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I have noticed that all halogenoalkanes of the form 2-Xpropane seem to have a mass spec peak at 27, implying that a $\ce{C2H3+}$ ion is being formed.

Forming a $\ce{C2H3+}$ ion, however, would mean breaking three bonds and, given its abundance, that wouldn't make sense - especially in the case of 2-chloropropane where it has an abundance of 25.36 according to this website.

Why exactly does the $\ce{C2H3+}$ ion form?

EDIT: You can also see this peak in the mass spectrum of 2-bromopropane, 2-iodopropane and 2-fluoropropane, but it is less pronounced.

EDIT 2: as pointed out by @PLD, it appears that other molecules of this form have peaks at 27 for a different reason to that of 2-chloropropane, so my previous edit is irrelevant.

Mass spectrum

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    $\begingroup$ Maybe alpha cleavage with H2 elimination or whatever; can't see unlikeliness in here. $\endgroup$
    – Mithoron
    May 4 '17 at 21:49
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    $\begingroup$ Nice question. I am not sure that 2-bromopropane and 2-iodopropane m/z 27 share the same originating process though. Referring to Rosenstock et al. (pubs.acs.org/doi/pdf/10.1021/ja00373a001) and Ikuta et al. ( journal.csj.jp/doi/pdf/10.1246/bcsj.49.66) the m/z 27 ion is described as a fragment of either m/z 41 (allyl cation) or m/z 29 (ethyl cation), but neither of these ions do appear in the fragmentation spectrum of 2-Chloropropane, nor do these two articles reference 2-chloropropane. I did not find any studies specifically on this issue. $\endgroup$
    – PLD
    May 6 '17 at 7:35
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    $\begingroup$ The m/z 42, 41, 39, 27 pattern looks like that of the EI spectrum of propene / cyclopropane. This suggests loss of HCl as first step, followed by fragmentation of m/z 42. Why so abundant? According to Maccoll & Mature (onlinelibrary.wiley.com.proxy.scd.u-psud.fr/doi/10.1002/…) : the appearance energy for m/z 42 (ionized cyclopropane or propene) from 1-chloropropane is lower (about 2.7 eV) than that for 2-chloropropane. Thus the m/z 42 ion would be formed with much higher energy, leading to more fragmentation in the latter case. I am not completely convinced though. $\endgroup$
    – PLD
    May 11 '17 at 7:13
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Here is a little mechanism I made up about it. Not quite sure if its correct, seems to fit the mass spectrum though.

enter image description here

Definitely have a look at McLafferty classical textbook. Maybe you will find the answer there.

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  • $\begingroup$ I guess that the ethene cation is higher in energy than the allyl cation. But maybe, it is just about in range to be formed in equilibrium, and as the neutral methylene is removed from the equilibrium, the reaction is effectively drawn towards the side of the ethene cation. Maybe two methylene units even recombine to ethene, though I am not convinced that this could be possible in the HV of the mass spectrometer. $\endgroup$
    – logical x 2
    May 12 '17 at 12:57

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