I think tertiary carbocation must be more stable due to Hyperconjugation and +I effect....

  • 1
    $\begingroup$ That wouldn't be homolysis, to begin with. $\endgroup$ Commented Apr 21, 2020 at 6:55
  • $\begingroup$ The comparison is not useful. Carbocations and radicals are produced in different ways, react with different types of partners, and cannot interconvert. $\endgroup$
    – Karl
    Commented Apr 21, 2020 at 8:15
  • $\begingroup$ It's about homolysis, not heterolysis. If you start wit a neutral molecule, product has to be radicals. $\endgroup$ Commented Apr 21, 2020 at 9:13

1 Answer 1


The magnitude of the effects needs to be considered. To form a radical from isobutane, you break a C-H bond. To form t-butyl radical requires about 93 kcal/mol (390 kJ/mol) https://en.wikipedia.org/wiki/Carbon–hydrogen_bond. To form a cation, you break the C-H bond AND ionize the radical, which costs you another 6.7 +/- eV (154 kcal/mol or 646 kJ/mol http://www.colby.edu/chemistry/PChem/Hartree.html) https://webbook.nist.gov/cgi/cbook.cgi?ID=C1605738&Mask=20

So, while the t-butyl carbocation is indeed stabilized by the indictive effect, and hyperconjugation, the energy level at which this is occurring is much higher than the energy level of the radical.

Hyperconjugation is a qualitative concept, with a possible energy of about 5 kcal/mol. Like a fudge factor, which explains all possible effects if you need it and apply it correctly, but in this comparison, its effect is negligible compared to bond-breaking and ionization. https://en.wikipedia.org/wiki/Hyperconjugation


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