# Why is chlorine and not fluorine labelled as the most reactive halogen for halogenation?

The reactivity of halogenation in sunlight is as follows:

$$\ce{F2}>\ce{Cl2}>\ce{Br2}>\ce{I2}$$

So why does we often label chlorine as most reactive halogen towards halogenation reaction?

• Because $\ce{F2}$ is so reactive and the $\ce{C-F}$ bond is so stable, it's not considered useful in most synthetic pathways. Jan 15 '17 at 8:54
• It’s not fully clear what you’re actually asking. The halogenation is, indeed, most reactive with fluorine. However, the halogen radical (which is what sunlight does) is most easily generated with iodine. I suspect you mean the former but could you clarify?
– Jan
Jan 18 '17 at 0:43
• Why did you decide to roll back ron’s edits which were improving your post’s grammar and title?
– Jan
Jan 19 '17 at 12:49

The reactivity of halogens in sunlight

OK, this tells us that we are dealing with a free radical reaction. Sunlight breaks the halogen bond ($\ce{X-X}$) to generate halogen free radicals.

$$\ce{X2 ->C[{h\nu}]\ 2X.}$$

In the next step the halogen radical abstracts a hydrogen atom from any hydrocarbon present to generate $\ce{HX}$ and a hydrocarbon free radical (note that multiple hydrocarbon radicals may be generated depending on the structure of our hydrocarbon).

$$\ce{X. + R-H -> HX + R.}$$

Finally, the hydrocarbon radical can react with the halogen free radical to produce a halogenated product.

$$\ce{R. + X. -> R-X}$$

Let's now consider the energetics of these steps. The first step, generation of the halogen free radical, uses the sunlight to break the $\ce{X-X}$ bond, so no additional energy is required for this step. Then we break a $\ce{C-H}$ bond (so we must add energy) and make an $\ce{H-X}$ bond and a $\ce{C-X}$ bond (so energy is given off). The various bond strengths are contained in the following Table.

\begin{array}{|c|c|c|c|c|c|} \hline \ce{C-H}~ \text{Bond Strength} & \ce{H-X} & \text{Bond Strength}& \ce{C-X} & \text{Bond Strength} & \text{Overall} \\ \text{(kcal/mol)}\ & & \text{(kcal/mol)} & & \text{(kcal/mol)} & \text{(kcal/mol)} \\ \hline \ 99 & \ce{H-F} & -135 & \ce{C-F} & -116 & -152\\ \hline \ 99 & \ce{H-Cl} & -103 & \ce{C-Cl} & -81 & -85\\ \hline \ 99 & \ce{H-Br} & -88 & \ce{C-Br} & -68 & -57\\ \hline \ 99 & \ce{H-I} & -71 & \ce{C-I} & -51 & -23\\ \hline \end{array}

We can clearly see that all of these free radical reactions will occur (e.g. they are exothermic) in the order you suggested, with fluorine being, by far, the most reactive.

$$\ce{F2 >> Cl2 > Br2 > I2}$$

Saying that chlorine is the most reactive is incorrect.

Note: 99 kcal/mol was used as a typical $\ce{C-H}$ bond strength. We could refine our model by using more exact $\ce{C-H}$ bond strengths (see this Wikipedia table) and calculate the relative rates of halogenation at the various carbon atoms in a complex hydrocarbon.

Reaction with fluorine are strongly exothermic and mostly during reaction explosion takes place and reaction generally not used for synthetic purpose.