Subtitles section Play video Print subtitles >> All right so listen today what I want to talk about is something that's more conceptual. We've been sort of talking nuts and bolts and today practical stuff, interpretive stuff. Today I want to talk a little bit more about how some stuff works in NMR spectroscopy. It's not going to be anything too fancy and it gives you a little perspective on my take and understanding of NMR experiments, which is very different say than Professor Shaka's [phonetic] take or interpretation or Professor Martin's take or interpretation. These are NMR development people and my group and you will be users of NMR spectroscopy as a research tool, but one of the things is you can't just take NMRs as a black box. As you're already seeing when you're going down using the instrument you kind of have to keep your head about you, oh, what am I doing, why am I locking it, why am I shimmying it, what's the shimmy doing, what does it do if I don't do this? And what I would like to be doing is giving you a feel for some parameters and things that you may not be understanding in spectra. So we've already, for example, had instances where people look at their own DEPT spectra and say wait a second why do I see in the DEPT 90 a little peak for say a methyl group or a little peak for a CH2 group? A lot of this has to do with parameters and so that's what I want to give you a feeling about. We're going to start with 1 experiment that's kind of like a DEPT but a lot less fancy and I think it's one that's easy to understand. It's called the APT. So it was a precursor. That stands for attached proton test. [ Writing on board ] And the more NMR spectroscopic name for the technique which actually is going to make a lot of sense to you when we start to talk about how it works is called the J modulated spin echo technique. [ Writing on board ] And then we'll talk a little bit about DEPT. We won't go, and understand it as well, but you'll see how the parameters that we talked about for the APT relate to the DEPT experiment. DEPT if you don't know is Distortionless Enhancement Bipolarization Transfer. [ Writing on board ] From a practical point of view what the depth does as you all know is allows you to identify the CHs, CH2s and CH3. So in other words, the methines, the methylenes and the methyl groups. What the APT does is it's a less sophisticated experiment. It gives you your Cs, your quats and CH2s and distinguishes them so I'll say distinguishes let me maybe put it that way. [ Writing on board ] Quat and CH2 from CH from your methine and methyl. So in a way, the DEBT experiment is more of a Litmus test experiment. So we're going to start to get a feel for complex pulse sequences. I'm going to introduce some concepts like the rotating frame and the effect of various pulses. We talked a little bit about this when we introduced NMR spectroscopy and we'll see a spin echo experiment. So I just want to review some of the spin physics that we learned when we started to talk about NMR spectroscopy. So remember you start and you've got this difference in population between the alpha and beta states and that leads to a net magnetic vector along the X axis and remember that vector is a bunch of vectors that are precessing and we talked before about the concept of applying a 90-degree pulse. So remember 90-degree X pulse and you think about the right hand rule that means you have a quail along the X axis you're applying essentially a 4 sign to the vector, which is bringing the vector down into the X, Y plane. That brings it along the Y axis. Now remember what that means. Having the net magnetization in the X, Y plane means that we've now equalized the alpha and beta state populations by having a differentiated population of alpha and beta states you have net magnetization along the Z axis, but when the alpha states are equal to the beta states, then you have no net magnetization along the Z axis. What you do have is magnetization on the X, Y plane and because you've done this as a pulse all of your vectors start out initially along the Y axis and then remember you have your precession that leads to motion in the X, Y plane of the vectors and that gets picked up by the coil and gives you a cosine wave that when you Fourier transform gives you a peak. So the other pulse I'll give you an idea on is just remember if we apply instead of a 90-degree pulse, 180-degree pulse along the X axis, now that flips the population of the alpha and beta states. So, instead of dragging the magnetization down into the X, Y plane you dragged it all the way along to the negative Z axis. I'm trying to represent things in 3 dimensions here. So that brings our magnetization to the negative Z axis. If you think about it, this means now that you've inverted the population of alpha and beta states so