# Why mass spectra of 2-methylbutane do not show big peak at 71 coming from tertiary carbocation?

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.

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

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.
• 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. – logical x 2 Oct 13 '17 at 15:07