lex asked in Science & MathematicsChemistry · 12 months ago

Need help with organic chemistry mass spectrometery questions?

1.Identify the important (diagnostic) peaks from the IR spectrum. List the cm functional group that corresponds to each peak. 2. List all possible classes of compounds that your unknown could be. 3. From the MS, identify the parent ion and determine the molecular weight. Identify the base peak and determine the molecular weight of the fragment. Label both the parent ion and the base peak on the MS spectrum. 

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2 Answers

  • 12 months ago
    Favourite answer

    You could start by concluding that the M+ = 116 m/z. This is the parent ion. Parent ion is the molecular ion, also assigned as M+. Even numbers often mean that this is an organic compound containing no N atoms. As there are no isotope patterns and the missing masses do not correspond with any halogens, I would assume that this molecule consists only of C, H and O. It could of course be that this molecule contains even numbers of N atoms (2,4,6,etc) but from my experience, it is unlikely that this is the case.

    If you look at the biggest peak in the spectrum: that is the base peak. The base peak (m/z=60) could be the [CH3-(C=O)-OH] fragment, which you could regard as a regular, protonated acetic acid molecule. This could be the result of ester cleavage of the original molecule. I would therefore say that this molecule might be some kind of ester, with an alkyl moiety on the end.

    The peak at m/z=73 suggests that during one of the fragmentation patterns the molecular ion has lost a fragment with a mass of 43. 43 is found often in mass spectrometry. It is the mass of either an 'acetyl' fragment (CH3-C=O) or an (iso)propyl fragment. This agrees with the alkyl ester hypothesis. We'll leave the m/z=73 peak for now.

    An acetyl ester group in a molecule has a mass of 59. Assuming your molecule contains one of these groups, the rest of your molecule has a mass of 57. Normaly for linear alkanes, you can clearly see peaks at 15 and 29, resulting from methyl and ethyl fragmens, respectively. In this mass spectrum, these are not so evident.

    On the other hand, there is a large peak at 43 which might as well be evidence for a (iso)propyl radical. isopropyl radicals are more stable than propyl radicals, so that could explain why there is less evidence of methyl and ethyl radicals. Assuming there is an isopropyl-group (43) in the molecule and a acetyl ester group (59), the total mass balance misses 14 mass units. This could be a CH2 fragment. I therefore would say that this is in fact not an isopropyl but an isobutyl group (m/z=43+14 = 57). Back to m/z=73. If we take the isobutyl group and add 1 O atom, the total mass is 73.

    My GUESS (I have to emphasize this, since there is little data to work with) would be that this MS spectrum corresponds to isobutyl acetate. I don't know if my answer is right or wrong, and maybe this answer is more than you asked for, but I hope you see the way you analyze these spectra and what you can do with it. I did a complete analysis to end up with the whole molecule, so that you can then explain all the fragments that you see. Draw out the isobutyl acetate molecule if you will, cleave any bond, determine the m/z of the fragment and see if it agrees with the MS spectrum. That way, you can really check if the answer is right of wrong. I have to say, I did a lot of assumptions and in general, that is not the right way to analyze your scientific data.

  • 12 months ago

    What IR spectrum?

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