just the right balance of they're really reactive. Acid fluoride is not. But it's not so reactive that it just falls apart when you put it in the bottle. And next week you come back and half of it's decomposed. Okay, so I want to come back and talk about this sort of relationship. I talk about why is it that acid fluoride is so much less reactive than an acid chloride? You can't explain that with electronegativity. Fluorine is way more electronegative than carbon. Let's just write out the electronegativity. So the electronegativity of fluorine is 4. That's the standard, it's the most electronegative atom in the periodic table. Chlorine when you drop down a row in the periodic table is only 3.2. So it's not as electronegative. And when you get to bromine, the drop is not as big but it's even less electronegative than chlorine. Idyne has about the same electronegativity. I'm not even talking about acyliodides. That would be crazy reactive. But that has about the same electronegativity as carbon, iodine does. Okay, so let's go ahead and talk about why. It's not because of electronegativity because you'd predict the acid fluoride would have, would make nucleophiles out of the carbonyl faster. And so why is it? So what I want to do is I want to look at the carbon halogen bond lengths in these compounds because that's the secret to understanding this order of reactivity. It's not the electronegativity. So if I come over here and I draw the acid fluoride like this and I look at the bond length. And an acid fluoride, fluorine is a second row atom just like carbon, just like oxygen, just like nitrogen. It's got a very typical, well maybe a little bit shorter. But it's got a nice kind of second row bond length for that carbon fluorine bond. And so in comparison, if I draw an acid fluoride, that's the secret to understanding why acid chlorides are so reactive. I'm hugely exaggerating because I worry that if I don't exaggerate, you won't remember it. So I'm going to exaggerate there. And let me come in and draw the acid bromide. And it's not as, right, now I'm really exaggerating and it's an over exaggeration. But that's why there's this difference. It has to do with bond lengths. When you drop down from the second row with carbon, nitrogen, oxygen, and fluorine to the third row or the fourth row of the periodic table, you end up with longer bonds. And what the bond length affects is it affects the ability of these electrons to donate back into pi star. Now with these misshapen huge orbitals on chlorine, so big and so far away, you don't get effective overlap back into that carbonyl. And it's that overlap, it's the donation into the carbonyl that makes this carbonyl less reactive. That's why esters are less reactive than ketones. So this fluorine atom makes acid fluorides. Even though fluorine is super electronegative, that's helping. But you're screwing up the reactivity by having the lone pairs donate back in. And you don't get that really with chlorine and I'm not even going to draw with the bromine, so forget that. So with those two, all you have is the light electronegativity effect. Well, let me just try to convince you here that this donation of fluorine is important because it may not be so obvious to you. So, it's like what's some evidence that this is an important feature that, right? If I've got a lone pair here, if you're a nucleophile trying to collide with that carbonyl group, how are you going to compete with a lone pair that's always right there? You're not going to compete with that. You might say, oh, I don't like that F plus in that resonance structure. Let me show you a little comparison here that'll help you understand this balance between the fact that, okay, fluorine is electronegative and should make it more positive here on carbon. Fluorine's electronegative. Delta minus delta plus. If I were a self-respecting nucleophile that had any negative charge, I would definitely want to attack that carbonyl except for the lone pair on fluorine getting in the way. Okay, so let's try to get straight the effects of fluorine, this contrasting effect of electronegativity which makes that more reactive versus donation which makes that less reactive so we can put it into perspective. I'm going to draw three different cations for you and I feel like I've shown you part of this comparison before. But let's imagine a fluorine substituted carbocation. What's the effect of fluorine? You know that fluorine is electronegative, that ought to really hurt, right? Fluorine's electronegative, ouch. There's a carbocation there with a positive charge. It turns out that a fluorine substituted, so I'm just giving you relative stability of these carbocations, a fluorine substituted carbocation is more stable than one that has no fluorine on there. So yes, fluorine is electronegative. Yes, okay. But it's still so good to have this donation that overall it's better to have a fluorine on a carbocation even though the fluorine's electronegative, that donation is so stabilizing it overcomes that. You couldn't have known that. You couldn't have known that the donation was more important in stabilizing than the electronegativity is in destabilizing, so that's why I'm telling