 Okay, so we're still on lecture number 11, I think, talking about eliminations. Lecture 11 continued. Now, we just have a little bit left to talk about, but very important stuff to talk about with lecture 11. So I've tried to convince you that it's super important to see anti-periplanar relationships between bonds and organic chemistry. And that's why I will push you to draw things using perfect chairs with perfect axial substituents and perfect equatorial substituents because those confirmations correctly drawn allow you to see anti-periplanar relationships. So it's this kind of anti-periplanar relationship that makes these types of grignards or alkalithium reagents unstable. You cannot have beta-alcoxy groups because these kinds of nucleophilic bonds will never make it to some carbonyl or some other electrophile, they're going to donate into the antibonding orbital and push out that visceral leaving group, which would be alcoxy. So you can use this reaction. So for example, this is the way you remove a trichloroethyl, sorry, trichloroethoxy carbonyl. This is a trach protecting group. And the way you trigger its release and removal is you typically use zinc. So hypothetically, you could use magnesium. It's just that zinc is a more tame metal to use. So like a grignard reagent, like magnesium, zinc will end up doing an oxidative addition to make a chlorometal alkyl species. So you can imagine the zinc doing a reduction basically in order to make that nucleophilic carbon zinc bond. And when you do that, this is now set up to push out the leaving group, which would now be this carbamate and ion. And then of course that decarboxylate. So ultimately, when you do these reactions in water, so zinc reacts slowly with water, but very quickly with alkyl halides, I won't show all the steps that are involved here. Eventually, this will release carbon dioxide to give you the free amine. So you can take advantage of these types of nucleophilic carbon metal bonds pushing at alkoxy groups with this kind of a protection scheme. Truck, I might have drawn that. Yeah, trichloroethoxy carbonyl is the truck protecting group. So we've talked about silicon carbon bonds being nucleophilic enough if it's a silicon 8 to push out alkoxy leaving groups that are beta or leaving groups that are beta. We just talked about carbon metal bonds, carbon magnesium bonds, carbon zinc bonds being nucleophilic enough to push out leaving groups. What do we need to do to make a carbon carbon bond nucleophilic enough to push out a leaving group? There's two things we need to do. First of all, we need to make the leaving group better. But second of all, we somehow need to make this bond so nucleophilic and so weak that it's capable of pushing out a leaving group. And the way you do that is you use good leaving groups like this and you need to weaken this bond. And one good way to weaken that bond is if you next door put a super strong donor like this, that can donate into this anti-bonding orbital for the CC bond and that will weaken this bond and make it more nucleophilic. Donation from the oxygen lone pair can make that bond nucleophilic enough where it can now push out the leaving group by donating into that sigma star. And particularly for a good leaving group like toslate. So we call these grobe fragmentations where you're fragmenting carbon-carbon bonds apart. So you can see that the product of this reaction is going to give you a carbonyl somewhere and an olefin and plus some sort of a leaving group. Now it takes a very, very special arrangement of bonds here in order to make this work. Bonding orbitals, bonds have to be perfectly aligned with anti-bonding orbitals for this to work. So let's go ahead and take a look at some systems where this, where you can see grobe fragmentations in action. So I'm going to draw two, this is called a decolin ring system when you have two six-membered rings and it's saturated. And I'm going to show you how this is a stereospecific reaction, how stereochemistry matters, how conformation matters, and how the alignment, how anti-paraplanar arrangements matter. So I'll show you two different consequences. You make this alkoxide by deprotonating the alcohol with T-butoxide. So the actual starting material is the alcohol. And they treat that with potassium T-butoxide. And when you do that, you end up with a grobe fragmentation that cleaves the bond in the middle of these two rings. And I'm going to draw a very funny shaped carbonyl group but you get the idea. There's just no room to draw the carbonyl in the center. So this would be a grobe fragmentation. So let me go ahead and draw the arrow pushing for that. So these electrons are donating into the antibonding orbital for this CC bond and weakening it. They're making this central CC bond weaker and more nucleophilic and those can push out the tosylate-leaving group. And that's what leads to that fragmentation of the middle bond. Okay, if I take an