Lecture 20. Phosphorus Chemistry
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Vranken, David Van
UCI Chem 201 Organic Reaction Mechanisms I (Fall 2012) Lec 20. Organic Reaction Mechanism - Phosphorus Chemistry Advanced treatment of basic mechanistic principles of modern organic chemistry. Topics include molecular orbital theory, orbital symmetry control of organic reactions, aromaticity, carbonium ion chemistry, free radical chemistry, the chemistry of carbenes and carbanions, photochemistry, electrophilic substitutions, aromatic chemistry.
 Okay, so this is our final lecture in this class. We just finished talking about sulfur chemistry. We dropped down from the second row to the third row. And we're going to finish up by talking about phosphorus chemistry. And I want to introduce you to some of the functional groups that we're going to talk about today. We can't cover everything. But we're going to focus on these four types of functional groups, phosphines. And that's where you have three alkyl or aryl groups bonded to phosphorus. So three carbon phosphorus bonds in a phosphine. When you switch to phosphorus heterodontal bonds, you change the name. And in particular with oxygen, we call this a phosphite. This kind of species where there's three oxygens on here. And more specifically, I would refer to this as a phosphite ester. But in general, I would simply say, oh, that phosphite over there. And whenever I see these top two, I'm focusing relentlessly on those lone pairs on phosphorus. So we'll talk about that in the chemistry. So it's very easy to oxidize phosphorus. Three, we'll talk about that. So this is phosphorus with an oxidation state of three. This has an oxidation state of five down here. This is a phosphine oxide. Let me contrast that with oxidation state three up above. So at the oxidation state of five, this is the most common functional group you'll see. It's a phosphine oxide where we have this oxo ligand sticking on top of phosphorus. And if you do the same for a phosphite, we don't call that a phosphite anymore. We call that a phosphate. And you should be familiar with this functional group because it's the backbone of DNA and RNA. And it's the key functional group on phosphorylated proteins. So let's go ahead and talk about what kinds of things change as we drop down from the third row, sorry, from the second row nitrogen to the third row with phosphorus. So let me remind you of some of the pKa's about the basicity of the lone pairs on nitrogen. And what we'll do is we'll look at the pKa of the protonated nitrogen compound. So triethylamine is a common base. And we can say something about the basicity of the lone pairs on triethylamine by looking at the acidity of the conjugate acid of triethylamine. So if you compare triethylamine with triphenylamine, as you might expect, triphenylamine is, it's not basic at all. If you look at the pKa for triphenylamonium, it's on the order of minus 5, whereas triethylamonium is on the order of 10 to 11. I'll use 11 as my approximate number there. It's extremely difficult to stick a proton on triphenylamonium. That's because the lone pairs on nitrogen are donating into, let me draw the lone pair up here, are donating into this pi system. And you can draw resonance structures all over the place. That's exactly why aryl amines are so reactive in electrophilic aromatic substitution because these nitrogen lone pairs are juicing up that pi system and making it more nucleophilic. So you have really good resonance here with nitrogens attached to benzene rings. If you look at the corresponding phosphorus compounds, just in terms of pKa, and I'm going to draw butyl groups here because that's the most common alkylphosphine, but it's about the same for triethylphosphine. If you look at the pKa, it's pretty similar. That's a pretty similar basicity. The desire to have a proton stuck on there is pretty similar for a phosphonium species. 10, 11, that's sort of within the same ballpark. If I look at triphenylphosphonium, protonated triphenylphosphine, now the pKa is significantly different from the nitrogen case. Instead of being way at the minus 5, in other words, this is a super strong acid. If you could figure out how to stick a proton on that nitrogen, that would be a very strong acid. But a protonated triphenylphosphine is not a super strong acid. Triphenylphosphine is not super duper acidic, and we can trace this back to the length of these carbon phosphorous bonds. So yes, there is a very nucleophilic lone pair on this phosphine. And hypothetically, there's some donation into this pi star system of the airing ring, but look how far away that, of course, I'm trying to exaggerate it so you'll remember, but it's so far away. You don't get good donation to form partial PC double bonds here. This goes back to this issue that, yes, the orbital is very big here on phosphorous, but it's so far away. You don't get effective donation, and that makes this lone pair more basic. It's spending less time donating into the pi system, so it has more room sort of to more left over reactivities to react with protons. Okay, so trialkylphosphines are more basic than triarylphosphines. And triarylphosphines are more basic than the corresponding second row atom system with nitrogen there. I want to return to a mechanism that you've probably, I hope you've seen before, or maybe not the mechanism, the reaction I hope you've seen before from sophomore organic chemistry. This description is not the reaction. This is a description of part of this process. So here's a reaction that you might have seen before in sophomore organic chemistry, and it involves taking PBR3, phosphorus tribromide, and using that as a reagent to substitute the hydroxyl groups of alcohols, primary alcohols, secondary alcohols. You could do tertiary alcohols. The stoichiometry looks like this. You take three equivalents of alcohol for each tribromo for each phosphorus tribromide, and that allows you to substitute the OHs. So that's a very useful substitution reaction because there's all kinds of things you can do without kilbromides. You can turn them into green yard reagents. You can substitute them now with nucleophiles. Okay, so here's a reaction I hope most of you have already seen before. Maybe something that was never mentioned to you before is that this reaction is acid catalyzed. Probably with every fiber of your being, you want to somehow use this oxygen, right? You have to make PO bonds in your product. Probably you want to do SN2 reactions on that, and I want to urge you to avoid that mistake. Let's talk about the mechanism for this reaction. The mechanism of the reaction starts off with protonation. Whoops. Starts off with protonation of the phosphorus lone pair. That's why acid is essential. You cannot buy PBR3 that doesn't have some acid in there. And once this reaction starts, you start generating acid from the get-go. And so the initial step here is that you protonate that phosphorus, and the second you put a proton on there, you've turned that phosphorus into a super-duper electrophile. So now that I've got a proton on there, look at that. If you want to make a bond of phosphorus, now it's easy to make a bond of phosphorus. Now it's easy to attack the phosphorus. Proton transfer reactions are generally very fast, and so now we've changed this to where it becomes very easy to attack. Let me go ahead and draw a nucleophile here like ethanol, E-T-O-H. Here's my lone pair, and there's that phosphorus, and so that those lone pairs on the ethanol, the alcohol, want to attack this. Stereoelectronic there's a preference to add anti to the most electronegative ligand. Proton is not the most electronegative ligand. You add anti to one of the bromines here. So you end up with a trigonal bipyramidal phosphorus. So here's my equatorial ligands, and here's my axial ligands on here. So here's that, and I'm trying to figure out whether I'm going to just skip this deprotonation step because I'm confident you can figure out that after I attack, yeah, you need to use deprotonate that proton from the ethoxy nucleophile. So you end up with this trigonal bipyramidal intermediate. That's not well drawn. So you end up with a trigonal bipyramidal intermediate. No charge on phosphorus. It's neutral. Don't freak out over the five bonds to phosphorus. This is not a second row atom. Perfectly happy as an intermediate with five bonds. Remember phosphate over there? That's got five bonds. Okay, so this is perfectly okay as an intermediate, but that's not to say that leaving groups don't want to leave. Now the axial leading group will prefer to leave. That's just a stereoelectronic preference. And so I get right back to a tetrahedral type intermediate. And so in that tetrahedral intermediate, I now have this ethoxy group attached. I now have this electrophilic phosphonium species. And so now, and that bromide is now floating around, and I'm going to draw it right over here so you can see that it's ready to attack carbon. Wow, that phosphorus and oxygen really want this pair of electrons here. And so they're simply going to take it. And I've got two choices for how I draw this reaction out. I can make the electrons form a P O double bond, or I can give the electrons to the oxygen and draw a resonance structure for that. And I'm simply going to draw that resonance structure. So here's my initial intermediate here. And here's my minus, and here's my plus. And so now to finish up this stage of the process, what I need to do is I need to stick a proton on the O minus using acid, and then pluck a proton off the phosphorus. I'm not going to draw those two steps for you. I'll draw the first step here. So I think you can see that after I do that, I can use the A minus. And so you just do this three more times. And that's what allows you to, sorry, I ran out of room there. And that's what allows you to replace those OHs on your alcohol with BR groups. So look at what we're doing. This was the key step right here. The fact that bromide came in and did an SN2 reaction in order to pop off a species that had this