CHEM 125b: Freshman Organic Chemistry II
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Freshman Organic Chemistry II
CHEM 125b - Lecture 24 - Higher-Order Effects, Dynamics, and the NMR Time Scale
Chapter 1. Hybridization and Splitting by C-13 [00:00:00]
Professor J. Michael McBride: So we’ve talked about the main events in NMR spectroscopy for determining structure. That is, about the chemical shift and about spin-spin splitting, that allows you to tell how many nearby nuclei there are. Today we’re going to talk about some complications that make the spectrum more complicated, and perhaps more informative, in terms of high-order effects, higher order effects in spin-spin splitting. And then about using NMR to study, not structure, but to study rates. And we’ll do some C-13 labeling stuff there.
So spin-spin splitting with other nuclei besides protons next door, and we actually saw some of this last time with deuterium splitting, with C-13 splitting. For example, if there’s a signal like this and you blow it way up by 30 times, then you can see that. What gives these things that are so far apart 126 Hz. Do you remember this from last– what are the size of splittings you normally see for protons that are on adjacent carbons? They’re the order of 7 Hz, right? Here we have got 126. What’s the big thing that’s going on? What makes coupling so big? It’s C-13-H. In fact, 99% of the sample, if you haven’t specifically put C-13s in there, are C-12, but about 1% are C-13. So you see that.
Now here’s a paper from 1959, which first noted something interesting about these C-13-H coupling constants, that they relate to the amount of s-character in the carbon atomic orbital that’s involved in the bonds. Remember, we said that the information has to be communicated through the electrons because the action of magnets through space, if you average over all angles of rotation in a liquid, goes to zero. So you communicate through the bonds. The nice thing about… that it’s so strong for JC-13-H, is that the C-13 has just one bond to get to the H. So it’s a very short path to communicate.
Now, but you remember, also, that to do that– so that it doesn’t average out as you tumble– you have to have the electrons in an s orbital. So the hydrogen uses its s orbital, that’s all it’s got essentially. But carbon can use s or p. So the hybridization, as they say here, relates to the coupling constant, Because to communicate that electron must be on the C-13 nucleus. So this is a table from that paper. It gives the coupling constant in units of per second. Those were then changed to be called Hz nowadays. So here are a bunch of different compounds and their JC-13-H coupling constants. If we make a histogram of them, you see that there’s a group between 120 to 140, one between 150 and 170, and another group down around 250. So there are different groups involved here. There are three different populations, we could draw the lines between them.
And then if we look to see what the carbons are, that bear these hydrogens that show that splitting, we could look first at the very big ones down there. And what’s special about those carbons that are splitting the hydrogens? They’re triple bonded. So sp hybridization, 50% s in the bond to hydrogen. The ones here, the green ones, are in aromatic systems, so they have sp2 hybridization. Except for cyclopropane. Why would it have such a high coupling constant? Why would it have a lot of s-character in the bond to hydrogen, the carbon that’s in the cyclopropane? Because its bonds normally are tetrahedral, sp3, but when you try to make a smaller angle, more p-character in the bonds to the other carbons, therefore more s-character coming out to the hydrogens. So we can actually see that hybridization change in the NMR. And then these other are normal methyl groups, sp3 hybridized carbons.
So here in NMR, we have direct evidence for this hybridization stuff we talked about last semester. So it’s sp, sp2, sp3.
Chapter 2. Higher-Order Effects: Why Methane Gives a Singlet [00:09:39]
Now here’s a subtle point. Perhaps we shouldn’t be spending time on it, but it’s so interesting, and I suspect the curious among you may have wondered about this. Because we talked about coupling goes through the bonds. So from H to C to C to H. How about H to C to H? That’s a shorter path. So methane, if you’re looking at one hydrogen, how many other hydrogens would you think it would be coupled to? Three. And they could be in 1:3:3:1. So you should see a quartet for a proton of CH4. But you don’t, you just see a single sharp peak. So where did the splitting go? And this is related to what we talked about last time, where a book said the splitting constant is zero. And I said that’s not true, the splitting constant is high, maybe 10. But it’s not observable. So why is it, why don’t you see quartets for CH4?