alternative, and this maybe is not so obvious, but if I take an alternative diastereomer where the hydroxy group at the bridge head, and the proton at the bridge head are sinned to each other on the same face of the molecule. And in this case, we've got the tosylate going down. So now when you fragment this in over 90% yield, you get a different product. And I've sort of steered the ring around. The important point is you get a cis-double bond. You don't get a trans-double bond. You get a cis-double bond. That's the important point. So I've moved the atoms around a little bit. But the top one gives you the e-isomer and the bottom one gives you the z-isomer. So let's try to draw these out so that we can understand why you get these differences. And these are examples of stereospecific reactions. You start off with this isomer. You get the e-isomer product. You start off with this alternative stereoisomer and you get the z-product. So let's draw these out using chair conformation. So we can see the relationships that lead to those different products. So I'm going to ask you to use your best chair drawing skills here so that we can see these really important anti-paraplanar relationships. So I'm going to draw the H at the bridgehead going up. This is that first starting material. And then I'll draw that alkoxide anion pointing down. Those are both axial substituents. If you've drawn them correctly, they should be perfectly parallel and anti-paraplanar to each other. And then finally, I look at that tosylate group. It's going up with a wedge. That means of the two groups here, there's an H and there's a tosylate. The tosylate is on top. That's the leaving group and the H is pointing down. I'm drawing the H there just so we can see that there's another substituent at that center. So this system is set up so that these lone pairs and there's lone pairs we can draw all over this thing are set up to donate into the antibonding orbital for this bond. The lone pairs on oxygen donate into and weaken that CC bond. And that CC bond is perfectly because of the rigidity of this ring system is held always anti-paraplanar to that tosyloxyl leaving group, the CO bond. So this system is perfectly set up for this grobe fragmentation in terms of orbitals. So again, this bond is perfectly parallel to that one and it's anti to that one. That's anti-paraplanar. And you need to see those relationships. Now when I draw the product, I'm simply going to draw the same two rings and then I'll erase the bond in the middle so, you know, you can just draw the middle bond very lightly if you can't, if you're writing in pen. But it's just to erase that bond. It's not there anymore in the product. The CO bond is here but now it's a double bond in the product. And the one thing I need to do in order to correctly draw the product is to see that there was an anti-relationship between these two protons here, between this CH bond and that CH bond. This CH bond was anti to the one on top here in the starting material. It will be anti to that in the product. And that's how I know for a fact that I'm going to get a trans double bond here like that. I still have that anti-arrangement. And if you didn't see those protons and their relationship, it might be hard for you to see why you get a trans double bond and only a trans double bond in the product. Let's go ahead and draw out the cis-decolon system, the two chairs. And this is, I mean, you just have to practice writing cis-decolon system so you know how to fuse the chairs together. We'll start off by drawing this equatorial substituent here. But instead of drawing another chair in a regular chair conformation, I'm going to draw two bonds going straight down. That's how you draw a cis-decolon system where the ring fusion is cis. And so now I can complete this by drawing another chair here like that. So that's called a cis-decolon ring system. And if I look at the ring fusion now, as I drew it up here, this hydrogen, this CH is still axial. This CH has to be on the top, or this O has to be on the top face as well. I'll draw that as an equatorial substituent relative to this first ring. And there's my O minus intermediate that I get when I deprotonate with t-butoxide. And now when I look at the orientation of the tosyloxy substituent, there's two substituents at this position here. There's a tosylate and there's an H. And there they are. The H is on the top face. The tosylate is going kind of in back. The H is coming more towards you. The tosylate is going farther away from you. Let me make this bond the same length as the other CC bonds. That wasn't essential. That was just bugging me. Okay, so let's draw these orbitals out again. Here's a non-bonding lone pair. It's donating again into this anti-bonding orbital for this CC bond. It's weakening that CC bond. That's the effect of having nucleophilic lone pairs donate next door to CC bonds. So these electrons are donating into the anti-bonding