if we had more nuclei in the alpha state than the beta state before then after more nuclei in the spin down state in the, if you've had more nuclei in the alpha state than the beta state, you flipped the population over here. So, that's basically all the spin physics that we need to get to a point to really think our way through an experiment that ultimately is going to allow us to distinguish our quats and our methylenes from our methyls and methines. So now I want to develop that idea. [ Writing on board ] So let's take this situation here and I'm just going to project the X, Y plane into the plane of the blackboard because it's going to help us to look down on the system rather than for me to try to keep making these horrible 3 dimensional drawings. So, okay, here we are in this situation where we've got our X axis, we've got our Y axis and now we're looking down the Z axis and we have our magnetization along the Y axis. Now, remember the magnetization along the Y axis is precessing and so if you just think of this as a single line, it's going to precess at an angular velocity omega. [ Writing on board ] Like if it's 500 megahertz you're going to have precession at 500 million times per second in the X, Y plane. That's this angular velocity if you're interested. It's just my way of representing that it's moving. So in other words as you wait, if we're precessing around in the X, Y plane as you wait, sometime T now when we look at the X, Y plane now your vector is over here and it's continuing to precess. Remember this is what gives rise to your signal that's your FID where you get this cosine wave over here that's basically the coil picking up your precessing vector and generating electricity, generating alternating current in the coil that ultimately you amplify and use the analog to digital converter on and then Fourier transform. All right what I want to do at this point though is to introduce a concept that NMR spectroscopists use to make their lives easier and the idea is the rotating frame and it's basically saying let's just have our axes precess for conceptual purposes actually works out for the detection purposes with various electronic techniques, but let's have our axes rotate at the angular velocity. So that means we'll have the axes rotate at omega and so we'll have our rotating frame and we'll call that X prime, Y prime. So now if you think about it if you're in the rotating frame, if your frame is rotating, your frame of reference is rotating with angular velocity omega and we wait time T from the rotating frame how does our magnetization look at time T? Exactly the same and that's the whole big concept of the rotating frame is simply we go ahead and we adjust so we're not spinning around it 5 million times per second. All right now we're ready to introduce the concept of the spin echo and that's going to be the basis for differentiating our methines and methylenes, our quats and methylenes from our methines and methyls in the DEBT experiment. Okay, so the idea behind the spin echo is as follows. We're going to be it the rotating frame and so we start with a 90-degree pulse and I'll call it a 90-degree X prime pulse just to show that we're working in the rotating frame. Imagine for a moment now that I have some vectors and they start along the Y prime axis, but now let's say for a moment that some of those vectors are moving a little bit faster than the Larmor frequency than the velocity omega and some of them are moving a little bit slower. So, in other words, we've got our frame rotating at 500 million cycles per second but some vectors are going a little faster and some vectors are going a little slower. So now if you imagine on this thought experiment that you wait time T, now what's going to happen after time T to after some time to those vectors? The ones that are rotating slower will appear to go clockwise and the ones that are going, right, clockwise and counterclockwise. So now our vectors are going to fan out with respect to the rotating frame and those that are going one direction are going to be going this way. I think it's actually counter clockwise in this axis, axis system, but the point is that they're diverging in velocity. So now we've got some that are headed this way and some that are headed this way. Now, imagine for a moment we now at this point after time T apply a 180-degree X prime pulse. In other words, now we do the same thing that we did before and we apply a 180-degree pulse. Now, when you do that, again just think right hand rule and remember any component of your magnetization that's along the X prime axis when you apply a pulse to it right hand rule, it doesn't do anything. Any component of your vector that's along the Y prime axis flips over 180 degrees. Does that make sense? So when we take this vector and we apply