[Violagirl]

The NMR spectrums below show overlaps of eugenol and acetyleugenol together. The acetyleugenol has two peaks that distinguish it from eugenol. It looked one peak was the found at around 2.3, which is representative of the acetyl group. However, I've been unable to determine where the other peak might be am thinking it possibly pertains to the methoxy group found on it. The other thing I am trying to figure out is the mole ratio of both compounds for purity/ratio calculations. I was thinking of using the peak found at 3.8, representative of eugenol and the peak at 2.3 for acetyleugenol to do it. I thought I was supposed to figure out the height of both peaks, divide by the number of H's in both and then divide the amount in cm of one compound to the number of the other when added to the height of the first peak and multiply it by 100. This did not seem to work as I took the height of the peak at 3.8, found it to be 1.50 cm, divided it by 2 for its number of H's and got .75 for an equivalent of 1 H. I did the same for the other peak, got a height of 5.5 cm, divided it by 3 H's and got 1.83 cm for the equivalent of 1 H Then I took .77 divided by 1.83 + .75 and multiplied by 100 to get a percentage of about 30%. This seems wrong however as eugenol has a higher presence then acetyleugenol and am not sure what I'm doing wrong or how to do it. Any help would be fantastic.

Edit: For the steam spectrum, for the way I had done it, I found that the Ha peak would have a measurement of .75 cm for one H and .17 cm for one H at the Hj peak. When calculating it, it appeared that eugenol would have have a purity percentage of 81%. Again, any help is appreciated!

[pfnm]

The protons that are different in the molecules will give you knowledge of their signals and where they should appear in the spectrum:
eugenol has an alcohol O-H which appears as a broad singlet in the spectrum
acetyleugenol has no alcohol O-H, but has an acetyl CH3 which will appear as a singlet.

Neither wll be coupled.

In the CO2, one of your structures is drawn incorrectly (the structure on the right, two Ha protons are bound to a carbon with three other bonds).

In all structures, you've drawn the Ha protons, but not the proton that undergoes 3J coupling with Hc and Hd (the hydrogens at the terminal end of the alkene). It is this missing proton, that splits Hc and Hd (to give overlapping doublet of doublets) and is split to give the complex pattern that you've labelled Hf.

I annotated your spectrum (CO2) and fixed the incorrect drawings.



The 5.4ppm peak is from the eugenol O-H. The 2.8ppm singlet is from the -CH3 on the acetyl functional group in acetyleugenol.

The hydrogens have already been integrated. A computer program in the NMR machine measures the integrals of the curves (the area under each curve) and prints this number at the bottom of the spectrum. The numbers are never exact, you have to assign integration comparatively within a spectrum:
once you find a peak representing a known number of hydrogens, you can use its integration as a basis to derive the integration of other peaks.

I don't understand this method of calculating the mole ratio of the hydrogens? The height of the integral does not offer you any useful information.

Are you sure this is the method you were given?

[Violagirl]

Thanks so much for your *delete me* That makes a lot more sense. It appears that the acetyl group was more apparent in the CO2 spectrum then in the steam spectrum.

On determining the purification ratios, I am not for sure if that is the correct method. I had followed an example lab report my professor had included with our manuel and it had looked like they had done some sort of measurement to determine the ratio of the compounds. However, I was not sure if they had measured the height of the peaks or if they had measured it in another way. I just know that they had calculations that derived from cm and then they had divided out the number of compounds from corresponding peaks to achieve the equivalent of one H for both compounds and then completed their calculations from there.

[Violagirl]

Ok so I spoke with my professor about it and she had to measure the lines of integration that lined up with each area of peaks. Then to divide the number of H's for both peak areas to gain the equivalent of the area of 1 H and then apply furthur calculations. I will try what she suggested and then upload what I get later on.

[Violagirl]

Ok so I obtained purity values. However, at the peak on the CO2 spectrum labled Hb, I found that the small peak corresponding to the large peak belonged to that of the methoxy group in acetyleugenol. In labeling my spectrum, would this be labled as corresponding to the same set of peaks or would it be different the H's are next to an acetyl group on acetyleugenol and next to an OH group on eugenol? Any help is great. 

[pfnm]

Hb produces both signals (one peak from the eugenol, one from the acetyleugenol). The substitution of the alcohol by the acetyl functional group does not  have a great effect on the position of the signal(s) for Hb.