See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Hydroboration-oxidation (BH3/THF followed by H2O2/OH-) is an anti-Markovnikov, syn addition of water across a double bond. The boron adds to the less substituted carbon and the overall process proceeds with syn stereochemistry. Step 1 - Identify the alkene: The starting material is 1-methylcyclohex-1-ene. The double bond is between C1 (bearing the methyl group) and C2. Step 2 - Regioselectivity: BH3 adds boron to the less hindered, less substituted carbon. In 1-methylcyclohex-1-ene, C1 bears a methyl substituent and is more substituted, while C2 is less substituted. Therefore, boron adds to C2 and hydrogen adds to C1 (anti-Markovnikov). After oxidation, the OH ends up at C2 (the less substituted carbon). Step 3 - Stereochemistry: Hydroboration occurs via a concerted syn addition (BH adds to the same face of the double bond). The oxidation step (H2O2/OH-) proceeds with retention of configuration at the carbon bearing boron. Thus the overall addition of H and OH is syn. Step 4 - Product structure: The OH is installed at C2 of the cyclohexane ring (anti-Markovnikov), and the H at C1. The methyl group remains at C1. The syn addition means the H (at C1) and OH (at C2) are on the same face of the ring. Step 5 - Stereochemical outcome at C1 and C2: In option (d), the product is drawn as trans-1-methyl-2-hydroxycyclohexane with the methyl and OH on opposite faces (the H atoms shown on wedges indicate specific relative configuration consistent with syn addition from one face). Option (d) shows CH3 on a dash and H on a wedge at one carbon, and OH on a dash and H on a wedge at the adjacent carbon, representing the correct syn addition product with the proper relative stereochemistry. Why other options fail: - Option (a): Shows the wrong relative stereochemistry (anti relationship between substituents, inconsistent with syn addition). - Option (b): Shows an enol/allylic alcohol; hydroboration-oxidation does not give an enol product and does not leave the double bond intact. - Option (c): Shows a tertiary alcohol at C1 (Markovnikov product), which would be the result of acid-catalyzed hydration, not hydroboration-oxidation. Therefore, the correct answer is D.