our 180-degree pulse the Y prime component comes on over to the negative Y prime like so and we do that to all of these different vectors and they've all flipped around the X prime axis and they've all gone to the negative Y prime axis and yet they are still precessing at the rotating frame but now they're headed inwards and that's the key to this whole thing. So because they're still going in the same direction these ones are a little bit faster, these ones a little bit slower than the rotating frame they converge inward and if we wait another time T, that same time increment so I'll say T again now what's our picture at this point. They're all on the negative Y axis so they are all like so along the negative Y axis. This is the basis for many, many sorts of NMR experiments. This is called the spin echo and one of the reasons that people invariably do spin echoes is in addition to what we're talking about here which I'll show you in a second which is J modulation, you've also got T2 relaxation where vectors fan out in the X, Y plane and what the spin echo experiment does, what the spin echo does is a refocusing of those vectors. So the ones that are inadvertently due to T2 effects moving a little faster and the ones that are moving a little slower by applying this 180-degree pulse halfway through they refocus and so almost every experiment you see, the ones I'm talking about now but also 2D experiments are going to involve some sort of thing that involves a weight and then an equal weight. So if I write out this simple pulse sequence, what it is is we apply a 90-degree pulse, we wait, we apply 180-degree pulse, we have an equal weight and at that point the spin is refocused. [ Writing on board ] And so if you've already started to notice as you're working some of the 2D experiments, there are various delays and you may see this in the handout about various delays, some of the delays you've learned about in the 1D experiments may be relaxation delays where it's just for your magnetization to return to the Z axis I think that's what's D0 or D1 on our instruments but then some of the other delays that you're choosing are specific parameters related to this. As I said, this now sets us up for the so-called J modulated spin echo experiment or APT. [ Writing on board ] All right so now we're going to bring in the idea of 2 different channels. Now the experiment here is a carbon detected experiment and so we're going to be doing stuff in the carbon channel. Remember, normally when you're running a carbon 13 NMR you're proton decoupling. The reason you're proton decoupling, remember those all coupling constants I talked about, the J1CH being like 125 hertz and the J2CH being like 10 hertz, if you took an undecoupled spectrum of ethanol it would be horrible because your CH2 group would appear as a quartet that was then split into a triplet. So it would be a quartet of triplets and your CH3 group would appear as, I'm sorry, your CH2 group would appear as a triplet of quartets and your CH3 group would appear as a quartet of triplets. So imagine how tiny your peaks would be, right, because a singlet is this tall, a doublet is half as tall, a quartet is 1 to 3 to 3 to 1 where the total height is the same as the height of the singlet and if you go now to a quartet of triplets everything is split and you've already seen when you collect your carbon 13 that you're fighting noise on this so you'd be horrible. So we always end up collecting these fully decoupled proton decoupled experiments, proton decoupled spectra. So the trick in the APT is we're going to turn on our proton decoupling partway through the experiment. So we do a 90-degree X pulse and I'll call it X prime pulse because we're in the rotating frame, and then we're going to wait 1 over J, 1 over the coupling constant, and then we're going to apply 180-degree pulse and meanwhile we're going to start to do stuff in the proton channel. So for this point we haven't been proton decoupling. We're going to at this point turn on the broadband decoupler. All right remember when you apply a broadband decoupler what you're doing is rapidly flipping the spin states of the proton. So that's going on at 500 megahertz. Meanwhile you're applying these pulses at 125 megahertz for the carbon. So now you're going to be rapidly flipping the spins and the protons that's why you don't see, why you normally get a singlet, but here we've done something first where we haven't immediately turned on the broadband decoupler. We've applied our 90-degree pulse, we're waiting 1 over our coupling constant, we're then going to go ahead and apply another, we're going to wait another 1 over the coupling constant and then we're going to observe all the time with our broadband decoupler going on. This is as I said what's going to lead to the CH3s and CH2 and