See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Hydroboration-oxidation (here protodeboronation via carboxylic acid) proceeds via syn addition. In hydroboration, boron adds to the less substituted carbon (anti-Markovnikov) and the hydrogen (or deuterium/tritium from the borane) adds to the more substituted carbon, both from the same face (syn). In the protonolysis step, the C-B bond is replaced by C-H (or C-D or C-T from the acid) with retention of configuration. For 1-methylcyclohex-1-ene: - The double bond is between C1 (bearing CH3) and C2. - Hydroboration: B adds to C2, the isotopic hydrogen (from BX3) adds to C1, syn addition. - Protonolysis: B at C2 is replaced by the isotope from the acid (retention at C2). So C1 ends up with CH3 + (H/D/T from borane), and C2 ends up with (H/D/T from acid). Reaction (a): BD3:THF then CH3CO2T - Step 1 (BD3): D from borane goes to C1, B goes to C2, syn addition. - Step 2 (CH3CO2T): B at C2 replaced by T (from tritiated acetic acid). - Result: C1 has CH3 and D; C2 has T → matches product (p): CH3 and D at C1, T at C2. - So a → p. Reaction (b): BT3:THF then CH3CO2D - Step 1 (BT3): T from borane goes to C1, B goes to C2, syn addition. - Step 2 (CH3CO2D): B at C2 replaced by D. - Result: C1 has CH3 and T; C2 has D → matches product (s): CH3 and T at C1, D at C2. - So b → s. Reaction (c): BD3:THF then CH3CO2H - Step 1 (BD3): D from borane goes to C1, B goes to C2, syn addition. - Step 2 (CH3CO2H): B at C2 replaced by H (protio acetic acid). - Result: C1 has CH3 and D; C2 has H → matches product (q): CH3 and D at C1, H at C2. - So c → q. Reaction (d): BH3:THF then CH3CO2D - Step 1 (BH3): H from borane goes to C1, B goes to C2, syn addition. - Step 2 (CH3CO2D): B at C2 replaced by D. - Result: C1 has CH3 and H; C2 has D → matches product (r): CH3 and H at C1, D at C2. - So d → r. Why other options fail: Each product differs only in which isotope (H, D, or T) is placed at C1 vs C2. The borane isotope always ends up at C1 (more substituted, Markovnikov carbon receives H from borane in anti-Markovnikov hydroboration — actually boron to less substituted C2, hydride/isotope to C1), and the acid isotope always ends up at C2. Mixing up these assignments would give the wrong product, ruling out alternative matches. Therefore, the correct answer is {"a": "p", "b": "s", "c": "q", "d": "r"}.