Mass spectra of 2-methylbutane show strong peak at 57 by loss of methyl radical to give secondary carbocation while a loss of hydrogen radical from 2-position would form tertiary carbocation with m/z = 71 peak.

Since rules suggest that it is the stability of carbocation that determines the fragmentation pattern.So this process must be more favorable process still we do not get base peak at 71 instead we get a small peak at 71.

Same pattern is observed in case of 2-methylpentane where M-1 peak is almost absent. Other compounds also do not favor loss of hydrogen radicals to give M-1 peak.

I had read about this somewhere that it had something to do with energy of overall process but not able to recall. Kindly help me to explain the reason behind this phenomenon.

Mass Spectra of 2-methylbutane

Mass spectra of 2-methylpentane

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    $\begingroup$ Possible duplicate of Why is the molecular peak not observed in the spectrum of 2,2-dimethylbutane? $\endgroup$ Aug 19, 2017 at 19:35
  • $\begingroup$ So tertiary carbocation gets fragmented further..plz do explain it further. My apology for not able to understand from linked question. $\endgroup$
    – aks0854
    Aug 19, 2017 at 19:45
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    $\begingroup$ @MelanieShebel Not dupey enough, I think, nice from you though :) $\endgroup$
    – Mithoron
    Aug 19, 2017 at 21:18
  • $\begingroup$ Basically, you have 2 pathways - 1) losing H atom 2) losing CH3 radical. Both give you comparable carbocations. $\endgroup$
    – Mithoron
    Aug 20, 2017 at 0:17
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    $\begingroup$ I don't see how the linked thread answers the regioselectivity question raised here. $\endgroup$ Oct 12, 2017 at 14:01

1 Answer 1


How do you know that losing a CH3 radical gives you a secondary carbocation? Maybe it will perform a hydride shift and you land at the tertiary carbocation. Always take this into account, because you can shoot yourself in the foot if you assume carbocations to be too stable. In EI, they rearrange all the time (e.g. benzylium <-> tropylium). However, seeing that the ion with $m/z~57$ subsequently looses $\mathrm{\text{:}CH_2}$, it seems reasonable to assume that the secondary carbocation is formed here.

The important point is the following: You propose that instead of loosing the methylradical, yielding a secondary carbocation (or a tertiary, considering the argument from above), there should be a loss of a hydrogen radical. However, the hydrogen radical is far less stable than the methyl radical, and I assume this is the reason for the regioselectivity, here.

Also keep in mind that you supply a lot of energy (70 eV). Hence, the induced fragmentation processes are highly endothermic, and should be under kinetic control.

If you really want to find out what is going on, you should probably have a look at McLafferty's/Turecek's classic textbook "Interpretation of Mass Spectra".

  • $\begingroup$ Come think of it, both homolytic bond breaks are effectively a $sp^3 \mapsto sp^2$ transition. For the secondary carbocation, only two hydrogens have to move in the plane to yield the $sp^2$ center. For the tertiary carbocation, all three neighbouring carbon atoms have to move into the plane of the new $sp^2$ center. This makes it very plausible that the carbocation is kinetically favored. $\endgroup$ Oct 13, 2017 at 15:07
  • $\begingroup$ Thanks a lot. I think unstability of hydrogen radical is behind this apart from planarity.. $\endgroup$
    – aks0854
    Oct 21, 2017 at 12:43

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