Question

Which of the following is obtained as major product when excess of methane is treated with limited chlorine in presence of UV light?

• A. Chloromethane
• B. Dichloromethane
• C. Trichloromethane
• D. Tetrachloromethane

Hint:

Excess Alkane $\rightarrow$ Mono-substituted product; Excess Halogen $\rightarrow$ Poly-substituted product.

Answer 1

The Correct Option is A

Step 1: Concept
Halogenation of alkanes is a free-radical substitution reaction.

Step 2: Analysis
- If chlorine is in excess, the reaction proceeds until all H-atoms are replaced ($CCl_4$). - If methane ($CH_4$) is in excess, the probability of chlorine radicals colliding with $CH_4$ is much higher than with $CH_3Cl$.

Step 3: Conclusion
Using excess methane limits poly-substitution and makes Chloromethane ($CH_3Cl$) the major product.
Final Answer: (A)