Addition of Organometallic Reagents to Carbonyls
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King, Susan
This is the third (and final) quarter of the organic chemistry series. Topics covered include: Fundamental concepts relating to carbon compounds with emphasis on structural theory and the nature of chemical bonding, stereochemistry, reaction mechanisms, and spectroscopic, physical, and chemical properties of the principal classes of carbon compounds. 
 All right. So we're still talking about irreversible additions to carbonyl compounds. And so we just started last time talking about organometallic reagents where we have carbon alkyl groups bonded to metals. And because we bond the alkyl groups to metals, we change the polarity of the bond. So this is what we're used to. Same groups like this where carbon's bonded to group that's more electronegative, so electrons flow this direction, which means carbon's partial positive. But if we change that to a metal, and the metals from the left-hand side of the periodic table are going to have electronegativities lower than this. And so the electrons now flow the other direction, making us have electron-rich carbon. And so there's a bunch of these compounds. We're only going to talk about three. So we're going to talk about compounds that have a carbon-lithium bond. We're going to talk about compounds that have a carbon-magnesium bond, and compounds that have a carbon-carp bond. So these are the only three that we're going to talk about. So let's check out electronegativities here and see if we can figure out which of these reagents is going to be more. Which of these bonds is going to be most polarized, and which one's going to be the most reactive? So carbon's going to be 2.5 all the way across for electronegativity. Lithium is 1.0, magnesium 1.2, and copper 1.8. So if we look at the electronegativity difference, we can see which of these bonds is the most polarized. So let's do electronegativity difference here on the left-hand side. The end difference. Here it's 1.5. Here it is 1.3. And here it is 0.7. So you can see that definitely the most polar bond is going to be the carbon-lithium bond. So it's polar because electronegativity difference is the largest. And so that also means that this is going to be the most electron-rich carbon for comparing all three of these species, most electron-rich carbon. So we have electrons flowing towards carbon here, but we have the most electron density flowing towards the carbon when we have a carbon-lithium bond. All right, I want to say a couple of things because there's a tendency to think of these bonds as ionic bonds or not ionic bonds. If you go back to chapter one, we sort of talked about the fact that once the electronegativity gets greater than about 1.7, 1.8, then electrons are transferred completely and we have an ionic compound. But this is not. This is the largest difference we have is 1.5. So that is considered to be a carbon-lithium covalent bond. And so these are actually quite different here. So about covalent carbon metal bonds, the reactivity towards electrophile depends on the polarity. So organolithium are going to be the most reactive, greater than organomagnetium, greater than organocopper. Now for the most part, we're going to be using organolithium compounds and organomagnetium compounds interchangeably, but there is one reaction where you need a super great electrophile that will work for organolithium but not work for organomagnetium compounds. And that's coming up. So let's label these certainly, and I want you to know this, that organolithium are the most reactive and organocopper least reactive. So we're going to have three different organometallic reagents. There's organolithium and organomagnetium. Organomagnetium have a special name and that's Grignard. So they're Grignard reagents. And this is how you make them. We have Rx, I don't know why that's cut off there, it looks like it's a little cut off, plus 2 lithium, and you get Rli plus Lix. So here's an example here. Okay, we're at that with lithium. Now I'm not going to be worrying about stoichiometry on the exam, you can just write lithium, you don't have to write 2 lithium. There's your organolithium product. And you can leave out stoichiometry and side products. So I don't care about the stoichiometry and I don't care about the side products. Alright so that's an example of organolithium compound. Grignard reaction has a little bit different structure. So we take Rx plus Mg and we get Rmgx. So the halogen from the Rx actually becomes part of the compound, so it looks a little bit different here. And it turns out that you must use an ether solvent for this reaction. Organolithium don't need to use ether, but you definitely need to for Grignard. Here's an example right here. Mg ether. And your product is Mgbr and this is a Grignard. You can also call it organomagnetium, but it's definitely much more commonly called a Grignard reagent. And the ether is there because you actually get something that looks like this. Where you have your Grignard and then you have the ether. The lone pairs on the oxygen of the ether are sort of donating electrons to the magnesium. Don't need to worry about this, but this is why you need the ether solvent. Alright, organocopper. I have a completely different structure also. So let's go, let's look at that on the next page. Organocopper reagents. So also known as cuprate or Gilman reagents. I will call them cuprates because that's how I learned it. But there's also these other names. They are prepared from organolithium reagents by reaction with a copper salt, usually copper iodide. So here's what it looks like. Here's what it looks like. Rx plus 2 Li gives you, there's your lithium reagent, right? So we did the same reaction we did on the previous page. And then you add copper iodide. Notice you have two R groups here. R2CULI. So the copper stays, the lithium stays. And that's a cuprate. So two steps. Number one, make organolithium. Number two, add a copper iodide. Okay, so that's how you make those. Let me give you an example. And you could use a vinyl chloride for this reaction. So this is what it would look like. You add the lithium, you don't need to put the two in front of the lithium. I'm not worried about that. Then you add copper iodide. And you get two R groups bonded to the copper. And then you have CULI. Just like that. All right, so those are the three reagents. Need to know how to make those three reagents. Grineer reagents are the form RMGX. And you can have bromine, chlorine, or iodide. No fluorides here. Organolithium is RLI. And organic copperate is R2CULI. Need to know that. They react as if they were carbonyl ions. They are not true carbonyl ions because they have covalent carbon metal bonds. However, we can think about them conceptually as carbonyl ions. So a carbonyl would be an ionic compound. And these aren't. This is a covalent bond. But that's the way they react. Compare this with sodium acetylide. And we have here carbon 2.5 and sodium 0.9. But this guy's actually ionic. So it's right on the borderline here. It's actually ionic. So conjugate base of a terminal alkyne. This is negatively charged. And then we have Na+. So they're going to behave in a lot of the same ways. But they're not exactly the same because this is an ionic compound. So here we have electronegativity differences 1.6. So somewhere around 1.6, 1.7 is where you start getting into ionic compounds. So far, anybody? All right. Let's talk about reactions of organometallic reagents. Here