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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--