CHs up in the CH2s and the CH2 and the Cs are down. [ Pause ] All right so let's think about what each of our carbons sees. So if you have a quat carbon, we can conceive of that quat carbon as being a peak at the line at the frequency omega. In other words, the quat carbon, the rotating frame as I said is partially an artifact, it's partially a conceptual tool, it's partially an electronic tool. So let us conceive of the rotating frame as being exactly tuned to the frequency of the carbon. So in other words, in the rotating frame you're going at omega which means here you're going to be going at 0. Okay, now imagine a methine and we're only going to be considering 1 bond coupling. So, a methine is going to be without decoupling it's going to be a doublet that's centered around this frequency omega. So you've got 1 line of the doublet is going to be going faster and think about it the J is the coupling constant so the separation is J so you're going to be going at omega plus J over 2 and here this other line is going at omega minus J over 2. Does that make sense? Now if you have a methylene, now you have without decoupling a triplet a 1 to 2 to 1 triplet and remember the separation of the lines is J so the center line is at omega and the line, the down field line, is at omega plus J and the up field line is at omega minus J. We'll do the same thing now for a methyl group and in a methyl group now you have 4 lines in a 1 to 3 to 3 to 1 ratio centered around omega and so what's the position of the line I just drew the arrow to? [ Inaudible response ] Omega plus one-half J and the position of the next line? [ Inaudible response ] Plus three-halves J, right. And the other big line? Minus J over 2 minus a half J. And the last small line? [ Inaudible response ] Minus three-halves J. All right, 3 J over 2. All right that sets us up now to start to consider the different scenarios and we'll just consider the quat, the methine, the methylene and I think by the point you get to the methyl it'll all sort of make sense. [ Pause ] All right so let's try this scenario for a quat carbon. We apply our 90-degree X prime pulse and so we're in the rotating frame. Our magnetization ends up along the Y prime axis. We wait 1 over J so if J is 125 hertz you would imagine waiting 1, 125th of a second. [ Writing on board ] How do things look after 1 125th of a second? >> Same. >> Same, yeah, because we don't have any J coupling in a rotating frame is going right at omega. So after 1 125th of a second after 1 over J nothing happens now we go ahead and apply our 180-degree X prime pulse and at the same time we turn on our broadband decoupler. If there had been coupling, in other words, if the proton, the carbons attached to a spin up proton were precessing faster and the carbons attached to a spin down proton were precessing slower now when we turn on the broadband decoupler you're flipping back and forth and they're both precessing at the same angular velocity omega, but here we don't have any attached protons so we don't even change that coupling thing. Okay, so at this point you go ahead and now your scenario looks like this where now your magnetization is along the negative Y prime axis and then you're again going to wait 1 over J so again we're waiting 1, 125th of a second if we're using this value of J is equal to 125 hertz. So now after 1 over J how do things look? >> Same. >> Same. So at this point our magnetization is along the negative Y prime axis. That means if you're not in the rotating frame your cosine wave, but remember how I've typically shown our cosine waves like this where we start positive? If your cosine wave starts negative, which is what happens when your magnetization is along the negative axis, the Fourier transform is a negative peak. So here when we Fourier transform we get a negative peak, a line down. All right remember the APT is going to be distinguishing our methines from our quats or should I say our methines and our methyls from our quats and methylenes. So, now let's try this exact same scenario but we're going to try it with a methine. [ Pause ] All right so now, and I'll just write things out exactly the same way. We apply our 90-degree X prime pulse, we end up with our magnetization, got to make my axes a little bit bigger here, we end up with our magnetization along the Y prime axis, but now we're precessing and remember we said one of them precessing at J over 2 relative to the rotating frame and the other of them is negative J over 2 so I guess our rotating frame rotates counterclockwise. Now we're going to wait 1 over J. So how do things look at this point after 1 over J? [ Inaudible response ] Forty-five degrees. Everyone agree? [ Inaudible response ] So let's put this in concrete numbers. Let's say we're at 125 hertz is J. In other words, our angular