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Homework Help: Mass spectrometry data analysis

  1. Jan 31, 2012 #1
    1. A liquid compound gave a mass spectrum showing a strong molecular ion at m/z = 156. The only fragment ions are seen at m/z = 127 and 29. Suggest a structure of this compound.

    2. A liquid compound gave a mass spectrum in which the molecular ion appears as a pair of equal intensity peaks at m/e = 122 and m/z = 124. Small fragment ion peaks are seen at m/z = 107 and 109 (equal intensity) and at m/z = 79, 80, 81 and 82 (all roughly the same size). Large fragment ions are seen at m/z = 43 (base peak), 41 and 39. Suggest a name for this compound.

    I suppose the first one would be C11H24 (MW = 156) which breaks down into ionic fragments C2H5+ (MW = 29) and C9H19+ (MW = 127). Am I correct? And I am totally clueless about the second question. Any guidance would be appreciated. Thanks.
    Last edited: Jan 31, 2012
  2. jcsd
  3. Jan 31, 2012 #2
    I haven't done these type of questions for a while. Do you have any more information to go on? Are you expecting it to be only hydrocarbon chains, etc.? Bromine, for example, will show two peaks of approximately equal intensity due to the approximately equal relative abundance of the two isotopes (incidentally, they will be found at 79 and 81 when they are on their own).
  4. Feb 1, 2012 #3
    It is not necessary for the compound to be a hydrocarbon. Its just that I had not thought of bromine. In that case, the first compound could also be ethyl iodide, since the MW of iodine is 127. Do you think I am on the right track? And what would could be the structure of the second compound, assuming that it does contain bromine?
  5. Feb 1, 2012 #4
    I don't know how much detail you have gone into on these spectra, but the question seems difficult to me: such that it seems a little misleading. You think you have a structure, and then its difficult to account for the 'extra' lines and the 'missing lines'.
    I thought it might contain bromine because of the two peaks of roughly equal magnitude separated by 2 mass units, as I mentioned. I subtracted away the bromine and you get a value of 43, which usually means a certain group; however, I found it difficult to see where the other peaks came from, I thought perhaps rearrangements, but I had no idea about the 80 and 82 m/z peaks.
    Anyway, to cut a long explanation short, it's not at all obvious what the other peaks correspond to, but you can find the molecule with a little searching. Finding what the other peaks correspond to and their names takes a little more work and a little effort. So, basically, what I am saying is it is up to you. If you want to try to work it out I can help; alternatively, you can search through some online sources yourself. I say all this as I don't want to give the impression I worked it out, since we never covered some of principles necessary to work out what is happening (at least I do not remember covering them).
  6. Feb 1, 2012 #5
    I finally found the questions here: (http://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/questions/Spectroscopy/masspec2.htm). The answer given for the first question is ethyl iodide while for the second one is 1-bromopropane or 2-bromopropane. However there is no explanation provided. The propyl group does have a mass of 43, but as you said, the other lines cannot be explained so easily. Hope you can shed some more light here. Thanks for your help so far!
  7. Feb 2, 2012 #6
    It wasn't easy to find, but I think I found, through various sources, the species responsible for the various m/z values. Firstly, the 80 and 82 appear to correspond to the production of HBr, one for each isotope.
    The other values were more difficult. I managed to find a paper for 3-Iodoprop-1-ene. Admittedly this is not iodopropane, which would be closer; however, the removal of iodine yields the allyl cation, which has an m/z = 41. Furthermore, the allyl radical can produce the cyclopropenyl cation, with an m/z = 39, or alternatively the propargyl cation with the same m/z.

    Allyl cation = C3H5+
    Cyclopropyl = c-C3H3+
    Propargyl = l-C3H3+
    You can google the structures no problem.

    C3H5 + hv -> c-C3H3+ + H2

    C3H5 + hv -> l-C3H3+ + H2

    Apparently, cyclopropyl is favoured over propargyl as it is more stable.

    The paper is:
    Ingo Fischer, Thomas Schüßler, Hans-Jürgen Deyerl, Mohamed Elhanine, Christian Alcaraz. Photoionization and dissociative photoionization of the allyl radical, C3H5.
    International Journal of Mass Spectrometry, Volume 261, Issues 2–3, 15 March 2007, Pages 227-233.

    As to how you get to the allyl radical, I don't know, it may involve the formation of propene in an initial step through an elimination reaction of bromopropane, and then proceed from there, maybe. It's difficult to find an explanation; as you can see, this is pieced together from various sources. If you want the other sources, if I get a spare five minutes I can find them, but the above is the paper for the m/z = 39 and 41 information.
  8. Feb 3, 2012 #7
    I guess then the 107 and 109 peaks would be due to CH3CHBr+ (with both isotopes of bromine of course). I think you are right about how the allyl radical is formed.

    Thanks for the great research! I'll go through the paper you mentioned. I am sure I would never have been able to figure this out by myself.
  9. Feb 4, 2012 #8
    Remember it depends on if you have 2-bromoporpane or 1-bromopropane, but yes the loss of the methyl group is the 109 and 107 values.

    I found another paper which I think is also useful:
    Dissociative ionization of hot C3H5 radicals.
    H. Fan, L. B. Harding, and S. T. Pratt.
    Molecular Physics. Volume 105, Issue 11-12, 2007.

    There are two radicals produced, the allyl radical and the 2-propenyl radical. As always, it ends up being more difficult than it should be, since some sources consider them to be the same thing. However, it appears that the allyl radical is H2C=CH-C°H2 <-> H2C°-CH=CH2, and the 2-propenyl radical is H2C=C°-CH3 (° denotes the electron). The allyl radical is resonance stabilised, the 2-propenyl radical isn't and is more likely to produce both forms of C3H3+.
  10. Feb 5, 2012 #9
    Thanks, I'll have a look at this paper as well. The problem is, I was given these questions on my term paper where I did not have access to all this literature you are citing. So could you tell me how do I even begin to solve such questions if I encounter them in the future? For example, peaks separated by 2 units may suggest bromine, odd mass molecular ion suggests presence of nitrogen (I read this somewhere) etc. Do you think there is some general purpose method of getting to the structure from the mass spectrum? I would be really grateful if you could help me here, or even suggest some paper to read regarding this.
  11. Feb 8, 2012 #10
    Peaks separated by two units doesn't necessarily suggest Bromine, it suggests isotopes of elements which differ by two neutrons. The clue it was bromine was 1) the relative abundance is approximately 50/50, and 2) the peaks at 79 and 81. Ignoring Cl 36: Cl 35 and 37 are in the ratio of approximately 75/25, this would also show peaks separated by two mass units, but the molecules would have a different relative abundance, and there should be peaks at 35 and 37.
    I don't know what a term paper is, but I don't think you'd be expected to know the other peaks, we weren't expected to, a 'good guess' using some main peaks might be all that is expected, you might want to ask. There are resources on line that go into some detail about how to analyse mass spect. Anyway, we were taught that you rarely do mass spec on it's own, its often used in conjunction with other techniques.
    Last edited: Feb 8, 2012
  12. Feb 9, 2012 #11
    Thanks! I guess I'll read up a bit more on this topic and then trouble you again if I come up against something!

    PS: Term paper means exam!
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