OK, so here’s one of these spectra– actually if you go up here and you can click on that and get taken to the website of Chem 220 where they have problems. I’ll work one problem later in the class, if we have time. But there are others you can practice on there. So this is a compound, and I’ll tell you its answer. It’s cinnamic acid. It’s related to… the adelhyde related to this acid, cinnamaldehyde is the smell of cinnamon. So what do we expect when we look at the proton spectrum– this is the proton spectrum of it– what do we expect? Well, notice that we can draw resonance structures, which will put positive charge on the carbon with that hydrogen, so remove electron density from that hydrogen. Or from that one, or that one, or that one. So those all should be shifted downfield relative to where they would otherwise be, relative to the ones that aren’t blue.
When we look at the compound, then, we see these really complicated peaks in here. We throw up our hands trying to understand that. And then we see here two doublets, really nice, sharp, clean things. So what do you think the two doublets are out here? Let’s ignore that bit for the time being and talk about these two doublets. Which protons do you think those come from? Which one is this? OK, let me start going around the compound. The OH is way down further. We’re not showing that. OK, so I go down. I come to this one. What do you think that is?
Student: The one near the COOH.
Professor J. Michael McBride: No, that’s the one up here. This is where double bonds normally come. Then you come here, and that one has a hydrogen on double-bonded carbon, but it’s down here because it has lost electron density. Remember, we drew a resonance structure where the blue ones were more positive. And then here, we have hydrogens that are on the benzene ring, shifted down this way because of the ring shift, the aromatic ring current that we talked about last time. And now there’s a group of three and a group of two. These are positive and are shifted down. That’s the group of two.
The group of three here is these two, plus this one. Which would be shifted down because it’s positive, but up because it’s far from this sp2 carbon. Let’s not worry about why that is. But at any rate, that’s the benzene ones. But that’s not what we’re going to talk about here. What we’re going to talk about is the clean signals, here and here.
Now this is measured with a magnetic field of 7.05 Tesla. That means it’s a 300 MHz spectrometer. So instead of using parts per million down here, we could change to Hz if we wanted to, change to a frequency scale. And in terms of frequency, if we call zero the middle, then that’s lower-frequency to the top, higher-frequency to the bottom. It turns out that the ratio of the chemical shift difference, ∂– measured in Hz, not in parts per million, we want to be comparing apples with apples here, Hz with Hz– so the chemical shift from here to here is much bigger than the coupling constant, J. It’s 24.4 times as large, and that’s the pattern we see.
Now suppose we used a smaller magnetic field. What’s going to happen? The chemical shift in parts per million will stay the same. But it’ll be the same fraction of a smaller frequency, if we use a smaller field, rhe precession frequency gets smaller. So if we measure in Hz, as we’re doing here, they’ll come closer together, but J will stay the same, because it doesn’t depend on the field. The magnets are not dependent on the field, they’re coupling. So what will happen? These two doublets will move together. So at 2.89, it turns out the ratio is 10. Or at 0.86 Tesla, it would be 3.
Notice, incidentally, that this would be a much cheaper spectrometer, because it would have only 1/10 as big the magnetic field as the other one. But also, those other peaks are going to be in the middle. Everything’s going to be getting on top of one another, and I’m just showing you what would happen to this double doublet.
There it is at 0.29. And finally, if we’re 100 times smaller at 0.2 for the ratio of chemical shift to J, measured in the same units, in Hz, you get the purple pattern. Now notice that the J is always the same, the splitting between the two peaks, but what do you notice between the doublet? What’s curious about this? I mean, I’m sure you all see. Some of the peaks are disappearing and others are getting stronger as they come together. The outer peaks are getting very weak. And in the limit where the chemical shift was zero, the difference– as in methane– then the outer peaks disappear altogether and you just see one peak.
Now notice that that part in the original spectrum was a mess, because all those hydrogens of the benzene ring have similar chemical shifts, and they’re all coupled with one another so it’s more of this same thing, and all on top of one another, and very complicated, so we’ll forget that.
Let’s think about what causes this change. This is a so-called higher-order effect, not the simple thing, that you have a chemical shift, count the neighbors, and get 1:2:1, 1:3:3:1; 1:4:6:4:1 kind of thing. So here we’re plotting the energy in a big applied field of different nuclear spin states. And we’re talking about just two nuclei. So they could both be with the field; that would be low in energy. Or one with and the other against, or one against and the other with. And then they could both be against the applied field. So you’d have these different energies, right? And that’s the 100 MHz, that’s the energy that it would take to get from here to here by flipping the red spin, would be 100 MHz.
