# α-cleavage of 1-pentene in EI-MS

I am studying the fragmentations in EI-MS, and this is the spectrum of 1-pentene.

The textbook says: The allyl carbocation (m/z=41) is an important fragment in the mass spectra of terminal alkenes and forms via an allylic α-cleavage.

First of all, at the initial ionization event, one electron draws off from π-orbital. In the below image, one can expect the 2nd carbon will bear the positive charge according to the Stevenson rule.

But in this case, I can't imagine the α-cleavage mechanism. Instead, if the first carbon bears a positive charge, I can draw the mechanism like this:

Is this the correct mechanism, although it violates the Stevenson rule, or is this an exception for the Stevenson rule?

• Is there space for a McLafferty rearrangement? Nov 1 '21 at 14:28
• @Buttonwood It is written that m/z= 42 is the peak for McLafferty rearrangement. Nov 1 '21 at 15:04

The main issue is that the structures that are named 1° carbocation and 2° carbocation in the question are not different structures. As a general rule in mass spectrometry, ionization is postulated to occur from the highest occupied molecular orbital, which is usually for alkenes the $$\pi$$ orbital. Removing an electron from a $$\pi$$ orbital leads to a singly occupied molecular orbital (SOMO) which is formally a singly-occupied $$\pi$$ orbital (imagine a double bond but only half occupied), for which there is no sensible Lewis representation. The usual way to represent in the Lewis formalism this $$\pi$$-type SOMO is to represent it as a resonant structure between the structures that are called 1° carbocation and 2° carbocation in the question, such as below:

So the short answer is that there is no reason, in general, to apply the Stevenson rule to these, as they are not two separate structures but only a single one.

There is nevertheless, for the specific case of 1-pentene an interesting twist. Leif A. Erikson et al. have shown through calculations and ESR experiments in cryo-matrices that the 2-pentene cation follows the general rule given above. On the other hand, the 1-pentene cation has an electronic structure for the SOMO which also involves delocalization of the orbital up to the adjacent sp3 carbon atom. This also translates in a shortening of the C(sp2)-C(sp3) bond upo,n ionization. In this instance, the Lewis formalism cannot be used to correctly represent the SOMO, but it helps in rationalizing the preferential break of the $$\alpha$$-bond.

The suggested formation of the allyl cation is not a contradiction of the Stevenson rule.

«What is the statement of Stevenson’s rule?

When a fragmentation takes place, the positive charge remains on the fragment with the lowest ionization energy.»

Jürgen H. Gross, public answer key for question 6.1c) on his companion site to Mass spectroscopy, 3rd edition (here).

As suggested by you, a plausible mechanism for the fragmentation of 1-pentene in a mass spectrometer is the homolysis yielding two radicals, one of them the allyl cation:

The parent for the allyl cation is the neuter allyl radical, $$\ce{C3H5^.}$$. In section Gas phase ion energetics data, NIST lists multiple entries for the reaction leading to $$\ce{C3H5^{.+}}$$. According to the entry by Kagramanov et al. (1983), this requires $$(8.18 \pm \pu{0.07) eV}$$ by EI (ref). This value corresponds to about $$\pu{789 kJ/mol}$$.

The alternative would is to depart from the neuter ethyl radical, $$\ce{C2H5^.}$$, to yield the radical cation $$\ce{C2H5^{.+}}$$. Here, the most recent entry for EI NIST lists is the by Lossing and Semeluk (1970) with $$(8.38 \pm \pu{0.05) eV}$$ (ref). This value corresponds to about $$\pu{806 kJ/mol}$$, which is greater than the one for the allyl radical.

Energy conversions $$\pu{eV} \rightarrow \pu{kJ/mol}$$: unitsconverters.com

• This answer is misleading : the reaction drawn is a fragmentation of neutral propene, whereas in mass spectrometry one deals with fragmentation of ionized propene, leading to the allyl cation and the neutral radical C2H5°. Second: the IE of the ethyl radical is around 8.2 eV (thus lower than 12.45 quoted in the answer), and the IE of the allyl radical is only slightly lower (around 8.1 eV). Accordingly and following the Stevenson rule, the m/z 29 (C2H5+) is also present, and only slightly less intense then m/z 41.
– PLD
Nov 4 '21 at 15:26
• @PLD «it is more likely that the allyl cation is formed» does not exclude that there is $\ce{C2H5^+}$ is formed. May you put your argument in an answer (well, counting the carbon atoms in my equation [5], it departs from neuter pentene), including an equation about the mechanism and quote of the source of the origin of the numbers you use for the discussion? Nov 4 '21 at 15:58
• @Buttonwodd. I posted an answer to the original question, but it does not contain elements of my comment. The values are from the NIST database as yours. For correct application of the Stevenson rule, one should consider the ioniozation energy between the neutral counterpart and the ion, which means that for cations (and not radical cations as usually stated for Stevenson's rule) one needs to start from the radical (allyl or ethyl) and not from the entire molecule.
– PLD
Nov 4 '21 at 17:02
• @PLD Based on your input, I revised data used and scheme. Does NIST equally list the enthalpy of formation for e.g., the neuter ethyl radical $\ce{C2H5^.}$ since webbook.nist.gov/cgi/… reads like $\Delta{}_fH (\text{ion})$, too? These data could be used for a back on the envelope estimation of $\Delta_rH$ (or, $\Delta_rG$) for the overall reaction. Nov 5 '21 at 9:53
• You have the $\Delta_{f}H$ of the neutral under the "Gas-phase thermochemistry data" section (webbook.nist.gov/cgi/…). But this value is likely derived from the ionization energy measurements (detailed in the page you refer to) and the formation enthalpy of the ion, so basically we talk of the same data.
– PLD
Nov 5 '21 at 10:02