See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Free radical halogenation is a chain reaction. When Cl2 reacts with an alkane (A) under hv (ultraviolet light), it proceeds via free radical substitution. The monochloro product formed can itself react further with Cl2 to give dichloro, trichloro, and polychloro products, reducing the yield of the monochloro product. Reasoning: To maximize the yield of the monochloro product, we need to minimize the probability that the monochloro product reacts again with Cl2. If reactant (A) is present in large excess relative to Cl2, then statistically Cl2 is far more likely to react with unreacted (A) molecules than with the monochloro product. This suppresses further chlorination and drives selectivity toward the monochloro product. Why other options fail: (a) Adding Cl2 in excess would increase the likelihood of the monochloro product reacting again with excess Cl2, leading to polychlorination and reducing monochloro yield. (c) Carrying out the reaction in the dark would prevent the free radical initiation step (since hv is required), so the reaction would not proceed at all under purely dark conditions. (d) Using an equimolar mixture does not sufficiently suppress further chlorination; excess (A) is needed to ensure Cl2 preferentially attacks unreacted substrate. Therefore, the correct answer is B.