sort of phosphine oxide character. And don't freak out over the five bonds to phosphorus that pentavalent, let me give you a name here so we know how to describe this. This is called a phosphorane. That's called a phosphorane intermediate. Trigonal bipyramidal, five bonds to phosphorus. That's like a tetrahedral intermediate when you substitute ester derivatives of carboxylic acids. Okay, I want to extend this idea that when you have phosphorus with lone pairs, you can protonate those. And so let's talk about phosphites, not PBR3, but phosphorus with some bonds on there. I want to, I'd like to talk about some patterns of substitution and I want you to compare just mentally. Please don't write this down because I'm assuming that this is stuff that you already know. Or if you don't know this, you don't have to. Which of these is easier to attack with alkoxide? The ester or the ether? Yeah, the ester. There is no reaction where you can substitute OR by having, this is very easy to attack a carbonyl. It is, I don't know of any way to get a hydroxide anion to attack an ether and do an SN2 reaction. Okay, but now let's talk about phosphorus because I want to urge you not to extend what you already know about carbon to phosphorus because it's not the same. Let's draw it a little table here. What I'll do is I'll now draw a phosphate ester instead of a carboxylic ester. And let's try to imagine the reaction of these two different species, a phosphate ester versus a phosphite ester with nucleophiles. And so what are the relative rates of hydrolysis of a phosphate versus a phosphite with hydroxide anion? If you look at the relative rates, it's exactly, it's not exactly opposite. It's, right, you can't even imagine attacking an ether oxygen but with phosphorus you get a thousand times faster rates of attack and substitution at phosphites. It's less reactive when there's an oxygen atom on top of there. And if that is in any way hard for you to understand, then go ahead and draw this resonance structure so that you can see that there's an O minus right next door that's already donating into that phosphorus. So now you can imagine why it's hard for hydroxide anion to attack a phosphate ester. Let's switch over to an acid catalyzed reaction. Suppose we use acid to catalyze the hydrolysis of these two esters, a phosphate ester versus a phosphite ester. The phosphate ester is still the loser here. It's still the least reactive of these pairs. If you want to make phosphorus oxygen bonds, you should use phosphites. And the relative rate difference is now, it's 10 to the 12. It won't matter if you write the right number of zeros. You get the idea. It is way, way, way more reactive to substitute this under acidic conditions. And I just showed you why that's true. I just showed you why. It's so much faster. When you protonate a phosphorus lone pair on a phosphate, it makes it super-duper-duper-duper fast to attack at that phosphonium species. And you can't do that with a phosphate ester. Protonating this, you still have lone pairs on that oxygen that are donating in. Okay, so phosphites are very fast to exchange. Okay. Let me give you one example of a reminder of how difficult it is to substitute phosphate esters. Whereas, phosphate esters substitute quickly. So if you look at the rates or if you look at the hydrolysis of trimethyl phosphate with alkoxide. Right? You know, the typical mechanism that you draw looks like this. And you kick the electrons up to the oxygen to give an O minus. You'll end up with this pentavalent phosphorus intermediate. Fifty percent of the substitution that takes this down to the product in this case. I guess I should, sorry, I should have hydroxide anion here. I apologize for that. Otherwise, there's no way to have a product. It would be. So this would be the product. Fifty percent of the product when you attack trimethyl phosphate with hydroxide anion comes from attack at carbon. Right? It is so hard to attack at phosphorus, you start to see attack at carbon. This is kind of like making a sulfonate ester. The reason you make sulfonate esters is not because you want to attack at sulfur, it's because you want to do SN2 reactions at carbon. So it's the same idea with phosphates. It's just very hard to attack here at this phosphorus of a phosphate ester. So if you're clever and you want to make phosphorus oxygen bonds, then you should use phosphates, not phosphates. Guess what they do in modern DNA synthesis? They don't use phosphates. They use phosphates. Let me go ahead and draw a typical phosphorus oxygen coupling reaction. This is going to be a picture of my DNA strand. And suppose I wanted to add one and I'm doing a chemical synthesis of DNA. I'm making a 50mer or a 60mer. And I'm now adding on my 50th subunit. The subunits that you use, the reagents that you use in modern DNA synthesis are basically, here let me draw this isopropyl correctly. So that's an isopropyl amino group, are basically phosphates. I don't want you to worry about what the protecting group is on that oxygen. So this is a subunit used for DNA synthesis. It's called a phosphoramidite. And there's a trital, a triphenyl methyl group there you take off with an SN1 reaction. And your other substituents on your DNA base, you've got some sort of a base there. Three of the four bases need protection on there. So when you want to make a phosphorus oxygen bond in DNA synthesis, you don't make phosphate esters initially, even though that's what you want. You have phosphate esters connecting DNA. What you do is you take this phosphoraminite reagent and you rely on the fact that if you put in an acid, it's now very fast for alcohols to attack that and substitute. And in this case, they substitute the nitrogen. So there's a super secret acid they use called tetrazole. You don't need to know that. The bottom line is there's an acid in there that accelerates the rate of oxygen nitrogen substitution. And we're not going to draw the whole mechanism because it can be more complex. Tetrazole can also substitute the nitrogen. Don't worry about that. The bottom line is in DNA synthesis. You have minute long reaction scales. When was the last time you did a reaction that took one minute or less? Okay, I'm done. Now, it doesn't work like that typically. But with DNA synthesis, it's so optimized and it's so fast to substitute at phosphites. Sorry, let me take off that plus charge there. But this is the way you make phosphorus oxygen bonds in DNA synthesis. You use phosphites. Now, nobody wants to make polyphosphites. Those bonds are now so sensitive to acid-catalyzed substitution, you've got to get rid of this. So what you do in DNA synthesis is you oxidize this now and the reagent combination that you use is iodine and water. And that converts and it's super fast. It's not the same mechanism as this other one. We'll talk about this. But it is super fast to oxidize phosphites and phosphines to phosphine oxides. And now those bonds are now stable, right? You do this initial ester exchange under acid-catalyzed conditions with a phosphite. But then when you want to stabilize these bonds, you convert that to a phosphate. And now those bonds will sit through round after round of DNA synthesis and be perfectly stable. So this is modern DNA synthesis. It takes advantage of the rapidity of phosphite ester exchange. Let's talk about phosphates and phosphine oxides in these P O double bonds. All these are the carbonarmeApplause. So we with you are simply going to talk about this now is super fast and the process can, it comes out like this. Many, many, many reactions in organic synthesis are driven by the formation of phosphorus-oxygen bonds or these things that we call, in other words, what do I call this? Do I call that a phosphorus-oxygen double bond? Well, that's not so clear. It's not so clear how I should draw this. This is just like a sulfur-oxygen double bond. There's another resonance structure that I can draw in which I'll bring these electrons up to the more electronegative oxygen atom. And I can draw this illid-like structure with P plus O minus. What you'll find is that I draw it like this and everybody else draws it like this because we're too darn lazy to draw it like this. There is no time ever in which it helps you predict reactivity by drawing it like this. You always can predict reactivity by drawing it like this. It just becomes burdensome to draw these charges around there. So because I'm lazy and because I always think of it like this, I continue to draw it like this. But I would urge you to draw it like the equivalent way because that way you're ready to use the O minus as a nucleophile when you need to. We're ready to rationalize why things will attack phosphorus. Okay, so it is very facile to oxidize. When you leave trialkyl phosphines exposed to air, they oxidize very rapidly. So if you open the top off your bottle of tributylphosphine, you can expect it to be very rapidly oxidizing the phosphine to the phosphine oxide. So in the case of trialkylphosphines, and this kind of correlates with basicity, alkylphosphines are more reactive than triarylphosphines toward acids. They're more basic. And they also attack oxygen faster. So tributylphosphine oxidizes very fast and very reactive. Whereas triphenylphosphine, you know, you can open the bottle and walk over to the balance kind of leisurely, scoop a little bit out to put in your reaction. And you don't have to worry about it all getting converted to the phosphine oxide. Okay, so it's very thermodynamically favorable to do this. Let me give you some bond energies here. In many, many reactions, you'll break bonds to things like carbon and oxygen. And let me just remind you that's about a 90 kcal per mole bond. And so if you can break bonds between carbon and oxygen and replace them with bonds between phosphorus and oxygen, and it doesn't matter how I draw that. It doesn't matter whether I draw this P plus or O minus. That bond, whether you draw P plus or O minus or P double bond O, is energetically very favorable. 