But all these look the same, whether you do a red line and flip the red arrow, or do a blue line and flip the blue arrow. They’re all the 100 MHz or whatever frequency, or 300 MHz in that spectrum we were looking at before. So we’re going to see just a single peak at first glance. But if we home in and magnify things so we can see small differences, so enlarge this center portion by a factor of 105 so we can see the difference is just a part per million in the difference between the environment of the blue and the environment of the red proton. Then as we expand it, we see that they’re not quite the same. It’s harder to flip the blue proton than it is to flip the red proton.
The blue proton is in a higher field because of the chemical shift difference. And if we call that difference ∂/2, now you’ll notice that that made the red lines shorter by ∂/2 and the blue lines longer by ∂/2, doing that. So now we’re going to see a chemical shift of ∂. One is smaller by ∂/2, the other larger by ∂/2, the difference is ∂, the chemical shift. So that’s the chemical shift.
And now suppose the spins interact not only with the big blue field, but also with each other. And just suppose that their coupling through these electrons is such that they prefer to be anti-parallel. You could have the opposite sign of the coupling constant, they could prefer to be parallel. But let’s suppose they prefer to be anti-parallel. So the ones where they’re anti-parallel will go down in energy, the ones where they’re parallel will go up in energy. So let’s do that. And now what happened to the red and blue peaks? Notice that this red transition went to lower energy, but this one went to higher energy. And this blue one went to lower energy, but this blue one went to higher energy. Everybody see that? So what’s going to happen? The red peak is going to split.
There’s going to be a lower and a higher red peak, there’s going to be a lower and a higher blue peak. So it’s going to look like that. And that’s going to be J. If these shifts are J /4, this transition got longer by J /2, this one got shorter by J /2. So the difference between them is J now. In terms of the energy of the things interacting with one another, that’s where we get chemical shift difference, and spin-spin coupling J.
OK, so that looks just fine, and that’s what’s called the first-order spectrum. That’s what we’ve been talking about.
But there are higher-order effects, and I think you’ll find this cute. Because J does something else besides just shifting these levels, it also mixes this state with this state. And it mixes them into a sum and a difference. And of course we’ve seen things just like this before. That’s what it looks like in the case of magnetism. The sum has the two pointing opposite directions along z, but not along x. And the difference has them pointing opposite directions altogether.
When you do that, when you mix this with this, you get the difference, which turns out to be a little lower in energy, and the sum, which is a little higher in energy. And of course, that’s just like mixing a p-orbital on carbon with a p-orbital on oxygen in a C=O double bond. You mix the two, their energies don’t match, so this one has more of that in it, this one has more of that in it. Oops, I wrote this wrong, it should be red and blue – OK, I’ll change that. Everybody with me?
So I didn’t tell you exactly why it couples to this, but it does. So this shifts those energies. Notice that that makes the reds a little bit smaller, both of the reds smaller. And both of the blues larger. So it shifted the reds to the right and the blues to the left. So ∂ isn’t exactly what you thought it was going to be. That is, the difference between the average of those two isn’t exactly the chemical shift, because of this higher-order effect.
But there’s something much more interesting that happens. That one is non-magnetic, obviously. They cancel one another. But how about the other one? It’s not magnetic along z, but it’s just as magnetic along x as either of those other states, the top one or the bottom one, is. So it has this resultant. So it as magnetic perpendicular to the main field as either the one that was where they’re both against the field, or the one where they’re both with the field. So this one is magnetic, and that one’s not magnetic. That is, this one can interact with the magnetic field of light, but this one can’t interact with the magnetic field of light. So if we look over here, we see that this one is weakened. It’s not interacting with light. This one is interacting even more strongly with light. Now what that means is this one is really strong, and this one is weak now, in terms of interaction with light.
So this influences the intensity of the transitions. Now what it does, is it makes the ones that involve this level– that’s this red one and this blue one get stronger, and these get weaker. Like that. So what’s going to happen over here? This one, which is the shortest of all, and this one which is the longest of all, this one and this one get weak. Pardon me, this is the longest of all. This is the shortest of all, this red one. So those get weak, but the inner ones get strong. So it changes to look like that. And the more you mix them,… They’re separated by ∂ or ∂/2, they’re mixed by J. So the bigger J is compared to ∂, the more they’ll mix, the more they’ll be exactly like this, which is when they’re completely mixed. And the stronger these will become, and the weaker these will become.
