Why is the reaction of alkanes with fluorine difficult to control?

Question

Madison Anderson · Accepted Answer

The reaction of alkanes with fluorine is difficult to control because the activation energy for hydrogen abstraction is so low.

The initiation step involves the homolytic cleavage of the F-F bond.

F-F → F· + ·F; #E_a# = + 155 kJ/mol

This is a high activation energy. But only a few F atoms need to form.

Each F atom can then initiate thousands of propagation steps.

CH₄ + ·F → CH₃F + H·

These steps are highly exothermic: ΔH° = -431 kJ/mol.

Hammond's postulate states that the structure of a transition state resembles that of the species nearest to it in energy.

In Case (a), the reaction is exothermic. The energy of the transition state is closer to that of the reactant than of the product. Therefore the transition state resembles the reactant.

In Case (c), the reaction is endothermic. The transition state resembles the product.

So the highly exothermic fluorination of an alkane has a very low activation energy.

Many molecules have enough energy to get over the barrier.

When they do so, they release large amounts of energy.

This raises the temperature, and the reaction goes even faster.

The chain reaction gets out of control and the mixture explodes.

The reactions with chlorine and bromine are less exothermic. They have higher activation energies, so they are easily controlled.

Jacob Anderson · Answer

undefined The reaction of alkanes with fluorine is difficult to control because fluorine is highly reactive and can undergo explosive reactions with alkanes even at low temperatures and pressures. Additionally, the high reactivity of fluorine makes it difficult to control the extent of the reaction and can lead to undesired side reactions or formation of multiple products.