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  • >> 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--

>> All right so listen today what I want to talk

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B2 中高級

化學203.有機光譜學。第20講。瞭解複雜的脈衝序列 (Chem 203. Organic Spectroscopy. Lecture 20. Understanding Complex Pulse Sequences)

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    Cheng-Hong Liu 發佈於 2021 年 01 月 14 日
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