What causes hydrogen abstraction in the radical chain mechanism of alkane halogenation?

Question

In the first propagation step of the radical chain mechanism of alkane halogenation, why does the halogen actually abstract hydrogen from the alkane? My organic chemistry textbook so far has went into details on MOs, hybridized orbitals, and general thermodynamics of reactions but has been lacking in the by and large why of reactions. Is the $\ce{C-H}$ bond broken by sufficient collisions from halogen free atoms and the hydrogen goes to the halogen as a result of $\ce{F}$/$\ce{Cl}$/$\ce{Br}$ being more electronegative than carbon, or what should cause this to happen?

Answer 1

why does the halogen actually abstract hydrogen from the alkane?

Look at the two tables of bond energies on this webpage. The second or bottom table shows the $\ce{C-H}$ bond energies for various carbon containing compounds. In the top table we see that the H-Cl bond strength is 103 kcal/m.

If we start with a chlorine radical, then if we break a C-H bond and make a Cl-H bond, the energetics of the reaction will be

$$\ce{\Delta H=bonds~ broken~-~bonds~made}$$ $$\ce{\Delta H=C-H ~bond~ broken~-~H-Cl ~bond~made}$$ $$\ce{\Delta H=C-H ~bond~ broken~-~103~ Kcal/m}$$

Look at the top row in the bottom table again. As long as our chlorine radical breaks any C-H bond that has a dissociation energy less than 103 kcal/m, our reaction will be exothermic, and the reaction will continue.

Edit

As I understand, the first propagation step in the free-radical monochlorination of methane is slightly endothermic but is driven to completion by the exothermic nature of the second propagation step

The first propagation step involves breaking a $\ce{C-H}$ bond (103 kcal/m) and making an $\ce{H-Cl}$ bond (103 kcal/m), so it is roughly thermoneutral according to the above link.

$$\ce{CH3-H + Cl{.} -> CH3{.} + H-Cl}$$

$$\ce{CH3{.} + Cl-Cl -> CH3-Cl + Cl{.}}$$

The second propagation step involves breaking a $\ce{Cl-Cl}$ bond (58 kcal/m) and making a $\ce{CH3-Cl}$ bond (81 kcal/m), so it is quite exothermic.

These steps are consistent with my earlier argument. You are correct that using free energies would be better (more accurate) than enthalpies, but the Tables I could find were based on enthalpies - I suspect they provide us with a reasonably accurate "back of the envelope" calculation and view of what is going on.

As to the actual mechanism of the reaction, once we've created a chlorine radical, we've created a reactive species. It is reactive because it doesn't have an octet of electrons. By undergoing the above-reactions it can regain its octet and become a stable entity.

Here is a diagram of a reaction coordinate (note we are using free energy as the Y-axis in this diagram). Generally the more exothermic (exergonic for free energy) a reaction, the lower the activation energy required to reach the transition state and pass over from reactants to products. So our reactive chlorine radical bounces around probing different parts of the reaction coordinate (e.g. different $\ce{C-H}$ bonds, different angles and approaches to these bonds) until it encounters a bond from the right direction, right angle, etc, such that the collision has enough energy to pass over the top of our reaction coordinate and become a product.

Answer 2

First mistake you do is speaking of "the radical chain reaction". There are many reactions with different mechanisms which are radical chains.

Re the cause: All reactions take place because there is some energy (more specific: delta G) gain. Any step in that radical chain has to fulfil this requirement.

Answer 3

The initiation step of all these radical halogenations is the homolytic cleavage of a halogen molecule: $\ce{Cl2 ->[{h \cdot \nu}] 2 Cl^{.}}$ These radicals rapidly move away from each other so immediate recombination is not an option. Also, the radical concentration is generally rather low, so it is unlikely for the radicals to find other possible recombination partners. That is what gets the chain started.

So the radical has the choice of capturing one of the other species present in the mixture which are the alkane $\ce{C_$n$H_{2$n$ +2}}$ or unreacted halogens $\ce{Cl2}$ (or previously reacted haloalkanes). In all cases, one bond would be cleaved and one formed, giving a new radical species. If the radical happens to meet $\ce{Cl2}$ then nothing apparantly happens, only if you colour one chlorine atom yellow, the second blue and the third red would you observe a change.

If the halogen radical meets an alkane and the Gibbs free energy change is favourable (refer to ron’s answer for numbers) it will react. It cannot directly bind to carbon, because that would induce a $\ce{H^{.}}$ radical which is much more unstable than anything else — $\Delta G$ is too high. Instead, it cleaves a $\ce{C-H}$ bond by abstracting hydrogen: $\ce{Cl^{.} + H-C_{$n$}H_{2$n$+1} -> Cl-H + ^{.}C_{$n$}H_{2$n$ + 1}}$. This new radical then goes looking for reaction partners and usually finds a $\ce{Cl2}$ molecule first according to what we write — but it could also find another alkane and transfer the radical from one alkane molecule to another.