The Robinson Annulation & Amines
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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. 
 Good afternoon. We're going to get started. Any questions? I did open up the sapling chapter 24 assignment. It's due on Wednesday. We don't want it due on Thursday night because then you're going to, human nature, right? You're going to procrastinate and then you're going to be more stressed out on Friday. Don't you think? Who thinks we should, who thinks we should have a due date of Wednesday night? Okay, so, and Thursday? Okay. You'll be glad. Alright. So here's how I know you guys are burned out and ready for summer to start because nobody emailed me to say, oh by the way the chapter 24 assignment's not up. Nobody emailed me. So, and we're finishing chapter 24 today so I can tell. Alright. So we left off. We're going to finish this today. We're going to start chapter 25. Two more chapters left and you're done. Alright. So we left off last time talking about the Michael followed by the Aldol also known as the Robinson Annulation. So we went through the Michael edition. So this is two mechanisms. So if I ask you a Robinson Annulation mechanism on the test, it's actually like me asking you two mechanisms, a Robinson, a Michael and an Aldol. Alright. So we do the Michael edition and we said without you you can stop at this Michael product here. So I will always add heat to let you know that you're going to keep going. So then we show we're showing the Aldol condensation. We kind of left off in the middle of that mechanism. So this is exactly the same as the less complex Aldol general mechanism that we showed already. So we're forming the enolate and then we're attacking the carbonyl. It's an intramolecular Aldol but that's no problem. And so what we need to do is we need to protonate this alkoxy group. So we're going to protonate. And then we're going to do a two step elimination. So because we're adding heat we're going to go all the way to the Aldol condensation product or the alpha-beta unsaturated carbonyl. Remember that is a two step process. What are the two steps? The two steps are we going to make the enolate and then we're going to kick off the hydroxyl in the second step. If you just directly kick off the hydroxyl you're going to miss some points on the test if you do that. Because it is not one step, it is two steps. So we're going to make the enolate by pushing the electrons all the way up onto the oxygen. And so I definitely going to need my lone pairs here on oxygen and then so that's our enolate. So again two step elimination we make the enolate first. You can make the enolate by pushing electrons onto oxygen or you can make the enolate by pushing electrons onto carbon just as long as you make the enolate first you're in good shape. These electrons are going to come down. We're going to move the electrons over and kick off hydroxyl. So notice I'm throwing lone pairs on the oxygen at the last minute. And there's your final product. Alphabet unsaturated carbonyl. Alright so that's the Robinson annulation. If I give you a compound and say what two starting materials would you use to make this by a Robinson annulation, we're going to take this apart and see what we would do. Your book shows you a little bit different way to do this that some students find easier. So by all means if you can use the book way to do that that's absolutely fine. So I'm going to do the, since the Robinson is a Michael followed by an Aldol, I'm going to, when I do retro synthesis I'm going to take apart the Aldol first and then the Michael. Alright so retro Aldol, same steps as we talked about before. Alpha, beta, we're going to break the alpha, beta bond. We're going to put a carbonyl in the beta position. So here's my carbonyl in the beta position. And enolate goes in the alpha position. Okay and a good idea to count carbons here. One, two, three, four, five, six, it's a six membered ring so we should have six carbons there. So that's the retro Aldol. And then the next thing we want to do is the retro Michael. And when we did the regular Michael, we took it apart two different ways when we're doing retro Robinson. Yes? The previous problem, if there's the minus the second last step, why can't we approach that with water? Are we talking about this right here? Yeah. Okay and that would be what you would do if it was acid catalyzed dehydration. But if you look at the pKa of a protonated alcohol, it's about negative five or negative six. You're not going to be forming that in a basic solution. Okay? It certainly would make a better leaving group but even if you formed it, what's the hydroxide going to do? In a shot it's going to deprotonate that. You know what I'm saying? So you can't form it and even if you did, it would instantly be deprotonated before it could leave. All right. So that would be what you would do for an acid catalyzed dehydration. So for the retro Michael, we want to only take it apart one way when we're doing retro Robinson. So for retro Robinson, always break at the bond hanging off the ring. So we're not going to take it apart two ways. We're going to take it apart one way. And so how you always break at the bond hanging off the ring is that you start counting from the chain hanging off the ring. All right. So that's how we're going to do that. So this guy is, well, we're going to go this way. Alpha, beta, gamma. Okay? We're going to break the beta gamma bond. We're going to put a double bond in the gamma position, in the alpha position. So there's the double bond right there. And the negative charge goes in the gamma position. So this is going to be a negative charge. And we're not going to show this deprotonated here. So this is what our two starting materials would be. That and this. I forgot to take away the negative charge from the retro aldol, but we would certainly do that also. So those would be our two starting materials. Is that going to work well? So we can take it apart, but is that actually going to work well in this reaction? What do we know about Michael reactions in the Michael donor? What do we want there? Don't we want to stabilize enolate? That's not a stabilized enolate. So that's not going to work very well. Okay. So not a stabilized enolate. Therefore, it won't work well in the Robinson annulation. And it won't work well in the Robinson annulation because the Michael reaction part of the step is not going to work well. So we talked about that last time, but let's talk about it again. Why won't an un-stabilized enolate work? And that's because the Robinson annulation involves a