137 kcal per mole. And for the tributylphosphine, 128 kcal per mole. I'm positive I have these reversed. But that's what's in my notes. I'll fix the notes later, but I'm super positive that I reversed those in my notes that I have drawn here. The bottom line is these are very much more favorable than any bond between oxygen and carbon. And you will find in many, many reactions where you see people dump in triphenylphosphine. They're trying to strip off an oxygen from carbon and end up with this. And so that provides a huge thermodynamic driving force. Let's come back and draw a mechanism for this iodine oxidation and DNA synthesis. Because this transformation is driven by formation of this P O, if you want to call it a double bond, that P O double bond. There's less than 50% pi character between the P and the O. It's kind of like a sulfur oxygen double bond. Is it really a double bond? There's not much pi character there. Let's go ahead and draw this mechanism for iodine oxidation. If you ever stain phosphines on your TLC plate by dipping it in an iodine chamber, this is what's going on. So if I have iodine in the presence of a phosphine, you should expect that that lone peron phosphine on phosphorous will very quickly displace that iodine. And it will give you this phosphonium intermediate. Let me just draw R here. And it doesn't matter whether the R's are alkoxy groups or alkyl or phenyl. So now I have this phosphonium. It's easy for things to attack, things like water. So if you have water present, it adds opposite the most electronegative group. That's a stereoelectronic preference to give you a trigonal bipyramidal phosphorane intermediate. I don't want to show the deprotonation of the water that comes after the attack. I think you guys can get the idea that after I attack with water I have to pull a proton off. And so now I've got this phosphorane. That's supposed to be an R. I just didn't do a good job of drawing that out. And so now I make something leave. And the leaving group that leaves best is iodide. And so the intermediate that you get. Looks like this. And now all I have to do is remove the proton. And I can see how I generate a P double bond. If that's the resonance structure I want to draw. You can see how I generate this P double bond O. So in DNA synthesis this is what you're doing. You're just treating your phosphite ester with iodine and water. And you end up with a phosphate ester that's very stable. If you ever wanted to convert a phosphine and an organic reaction to a phosphinoxide you could use any oxidant you want. And it will make the phosphinoxide. MCPBA, hydrogen peroxide, maybe in a less controlled manner if you just blew air through the reaction. Nobody does it that way. But okay. So phosphorus lone pairs attack things. That's the basic idea. So let's talk about that more explicitly in a classical, very old reaction. I'm sure it dates back to the early 1900s, maybe 1800s. Don't know. It's called the Arbusoff reaction. More specifically this is called the McHales Arbusoff reaction but it's not really important. So if you take a phosphite like this triethylphosphite would be a good one for this. And you mix that with an alkyl halide. You're going to get an SN2 reaction. There's no really acidic protons here but there's a good electrophile. That non-bonding lone pair orbital on phosphorus is very big and very nucleophilic. It's easy to overlap with sigma star here so you do an SN2 reaction. And the intermediate that you get, one of these ethoxys I'm going to draw with all the bonds there. So the intermediate that you get now has this phosphonium salt there and I'm going to take my bromide counter and move it to the other side so I can see how ready it is to cleave that carbon oxygen bond. Right? We want to generate a P double bond O type bond here. There's a huge thermodynamic driving force and kinetically that's very fast. Well, you still have to heat these reactions substantially but I don't want to say it's that fast. That's the slow step right there is cleaving that carbon oxygen bond. But ultimately what you get out of this is a phosphonate and this is what you use for a Vittig-like reaction. I'll say a little bit about Vittig reactions. But you can use this for olefination, something called the Horner-Wadsworth-Emmons olefination procedure. And you make these reagents for Horner-Wadsworth-Emmons reactions using this Arbusoff, Mikhailis Arbusoff reaction. This particular type of species where there's one carbon phosphorus, is it like a phosphate ester or is it a phosphene oxide? What is that? We call this a phosphonate. That own sound tells me phosphonate ester. Tells me that there's a carbon oxygen, sorry a carbon phosphor, it's not just a phosphate ester. Phosphate has three alkoxy groups on there. This is a phosphonate ester. Okay, so I'm trying to play up, oh gee, it's so hard to add to phosphorus, to phosphene oxides. It is hard to add to phosphates. And so I'll show you something ironic that maybe is not so ironic and that is that you can have phosphorus 5-oxo phosphorus species that are attacked very easily. And here's the reagent that's very fast. Everybody calls this Pockel III phosphorus oxychloride. And this is not like a regular phosphate ester. This is screaming hot reactive. Let me just show you where, let me just remind you of why this should be so reactive. Typically if you make a phosphate ester like this, there is some donation, it's not great, but there's some donation between the lone pairs into the phosphorus. And then make it hard for other nucleophiles to attack. If you're coming along with your lame, pathetic lone pair and there's already three, four sets of lone pairs donating in there, how are you going to wedge your way into that? But we come back over to phosphorus oxychloride and I'll try to exaggerate this. Okay, yeah, there's things with lone pairs, but those lone pairs are way, way, way, way away from the phosphorus. All right, chlorine's electronegative, but those lone pairs are not really deactivating the phosphorus. This is extremely reactive. This is nothing like a regular phosphate ester in terms of reactivity. So here's a typical place where you would take advantage of the super-duper electrophilicity of that phosphorus in PAKL3. I'll show you a type of electrophilic aromatic substitution called a Wilsmeyer reaction. So I hope you're familiar with some sort of electrophilic aromatic substitution reaction called Friedel-Kraftz isolation. You can add acyl groups to benzene rings, but there's one particular acyl group that you can't add and that's a formula group where you have an alde, where you create a new aldehyde. And the way you do that, the way you affect a Wilsmeyer reaction is you don't use aluminum trichloride or any of these typical Friedel-Kraftz reagents. You use this. This is called the Wilsmeyer reagent because it's the key intermediate in this reaction. If you want to add an aldehyde carbon bond to this benzene ring, this is the reagent that you're going to need. So it does electrophilic aromatic substitution. I'm not going to draw the mechanism for electrophilic aromatic substitution. I expect you guys to know that. The end result, not the end result, but before you work this up, you have now substituted. There was an H on the benzene ring. It's gone. So you spit off HCl as a byproduct. So you don't make an aldehyde initially. You make this a mini-amiand. But I don't know of anybody who works to keep that. You always work that up with water and it hydrolyzes the mini-amiand. So this is the way you add aldehydes to aromatic rings using electrophilic aromatic substitution. You use the Wilsmeyer-Hock reaction. So where do you get this chloroaminium species? That's not just a regular aminium ion. There's a leaving group attached to that aminium ion. The way you get it is you mix POCl3, Pocl3 with DMF. And that's typically used as the solvent for the reaction. So let's talk about that. Let's talk about the reaction of POCl3 with amides. Phosphorous oxychloride is so reactive that the lone pairs on amide oxygens will attack. So typically with the Wilsmeyer reaction, you use DMF as a common solvent for organic synthesis. You just take your pot, use DMF as your solvent, and you dump in POCl3. That's supposed to be a chloride right there. I just didn't do a good job of drawing that. So the lone pairs on the amide are very nucleophilic. If you'd like a reminder for why that's nucleophilic, here's a lone pair on nitrogen that's donating in. So I'm not going to draw that resonance structure. Let me just take my arrow away because I don't want to have too many arrows. Okay, so the lone pairs on the oxygen can attack, and they attack opposite the most electronegative group. Most electronegative groups are chlorine, not the oxygen on here. But if you want to understand the reactivity of this, you could draw it like this, right? So what's more electronegative? Oh, with an anionic charge on it or chlorine? Chlorine's more electronegative. So you attack opposite one of the three chlorines, and you generate an intermediate that looks like this. This is one of two possible resonance structures that I could draw. Or if you wanted to draw the double bond O, you could have kicked the electrons up to the oxygen. It's the same thing. Okay, so you end up with this phosphorane intermediate, this pentavalent phosphorane intermediate, and so now we'll pop off one of the leaving groups. And there's another resonance structure you could have drawn for that amide if you wanted. That would be the equivalent of what I'm drawing here. And so now that chloride ion that you popped out is just floating around looking for something to do. So it shouldn't surprise you that that chloride will eventually attack here at this carbonyl species. Or if you drew the aminium ion resonance structure, we still attacked the same group. You'd still draw it attacking the same carbon. And I don't want to draw the substitution mechanism here, but you'll add, well, I might as well. So you have this chlorophosphate leaving group ready to leave. And as soon as you add in to this species, it's now ready to push out