Question

When light is shown on a mixture of chlorine and chloromethane, carbon tetrachloride is one of the components of the final reaction mixture. Propose a series of mechanistic steps which explain this observation?

Answer 1

A free-radical substitution reaction is likely to be responsible for the observations. The reaction mechanism of a reaction like this can be grouped into three phases:

• Initiation; the "light" on the mixture deliver sufficient amount of energy such that the halogen molecules undergo homologous fission. It typically takes ultraviolet radiation to initiate fissions of the bonds.  
• Propagation; free radicals react with molecules to produce new free radicals and molecules.
• Termination; two free radicals combine and form covalent bonds to produce stable molecules. Note that it is possible for two carbon-containing free-radicals to combine, leading to the production of trace amounts of long carbon chains in the product.

Initiation

$\text{Cl}-\text{Cl} \stackrel{\text{UV}}{\to} \text{Cl}\bullet + \bullet\text{Cl}$

where the big black dot indicates unpaired electrons attached to the atom.

Propagation

$\text{CH}_3\text{Cl}+ \text{Cl}\bullet \to \bullet\text{CH}_2\text{Cl} + \text{HCl}$

$\bullet\text{CH}_2\text{Cl} + \text{Cl}_2 \to \text{CH}_2\text{Cl}_2 + \text{Cl}\bullet$

$\text{CH}_2\text{Cl}_2 + \text{Cl}\bullet \to \bullet\text{CHCl}_2 + \text{HCl}$

$\bullet\text{CHCl}_2+ \text{Cl}_2 \to \text{CHCl}_3 + \bullet \text{Cl}$

$\text{CHCl}_3 + \text{Cl}\bullet \to \bullet\text{CCl}_3 + \text{HCl}$

$\bullet\text{CCl}_3 + \text{Cl}_2 \to \text{CCl}_4 + \text{Cl}\bullet$

Termination

$\text{Cl}\bullet + \bullet\text{Cl} \to \text{Cl}-\text{Cl}$