are some examples. The organometallic reagents are powerful bases. Extremely powerful bases. They react rapidly with protons in an acid-base reaction. So this is what it looks like here. This is an acid-base reaction. And we'll draw out the reaction and then we'll see which way the equilibrium is favored. So there is going to come from the carbon magnesium bond. We're going to grab a proton and then we're going to break the hydrogen-oxygen bond. Very rapid reaction with water. So there's the hydrogen that we just added on. Let me mark that. That's the hydrogen right here. So you're going to get that plus HLMGBR. You can write it like that or you can write it like you get a hydroxide, you get an Mg2 plus, and you get a Br minus. So that's what happens when you hydrolyze a griniard. Let's look at the direction of equilibrium. So we're going to look at for conjugate acid-base pairs. And what do we have here? So we have water and we have hydroxide over here. Conjugate acid-base pairs right here. The one with the extra proton is the acid. So here's our acid on this side of the equation. And here's another conjugate acid-base pair. Here's our acid on this side of the equation. This has a pKa of about 15 and this has a pKa of about 50. So as you can see, equilibrium very strongly favored to the right. So long arrow to the right, tiny, tiny short arrow. Really there is no reverse reaction for this. But we'll just draw it out like that anyway. All right. So same thing with organolithium reagents. And so you've got to be careful with these guys so that you don't have proton sources around. If you have proton sources around, then you will hydrolyze. So proton sources can be water, could be acids, any acidic protons. So there we just put a hydrogen in. And then we get a carboxylate. All right so it's easy to forget about that, but that's the fastest thing that's going to happen with these guys. If there's a proton source, they will react with it. And direction of equilibrium, let's do that again here. So here's our acid on this side of the equation. Here's our acid on this side of the equation. This is pKa about 5. This is pKa about 45. So once again, very, very strongly favored to the right. So try not to forget that. It's really easy to forget when you're doing synthesis. We get so caught up in the carbonyl that we forget about these acidic hydrogens that kind of interfere with what we're trying to do. All right so this rapid reaction with water or any proteic acid means that Grineard reagents or Genolythium cuprates cannot be prepared from compounds that contain acidic groups. If you want to make a Grineard and your compound has an acidic group, you can't make a Grineard from it. So here's the acidic groups that interfere. Anything in here? There's an acidic hydrogen right here, a hydrogen on a nitrogen. Here's another acidic hydrogen. Hydrogen attached to sulfur, carboxylic acid, acidic hydrogen, even a terminal alkyne, hydrogen on a terminal alkyne, you can't make Grinears from these compounds. So don't make a Grineard. Make a Grineard or Organolythium compound from compounds that contain these groups. All right, so that's super important and very, very easy to forget. As I will tell you when I grade exams, I will see a lot of people forgetting about that. All right, so here's an example of a failed synthesis. This person is trying to make a Grineard from this compound here. So remember when we make the Grineard, the magnesium goes in between the carbon and the bromine bond. So we have here Mg, ether. You actually will make a little bit of this, but as soon as you make it, it's going to be protonated by any acidic hydrogens that are in the medium. So another molecule here, definitely. So you make that and then any other proton sources and any of these other compounds here. So air comes from the carbon, magnesium, bond, grabs the hydrogen and looks like that. And then so what you end up getting is ethanol, not what you wanted. You wanted to make a Grineard out of this compound. So you're going to get that and you're going to get deprotonated alcohol. So you can draw it like this or you can draw this like this with Br, Ch2, Ch2, Omg Br. You can write it either way, but that's what you're going to get. So that's a big problem. That's not a Grineard reagent. So these guys are the actual products. So this rapid reaction with water and any protochastic can be used to synthetically turn an alkyl halide into a hydrocarbon. If you use a labeled proton source like D2O, you can actually make deuterated compounds. So here's an example right here. We have no acidic hydrogens here, so we will be able to make this Grineard rather nicely here. So the magnesium ends up being going in between the carbon and the bromine. And then if you protonate this, if you actually deliberately add water, you will protonate that. So that's a way to actually get rid of this bromine in our molecule. So that's what you would get. If you use D2O instead or some sort of proton source that has deuterium, then you'll get CH2D. So there's the deuterium. Let's circle that deuterium from one of these bonds here. And now you've made it a deuterated compound. And that's just by protonating the Grineard. Once you form it, you protonate it. And if you use a deuterated proton source, you'll get a deuterium incorporated. Question so far. Anybody? All right. So gannomatocryogens are powerful nucleophiles. They're very important in organic chemistry. They can react with epoxides to form alcohols. The result is an extension of the carbon chain by two carbons. So what am I talking about? Let's keep an eye on this number here. So that's going to keep an eye on that. We're going to come back and circle those when we come back. All right. So we're making, again, a lithium compound. So alkyl lithium. All right. And now we're going to open up an epoxide. So this is going to attack an epoxide. It's going to be very similar to lithium-lunium hydride attacking an epoxide. Arrow's going to come from the carbon-lithium bond. If this was a substituted epoxide, it would attack on the least substituted side. Let's circle that original group that we started with. Here's our three carbons right here. Here's our three carbons right here. And you see how I've added on two carbons. I've added on a CH2, CH2O minus. Once we protonate that, it will be a CH2, CH2OH. And that's something you might want to tuck away. You might see that on a future exam, where you have to add on a two-carbon piece that looks like that. We're going to talk about other ways to add on two carbon pieces, but this will add on a CH2, CH2OH. Two carbon extension where the functional group is on the end. We had another two-carbon extension in Chapter 11, where we took acetylene, we deprotonated it, and we attacked an alkyl halide. That was also adding on a two-carbon piece. Just different functional groups. So good thing to tuck away. What I think would be a good idea right now is if you have an index card or a larger index card or a sheet where you start to now write carbon-carbon bond forming reactions. We have one from 51 B, and we're going to now have granules or, again, lithiums and cuprates here for 51 C. So those are great when you're trying to build up carbon skeletons. All right. Longer extensions can be made by using substituted epoxide. So here's another example here. Which side is this going to tuck on left or right? Well nucleophile takes on the least substituted side, right? This goes back to Chapter 9. All right. So that's what's going to happen here. It's going