you that. And I expect you to remember that. If you put two fluorines on, it's even better. That's even better. What I don't have on here and something that I just happened to know is if you put on three fluorines and I don't, what I don't know is where it falls in this sort of list. Three fluorines is worse. The third fluorine doesn't help you. And I, this one falls somewhere in this region. A third fluorine doesn't help you. You just get the electronegativity effect. So of this series, this is the most stable and this trifluorocompound cell falls somewhere in between. The way to think about it is that, well, you can only have so many fluorines donating into that carbocation at the same time. A third one's kind of like, well, there's nowhere left for me to donate as a sort of descriptive way to think about that. If you really want to, well, I'll just go ahead and leave it at that. Okay, so two fluorines. So fluorine is electronegative, yes, but the lone pairs on fluorine can donate into empty orbitals. So remember that. So I'm going to show you a special type of aldol condensation called the novenagle condensation. It's a very common way to make enone derivatives. It involves taking an aldehyde, a 1, 3-dicarbonyl compound. So I'll show you one of the most common ones, ethyl acetoacetate. So you know that the, I hope you know that the protons between the two carbonyls are acidic and so you add some sort of a secondary amine base to this. And don't worry about the easy, I'm not trying to make any comment about the stereoselectivity. You'd get a mixture of ENZ. The point is it does an aldol condensation, makes a new CC double bond in the reaction. If you go back and you look at all the novenagle condensations, almost always using pippuridine or diethylamine, not triethylamine, diethylamine. Not tertiary amines, secondary amines every time. And there's another weird thing about novenagle condensations. They usually add acid in there. What's that about? Right? If the mechanism is simply to have the base deprotonate here, why are they adding acid into the reaction? That can't possibly help that deprotonation. It can't possibly increase the amount of the enolate that's present in there. This is a classical recipe for a novenagle condensation in 90% of all cases. What's going on here is not what you think. It's the amine is not simply in there to form an enolate. The point of the amine and the acid is to make the aminium ion. And I'm going to draw the totomer, the enol totomer of this. So the ideal conditions for a novenagle condensation are not basic. They're actually just slightly acidic, a little below pH 7. Under those conditions, you have this super duper reactive aminium ion. Aminium ions are way more reactive than ketones. This is a classical novenagle condensation. And when you draw, I'm not going to draw up the mechanism. There's a five-step mechanism to show how you get from an aldide to an aminium ion. I expect you to know that mechanism for aminium ion formation. So when this adds, and we're going to talk a lot about enols as nucleophiles later. So I'm not going to draw the deprotonation step. I just wanted to get to this stage here where we, so you get to this stage where you have this intermediate and just trying to figure out which steps should I show you. So in this particular case, you'll have this R group on here and you'll have the amine. And so now you're set up to do an elimination reaction. This is going to be an E1CB type elimination. You're going to end up protonating this under the slightly acidic conditions of the reaction. And let me draw this intermediate out, otherwise I'll just be talking in total abstraction. So you can't claim that it's hard to form an enol because we started off our mechanism with an enol. So we're going to add this nucleophilic enol. And we said that the conditions are slightly acidic. So it makes total sense that we're going to have some of this Piperidinium group on there. And so this nucleophilic pi system here, this enol, will simply push out the leaving group. And that's what leads to the final enone in this product. So all of these steps are faster. The simple mechanism that you might think of would be to deprotonate this and make an enolate and have the enolate add to the aldehyde. And it wasn't until people studied the mechanism of this that they realized these are all going through aminium ions. This isn't the power of aminium ion chemistry. Okay, very powerful stuff. It dates back, these date back to the 1940s and 1950s, these react, the Novinagel reaction. It's a very powerful and general CC bond forming reaction. And there's tremendous insight in this idea that aminium ions are way more reactive than ketones or aldehydes. Aminium ions are more reactive than carbonyls. So let me just sort of be clear about what's going on here. If you look at the relative molecular orbitals energies for pi stars for a carbonyl, a pi star for an aminium ion is lower. So nitrogen is not as electronegative as oxygen. But when it's a positively charged nitrogen, then that orbital is even lower in energy. That's what's making that work. So this stuff was sitting in the literature for 50 years until Dave McMillan