So as the ∂ gets smaller and smaller, in terms of Hz as you change the magnetic field, the outer peaks shrink and the inner peaks grow. So that’s why methane has a single peak and you can’t see it splitting. Even though the coupling is still there, you just can’t see it.
So the radio frequency light loses its handle on the flanking peaks as ∂ becomes small compared to J, as shown here. So by increasing ∂, measured in Hz, that is by using an expensive, big field, you make things look more like first-order spectra that are easy to interpret. 1:2:1, 1:3:3:1; 1:4:6:4:1, and so on.
So that’s the subtle aspect.
Chapter 3. Averaging and the NMR Time Scale [00:15:57]
Now we’re going to talk about how you use NMR to study dynamics, to study rate, and what we mean by the NMR time scale, and how decoupling works. So we looked at this spectrum last time, this is a spectrum of ethanol measured, as it says over here, by your own TA, Phillip Lichtor. If we look at these three peaks, we see there the OH, the CH2, and the CH3, and we analyzed these last time and saw that this is a triplet, this is a triplet, both being split by this CH2, but this one is a quartet of doublets. Now that was measured in DMSO solvent for a special reason. Remember, that peak came from DMSO that had a little hydrogen in it, as well as deuterium. If we measured it in chloroform with a little bit of acid in there– or if Phil did, he measured this one as well– notice the peaks shift a little bit. Especially the OH peak shifts a lot, the others shift a little bit, this one not too much.
What’s much more interesting than these shifts is what the peaks actually looked like, in terms of their splitting. Now bear in mind this OH peak is here. It’s now over here, but I’m drawing it here. So instead of a triplet, it’s now a singlet. And this one, instead of being a quartet of doublets, is just a quartet. No changes over here. We’ve lost the splitting between the H and the CH2. It’s no longer split. It’s not split by the CH2, and the CH2 is not split by the H. So what’s going on when we did this? In fact, what happens if we don’t have the acid in there? If we don’t have the acid in there, this peak moves way over here. This one stays the same place, we’re still in chloroform solvent. And this one stays the same place, but the OH really changes. But what’s, again, much more interesting is what the patterns look like. They look like that now.
So this, instead of being either a triplet or a singlet, is a really broad, low peak. This one still looks a little bit like a quartet of doublets, but it’s much broadened. This one is the same as it was before. So something really funny is happening here.
Now let’s look at the integrals. So within experimental error, it’s 1 to 2 to 3. It’s 0.96 to 2, to 3.04 in DMSO. In CDCl3, it’s 1.00 to 2.00 to 3.05, again with an experimental error 1, 2, 3. But the one that had a little bit of acid in it is much more interesting. It’s 2 to 3, that’s fine. But this is bigger, 1.14. So that peak – this sharp peak here, the one that has acid in it, is bigger. There are more protons there than just one. How did that happen? It’s an average of both the OH that’s on the alcohol and the H+.
Now let’s see, how did that happen? Well, H+ can go onto the alcohol, and then the H that was on the alcohol can come back into the solution. So we’ve exchanged those two H’s. And if that happens rapidly, there are H’s moving all over. Most of the time they’re on alcohol but some time– 14%– there as H+ and what you’re seeing is the average of those, in this case. So this is averaging. The H’s are zipping around not only between acid and alcohol, but from one alcohol to another alcohol, by way of the acid. And what that means is, remember, the reason we see splitting is that there are these spin isomers. There are different molecules in there. Some molecules have the neighboring CH2 both up, some have them both down, some have one up and one down. But if you’re zipping from molecule to molecule very rapidly, you see the average, which is nothing. So when we have acid in there and it’s averaging very fast by going on and off and then going to different molecules, then we see a sharp peak here, the splitting disappears.
Here, what’s going on? Well how about the ROH chemical shift, that thing? It’s the average of many hydrogen bonded structures, because the hydrogen bonds keep getting made and broken very rapidly. So you average among many different hydrogen bonds, but the H stays with the original molecule. It’s not going from molecule to molecule. It’s just breaking hydrogen bonds and going back and forth. And how fast you do that and how much you do it depends on concentration, on what the solvent is, on the temperature, and so on.
