See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: In hydroboration, BH3 reacts with alkenes by sequentially adding across double bonds. Each BH3 molecule has three B-H bonds, meaning it can react with three molecules of alkene (one alkene per B-H bond), forming a trialkylborane (R3B). Reasoning: - BH3 + alkene → R-BH2 (first addition, one B-H bond used) - R-BH2 + alkene → R2-BH (second addition, second B-H bond used) - R2-BH + alkene → R3-B (third addition, third B-H bond used) - Therefore, 1 mole of BH3 reacts with 3 moles of alkene. - To react with 2 moles of 1-pentene: moles of BH3 = 2/3 mole. Why other options fail: - (a) 2 mole: This would be the answer if BH3 reacted 1:1 with alkene, ignoring its trifunctional nature. - (b) 3 mole: This inverts the ratio (3 moles BH3 per 1 mole alkene), which is incorrect. - (d) 3/2 mole: This assumes BH3 reacts with only 2 alkenes per molecule (dialkylborane scenario), which is not the standard hydroboration stoichiometry. The correct stoichiometry is 1 mole BH3 per 3 moles alkene, so for 2 moles of 1-pentene, 2/3 mole of BH3 is required. Therefore, the correct answer is C.