a leaving group. It's okay if it pushes out the chloride and goes back, right? But eventually it's going to push out the chlorophosphate leaving group. And that gives you your byproduct for this reaction. The byproduct for this reaction now has this O minus attached to phosphorus, which helps to satisfy some of its desire for electrons. And then the other intermediate is this Vilsmeyer reagent that can now go attack stuff. Very reactive Vilsmeyer reagent. Yeah? And if you add the phosphate product go again, and try to find another reagent. Typically you add one equivalent of this. And if you have this floating around in the presence of this, that 1 O minus makes it way less reactive. Okay, so that's the Vilsmeyer-Hawk reaction. Okay, so let's talk more about phosphene oxides as leaving groups. And I'm not going to title that. I'll simply call this the Appel reaction. This is a reaction that's useful for substituting OHs with bromides under mild conditions. So that PBR3 thing, nobody uses that in typical organic chemistry labs. They use the Appel reaction because it's ultra mild. And here's the reagents that you use for the Appel reaction. You use bromiform. And I'm going to draw, sorry, carbon tetra bromide. And I'm not going to draw a carbon tetra bromide like this. I'm going to draw one of those carbon bromine bonds. So you can see that we're going to break that right off the bat. And we throw in triphenylphosphine with that. And that's the magic system. And a thousand years, a million years, I don't think you could have guessed what's going to happen. Maybe you could. You're going to attack this into a substitution like this. Right? I know you want to attack carbon. You want to attack carbon so badly. That's so logical. Everything you've ever done is attacking carbon and making bromide act as a leaving group. And it doesn't do that. It makes this. Who would have thought that? Oh, let's not be our minus. Bosvonium. That's the hardest part about getting the mechanism right for the Appel reaction is remembering to attack the bromine and know don't attack the carbon atom. Now you've got this basic. This is the conjugate base from bromiform. You've got this basic anion here. And that will very quickly deprotonate in order to give you bromiform byproduct. Kind of like chloroform that you use for a solvent. This is called bromiform. I don't know who invented that name back in the 1800s, but that's what it's called. And so now you've got this alkoxy minus species. And you've got that phosphonium plus species. And so of course they're going to react with each other. It bothers me that I'm not showing the oxygen attacking from the backside of the, but whatever. So the oxygen will attack opposite the bromide. And so you get this phosphorane intermediate. Pentevalent phosphorus. It's okay. No charge on five bonds to phosphorus. So you can kind of see what's going to happen here. We've got this great leading group now that's bromide. It's going to leave. And once the bromide leaves, it's ready to come back in and attack something. So when that bromide leaves, we now have, you can see the phosphine oxide. Component of this reaction. Just ready to take off. So that's what happens when you eject a bromide. Yeah. What do I mean to have that arrow happen? What do I do? It kind of looks a-send to you, but yeah, in the middle there. Oh, no. I did not mean to do that. Thank you. I'm going to go ahead and get the other one. I'm going to get the other one. I'm going to get the other one. I'm going to get the other one. Well, thank you. I don't know what-whatever. Sorry. It sickenes me that I did that. Oh. Okay, there's the bromide anion. And now it just comes right in and doot. Or you could make the P-O double bond if you want. And so that's what leads to the phosphine oxide. That's what gives you a product where you've now substituted the alcohol and your byproduct as in most triphenylphosphine reactions is triphenylphosphine oxide. So that's the problem. The Mitsunobu reaction, it does something very similar. The Mitsunobu reaction generates a leaving group like this. And in the Mitsunobu reaction, a carboxylate anion displaces. And I'm not going to talk about the Mitsunobu reaction. It's not really important. Okay, we've got just a couple of minutes here. So let me finish up by talking about the Vittig reaction. I expect that most of you, I expect you've covered the Vittig reaction in sophomore organic chemistry. But I'll just remind you of what that's all about. So let's just suppose you've got some sort of aldehyde. And what do you do with aldehydes at UC Irvine or anywhere else? You attack them with things. And one powerful way to attack them is a way that makes double bonds. So if you take this Vittig reagent, and I've got a triphenylphosphine with a phosphorus with three phenyl groups, so there's now five bonds to phosphorus. Let me draw this out more explicitly just so you can see. You've got five bonds to phosphorus. That's okay. Phosphate esters have five bonds to phosphorus. And that's your reagent. When you mix in this Vittig reagent, this phosphonium illid species, it allows you to make carbon-carbon double bonds. And the byproduct for this reaction, as you might expect, is triphenylphosphine oxide. So huge driving force for a generation of that byproduct. And so let's go ahead and talk about the mechanism for this. I'm going to draw a resonance structure for this in which I give the electrons to carbon. So let me draw this resonance structure here for this, because I call it a phosphonium illid. Why do I call it an illid? Because I think of it like this. I think of it like a carbanion that's stabilized by a neighboring phosphonium group. And so when you put this in the presence of oxygen, of carbonyl compounds, particularly aldehydes and ketones, esters are not so reactive, then this ends up doing a reaction that looks like this. And I'll talk about the transition state for this in just a moment. Let me put a prime here so I don't have the same labels. And it might bother you to have these five bonds to phosphorus. Don't be bothered by that. That's OK. We're not trying to isolate this. I'm not saying it's stable, but there's no problem having intermediates with five bonds to phosphorus, a phosphane like that. So now I get to choose how I pop this group off. I'm going to do it so it makes a P0 double bond directly, but you could simply give the electrons to oxygen and have P plus O minus. And so this is what allows you to make these two by products here. And I want to talk about this step right here. There were arguments for, I feel like, decades about, is this really concerted? Is that really a 2 plus 2 reaction? I feel like this was finally resolved with some good ab initio calculations. That show that, yes, it is concerted. If you look at the transition state for this Vittig reaction, the transition state looks like this. In the transition state, you've got this nucleophilic species here with this carbon atom and the phosphonium coming very close to the aldehyde. And as it gets closer and closer and closer and closer and closer, you start to attack and build up negative charge here on this oxygen. And that negative charge starts to come back in and attack back on the phosphorous at the same time. It is not equal. In the transition state, you've got about 44% of a carbon-carbon bond here. But in that same transition state, there's only 1% of a P0 bond. It's not equal. By the time you're over the energetic hump, you're starting to form a little bit of a bond here, but not a lot. But that's important. Even that 1% means that you are not going through a transition state that looks like this. The fact that there's 1% bond says the transition state does not look like this. The transition state does not look like you're simply, let me draw it delta minus and delta plus. The transition state does not look like this carbanion. It's simply attacking and then this is freely floating around. There's partial bonding. So you should draw this as a concerted, I'll call that a 2 plus 2 reaction. So it's not what we call paracyclic, but it is concerted. If you take Chem 202, you'll talk about paracyclic reactions. Okay, so that's it. Let me talk about how you make these illid reagents if you don't remember that. So you use phosphines as nucleophiles. Triphenylphosphine is not great, just like it's not a great base. But if you heat them enough or if you have a reactive enough electrophiles, you can do S and 2 reactions. So in this reaction, it takes 2 days at room temperature. So it's not. Triphenylphosphine is nowhere near as nucleophilic as trialkylphosphine. And the reason you want phenyls on here is because as long as you've got phenyls, that means there's only one group that can be deprotonated and that's the methyl. If you use tributylphosphine, now it's like, well, where are you going to deprotonate if there's 3 alkyl groups on there? And now you need a very, very, very strong base to pull a proton off of a phosphonium salt. So a common one is just straight-up alkyl lithium. And I know maybe you're going to want to say to try to attack the phosphorus with the alkyl lithium, but that's not what happens. It's faster for the alkyl lithium to deprotonate. And so that's how you make illids. You do SN2 reactions to make phosphonium salts. And then in a second step, you use super powerful bases to pluck those protons off. And that's what you throw in with your aldehyde. Okay, so that's it for phosphorus chemistry and that's it. I'll see you at, there's a discussion section tonight at 5 o'clock. It's optional. You don't have to come, but I'd like you to be there just so we can go over some of this phosphorus and sulfur chemistry. And the exam is at, somebody know? It's in this room. We will post the answer key today. Yeah, for all the problem sets by today. And these lectures are online.
