[Rutherford]

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?

[discodermolide]

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.

[Rutherford]

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.

[discodermolide]

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?

[Rutherford]

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.

[discodermolide]

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

[nox]

^ethyl acetate

[discodermolide]

^ethyl acetate

Thanks for the correction

[Rutherford]

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?

[discodermolide]

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.

[Rutherford]

I suppose that without the red line the intensities can't be determined just by looking. Thanks for helping.

[AWK]

Integration is a standard procedure during H-NMR measurement
http://www.stolaf.edu/people/hansonr/nmrtalk/ethylacetate/spectrum.gif

[discodermolide]

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"

[nox]

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.

[discodermolide]

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!