[Woopy]

Hello,

I am having trouble figuring out why the relationship between C-H bond ease of abstraction and radical stability are opposites.

As in, a tertiary hydrogen is extracted easier than a primary hydrogen for a radical reaction. Is this just because the tertiary hydrogen has a more stable radical, and to get that more stability it needs less energy to break, since the less energy there is involved in the bond breakage, the more stable?

[discodermolide]

Surely most C-H bond energies are roughly the same? The radical that comes along doesn't stop a minute and make choices as to which H to abstract and the molecule doesn't say "here take my tertiary hydrogen because it gives me a more stable intermediate".
So there is something else going on here. What about steric factors do they play a role?

[Woopy]

In radical reactions, I don't think they play a role. I think electronic factors weigh in more than steric factors, which is why bromine (large molecule) would choose a tertiary hydrogen over a primary, right?

[discodermolide]

As I said bromine radicals do not know if a H is primary or tertiary. They will extract the most accessible H, primary. This primary radical may then re-arrange to a "more stable" position.

[Woopy]

bromine radicals are more selective, how can it rearrange to a more stable carbon if there is nothing analogous to hydride shifts for radicals?

[discodermolide]

Of course radicals can rearrange, just like carbocations.
A primary radical, the least stable can abstract a H radical from a neighbouring group. If that group is t-Butyl then you will get a tertiary radical.

[Woopy]

Ok then why do I have a question in my textbook asking ''Which of these are the easiest to abstract from the C-H bond?'' and the difference between all the hydrogens is their degrees

[discodermolide]

No idea, I don't have your textbook.
What does it mean by degrees?

Have a look here: http://www.chem.ucla.edu/harding/notes/notes_14D_radicals02.pdf

[orgopete]

I am having trouble figuring out why the relationship between C-H bond ease of abstraction and radical stability are opposites.

As in, a tertiary hydrogen is extracted easier than a primary hydrogen for a radical reaction. Is this just because the tertiary hydrogen has a more stable radical, and to get that more stability it needs less energy to break, since the less energy there is involved in the bond breakage, the more stable?

Ah, the why question. That is what makes the question harder.

First, I don't agree with Disco to this point. Second, easy to break, hence more stable is beguiling, but I think you have to be careful here. Flourine-flourine bonds are very easy to break, but that does not mean flourine radicals are exceptionally stable. They are very reactive. Iodine radicals are more stable and hence less reactive. (When I suggest this point, I seem to be in a very distinct minority. I have made a similar argument vis a vis an acid and it's resonance stabilized anion and have picked up many flags for this argument.)

Let us analyze the reaction, and try to draw some inferences as to how it may be taking place, or how radical stability may affect radical formation. We know halogen radical stability falls F>>Cl>Br>>I. Hence flourine reacts with any C-H bond and iodine virtually not at all. I speculate that a complex forms between the various C-H bonds and the radical. Decomposition of this complex should give a more stable or most stable radical. If this were correct, then e could argue the closer a carbon radical is to a halogen radical, the more selective the formation may become. For example, of a flourine radical is much less stable than any possible carbon radical, the complex should always decompose to form a carbon radical. Similarly, if in iodine radical is more stable, they should always reform iodine radicals. If bromine radicals are very similar to carbon radical stability, then formation of the more stable carbon radical should occur more often and result in the formation of the more stable radical.

I further think that reactivity and stability can play a role in the equilibrium of the complex. If a tertiary C-H bond is more reactive than a primary bond, it can form more complexes and hence more product.

This is all simple speculation. You need not give me a brace of flags for this. I am not saying it is true. I am saying you could use this to rationalize the result.

[Woopy]

Actually, I need to reword my question

So I was looking at a molecule 2-methylbutane. It says to rank the C-H bonds in order of increasing bond strength.

Then it says rank the C-H bonds in order of increasing ease of H abstraction in a radical halogenation reaction.

The answers are opposite of each other, why?

answer for part A is b < c< a
for part D a < c< b