to come in attack on the least substituted side. We're going to kick electrons up onto oxygen. That's at the very end of Chapter 9 if you want to take a look at that. The after protonation. And there's our phenyl ring. So we've added on actually a four-carbon piece where the hydroxyl is in the two position. So that's just something to keep in mind here. We've added on a one, two, and then this is three and four if we drew out that ethyl. Added on a four-carbon extension. And so just remember that this is, of course, a strong nucleophile. That's at the least substituted side of the epoxide. All right. So grignards can be protonated. They can attack epoxides. But the most important reactions that we see with grignards are grignards attacking carbonyls. Okay. So let's look at some examples of that. Let's look at some examples with type one and type two carbonyls. So here's a typical reaction with the type one carbonyls. So we have ethyl, magnesium, bromide, or ethyl-grignard. Arrow's going to come from the carbon-magnesium bond. These are going to look just like lithium-luminum hydride. Okay. So we're going to do a very similar way. Arrow's going to come from the carbon-magnesium covalent bond. We're going to attack the carbonyl carbon. We're going to kick electrons up onto oxygen. So this will look really familiar to you. It looks just like the mechanism for the lithium-luminum hydride with the type one carbonyl, except rather than transferring a hydride, we're actually transferring an alkyl group. So we're building up carbon skeletons. But still the same thought process in that we have tetrahedral intermediate with no leaving group. Therefore it stops here until you add acid in a second step. So really key here is that we want to have the water in a second step. What happens if we mix the grignard and the water? What's going to happen? We're going to kill our grignard reagent. A word that you'll hear used is we're going to quench it. We're going to kill that because the acid-base reactions are going to be faster. And so then our grignard will be gone and then we don't get anything that we want. So add water in a second step. So mechanistically very similar to the lithium-luminum hydride. All right, question about reactions with type one carbonyls. Anybody? All right, so we can agree, organolithium is the same. Like I said, most of the time we're using greenery or xanoganyl lithium interchangeably except for one reaction. So same thing happens here with organolithium. These guys are more powerful, but in this instance it would do the exact same thing. And then of course we're going to add water in a second step. And same thing as when you're using lithium-luminum hydride. You cool the reaction vessel down in an ice bath and then you cautiously add a drop of water. You wait for the exothermic reaction to go away and then you add more. So it's really very, very extremely strong base here. All right, so that's type one carbonyls, organomagnetium, organolithium cuprates. We're going to talk about it a little bit later because they do different chemistry. They're sort of their own category. Sodium acetylide, we've talked about sodium acetylide reagents. We know that they can open up epoxides. They can also attack carbonyls. So familiar reagent, unfamiliar reaction, but you probably predicted that something like this would happen. I'm not worrying about stereochemistry because we have achyro reagents reacting with achyro reagents. So it's going to look like that. Okay, and then we add water in a second step. And notice something also, when we use acetaldehyde, we're adding on a two-carbon piece. But it's different than the two-carbon piece we do with an epoxide. So we've also added on two carbons. So let me circle what we started with. So this is one, two. This also adds on a two-carbon piece. What's the difference between this two-carbon piece and the one we did with an epoxide? They're not the same, right? Anybody have an idea? Excuse me? No, we're at, well, we started with this. We're adding two carbons onto this. It depends on the perspective. You're looking from the perspective of the acetaldehyde. Yes, we're adding on three pieces. But if we're starting with this and we're adding on two pieces right here. So the only difference here between this is, remember when we added on the two-carbon piece with an epoxide, it was CH2, CH2OH. The OH was on the end. Here the OH is right where we bonded directly to the carbon of our nucleophile. So the only difference is the location of the OH. Sometimes you want to use an epoxide. It depends on what you're doing afterwards. It depends on what you're trying to synthesize. But I want you to point out that when you use acetaldehyde, you're also adding on a two-carbon piece. It's just the only difference is where the hydroxyl is. So I'm just trying to point these things out to you because they're going to be relevant when you do start doing synthesis. All right, let's look at a typical reaction with a type II carbonyl. All right, and I'm not going to show you the product because we're going to work through what the product is. Another possible mechanism for midterm one. Aero's going to come from the methyl carbon magnesium bond. We attack the carbonyl carbon. We kick electrons up onto oxygen. So let's just kind of follow it along. We know leaving groups. We know things that can be leaving groups in this reaction. We know things that cannot be leaving groups in this reaction. So here's our tetrahedral intermediate. Does it have a leaving group? Yes, tetrahedral intermediate has leaving group. If it has a leaving group, it's going to leave. So electrons on oxygen are going to come down, and we're going to kick off our leaving group. Now we're going to look at the product that we get, and we're going to make a decision of something more is going to happen or not. All right, so our product is acetone. So it's a ketone. Is a ketone more reactive or less reactive than an ester? It's more reactive than an ester. So that means the griniere is going to add again. Can't stop here. Why can't you stop there? Ketones are more reactive than esters. All right, so what that means is that the griniere is going to add again. All right, arrow comes from the carbon magnesium bond. Now we're attacking a type 1 carbonyl. We know what's going to happen next. We know that since it's a type 1, we are not going to have a leaving group. Absolutely no possible leaving groups here. All right, so our tetrahedral intermediate has no leaving group. Therefore, it's going to stop here until we add a proton source, and we're going to add water first. So conceptually, the same thing is happening here. As was happening with lithium-aluminum hydride. We're doing the same thing, we're adding more than once here. And so we certainly have to add H2O in a second step. And then when we do that, we get our alcohol. So this is also a synthesis of alcohols. So there's our product. We figured it out ourselves here. And so the way that I want you to approach these problems is rather than just memorizing products, I want you to go through this thought process and figure out what the product is yourself. So it turns out that you must use two equivalents of Grignard. If you think you can trick it and only use one equivalent and get that product cleanly, you're wrong. What you will get is that product and unreacted starting material. OK, so if you only use one equivalent, you will get 50% conversion. All right, now I don't remember specifically what kind of sapling problem, but there was