But how about the spin-spin splitting between the CH2 and the OH? That can be the average involving H-exchange among many molecules. It’s fast if you have H+. It’s slow in DMSO. So in DMSO, you saw H’s staying with their molecule, and you see they’re splitting. In H+, they go from molecule to molecule fast. You see the average. And what happens if you have chloroform, but you don’t have the H+? Some kind of averaging is beginning to happen, it looks like. So it’s clear that if you have H+ in there, it can zip around.
What happens if you’re just making and breaking hydrogen bonds? Well, you could have three alcohols hydrogen bonded together in a circle. And they could trade which hydrogen, so this is a mechanism for exchanging hydrogens from one alcohol to another. And then as the trimer comes apart and the molecules become another trimer and do this exchange, another, another, another, this is a mechanism – a slow mechanism, much slower than this H+ thing– which can allow slow averaging of the Hs. But if you have DMSO as the solvent, this doesn’t happen. Why? Because DMSO forms very strong hydrogen bonds with the OH, and it doesn’t form these trimers in which the hydrogens can shift from one molecule to the other. If DMSO were a strong enough base, it could pull the H off and be like an acid catalyst, it would be a base catalyst. Pull it off one molecule, put it on another one. But it’s not that strong a base, it’s just strong enough to forbid forming those hydrogen bonded trimers. So we can get this averaging.
But wait a second, we looked at this one a couple lectures ago when we were looking at IR. And we said that even though this was said to be an anhydrate, it had these water peaks. And we said there were two kinds of water peaks. There are the ones that are hydrogen bonded here, and the ones that are not hydrogen bonded. That’s two different peaks. But shouldn’t those average the same way they do in NMR, where as you go from molecule to molecule, as the molecules come apart and come together in hydrogen bonding, you quickly average those? Why don’t you get averaging in the IR, when you do get averaging in the NMR? Another question is, what’s fast? How fast does it have to be to see a blur, or to see the average as a sharp average, rather than to see the two distinct things?
This particular sample is a solid, so you could imagine that these are held in place by the surrounding molecules. But you see the same effect in solution, so that’s not the difference.
Now notice that in IR, we have wave numbers here to tell us how fast this is vibrating and how fast this is vibrating. And this one is 110 x 1010 Hz, and this OH bond is 102 x 1010 Hz. That is, the difference between them is 1011 Hz, whereas the difference we were looking at in the alcohol things were 100 Hz or a few hundred Hz, at most. So this is a much bigger frequency difference here in IR than it was in NMR. What does that have to do with anything?
How long does it take to measure a frequency precisely? How long is long for purposes of an average? Well suppose you have two different swings. I’m going to put these two guys on swings up to the ceiling and they’ll start swinging. And they swing at a certain rate, but they won’t swing at exactly the same rate if they’re a little bit different, the two swings. This is pretty fast swinging. One of them swings a 20 Hz, and the other swings at 21 Hz. And I’m going to try to push them, but I’m not going to adjust the frequency I’m pushing at. I’m just going to sit here and go bum bum bum bum bum bum bum bum bum. And I’m going to do it a 20 Hz, bum bum bum bum bum. So the guy that’s swinging naturally at 20 Hz, I always push him every time he comes, just right. So the light field in the IR case, or the NMR case, matches perfectly for 20 Hz but it doesn’t match for 21 Hz, of course.
If I just did it for a little while on the 21 Hz, just for a little while, it would be almost in phase, and I’d be pushing them. But after a little while, we’d get out of phase, and I’m pushing when he’s coming and then pulling when he’s going and so on. So it’s going to get out of whack.
How long does it take to get out of whack? How long do I have to be doing this in order to find out that I’m not really doing anything with him? Well notice that a one-second pulse samples the full range of phases, from being completely in phase to coming back to being completely in phase again. So if I do it for one second, I get nothing. It just cancels. On this one, everything was golden. But obviously, then, there’s no net interaction with the light in that one over the course of one second. So a one second light pulse can clearly distinguish between 20 Hz and 21 Hz. Everybody with me?
That means… Suppose we have 22 Hz, a swing that I’m trying with my light to hit, but I’m going at 20 Hz. So that goes through two complete cycles there. In fact, if it goes through one complete cycle, you can tell it’s not working after only half a second. That is, 1 over the difference in λ. That’s long enough to sample the full range, favorable and unfavorable. So I can more quickly tell the difference when the frequencies are more different. It depends on 1 over the frequency difference, that’s how long I have to go to know that I’m not going to get any action out of it.
