Mass Spectrum of Paracetemol

[aj47]

I am used to reading mass spectrums of elements with giant atomic stuctures e.g. copper and zinc so im finding this quite hard.

At the moment I am looking at a mass spectrum for paracetemol (4-Hydroxy-(N-ethanoyl-aminobenzene)) which has four main m/z peaks at around 43, 80, 110 and 151.

I know fragmentation will occur but i cannot account for all of them. The peak at 151 can be accounted for the unfragmented molecule and 43 for a C2H3O group but what about the rest. I would have thought the c-c bond would be most easily broken but there is no evidence for any methyl radicals?

Help would be much appreciated

[akcapr]

i dont think many ppl kno much about spectroscopy here

[jdurg]

The peak at around 110/108 is the other fragment you get when you rip off the C2H3O group from the molecule. Taking the mass of that fragment away from the main molecule's molar mass leaves you with a value of 108. Values in and around that ideal number are typicaly explained by high energy collisions which results in rearrangments in the fragments. Your peak at 80, however, is most likely a contamination in your sample, as the paracetamol molecule can't really fragment in any repeatable manner and leave you with a fragment of 80. Also, all the mass specs of paracetamol that I've ever seen don't have a peak anywhere near 80.

 

What solvent did you use and what is column made of?

[aj47]

I first thought the 80 peak would be from the benzene fragment but it would have a peak at 78 not 80. Is there any chance two more hydrogens could attach to it during fagmentation to form C6H8?

[jdurg]

That may be possible if you use a VERY high energy beam to fragment the molecule, but then you'd also have to have the -OH group fragmented off so you'd see some peaks showing up pretty intensely at around 17. You wouldn't happen to have the actual spectrum that you could show, do you?

[akcapr]

guess i was wrong

[jdurg]

guess i was wrong

 

Hehe. My internship my senior year of college was working with a GC/MS/MS system for urine sample analysis. I worked on those machines 8 hours a day, five days a week for a good two or three months. Had to take them apart, put them back together, clean them, etc. etc. Biggest project was when the person I was working for gave me an unidentified urine sample and asked me to determine the sex of the donor, what they had in their system at the time of the test, and any other unique items. So I had to pretty much think of every solvent that could be used to test for all types of compounds, what settings to use, and a bunch of other stuff. It was pretty neat.

 

In case you're wondering, the sample I had was of a female who had caffeine and nicotine and their metabolites, acetaminophen and its metabolites (acetaminophen is a.k.a. Paracetamol), she had eaten a poppy seed concoction based on the ratio and amount of opiate materials in there, and a few other things. It was really cool figuring all of that out. Makes me wish that I got a job in the field upon graduation, but there just weren't any out there.

[aj47]

No i don't seeing as its a print out. But seeing as you seem to know quite a lot about analytical chemistry I have another question (quite long though). I'm now looking at the low resolution NMR spectrum print out for paracetemol and I'm a little confused. (You can see paracetemol structure here)

http://www.ch.ic.ac.uk/rzepa/mim/drugs/html/paracet_text.htm

 

The first and highest peak has a chemical shift value of 2 and is labelled 3H which I can identify as the hydrogen on the hydroxy group due to oxygens electronegative behavior.

However there are 4 more peaks with two at 6.5 and 7.5 labelled '2H' (equal height) and the final two at 9 (high peak) and 9.6 (lower peak).

Does this indicate that the final four hydrogens are all in different enviroments or only in two. I would have thought there would be only two more peaks with one representing the three hydrogens on the methyl group and the other for the Hydrogen on the Nitrogen or does the oxygen on the carboxyl group withdraw electron density from one of the hydrogens on the methyl giving rise to one hydrogen in a different enviroment? Hope you can vaugly understand my question. thanks

[jdurg]

Unfortuneately, NMR is not an area that I'm very experienced in. I did maybe one or two analyses on an NMR so I can't really give any confident answers there.

[aj47]

ah well thanks anyway

[greg1917]

Not sure i understand your question about the proton NMR, but having had a quick glimpse at the structure of paracetamol, it should have 5 distinct peaks.

 

Forgive me if this is too basic for you. There are 5 separate proton environments in the structure - the methyl group as you point out, the OH proton, the NH proton and then the 4 protons on the the phenol ring. These last 4 protons will only give 2 peaks.

 

The methyl group should be a large singlet because the nearest proton is the NH proton, and is unlikely to have a splitting effect. The height of the peak is unimportant - its the peak integral you're after, which should be 3, 1, 1, 2 and 2, not in that order.

 

The OH proton peak should (i think) be a triplet as its close to two other hydrogens, and the two phenyl peaks should be 2 multiplets, possibly doublets of doublets. Although if your NMR spectrum is low resolution the splitting patterns may not appear anyway. The coupling constants would allow you to further confirm the structure.

[aj47]

Ah thanks, yes me being a retard completly forgot about the hydrogens on the phenol group and no there is no splitting of peaks due to it being low resolution.

One thing though, like you said, the two peaks at 6.5 and 7.5 are the phenol hydrogens but why are they split into two peaks. Surely the hydroxy group and the nitrogen are too far away to withdraw electron density as they are already delocalised?

[greg1917]

the reason the phenyl hydrogens are displayed as two peaks has nothing to do with the NH or OH protons whose proximity would only affect the splitting pattern, not the peak integral value.

 

Look at the four hydrogen atoms coming off the ring. If you draw an imaginary line down the ring, going from the OH group to the NH group, you'll see the ring is symmetrical. Ive drawn a (crude) picture to show this: protons A and C are equivalent, as are protons B and D. Thus one of the peaks in the NMR spectrum will correspond to protons A and C, the other peak to protons B and D, both with integral intensity 2 (two protons). This is quite common in para-substituted phenol rings.

[greg1917]

Apologies that took so long for me to reply, i didnt realise Terminator 3 was on channel 5