Michael reaction. The Michael reaction does not work well with un-stabilized enolates because un-stabilized enolates are going to do a lot of one-two addition. We talked about that last time, so let's do it again. And I believe this is page 32 of the notes, maybe. I don't know. You can look. 31 or 32. I think that number might be off. Remember, the stronger the nucleophile, the more one-two addition. So if we compare, this pKa of the conjugate acid is about 20, and pKa for active methylene compounds, if we round to the nearest 5, is about 10. So if we have an un-stabilized enolate, it's 10 to the 10th times more acidic. So this here is a stronger base. Therefore, it's a stronger nucleophile. How do we know it's a stronger nucleophile? Well, we do know some trends from 51A and 51B. Here's the trend that would tell us that. And that is CnOF. And we're going across. And I won't even bother to go down, but if we're going across this way, increasing base strength, increasing base strength that direction. And we also will have increasing nucleophile strength in the same direction. Increasing nucleophile strength in the same row of the periodic table as the base strength increases, the nucleophile strength increases. So stronger base, therefore stronger nucleophile, the stronger the nucleophile, the more one-two addition. And that's why it's not going to work very well in the Michael reaction because we're going to get competition, a lot more competition from one-two addition. All right. So, and that's where I should have drawn this down here where I had more room. That's what we're talking about here. Base strength correlates with nucleophile strength. And so conditions for this Robinson annulation that we just drew, we just took this thing apart and saw what we would use to make it and realized that it wasn't going to work very well. Not only do we get competition with one-two and one-four addition, we also have two enolates that it can competitively form. All right. So we've got, what do we have here? We have enolate conform here. It can also form on the other side. That would give you the same enolate. But we also have this one here, enolate conform here also, right? So when you mix the two together with enolate, I'm putting this, I'm drawing this in the wrong spot. I don't know why I'm doing that. NaOH, H2O, and water. Okay. When you do that, you're going to get one-two and one-four addition in addition to two different enolates. I'm drawing everything in the wrong section here. I'll fix that in just a minute here. In addition to two enolates forming. All right. So that's going to be a big mess. So really, what we really want to do is when we're doing a Michael, we use a stabilized enolate. And so we can use a stabilized enolate or we can use a stork eno- what's called the stork enamine synthesis. So instead of making the enolate here, we can use an enamine. And we talked about that before in a different context, but that also works really well here. So this guy is supposed to be here. And now I'm going to get on the right page here in the right column. So stork enamine synthesis will make this reaction work a lot better. So it's going to be a great alternative. So we make the enamine. And the enamine works really well. So it's much more well behaved than an enolate ion. It's going to do one for addition. All right. So there that would be one for addition. We add H3O plus in water. We're going to hydrolyze that in a minimum ion to a ketone. We're going to protonate the enolate ion to make a ketone also. And then we can do intermolecular aldol. You could keep it in acid. It could be acid catalyzed or it can be base catalyzed. So if we just keep it in acid and heat it up, we're going to go all the way to the product, a single product here. Cleanly. If we did it without making any enamine, it's not going to be as clean. All right. So that's the Robinson annulation. Any questions on that? Anybody? So as much as possible, we want to use a stabilized enolate if we can. If we can't, then this is a good alternative to the Storkenamine synthesis. All right. Let's do some a little bit more organic synthesis before we move on to chapter 24. In this chapter, we've learned to synthesize beta hydroxycarbonyls, alpha beta unsaturated carbonyls, beta keto carbonyls, and 1,5 dicarbonyls. So when a target molecule, you're asked to synthesize a target molecule, has these key structural features, you should be able to recognize those features and know how to make it. But what if we had those key structural features and maybe we reduced? And so the carbonyls would be alcohols, wouldn't be able to as easily recognize those structural features. So we want to be able to be ready to look for that. So it's not only going to be a 1,5 dicarbonyl. If we reduce that 1,5 dicarbonyl with lithium-lunum hydride, we'd have a 1,5 diol. So not obvious that you would make that biomichael, but if you remember that that 1,5 diol could come from a 1,5 dicarbonyl, then you're going to be in good shape. So you got to get yourself trained to look for these structural features. All right, so here's an example right here. This is a beta hydroxy ketone. So here's alpha, here's beta. So how do we make a beta hydroxy ketone? What reaction? What reaction gives us a beta hydroxy ketone? Aldol addition. So that we can recognize or we'll be able to recognize by the test on Friday, next Friday. Okay. But what if we did sodium borohydride here and we reduced that alcohol now no longer looks recognizable. So no longer has one of these recognizable features. So you have to be ready for this. You have to be ready to look for things that are, if you want to call it one degree of separation from the patterns that we recognize. You have to be able to look for that. So let's say, for example, I asked you to synthesize this compound. And so you're thinking, okay, well I just saw this example. I could make that by an aldol and then followed by sodium borohydride. So that would mean I could make this from one of these two compounds here. This to match the one above. But assuming I hadn't seen what I started with here. Or I could do it the other way, right? Have one of those. Okay. So either one of those, if I reduced it with sodium borohydride, would give me that target molecule. Let's call this first one A. And then let's call the second one B. And we're going to see which one actually ends up working better. So I'm going to make this target molecule by doing an aldol reaction followed by sodium borohydride to reduce it. But we want to make sure, let's take these guys apart using the retro aldol and see which