See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Concept: Hydroboration-oxidation is an anti-Markovnikov, syn-addition of water across a double bond. The boron (electrophilic) adds to the less substituted carbon of the alkene, and the hydrogen adds to the more substituted carbon. Subsequent oxidation with H2O2/OH^- replaces the B with OH with retention of configuration. Step 1: Identify the substrate. Methylenecyclohexane has an exocyclic double bond: the cyclohexane ring bears a =CH2 group. The two carbons of the double bond are: (i) the exocyclic =CH2 (less substituted, terminal carbon) and (ii) the ring carbon bearing the =CH2 (more substituted, secondary carbon attached to two ring carbons). Step 2: Apply hydroboration regioselectivity. In hydroboration, BH3 adds boron to the less hindered (less substituted) carbon. Here, the terminal =CH2 carbon is less substituted, so boron attaches to -CH2- (the exocyclic carbon), and hydrogen goes to the ring carbon. Step 3: After oxidation (H2O2/OH^-), the C-B bond is converted to C-OH. Since boron was on the exocyclic CH2, the product is cyclohexyl-CH2OH, i.e., (cyclohexylmethanol), a cyclohexane ring with a -CH2OH group — anti-Markovnikov alcohol. Step 4: Evaluate options. - (a) cyclohexyl-CH2OH: This is the anti-Markovnikov product with -OH on the terminal (exocyclic) carbon. CORRECT. - (b) 1-methylcyclohexan-1-ol: This would be the Markovnikov product (acid-catalyzed hydration), not hydroboration-oxidation. - (c) cyclohex-2-en-1-one: This is an oxidation product unrelated to this reaction. - (d) cyclohexane-1-carboxylic acid: This requires oxidation beyond an alcohol, not the result of hydroboration-oxidation. Therefore, the correct answer is A.