See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Free radical halogenation of methane is a chain reaction that can proceed through multiple substitution steps. Once CH3Cl is formed, it can react further with Cl2 to give CH2Cl2, CHCl3, and CCl4. Reasoning: To maximize the yield of the monohalogenated product CH3Cl and minimize further chlorination, there must be a large excess of CH4 relative to Cl2. When CH4 is in large excess (high CH4:Cl2 ratio), the probability that Cl• radicals encounter CH4 molecules is much greater than the probability of encountering CH3Cl molecules. This statistical excess suppresses the secondary, tertiary, and quaternary chlorination steps. Conversely, if the ratio is low (more Cl2 relative to CH4), the formed CH3Cl is likely to undergo further chlorination, reducing selectivity for the monosubstituted product. Why other options fail: - (b) low: A low ratio means excess Cl2, which promotes polychlorination of CH3Cl to CH2Cl2, CHCl3, and CCl4, reducing CH3Cl yield. - (c) equal: A 1:1 ratio still allows significant polychlorination because as CH4 is consumed the relative proportion of CH3Cl increases, making over-chlorination likely. - (d) can't be predicted: The outcome is well understood from free radical mechanism principles; the ratio clearly affects selectivity. Therefore, the correct answer is A.