CHEM 125b: Freshman Organic Chemistry II
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Freshman Organic Chemistry II
CHEM 125b - Lecture 8 - Solvent, Leaving Group, Bridgehead Substitution, and Pentavalent Carbon
Chapter 1. Nucleophilicity and the Influence of Solvent [00:00:00]
Professor Michael McBride: So we’re studying nucleophilic substitution, in part at least, as a means of seeing how one goes about establishing a mechanism, proving a mechanism. Remember that the SN2 nucleophilic substitution has a number of components here, and the logic is to vary them, see how the rate changes, and see if you can disprove some of the mechanisms. And last time we looked at kinetic order and at stereochemistry, and began looking at how the different components influence the rate constant.
So at the end of last time we looked at the nucleophile, and saw that it was unusual. The reactivity with a carbon in the SN2 transition state didn’t seem to parallel the reactivity with the proton. If you went across the first row it was fine, but when you went down it looked funny.
But then we saw that the solvent had something to do with it. That although in water it looks funny, it’s backwards from what you would expect. In acetone it’s sensible. So we concluded that the solvent makes a difference. The solvent hydrogen-bonds in the case of water to the nucleophile, making it less reactive. The smaller, the more reactive the nucleophile, the more it deactivates it–the stronger the hydrogen bonding. And you don’t have that factor in acetone. But that’s just one example of a solvent effect.
There’s a very general kind of solvent effect where a polar solvent stabilizes ions. We saw this in the case of water, the gas phase ions, how much they were stabilized by having water come around them to the order of 100 kilocalories per mole, compared to the gas phase.
So whenever you generate ions, or increase the localization of charge, you get stronger bonding with polar solvents. So polar solvents accelerate reactions that generate or concentrate charge, because they bind more strongly to it, and vice versa. If charge is being destroyed in the reaction, than a polar solvent will stabilize the starting materials more than it does the transition state, and the reaction will slow down. So that’s an important consideration in choosing a solvent for a reaction.
Chapter 2. Leaving Groups & Bridgehead Substitution [00:02:34]
Now let’s look at the leaving group. Fortunately here things make sense. You notice that the stronger the acid, that is the more weakly that the particular group bonds to a proton, the better it is as a leaving group. That is it can give up its bond to carbon, the same way it gives up its bond to protons. And there’s a strict parallel between how good a nucleophile [correction: leaving group] is and how acidic and what its pKa is. So weak bases are good leaving groups. That is, stable anions form easily. Or, those that don’t hold tightly to H+ don’t hold tightly to the carbon in the transition state of the SN2 reaction. So that makes perfect sense.
We looked last time at the work of Bartlett and Knox, this classic work of using a bridgehead reactivity, this bridgehead chloride, in order to test some possibilities about the mechanism of nucleophilic substitution. You can’t get at the backside, so that’s out. You can’t form a double bond at the bridgehead, so you don’t have to worry about elimination. The question is whether you can have a non-planar cation, or whether you could attack from the front, instead of from the back. And by showing that this compound was inert to such substitutions, they were able to show that those inferences were correct.
So they specifically designed and prepared this molecule to test these mechanistic questions. Incidentally, since last time I got hold of this picture of Lawrence Knox. It’s his graduation picture from Bates College. Now he had to prepare the molecule. How did he prepare it? Because what we’re saying is it’s very difficult to prepare things at the bridgehead, because you can’t do nucleophilic substitution. That’s what their experiment showed. So how did they do it?
Well they started with this amine. Actually, they started with a product that already had the carboxylate group in there. And they converted that to the amine by a reaction we’ll discuss later in the last quarter of the semester. But they had this compound available. How do they replace NH2 by Cl? NH2 is a crummy leaving group, and even a better leaving group like chloride, what they showed was that it doesn’t undergo substitutions. So how do you put chloride in place of the NH2? You need some fabulous leaving group if you want to be able to do something at the bridgehead. So what they did was treat the amine –notice that the amine itself is not a leaving group. The pKa is 34. So the higher the pKa, the worse it is as a leaving group, as we just saw. If you protonated it, it would make it a much better leaving group because protonated NH3 the ammonium ion has a pKa that’s 25 powers of ten better. But still, that’s not nearly good enough to make a bridgehead cation in order to do substitution at the bridgehead.