one is going to be more feasible. So that was, we'll do that on the next page. Let's do A first. Alright, so we could start counting from the ketone alpha beta. Break the alpha beta bond. Beta position gets the carbonyl, alpha position gets the negative charge. So it's that enolate. And this would be an aldehyde. So beta position gets the ketone. And so that's what my two starting materials would be. Alright, let's draw the other one and then we'll take a look and see which we think is going to be better. So if V was my strategy. Alright, now we're going to, now we have the opposite direction. So this is alpha, this is beta. We're going to break right here. So beta position gets the carbonyl. So this would be acetaldehyde. And alpha position is the enolate. Alright, so let's look at these. I don't have my laser pointer today, so we'll have to do the best we can. So let's look at the first one. Are we going to be able to make that enolate, well we've got more than one enolate to conform. This only has one enolate, right? If we mix the two of these. Only one enolate. That conform. But this, we could also form the enolate on the second carbon here. So in other words, we would have to use directed aldol here. Can also form enolate here. Therefore we need directed aldol. So remember directed aldol, we form the enolate completely using LDA and then we add our second compound. Let's look at the second one and see if this is going to work better. We have an aldehyde here. We can form the enolate. This is not going to go well here. So enolate here, enolate here, enolate here. We have three different enolates that conform. So what we really want to do is we definitely want to make, we want to make this enolate here so we could use LDA, THF, you know, minus 78 to form the kinetic enolate. And then we'd also need directed aldol here. But we expect to see some contamination with the thermodynamic enolate here. Because that's got two different ones that conform. So we would also need directed aldol. The good news is if we use the directed aldol in either one of these cases, it's going to add selectively to the aldehyde. So that part of it will be clean. It's just a matter of how many enolates we can form. This one here, because it's symmetrical, there's only one enolate that can form. We're going to form that enolate first and then we add the aldehyde. It seems to me like A is better, don't you think? Just because of the complication with this guy right here, that guy right there is going to be more complicated. So I think that A is the best strategy. This one's best. Alright, so let's draw out the synthesis. Aldea, THF. Minus 78 degrees centigrade. Completely formed the enolate ion and then step one, we add our aldehyde. Step two, acid or water here? Definitely water. If we use acid, we're going to tend to get the alpha, beta, and saturated carbonyl and that's not what we want to synthesize. Okay, so then that will give us, what did I just, did I say something? What did I say? Oh my gosh. That's awesome. Okay. Alright, H2O. Okay, so we're in good order. We have to do a sodium borohydride in ethanol. And that will give us the product that we're trying to synthesize. Okay, kind of an elaborate thing here. Alright, so another crazy synthesis here on the last one in this part. Again, we're trying to provide a synthesis of the falling compound from acyclic precursors. So we're thinking about doing intramolecular aldols or any sort of cyclization intramolecular is going to help us make this compound here. Alright, so what I'm thinking is, if I replace that bromine with a hydroxyl, I could use this strategy. What are you thinking here? What does this look like? Are you thinking of a strategy now that you see it like that? Does it seem like you can see something now? If you look here, we've got, this is a substituted acetone, right? Substituted ketone. So maybe you think acetoacetic ester synthesis? Alright, so if I made that by acetoacetic ester synthesis, then I would go here. And then I would have this group on here. I would do alkylation to get this group on here. See that purple group on there? And then I could take this apart further. And that would be this. So and each time we do this, we're thinking of ways to make these compounds. If I did this, and then I had propybromide, propyl iodide, propyl chloride, your choice. Are you seeing another recognizable structural feature there? In the carbonyl-containing compound, what does that remind you of? How can we, how can you make that compound that's right there? We've made that already. 1,3-dicarbonyl or beta-ketoester, right? What is that? Beta-ketoester. This is a product of a clasing condensation. How to know that? And so then how, that's how we're going to take this apart by acyclic precursors. Actually, what was an intramolecular clasing called? A Dieckman condensation, right? I don't know if it's 1n or 2n, Stephen. That's close enough. Actually, a Dieckman condensation. A dieckman condensation is a Dieckman condensation. All right, so that means that this is from, we'll keep going here. And remember how we take apart a clasing, start counting from the ester. So let's, I'm going to label that, start counting from ester. If you don't start counting from the ester, you're going to get, you're not going to get acyclic precursors. So classic clasing, you start counting from the ester. This is alpha-beta. We put an ethoxy group in the beta position. Alpha position becomes the enolate. So let's take this apart further. And there, and so in the alpha position gets the negative charge, so that would be what we would cycleize with. Let's double check that we have not made a mistake in counting carbons here. This would be 1, 2, 3, 4, 5. And yes, we have 1, 2, 3, 4, 5. We have a 5-membered ring here. So that's our acyclic precursor. All right, so what would our, what would our synthesis look like? Writing our synthesis out in the forward direction. We start off with our acyclic precursor, our diester. We choose our base to match our ester, so we don't get any transesterification. We certainly would not use sodium hydroxide because we're going to get saponification. We want to do the clasing condensation. So remember what drives the equilibrium in a clasing condensation is forming the deprotonated active methylene compound. So that's what we're going to form. We don't need to protonate because we actually need to alkylate. So we're going to put this on there here. SN2 reaction to alkylate that position. So it looks like that at that point. And with our group here. Now we want to get rid of that ester group. We're going