See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The starting material is an indene derivative where the five-membered ring is fused to a benzene ring. The five-membered ring contains a double bond (C2=C3 in the cyclopentene portion) and an OR group on C1 (the carbon directly attached to the benzene ring at the ring junction adjacent position). Step 2 - Understand hydroboration-oxidation: BH3 adds to the double bond in an anti-Markovnikov, syn fashion. The boron (electrophilic) adds to the less substituted carbon, and the hydrogen adds to the more substituted carbon. The subsequent H2O2/HO- oxidation replaces B with OH with retention of configuration. Step 3 - Apply the hint about positive charge stabilization: During hydroboration, BH3 acts as an electrophile approaching the double bond. The transition state has partial positive charge developing on one carbon. The carbon adjacent to the OR-bearing carbon (C1, which bears an oxygen substituent) can stabilize positive charge through lone pair donation from oxygen (oxocarbenium-like stabilization). This means the partial positive charge preferentially develops on C2 (the carbon closer to C1-OR), so boron (electrophilic) adds to C3 (the carbon farther from OR), and H adds to C2. Step 4 - Determine regiochemistry: Boron adds to C3 (exo carbon, away from OR), hydrogen adds to C2 (carbon adjacent to C1-OR). After oxidation, OH replaces B at C3... Wait, re-examining: The OR is at C1 (ring junction carbon adjacent to benzene). The double bond is C2=C3. The developing positive charge is stabilized at C2 by the adjacent C1-OR group (oxygen lone pair donation). Therefore, electrophilic boron adds to C2, and H adds to C3. After oxidation, OH is placed at C2 (adjacent to C1-OR). Step 5 - Product structure: The product has the indane skeleton with OR at C1 and OH at C2, both on the same face (syn addition). This corresponds to option (b): indane with OR at C1 (adjacent to benzene ring) and OH at C2 of the five-membered ring. Step 6 - Why other options fail: Option (a) would require OH at C3 (Markovnikov-like addition, wrong regiochemistry). Options (c) and (d) show incorrect regiochemistry or additional substituents not consistent with simple hydroboration-oxidation. Therefore, the correct answer is B.