realized, well gee, maybe I can use that for other reactions. So if you take Chem 202, we'll talk about paracyclic reactions like the Diels-Alder reaction. And one of the first things that you learn about the Diels-Alder reaction is they work better when there's a carbonyl on the dienophile. And the reason why carbonyls make this faster is because it lowers the LUMO energy for this double bond. In other words, the way to think about a Diels-Alder reaction is you think about the diene as the nucleophile attacking the, sorry, as the diene attacking the dienophile. So if you take this reaction, which may not be particularly fast, and you add a secondary amine to this, so I'll just write catalytic. And importantly, if it's a chiral amine, there will be this fast equilibrium formation of a small amount of this chiral aminium ion. And that is now more reactive in Diels-Alder reactions or in nucleophilic addition reactions. It's now more reactive than the starting cinnamaldehyde starting material. So nucleophiles, like the diene, will attack that faster. And of course, this now acts as a nucleophile and comes back in and adds. And that's the mechanism for a paracyclic Diels-Alder reaction 4 plus 2. And if this was chiral, you can imagine it will influence which phase of this dienophile that you add to. And that controls the stereochemistry. You couldn't have known. You couldn't have guessed that this equilibrium would be so fast. This is why Dave McMillan had to run the experiment. He was a graduate student in her department. That's why he had to run the experiment to see, is this faster than the, is the aminium ion faster than the simple Diels-Alder reaction? And so in this case, it turns out to be faster. So the initial product has this aminium ion still on there, a aminium ion formation is reversible. So let me draw the double bond down here on this nor-bornening ring system. And so a aminium ion formation is fast and reversible. And so you very quickly hydrolyze this. And so ultimately what you get out of the reaction at the end is not the aminium ion. It's the aldehyde. You can't even tell there was ever an aminium ion in this reaction mixture. You're only adding a catalytic amount of the amine. So this is fast. This hydro, this formation of, of aminium ions and the hydrolysis of aminium ion is fast under many, many different types of reaction conditions. So, so anytime you want to make some sort of a carbonyl group or an enone, more reactive, you can simply throw in a catalytic secondary amine. And, and it will work best if the conditions are not really basic but slightly acidic. You'll get the fastest rates for aminium ion formation. Okay. So that's the whole field of organo catalysis. It was, was that one simple insight that aminium ions are more reactive than, than carbonyls. Okay. So the last thing we're going to talk about is conjugate addition reactions. In other words, when you have an enone like this, you can imagine nucleophiles adding either to the carbonyl or adding to this beta carbon right here, not alphabet beta. And so let's talk about that. In order to set us up to understand the, the distribution of products, the regioselectivity in these reactions, there's a, a, a couple of pieces of information that we have to put together. So would that make sense? And so I want to talk about, first I want to talk about the effects of conjugation on reactivity to remind you of something we said already. And then I want to talk about the effects of substituting carbon atoms with more electronegative atoms. So let me go ahead and remind you of the effect of conjugation. I'm going to draw some sort of MO diagram where I want to compare the, the react, the energies of the homos and lumos. And I'm going to start off with ethylene. So here's the pi orbital for ethylene. It's down here. Not very nucleophilic. Here's pi star. It's not very electrophilic. It's not very, if you throw in, if you throw in an amine it doesn't just attack ethylene. It's not a fast reaction. If you wanted to add HBR across this, that's not a very reactive alkene unfortunately. But if I take a dyne, if I add conjugation to this, and it doesn't matter that it's another double just simple olefin, it could be a benzene ring that I attach on there, same idea. If I add conjugation to this, what it does is it soups up the reactivity of this as a nucleophile so pi now gets higher in energy and it makes it easier for nucleophiles to add back to this. So it's a better electrophile. So pi gets higher, pi star gets lower. That's the effect of conjugation. The more conjugation you have, let me just draw some sort of central line here, the more I see the homo raised and the more I see the lumo raised, add infinitum. So we've already talked about that. And of course if you go out to make this in infinitely long, that's molecular wires. That was the 2000 Nobel Prize. So somebody just extrapolated that to very long lengths and realized they could make molecular wires out of that. So that's one piece of information I expect you to know. Okay, so let's talk about a second effect. It's the idea of replacement. So if I take a dyne like this and I draw out these orbitals for pi and pi star for this