[LeighVic]

I'm trying to understand why exactly a tertiary radical is considered more stable than either of secondary or primary.

In my textbook, it states that "alkyl groups are electron donating and hence, donate electron density into the orbital containg the unpaired electron. This trend is then supported by a comparison of bond dissociation energies (BDE)".

I don't understand that statement quite well, so would someone be able to explain it to me? As well as this, I don't really understand what a "bond dissociation energy is".

Thanks

[phth]

What type of a function is an orbital?  BDE is the energy required to break a bond i.e. R-R' R·+R'· OR R-R' R:+R'⁺; one electron stays or two stay.

[Babcock_Hall]

@OP, Is the atom upon which the radical is centered electron-rich or electron-poor?  With respect to bond dissociation energies, I have always used that term to refer only to homiletic cleavages.

[LeighVic]

Thanks for the replies.

@Babcock_Ha.

I was copying that quote word for word, so the example organic compounds were all hypotheticals. (Example. R-X-R)

In saying that, here's an example of a picture I'm referring to.

http://jamesash.wpengine.netdna-cdn.com/wp-content/uploads/2013/08/2-bdes-1-2-3-4.png

So, I assumed that if a compound was more stable, it would require more energy to break bonds. But in the picture above, this isn't the case, as BDE is lowest for most stable radicals.

Thanks.

[orgopete]

Let me pose a hypothetical instance. If the radicals were derived from hydrocarbons of equal stability, which radicals would form most easily?

This is a different question from comparing the bond energy of a RO-OR and a RCH2-CH2R bond. In this case the O-O bond would break more easily as it is the most reactive.

[Babcock_Hall]

In saying that, here's an example of a picture I'm referring to.

http://jamesash.wpengine.netdna-cdn.com/wp-content/uploads/2013/08/2-bdes-1-2-3-4.png

So, I assumed that if a compound was more stable, it would require more energy to break bonds. But in the picture above, this isn't the case, as BDE is lowest for most stable radicals.

Thanks.

Yet we are not asking which compound is most stable, we are asking which radical is most stable (I think your reasoning might work for reactants, but not for products).  The bond dissociation energy is greatest for the radical which is least stable.  Let's put this idea to a test.  Do you agree that the benzyl radical, Ph-CH₂· is relatively stable.  If so, what do you predict (qualitatively) about the BDE for toluene. Ph-CH₃?