I was wondering the same thing… my guess is this is what James was referring to in the notes of a prior post, “Bond Strength and Radical Stability” where he writes:

“the general trends in this post are valid because we discuss bonds to H, but use caution when comparing any other type of bond other than hydrogen…. Using the bond strengths (BDE’s) of unstrained bonds to hydrogen is a reasonable method for discerning trends in radical stabilities, as discussed in this post. However, BDE’s in and of themselves are not reliable for discerning absolute radical stabilities in cases where the bond may be weakened by strain, repulsion between lone pairs, or other factors. For example the BDE for hydrogen peroxide is 51 kcal/mol, which does NOT imply that the HO• radical is stable, but rather that the O–O bond is destabilized by repulsion between the lone pairs.”

Now when I look at the BDE table in Wade’s Organic Chemistry the only cases I find where tertiary BDE are significantly lower than primary BDE was in breaking of C-H or C-C bonds. There was little BDE difference between tertiary and primary halides (C-F, C-Cl, C-Br, C-I) or C-O bonds.

Since breaking of the C-X halide bond still produces a C* radical (and the halide is now gone), that C* radical ITSELF must be more stable if it is tertiary rather than primary. To me, that suggests the only way BDE can be equal (for tertiary and primary C-X halides) is if the tertiary C-X halide bond is more stable than a primary C-X halide bond. But why would that be?

My guess (only a guess): the CH3- groups are electron donating and the halide is electron withdrawing, so perhaps that stabilizes the C-X halide bond in tertiary carbons (halides like electrons!) and make the halide more loathe to leave as a radical (where it only takes a single electron from the bond) … but I’m not a chem-major… so it would be good if James weighed in…

So what says you James?!?

Thanks!

Great quote, and very relevant to this blog post. Was this a question?

Iodine is not reactive at all in free-radical halogenation of alkanes. The H-I bond strength is too weak.

March, Advanced Organic Chemistry, 5th ed. is my standard reference.

Thanks for the correction on cal vs kcal. Fixed!

The first propagation step is the key step because it results in formation of a radical at the alkane – this is the slow step (breakage of C-H). The second step is fast because a considerably weaker bond is being broken (Cl-Cl) .