velocity is going around for J is going around 125 times a second, but we're going around at J over 2 so we're going around at half that angular velocity. So in other words, if you wait 125th of a second, where do you get to? >> One-hundred and eighty. >> One-hundred and eighty. So you get to the negative Y prime axis and that's the main point over here is now your 2 vectors both the one attached to the, the one precessing faster due to being attached to a spin up proton and the one precessing slower due to being attached to a spin down proton, they've come by on a negative Y axis. Now at this point we go ahead we apply our 180-degree X prime pulse and we turn on our broadband decoupler. So after you've applied your pulse, your magnetization is along the Y prime axis, but you've just turned off your J coupling because you've turned on your proton decoupler. So now you don't have 2 separate vectors, 1 precessing faster and 1 precessing slower, they're both precessing at the Larmor frequency. In other words, they're staying along the positive Y axis. So now we go ahead and we do this whole shebang again of we wait 1 over, I guess where am I in my drawing? Okay, so we wait 1 over J but nothing happens except remember what I said about T2, remember what I said about defocusing, the reason you're doing this is to refocus any inhomogeneity in the spins due to T2 relaxation but nothing changes in regards to this primary picture. Now, you're decoupling your magnetization is still along the positive Y axis and so at this point when you observe what do you get? You get a positive peak. [ Writing on board ] This translates into this situation here where you have a positive cosine wave instead of a negative cosine wave and the Fourier transform of a positive cosine wave is a positive peak. So you can see how we've differentiated our methines from our methylenes by way of this spin echo experiment. Our methines from our quats by way of the spin echo experiment. Let's move on to our methylenes at this point. [ Pause ] So our methylenes now you can think of as being composed of 3 different vectors. So you've got a vector that's going on omega and a vector that's going at omega plus J over 2, plus J, and a vector that's precessing at omega minus J. So we apply our 90-degree X prime pulse. You end up with a situation where you can think of this as 3 vectors comprising your triplet. One of these is going at omega, one of them is going at omega plus J and the other of them is going at omega minus J. Of course, in the rotating frame you just forget about the omega because the whole system is going at omega. So you've got 1 of them going at plus J, 1 of them going at minus J. Maybe I'll do parentheses, I'll put my parentheses around the omega here. All right so again we waiting 1 over J and if you can think about it what does it mean if you wait 1 over J? [ Inaudible response ] They're all on the positive Y because you've waited your 125th of a second or whatever it is and everything has come back over to here and at this point you go ahead and you apply your broadband decouple and your 180-degree X prime pulse. That flips your magnetization over to your negative Y axis and recombines it. Again, you wait 1 over J and now nothing happens because you've turned on your broadband decoupler except any refocusing of the T2 relaxation that's occurred. So after 1 over J your scenario is still a negative peak and so you get a negative peak over here. [ Inaudible response ] Spin lattice relaxation. So, invariably you're always fighting spin lattice relaxation. Remember that's the defocusing in the X, Y plane. So a lot of the pulse sequences incorporate these spin echoes to bring up the intensity and sharpen things back up. So that's why because if you looked at this and said why are you doing the second half of the experiment? Why don't we just turn on the broadband decoupler and observe? The short answer is, well, yeah, it would work, it just wouldn't work as well. Your peaks would be less intense, you'd have more artifacts and so forth so the spin echo gets incorporated and it gets incorporated into all of the experiments that we won't go into a lot of depth on like the HMBC experiment where we're going to use it but not necessarily understand it quite as well. All right let me show you one of these experiments. The APT is written up in Phil's handbook so you can actually use it, but in practice you'd have no reason to use it versus the more modern depth experiment. I just like this because I think it's something that we can easily wrap our heads around for starters without too much mental gymnastics. [ Pause ] All right so this is one I pulled from the supplemental book [inaudible] and it's just an example. It's a sugar molecule. This particular sugar they ran an APT experiment. So here you have