# Why is the chlorine radical used in halogenation reactions? What are the steps of reaction with organic molecules?

Feb 5, 2017

The chlorine radical is exquisitely reactive...........

#### Explanation:

Normally, chlorine is a diatomic molecule, $C {l}_{2}$. Under certain conditions, however, for instance under ultraviolet light, homolytic fission (i.e. breakage) of the $C l - C l$ bond can occur to give chlorine radical $\dot{C} l$. This is a 7 electron species.

And we can represent this process by the given chemical reaction:

$C {l}_{2} + h \nu \rightarrow 2 \dot{C} l$ $h \nu$ represents an ultraviolet light source.

Now $\dot{C} l$, is a high energy radical species, and will react with other chemical bonds. However, given that by reaction it generates ANOTHER radical species, the newly generated radical can continue the chain of chemical reaction. This is what they refer to as the propagation step in radical chain mechanisms: the actual bond making step generates another radical to continue the chain of reaction.

$\left(i\right)$ $C {l}_{2} + h \nu \rightarrow 2 \dot{C} l$ $\text{INITIATION}$

$\left(i i\right)$ $\dot{C} l + C {H}_{4} \rightarrow H - C l + \dot{C} {H}_{3}$ $\text{PROPAGATION}$

$\dot{C} {H}_{3} + C {l}_{2} \rightarrow C l - C {H}_{3} + \dot{C} l$

$\left(i i i\right)$ $\dot{C} l + \dot{C} {H}_{3} \rightarrow C l - C {H}_{3}$ $\text{TERMINATION}$