to hydrolyze and decarboxylate H3O plus H2O heat, which is going to hydrolyze the ester. Then we're going to decarboxylate. That gets us a substituted acetone derivative. And then we're going to reduce. You could do lithium aluminum hydride or you could do sodium borohydride. I'm going to use lithium aluminum hydride this time. And then PBR3-pearding in our last step. And that will take us to their target compound. So you can get very complicated, very fast. But all the reactions in this chapter allow us to build up carbon skeletons quickly and easily. Questions on Chapter 24. That's carbonyl chemistry. We're going to change gears and we're going to start on amines. So I'll very briefly start on them today. And so recommendation to you is that when we get into the meat of this chapter where we're doing a lot of synthesis, you're going to review Chapter 18, right? I know you're not going to do it now because we have a test coming up. But after the test, you're going to want to review Chapter 18. Any questions before we get started on Chapter 25? Anybody? Up at the top. Really loud. You're going to have to show me because now that it's closed, I'm not going to be able to. Can you come up afterwards and show me? Okay. All right. Pretty exciting. You guys only have two more chapters and you're going to be done with OCAM. Kind of bittersweet, huh? I do have a lot of students that come up to me afterwards and they say that next year, you know, you got some of you might come up to me and say, I miss OCAM so much. You know, it's cool. All right. Okay. So amines super important. Many plants that synthesize complex amines called alkaloids and some which have medicinal or poisonous property. And a lot of amines are very bioactive in the brain. I have a book that's called Drugs in the Brain and there's only one compound in that entire book that doesn't have an amine in it. So they're very active in the brain. I'm going to give you some examples here and we're going to look for structural features in common with all these compounds. This first compound here, anybody know the name of it? That's good that you don't know the name because that means you're not making drugs on the side which is awesome. This is methamphetamine. Okay. I had a student one time when I, we used to have the, I do the honors lab now but we used to have the honors lecture and there was a guy in the back and I made a mistake on one of these compounds and he corrected me on that, you know. He never ended up passing the class. He just kind of disappeared. So anyway, I don't know, I'm just saying. This is amphetamine. So you see the difference between methamphetamine and amphetamine? So it changes the addictive potential dramatically when you just simply take that hydrogen from amphetamine and replace it with a methyl. This is amphetamine. This is also the active ingredient in Adderall just in case you were wondering. This is dopamine. This is dopamine. This is a, a fedrin also known as adrenaline. This is norephedrin or I'm sorry, norepinephrine. This is riddlin. Oops. And what do we have here? Pseudofed. Pseudofedrin used to be over the counter. Now you have to go, go to the pharmacy and show your driver's license to get this compound here which, that, that was pretty recent that that happened. So this is Pseudofed. And some people, when they take Pseudofed they have to take it in the morning. It's for decongest, it's a decongestant but when they take it they have to take it in the morning because it gets you kind of wired. And so you can see the relationship here. Are you starting to see some things in common here? This is ecstasy. Let's stop right here and look at what these all have in common. These are all. Beta-phenethylamines. Let me start with, tell you what that means. Beta-phen, I spelled that wrong. Dog on it. Beta-phenethylamines. So all our beta-phenethylamines, here's the structural skeleton of a beta-phenethylamine. We have an aromatic ring. So that's the phen for the phenyl. And then we have an ethyl group. And then we have an amine. And that could be, that could be hydrogens or R groups. So there's the phenyl for the phenethyl and here's the ethyl right here. And here's the amine. So you can see that in all of these. Here's your ethyl. Here's your amine. There's your ethyl group. Here's your amine. So there can be other substituents but these all have that same thing in common. There's your ethyl. There's your amine. And we can go through and do that. And there's two types. There's the type that have no hydroxyl in the alpha position. So this is alpha. This is beta. There's no hydroxyl in the alpha position or a hydroxyl in the alpha position. So everybody knows why you can't get pseudofed, right? Without showing your license. Does everybody know why that is? Okay, you can take pseudofed and convert it into methamphetamine by getting rid of that hydroxyl, right? That's also why you have to go when you get acid, when you get nail polish remover you have to show your driver's license now. Isn't that crazy? Because people take the acetone and they use it as a solvent to take pseudofed and convert it into meth. So that's why you have to show your license. Okay, so. Guys are so funny. So this guy here, I think this is gone now but there's designer drugs out there where people are like, okay, I'm going to make something that looks a lot like this but it'll be slightly different. And this came out, this is methadone. So what they did is they oxidized that hydroxyl in the alpha position. We have some other substituents here. This was a designer drug that was in bath salts. It was sold in stores. If you heard about that, it was called bath salts and people got really, really bad, bad on this. They had guys, they talked about this guy that was arrested and he just makes people extremely angry and he like, with his teeth he pulled off the knob in the cop car just like, you know, like that. It's crazy. And these people would come to the emergency room and they couldn't sedate them. They had to just tie them down because they were so whatever. Anyway, two more to show you. This is a fenfloramine and this is fenteramine. This was a combination that was used for a diet aid, fenfen. And they had to pull it off the market because it caused heart valve defects. So now I think it was the fenfloramine that was causing the heart valve defects. Now you can buy fenteramine by itself and this of course is fentanyl. These are all beta hydroxy amines and we'll talk more about those on Monday.
