Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Organic Spectroscopy => Topic started by: Rutherford on July 22, 2012, 11:19:51 AM
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How to determine the relative intensities of the signals in a NMR spectre? Why is it 2:3:3 in the following picture: http://imageshack.us/photo/my-images/853/nmrethyleth.gif/ when the middle one is higher than the other two together?
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How to determine the relative intensities of the signals in a NMR spectre? Why is it 2:3:3 in the following picture: http://imageshack.us/photo/my-images/853/nmrethyleth.gif/ when the middle one is higher than the other two together?
The intensities follow a binomial expansion, (Pascal's Triangle), i.e. 1, then 121, then 1331, then 14641 and so on.
Your signal is a triplet, intensity 121, the other signal is a quartet, 1331.
In the case 121, the middle signal is twice the height of the outer signal.
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Didn't mean the intensities of the multiplet lines, I meant the intensities of the signals (that are made from those multiplet lines). There are exactly 3 signals and above every signal, its relative intensity is written. I am asking why are those numbers (2,3,3) their relative intensities, because the middle signal is significally taller that the last signal, but they are both marked as 3.
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Didn't mean the intensities of the multiplet lines, I meant the intensities of the signals (that are made from those multiplet lines). There are exactly 3 signals and above every signal, its relative intensity is written. I am asking why are those numbers (2,3,3) their relative intensities, because the middle signal is significally taller that the last signal, but they are both marked as 3.
Sorry misunderstood.
Those numbers correspond to the peak integration (area under the peaks), i.e. the number of protons.
You have the following, 2, 3, & 3 protons.
Methyl acetate?
p.s. May I ask what level of chemistry you are studying?
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I am in high school, we didn't study this yet, but I like chemistry so I study more. It is ethyl formiate (there is a CH2 group, the 4-4,35ppm quartet).
I still don't understand how to determine the area under the peak if it is not given. Here is given 2,3,3 but why is it so? The two signals that have integration rates 3, don't occupy the same area, but they still have the same integration rates.
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I am in high school, we didn't study this yet, but I like chemistry so I study more. It is ethyl formiate (there is a CH2 group, the 4-4,35ppm quartet).
I still don't understand how to determine the area under the peak if it is not given. Here is given 2,3,3 but why is it so? The two signals that have integration rates 3, don't occupy the same area, but they still have the same integration rates.
If it was ethyl formiate there would be an aldehyde signal at about 9.5ppm, and no methyl group at 2.1. It is methyl ethyl acetate, CH3C(O)-CH2CH3, 2.1 = CH3, 4.1 =OCH2, 1.5 = OCH2-CH3.
Anyway. The old fashioned way was to cut out the peaks and weigh them.
As I said the area under the peaks is proportional to the number of protons. So the peaks showing 3 have the same area and indicate the same number of protons. Here is a link:
http://www.wfu.edu/~ylwong/chem/nmr/h1/integration.html (http://www.wfu.edu/~ylwong/chem/nmr/h1/integration.html)
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^ethyl acetate
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^ethyl acetate
Thanks for the correction ;D
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Yes, my mistake, I didn't count the C atom that has oxygen atoms bounded to it.
I see that the red line is used for determining the intensity ration, but where does the red line come from? Could I determine the intensity ratio without it?
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Yes, my mistake, I didn't count the C atom that has oxygen atoms bounded to it.
I see that the red line is used for determining the intensity ration, but where does the red line come from? Could I determine the intensity ratio without it?
The red line is produced by the printer! The NMR machine determines the intensities.
As I said above you can cut out the peaks and weigh the paper. Usually the spectrum is printed on squared paper, so you can count the squares.
I hope I understood you correcly.
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I suppose that without the red line the intensities can't be determined just by looking. Thanks for helping.
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Integration is a standard procedure during H-NMR measurement
http://www.stolaf.edu/people/hansonr/nmrtalk/ethylacetate/spectrum.gif
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I suppose that without the red line the intensities can't be determined just by looking. Thanks for helping.
No they cannot be determined by "just looking"
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This brings up an interesting question though, how did you determine the integration in the old days when fancy software and powerful computers weren't around? For chromatograms, I've heard you literally cut out the peak and measure the weight against some standard, but surely that would be impractical for a reasonably complicated proton NMR.