So what they did was treat the amine with NOCl. Now let’s think about how those would react with one another. What makes the NH2 reactive? Antonia, what do you say?
Student: The high HOMO?
Professor Michael McBride: And the high HOMO is–
Student: The lone pair on the nitrogen.
Professor Michael McBride: Right, the lone pair on the nitrogen. So what’s it going to attack? Antonia, keep us going.
Student: The N-O p*.
Professor Michael McBride: OK, there’s a p* bond of N-O. So it could form a bond to nitrogen. But, if you have that negative charge on the oxygen–that’s perfectly good, and it’s almost certainly the way it starts. But that negative charge can come back in, reform the bond, and lose chloride, which is a good leaving group. So I’m not going to draw the intermediate step, which is probably correct, that you suggested. But just say that it displaces the chloride. It probably is because of the p* that does it so easily, although it could do it directly.
So then we’re going to have chloride displaced and a positive charge on this nitrogen. What will happen next? How do you get rid of positive charges on nitrogen like that? Po-Yi? We have an ion. Generally there are other things around that are better at having positive charge than this. Can you think of some way you can get rid of that positive charge? What did we just say about protonated nitrogen? Remember? Remember its pKa is 9. It’s 25 powers of 10 better as a leaving group–that is to give up a proton–than NH3 is to give up a proton and form NH2–. Much better to have NH3 as the leaving group–trivalent nitrogen–rather than the anion. So it’s acidic. It gives up a proton.
The next thing you need to do to get toward the product–I mean obviously this particular reaction wasn’t figured out on paper by saying what’s the HOMO and what’s the LUMO, and so on. They mixed it and found out what happened. But we can predict what’s going to happen. But fortunately I have some insight into what does happen, so I know what has to happen next. What you have to do is have what’s called an allylic rearrangement, where are you shift the double bond and the H, so the H goes from N to O. The double bond goes between the two Ns, from one end to the other, of a three atom allylic system. How do we go about that?
Well first we can put the proton on that shared pair of oxygen. That’s pretty easy. And now there’s a resonance structure of that involving the unshared pair on nitrogen again. So we can go in, and the p electrons can go on to oxygen in the resonance structure we’re talking about. So now we’ve done the allylic rearrangement, except it’s still got a positive charge on the nitrogen. Po-Yi, what do you do now? To get rid of the positive charge on the nitrogen?
Student: The proton leaves?
Professor Michael McBride: Right. You lose the proton again. Now you do the same thing again. Watch this. So you protonate the water, and now use the unshared pair of the nitrogen. And we’re going to draw two arrows. So you form a triple bond between the two nitrogens, and water is the good leaving group. So we have that. And now that truly is a fabulous leaving group. What is the leaving group? What’s going to leave? Ruoyi?
Professor Michael McBride: N2. I looked. I couldn’t find a pKa for protonated N2. It just has no affinity for a proton on at all. I couldn’t find it. But it’s certainly a very, very strong acid. Protonated nitrogen gas. So it’s a very good leaving group. And you get N2, and it as a leaving group can generate even a bridgehead cation, the thing that you can’t do with chloride. Even with the help of silver ion as Bartlett and Knox found.
So once you have that, remember we already got the chloride from the first reaction. So you just put that in, and that’s how they made it. So there are leaving groups even better than these. N2 is a fabulous leaving group, and can even make the bridgehead cation.
Chapter 3. Making OH a Leaving Group [00:11:12]
Now this idea of protonating or charging things in order that they leave as neutral molecules like N2, is also present in molecules in the list we already have here. Like ROH. That’s a very bad leaving group, OH–. However, if you protonate it, then you have water as the leaving group and that’s good. You’ve changed the pKa by 17.5 powers of 10. You’ve changed the pKa by 17.5 units. That is the equilibrium constant by that many powers of 10. So that becomes very good.