 Good afternoon. We're going to get started. Any questions? I did open up the sapling  chapter 24 assignment. It's due on Wednesday. We don't want it due on Thursday night because  then you're going to, human nature, right? You're going to procrastinate and then you're  going to be more stressed out on Friday. Don't you think? Who thinks we should, who thinks  we should have a due date of Wednesday night? Okay, so, and Thursday? Okay. You'll be glad.  Alright. So here's how I know you guys are burned out and ready for summer to start because  nobody emailed me to say, oh by the way the chapter 24 assignment's not up. Nobody emailed  me. So, and we're finishing chapter 24 today so I can tell. Alright. So we left off. We're  going to finish this today. We're going to start chapter 25. Two more chapters left and  you're done. Alright. So we left off last time talking about the Michael followed by  the Aldol also known as the Robinson Annulation. So we went through the Michael edition. So  this is two mechanisms. So if I ask you a Robinson Annulation mechanism on the test,  it's actually like me asking you two mechanisms, a Robinson, a Michael and an Aldol. Alright.  So we do the Michael edition and we said without you you can stop at this Michael product here.  So I will always add heat to let you know that you're going to keep going. So then we  show we're showing the Aldol condensation. We kind of left off in the middle of that  mechanism. So this is exactly the same as the less complex Aldol general mechanism that  we showed already. So we're forming the enolate and then we're attacking the carbonyl. It's  an intramolecular Aldol but that's no problem. And so what we need to do is we need to protonate  this alkoxy group. So we're going to protonate. And then we're going to do a two step elimination.  So because we're adding heat we're going to go all the way to the Aldol condensation  product or the alpha-beta unsaturated carbonyl. Remember that is a two step process. What  are the two steps? The two steps are we going to make the enolate and then we're going to  kick off the hydroxyl in the second step. If you just directly kick off the hydroxyl  you're going to miss some points on the test if you do that. Because it is not one step,  it is two steps. So we're going to make the enolate by pushing the electrons all the way  up onto the oxygen. And so I definitely going to need my lone pairs here on oxygen and then  so that's our enolate. So again two step elimination we make the enolate first. You can make the  enolate by pushing electrons onto oxygen or you can make the enolate by pushing electrons  onto carbon just as long as you make the enolate first you're in good shape. These electrons  are going to come down. We're going to move the electrons over and kick off hydroxyl. So  notice I'm throwing lone pairs on the oxygen at the last minute. And there's your final  product. Alphabet unsaturated carbonyl. Alright so that's the Robinson annulation. If I give  you a compound and say what two starting materials would you use to make this by a Robinson annulation,  we're going to take this apart and see what we would do. Your book shows you a little  bit different way to do this that some students find easier. So by all means if you can use  the book way to do that that's absolutely fine. So I'm going to do the, since the Robinson  is a Michael followed by an Aldol, I'm going to, when I do retro synthesis I'm going to  take apart the Aldol first and then the Michael. Alright so retro Aldol, same steps as we talked  about before. Alpha, beta, we're going to break the alpha, beta bond. We're going to  put a carbonyl in the beta position. So here's my carbonyl in the beta position. And enolate  goes in the alpha position. Okay and a good idea to count carbons here. One, two, three,  four, five, six, it's a six membered ring so we should have six carbons there. So that's  the retro Aldol. And then the next thing we want to do is the retro Michael. And when  we did the regular Michael, we took it apart two different ways when we're doing retro  Robinson. Yes?  The previous problem, if there's the minus the second last step, why can't we approach  that with water?  Are we talking about this right here?  Yeah.  Okay and that would be what you would do if it was acid catalyzed dehydration. But if  you look at the pKa of a protonated alcohol, it's about negative five or negative six.  You're not going to be forming that in a basic solution. Okay? It certainly would make a  better leaving group but even if you formed it, what's the hydroxide going to do? In a  shot it's going to deprotonate that. You know what I'm saying? So you can't form it  and even if you did, it would instantly be deprotonated before it could leave. All right.  So that would be what you would do for an acid catalyzed dehydration. So for the retro  Michael, we want to only take it apart one way when we're doing retro Robinson. So for  retro Robinson, always break at the bond hanging off the ring. So we're not going to take it  apart two ways. We're going to take it apart one way. And so how you always break at the  bond hanging off the ring is that you start counting from the chain hanging off the ring.  All right. So that's how we're going to do that. So this guy is, well, we're going to  go this way. Alpha, beta, gamma. Okay? We're going to break the beta gamma bond. We're  going to put a double bond in the gamma position, in the alpha position. So there's the double  bond right there. And the negative charge goes in the gamma position. So this is going  to be a negative charge. And we're not going to show this deprotonated here. So this is  what our two starting materials would be. That and this. I forgot to take away the negative  charge from the retro aldol, but we would certainly do that also. So those would be our two starting  materials. Is that going to work well? So we can take it apart, but is that actually  going to work well in this reaction? What do we know about Michael reactions in the Michael  donor? What do we want there? Don't we want to stabilize enolate? That's not a stabilized  enolate. So that's not going to work very well. Okay. So not a stabilized enolate. Therefore,  it won't work well in the Robinson annulation. And it won't work well in the Robinson annulation  because the Michael reaction part of the step is not going to work well. So we talked about  that last time, but let's talk about it again. Why won't an un-stabilized enolate work? And  that's because the Robinson annulation involves a Michael reaction. The Michael reaction does  not work well with un-stabilized enolates because un-stabilized enolates are going to  do a lot of one-two addition. We talked about that last time, so let's do it again. And  I believe this is page 32 of the notes, maybe. I don't know. You can look. 31 or 32. I think  that number might be off. Remember, the stronger the nucleophile, the more one-two addition.  