a sapling problem with either this or lithium-lunar hydride. And they asked for all the organic products. And so you just got to remember that right here, I just kicked off a methoxy group. So I'm also going to get methanol as a product, aren't I, when I start with an ester. So you want to keep that in mind. So you're going to get this plus this methoxy group that you made is going to get protonated. Now I don't normally worry about that side product, but it could be something that you would have to worry about if you were actually doing the reaction. So you're going to get that product and you're going to get methanol. So two products. Questions on inability to stop at the ketone. Anybody have trouble with that? I think one of the common misconceptions that I see in organic chemistry is people think, well, if I just use one equivalent and everything reacts simultaneously, then it should be able to stop at what I want. And it's not that way because these reactions are not happening simultaneously. So in other words, if I lined up all the esters in a row and then I lined up all the griniards right next to them in a long line and I said, OK, on your mark, it said go. Then they all reacted simultaneously. Then I would be able to stop at the ketone, but you and I know that's impossible to do. They're reacting a little bit at a time. And as soon as you make a more reactive product, it's going to react faster with the griniar that hasn't reacted yet. So I know that's tricky, but that's just the way it works. All right, questions on that aspect? Anybody? We like to think we're in charge of these molecules, but we're really not. They're going to do what they're going to do. As any grad student will tell you. All right. OK, let's do an example with an acycloid and organolithium reagent. Same idea here. Let's go through the mechanism. So another possible mechanism for midterm one. We've got a lot here. Do you remember to put lone pairs on all your reacting atoms when you're doing a mechanism? So let's see what we get here. So rather than just writing what the answer is, I like to go through it so you can think your way through to the answer. All right, tetrahedral intermediate has leaving group. So electrons on oxygen come down and kick off that fantastic leaving group. Chloride's a fantastic leaving group. All right, so then let's look at the product and let's make some decisions here. What's more reactive, an acycloid or a ketone? Acycloid. So theoretically, we should be able to stop here. The problem is the same problem we had with lithium-luminum hydride. It's too reactive to stop there. It's just way too reactive. So it's going to add again, unfortunately. Or maybe fortunately, if you want that product. So it's going to add again. So I will label that very clearly. You should be able to stop here because an acycloid is more electrophilic than a ketone by far. The problem is that granules are too reactive. All right, so you can't stop here. So these powerful reagents are just too powerful to stop. And so it's going to add again. And so what do we get from that step when it adds again? Gosh, we get the same thing that we get with an ester. We get the same exact thing. Oops, I skipped a step, didn't I? Let's fix that. So you get an alkoxide. Same alcohol you get with an ester. Tetrahedral intermediate has no leaving group. Therefore it stops here. And then when you add water, we protonate and we get the same exact product. All right, so there's two examples of reaction with type 2 carbonyls. Yes? Yeah, right. How about that? Organolythium. Thank you for catching that. They're so interchangeable. I'm interchanging the names and I shouldn't do that. More questions? Anybody? All right, so granules will also add twice to acid chlorides because they're two reactive cuprates which are the least reactive organometallic reagents will add only one. So cuprates do their own chemistry. Think of them in their own special category. Okay? This reaction only works with acid chlorides. Only works with acid chlorides. No other carbonyls. Okay, so cuprates having copper. Now copper is a transition metal and so they do have their own different chemistry that's completely different than granules and again, allithium compounds. What they do, they will react with acid chlorides to make ketones. Now I'm going to draw arrows for this. This isn't really how this reaction takes place but this is a way you can think about it. Cuprates, again, they do their own chemistry and it involves radical reactions but I'm not going to show you that. I'm just trying to give you a conceptual basis here but the idea here is that these are going to attack the carbonyl carbon and they, remember these were the least reactive of the three so they will only react with acid chlorides. And since they only react with acid chlorides, when we form the ketone here it's not going to react again. So we have tetrahedral intermediate has leaving group. So leaving group is going to leave, electrons come down, kick off the leaving group and since cuprates don't attack any other carbonyls but acid chlorides, it stops here. So if we want to stop at the ketone we have to use an acid chloride with a cuprate. If we use an acid chloride with a granule or an allithium it will add twice and give you the alcohol. So that's the way to do this. Alright and let's just label this when you go back. I won't ask for cuprate mechanisms. This is not the true mechanism. Way too complicated the true mechanisms, not completely 100% understood and way too complicated for this class. Okay yes. That compound there with an aldehyde on one side and an acid chloride those are very unstable so I don't know if you can actually do the cuprate reaction. Do you know? They're very unstable. They decompose into carbon monoxide and hydrochloric acid so I don't know whether, I don't know the answer to that. Yeah, yeah. Yes. Okay. What about carboxylic acids? Will they work in this reaction? Well remember what we don't want to forget. We're soaked tapped into the carbonyl. We don't want to forget about the acidic hydrogen. Don't forget acidic hydrogen. Alright, so the first thing that's going to happen these guys are extremely powerful bases. We're going to deprotonate that acidic hydrogen. Alright and we know something about carboxylates. We know that they're the least reactive of all the carbonyls. So the question is are carboxylates, and you look at a carboxylate you normally think of an electrophile as having a partial positive charge or a full positive charge, right? This has a negative charge so it's kind of strange to think of it as an electrophile anyway. But the question is a grignard, our side product is methane by the way, that's our side product, that's methane gas which is going to bubble off the reaction mixture. And the question is if we have an additional equivalent grignard, is it strong enough to attack a carboxylate? And the answer is no. So methyl magnesium bromide is not powerful enough, so it's powerful but we've come to the limit of its strength, not powerful enough to attack a carboxylate. Okay so if it's not powerful enough to attack a carboxylate, which is what is powerful enough to attack a carboxylate? Organolethium. So this is the one reaction where they do have a different something happening. Organolethium is more powerful so if we want to have that happen again we'll talk about what happens next time. So Steve you could try to figure it out on your own.