And in fact there’s another good way of doing it. So that’s acid catalysis to help lose OH, that is you lose it as water not as OH–. And another way is to convert it into a sulfonate ester, which is a good leaving group. Did you ever see that before? To have that complicated thing leave instead of OH–?
Because of the new dispensation that the exam is going to be on Friday rather than Wednesday, you have a whole week to learn that. So that was what Kenyon and Phillips did, remember, in order to study the inversion. So there are leaving group tricks, and there’s lore about that, of what you could do to help make OH a leaving group. You can protonate it and have the leaving group be H2O. So that could be the leaving group, OH2+, and then it leaves as H2O. So we can look at an example of that.
So here’s bromide going to displace hydroxide. So we can imagine it coming on, and hydroxide going off. But there’s a problem with this. The pKa of water is 16, roughly. And the pKa of HBr is –5. So one way to say it is that bromide is a much better anion than hydroxide is. When one of them is going to be protonated. What does that mean about this reaction? That you’d rather have bromide than have hydroxide. Luke, what does that say about this reaction? Going to make a lot of money doing that reaction?
Student: Probably not.
Professor Michael McBride: Why not?
Student: Well, that probably wouldn’t happen.
Professor Michael McBride: I can’t hear very well.
Student: That probably wouldn’t happen.
Professor Michael McBride: What would happen?
Professor Michael McBride: You’d go the other direction. So it would go that way. And so you’d start with an expensive starting material and make cheap product if you want to use that reaction. But if you put HBr in there, you protonate it because HBr has a pKa of 5 [correction: –5]. But water or alcohol, has a pKa of –1.7. So there’s a difference between that of over three. So it’s more than a thousand times better to protonate the oxygen, than it is to protonate Br–. So this reaction will certainly go in the right direction. And now you have a great leaving group. So bromide is a good nucleophile and will displace water. So if you just tried using sodium bromide as the source of bromide, it wouldn’t work. But if you use hydrobromic acid, then you zip up the starting material and now it’s downhill to the product.
And you can cleave ethers. You know ether is a common solvent. I suspect you’ve used ether in lab. Because it doesn’t react well with–you can have lots of nucleophiles in there, but you don’t displace OR–. That’s not a good leaving group. It’s like OH–. But if you use HBr like this example here, excess 47% HBr, treated for eight hours. It protonates. And now you have a good leaving group. So you have a good nucleophile, and a good leaving group. The bromide can attack the carbon, and the alcohol is the leaving group. And then we already saw that the alcohol reacts with HBr to give bromide. So you convert the ether completely to the bromide if you add HBr. So HBr can cleave ethers. So that’s one trick to make OH– into a leaving group, to make OH into OH2+.
Another one is the one we already alluded to and talked about last time. Which is to convert the alcohol into what’s called a sulfonate ester. A tosylate. The reason it’s called tosylate, is that acid is called toluenesulfonic acid. So it’s abbreviated tosic acid. So tosylate is the leaving group. The pKa of that acid is –3. So it’s not surprising it’s a good leaving group. It worked well for Kenyon and Phillips in 1923. That’s a very common way to make alcohols into leaving groups.
Another way is talked about in the Jones text, which is OSOCl. So the way to get that is to start , as they say here, with thionyl chloride, SOCl2, which reacts with the alcohol, to give an alkyl chloride plus SO2 and HCl. And there’s an example shown here of a particular acid being converted with thionyl chloride, chloroform, 100 degrees, two hours, into an 80% yield, they say of this, plus SO2 and HCl. 80% is not such a great yield for this reaction. This is one of the reactions I’ve done a lot, and it really gets very, very good yields typically. One of the reasons it’s so handy is that the byproducts are gases. So you don’t have to do any separation, they just go away. You can appreciate from lab how handy it would be if you didn’t have to do any purification of the product at the end, everything just went up the hood.