orbital for this CC bond. It weakens it, makes it more nucleophilic. And that more nucleophilic bond is now pushing out the leaving group by donating into the sigma star. And again, I want to look at the relationship between those two hydrogen atoms. So when I draw the product, I'm going to draw everything the same except I'm going to erase my middle bond there. I'm just going to erase this middle bond. If you wanted, you could have drawn that in very lightly or something. So if you're using a pen, if you're using a pencil, it's easy to erase. So this 6-6 ring system we call decalin. When it's got all single bonds in it, that's called decalin. Sometimes it's used as a solvent, a high boiling solvent. This particular one we call transdecolin because it has a trans ring fusion. One up, one down at the ring junction. This we call cisdecolin, a 6-6 ring system that has cis substituents at the ring junction. Okay, so let me draw out the oxygen. It's still sort of equatorial here. Of course, now it's a double bond in the product so it's a carbonyl. I didn't draw that really centered correctly. So there's the carbonyl you get as those lone pairs on oxygen kick around. Let me draw the CH that was pointing up. It's still pointing up. It started off cis, syn, it started off syn to the CH over here and so it's also that CH is going to stay cis to that or syn to that. And so that's why you get the cis-double bond. Okay, so grobe fragmentations are a way for me to test whether you see these anti-periplanar relationships. So of course on the problem set I'm going to ask you all kinds of grobe fragmentations. And no, I'm not going to draw it in the right conformation for you. You're going to have to do that. And yes, I love to put these kinds of problems on exams because it's a test of whether you're practicing drawing chairs and seeing anti-periplanar relationships. Okay, so that's it for eliminations. So practicing anti-periplanar relationships is about all I can say. So we're going to majorly switch gears to probably the second major theme of organic chemistry. We talked a lot about carbocation chemistry. Let's switch over to carbonyl additions. So let me just sort of get things situated here about what we've been talking about so far. In this class we've been talking about the addition of some sort of a filled orbital which could be lone pairs, pi bonds, or sigma bonds donating into empty orbitals. We started off by talking about carbocations, empty p orbitals. Those are the most reactive things you can attack. Those are the lowest energy orbitals, the empty orbitals that you can put electrons into. And so then we talked about sigma star orbitals. And SN2 reactions, elimination reactions. And now we're going to talk about things attacking pi bonds. Things attacking, donating into pi star orbitals. So when you attack CC double bonds. And that'll sort of complete our little tour through things that you can attack. And I want to start off by comparing pi bonds because all pi bonds are not equal most certainly. So I'd like to start off by contrasting two different reactions and we'll ponder on the relative facility we expect these two reactions to have when you set them up. So if you take some sort of alkyl lithium, I'm too lazy to draw butyl lithium so I'll draw ethyl lithium. It's the same idea. And so let's just imagine the reaction of ethyl lithium with carbonyl. That's not a bad reaction. I hope you all know that reaction from some earlier version of organic chemistry. But what about this? Aero pushing is exactly the same, right? I can push the same kinds of arrows. I just finished telling you how glorious pi star is at being attacked. Look at that. It's lower in energy. And so I don't know about you but I don't feel so good about this bottom reaction, right? I feel really good about the top reaction. I don't feel so good about the bottom reaction. I think you know that these kinds of additions of nucleolive alls of carbonyls, especially by alkyl lithiums and green yard reagents, those are always fast. There's no cases when those aren't, when this kind of addition that I've shown is not fast. You can add to simple CC, not substituted enone, just simple alkenes under certain conditions. And I'll tell you those conditions. When is this fast? Because intrinsically it's generally not something you worry about. It's fast if you have very nucleophilic alkyl lithiums. So it's not fast with ethyl lithium. That's a primary alkyl lithium. The same way the butyl lithium is primary. If you look at the product of this reaction, if you start off with a primary alkyl lithium, ethyl lithium, and you add to a simple olefin, get another primary alkyl lithium just thermodynamically. You haven't done a whole lot for this system. It just cost you something, an entropy, to bring these, a mole of these next to a mole of those. So you paid an entropic price and you didn't get anything out. You started with a primary alkyl