The course introduces the three key spectroscopic methods used by chemists and biochemists to analyse the molecular and electronic structure of atoms and molecules. These are UV/Visible , Infra-red (IR) and Nuclear Magnetic Resonance (NMR) spectroscopies. The content is presented using short focussed and interactive screencast presentations accompanied by formative quizzes to probe understanding of the key concepts presented. Numerous exercises are provided to facilitate mastery of each topic. A unique virtual spectroscopic laboratory is made available to enable students to measure and analyse spectra online. Assessment is via summative quizzes completed during the course period.
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来自 University of Manchester 的课程
Introduction to Molecular Spectroscopy
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从本节课中
Nuclear Magnetic Resonance (NMR) Spectroscopy
This week we concentrate on Nuclear Magnetic Resonance (NMR) spectroscopy. Here a magnetic field is used to create energy levels for magnetic nuclei present in a molecule. Transition between these energy levels occurs in the radiofrequency region of the electromagnetic spectrum. The positions of the bands in the observed spectrum is dependent on the shielding of the nuclei by the local electronic structure, giving rise to a parameter known as chemical shift. Bands also display fine structure caused by spin-spin coupling with neighbouring nuclei. Examples on the analysis of NMR spectra for structure determination will be given. As the final activity in this module you are given a link to view how to obtain an NMR spectrum in the laboratory. Don't forget to compete this end of week laboratory quiz which contributes to your final mark.
与讲师见面
Patrick J O'Malley, D.ScReader
Chemistry
Okay, so we've developed that point a little bit about deshielded,
high frequency, shielded, low frequency.
And again, you'll always come across these terms and in a more upfield and downfield.
Right, so just we talked about,
the idea that all electrons tend to shield the nucleus.
I'm just going to mention that, that's always not the case.
And some will actually deshield the nucleus.
And that means that they'll add to the magnetic field, so
that the field that you feel at the nucleus position will be enhanced,
if you like, compared with the external field you're using.
And the most common ones that you come across are benzene and ethylene.
And the idea here is that these aromatic ring system here, and
the pi bonds here, generate a magnetic field.
And the magnetic field can be shown to be in this direction.
Here you have applied magnetic field in this direction.
And you can see the currents generated here at the protons are in
the same direction as the magnetic field.
So that means that this will field an actual field
that's larger than the applied.
And the same thing for the protons.
So not all electrons shield the nucleus.
Sometimes in systems like this, benzene and
ethylene they actually deshield the nucleus.
Or they make the applied field increase in magnitude at the actual position.
Right, so again you represent that this way you have your outside external field.
You add to it and the effective field at the proton and nucleus is larger.
Right, okay, so,
When you measure we talked about let's go back again to the,
So our demonstration here.
We said we had these,
Different shielding positions and we had a spectrum we plotted against new,
to frequency and we get three different values.
And these are absolute values of the frequency.
You could say they're in megahertz.
But in practice you don't actually do that.
What you usually do is you put in a, what's known as a standard
And the usual standard we talk about.
We talk about this in the notes in a minute, it's called tetramethylsilane.
And you measure the shifts in your molecule, that you're looking at,
you put that along in with your sample, in the sample tube.
And you measure.
You get the spectrometer to measure the shifts relative.
So what you do is you measure relative shifts.
So your typical spectrum then will contain something.
Say this is our molecule with three different, say, proton positions.
You'd also get a peak down here for the TMS.
So that's TMS there.
And then you measure these relative to that.
So in effect what you do is you say this is zero.
And then you measure the position of these other peaks relative to TMS.
So what this is called when you do it like this it's called the delta scale.
And the exact definition is the frequency of, say,
a sample minus the frequency of tetramethylsilane divided
by the tetramethylsilane multiplied by 10 to the 6.
And that's because you multiply by 10 to 6 it's called parts,
PPM or parts per million.
So generally when you see anymore spectra,
you're not comparing that the absolute frequency is new.
But you're using a scale, the delta, the delta scale.
Okay, all right, so this is just a, so
the chemical shift is measured in what they call a delta scale.