And furthermore, there seems to be a little bit of a problem here. Because they said they did this in chloroform at 100°. It could be that the bath was at 100°, but if so the solution wasn’t at 100°. Does anybody have an idea why that might be? Because the boiling point of chloroform is 61°, and the boiling point of thionyl chloride is 75°, so it’d be very difficult to get a solution like that heated to 100°.
So the reason I point this out on this page from the book, is that Professor Jones is a great friend of mine, and a very good physical organic chemist. And he’s taken the care in his book, as in most good books, to actually put in real examples, not just theoretical examples made up out of your head. There was a very famous professor in the in the first half of the twentieth century at MIT whose occupation in the summer was to write textbooks. But he did it by going up to his cabin in Maine, and writing the textbooks there, where he had no access to a library. So everything he wrote, he made up out of his head in there.
Much better to have an example like this where they give exactly what the yield is, 80%, and what the conditions were. But even in this case, I doubt that that really was the condition. Although that might have been what the heating mantle was, at a 100°. But the solution couldn’t possibly have been at 100°. Even though when you mix things the boiling point is elevated, but it wouldn’t be that much elevated I don’t think. But the reason I mention that is not just to point out that you have to take things with a grain of salt and look as much as you can at original literature, if you really want to repeat some process, but to point out that thionyl chloride is not a very high boiling liquid. So in fact all you do is take the alcohol–You don’t need chloroform, at least I never used a solvent. This is a liquid. This’ll dissolve this. And the product chloride is a liquid even with bigger molecules where the alcohol might be a solid. So you just heat it up for a while, reflux it, and then distill out. And this SOCl2 distills out, the SO2 and HCl distills out, and all you’re left with is the alkyl chloride. It’s a beautiful reaction.
OK and here’s an example that shows the mechanism of this. The crucial part is this. The alcohol displaces at sulfur. So the chloride goes away, and you get R-O, and this is the new bond, SOCl.
Oh pardon, me. I said the wrong thing at first. You displace one of these chlorides with the O. So you get ROSOCl. But now the neat reaction is that you generated chloride in that process, when you eliminated a chloride here. That chloride can do an SN2 displacement on the R. These electrons go in here, but at the same time these electrons go on to the chloride. So you directly make SO2 and Cl. So it’s an unusual kind of leaving group, because it breaks apart as it leaves.
Now another way to make the OH into a leaving group is to have phosphorus on there with electron-withdrawing groups on it. Because as it turns out, the double bond between phosphorus and oxygen is quite a strong one. Remember that phosphorus has d orbitals that are vacant. So they can stabilize the electrons on the adjacent oxygen when you make a negative charge there. And if X is something that’s electronegative this is a very good leaving group.
So here’s an example of an alcohol treated with PBr3 and it gives a 58% yield they said here with specific conditions, and phosphoric acid is a byproduct. And here’s another example, using PCl5. It says 100%. Beware of things that say the yield is 100%, that’s a sign that the person probably wasn’t very careful. Have you ever gotten 100% yield in lab? These people maybe have more experience than you, but they’re not that much better. Very rarely do you truly get 100% yield.
Now how does this work? Well, first you do a substitution at phosphorus. So the O of the OH attacks phosphorus and displaces chloride. Probably that’s A/D, association and dissociation. Probably you first form a bond to phosphorus, remember it has vacant d orbitals, and then the chloride leaves.
So you have this with these electronegative groups on it, and now that can be a good leaving group. So you’ve converted OH into this good leaving group. Why is that particular one of interest in context of what we’ve been talking about lately? What’s interesting about that particular R group to do a substitution at that carbon? Linda?
Student: You can’t attack the back.
Professor Michael McBride: I can’t hear very well.
Student: You can’t do a backside attack.