lithium. You ended up with one. There is a trade-off for pi bond. That's worth a lot actually. But in the end, that's, even if it's slightly thermodynamically favorable, you really need to have a super nucleophilic alkyl lithium. We'll talk more about those, secondary and tertiary and more nucleophilic than primary alkyl lithiums. The other way to make it fast is use the most unsubstituted olefin on the planet and that would be ethylene. Now you're rarely going to bubble ethylene gas into your reaction, so, but if you're a polymer chemist, you might do that. You can make it fast by putting in super secret additives that have nucleophilic lone pairs on them. Trialkyl amines, those can poke their electrons into lithium and it makes this lithium carbon bond more nucleophilic. We'll talk more about that. And then finally you can use solvents present in very high concentrations. It's the same idea. There's just oxygen, lone pairs aren't very nucleophilic, but if it's the solvent, there's a lot of it. And those can poke their electrons into lithium and make the carbon lithium bond more nucleophilic. So that's what it takes in order to get you to add alkyl lithiums to simple C-C double bonds. It takes all this extraordinary stuff. Generally you don't worry about alkyl lithiums attacking simple olefins. So let's kind of get this out of our mind. This is special stuff, right? If you don't see those special circumstances, don't worry about alkyl lithiums reacting with simple C-C double bonds. Let me give you one example of, this seems, I would say, synthetically useful to me, how you can use entropy to help make these work. So I don't want to say, so it's not fair for me to say you can never see alkyl lithiums, which are about as nucleophilic as we can imagine, adding to simple double bonds. Here's a special case where somebody showed that you can take allyl alcohol and you can get alkyl lithiums to add to the C-C double bond. So what did they do? They used two equivalents of isopropyl lithium. So they're starting with a secondary alkyl lithium. So I think many of you have used butyl lithium before. It's not that common. I would doubt that any of you have ever used isopropyl lithium. You might have used sec butyl lithium. It's easier to get your hands on. But here's an example of one of those tricks that I showed you as switching to a secondary alkyl lithium. Of course, it's fast for the alkyl lithium to deprotonate the alcohol. So you're going to have in this solution a lithium alkoxide and there's a second equivalent of isopropyl lithium in there. And of course, lithium wants eight electrons, not just two, so it shouldn't be surprising to you that oxygen lone pairs will be donating into lithium atoms. I have to conserve my charges. If I use a lone pair on oxygen to attack lithium, if I have three bonds to this, I need to put a positive charge there. And if I have two bonds to lithium, I need to put a negative charge on there. And by drawing this Lewis structure correctly, it reminds me that this lithiate here, this negative charge on lithium soups up the nucleophilicity of all the bonds to that lithium. That negative charge tells me that all those bonds are more nucleophilic. And in particular, this bond right here is more nucleophilic. That's what that negative charge tells me. That's why it's so important when I say lithiate, cuprate, pallidate, I'm listening for that H sound. And so when this is coordinated, not when it's floating around in solution, but when it is coordinated, this bond gets souped up in nucleophilicity and now five atoms away is that olefin just dangling there, waiting to be attacked. I mean, that's the best kind of transition state. The number of collisions between this bond and that C-C double bond, they collide so much more frequently when they're five atoms away. That's what makes this reaction work. So normally you can't add, I don't want you to believe that you can normally add alkyl lithiums to olefins, but you see all of this stuff that's built into this system, that's what makes it happen. And of course, there's nothing else in the reaction mixture that can react. Pretty much anything else would react faster than the C-C double bond. Okay, so this ultimately swings around and forms a chelate, but that's not really important for the reaction, but I'll draw that. So that oxygen lone pair, I'll draw it there, just swings around. Okay, so that wasn't important. The important thing is that sort of intramolecularity effect, the fact that this brings in this very nucleophilic double bond. Helps it to collide many, many times more frequently. That's the effect of entropy. Okay, so let's go ahead and take a look at stereochemistry. And we're going to start off with the probably the most basic issue one can think of, and that is trajectories for addition to carbonyl groups. So I want to start off by saying something that I think is totally obvious, and that is carbonyl groups are