 All right.  So we're still talking about irreversible additions to carbonyl compounds.  And so we just started last time talking about organometallic reagents where we have carbon  alkyl groups bonded to metals.  And because we bond the alkyl groups to metals, we change the polarity of the bond.  So this is what we're used to.  Same groups like this where carbon's bonded to group that's more electronegative, so  electrons flow this direction, which means carbon's partial positive.  But if we change that to a metal, and the metals from the left-hand side of the periodic  table are going to have electronegativities lower than this.  And so the electrons now flow the other direction, making us have electron-rich carbon.  And so there's a bunch of these compounds.  We're only going to talk about three.  So we're going to talk about compounds that have a carbon-lithium bond.  We're going to talk about compounds that have a carbon-magnesium bond, and compounds that  have a carbon-carp bond.  So these are the only three that we're going to talk about.  So let's check out electronegativities here and see if we can figure out which of these  reagents is going to be more.  Which of these bonds is going to be most polarized, and which one's going to be the most reactive?  So carbon's going to be 2.5 all the way across for electronegativity.  Lithium is 1.0, magnesium 1.2, and copper 1.8.  So if we look at the electronegativity difference, we can see which of these bonds is the most  polarized.  So let's do electronegativity difference here on the left-hand side.  The end difference.  Here it's 1.5.  Here it is 1.3.  And here it is 0.7.  So you can see that definitely the most polar bond is going to be the carbon-lithium bond.  So it's polar because electronegativity difference is the largest.  And so that also means that this is going to be the most electron-rich carbon for comparing  all three of these species, most electron-rich carbon.  So we have electrons flowing towards carbon here, but we have the most electron density  flowing towards the carbon when we have a carbon-lithium bond.  All right, I want to say a couple of things because there's a tendency to think of these  bonds as ionic bonds or not ionic bonds.  If you go back to chapter one, we sort of talked about the fact that once the electronegativity  gets greater than about 1.7, 1.8, then electrons are transferred completely and we have an  ionic compound.  But this is not.  This is the largest difference we have is 1.5.  So that is considered to be a carbon-lithium covalent bond.  And so these are actually quite different here.  So about covalent carbon metal bonds, the reactivity towards electrophile depends on the polarity.  So organolithium are going to be the most reactive, greater than organomagnetium, greater  than organocopper.  Now for the most part, we're going to be using organolithium compounds and organomagnetium  compounds interchangeably, but there is one reaction where you need a super great electrophile  that will work for organolithium but not work for organomagnetium compounds.  And that's coming up.  So let's label these certainly, and I want you to know this, that organolithium are the  most reactive and organocopper least reactive.  So we're going to have three different organometallic reagents.  There's organolithium and organomagnetium.  Organomagnetium have a special name and that's Grignard.  So they're Grignard reagents.  And this is how you make them.  We have Rx, I don't know why that's cut off there, it looks like it's a little cut off,  plus 2 lithium, and you get Rli plus Lix.  So here's an example here.  Okay, we're at that with lithium.  Now I'm not going to be worrying about stoichiometry on the exam, you can just write lithium, you  don't have to write 2 lithium.  There's your organolithium product.  And you can leave out stoichiometry and side products.  So I don't care about the stoichiometry and I don't care about the side products.  Alright so that's an example of organolithium compound.  Grignard reaction has a little bit different structure.  So we take Rx plus Mg and we get Rmgx.  So the halogen from the Rx actually becomes part of the compound, so it looks a little  bit different here.  And it turns out that you must use an ether solvent for this reaction.  Organolithium don't need to use ether, but you definitely need to for Grignard.  Here's an example right here.  Mg ether.  And your product is Mgbr and this is a Grignard.  You can also call it organomagnetium, but it's definitely much more commonly called  a Grignard reagent.  And the ether is there because you actually get something that looks like this.  Where you have your Grignard and then you have the ether.  The lone pairs on the oxygen of the ether are sort of donating electrons to the magnesium.  Don't need to worry about this, but this is why you need the ether solvent.  Alright, organocopper.  I have a completely different structure also.  So let's go, let's look at that on the next page.  Organocopper reagents.  So also known as cuprate or Gilman reagents.  I will call them cuprates because that's how I learned it.  But there's also these other names.  They are prepared from organolithium reagents by reaction with a copper salt, usually copper  iodide.  So here's what it looks like.  Here's what it looks like.  Rx plus 2 Li gives you, there's your lithium reagent, right?  So we did the same reaction we did on the previous page.  And then you add copper iodide.  Notice you have two R groups here.  R2CULI.  So the copper stays, the lithium stays.  And that's a cuprate.  So two steps.  Number one, make organolithium.  Number two, add a copper iodide.  Okay, so that's how you make those.  Let me give you an example.  And you could use a vinyl chloride for this reaction.  So this is what it would look like.  You add the lithium, you don't need to put the two in front of the lithium.  I'm not worried about that.  Then you add copper iodide.  And you get two R groups bonded to the copper.  And then you have CULI.  Just like that.  All right, so those are the three reagents.  Need to know how to make those three reagents.  Grineer reagents are the form RMGX.  And you can have bromine, chlorine, or iodide.  No fluorides here.  Organolithium is RLI.  And organic copperate is R2CULI.  Need to know that.  They react as if they were carbonyl ions.  They are not true carbonyl ions because they have covalent carbon metal bonds.  However, we can think about them conceptually as carbonyl ions.  So a carbonyl would be an ionic compound.  And these aren't.  This is a covalent bond.  But that's the way they react.  Compare this with sodium acetylide.  And we have here carbon 2.5 and sodium 0.9.  But this guy's actually ionic.  So it's right on the borderline here.  It's actually ionic.  So conjugate base of a terminal alkyne.  This is negatively charged.  And then we have Na+.  So they're going to behave in a lot of the same ways.  But they're not exactly the same because this is an ionic compound.  So here we have electronegativity differences 1.6.  So somewhere around 1.6, 1.7 is where you start getting into ionic compounds.  So far, anybody?  All right.  Let's talk about reactions of organometallic reagents.  Here are some examples.  The organometallic reagents are powerful bases.  Extremely powerful bases.  They react rapidly with protons in an acid-base reaction.  So this is what it looks like here.  This is an acid-base reaction.  And we'll draw out the reaction and then we'll see which way the equilibrium is favored.  So there is going to come from the carbon magnesium bond.  