So if we compare, this pKa of the conjugate acid is about 20, and pKa for active methylene  compounds, if we round to the nearest 5, is about 10. So if we have an un-stabilized enolate,  it's 10 to the 10th times more acidic. So this here is a stronger base. Therefore, it's  a stronger nucleophile. How do we know it's a stronger nucleophile? Well, we do know some  trends from 51A and 51B. Here's the trend that would tell us that. And that is CnOF.  And we're going across. And I won't even bother to go down, but if we're going across this  way, increasing base strength, increasing base strength that direction. And we also  will have increasing nucleophile strength in the same direction. Increasing nucleophile  strength in the same row of the periodic table as the base strength increases, the nucleophile  strength increases. So stronger base, therefore stronger nucleophile, the stronger the nucleophile,  the more one-two addition. And that's why it's not going to work very well in the Michael  reaction because we're going to get competition, a lot more competition from one-two addition.  All right. So, and that's where I should have drawn this down here where I had more room.  That's what we're talking about here. Base strength correlates with nucleophile strength.  And so conditions for this Robinson annulation that we just drew, we just took this thing  apart and saw what we would use to make it and realized that it wasn't going to work  very well. Not only do we get competition with one-two and one-four addition, we also  have two enolates that it can competitively form. All right. So we've got, what do we  have here? We have enolate conform here. It can also form on the other side. That would  give you the same enolate. But we also have this one here, enolate conform here also,  right? So when you mix the two together with enolate, I'm putting this, I'm drawing this  in the wrong spot. I don't know why I'm doing that. NaOH, H2O, and water. Okay. When you  do that, you're going to get one-two and one-four addition in addition to two different enolates.  I'm drawing everything in the wrong section here. I'll fix that in just a minute here.  In addition to two enolates forming. All right. So that's going to be a big mess. So really,  what we really want to do is when we're doing a Michael, we use a stabilized enolate. And  so we can use a stabilized enolate or we can use a stork eno- what's called the stork  enamine synthesis. So instead of making the enolate here, we can use an enamine. And we  talked about that before in a different context, but that also works really well here. So this  guy is supposed to be here. And now I'm going to get on the right page here in the right  column. So stork enamine synthesis will make this reaction work a lot better. So it's going  to be a great alternative. So we make the enamine. And the enamine works really well. So it's  much more well behaved than an enolate ion. It's going to do one for addition.  All right. So there that would be one for addition. We add H3O plus in water. We're  going to hydrolyze that in a minimum ion to a ketone. We're going to protonate the enolate  ion to make a ketone also. And then we can do intermolecular aldol. You could keep it  in acid. It could be acid catalyzed or it can be base catalyzed. So if we just keep it  in acid and heat it up, we're going to go all the way to the product, a single product  here. Cleanly. If we did it without making any enamine, it's not going to be as clean.  All right. So that's the Robinson annulation. Any questions on that? Anybody? So as much  as possible, we want to use a stabilized enolate if we can. If we can't, then this is a good  alternative to the Storkenamine synthesis. All right. Let's do some a little bit more  organic synthesis before we move on to chapter 24. In this chapter, we've learned to synthesize  beta hydroxycarbonyls, alpha beta unsaturated carbonyls, beta keto carbonyls, and 1,5 dicarbonyls.  So when a target molecule, you're asked to synthesize a target molecule, has these key  structural features, you should be able to recognize those features and know how to make  it. But what if we had those key structural features and maybe we reduced? And so the  carbonyls would be alcohols, wouldn't be able to as easily recognize those structural features.  So we want to be able to be ready to look for that. So it's not only going to be a 1,5  dicarbonyl. If we reduce that 1,5 dicarbonyl with lithium-lunum hydride, we'd have a 1,5  diol. So not obvious that you would make that biomichael, but if you remember that that  1,5 diol could come from a 1,5 dicarbonyl, then you're going to be in good shape. So  you got to get yourself trained to look for these structural features.  All right, so here's an example right here. This is a beta hydroxy ketone. So here's  alpha, here's beta. So how do we make a beta hydroxy ketone? What reaction? What reaction  gives us a beta hydroxy ketone? Aldol addition. So that we can recognize or we'll be able  to recognize by the test on Friday, next Friday. Okay. But what if we did sodium borohydride  here and we reduced that alcohol now no longer looks recognizable. So no longer has one of  these recognizable features. So you have to be ready for this. You have to be ready to  look for things that are, if you want to call it one degree of separation from the patterns  that we recognize. You have to be able to look for that. So let's say, for example,  I asked you to synthesize this compound. And so you're thinking, okay, well I just saw  this example. I could make that by an aldol and then followed by sodium borohydride.  So that would mean I could make this from one of these two compounds here. This to match  the one above. But assuming I hadn't seen what I started with here. Or I could do it  the other way, right? Have one of those. Okay. So either one of those, if I reduced it with  sodium borohydride, would give me that target molecule. Let's call this first one A. And  then let's call the second one B. And we're going to see which one actually ends up working  better. So I'm going to make this target molecule by doing an aldol reaction followed by sodium  borohydride to reduce it. But we want to make sure, let's take these guys apart using the  retro aldol and see which one is going to be more feasible. So that was, we'll do that  on the next page. Let's do A first. Alright, so we could start counting from the ketone  alpha beta. Break the alpha beta bond. Beta position gets the carbonyl, alpha position  gets the negative charge. So it's that enolate. And this would be an aldehyde. So beta position  gets the ketone. And so that's what my two starting materials would be. Alright, let's  draw the other one and then we'll take a look and see which we think is going to be better.  