I've noticed that in the mass spectrum of 3-chloropropene, there is a peak at m/z = 39, which to me, indicates the formation of the C3H3 ion.

However, given that 3 bonds would have to be broken to form this ion, how can the frequency of this cleavage be explained?

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1 Answer 1


It would indeed be $\ce{C3H3^+}$ assuming predominant isotopic species. Although there can be other isomers, the cyclic isomer, the cyclopropenyl cation is strongly stabilized by aromaticity.

As PLD comments, there are really only two steps, not three, to forming this cation. The allyl cation ($\ce{C3H5^+},m/z=41)$ is first obtained by loss of the chlorine atom plus an electron. The aromatic cyclopropenyl cation noted above is then obtained by loss of $\ce{H2}$.

  • $\begingroup$ That makes sense, but why would it form such a predominant peak given that it seems quite unlikely that such a cleavage would occur? $\endgroup$ Commented Mar 8, 2020 at 1:35
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    $\begingroup$ Probably multiple-bond cleavage does occur, but normally we would not see it. Usually if something breaks three bonds it immediately gets one or two back. However, if the ion is stabilized it might last long enough to show up in the detector. $\endgroup$ Commented Mar 8, 2020 at 1:41
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    $\begingroup$ It is a consecutive process: homolytic bond cleavage leads to m/z 41. And for C3H5+, the lowest energy pathway for fragmentation is loss of H2 with the formation of the cyclopropenium ion (m/z 39). It is not so unlikely if one considers that EI can introduce quite a lot of energy within the ionized molecule. $\endgroup$
    – PLD
    Commented Mar 8, 2020 at 7:50

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