polarized. Oxygen is electronegative, more electronegative than carbon. That bond is polar. And if I draw that resonance structure, it reminds us of why carbonyl groups are more electrophilic than CCN. In fact, let me make a point of this. Let me come back over here. So why are carbonyl groups more electrophilic? I'll just say more reactive, but I mean more reactive towards nucleophiles than just simple CC double bonds. Why is it? So the first reason has to do with this polarization. So there's two reasons, and they're not the same reason. Reason number one is a colomic effect relative to this. So if I compare the charges on these two types of pi systems, there's a substantial amount of partial positive charge on that carbon atom. There's also partial negative on oxygen. So from a strictly colomic point of view, if my nucleophile has any partial negative charge than they always do, the good ones do, then I should expect it to attack faster here. That's a colomic effect. Totally separate from that is reason number two. If I sketch out the energies of the molecular orbitals, the canonical orbitals here, for some sort of nucleophilic filled orbital, it can be like an alkylithium bond or some lone pair. It doesn't matter. And I compare pi star orbitals, for example, for a CC double bond. What happens if I replace a carbon atom in a CC double bond with a more electronegative atom? I should expect the energies of that orbital to drop. And if I go all the way down to oxygen, which is even more electronegative than nitrogen, I expect pi star to be lower in energy. If you were a pair of electrons, wouldn't you want to donate into this lower energy orbital more than the higher energy orbital? Of course you would. So there's two completely different reasons why carbon yields are so much more reactive than CC double bonds. One issue is charge and one issue is the energies of the frontier orbitals or lowers. Those are two completely different issues. You can guesstimate the relative proportions, the contributions of these two resonance structures by looking at dipole moments. So dipole moment, and I'm not sure what kind of technique they use to measure this, dipole moment is measured, is defined as a separation between two charges. So it's a distance times charge differential between charges. And if you measure the dipole moment for acetone, it's 2.7 Debye units. I have no idea what the heck of Debye, it's whatever. It's some charge unit times distance in angstroms. What you can, what that allows you to do is it allows you to say that there's the equivalence of plus or minus 0.23 charges on acetone. And that's a number you should keep in mind. Essentially what it's telling you is that if you think about the contributions of these two resonance structures, it's like 75, 25. It's like 25% of a positive charge on the carbonyl carbon of acetone. So that's one reason, one of two different reasons why nucleophiles are so adept at adding to acetone. So keep that charge issue in mind for acetone and you can extrapolate it to other types of carbonyl groups. Okay, I want to look at, so that's the charge issue. Let's come back and look at the issue of this pi star orbital. And what should we expect it to look like? And I want to, this will give us a chance to talk about the effects of electronegative atoms, not on the energies of orbitals, but on the shape of the orbitals. And we haven't spent any time talking about this, so this is going to be kind of new stuff. These are the frontier orbitals that we care about, the highest occupied molecular orbital, the homo, the lumo, things like sigma star and pi star. And so electronegative atoms distort the shape of the orbital's orbitals. And I'll show you how it does that for a carbonyl in just a minute. So I'm going to fall back on a term, and so let me explain that term because I'll find myself using it over and over and it will seem sort of mysterious. I'm going to use a term from molecular orbital theory. Maybe at some point in time you learned about something called linear combinations of atomic orbitals. It basically said take two p orbitals, you mix it together, you get a pi star molecular orbital out. If you take a 50-50 mixture, a 50% of one pi, a p orbital and another p orbital, you can make pi orbitals out of that. So coefficients, these are these things from linear combinations of atomic orbitals. What is a coefficient? It's like 50% p on this atom plus 75% s character from another atom. Those are the coefficients, the 50% and the 75%. So coefficients tell you about the relative size of pi and pi star on each atom. And I'll draw these out in just one moment and then we'll draw out a conclusion. So I want to draw out a diagram here, an MO diagram. And I'll draw all the energies. We've been dealing a lot with energies. I've been talking endlessly about the energies of unfilled orbitals and filled orbitals. And now I want to talk about the shape of these orbitals. So if I