dyne, what would happen if I replace this carbon atom with oxygen with, not without, but with O. If I replace that carbon atom with oxygen and I look at the energy of this pi orbital and I look at the energy of pi star, the effect of replacing carbon with a more electronegative atom is that both of these should now drop. Pi should drop, pi star should drop, everything should drop, sigma, any sigma bond that involves that oxygen, so now drop in energy. So that's the effect of replacement. So if I draw out the energies of an enone versus a dyne, what I expect is that both of these should now be lower in energy and you kind of knew that. You kind of knew that this enone would be less nucleophilic than a dyne. You kind of knew that. You kind of knew that ketone pi bonds aren't as nucleophilic as C-C pi bonds. You knew that already. You kind of knew that it's faster to add to carbonyls than it is to add to bi bonds. So you already knew this was lower. So you knew this stuff already. Okay, so that I think I expect you to know. Here's what you couldn't know is the shape of pi star in a dyne. So I'm going to have to tell you that. So I want you to try to think about what pi and pi star should look like for a simple dyne. And I'm going to draw out a simple dyne like this. And in order to make pi and pi star, what you have to imagine is mixing some p orbitals together. So I want you to imagine looking at some p orbitals top down, all you can see is the tops of these p orbitals. And I'm going to start by drawing these p orbitals and then we'll draw in some phasing here. And so what I'm interested in is the lumo. If I were to attack a dyne, if I were to add a nucleophile to a dyne, what would I expect for the regio selectivity? It turns out that this is the shape, the rough shape of the lumo for a dyne. These carbon atoms on the ends contribute more p orbital character to pi star. So pi star, the lumo in this case. Then the two inner carbons do. You could not have known that. Now let me show you how you kind of knew this but you didn't know it. So if you take a nucleophile and you add it to but a dyne, where would you add? Where would you prefer to add? You already knew that you prefer to add to the end. And maybe you made up some ridiculous argument that oh, I want to generate a resonance stabilized carbene ion. Like that's not why you add to the ends. It's not because of sterics. The reason why it's faster to add to the ends of a dyne is because the lumo is bigger there. If I have a filled orbital, I'm trying to figure where can I overlap with? You're going to overlap better with the ends, the carbons at the ends. That's why nucleophiles attack. Dyne's at the end. So you need to just remember this. If you do a five second calculation, a five microsecond calculation, it will show you that the pi star is bigger on the ends. Okay, so now let's try to map this picture on top of this picture that I showed you over here for pi star for an enone. Let's go ahead and try to combine those two effects together. So what I want to try to do is I want to try to draw an edge on depiction and I feel like I'm going to need a new board for this. I want to try to depict an enone taking this sort of orbital thing into account. I want to try to draw pi star not for a dyne but for an enone. So if I'm a nucleophile and I want to attack an enone, let me draw this enone edge on here. I'm trying to draw it looking sort of at the side of that for the edge because I want to see the p orbitals now. And remember what I told you. I told you that the orbitals are bigger on the ends for a dyne and I don't feel like I'm doing this justice here. What I'm going to do is I'm going to give you some numbers called coefficients that will more accurately express the sizes of these. So in other words, how much p character got mixed in? Was it a full p orbital that got mixed in to make pi star here? Each one of these atoms contributed some p character. This carbon atom contributed some of the p orbital to make this pi star orbital. This carbon atom contributed some p orbital but how much did they contribute? They don't all contribute equally. And so what we do is we look at the coefficients, tell you how big these are. This comes from linear combinations of atomic orbitals, LCAO theory. You might have remembered that from P chem. So if I draw out the coefficients for this, here's the coefficients, 0.62 for the carbon at the end here and it's plus. So let me just arbitrarily put phasing on here. And then I look at the coefficient for this orbital here and it's minus 0.43. Minus means I phase at the opposite direction. And then I look at this one over here, it's minus 0.53 is the coefficient size. And then I look over here and it's plus 0.60. So I phase that the opposite direction. So in other words, a nucleophilic comes in, doesn't want to come in from this side because it doesn't want to have a destructive interaction. With this, you'll come in from the Berge Dunn's angle over here. So I've got a clear picture here for compounds that have nucleophiles that have big occupied orbitals, nucleophilic orbitals, will tend to want to come in either on the carbon or on the oxygen. That's what this picture