 Okay, so this is our final lecture in this class.  We just finished talking about sulfur chemistry.  We dropped down from the second row to the third row.  And we're going to finish up by talking  about phosphorus chemistry.  And I want to introduce you to some of the functional groups  that we're going to talk about today.  We can't cover everything.  But we're going to focus on these four types  of functional groups, phosphines.  And that's where you have three alkyl  or aryl groups bonded to phosphorus.  So three carbon phosphorus bonds in a phosphine.  When you switch to phosphorus heterodontal bonds,  you change the name.  And in particular with oxygen, we call this a phosphite.  This kind of species where there's three oxygens on here.  And more specifically, I would refer to this  as a phosphite ester.  But in general, I would simply say, oh,  that phosphite over there.  And whenever I see these top two,  I'm focusing relentlessly on those lone pairs on phosphorus.  So we'll talk about that in the chemistry.  So it's very easy to oxidize phosphorus.  Three, we'll talk about that.  So this is phosphorus with an oxidation state of three.  This has an oxidation state of five down here.  This is a phosphine oxide.  Let me contrast that with oxidation state three up above.  So at the oxidation state of five,  this is the most common functional group you'll see.  It's a phosphine oxide  where we have this oxo ligand sticking on top of phosphorus.  And if you do the same for a phosphite,  we don't call that a phosphite anymore.  We call that a phosphate.  And you should be familiar with this functional group  because it's the backbone of DNA and RNA.  And it's the key functional group on phosphorylated proteins.  So let's go ahead and talk about what kinds of things change  as we drop down from the third row, sorry,  from the second row nitrogen to the third row with phosphorus.  So let me remind you of some of the pKa's about the basicity  of the lone pairs on nitrogen.  And what we'll do is we'll look at the pKa  of the protonated nitrogen compound.  So triethylamine is a common base.  And we can say something about the basicity of the lone pairs  on triethylamine by looking at the acidity of the conjugate acid  of triethylamine.  So if you compare triethylamine with triphenylamine,  as you might expect, triphenylamine is,  it's not basic at all.  If you look at the pKa for triphenylamonium,  it's on the order of minus 5,  whereas triethylamonium is on the order of 10 to 11.  I'll use 11 as my approximate number there.  It's extremely difficult to stick a proton on triphenylamonium.  That's because the lone pairs on nitrogen are donating  into, let me draw the lone pair up here,  are donating into this pi system.  And you can draw resonance structures all over the place.  That's exactly why aryl amines are so reactive  in electrophilic aromatic substitution  because these nitrogen lone pairs are juicing up that pi system  and making it more nucleophilic.  So you have really good resonance here  with nitrogens attached to benzene rings.  If you look at the corresponding phosphorus compounds,  just in terms of pKa, and I'm going to draw butyl groups here  because that's the most common alkylphosphine,  but it's about the same for triethylphosphine.  If you look at the pKa, it's pretty similar.  That's a pretty similar basicity.  The desire to have a proton stuck on there is pretty similar  for a phosphonium species.  10, 11, that's sort of within the same ballpark.  If I look at triphenylphosphonium,  protonated triphenylphosphine,  now the pKa is significantly different  from the nitrogen case.  Instead of being way at the minus 5,  in other words, this is a super strong acid.  If you could figure out how to stick a proton  on that nitrogen, that would be a very strong acid.  But a protonated triphenylphosphine is not a super  strong acid.  Triphenylphosphine is not super duper acidic,  and we can trace this back to the length  of these carbon phosphorous bonds.  So yes, there is a very nucleophilic lone pair  on this phosphine.  And hypothetically, there's some donation  into this pi star system of the airing ring,  but look how far away that, of course,  I'm trying to exaggerate it so you'll remember,  but it's so far away.  You don't get good donation  to form partial PC double bonds here.  This goes back to this issue that, yes,  the orbital is very big here on phosphorous,  but it's so far away.  You don't get effective donation,  and that makes this lone pair more basic.  It's spending less time donating into the pi system,  so it has more room sort of to more left  over reactivities to react with protons.  Okay, so trialkylphosphines are more basic  than triarylphosphines.  And triarylphosphines are more basic  than the corresponding second row atom system  with nitrogen there.  I want to return to a mechanism that you've probably,  I hope you've seen before, or maybe not the mechanism,  the reaction I hope you've seen before  from sophomore organic chemistry.  This description is not the reaction.  This is a description of part of this process.  So here's a reaction that you might have seen before  in sophomore organic chemistry,  and it involves taking PBR3, phosphorus tribromide,  and using that as a reagent  to substitute the hydroxyl groups of alcohols,  primary alcohols, secondary alcohols.  You could do tertiary alcohols.  The stoichiometry looks like this.  You take three equivalents of alcohol for each tribromo  for each phosphorus tribromide,  and that allows you to substitute the OHs.  So that's a very useful substitution reaction  because there's all kinds of things you can do  without kilbromides.  You can turn them into green yard reagents.  You can substitute them now with nucleophiles.  Okay, so here's a reaction I hope most  of you have already seen before.  Maybe something that was never mentioned to you before is  that this reaction is acid catalyzed.  Probably with every fiber of your being,  you want to somehow use this oxygen, right?  You have to make PO bonds in your product.  Probably you want to do SN2 reactions on that,  and I want to urge you to avoid that mistake.  Let's talk about the mechanism for this reaction.  The mechanism of the reaction starts off with protonation.  Whoops. Starts off with protonation  of the phosphorus lone pair.  That's why acid is essential.  You cannot buy PBR3 that doesn't have some acid in there.  And once this reaction starts,  you start generating acid from the get-go.  And so the initial step here is that you protonate that phosphorus,  and the second you put a proton on there,  you've turned that phosphorus  into a super-duper electrophile.  So now that I've got a proton on there, look at that.  If you want to make a bond of phosphorus,  now it's easy to make a bond of phosphorus.  Now it's easy to attack the phosphorus.  Proton transfer reactions are generally very fast,  and so now we've changed this  to where it becomes very easy to attack.  Let me go ahead and draw a nucleophile here like ethanol, E-T-O-H.  Here's my lone pair, and there's that phosphorus,  and so that those lone pairs on the ethanol,  the alcohol, want to attack this.  Stereoelectronic there's a preference to add anti  to the most electronegative ligand.  Proton is not the most electronegative ligand.  You add anti to one of the bromines here.  So you end up with a trigonal bipyramidal phosphorus.  So here's my equatorial ligands,  and here's my axial ligands on here.  So here's that, and I'm trying to figure  out whether I'm going to just skip this deprotonation step  because I'm confident you can figure out that after I attack,  yeah, you need to use deprotonate that proton  from the ethoxy nucleophile.  So you end up with this trigonal bipyramidal intermediate.  That's not well drawn.  So you end up with a trigonal bipyramidal intermediate.  No charge on phosphorus.  It's neutral.  Don't freak out over the five bonds to phosphorus.  This is not a second row atom.  Perfectly happy as an intermediate with five bonds.  Remember phosphate over there?  That's got five bonds.  Okay, so this is perfectly okay as an intermediate,  but that's not to say that leaving groups don't want to leave.  Now the axial leading group will prefer to leave.  That's just a stereoelectronic preference.  And so I get right back to a tetrahedral type intermediate.  And so in that tetrahedral intermediate,  I now have this ethoxy group attached.  I now have this electrophilic phosphonium species.  And so now, and that bromide is now floating around,  and I'm going to draw it right over here so you can see  that it's ready to attack carbon.  Wow, that phosphorus and oxygen really want this pair  of electrons here.  And so they're simply going to take it.  And I've got two choices for how I draw this reaction out.  I can make the electrons form a P O double bond,  or I can give the electrons to the oxygen  and draw a resonance structure for that.  And I'm simply going to draw that resonance structure.  So here's my initial intermediate here.  And here's my minus, and here's my plus.  And so now to finish up this stage of the process,  what I need to do is I need to stick a proton  on the O minus using acid,  and then pluck a proton off the phosphorus.  I'm not going to draw those two steps for you.  I'll draw the first step here.  So I think you can see that after I do that,  I can use the A minus.  And so you just do this three more times.  And that's what allows you to, sorry,  I ran out of room there.  And that's what allows you to replace those OHs  on your alcohol with BR groups.  So look at what we're doing.  This was the key step right here.  The fact that bromide came in and did an SN2 reaction in order  to pop off a species that had this sort  of phosphine oxide character.  And don't freak out over the five bonds to phosphorus  that pentavalent, let me give you a name here  so we know how to describe this.  This is called a phosphorane.  That's called a phosphorane intermediate.  Trigonal bipyramidal, five bonds to phosphorus.  That's like a tetrahedral intermediate  when you substitute ester derivatives  of carboxylic acids.  Okay, I want to extend this idea that when you have phosphorus  with lone pairs, you can protonate those.  And so let's talk about phosphites, not PBR3,  but phosphorus with some bonds on there.  