take a simple carbonyl group like acetone or some aldehyde, it doesn't matter. What would I expect? I'm going to have to have two different pictures of the orbitals here. So let me start off by drawing, and I don't care that there's a double bond there. I'll draw in the orbital in a minute. So I'm going to draw two pictures of acetone. And I'm trying to draw this kind of edge on. I don't know if you can tell and kind of distorting it because I want to see sort of side on this pi and pi star orbital. So if I depict a pi orbital as arising from a contribution of two p orbitals mixing together, when the orbitals mix together in phase, I get a pi bond. That's where electrons want to be. They want to be in a bond. And here's what I would see if I looked at those two p orbitals that I'm running out of room to draw here. So you may have, I'll use that bottom there. Here we go. So here's the contribution from the pi orbital on the carbon atom, sorry, p orbital. And there's another p orbital in oxygen that it combines with in order to make that p bond. And what I would find is, and I apologize, you can change your drawing appropriately. I made that too big. There we go. And I'll phase that. What I would find is something that if you think about it is totally intuitive. And that is that if you look at the bonding orbital, the coefficient, the size of the pi orbital is bigger on the more electronegative atom. Now why is that? If I were a pair of electrons, I would want to spend my time closer to the electronegative atom. There's more protons in the nucleus of oxygen. It makes total sense. That's totally intuitive to me that if I were a pair of electrons, I'd want to spend more time over the oxygen. So that's the shape of a pi bonding orbital on a carbonyl. It's distorted to be bigger on oxygen. Here's what's not obvious. I don't think it's obvious. If you look at pi star, now when you combine orbitals together, two different p orbitals together to make an antibonding combination, we change the phasing. So now I'll phase this one downward. And what's not so obvious is now in the antibonding orbital, electrons don't want to be on the electronegative atom. It's the opposite of everything you think about bonding for the pi star orbital. It's the opposite. The phasing is opposite and the coefficient sizes are opposite. And there's powerful, powerful implications to this drawing. The implications are this. The implications are if you are some sort of a nucleophile trying to add into a carbonyl. And let me draw a pi star. If I've got a pair of electrons and I'm trying to donate into an orbital, no, don't donate here. It's filled. Donate into the empty orbital. I'm going to want to try to overlap with my filled orbital somehow with this picture. And what's the area that I can overlap with? Well, I want to overlap where the coefficients are big, where the orbital looks big. But phasing is important. Correct phasing leads to a bonding combination. Incorrect phasing leads to a non-anti-bonding configuration. So just imagine if you will a nucleophile coming in. Let's make it come in right here. Now let's suppose I make the phasing hashed. Well, that's going to suck. Here's my nucleophile. It can be like a lone pair, it doesn't matter. Because I'll have this anti-bonding common. I'm not going to form a bond like that. Right? I'm going to end up with an anti-bond. You need to make your phasing match. So if I just happen to some extent to this is arbitrary, how I chose to phase the nucleophile. According to my depiction, what I want to do is I want to come in like this. I don't want to come in over here because then I'll be overlapping with this and have an anti-bonding interaction with the oxygen. I want to steer away from that. And I don't want to come in from the back end because then I'll be interacting with this other unhashed lobe. The ideal trajectory is to come in from an angle so that I can skip the anti-bonding interactions with these two lobes. I'll just write here that's bad. That stuff leads to an anti-bonding combination and this is a good interaction. And so what's the preferred angle here for attack? And this was no longer, the preferred angle for attack was figured out long before people understood the molecular orbital origins of this trajectory. Jack Dunnitz is a famous Swiss crystallographer who worked closely with chemist Hans Berge. And Berge made lots and lots of organic compounds, lots of 9, 10-membered ring compounds. And Dunnitz would go get crystal structures of these. And Berge tended to make compounds that had all kinds of atom strain in them where atoms were just uncomfortably close to each other. Some of them looked like this and maybe I'm not doing a good job here. But this would be an example of a way to design a compound where atoms are just uncomfortably close to each other. They're getting in each other's face. And what Dunnitz noted because all he had were crystal structures to look