Professor Michael McBride: Ahah! You can’t do a backside attack. This is like the Bartlett-Knox thing. It’s set up so you can’t get a nucleophile to the backside. The alternative, remember, was to have it leave, and leave a cation. Just leave without the help from being pushed backside. But, what we learn from this, the fact that we got 100% yield, is that we can do that. So these rings are bigger. Remember the one that Bartlett and Knox made. The bridges were two carbons, two carbons, and one carbon. Here the rings are much bigger, so it’s easier to distort the angles in order to flatten the rings. So we can do substitution involving bridgehead cations, just not when the rings are very small. But what you learn here is that this leaving group is good enough to do that, even though it’s a bridgehead.
Another trick is this one called the Appel reaction. I took this figure from Wikipedia. The reagents that you add to the alcohol are triphenylphosphine and carbon tetrachloride, an unusual reagent. You more often would think of that as a solvent then as a reagent. But as you see, what happens is the unshared pair on phosphorus does a substitution on carbon tetrachloride. But the phosphorus attacks chloride, not carbon. So it attacks the chloride and CCl3– is the leaving group. Because the three chlorines are sufficiently electron attractive, that the anion is not a bad anion. The pKa of chloroform is 24, so it’s pretty easy to generate that anion and lose a proton. Now notice at this stage when you add the alcohol, its pKa is about 17, so the trichloromethyl anion, although it’s unusually stable for a carbon anion, is much more avid to get a proton, than the O of the alcohol is.
So this base that you’ve made in the first reaction, the leaving group takes the proton off the alcohol and now you have OR–. But now you have the possibility for chloride to attack this carbon where you’ve made this new bond here. That’s probably an association followed by dissociation. But notice the substitution is at phosphorus, because the O attacked the phosphorus and chloride came off. So here’s the chloride. It does a nucleophilic substitution now at carbon, and Ph3PO, triphenylphosphine oxide, is the leaving group. And that’s a super-leaving group. This is a very strong bond, the P-O double bond there. So this converted what would have been an OH leaving group into this triphenylphosphine cation, which leaves to make triphenylphosphine oxide. So again you’ve made a very good leaving group out of an oxygen.
So what this shows is that they’re lots of ways to skin a cat. OH– is a terrible leaving group, but we have lots of different ways of going at it. You can protonate it. You can use SOCl2. You can use the phosphorus chlorides or bromides. Phosphorus halides. And you can do this funny trick in the Appel Reaction using triphenylphosphine and carbon tetrachloride.
Here’s an example from the Jones book, where the OH is made into a leaving group by triphenylphosphine and carbon tetrachloride. It says it was done over a day at room temperature, and here’s the chloride. Notice it inverted. The presence of deuteriums here allows you to know whether the configuration is retained or inverted. And as you expect with an SN2 reaction, it’s inverted.
Chapter 4. Accelerating SN2 to Support PET Scanning [00:27:48]
So we’ve looked a lot at this mechanism. Let’s see how you can use knowledge of the mechanism to do something useful. In this particular case it was to maximize synthetic speed in order to make PET scanning possible. Do people know what PET scanning is? PET, what the P stands for?
Professor Michael McBride: The P is positron. And what is a positron? It’s something you don’t encounter very often. You can’t buy a bottle of it.
Professor Michael McBride: Pardon me?
Student: It’s antimatter.
Professor Michael McBride: It’s antimatter. It’s a positive electron. It’s an electronic that has a positive charge instead of a negative charge. So that’s a positron. Now where do you get positrons?
Professor Michael McBride: You get them by nuclear decay. Unstable isotopes decay to give a number of particles, especially in the case of F-18. F-18 can have one of its protons in the nucleus become a neutron. That is it’s the same particle, but it loses a positive charge. So a proton becomes a neutron. So instead of fluoride, it’s now oxygen. So you’ve made a new element. And it’s lost a positive charge. So if it was F–, it’s oxygen double minus. But it’s the same mass, 18, because all you lost was this positive electron.