And as I defined it on the last overhead,
it's nu minus nu reference.
Now the reference for protons in carbon
13 is almost always tetramethylsilane and you divide it then.
So the reason you divide it is because sometimes you can use, and
NMR you have spectrometers with different frequencies.
So if you didn't divide it by the reference,
then you'd get different values for each spectrometer and that would be confusing.
If you divide it by the reference, then it'll be the same for all spectrometers.
And you multiply it by 10 to the 6 because you're
going to get a very small number here if you don't multiply it by 10 to the 6.
So when you multiply by 10 to the 6 you get ppm and
they usually range from about 0 up to 12.
All right, so this is the way you
monitor an NMR spectrum.
So as we say here, the delta scale uses a reference compound.
And it records the absorption peak as a shift from the reference.
And then by dividing by the reference frequency,
you come independent of the actual field, and it's universal for all NMR machines.
All right.
And this goes to parts per million and
multiplied by 10 to 6th because the shifts are very tiny.
Okay, so as I said in the overhead,
the reference compounds used is usually tetramethylsilane.
This molecule here.
You have 12 protons.
They're all the same, same environment.
So you get a strong single peak.
And it's also quite upfield.
These are highly shielded protons because the silicon is responsible for
putting a lot of electron density near the hydrogen.
So they're highly shielded, or again, this upfield, low frequency.
And the vast majority of organic compounds that you'll be looking at
are compounds have downfield or the deshielded.
Compared with TMS and by default the delta,
the chemical shift for delta on the delta scale is equal to 0.
Okay so again I emphasize this bit here that you'll come across.
Here you have your frequency scale.
Here you have your delta scale.
They're going the same way.
So peaks up this region spectrum, large delta values high frequency.
Down field, deshielded protons.
Protons in electron-poor environments.
On this region, up this side of an NMR spectrum, protons in electron-dense
environments, shielded protons, upfield, low frequency, small delta values.
So you really at this stage should be getting this into your head.
And also getting a good understanding of why these are the terms used for
the different regions of the spectrum.
Okay, so just to get this delta scale into your heads
a little bit more we'll do an example here.
So you're asked what is the shift of the resonance from tetramethylsilane which is
a standard, if a group of protons.
You measure it, the delta is equal to 6.33 three parts per million for
a polypeptide in a spectrometer operating at 420 megahertz.
So how would you work that out?
So if we go up here, so I have the same question here.
So you remember that delta is equal to the frequency
of your sample minus the frequency of tetramethylsilane.
Divided by frequency of tetramethylsilane, by 10 to the 6.
So what you want is you want the shift of the resonance of TMS, so
you want that nu, what is nu minus nu TMS.
So you just rearrange that equation.
nu minus nu TMS.
Is equal to delta times nu TMS.
Sorry.
So delta, we're given a 6.33.
Sorry I left out a 10th of a 6th haven't I, sorry.
So it's delta times,
New tier, delta times your tier mass,
and you have to divide it, because you're dividing it across by 10ths of a 6th.
So if you work that out.
You're going to get delta 6.33.
The spectrometer's frequency which is the new TMS
value is 420 by 10 to the 6 seconds minus 1.
And then you divide it by 10 to the 6th.
So what you will get is the shift from the TMS.
It's 2.66 by 10 to the 3 seconds minus 1, or hertz.
Okay, right, so what you can do now is you can,
we said that before we saw the spectra.
And these are some typical example spectra that you might come across.
So you can see for them all, so you also take the first one here.
We have the deltas on the delta scale, and it goes from 0, this one, up to 10.
Usually, for organic compounds, it goes from about 0 to 10.
And the delta scale is increasing in that direction,
that's the convention for explaining the NMR spectrum.
And look at this molecule here.
This is a paraxylene.
So what you have here, your NMR spectrum.
You have the TMS peak of 0.
And then you have two peaks here.
You have this peak at about 2.4, 2.5.
And then you have another peak equal to the delta value of 7.
Now what you find here these lines these are called integration lines.
And basically they're measuring the area of the peaks.
because one of the things about in the NMR spectra is that
the number of protons governs the intensity of the lines.