at, crystal structure after crystal structure after crystal structure is when he looked at the distances between the nitrogen and the carbonyls in these compounds, what he noted was that in crystal structures were the nitrogen and the carbonyl carbon were close to each other that the carbonyls were no longer planar. He noted that there was a distortion in the carbonyls towards tetrahedral geometries. And he plotted these out. What he plotted out was the angle, this NCO angle here. And what he was able to show is that the closer and closer the nitrogen and the carbon got to a bonding distance, the closer that reached to 109 degrees. And this is now referred to as the Berge Dunnitz angle. You'll recognize this as the angle between bonds in methane or in any tetrahedral structure. So nucleophiles don't want to come in and attack carbonyls from the back end or directly from the top. They prefer to come in from, there's my, the bonding angle. They prefer to come in from an angle of 109 degrees. That's the Berge Dunnitz angle. If you want to analyze trajectories that lead to stereoselective additions, you have to recognize that all nucleophiles have a preference to attack carbonyls, pi star in general. They have a preference to attack pi star from 109 degree angles like this. So that's the Berge Dunnitz angle. Okay, so let's see how that plays out in analyzing stereoselective additions. So I'm going to show you how to predict stereoselective additions. And let me start out by giving you an example of a stereoselective addition. Let's just imagine taking a nucleophile and adding it to this carbonyl, but a carbonyl where there's a stereogenic center right next door. And it doesn't have to be next door, but that's where you'll see the biggest effects. Okay, so there's a stereocenter next door. That stereocenter is going to influence which face of the carbonyl you add to. And this is a problem that people first started to deal with in the 1950s. So if I take lithium aluminum hydride that donates hydridogrups, it's going to form a new CH bond to the carbonyl. You don't get equal addition from both faces. So some of the hydride groups add, this is after workup, from the front face, and then some of the hydride groups add from the back face. And it's about a 3 to 1 mixture. It's like what? 0.5, 0.6 kcals per mole difference in transition state energies. So 2 to 1 would be 0.4 kcals per mole. So this is 3 to 1. Okay, so there's something that's distinguishing the facial selectivity. Don Cram at UCLA was the first to come up with a rule that explained almost all the additions that were known. Let me not say almost all the additions. I'll say over 80% of the examples out there. So what he did is he tried to analyze so there's a stereogenic center here. What Cram said is just orient yourself like this so you can view down this bond. And that way if you imagined looking down that bond at a Newman projection you could imagine how these three different groups disposed on different sides of the carbonyl might influence one trajectory versus the other. And so the Cram rule said, or the Cram model said, assume that if you draw a Newman projection that the medium size group that would be methyl in this case is sticking on one side. Well the important part of his rule is to say assume that the big group, in this case the phenyl is sticking away from the carbonyl. So assume that the big group phenyl is flipped downward away from the carbonyl and that leaves you with two other groups, a medium and a small one. And according to Cram's model that medium group and the small group would now distinguish addition from one face or the other. That if you put the big group down away from the carbonyl that now you would expect the nucleophile to attack on the same face as the small group because there would be less of a steric interaction. And that explained most of the cases, 80% of the cases. But obviously if it doesn't explain all the cases then there's something wrong with the rule. The rule doesn't work every case. And so Cram's rule is now known to be incorrect. This is not the ideal conformation in the transition state. And so let's talk about the transition states and how to think about them so you can correctly predict. Which face is preferred for addition? And so here's what, here's, there's four different factoids that you need to know in order to understand how to analyze this. And so let's start off with factoid number one. And this is really came from Cram. So this is what Cram got right. What Cram got right and which was almost a revolution for him to be able to predict 80% of the cases was he said stop looking at the products. Stop drawing out Newman projections that look like this with an oxygen atom here and a nucleophile here and some group there and small. You can look at the products as much as you want and you will never be able to predict by looking at the products what's the favored direction of attack. The