Question

Chlorination of methane does not occur in dark because:
A. methane can form free radicals in presence of sunlight only
B. to get chlorine free radicals from $Cl_2$ molecules energy is required. it cannot happen in the dark.
C. substitution reactions can take place only in sunlight and not in dark.
D. termination step cannot take place in dark it requires sunlight.

Answer 1

Hint: Chlorination of methane is a free radical substitution reaction. Chlorine is not able to convert into free radicals in the dark, so the reaction doesn’t happen. Hence, presence of sunlight is must for the reaction to occur.

Complete step-by-step answer:
Methane $\left( {C{H_4}} \right)$ doesn’t react with chlorine in dark because chlorine atoms are pertained in the presence of light.
Let us understand this in brief with the help of terms and reactions.
Free Radical: Molecule that contains at least one unpaired electron.
Example - Hydroxyl radical (HO.)
Free Radical Substitution reaction -
$ \Rightarrow \,\,C{H_4}\, + \,C{l_2} \to \,C{H_2}Cl\, + \,HCl$

Initiation -
- Homolytic fission bond breaks by radical form
- Covalent bond splits and after that each atom contains one unpaired electron.
-UV needed
   $ \Rightarrow \,\,C{l_2}\xrightarrow{{h\nu }}2Cl \bullet $
Hence, the answer is (C) - Substitution reaction can take place only in sunlight and not in dark.
Now let us study some other aspects.

Propagation -
-Radical react with molecule
$Cl \bullet + \,C{H_4} \to \,HCl\, + \,C{H_3}$
$C{H_3} \bullet \, + \,C{l_2} \to \,C{H_3}Cl\, + Cl \bullet $

Termination -
-Radical + radical $ \to $ molecule
$Cl \bullet \, + Cl \bullet \to C{l_2}$
$C{H_3} \bullet \, + C{H_3} \bullet \to C{H_3} - C{H_3}$
$Cl \bullet + \,C{H_3} \bullet \to C{H_3}Cl$

Note: High intensity, broad spectrum UV system decreases both free chlorine and combined chlorine compounds into easily removed byproduct. Wavelengths of 180 and 400 nm, UV light produces photochemical reactions that dissociate free chlorine to make hydrochloric acid.