Now when you have a positive electron, it goes away instantly because the antimatter meets it’s corresponding matter, an electron right where it lives. It’s made right there, but they’re a zillion electrons around, even on the same atom there are electrons. So when the electron and a positron meet one another, they annihilate. They go away, and what comes out is two photons, γ particles, which as you saw, flew off antiparallel to one another along the same line. So if you have a detector you see a flash here and a detection over there as well.
Now you can do that again and again. You can get, bingo, those two happen at the same time. Because they came from the same decay. Or those two. So if you surround the thing with the detector that can tell when things hit it, when two things hit it at the same time, they came from the same nuclear decay.
Now do you see what you could do with respect to imaging? How can you tell where the sample that had F-18 was? All you know is where these things hit your detector. You see what you do to figure out where they came from? Yigit?
Student: Trace the lines–
Professor Michael McBride: Aha, yes. You trace lines. So if you connect those, where they intersect is where the fluorine-18 was. Each line comes from one fluorine-18, but the sample had a number of fluorine-18s in it. So you find out where the sample was. If you want to use this imaging to find tumors, you have to find some way to get fluorine-18 to where the tumor is. And then if you surround the patient with this detector and measure these simultaneous flashes, or detection events, then you know where that substance was. So if you can make that substance be in the tumor, then you’re home free. You can read about it, as I did, in Wikipedia. But you have to do this really quickly after you’ve prepared the element. So you prepare the element, F-18, but its half-life is only 110 minutes. So you don’t have much time to make it into something that will go where the tumor is. So speed is of the essence in doing whatever kind of synthesis you’re going to get.
Now where do you get F-18? You get it from O-18, the same as the product by adding a proton and losing a neutron. So you don’t change the mass, but you put a positive charge into the nucleus so you get fluorine instead of oxygen. OK fine if you have O-18–you buy O-18, but where are you going to get a proton that’s going really fast? Seven million electron volts. You have to get it from an accelerator. From a cyclotron. Now you can also do it in place of fluorine you could have carbon, or nitrogen, or oxygen, but they have even shorter half-lives. So you have to work even quicker in order to have any there at the time to get it where you want it to be detected.
Now if you go on to the petcenter.yale.edu, you see that they do this positron emission tomography. And you see what they have is a cyclotron that generates these high-energy protons, that can convert one element into one of these radioactive elements that will generate positrons. And you see they have a water target so that you can produce fluoride. They also have targets to produce those other elements I talked about. And they say they also have two fluorine-18 synthesis modules–one for nucleophilic substitution, and one for electrophilic substitution. So they have the facility for making this stuff and then doing reactions quickly with it, but the question is what are you going to synthesize so that it will go to a tumor, so you can find where the tumor is? And you have to do it really quickly.
Well glucose is the basis of metabolism, and tumors are growing rapidly, so they really snarf up glucose. So if you had F-18 in glucose that might do the trick. If you could make 2-fluoroglucose, with an F-18 in it, maybe that would be sucked up by the tumors the same way that glucose is. So we need to make that molecule as quickly as possible from glucose. Well, let’s just do a nucleophilic substitution. Fluoride in, hydroxide out. Does that sound good? Or might there be problems with that in doing the SN2 reaction? Can you think of a problem doing this reaction? Fluoride comes in, displaces OH to give the compound on the right, which has the same shape as glucose. Lauren, what’s the funny problem you see?
Student: OH isn’t a good leaving group.
Professor Michael McBride: The hydroxide is a crummy leaving group. Whoops, and it was supposed to come up there, but it didn’t. I had anticipated you. I think I did. Let’s see what the next one is. Yes. But it could be that the wrong carbon gets attacked. There are lots of OH’s here. Even if it could be attacked, the others could be just as good. And if you did SN2 displacement at the oxygen, even if you attack that one, it would give the wrong fluorine-18 arrangement. There’s Lauren’s. It’s a terrible leaving group.
That’s not all that’s wrong with this. A fluoride is a horrid nucleophile as we saw. It’s tied up by hydrogen-bonding. So you have to break the hydrogen bonds. And it’s very small and highly charged, so it has very strong hydrogen bonds. And furthermore, it’s tied up by the potassium cation, which makes it less reactive. So you’ve got lots and lots of problems, so that doesn’t look like it’s going to do the trick.