We're going to grab a proton and then we're going to break the hydrogen-oxygen bond.  Very rapid reaction with water.  So there's the hydrogen that we just added on.  Let me mark that.  That's the hydrogen right here.  So you're going to get that plus HLMGBR.  You can write it like that or you can write it like you get a hydroxide, you get an Mg2  plus, and you get a Br minus.  So that's what happens when you hydrolyze a griniard.  Let's look at the direction of equilibrium.  So we're going to look at for conjugate acid-base pairs.  And what do we have here?  So we have water and we have hydroxide over here.  Conjugate acid-base pairs right here.  The one with the extra proton is the acid.  So here's our acid on this side of the equation.  And here's another conjugate acid-base pair.  Here's our acid on this side of the equation.  This has a pKa of about 15 and this has a pKa of about 50.  So as you can see, equilibrium very strongly favored to the right.  So long arrow to the right, tiny, tiny short arrow.  Really there is no reverse reaction for this.  But we'll just draw it out like that anyway.  All right.  So same thing with organolithium reagents.  And so you've got to be careful with these guys so that you don't have proton sources  around.  If you have proton sources around, then you will hydrolyze.  So proton sources can be water, could be acids, any acidic protons.  So there we just put a hydrogen in.  And then we get a carboxylate.  All right so it's easy to forget about that, but that's the fastest thing that's going  to happen with these guys.  If there's a proton source, they will react with it.  And direction of equilibrium, let's do that again here.  So here's our acid on this side of the equation.  Here's our acid on this side of the equation.  This is pKa about 5.  This is pKa about 45.  So once again, very, very strongly favored to the right.  So try not to forget that.  It's really easy to forget when you're doing synthesis.  We get so caught up in the carbonyl that we forget about these acidic hydrogens that kind  of interfere with what we're trying to do.  All right so this rapid reaction with water or any proteic acid means that Grineard reagents  or Genolythium cuprates cannot be prepared from compounds that contain acidic groups.  If you want to make a Grineard and your compound has an acidic group, you can't make a Grineard  from it.  So here's the acidic groups that interfere.  Anything in here?  There's an acidic hydrogen right here, a hydrogen on a nitrogen.  Here's another acidic hydrogen.  Hydrogen attached to sulfur, carboxylic acid, acidic hydrogen, even a terminal alkyne, hydrogen  on a terminal alkyne, you can't make Grinears from these compounds.  So don't make a Grineard.  Make a Grineard or Organolythium compound  from compounds that contain these groups.  All right, so that's super important and very, very easy to forget.  As I will tell you when I grade exams, I will see a lot of people forgetting about that.  All right, so here's an example of a failed synthesis.  This person is trying to make a Grineard from this compound here.  So remember when we make the Grineard, the magnesium goes in between the carbon and the  bromine bond.  So we have here Mg, ether.  You actually will make a little bit of this, but as soon as you make it, it's going to  be protonated by any acidic hydrogens that are in the medium.  So another molecule here, definitely.  So you make that and then any other proton sources and any of these other compounds here.  So air comes from the carbon, magnesium, bond, grabs the hydrogen and looks like that.  And then so what you end up getting is ethanol, not what you wanted.  You wanted to make a Grineard out of this compound.  So you're going to get that and you're going to get deprotonated alcohol.  So you can draw it like this or you can draw this like this with Br, Ch2, Ch2, Omg Br.  You can write it either way, but that's what you're going to get.  So that's a big problem.  That's not a Grineard reagent.  So these guys are the actual products.  So this rapid reaction with water and any protochastic can be used to synthetically turn an alkyl halide  into a hydrocarbon.  If you use a labeled proton source like D2O, you can actually make deuterated compounds.  So here's an example right here.  We have no acidic hydrogens here, so we will be able to make this Grineard rather nicely  here.  So the magnesium ends up being going in between the carbon and the bromine.  And then if you protonate this, if you actually deliberately add water, you will protonate  that.  So that's a way to actually get rid of this bromine in our molecule.  So that's what you would get.  If you use D2O instead or some sort of proton source that has deuterium, then you'll get  CH2D.  So there's the deuterium.  Let's circle that deuterium from one of these bonds here.  And now you've made it a deuterated compound.  And that's just by protonating the Grineard.  Once you form it, you protonate it.  And if you use a deuterated proton source, you'll get a deuterium incorporated.  Question so far.  Anybody?  All right.  So gannomatocryogens are powerful nucleophiles.  They're very important in organic chemistry.  They can react with epoxides to form alcohols.  The result is an extension of the carbon chain by two carbons.  So what am I talking about?  Let's keep an eye on this number here.  So that's going to keep an eye on that.  We're going to come back and circle those when we come back.  All right.  So we're making, again, a lithium compound.  So alkyl lithium.  All right.  And now we're going to open up an epoxide.  So this is going to attack an epoxide.  It's going to be very similar to lithium-lunium hydride attacking an epoxide.  Arrow's going to come from the carbon-lithium bond.  If this was a substituted epoxide, it would attack on the least substituted side.  Let's circle that original group that we started with.  Here's our three carbons right here.  Here's our three carbons right here.  And you see how I've added on two carbons.  I've added on a CH2, CH2O minus.  Once we protonate that, it will be a CH2, CH2OH.  And that's something you might want to tuck away.  You might see that on a future exam, where you have to add on a two-carbon piece that  looks like that.  We're going to talk about other ways to add on two carbon pieces, but this will add on  a CH2, CH2OH.  Two carbon extension where the functional group is on the end.  We had another two-carbon extension in Chapter 11, where we took acetylene, we deprotonated  it, and we attacked an alkyl halide.  That was also adding on a two-carbon piece.  Just different functional groups.  So good thing to tuck away.  What I think would be a good idea right now is if you have an index card or a larger index  card or a sheet where you start to now write carbon-carbon bond forming reactions.  We have one from 51 B, and we're going to now have granules or, again, lithiums and  cuprates here for 51 C. So those are great when you're trying to build up carbon skeletons.  All right.  Longer extensions can be made by using substituted epoxide.  So here's another example here.  Which side is this going to tuck on left or right?  Well nucleophile takes on the least substituted side, right?  This goes back to Chapter 9.  All right.  So that's what's going to happen here.  It's going to come in attack on the least substituted side.  We're going to kick electrons up onto oxygen.  That's at the very end of Chapter 9 if you want to take a look at that.  The after protonation.  And there's our phenyl ring.  