So if V was my strategy. Alright, now we're going to, now we have the opposite direction.  So this is alpha, this is beta. We're going to break right here. So beta position gets  the carbonyl. So this would be acetaldehyde. And alpha position is the enolate. Alright,  so let's look at these. I don't have my laser pointer today, so we'll have to do the best  we can. So let's look at the first one. Are we going to be able to make that enolate,  well we've got more than one enolate to conform. This only has one enolate, right? If we mix  the two of these. Only one enolate. That conform. But this, we could also form the enolate on  the second carbon here. So in other words, we would have to use directed aldol here.  Can also form enolate here. Therefore we need directed aldol. So remember directed aldol,  we form the enolate completely using LDA and then we add our second compound. Let's look  at the second one and see if this is going to work better. We have an aldehyde here.  We can form the enolate. This is not going to go well here. So enolate here, enolate  here, enolate here. We have three different enolates that conform. So what we really want  to do is we definitely want to make, we want to make this enolate here so we could use  LDA, THF, you know, minus 78 to form the kinetic enolate. And then we'd also need directed  aldol here. But we expect to see some contamination with the thermodynamic enolate here. Because  that's got two different ones that conform. So we would also need directed aldol. The  good news is if we use the directed aldol in either one of these cases, it's going to  add selectively to the aldehyde. So that part of it will be clean. It's just a matter of  how many enolates we can form. This one here, because it's symmetrical, there's only one  enolate that can form. We're going to form that enolate first and then we add the aldehyde.  It seems to me like A is better, don't you think? Just because of the complication with  this guy right here, that guy right there is going to be more complicated. So I think  that A is the best strategy. This one's best. Alright, so let's draw out the synthesis.  Aldea, THF. Minus 78 degrees centigrade. Completely formed the enolate ion and then step one,  we add our aldehyde. Step two, acid or water here? Definitely water. If we use acid, we're  going to tend to get the alpha, beta, and saturated carbonyl and that's not what we  want to synthesize. Okay, so then that will give us, what did I just, did I say something?  What did I say? Oh my gosh. That's awesome. Okay. Alright, H2O. Okay, so we're in good  order. We have to do a sodium borohydride in ethanol. And that will give us the product  that we're trying to synthesize. Okay, kind of an elaborate thing here. Alright, so another  crazy synthesis here on the last one in this part. Again, we're trying to provide a synthesis  of the falling compound from acyclic precursors. So we're thinking about doing intramolecular  aldols or any sort of cyclization intramolecular is going to help us make this compound here.  Alright, so what I'm thinking is, if I replace that bromine with a hydroxyl, I could use  this strategy. What are you thinking here? What does this look like? Are you thinking  of a strategy now that you see it like that? Does it seem like you can see something now?  If you look here, we've got, this is a substituted acetone, right? Substituted ketone. So maybe  you think acetoacetic ester synthesis? Alright, so if I made that by acetoacetic ester synthesis,  then I would go here. And then I would have this group on here. I would do alkylation  to get this group on here. See that purple group on there? And then I could take this  apart further. And that would be this. So and each time we do this, we're thinking of  ways to make these compounds. If I did this, and then I had propybromide, propyl iodide,  propyl chloride, your choice. Are you seeing another recognizable structural feature there?  In the carbonyl-containing compound, what does that remind you of? How can we, how can  you make that compound that's right there? We've made that already. 1,3-dicarbonyl or  beta-ketoester, right? What is that? Beta-ketoester. This is a product of a clasing condensation.  How to know that? And so then how, that's how we're going to take this apart by acyclic  precursors. Actually, what was an intramolecular clasing called? A Dieckman condensation, right?  I don't know if it's 1n or 2n, Stephen. That's close enough. Actually, a Dieckman condensation.  A dieckman condensation is a Dieckman condensation. All right, so that means that this is from,  we'll keep going here. And remember how we take apart a clasing, start counting from  the ester. So let's, I'm going to label that, start counting from ester. If you don't start  counting from the ester, you're going to get, you're not going to get acyclic precursors.  So classic clasing, you start counting from the ester. This is alpha-beta. We put an ethoxy  group in the beta position. Alpha position becomes the enolate. So let's take this apart  further. And there, and so in the alpha position gets the negative charge, so that would be  what we would cycleize with. Let's double check that we have not made a mistake in counting  carbons here. This would be 1, 2, 3, 4, 5. And yes, we have 1, 2, 3, 4, 5. We have a  5-membered ring here. So that's our acyclic precursor. All right, so what would our, what  would our synthesis look like? Writing our synthesis out in the forward direction. We  start off with our acyclic precursor, our diester. We choose our base to match our ester, so  we don't get any transesterification. We certainly would not use sodium hydroxide because we're  going to get saponification. We want to do the clasing condensation. So remember what  drives the equilibrium in a clasing condensation is forming the deprotonated active methylene  compound. So that's what we're going to form. We don't need to protonate because we actually  need to alkylate. So we're going to put this on there here. SN2 reaction to alkylate that  position. So it looks like that at that point. And with our group here. Now we want to get  rid of that ester group. We're going to hydrolyze and decarboxylate H3O plus H2O heat, which  is going to hydrolyze the ester. Then we're going to decarboxylate. That gets us a substituted  acetone derivative. And then we're going to reduce. You could do lithium aluminum hydride  or you could do sodium borohydride. I'm going to use lithium aluminum hydride this time.  