Now can we address these problems on the basis of what we’ve been talking about? Can we figure out some way of doing it so we can do this transformation really rapidly after the cyclotron generates this stuff, that half of it’s going away every couple hours. How about the SN2 inversion gives the wrong configuration? What could we do to get the right configuration if we did an inversion? Use a different starting material. And it turns out that mannose, which has this OH axial, pointing up, is readily available. It’s a different sugar. And if you did substitution on that, you’d get the equatorial thing that you wanted. So that’s no problem.
Now we have the problem that the wrong CO could be attacked. There are one, two, three, four, five different C-OH’s in here. So what you have to do is make one of them different from the others. The one you want to attack has to be different from the others. And the way you do it has been known a long time in sugar chemistry. Which is, you can protect some of the groups and not another one. So what’s done here is to put acetyl groups on all the other four, but not on this particular OH. How that’s done we won’t go into now. That’s a topic in sugar chemistry. But you know how to do it.
But there’s a problem here still–that OH– is a horrid leaving group. It’s now different from the others, but it’s a horrid leaving group and acetate is a better leaving group. It’s a more acidic compound–acetic acid–than water. Better anion. So what are we going to do now? Lauren, this was your problem. That the OH– was a bad-leaving group. What can we do?
Student: Try to protonate it.
Professor Michael McBride: We could protonate it, but you could protonate the acetates too and make them better leaving groups, so that won’t do the trick. What they did, was convert it into a sulfonic acid. The one that Kenyon and Phillips did had a phenyl group here, or toluene actually–a benzene with a methyl on the edge of it. This is CF3. Why might CF3 be better than just a regular old hydrocarbon like toluene? It’s got the fluorines in it that are electron withdrawing. So it makes the anion that you’re going to want to leave–this OSO2CF3 is a much stronger acid, a much more stable anion, and a better leaving group. The pKa of the corresponding acid here is –14. It’s called a super acid. Remember that the strongest that we’ve talked about so far, had a pKa of –11. This is a 1,000 times stronger than HI. So we put on a really super leaving group. It’s trifluormethansulfonate, and called triflate.
So now we have a compound that’s as good as we can make it for being attacked at this center, not the others. For having a great leaving group, and for having the proper stereochemistry so that we’ll get the fluoride where we want it. And now we have to do the reaction with fluoride.
Notice that all this can be done ahead of time. We can have the bottom right molecule, take however much time we want to make it. Now we have it in a bottle. We go to the cyclotron at the medical school, get fluoride-18, and try to do our substitution. Now we have to introduce the fluoride-18. So we have to go from this protected triflate to this compound as soon as we can, because the fluoride’s getting off these positrons all the time.
So let’s react it with potassium fluoride, with fluoride-18. What’s the problem? The fluoride is tied up by hydrogen bonding. It’s not a good nucleophile. How could we get around that problem with the hydrogen bonding deactivating fluoride? Any ideas on that one? How can you do that? Lauren?
Student: Could changing the solvent help?
Professor Michael McBride: Change the solvent to something that can’t make hydrogen bonds. So they use acetonitrile, methyl cyanide, which doesn’t have any hydrogen bonding protons in it. It has CH protons, but those aren’t involved in hydrogen bonding, so that’s no problem.
But we still have this last problem. The potassium cation is going to be tightly associated with the fluoride, and we’re going to have to break it away to make the fluoride more reactive. How can we get rid of the potassium cation? Remember the problem is that a single potassium atom is rather small. If it were big then it wouldn’t attract by Coulomb’s law, the fluoride so tightly. Then you wouldn’t have to worry about it. How could we make potassium big? Arvind, did you have an idea?
Student: Don’t atoms get bigger if you add an electron?
Professor Michael McBride: Well, if you add an electron to the potassium cation, it’s not a cation anymore. It’s not the thing that’s associated. It has to still be a cation, but you want to make it bigger. How do you make a potassium cation big? Nate?
