[theanonymous]

I have in my notes: The larger the energy cost (to break and form a bond), the larger the bond, the least stable.

and then I drew what my teacher wrote down on the board:

[theanonymous]

Ok, I Googled it and this is what I found:




The general stability order of simple alkyl radicals is:
(most stable) 3° C (tertiary alkyl) > 2° C  (secondary alkyl) > 1° C (primary alkyl) > methyl (least stable)

But why???
*delete me*

[theanonymous]

Someone else had a similar question like this in 2010 apparently (on this site) - they said their book said this:

"The more highly substituted the carbon atom, the less energy is required to form the free radical. We conclude that free radicals are more stable if they are more highly substituted."

Would that be the answer to my question (Why the radical stabilities is in that order)?

[discodermolide]

Have a look at hyper conjugation effects.
http://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/12/oc/vlu_organik/radikale/radikale_einfuehrung.vlu/Page/vsc/en/ch/12/oc/radikale/stabilitaet_radikale/stabilitaet_radikale.vscml.html

[theanonymous]

I get this: The more alkyl substituents a radical carbon atom possesses, the more stabilized it becomes from hyperconjugation.

But I don't get the second paragraph: The interaction of the double-occupied C-H σ bonding orbital with the single-occupied, non-bonding p orbital of the radical carbon atom is comparable to the stabilization by hyperconjugation in carbenium ions. However, they differ greatly in one important factor. The stabilization of carbenium ions, for example, is the result of the overlapping of a double-occupied C-H bonding orbital with an unoccupied, non-bonding 2p orbital. In radicals, on the other hand, this stabilization is obtained by the overlapping of a C-H bonding orbital with a single-occupied, non-bonding 2p orbital.

??

[theanonymous]

Sooo...
I suppose I have to  explain/show why the degree of hyperconjugation increases...

Oh boy.

[theanonymous]

Anyone there?

[Babcock_Hall]

But I don't get the second paragraph: The interaction of the double-occupied C-H σ bonding orbital with the single-occupied, non-bonding p orbital of the radical carbon atom is comparable to the stabilization by hyperconjugation in carbenium ions. However, they differ greatly in one important factor. The stabilization of carbenium ions, for example, is the result of the overlapping of a double-occupied C-H bonding orbital with an unoccupied, non-bonding 2p orbital. In radicals, on the other hand, this stabilization is obtained by the overlapping of a C-H bonding orbital with a single-occupied, non-bonding 2p orbital.

In a carbenium ion, the carbon with the positive charge has only six electrons.  Therefore the 2p orbital has no electrons.  In a radical, the carbon has seven electrons.  The seventh electron is in the 2p orbital.