Why are radical intermediates are more stable on tertiary carbons?

Question

Is this because there's more electron density? More carbons around means the tertiary carbon "has more electrons around it"?

Answer 1

Before thinking about why tertiary radicals are 'stable', you need to have some kind of understanding about the structure of radicals.

Carbon centered radicals adopt similar geometries to carbocations: three coordinate and planar. Whereas carbocations have an empty p-orbital, radicals have a singularly occupied p-orbital. Overall the geometry is usually described as sp2 if we're thinking about hybridisation. (*)

With the structure of the radical now defined, we can think about what gives the tertiary radical such stability, and to do that we need to discuss what hyperconjugation is.

A good place to start (other than a textbook) might be the IUPAC definition from the Gold book:

...The concept of hyperconjugation is [...] applied to carbenium ions and radicals, where the interaction is now between σ-bonds and an unfilled or partially filled π- or p-orbitals...

Essentially, in the sp2 geometry, adjacent C-H bonds are able to interact with the vacant p-orbital, donating a small amount of electron density in order to stabilise the electron deficient p-orbital.

Individual C-H bonds to not exert a particularly strong stabilising effect, however, in a t-butyl cation/radical, there are 9 adjacent C-H bonds, each donating a very small amount of electron density, the sum of which amounts to a net overall stabilisation relative to primary radicals.

Whilst this explains the stability of the tert-butyl radical, there are other mechanisms of stabilisation for radicals which carbocations are not able to participate in. Due to the single electron in the p-orbital, radicals can also be stabilised by resonance, in a similar (but not identical) way to lone pairs or carbanions. A classic example is a radical generated next to a C=O group or C=C group.

Both of these effects discussed (hyper conjugation and resonance) are thermodynamic effects. The stabilisation is lowering the energy, making the radical less reactive.

(*) In the case of radicals, this is a bit misleading, theres a lot of spectroscopic and computational evidence to suggest that radicals deviate from planarity towards a pyramidal structure (the idea of hybridisation and sp2 doesn't hold up well)- especially when electron withdrawing groups such as fluorine are bonded to the radical centre. From the point of view of undergraduate chemistry this is largely irrelevant, however, given that the planar model explains the features of radical reactions that we're usually interest in, as organic chemists rather than theoreticians (EPR studies are one place in which the planar model falls down, the measured values don't correlate well to a completely flat carbon centre).

Answer 2

Radical intermediate's stability is same as that of carbocation's stability. It is stabilised by hyperconjugation and +I effect of alkyl groups.