So we've added on actually a four-carbon piece where the hydroxyl is in the two position.  So that's just something to keep in mind here.  We've added on a one, two, and then this is three and four if we drew out that ethyl.  Added on a four-carbon extension.  And so just remember that this is, of course, a strong nucleophile.  That's at the least substituted side of the epoxide.  All right.  So grignards can be protonated.  They can attack epoxides.  But the most important reactions that we see with grignards are grignards attacking carbonyls.  Okay.  So let's look at some examples of that.  Let's look at some examples with type one and type two carbonyls.  So here's a typical reaction with the type one carbonyls.  So we have ethyl, magnesium, bromide, or ethyl-grignard.  Arrow's going to come from the carbon-magnesium bond.  These are going to look just like lithium-luminum hydride.  Okay.  So we're going to do a very similar way.  Arrow's going to come from the carbon-magnesium covalent bond.  We're going to attack the carbonyl carbon.  We're going to kick electrons up onto oxygen.  So this will look really familiar to you.  It looks just like the mechanism for the lithium-luminum hydride with the type one carbonyl, except  rather than transferring a hydride, we're actually transferring an alkyl group.  So we're building up carbon skeletons.  But still the same thought process in that we have tetrahedral intermediate with no leaving  group.  Therefore it stops here until you add acid in a second step.  So really key here is that we want to have the water in a second step.  What happens if we mix the grignard and the water?  What's going to happen?  We're going to kill our grignard reagent.  A word that you'll hear used is we're going to quench it.  We're going to kill that because the acid-base reactions are going to be faster.  And so then our grignard will be gone and then we don't get anything that we want.  So add water in a second step.  So mechanistically very similar to the lithium-luminum hydride.  All right, question about reactions with type one carbonyls.  Anybody?  All right, so we can agree, organolithium is the same.  Like I said, most of the time we're using greenery or xanoganyl lithium interchangeably  except for one reaction.  So same thing happens here with organolithium.  These guys are more powerful, but in this instance it would do the exact same thing.  And then of course we're going to add water in a second step.  And same thing as when you're using lithium-luminum hydride.  You cool the reaction vessel down in an ice bath and then you cautiously add a drop of  water.  You wait for the exothermic reaction to go away and then you add more.  So it's really very, very extremely strong base here.  All right, so that's type one carbonyls, organomagnetium, organolithium cuprates.  We're going to talk about it a little bit later because they do different chemistry.  They're sort of their own category.  Sodium acetylide, we've talked about sodium acetylide reagents.  We know that they can open up epoxides.  They can also attack carbonyls.  So familiar reagent, unfamiliar reaction, but you probably predicted that something like  this would happen.  I'm not worrying about stereochemistry because we have achyro reagents reacting with achyro  reagents.  So it's going to look like that.  Okay, and then we add water in a second step.  And notice something also, when we use acetaldehyde, we're adding on a two-carbon piece.  But it's different than the two-carbon piece we do with an epoxide.  So we've also added on two carbons.  So let me circle what we started with.  So this is one, two.  This also adds on a two-carbon piece.  What's the difference between this two-carbon piece and the one we did with an epoxide?  They're not the same, right?  Anybody have an idea?  Excuse me?  No, we're at, well, we started with this.  We're adding two carbons onto this.  It depends on the perspective.  You're looking from the perspective of the acetaldehyde.  Yes, we're adding on three pieces.  But if we're starting with this and we're adding on two pieces right here.  So the only difference here between this is, remember when we added on the two-carbon  piece with an epoxide, it was CH2, CH2OH.  The OH was on the end.  Here the OH is right where we bonded directly to the carbon of our nucleophile.  So the only difference is the location of the OH.  Sometimes you want to use an epoxide.  It depends on what you're doing afterwards.  It depends on what you're trying to synthesize.  But I want you to point out that when you use acetaldehyde, you're also adding on a  two-carbon piece.  It's just the only difference is where the hydroxyl is.  So I'm just trying to point these things out to you because they're going to be relevant  when you do start doing synthesis.  All right, let's look at a typical reaction with a type II carbonyl.  All right, and I'm not going to show you the product because we're going to work through  what the product is.  Another possible mechanism for midterm one.  Aero's going to come from the methyl carbon magnesium bond.  We attack the carbonyl carbon.  We kick electrons up onto oxygen.  So let's just kind of follow it along.  We know leaving groups.  We know things that can be leaving groups in this reaction.  We know things that cannot be leaving groups in this reaction.  So here's our tetrahedral intermediate.  Does it have a leaving group?  Yes, tetrahedral intermediate has leaving group.  If it has a leaving group, it's going to leave.  So electrons on oxygen are going to come down, and we're going to kick off our leaving group.  Now we're going to look at the product that we get, and we're going to make a decision  of something more is going to happen or not.  All right, so our product is acetone.  So it's a ketone.  Is a ketone more reactive or less reactive than an ester?  It's more reactive than an ester.  So that means the griniere is going to add again.  Can't stop here.  Why can't you stop there?  Ketones are more reactive than esters.  All right, so what that means is that the griniere is going to add again.  All right, arrow comes from the carbon magnesium bond.  Now we're attacking a type 1 carbonyl.  We know what's going to happen next.  We know that since it's a type 1, we are not going to have a leaving group.  Absolutely no possible leaving groups here.  All right, so our tetrahedral intermediate has no leaving group.  Therefore, it's going to stop here until we add a proton source, and we're going to add  water first.  So conceptually, the same thing is happening here.  As was happening with lithium-aluminum hydride.  We're doing the same thing, we're adding more than once here.  And so we certainly have to add H2O in a second step.  And then when we do that, we get our alcohol.  So this is also a synthesis of alcohols.  So there's our product.  We figured it out ourselves here.  And so the way that I want you to approach these problems is rather than just memorizing  products, I want you to go through this thought process and figure out what the product is  yourself.  So it turns out that you must use two equivalents of Grignard.  If you think you can trick it and only use one equivalent and get that product cleanly,  you're wrong.  What you will get is that product and unreacted starting material.  OK, so if you only use one equivalent, you will get 50% conversion.  All right, now I don't remember specifically what kind of sapling problem, but there was  a sapling problem with either this or lithium-lunar hydride.  And they asked for all the organic products.  