I want to, I'd like to talk about some patterns  of substitution and I want you to compare just mentally.  Please don't write this down because I'm assuming  that this is stuff that you already know.  Or if you don't know this, you don't have to.  Which of these is easier to attack with alkoxide?  The ester or the ether?  Yeah, the ester.  There is no reaction where you can substitute OR by having,  this is very easy to attack a carbonyl.  It is, I don't know of any way to get a hydroxide anion  to attack an ether and do an SN2 reaction.  Okay, but now let's talk about phosphorus  because I want to urge you not to extend what you already know  about carbon to phosphorus because it's not the same.  Let's draw it a little table here.  What I'll do is I'll now draw a phosphate ester instead  of a carboxylic ester.  And let's try to imagine the reaction  of these two different species, a phosphate ester  versus a phosphite ester with nucleophiles.  And so what are the relative rates of hydrolysis  of a phosphate versus a phosphite with hydroxide anion?  If you look at the relative rates, it's exactly,  it's not exactly opposite.  It's, right, you can't even imagine attacking an ether  oxygen but with phosphorus you get a thousand times faster rates  of attack and substitution at phosphites.  It's less reactive when there's an oxygen atom on top of there.  And if that is in any way hard for you to understand,  then go ahead and draw this resonance structure  so that you can see that there's an O minus right next door  that's already donating into that phosphorus.  So now you can imagine why it's hard for hydroxide anion  to attack a phosphate ester.  Let's switch over to an acid catalyzed reaction.  Suppose we use acid to catalyze the hydrolysis  of these two esters, a phosphate ester  versus a phosphite ester.  The phosphate ester is still the loser here.  It's still the least reactive of these pairs.  If you want to make phosphorus oxygen bonds,  you should use phosphites.  And the relative rate difference is now, it's 10 to the 12.  It won't matter if you write the right number of zeros.  You get the idea.  It is way, way, way more reactive to substitute this  under acidic conditions.  And I just showed you why that's true.  I just showed you why.  It's so much faster.  When you protonate a phosphorus lone pair on a phosphate,  it makes it super-duper-duper-duper fast to attack  at that phosphonium species.  And you can't do that with a phosphate ester.  Protonating this, you still have lone pairs on that oxygen  that are donating in.  Okay, so phosphites are very fast to exchange.  Okay. Let me give you one example of a reminder  of how difficult it is to substitute phosphate esters.  Whereas, phosphate esters substitute quickly.  So if you look at the rates or if you look at the hydrolysis  of trimethyl phosphate with alkoxide.  Right? You know, the typical mechanism  that you draw looks like this.  And you kick the electrons up to the oxygen to give an O minus.  You'll end up with this pentavalent phosphorus intermediate.  Fifty percent of the substitution that takes this down  to the product in this case.  I guess I should, sorry, I should have hydroxide anion here.  I apologize for that.  Otherwise, there's no way to have a product.  It would be.  So this would be the product.  Fifty percent of the product when you attack trimethyl phosphate  with hydroxide anion comes from attack at carbon.  Right? It is so hard to attack at phosphorus,  you start to see attack at carbon.  This is kind of like making a sulfonate ester.  The reason you make sulfonate esters is not because you want  to attack at sulfur, it's because you want to do SN2 reactions  at carbon.  So it's the same idea with phosphates.  It's just very hard to attack here at this phosphorus  of a phosphate ester.  So if you're clever and you want  to make phosphorus oxygen bonds,  then you should use phosphates, not phosphates.  Guess what they do in modern DNA synthesis?  They don't use phosphates.  They use phosphates.  Let me go ahead and draw a typical phosphorus oxygen  coupling reaction.  This is going to be a picture of my DNA strand.  And suppose I wanted to add one  and I'm doing a chemical synthesis of DNA.  I'm making a 50mer or a 60mer.  And I'm now adding on my 50th subunit.  The subunits that you use, the reagents that you use  in modern DNA synthesis are basically,  here let me draw this isopropyl correctly.  So that's an isopropyl amino group, are basically phosphates.  I don't want you to worry  about what the protecting group is on that oxygen.  So this is a subunit used for DNA synthesis.  It's called a phosphoramidite.  And there's a trital, a triphenyl methyl group there you take  off with an SN1 reaction.  And your other substituents on your DNA base,  you've got some sort of a base there.  Three of the four bases need protection on there.  So when you want to make a phosphorus oxygen bond  in DNA synthesis, you don't make phosphate esters initially,  even though that's what you want.  You have phosphate esters connecting DNA.  What you do is you take this phosphoraminite reagent  and you rely on the fact that if you put in an acid,  it's now very fast for alcohols to attack that and substitute.  And in this case, they substitute the nitrogen.  So there's a super secret acid they use called tetrazole.  You don't need to know that.  The bottom line is there's an acid in there  that accelerates the rate  of oxygen nitrogen substitution.  And we're not going to draw the whole mechanism  because it can be more complex.  Tetrazole can also substitute the nitrogen.  Don't worry about that.  The bottom line is in DNA synthesis.  You have minute long reaction scales.  When was the last time you did a reaction  that took one minute or less?  Okay, I'm done.  Now, it doesn't work like that typically.  But with DNA synthesis, it's so optimized and it's so fast  to substitute at phosphites.  Sorry, let me take off that plus charge there.  But this is the way you make phosphorus oxygen bonds  in DNA synthesis.  You use phosphites.  Now, nobody wants to make polyphosphites.  Those bonds are now so sensitive  to acid-catalyzed substitution, you've got to get rid  of this.  So what you do in DNA synthesis is you oxidize this now  and the reagent combination that you use is iodine and water.  And that converts and it's super fast.  It's not the same mechanism as this other one.  We'll talk about this.  But it is super fast to oxidize phosphites  and phosphines to phosphine oxides.  And now those bonds are now stable, right?  You do this initial ester exchange  under acid-catalyzed conditions with a phosphite.  But then when you want to stabilize these bonds,  you convert that to a phosphate.  And now those bonds will sit through round after round  of DNA synthesis and be perfectly stable.  So this is modern DNA synthesis.  It takes advantage of the rapidity  of phosphite ester exchange.  Let's talk about phosphates and phosphine oxides  in these P O double bonds.  All these are the carbonarmeApplause.  So we with you are simply going to talk  about this now is super fast and the process can,  it comes out like this.  Many, many, many reactions in organic synthesis are driven  by the formation of phosphorus-oxygen bonds  or these things that we call, in other words,  what do I call this?  Do I call that a phosphorus-oxygen double bond?  Well, that's not so clear.  It's not so clear how I should draw this.  This is just like a sulfur-oxygen double bond.  There's another resonance structure that I can draw  in which I'll bring these electrons  up to the more electronegative oxygen atom.  And I can draw this illid-like structure with P plus O minus.  What you'll find is that I draw it like this  and everybody else draws it like this  because we're too darn lazy to draw it like this.  There is no time ever in which it helps you predict reactivity  by drawing it like this.  You always can predict reactivity by drawing it like this.  It just becomes burdensome to draw these charges around there.  So because I'm lazy and because I always think  of it like this, I continue to draw it like this.  But I would urge you to draw it like the equivalent way  because that way you're ready to use the O minus  as a nucleophile when you need to.  We're ready to rationalize why things will attack phosphorus.  Okay, so it is very facile to oxidize.  When you leave trialkyl phosphines exposed to air,  they oxidize very rapidly.  So if you open the top off your bottle of tributylphosphine,  you can expect it to be very rapidly oxidizing the phosphine  to the phosphine oxide.  So in the case of trialkylphosphines,  and this kind of correlates with basicity,  alkylphosphines are more reactive  than triarylphosphines toward acids.  They're more basic.  And they also attack oxygen faster.  So tributylphosphine oxidizes very fast and very reactive.  Whereas triphenylphosphine, you know, you can open the bottle  and walk over to the balance kind  of leisurely, scoop a little bit out to put in your reaction.  And you don't have to worry  about it all getting converted to the phosphine oxide.  Okay, so it's very thermodynamically favorable to do this.  Let me give you some bond energies here.  In many, many reactions, you'll break bonds to things  like carbon and oxygen.  And let me just remind you that's  about a 90 kcal per mole bond.  And so if you can break bonds between carbon and oxygen  and replace them with bonds between phosphorus and oxygen,  and it doesn't matter how I draw that.  It doesn't matter whether I draw this P plus or O minus.  That bond, whether you draw P plus or O minus  or P double bond O, is energetically very favorable.  137 kcal per mole.  And for the tributylphosphine, 128 kcal per mole.  I'm positive I have these reversed.  But that's what's in my notes.  I'll fix the notes later, but I'm super positive  that I reversed those in my notes that I have drawn here.  The bottom line is these are very much more favorable  than any bond between oxygen and carbon.  