Student: Move the electrons to higher energy orbitals.
Professor Michael McBride: That’s an interesting idea, but I don’t think it’s practical, because you couldn’t have enough light to make a significant number of them bigger all at the same time, without just blowing everything apart. Matt?
Student: Could you separate it using a crown ether?
Professor Michael McBride: Aha! Crown ether. Or even better, a cryptate. So you make it big that way. And now the fluoride is a good nucleophile. There’s no hydrogen bonding. The cation is really big. And now you can do that. So that’s how they do it. So it’s all using these ideas to figure out how to tune the SN2 reaction to make it as efficient as possible. And of course there’s one more step. You have to then take off these acetate groups. But you can do that just by treating it with water and acid.
Here’s an example of a PET scan image that’s measured one hour after administering. So you not only have to make the stuff, you have to give it to the patient, and the patient has to process it enough that compound gets to where the tumor is. It just goes to where glucose is being consumed. The 2-fluoroglucose. One place that a lot of glucose is being consumed is up there. Where’s that? Is it happening for you? If I had given you 2-fluoroglucose, and was looking at your positrons, would I see them coming from here? That’s where a lot of metabolism–but unfortunately also there. And if we slice through there, and tip it up you can see that that’s the tumor that was located this way. So the metabolism is in the brain, but also in the cancerous lymph node. Understanding how nucleophilic substitution works, and how to control it was crucial in designing this technique.
Chapter 5. Using Theory to Investigate the Possibility of a Pentavalent Carbon Intermediate [00:44:57]
Now one last topic, which I’ll just introduce here and then we’ll complete next time. We’ve talked about stereochemistry, rate law, and rate constant. But can you do actual structural work? Remember how we talked about the mechanism of a nucleophile attacking a carbonyl group? And you could do it by X-ray? You could get a bunch of different compounds and add nitrogens in different positions relative to the carbonyl, and you could put them together in sort of a movie in the Bürgi-Dunitz material that we did last semester and see the stage of the reaction. Might you be able to do that for nucleophilic substitution? Might you be able to use X-ray, and if not X-ray at least use quantum mechanics, to see what the structure of this intermediate is. Because so far we’ve mostly been beating up on the dissociation/association mechanisms. Most of the things we’ve disproven are things that involve that mechanism with the trivalent carbon intermediate. Although it can happen, and we’re going to talk about it next–go through the intermediate cation. So there are some cases where it in fact applies.
But the tougher problem is to distinguish between concerted and association/dissociation with a very weakly stabilized pentavalent intermediate. There’s stuff on the course website about that.
But the question is, might there be a pentavalent intermediate instead of the concerted transition state? So here’s the transition state, which is an energy maximum. Here’s the pentavalent intermediate. So there would be nothing to see at the transition state here. It wouldn’t last any time. It’s just passing through. But if there were a little dimple up at the top, there could be some stability. That is, that is a stable pentavalent carbon. But does it exist? You can see that it would merge smoothly into one if it just had a teeny tiny dimple, or a little bit bigger dimple. If it were big enough, you might be able to prepare it under some circumstances where you could actually see it, say an X-ray. But if it’s too small, you’re never going to see it. You might be able to try to calculate it by quantum mechanics. So the question is, is there an example of a pentavalent carbon? So by quantum mechanics, they said that for water, displacing H2O, that cation, that symmetrical compound there is actually a transition state, not an intermediate. It goes downhill if you lengthen one bond and shorten the other.
The same is true in the anion. These are calculations of isolated materials, so gas phase. If you have OH– come in and displace OH–, that one quantum mechanics also says is transition state. But neither reaction is practical in the laboratory. So the question is, what does experiment say? Can you do X-ray studies and actually see whether there might be a stable pentavalent–maybe very transiently stable, very weakly stabilized, but might that pentavalent compound be stable? So that’s what we’ll talk about with the X-ray material next time.
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