a regular C13 and then you have, so this is, of course, your H1 decoupled, you always run H1 decoupled these days, C13 NMR, and here's your APT experiment on the sugar. So you'll notice in the sugar that carbon 2, which is this quat, whoops, carbon 2 is down over here and this carbon here, which is the methylene out here it's carbon 9. That one is down over here, and then you have 1 more down and that 1 more down if you notice there's this methylene hiding out at position 3. So you have 2, 9 and 3, the methines and methylenes are all down. [ Pause ] All right let's at this point let me raise the 180-pound gorilla in the room that should be, should be bothering you and you've seen this and some of you have found it confusing on the exam and that's the issue of what is J. Because the problem is, of course, you're going to do one experiment and so you have to choose an average J. So as I mentioned before for an SP3CH, J is typically, J1CH, is typically about 125 hertz. For an SP2CH remember the degree of coupling is intimately linked to the hybridization of the carbon because that tells you how much time the electrons spend near the nucleus, right? They have more S character if you have 33% S character those electrons are spending more time near the carbon nucleus than if you have only 25% S character. So you get a bigger coupling constant. So the J1CH is about 160 hertz. So normally you go ahead and you say, all right, you compromise and you use a middle value of 100, let's say 140 or 145 hertz, and you say all right that's pretty good. You can get away with that largely, but now imagine a situation where you have something unusual so I'll say but what does that do for something where you have a J1CH of 250 hertz like an alkine? In an alkine, you'll look and you say wait a second everything is going to be screwed up because while you're waiting your 125th of a second or 140th of a second and your CHs, your SP3s and your SP2CHs are doing exactly what you expect, your alkine CH is zipping around and precessing twice as fast. So what does that mean when you're looking at your data that means you look at the alkine CH and you come up with exactly the wrong conclusion about, or completely ambiguous thing where you see a peak in one spectrum and a peak, you know, peak up, well, we'll get to the DEBT spectrum which is where it really counts, but in other words, the APT if it's going around twice as fast you'll come to the wrong conclusion. So alkines end up being oddballs in many, many of these experiments. Not just in the APT but also in the DEBT and then in the HMQC as well, which also relies on these types of things and that's an important take home message. The other one somebody, I forget who raised this question, it was a good question over the homework was saying well do you ever see and it was something, it may have been DEBT, do you ever see J2CH? Well, no, usually they're very small. It's like 20 hertz and 20 hertz is nothing. Well, so we had, I don't know if people, I guess about 7 people took this on the exam and took this problem and sure enough I put an X over it but you may remember in the DEBT this peak actually showed up and I'm like what the heck is going on? I just put an X on it to help you out on this, but I was just looking and it turns out in this situation J2, right, you've got all of this SP character. J2CH is equal to 50 hertz for alkine. So, again, you're going to get a nominalist coupling. In other words, that's this coupling from this hydrogen over to this carbon. So none of these experiments are Litmus tests that you can use without really engaging your mind and thinking about what's going on. I guess another one I mentioned earlier is chloroform because electronegative atoms can have a big effect on coupling constant too and in chloroform J1CH is equal to 208 hertz. So in other words even though the carbon in chloroform is formerly SP3 hybridized because you've got 3 electronegative atoms attached to that carbon, your J1CH is a lot bigger than 125 hertz; it's 208 hertz. [ Pause ] So the DEBT experiment ends up having the same types of issues here. Because you have to use an average J value your DEBT experiment is not necessarily going to be perfectly clean. So you're going to be looking and saying okay, well, I see this peak as having a positive peak in the DEBT 135 and a teeny positive peak in the DEBT 90, but I know that that's not a real methine because it's just a little bit of the inaccuracy of choosing an average J value. So you're always making these judgment calls. Let me take a moment. I'm going to show you the DEBT pulse sequence and tell you what it does without necessarily walking through this physics. So as I said it's distortionless enhancement bipolarization transfer. The polarization transfer really is the key part. So remember