And you will find in many, many reactions  where you see people dump in triphenylphosphine.  They're trying to strip off an oxygen from carbon  and end up with this.  And so that provides a huge thermodynamic driving force.  Let's come back and draw a mechanism for this iodine  oxidation and DNA synthesis.  Because this transformation is driven by formation of this P O,  if you want to call it a double bond, that P O double bond.  There's less than 50% pi character between the P and the O.  It's kind of like a sulfur oxygen double bond.  Is it really a double bond?  There's not much pi character there.  Let's go ahead and draw this mechanism for iodine oxidation.  If you ever stain phosphines on your TLC plate by dipping it  in an iodine chamber, this is what's going on.  So if I have iodine in the presence of a phosphine,  you should expect that that lone peron phosphine  on phosphorous will very quickly displace that iodine.  And it will give you this phosphonium intermediate.  Let me just draw R here.  And it doesn't matter whether the R's are alkoxy groups  or alkyl or phenyl.  So now I have this phosphonium.  It's easy for things to attack, things like water.  So if you have water present, it adds opposite the most  electronegative group.  That's a stereoelectronic preference  to give you a trigonal bipyramidal phosphorane intermediate.  I don't want to show the deprotonation of the water  that comes after the attack.  I think you guys can get the idea that after I attack  with water I have to pull a proton off.  And so now I've got this phosphorane.  That's supposed to be an R. I just didn't do a good job  of drawing that out.  And so now I make something leave.  And the leaving group that leaves best is iodide.  And so the intermediate that you get.  Looks like this.  And now all I have to do is remove the proton.  And I can see how I generate a P double bond.  If that's the resonance structure I want to draw.  You can see how I generate this P double bond O. So  in DNA synthesis this is what you're doing.  You're just treating your phosphite ester  with iodine and water.  And you end up with a phosphate ester that's very stable.  If you ever wanted to convert a phosphine  and an organic reaction to a phosphinoxide you could use any  oxidant you want.  And it will make the phosphinoxide.  MCPBA, hydrogen peroxide, maybe in a less controlled manner  if you just blew air through the reaction.  Nobody does it that way.  But okay.  So phosphorus lone pairs attack things.  That's the basic idea.  So let's talk about that more explicitly  in a classical, very old reaction.  I'm sure it dates back to the early 1900s, maybe 1800s.  Don't know.  It's called the Arbusoff reaction.  More specifically this is called the McHales Arbusoff reaction  but it's not really important.  So if you take a phosphite like this triethylphosphite would  be a good one for this.  And you mix that with an alkyl halide.  You're going to get an SN2 reaction.  There's no really acidic protons here  but there's a good electrophile.  That non-bonding lone pair orbital on phosphorus is very big  and very nucleophilic.  It's easy to overlap with sigma star here  so you do an SN2 reaction.  And the intermediate that you get, one of these ethoxys I'm going  to draw with all the bonds there.  So the intermediate that you get now has this phosphonium salt  there and I'm going to take my bromide counter and move it  to the other side so I can see how ready it is  to cleave that carbon oxygen bond.  Right? We want to generate a P double bond O type bond here.  There's a huge thermodynamic driving force  and kinetically that's very fast.  Well, you still have to heat these reactions substantially  but I don't want to say it's that fast.  That's the slow step right there is cleaving  that carbon oxygen bond.  But ultimately what you get out of this  is a phosphonate and this is what you use  for a Vittig-like reaction.  I'll say a little bit about Vittig reactions.  But you can use this for olefination,  something called the Horner-Wadsworth-Emmons  olefination procedure.  And you make these reagents for Horner-Wadsworth-Emmons  reactions using this Arbusoff,  Mikhailis Arbusoff reaction.  This particular type of species  where there's one carbon phosphorus, is it like a  phosphate ester or is it a phosphene oxide?  What is that?  We call this a phosphonate.  That own sound tells me phosphonate ester.  Tells me that there's a carbon oxygen,  sorry a carbon phosphor, it's not just a phosphate ester.  Phosphate has three alkoxy groups on there.  This is a phosphonate ester.  Okay, so I'm trying to play up, oh gee, it's so hard  to add to phosphorus, to phosphene oxides.  It is hard to add to phosphates.  And so I'll show you something ironic that maybe is not  so ironic and that is that you can have phosphorus 5-oxo  phosphorus species that are attacked very easily.  And here's the reagent that's very fast.  Everybody calls this Pockel III phosphorus oxychloride.  And this is not like a regular phosphate ester.  This is screaming hot reactive.  Let me just show you where, let me just remind you  of why this should be so reactive.  Typically if you make a phosphate ester like this,  there is some donation, it's not great, but there's some  donation between the lone pairs into the phosphorus.  And then make it hard for other nucleophiles to attack.  If you're coming along with your lame,  pathetic lone pair and there's already three,  four sets of lone pairs donating in there,  how are you going to wedge your way into that?  But we come back over to phosphorus oxychloride  and I'll try to exaggerate this.  Okay, yeah, there's things with lone pairs,  but those lone pairs are way, way, way,  way away from the phosphorus.  All right, chlorine's electronegative,  but those lone pairs are not really deactivating the  phosphorus.  This is extremely reactive.  This is nothing like a regular phosphate ester  in terms of reactivity.  So here's a typical place where you would take advantage  of the super-duper electrophilicity  of that phosphorus in PAKL3.  I'll show you a type of electrophilic aromatic  substitution called a Wilsmeyer reaction.  So I hope you're familiar with some sort  of electrophilic aromatic substitution reaction  called Friedel-Kraftz isolation.  You can add acyl groups to benzene rings,  but there's one particular acyl group that you can't add  and that's a formula group  where you have an alde, where you create a new aldehyde.  And the way you do that, the way you affect a Wilsmeyer  reaction is you don't use aluminum trichloride  or any of these typical Friedel-Kraftz reagents.  You use this.  This is called the Wilsmeyer reagent  because it's the key intermediate in this reaction.  If you want to add an aldehyde carbon bond  to this benzene ring, this is the reagent  that you're going to need.  So it does electrophilic aromatic substitution.  I'm not going to draw the mechanism  for electrophilic aromatic substitution.  I expect you guys to know that.  The end result, not the end result,  but before you work this up, you have now substituted.  There was an H on the benzene ring.  It's gone.  So you spit off HCl as a byproduct.  So you don't make an aldehyde initially.  You make this a mini-amiand.  But I don't know of anybody who works to keep that.  You always work that up with water  and it hydrolyzes the mini-amiand.  So this is the way you add aldehydes to aromatic rings  using electrophilic aromatic substitution.  You use the Wilsmeyer-Hock reaction.  So where do you get this chloroaminium species?  That's not just a regular aminium ion.  There's a leaving group attached to that aminium ion.  The way you get it is you mix POCl3, Pocl3 with DMF.  And that's typically used as the solvent for the reaction.  So let's talk about that.  Let's talk about the reaction of POCl3 with amides.  Phosphorous oxychloride is so reactive  that the lone pairs on amide oxygens will attack.  So typically with the Wilsmeyer reaction,  you use DMF as a common solvent for organic synthesis.  You just take your pot, use DMF as your solvent,  and you dump in POCl3.  That's supposed to be a chloride right there.  I just didn't do a good job of drawing that.  So the lone pairs on the amide are very nucleophilic.  If you'd like a reminder for why that's nucleophilic,  here's a lone pair on nitrogen that's donating in.  So I'm not going to draw that resonance structure.  Let me just take my arrow away  because I don't want to have too many arrows.  Okay, so the lone pairs on the oxygen can attack,  and they attack opposite the most electronegative group.  Most electronegative groups are chlorine,  not the oxygen on here.  But if you want to understand the reactivity  of this, you could draw it like this, right?  So what's more electronegative?  Oh, with an anionic charge on it or chlorine?  Chlorine's more electronegative.  So you attack opposite one of the three chlorines,  and you generate an intermediate that looks like this.  This is one of two possible resonance structures  that I could draw.  Or if you wanted to draw the double bond O,  you could have kicked the electrons up to the oxygen.  It's the same thing.  Okay, so you end up with this phosphorane intermediate,  this pentavalent phosphorane intermediate,  and so now we'll pop off one of the leaving groups.  And there's another resonance structure you could have drawn  for that amide if you wanted.  That would be the equivalent of what I'm drawing here.  And so now that chloride ion that you popped  out is just floating around looking for something to do.  So it shouldn't surprise you  that that chloride will eventually attack here  at this carbonyl species.  Or if you drew the aminium ion resonance structure,  we still attacked the same group.  You'd still draw it attacking the same carbon.  And I don't want to draw the substitution mechanism here,  but you'll add, well, I might as well.  So you have this chlorophosphate leaving group ready to leave.  And as soon as you add in to this species,  it's now ready to push out a leaving group.  It's okay if it pushes out the chloride and goes back, right?  