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This brings up an interesting question though, how did you determine the integration in the old days when fancy software and powerful computers weren't around? For chromatograms, I've heard you literally cut out the peak and measure the weight against some standard, but surely that would be impractical for a reasonably complicated proton NMR.
In those days we did not have very high field machines, so the spectra were, to put it mildly "simple".
I did cut out the peaks and weigh them, but I won't tell you either how old I am or the result I got!
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I cannot say that I am older than discodermolide, but the first NMR I used was a Varian A60 and it could plot an integral. I would have bet all commercial NMRs could.
GCs were a lot more rudimentary and I triangulated, weighed, and used the integral gizmo that gave counts for area above baseline. After that, I used a digital plotter that calculated integrals. I was in synthesis, so areas really didn't have the same importance as analytical chemists.
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My experience was the same as Orgopete's. The first NMR I used on a hands-on basis, back in 1984, was a Varian A60. These were scanning analog machines - the radio frequency signal projected through the sample would gradually scan from 10ppm to 0ppm (or whatever boundaries you set) and a chart printer indicated the absorption as the machine scanned the sample. To get better resolution, you just scanned slower. When you had scanned the sample once in a frequency readout mode, you went back and scanned it again in an integral mode. Essentially the same, but it didn't return to zero, it just kept accumulating signal. You would manually zero it after each set of peaks. I heard at the time that previous instruments lacked an integral mode and that you could Xerox the spectrum, zooming in on your peaks (there was one Xerox machine in the building that actually had a zoom function!) and then cut them out and weigh them to get the integral, but I never had to do that for an NMR.
Also, like Orgopete, I did have to do that for GC and later for LC traces which also had simple paper traces. It took much longer to get peak integration for those systems, and I never had access to a digital integrator. For simple traces where the peaks were reasonably triangular, however, it was faster and sufficiently accurate to use the height of the peak and width at half-height to calculate the peak area.
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For simple traces where the peaks were reasonably triangular, however, it was faster and sufficiently accurate to use the height of the peak and width at half-height to calculate the peak area.
I am interested in this, could you explain it?
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For simple traces where the peaks were reasonably triangular, however, it was faster and sufficiently accurate to use the height of the peak and width at half-height to calculate the peak area.
I am interested in this, could you explain it?
This is just the area of a triangle, half the base length times the height.
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For simple traces where the peaks were reasonably triangular, however, it was faster and sufficiently accurate to use the height of the peak and width at half-height to calculate the peak area.
I am interested in this, could you explain it?
This is just the area of a triangle, half the base length times the height.
But since chromatographic peaks tend to have shallow slopes and shoulders at the base making it hard to determine base length, you usually use the form "width at half-height times the height". The width of a triangle measured at half its height is half of the width of the base of the triangle.
And since the peak is rounded at the top too, it can be a little tricky measuring the height as well. As I said, however, for simple peaks it was usually a close enough approximation.
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Okay, I tried it and it was very tricky. Didn't give me the correct result, maybe because the picture I had was to small, so for precise measurement I need something that measures micrometers maybe.
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I cannot say that I am older than discodermolide, but the first NMR I used was a Varian A60 and it could plot an integral. I would have bet all commercial NMRs could.
Dear Orgopete,
You make it sound like I am 100 years old, :o
When I was an undergrad. The University was so poor it could not afford anything, we had to make it all ourselves. I think our physics/metalworking/electronic departments made the NMR machine although I can't be sure about that. I do remember getting a NMR copy that had no integration, it may have come from that machine!
So of cutting we went, although my scissors were not sharp and I ended up ripping the paper and having to estimate the area/weight.
The delights of Organic chemistry.
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This brings up an interesting question though, how did you determine the integration in the old days when fancy software and powerful computers weren't around? For chromatograms, I've heard you literally cut out the peak and measure the weight against some standard, but surely that would be impractical for a reasonably complicated proton NMR.
I did personally integrated spectra (as operator) on Perkin-Elmer 60 MHz with solid magnet in 1972 yr. Integration device did not need computer if you do not use a Fourier transform.
Those were days!