Now here’s the interesting thing. Suppose the frequency is changing in time. It’s averaging, a proton is jumping from one position to another and changing its frequency. Suppose that it goes at 22 Hz for a while, then at 20 Hz for a while– it’s in a different environment. Then at 22 Hz, then at 20 Hz, then at 22 Hz, then at 20 Hz for a while. It’s jumping back and forth randomly at a certain rate. Now obviously, that’s not going to work with the 22 Hz light. It goes in and out of phase, so we won’t get any absorption of 22 Hz light. Suppose we try 20 Hz light, so we’ll move it up there. Again, it goes in and out of phase. No profit in that one. But suppose I try 21 Hz. Now it stays pretty much in phase, so it’s averaging properly. And so what we will see is we have a very good match at 21 Hz, the average frequency, and we’ll see a single sharp peak. 21 Hz will work, 20 Hz won’t, 22 will [correction: won’t]. We’ll see an average peak.
What’s determining? It’s how long the pulse is compared to the difference in frequency. If the pulse is long enough, as long as the difference in frequency, then we can tell it. So when do peaks average? When atoms don’t stay put long enough to tell the difference in frequency between the two sites that they’re changing between. So if two peaks differ by 100 Hz, you have to count for a hundredth of a second, roughly, to tell them apart. But in the IR, they differed by 1011 Hz. So the exchange of position would have to be 1011 times a second in order to be able to see separate peaks. That’s as fast as atoms vibrate. They can’t change their partners that rapidly. So that’s why in the IR you don’t see averaging when you do see it in the NMR. Because the NMR differences in frequency are small, just a few Hz or a few hundreds of Hz. So if something happens in a hundredth of a second, you can see separate peaks for it. But you can’t do that in the IR, because the differences in frequency are so large.
So this gives rise to what’s called the NMR time scale. Now suppose you had cyclohexane-d11. It gives this spectrum. It gives a spectrum for axial and equatorial hydrogens. Incidentally, notice that that’s sort of as if this were like benzene with the ring current. It gets shifted down when it’s out the side, and shifted up when it’s along the axis. At least, that’s a way to remember it. I’m not really sure that that’s the explanation. I’m not really sure that ring current is true. But anyhow, you can remember which is which if you want to. Now notice this is a special molecule, d11, just one proton. And the reason is that you would get very complicated spin-spin splitting if these are all hydrogen. Because the axial will be split by that, this equatorial will be split by that axial one. Because they don’t have the same chemical shift, so they’re going to split one another. And then they’ll also be split by these with different coupling constants and so on. So if we make all those deuterium, we get sharper peaks so we can study it. So it’s d11.
But notice this is at minus 89°C. And watch what happens if you start warming up. The difference in chemical shift is 0.48 ppm, and this was measured in an old spectrometer in 1964 by Frank Bovey at Bell Labs. And the difference here is 0.48 ppm at 60 MHz, or it’s 29 Hz. We have to put it in frequency, if we want to talk about this kind of thing. So 29 Hz difference in frequency. Now we go from -89° to -68°. The peaks get broad. To -63°, broader still and moving in. At 60 they coalesce and become a single peak at -60°. At 57°, it starts sharpening up, and sharpened up still more at -49°. And what do you think it looks like at room temperature? A single, really sharp peak, as sharp as these or sharper. So it’s a single sharp peak up there.
Even if you have C6H12, because it’s flipping back and forth rapidly enough that it’s averaging the chemical shifts. And once the chemical shifts become the same on average, then you can’t see the splitting anymore. So cyclohexane gives a single sharp peak. But this allows you to tell how fast, because you know this frequency difference. So it turns out that they coalesce to give a single peak when it’s at about that same– the lifetime is about the same as that. It’s actually four times that, but for qualitative purposes it’s the same at the time they coalesce. So it’s about 100 times a second that it’s flipping back and forth at -60°. But you can measure this, you can fit these lines to theoretical calculations for rates at other temperatures. Or you can go to a different spectrometer which has the same chemical shift difference, but a bigger frequency difference. And then get at what temperature you’d have that frequency, or four times that frequency.
So that’s a great way to measure the rates, and by measuring the rates at different temperatures, you can get what the barrier is. And that’s where we got the number from last semester, when we were talking about conformational analysis. We said that the barrier for flipping one chair to another in cyclohexane is 10 10.5 kcal/mole, and that’s where you get it. It’s not easy to measure something like that, but NMR allows you to do it. So this is a handy way to measure really fast rates and very small barriers that correspond to these fast rates. So the NMR application in dynamics is very important for studying fast rate processes. Let me see how much time.