products don't tell you what the transition states look like when you add strong nucleophiles like alkylithiums or a lithium aluminum hydride the transition states are early. They look like carbonyl groups, not like tetrahedral atoms. So that was a revolution. So second factoid that helps you to predict things was that the big group that's alpha to the carbonyl doesn't want to be opposite the carbonyl group, it wants to be on the, as far away as possible from the nucleophile. So I'm going to go ahead and draw out two different Newman projections. And this is the way you should always analyze stereoselective conditions. So what, the correct way to analyze this is that the big group wants to be on the opposite face. Ninety degrees from the carbonyl, not down here, the big group wants to be on the opposite face. There's two different ways that the big group can be on opposite faces of the carbonyl. One would be if it was swung over on this side. So one would be if it was swung over on the left and the other would be if it was swung over on the right. So these are the two confirmations, the reactive confirmations that put the big group on the completely opposite face of the carbonyl from where the nucleophile is coming in. So that's number two. Number three, as you don't attack at 90 degree angles directly away from the carbonyl, you attack at the Berge-Dunnitz angle. You don't come in directly opposite the big group. You come in at a Berge-Dunnitz angle. Let's draw that trajectory in. So in other words, the nucleophile is going to come in and start forming a bond with this 109 degree angle. There's 109 degrees. It's got a kind of a lot of lines going on there. But you see the idea, and so that means the nucleophile is going to come in, if the big group's over there, the nucleophile wants to come in at a 109 degree angle. Those are the two competing transition states. And now it becomes very easy to distinguish these because if I fill in this medium and small related to those cases that I showed above, now I've totally distinguished the good from the bad transition states. So if you were a nucleophile, would you rather be eclipsing almost directly on top of the small group? Or would you rather be coming in directly on top of the medium group? It's the small one. I mean it's almost totally eclipsing with this where they're bumping into each other as bad as you can possibly imagine. And so this is the preferred transition state. Let me just put little explosion lines here so you can see them knocking into each other. So this is the bad transition state. And this is the good transition state. And so the last part of the analysis, you want to minimize eclipsing interactions. So if I ask you, and I will ask you, to predict the stereoselectivity every single time, I want you to draw two different Newman projections. Don't just draw one, draw two. Draw the good one and the bad one. Label one is good, label one is bad. You can use any term you want, crappy, great, whatever, but draw two Newman projections. Okay, so that's how you analyze stereoselective additions. There's one last factoid that's not obvious. Sorry, that's a permanent marker up there. We can't erase that. And this is kind of like rule number five. So if you have a polar group on your alpha position, it acts like the big group. And let's talk about why. So in other words, if you have an alkoxy group at your alpha position, your nucleophile wants to add opposite the alkoxy group. Even though the alkoxy might not be as big as an ethyl, the important point is that it's polar. Treat polar groups like methoxy, chloro. Treat those like they're big groups. So in this case, the preferred trajectory of attack. And I don't have where you can draw the other Newman projection yourself, but in this case, this would be the better Newman projection. That's the preferred trajectory. And the least preferred trajectory would be if the nucleophile attacks the opposite side. Okay, let me try to explain why that's true. Now I'm going to draw all this out so that you can see anti-pyrriplanar relationships. As I'm coming in with my nucleophile, to add into this carbonyl group. If I have this polar group here, it could be something like chloro or methoxy. The important point is my nucleophile is coming in with its filled orbital. If you can maximize overlap between this polar bond, carbon chlorine, carbon oxygen, if there's an empty orbital here, the nucleophile wants to interact with that. And it lowers the transition state energy. It's not a steric interaction, it's the interaction between filled and unfilled orbitals. Yes, you're donating into this unfilled orbital. If you can also donate into this unfilled orbital, super important. That's the preferred transition state. Okay, so that's how to predict stereoselective additions. I'll give you one last type of stereoselective addition when we come back. And lots of problems for you to work practicing these tools.