Question

When methane is mixed with excess chlorine and the mixture is exposed to sunlight, the final product is:

• A. $C{H}_{3}Cl$
• B. $C{H}_{2}C{l}_2$
• C. $CHC{l}_3$
• D. $CC{l}_4$

Answer 1

The correct option is D $CC{l}_4$

When a mixture of methane and chlorine is exposed to ultraviolet light - typically sunlight - a substitution reaction occurs and the organic product is chloromethane.
$C{H}_4+C{l}_2\to C{H}_{3}Cl+HCl$
However, when excess chlorine is present, the reaction doesn't stop there, and all the hydrogens in the methane can in turn be replaced by chlorine atoms. That means that you could get any of chloromethane, dichloromethane, trichloromethane or tetrachloromethane.
$C{H}_4+C{l}_2\to C{H}_{3}Cl+HCl$
$C{H}_{3}Cl+C{l}_2\to C{H}_{2}C{l}_2+HCl$
$C{H}_{2}C{l}_2+C{l}_2\to CHC{l}_3+HCl$
$CHC{l}_3+C{l}_2\to CC{l}_4+HCl$