And then PBR3-pearding in our last step. And that will take us to their target compound.  So you can get very complicated, very fast. But all the reactions in this chapter allow  us to build up carbon skeletons quickly and easily. Questions on Chapter 24. That's carbonyl  chemistry. We're going to change gears and we're going to start on amines. So I'll very  briefly start on them today. And so recommendation to you is that when we get into the meat of  this chapter where we're doing a lot of synthesis, you're going to review Chapter 18, right? I  know you're not going to do it now because we have a test coming up. But after the test,  you're going to want to review Chapter 18. Any questions before we get started on Chapter  25? Anybody? Up at the top. Really loud. You're going to have to show me because now that  it's closed, I'm not going to be able to. Can you come up afterwards and show me? Okay.  All right. Pretty exciting. You guys only have two more chapters and you're going to be done  with OCAM. Kind of bittersweet, huh? I do have a lot of students that come up to me  afterwards and they say that next year, you know, you got some of you might come up to  me and say, I miss OCAM so much. You know, it's cool. All right. Okay. So amines super  important. Many plants that synthesize complex amines called alkaloids and some which have  medicinal or poisonous property. And a lot of amines are very bioactive in the brain.  I have a book that's called Drugs in the Brain and there's only one compound in that entire  book that doesn't have an amine in it. So they're very active in the brain. I'm going  to give you some examples here and we're going to look for structural features in common  with all these compounds. This first compound here, anybody know the name of it? That's  good that you don't know the name because that means you're not making drugs on the  side which is awesome. This is methamphetamine. Okay. I had a student one time when I, we  used to have the, I do the honors lab now but we used to have the honors lecture and  there was a guy in the back and I made a mistake on one of these compounds and he corrected  me on that, you know. He never ended up passing the class. He just kind of disappeared. So  anyway, I don't know, I'm just saying. This is amphetamine. So you see the difference  between methamphetamine and amphetamine? So it changes the addictive potential dramatically  when you just simply take that hydrogen from amphetamine and replace it with a methyl.  This is amphetamine. This is also the active ingredient in Adderall just in case you were  wondering. This is dopamine. This is dopamine. This is a, a fedrin also known as adrenaline.  This is norephedrin or I'm sorry, norepinephrine. This is riddlin. Oops. And what do we have  here? Pseudofed. Pseudofedrin used to be over the counter. Now you have to go, go to the  pharmacy and show your driver's license to get this compound here which, that, that was  pretty recent that that happened. So this is Pseudofed. And some people, when they take  Pseudofed they have to take it in the morning. It's for decongest, it's a decongestant but  when they take it they have to take it in the morning because it gets you kind of wired.  And so you can see the relationship here. Are you starting to see some things in common  here? This is ecstasy. Let's stop right here and look at what these all have in common.  These are all. Beta-phenethylamines. Let me start with, tell you what that means. Beta-phen,  I spelled that wrong. Dog on it. Beta-phenethylamines. So all our beta-phenethylamines, here's the  structural skeleton of a beta-phenethylamine. We have an aromatic ring. So that's the phen  for the phenyl. And then we have an ethyl group. And then we have an amine. And that  could be, that could be hydrogens or R groups. So there's the phenyl for the phenethyl and  here's the ethyl right here. And here's the amine. So you can see that in all of these.  Here's your ethyl. Here's your amine. There's your ethyl group. Here's your amine. So there  can be other substituents but these all have that same thing in common. There's your ethyl.  There's your amine. And we can go through and do that. And there's two types. There's  the type that have no hydroxyl in the alpha position. So this is alpha. This is beta.  There's no hydroxyl in the alpha position or a hydroxyl in the alpha position. So everybody  knows why you can't get pseudofed, right? Without showing your license. Does everybody  know why that is? Okay, you can take pseudofed and convert it into methamphetamine by getting  rid of that hydroxyl, right? That's also why you have to go when you get acid, when you  get nail polish remover you have to show your driver's license now. Isn't that crazy? Because  people take the acetone and they use it as a solvent to take pseudofed and convert it  into meth. So that's why you have to show your license. Okay, so. Guys are so funny.  So this guy here, I think this is gone now but there's designer drugs out there where  people are like, okay, I'm going to make something that looks a lot like this but it'll be  slightly different. And this came out, this is methadone. So what they did is they oxidized  that hydroxyl in the alpha position. We have some other substituents here. This was a designer  drug that was in bath salts. It was sold in stores. If you heard about that, it was called  bath salts and people got really, really bad, bad on this. They had guys, they talked about  this guy that was arrested and he just makes people extremely angry and he like, with his  teeth he pulled off the knob in the cop car just like, you know, like that. It's crazy.  And these people would come to the emergency room and they couldn't sedate them. They had  to just tie them down because they were so whatever. Anyway, two more to show you. This  is a fenfloramine and this is fenteramine. This was a combination that was used for a  diet aid, fenfen. And they had to pull it off the market because it caused heart valve  defects. So now I think it was the fenfloramine that was causing the heart valve defects.  Now you can buy fenteramine by itself and this of course is fentanyl. These are all  beta hydroxy amines and we'll talk more about those on Monday.
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