carbon has a quarter of the magnetogyric ratio of proton, which means the Boltzmann differential and population is about a quarter as big for carbon as it is for proton. Carbon is a very insensitive experiment. You've got 1% C13, 99% C12, which is spin inactive. There is nothing you can do about that. You're only detecting what's associated with your C13, but then you're damned and damned and damned again by the fact that the magnetogyric ratio of carbon is a quarter of that of proton because your Boltzmann distribution is a quarter as big, your vector is a quarter as big, it's processing a quarter as fast and each of those goes to a quarter of a quarter of a quarter or the signal that you get for proton and then you throw that on top of the 1% and you're down at like 1, 5,700th a signal. Okay, so what the polarization transfer does is it flips the population of the proton and the carbon. So the carbon in the end through this pulse sequence gets the bigger Boltzmann distribution of the protons. Let me just show you the pulse sequence and tell you the things that it does. All right. So here's in your proton channel you actually start out you apply a 90-degree X prime pulse, you wait 1 over 2J and you apply 180-degree X prime pulse. In the meantime, you start doing things in your C13 channel so at that point concurrent with applying 180-degree X prime pulse in the proton channel so that's at like 500 megahertz, you're applying a 90-degree X prime pulse in the carbon channel, which is like 125 megahertz, now you wait another 1 over 2 J. now you apply a pulse which I'll show you in a second we'll call that a theta Y pulse, I guess I'll call it theta Y prime pulse. That one is going to vary. You'll do 2 or 3 experiments. So I'll say 2 or 3 experiments and put that in parentheses for or 3. So while you're applying the theta Y prime pulse you'll apply 180-degree pulse in your carbon channel. It doesn't matter if it's X or Y that you're applying, you again wait 1 over 2J and now you go ahead and finally turn on your broadband decoupler and you acquire. [ Writing on board ] This sequence of pulses leads to both the polarization transfer and it's also going to lead to differentiation of our quats, of our methines, methylenes and methyl groups. So if you apply a theta Y is equal to 135 degrees. Your CHs and CH3s end up being positive and your CH2s are negative. If you apply a theta Y prime pulse of 90 degrees, then your CHs are positive and some people will do an experiment s this is, of course, called the DEBT 135, and this is called a DEBT 90, and some people will do a third DEBT experiment where they do a theta Y prime pulse of 45 degrees and in that experiment you end up with your CHs, CH2s and CH3s all positive and you'll look at that and you say what's there to gain because all of that information in the DEBT 90 and the DEBT 135 is redundant with the information, the DEBT 45 or to put it another way the DEBT 45 is redundant but what you can do is you can add and subtract your spectra to get subspectra. So you can do what's called spectral editing to give your CH, CH2 and CH3 subspectra. Let me just show you what I mean. Imagine for a moment that we go ahead and take our DEBT 45 in which all of these guys are positive, the CH3s, the CHs, the CH2s and the CH3s, and we subtract from that the DEBT 135 in which the CHs are positive and the CH3s are positive and the CH2s are negative. Well, you subtract the positive number from a positive number you get 0, you subtract a positive number from a positive number you get 0. You subtract a negative number from a positive number and you get a bigger positive number. So you go ahead and when you do that subtraction you get a spectrum in which your CH2s are the only thing that shows up, you do other subtractions. You can get 1 that you have just your CH3s or just your CHs. In practice, it doesn't work so well and it's messy and given the fact that it's really no big deal to stare at your DEBT 135 and stare at your DEBT 90, people generally don't do it but you can do it as well. Anyway that pretty much sums up what I want to say. So the main take home message is in all of these pulse sequences you're making choices about delays that are based on coupling constants and if those choices, there are always compromises, they're always estimate, if those choices don't match your molecule, you will get confusing results which means always be careful interpreting your spectra and when you're collecting your spectra pay attention to those parameters that Phil talks about because they actually have meaning. ------------------------------e2ae5245219b--
B2 pulse axis omega prime rotating spin Chem 203. Organic Spectroscopy. Lecture 20. Understanding Complex Pulse Sequences 48 2 Cheng-Hong Liu posted on 2015/02/02 More Share Save Report Video vocabulary