And so you just got to remember that right here, I just kicked off a methoxy group.  So I'm also going to get methanol as a product, aren't I, when I start with an ester.  So you want to keep that in mind.  So you're going to get this plus this methoxy group that you made is going to get protonated.  Now I don't normally worry about that side product, but it could be something that you  would have to worry about if you were actually doing the reaction.  So you're going to get that product and you're going to get methanol.  So two products.  Questions on inability to stop at the ketone.  Anybody have trouble with that?  I think one of the common misconceptions that I see in organic chemistry is people think,  well, if I just use one equivalent and everything reacts simultaneously, then it should be able  to stop at what I want.  And it's not that way because these reactions are not happening simultaneously.  So in other words, if I lined up all the esters in a row and then I lined up all the griniards  right next to them in a long line and I said, OK, on your mark, it said go.  Then they all reacted simultaneously.  Then I would be able to stop at the ketone, but you and I know that's impossible to do.  They're reacting a little bit at a time.  And as soon as you make a more reactive product, it's going to react faster with the griniar  that hasn't reacted yet.  So I know that's tricky, but that's just the way it works.  All right, questions on that aspect?  Anybody?  We like to think we're in charge of these molecules, but we're really not.  They're going to do what they're going to do.  As any grad student will tell you.  All right.  OK, let's do an example with an acycloid and organolithium reagent.  Same idea here.  Let's go through the mechanism.  So another possible mechanism for midterm one.  We've got a lot here.  Do you remember to put lone pairs on all your reacting atoms when you're doing a mechanism?  So let's see what we get here.  So rather than just writing what the answer is, I like to go through it so you can think  your way through to the answer.  All right, tetrahedral intermediate has leaving group.  So electrons on oxygen come down and kick off that fantastic leaving group.  Chloride's a fantastic leaving group.  All right, so then let's look at the product and let's make some decisions here.  What's more reactive, an acycloid or a ketone?  Acycloid.  So theoretically, we should be able to stop here.  The problem is the same problem we had with lithium-luminum hydride.  It's too reactive to stop there.  It's just way too reactive.  So it's going to add again, unfortunately.  Or maybe fortunately, if you want that product.  So it's going to add again.  So I will label that very clearly.  You should be able to stop here because an acycloid is more electrophilic than a ketone  by far.  The problem is that granules are too reactive.  All right, so you can't stop here.  So these powerful reagents are just too powerful to stop.  And so it's going to add again.  And so what do we get from that step when it adds again?  Gosh, we get the same thing that we get with an ester.  We get the same exact thing.  Oops, I skipped a step, didn't I?  Let's fix that.  So you get an alkoxide.  Same alcohol you get with an ester.  Tetrahedral intermediate has no leaving group.  Therefore it stops here.  And then when you add water, we protonate and we get the same exact product.  All right, so there's two examples of reaction with type 2 carbonyls.  Yes?  Yeah, right.  How about that?  Organolythium.  Thank you for catching that.  They're so interchangeable.  I'm interchanging the names and I shouldn't do that.  More questions?  Anybody?  All right, so granules will also add twice to acid chlorides because they're two reactive  cuprates which are the least reactive organometallic reagents will add only one.  So cuprates do their own chemistry.  Think of them in their own special category.  Okay?  This reaction only works with acid chlorides.  Only works with acid chlorides.  No other carbonyls.  Okay, so cuprates having copper.  Now copper is a transition metal and so they do have their own different chemistry that's  completely different than granules and again, allithium compounds.  What they do, they will react with acid chlorides to make ketones.  Now I'm going to draw arrows for this.  This isn't really how this reaction takes place but this is a way you can think about  it.  Cuprates, again, they do their own chemistry and it involves radical reactions but I'm  not going to show you that.  I'm just trying to give you a conceptual basis here but the idea here is that these are going  to attack the carbonyl carbon and they, remember these were the least reactive of the three  so they will only react with acid chlorides.  And since they only react with acid chlorides, when we form the ketone here it's not going  to react again.  So we have tetrahedral intermediate has leaving group.  So leaving group is going to leave, electrons come down, kick off the leaving group and  since cuprates don't attack any other carbonyls but acid chlorides, it stops here.  So if we want to stop at the ketone we have to use an acid chloride with a cuprate.  If we use an acid chloride with a granule or an allithium it will add twice and give  you the alcohol.  So that's the way to do this.  Alright and let's just label this when you go back.  I won't ask for cuprate mechanisms.  This is not the true mechanism.  Way too complicated the true mechanisms, not completely 100% understood and way too complicated  for this class.  Okay yes.  That compound there with an aldehyde on one side and an acid chloride those are very unstable  so I don't know if you can actually do the cuprate reaction.  Do you know?  They're very unstable.  They decompose into carbon monoxide and hydrochloric acid so I don't know whether, I don't know  the answer to that.  Yeah, yeah.  Yes.  Okay.  What about carboxylic acids?  Will they work in this reaction?  Well remember what we don't want to forget.  We're soaked tapped into the carbonyl.  We don't want to forget about the acidic hydrogen.  Don't forget acidic hydrogen.  Alright, so the first thing that's going to happen these guys are extremely powerful bases.  We're going to deprotonate that acidic hydrogen.  Alright and we know something about carboxylates.  We know that they're the least reactive of all the carbonyls.  So the question is are carboxylates, and you look at a carboxylate you normally think of  an electrophile as having a partial positive charge or a full positive charge, right?  This has a negative charge so it's kind of strange to think of it as an electrophile  anyway.  But the question is a grignard, our side product is methane by the way, that's our side product,  that's methane gas which is going to bubble off the reaction mixture.  And the question is if we have an additional equivalent grignard, is it strong enough  to attack a carboxylate?  And the answer is no.  So methyl magnesium bromide is not powerful enough, so it's powerful but we've come to  the limit of its strength, not powerful enough to attack a carboxylate.  Okay so if it's not powerful enough to attack a carboxylate, which is what is powerful enough  to attack a carboxylate?  Organolethium.  So this is the one reaction where they do have a different something happening.  Organolethium is more powerful so if we want to have that happen again we'll talk about  what happens next time.  So Steve you could try to figure it out on your own.
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