is telling me. If charge was irrelevant, if coulombic interactions were totally irrelevant, it might be almost equally probable to add to the carbon or the oxygen. But coulombic interactions are relevant. They're relevant. No nucleophile, all good nucleophiles have partial negative charge, they would never want to add on oxygen. So if you pay attention to coulombic interactions, they'll tell you carbon, well, sorry, if you pay attention to where pi star is big, you would predict either end. But if you pay attention to charge, then there's partial positive here and partial negative here. And we'll see how much in just a second. So this simple picture of a dyne predicts that you'll react at either end of an enone. But when you think about charge, it doesn't predict that. Let's take a look at what charge predicts for an enone. And this comes from calculation. So the Lumo picture tells us one thing. If I look at the charges from calculations, I'll draw two resonance structures for an enone. So here's one resonance structure for an enone, where I draw this charge separated resonance structure. And this reminds me, yeah, nucleophiles want to attack carbonyls. You knew that already. Let me draw another resonance structure. I could swing this double bond over and draw it like this. This is a different charge separated structure. And both of those resonance structures are important. That the resonance structures predict that there will be partial positive charge at the carbonyl carbon and partial positive charge. Let me label these. This is the alpha carbon and this is the beta carbon. Alpha carbon, beta carbon. Whoops. Alpha carbon, beta carbon. There will be partial positive charge at the beta carbon. But how much charge? It turns out from calculations you can tell that the relative charges are 1.25 at the carbonyl and only 0.21 at the beta position. There's much less partial positive charge at the beta position. If you're a nucleophile that has lots of negative charge, you're going to tend to want to attack at the carbonyl. If you're a nucleophile that has a very nucleophilic homo and you're looking for good overlap, you're not going to want to attack here. You're going to want to attack there. Our expectation is that highly charged anionic nucleophiles will tend to attack here whereas nucleophiles that have big orbitals, filled orbitals will tend to attack there. And this explains the differences in reactivity. Yeah. That's, it's related to hard, it's related to hard soft. Hard soft is just this. This is hard, but I don't, again hard soft doesn't, I don't know how to interpret that in terms of coulomic interactions in Coulomb's equation. So that's why I break it down to Coulomb's law. Okay, so let's go ahead and draw some, just draw out these sort of conclusions here. So if you want 1, 2 addition, here would be my champion nucleophiles for 1, 2 addition. If I wanted to add 1, 2 at the carbonyl, so for example something like this. If I did not want to add at the beta carbonyl, I would want to have something that's super charged that has all, has all the negative charge concentrated in some little point. Fluoride would be a perfect example. It's just like a naked little pinpoint of negative charge. If you had fluoride, you would expect vastly faster rates of addition here where the positive charge is closer to 1.25, I think I erased that. Then addition here to the beta carbon. Similarly, if I take an alkoxide and put a non-coordinating or poorly-coordinating counter ion there, I would also expect that very naked alkoxide. Then the more naked, the better. Lithium not so good, cesium, naked alkoxide and ion. That'll go 1, 2. So if you want to make things go 1, 2, use things that have all the charge concentrated in a single point so that as this gets closer and closer and closer to that partial positive charge, Coulomb's law predicts that that's going to be an attractive interaction. Now, let's contrast that. So we call this 1, 2 addition because we normally label these 1, 2, 3, 4. So these are good nucleophiles for 1, 2 addition. If you want 1, 4 addition, addition to the beta carbon, here would be some great nucleophiles. An enolate. And more importantly, a 1, 3-dicarbonyl enolate. Where's the negative charge in this case? Well, it's not on the nucleophilic carbon, it's way out here in these oxygens. So now, when you have some sort of electrophile in this case, since there's not a lot of negative charge on that carbon, let me not put a circle in there, it turns out that you end up attacking 1, 4. If I took an enamine, very little negative charge on that nucleophilic carbon, it turns out that that attacks 1, 4. If I take a thiolate anion, which very big atom, all that negative charge is spread all over the place. It's not concentrated in a little point. You end up with 1, 4 addition. If you take a phosphine, again a third row atom, you expect vastly faster rates for 1, 4 addition over 1, 2. Okay, so charged, super pinpoint charges like to attack the carbonyl carbon, things without a lot of negative charge tend to attack out here on the beta carbon. Okay, I think we're done with pi star for now. We've got problems on the problem set, and sorry I went over.