Question

Convert Methane into Chloroform.

Answer 1

In presence of diffused sunlight, Methane$\left(C{H}_4\right)$ undergoes substitution reaction with Chlorine$\left(C{l}_2\right)$ to give various products depending on the intensity of light as a chain reaction. The compounds formed are namely Methyl chloride$\left(C{H}_{3}Cl\right)$, Methylene chloride$\left(C{H}_{2}C{l}_2\right)$, Chloroform$\left(CHC{l}_3\right)$, and Carbon tetrachloride$\left(CC{l}_4\right)$. In every step, one or more hydrogen atoms get substituted.

Step-1: Methane reacts with chlorine to form Methyl chloride

$\underset{\left(methane\right)}{C{H}_4}+\underset{\left(chlorine\right)}{C{l}_2}\stackrel{sunlight}{\to }\underset{(Methylchloride)}{C{H}_{3}Cl}+\underset{(Hydrochloricacid)}{HCl}$

Step-2: The product formed (chloromethane) further reacts with chlorine producing dichloromethane.

$\underset{(Methylchloride)}{C{H}_{3}Cl}+\underset{\left(Chlorine\right)}{C{l}_2}\stackrel{sunlight}{\to }\underset{(Methylenechloride)}{C{H}_{2}C{l}_2}+\underset{(Hydrochloricacid)}{HCl}$

Step-3: This further produces trichloromethane or chloroform

$\underset{(Methylenechloride)}{C{H}_{2}C{l}_2}+\underset{\left(Chlorine\right)}{C{l}_2}\stackrel{sunlight}{\to }\underset{\left(Chloroform\right)}{CHC{l}_3}+\underset{(Hydrochloricacid)}{HCl}$

The substitution products so formed can be separated by fractional distillation due to their appreciable difference in boiling point. Hence Chloroform can be obtained easily by this method.