But eventually it's going to push  out the chlorophosphate leaving group.  And that gives you your byproduct for this reaction.  The byproduct for this reaction now has this O minus attached  to phosphorus, which helps to satisfy some  of its desire for electrons.  And then the other intermediate is this Vilsmeyer reagent  that can now go attack stuff.  Very reactive Vilsmeyer reagent.  Yeah?  And if you add the phosphate product go again,  and try to find another reagent.  Typically you add one equivalent of this.  And if you have this floating  around in the presence of this,  that 1 O minus makes it way less reactive.  Okay, so that's the Vilsmeyer-Hawk reaction.  Okay, so let's talk more  about phosphene oxides as leaving groups.  And I'm not going to title that.  I'll simply call this the Appel reaction.  This is a reaction that's useful for substituting OHs  with bromides under mild conditions.  So that PBR3 thing, nobody uses that  in typical organic chemistry labs.  They use the Appel reaction because it's ultra mild.  And here's the reagents that you use for the Appel reaction.  You use bromiform.  And I'm going to draw, sorry, carbon tetra bromide.  And I'm not going to draw a carbon tetra bromide like this.  I'm going to draw one of those carbon bromine bonds.  So you can see that we're going to break that right off the bat.  And we throw in triphenylphosphine with that.  And that's the magic system.  And a thousand years, a million years,  I don't think you could have guessed what's going to happen.  Maybe you could.  You're going to attack this into a substitution like this.  Right? I know you want to attack carbon.  You want to attack carbon so badly.  That's so logical.  Everything you've ever done is attacking carbon  and making bromide act as a leaving group.  And it doesn't do that.  It makes this.  Who would have thought that?  Oh, let's not be our minus.  Bosvonium.  That's the hardest part about getting the mechanism right  for the Appel reaction is remembering to attack the bromine  and know don't attack the carbon atom.  Now you've got this basic.  This is the conjugate base from bromiform.  You've got this basic anion here.  And that will very quickly deprotonate in order  to give you bromiform byproduct.  Kind of like chloroform that you use for a solvent.  This is called bromiform.  I don't know who invented that name back in the 1800s,  but that's what it's called.  And so now you've got this alkoxy minus species.  And you've got that phosphonium plus species.  And so of course they're going to react with each other.  It bothers me that I'm not showing the oxygen attacking  from the backside of the, but whatever.  So the oxygen will attack opposite the bromide.  And so you get this phosphorane intermediate.  Pentevalent phosphorus.  It's okay.  No charge on five bonds to phosphorus.  So you can kind of see what's going to happen here.  We've got this great leading group now that's bromide.  It's going to leave.  And once the bromide leaves, it's ready to come back  in and attack something.  So when that bromide leaves, we now have,  you can see the phosphine oxide.  Component of this reaction. Just ready to take off.  So that's what happens when you eject a bromide.  Yeah.  What do I mean to have that arrow happen?  What do I do?  It kind of looks a-send to you, but yeah, in the middle there.  Oh, no. I did not mean to do that.  Thank you.  I'm going to go ahead and get the other one.  I'm going to get the other one.  I'm going to get the other one.  I'm going to get the other one.  Well, thank you.  I don't know what-whatever.  Sorry.  It sickenes me that I did that.  Oh.  Okay, there's the bromide anion.  And now it just comes right in and doot.  Or you could make the P-O double bond if you want.  And so that's what leads to the phosphine oxide.  That's what gives you a product where you've now substituted  the alcohol and your byproduct as in most  triphenylphosphine reactions is triphenylphosphine oxide.  So that's the problem.  The Mitsunobu reaction, it does something very similar.  The Mitsunobu reaction generates a leaving group like this.  And in the Mitsunobu reaction, a carboxylate anion displaces.  And I'm not going to talk about the Mitsunobu reaction.  It's not really important.  Okay, we've got just a couple of minutes here.  So let me finish up by talking about the Vittig reaction.  I expect that most of you, I expect you've covered the Vittig  reaction in sophomore organic chemistry.  But I'll just remind you of what that's all about.  So let's just suppose you've got some sort of aldehyde.  And what do you do with aldehydes at UC Irvine or anywhere else?  You attack them with things.  And one powerful way to attack them is a way that makes double  bonds.  So if you take this Vittig reagent, and I've got a  triphenylphosphine with a phosphorus with three phenyl  groups, so there's now five bonds to phosphorus.  Let me draw this out more explicitly just so you can see.  You've got five bonds to phosphorus.  That's okay.  Phosphate esters have five bonds to phosphorus.  And that's your reagent.  When you mix in this Vittig reagent, this phosphonium  illid species, it allows you to make carbon-carbon double  bonds.  And the byproduct for this reaction, as you might expect,  is triphenylphosphine oxide.  So huge driving force for a generation of that byproduct.  And so let's go ahead and talk about the mechanism for this.  I'm going to draw a resonance structure for this in which I  give the electrons to carbon.  So let me draw this resonance structure here for this, because  I call it a phosphonium illid.  Why do I call it an illid?  Because I think of it like this.  I think of it like a carbanion that's stabilized by a  neighboring phosphonium group.  And so when you put this in the presence of oxygen, of  carbonyl compounds, particularly aldehydes and ketones,  esters are not so reactive, then this ends up doing a  reaction that looks like this.  And I'll talk about the transition state for this in just  a moment.  Let me put a prime here so I don't have the same labels.  And it might bother you to have these five bonds to  phosphorus.  Don't be bothered by that.  That's OK.  We're not trying to isolate this.  I'm not saying it's stable, but there's no problem having  intermediates with five bonds to phosphorus, a  phosphane like that.  So now I get to choose how I pop this group off.  I'm going to do it so it makes a P0 double bond directly,  but you could simply give the electrons to oxygen and have  P plus O minus.  And so this is what allows you to make these two by  products here.  And I want to talk about this step right here.  There were arguments for, I feel like, decades about,  is this really concerted?  Is that really a 2 plus 2 reaction?  I feel like this was finally resolved with some good  ab initio calculations.  That show that, yes, it is concerted.  If you look at the transition state for this  Vittig reaction, the transition state looks like this.  In the transition state, you've got this nucleophilic  species here with this carbon atom and the  phosphonium coming very close to the aldehyde.  And as it gets closer and closer and closer and closer  and closer, you start to attack and build up  negative charge here on this oxygen.  And that negative charge starts to come back in and  attack back on the phosphorous at the same time.  It is not equal.  In the transition state, you've got about 44% of a  carbon-carbon bond here.  But in that same transition state, there's only 1%  of a P0 bond.  It's not equal.  By the time you're over the energetic hump, you're  starting to form a little bit of a bond here, but not a  lot.  But that's important.  Even that 1% means that you are not going through a  transition state that looks like this.  The fact that there's 1% bond says the transition state  does not look like this.  The transition state does not look like you're simply,  let me draw it delta minus and delta plus.  The transition state does not look like this carbanion.  It's simply attacking and then this is freely floating  around.  There's partial bonding.  So you should draw this as a concerted, I'll call that a  2 plus 2 reaction.  So it's not what we call paracyclic, but it is concerted.  If you take Chem 202, you'll talk about paracyclic  reactions.  Okay, so that's it.  Let me talk about how you make these illid reagents if  you don't remember that.  So you use phosphines as nucleophiles.  Triphenylphosphine is not great, just like it's not a  great base.  But if you heat them enough or if you have a reactive  enough electrophiles, you can do S and 2 reactions.  So in this reaction, it takes 2 days at room temperature.  So it's not.  Triphenylphosphine is nowhere near as nucleophilic as  trialkylphosphine.  And the reason you want phenyls on here is because as long  as you've got phenyls, that means there's only one group  that can be deprotonated and that's the methyl.  If you use tributylphosphine, now it's like, well, where are  you going to deprotonate if there's 3 alkyl groups on there?  And now you need a very, very, very strong base to pull a  proton off of a phosphonium salt.  So a common one is just straight-up alkyl lithium.  And I know maybe you're going to want to say to try to  attack the phosphorus with the alkyl lithium, but that's not  what happens.  It's faster for the alkyl lithium to deprotonate.  And so that's how you make illids.  You do SN2 reactions to make phosphonium salts.  And then in a second step, you use super powerful bases to  pluck those protons off.  And that's what you throw in with your aldehyde.  Okay, so that's it for phosphorus chemistry and that's  it.  I'll see you at, there's a discussion section tonight  at 5 o'clock.  It's optional.  You don't have to come, but I'd like you to be there just  so we can go over some of this phosphorus and sulfur  chemistry.  And the exam is at, somebody know?  It's in this room.  We will post the answer key today.  Yeah, for all the problem sets by today.  And these lectures are online.
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