See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Hydroboration-oxidation with isotopic reagents. In this reaction, the alkene CH3-CH=CH2 undergoes hydroboration using BD3 (borane with deuterium) in THF, followed by protodeboronation (protonolysis) using CH3CO2T (tritiated acetic acid, where T = tritium, 3H). Step 1 - Hydroboration with BD3: BD3 adds across the double bond in a syn (cis) addition following anti-Markovnikov regioselectivity. The boron attaches to the less substituted carbon (terminal carbon, C1 = CH2 end) and deuterium (D) attaches to the more substituted carbon (C2 = CH carbon). This gives: CH3 — CHD — CH2 — B (organoborane intermediate), where D is on C2 and B is on C1. Step 2 - Protonolysis with CH3CO2T (tritiated acetic acid): The C-B bond is cleaved by the tritiated acid. The boron is replaced by tritium (T) from CH3CO2T. This places T on C1 (the terminal carbon where B was attached). Result: CH3 — CHD — CH2T - C3 (methyl): CH3 (unchanged) - C2: CHD (got D from BD3 hydroboration) - C1: CH2T (got T from protonolysis with CH3CO2T) Why other options fail: - (a) CH3-CHD-CH2D: This would result if both steps used deuterium, but the second step uses tritiated acid (CH3CO2T), so T not D is incorporated at C1. - (b) CH3-CHT-CH2T: This would result if both steps used tritium, but the first step uses BD3 (deuterium), so D not T is incorporated at C2. - (d) CH3-CHT-CH2D: This would require T on C2 and D on C1, which is the reverse of what happens; BD3 delivers D to C2 and CH3CO2T delivers T to C1. Therefore, the correct answer is C.