Chapter 4. Predicting an NMR Spectrum [00:25:04]
Let me just go very quickly, I’m going to go really quickly through this so we can get to a little bit of something else. But this was the kind of question that I might ask on an exam. The first part, Question 1, had to do with making this molecule. We’ve talked about making hemiacetals before. But the relevant part here is Question 2, which was to sketch what you’d expect for the NMR spectrum of that compound. So let’s look at the different groups and you predict what we’re going to get in terms of chemical shift. First, (a). How about its chemical shift, where do you expect it on the spectrum? The OH?
Professor J. Michael McBride: It’ll be downfield because of the O, but it’ll be almost any place because of the hydrogen bonding and so on, that has a big range. And how about the splitting? Should it be a singlet, a doublet, a triplet? quartet? quintet? 1:4:6:4:1? What determines the splitting of this? How many protons are on the next adjacent atom. So we couple through this bond, this bond, to here. These are too far away. Those are removed by one, two, three, four bonds. But to here, it’s only one, two, three bonds. So it should be a doublet, right? Everybody agree it should be a doublet? Sebastian, you shake your head slightly.
Student: Well, it depends on what the solvent is.
Professor J. Michael McBride: Ah, in DMSO, it would be a doublet. But mostly, you don’t use that. And in chloroform, it would be exchanging so you wouldn’t see the splitting. So it could be almost any place between 1 and 6 for hydrogen bonding, and the splitting is… probably it would be a broad singlet, because of rapid OH exchange.
OK, now let’s go to (b). Where should that be, chemical shift? Way at the right, way at the left, in the middle, almost to the right?
Student: Way at the right.
Professor J. Michael McBride: So it’s hydrogens that are on a carbon and the next atom is carbon too. That’s up at the right. But it has a couple oxygens here. So it’ll be shifted down a little bit. Slightly deshielded by two oxygen atoms on the neighboring carbon. So one might predict someplace between 1 and 2, and there’s a compound that’s very like that, that is at 1.3.
OK, how about the splitting? This one? Now it has three bonds to get from here to here. So that should be a splitting, there’s no exchanging there. So it should be a 1:1 doublet, split by– and the J should be about 7 Hz.
OK, now how about (c), what should it look like? What should its chemical shift be, relative to, say, these hydrogens?
Student: A doublet of quadruplets.
Professor J. Michael McBride: Derek, what did you say?
Student: I said a doublet of quadruplets.
Professor J. Michael McBride: Not the splitting, where should the chemical shift be first?
Student: In the middle.
Professor J. Michael McBride: It’s attached to a carbon. What’s that carbon attached to. Amy?
Professor J. Michael McBride: They are oxygens. So it’ll be shifted down more than this one was. Because it’s closer to the oxygens. So it’s deshielded by two oxygen atoms on the same carbon. So it would be between 4.5– there’s an analog that’s at 4.7. Now how about, Derek, now you tell us about the splitting of this one.
Student: I think it’s a doublet of quadruplets– or a quadruplet of doublets. I don’t know how to–
Professor J. Michael McBride: It would be – if it were split by this, that makes it a quartet.
Professor J. Michael McBride: And this would make it a doublet, except we already said that that’s probably exchanging. It would depend on the solvent. So probably it would just be a 1:3:3:1 quartet. J about 7 Hz.
OK, let’s keep going. How about this one? What’s that going to be like? Jack? Chemical shift for that CH3?
Professor J. Michael McBride: Maybe about 2, because it’s attached to a carbon which bears oxygen. So slightly deshielded by the oxygen. That’s rather like this one, which is attached to a carbon which bore two oxygens. So not as far down as that. 1.1. And how about the splitting of that one, Jack?
Student: It would be a doublet.
Professor J. Michael McBride: Just a doublet, because these are too far away. 1:1 doublet, 7 Hz.
OK, now let’s go to (e). What’s it going to look like? How about its chemical shift? Debby?
Student: It would be like 4?
Professor J. Michael McBride: Yeah, because it’s got an oxygen on the same carbon. So it’s 3.8. Not bad. And how about its multiplicity? How many hydrogens splitting it three bonds away? I can’t hear very well.
