The final products of the reaction between methane and chlorine in the presence of ultraviolet light are hydrogen chloride and?

Analysis of the Question

This question tests your understanding of free-radical substitution reactions in alkanes. Specifically, it asks for the final product of the reaction between methane ($CH_4$) and chlorine ($Cl_2$) under ultraviolet (UV) light.

Concept Explanation

When an alkane like methane reacts with a halogen (like chlorine) in the presence of UV light, a series of substitution reactions occurs. The UV light provides the energy to break the chlorine molecule into highly reactive chlorine radicals. These radicals then replace the hydrogen atoms in the methane molecule one by one:

1. $CH_4 + Cl_2 \rightarrow CH_3Cl + HCl$ (Chloromethane)
2. $CH_3Cl + Cl_2 \rightarrow CH_2Cl_2 + HCl$ (Dichloromethane)
3. $CH_2Cl_2 + Cl_2 \rightarrow CHCl_3 + HCl$ (Trichloromethane)
4. $CHCl_3 + Cl_2 \rightarrow CCl_4 + HCl$ (Tetrachloromethane)

The reaction continues until all hydrogen atoms are replaced by chlorine atoms. The "final product" refers to the stage where no more hydrogen atoms remain to be substituted.

Step-by-Step Solution

1. Identify the reactants: Methane ($CH_4$) and Chlorine ($Cl_2$).
2. Understand the process: The reaction is a chain reaction. As long as there is an excess of chlorine, the substitution will proceed until all four hydrogen atoms in the methane molecule are replaced by chlorine atoms.
3. Determine the final state: After the fourth substitution, the molecule becomes $CCl_4$.
4. Name the product: $CCl_4$ is known as tetrachloromethane (also commonly called carbon tetrachloride).

Conclusion

Since the reaction proceeds until all hydrogen atoms are replaced, the final product is tetrachloromethane.

Exam Tip: Remember the sequence of substitution: Mono- (one Cl), Di- (two Cl), Tri- (three Cl), and Tetra- (four Cl). If the question asks for the "final product," always look for the option with the highest number of chlorine atoms (the "tetra" prefix).