So what you see here is you have two CH3 groups, that's six protons.
But they're what we call equivalents.
There's no difference between them they're in the exact same chemical environment.
And we've also got four on the ring and
they're exactly in the same chemical environment.
So what you should notice is that so you'd expect that the 6 protons,
the ratio of these lines should be 6 to 4.
And if you did the integration you'd find that.
So the ration is 6 to 4.
So, from that, you would know that this peak here, at about 2,
corresponds to the methyl groups.
And this peak here, at about 7, corresponds to the protons on the ring.
And this value here from TMS,
you measure the chemical shift
relative to the tetramethylsilane.
So let's move on.
We could talk about the chemical shift.
I just want to finish this off.
So here you have this other compound, and if you look at it
in a NMR spectrum, you've got the TMS signal here at 0 ppn.
Then you've got a peak here at about 1.5 delta value,
and another peak here up about 4.
And you can see that this one is more intense.
It's about with the integration it's about three times more intense.
So here you have two CS trig routes.
They're in the same environment so you're going to have 6 protons for that and
here you have 2 protons in the CH2 group and again they're equivalent.
So again as I said, you have 6 versus 2.
So again you measure exactly, find this is about one third the intensity of that one.
So you assign this peak here to the CH3 groups and
you assign this peak here to the CH2 groups.
Right, so we'll move on then to this one here.
We're just measuring proton NMR, so we're just looking at the hydrogens.
Here you have three equivalent hydrogens and
here you have three equivalent hydrogens.
And they're in two different environments so you expect two peaks.
because they're going to have slightly,
electronic environment around them is going to be slightly, slightly different.
Now, from the integration this time you can see that the peak heights
here are the same.
And that's what you'd expect because you've got three protons for
each distinct group.
So you have to assign this.
I don't know which one is which, but if you can see here that this CH3 group is
close to the oxygen, it's bound to an oxygen.
Whereas this CH3 group is bound to a carbon.
Now you should know that oxygen is more electro-negative than carbon.
So you'd expect the oxygen would pull the electron density to itself.
So it will pull electron density from the CH3 groups.
It'll pull electron density away from that metal group.
So you'd expect that to be more deshielded than this one.
So we know that our shield is up this side.
Deshielded is this side.
So you'd probably say that this one corresponds to that and
this one corresponds to that one there.
Okay, so let's look at this one here.
It's an aldehyde.
So here you have C and 3CH3 groups and they're all going to be equivalent.
They're all in the same chemical environment.
And then you have the H of the aldehyde group.
So you expect peaks.
You expect two peaks, you have two distinct chemical environments so
you expect two peaks and the ratio of a 9 to 1.
And here actually in this spectrum when it was done it's recorded
in the integration that measured the peak intensities.
And you can see here the peak intensity is 17.3 to 255.4 which is about 9 to 1.
So you can initially say this is due to the nine protons form the methyl groups.
And this one here, which is widely deshielded,
is due to the aldehydic H group and so forth.
Let's do the last ones.
Here you have this is a nitine group here.
So, you have a methyl groups.
These are equivalents.
So, there are six there.
There's one there and there's a one that's attached to the triple bond.
Carbon, carbon triple bond.
So, you'd expect three environments and as you can see, you do.
Here you have nine and here you have one each.
So this large peak here is due to the nine protons.
Sorry that's six.
CS3, so six.
And these ones here are due to the H1s of the oxygen and the h1s of the C.
And again they're difficult to assign these, but
again you have the more likelihood of negative oxygen.
So you might expect that the more deshielded one,
the one to higher chemical shifts is due to this and
the other one, the less shielded one is due to this proton here.
And then lastly you have this one here.
And you have CH3 group here,
expect three protons for that, two protons here.
And then you have actually the two protons on the metal group are going
to be equivalent and you have six protons here and one there.
So if you go again you're given the integration values here on the bottom.
And this will correspond to the six protons in the metal group and
then you have the three one, this one here.
And then you have two, which corresponds to this one here.
And then you have one which will correspond to the Group.