Student: It would be…I don’t know.
Professor J. Michael McBride: Well, we already said that this one is going to be a doublet split by that. If that’s true– if this one splits these, then these split that.
Student: So do the top ones.
Professor J. Michael McBride: So do the top ones. So how many altogether?
Student: So it’s six hydrogens, so its–
Professor J. Michael McBride: So it’ll be seven lines in the binomial intensity. Seven line multiplet from the six atoms on the neighboring carbons. 7 Hz. So that’s going to be really, a whole bunch of them. Seven lines.
OK, now how about (f)?
Student: Like (d).
Professor J. Michael McBride: Chris?
Student: Same as (d).
Professor J. Michael McBride: Ah, same as (d). It’s got the same connectivity as (d), same hydrogen coupling and so on. But it’s slightly deshielded by the oxygen atom there– so I copied that from here. And I copied this from here, and that’s what you said.
But it’s different from (d) because here is a chiral center. So one of those is pro-R and the other is pro-S. They’re different with another chiral center. So in fact, when you look at a situation like that, there are two that are really close to one another. Two doublets. They’re about the same, but I caught you.
Student: Can you see that?
Professor J. Michael McBride: Yeah, you do see that. Not in every molecule. Notice it’s an isopropyl group, two CH3s on a CH attached to something. If the rest of the molecule is chiral, those will be different.
Student: By how much?
Professor J. Michael McBride: That depends on what the molecule is. It’s sometimes hardly observable. But very often, you can see it. I saw one, when I was a graduate student I got a spectrum like that, and I couldn’t… I said, what the heck is going on? So that was a standard question I would ask on exams down here of graduate students at that time, after I learned it myself.
Chapter 5. Proton Decoupling [00:42:32]
OK now, proton decoupling. Let me just start this, and then we’ll show the example next time. So we have JC-13-H here. Now notice that when we put this in a certain magnetic field, the proton will precess at 100 MHz. How about the C-13? Will it precess at 100 MHz? No, because it’s a much weaker magnet. It doesn’t feel as much torque, it doesn’t precess as fast.
So the H has a steady field in that direction. Or there are other ones that have exactly the same steady field in the opposite direction. Now C-13 is 25 MHz, it’s only a quarter as strong as a magnet. So it precesses much more slowly, it’s a smaller magnet, and it can be either up or down. So both these things are happening when we put it in a magnetic field. And here’s the NMR spectrum, and we’ve seen that, and saw the satellites– this peak comes from C-12s. This is from C-13s when they’re down, and this is the proton precession when the C-13 is up, reinforcing the applied field.
And at the same time, if we looked at the C-13 spectrum, notice that essentially all of hydrogen is protium. Only 0.015%, very little, is deuterium. So you don’t see that big peak in the middle here, because all the hydrogen is either this way or this way. And the magnitude of the splitting in both cases is 125 Hz, that’s fine.
So now, suppose we do something really cute. Suppose we put two radio frequencies in. One at 100 MHz, one at 25 MHz. But what we’re going to do – the way we’re going to do it is to irradiate 100 MHz all the time. But just do a pulse, a 90° pulse, of the C-13 to twist it down into the plane, and then allow it to precess. And that’s what we’re going to listen to see if we got a signal. We’re going to listen to the C-13 while irradiating the H.
Now what effect does that have, then? If you were in a frame rotating at 100 MHz, you’d see that this little 100 MHz field is going to cause the hydrogens to precess like this. To go up, down, up, down, up, down. What determines how fast they go up and down? How fast they precess? What determines that? It’s how big this field is. How strong the field is. The stronger the field, the faster the precession.
So that means H is going to be up, then down, then up, then down. And how fast it’s going up, down, up, down, up, down depends on how much power we put into our radio. If we do it fast enough, we see the average of the C-13 and decouple the spectrum. We can see the C-13 spectrum without the H splitting it, by making the H’s quickly go up and down. How fast do they have to go up and down in order to average the 125 Hz splitting to be nothing, to just be a single peak? How fast? Once you’ve had time to review the lecture, you’ll be able to answer that.
Professor J. Michael McBride: About 125 times a second, right? The frequency difference is how fast you have to flip to average. So if you put enough power in on the hydrogen, you remove the splitting from the C-13 spectrum, which can be handy, as we’ll show you next time. So the decoupling power determines the rate of this precession. And this, we’ll see next time.
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