See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Hydroboration-oxidation (1. B2H6; 2. H2O2, HO-) is an anti-Markovnikov, syn-addition of water across a double bond. The boron adds to the less hindered (less substituted) carbon, and after oxidation with H2O2/OH-, the OH ends up on the less substituted carbon. Step 1: Identify the starting material. The alkene shown is 2-methylbut-2-ene: (CH3)2C=CHCH3. It has two carbons in the double bond: C2 (bearing two methyl groups, more substituted) and C3 (bearing one methyl and one H, less substituted). Step 2: Apply hydroboration regioselectivity. Boron (electrophilic, bulky) adds to the less substituted carbon (C3), placing H on C2. After oxidation, OH replaces B at C3. Step 3: Determine the product. OH ends up on C3 of the original alkene: (CH3)2CH-CH(OH)-CH3, which is 3-methylbutan-2-ol. This matches option (b): a saturated alcohol with OH on the internal less-substituted carbon. Step 4: Eliminate other options. - Option (a) retains the double bond and places OH on the more substituted carbon - this would be a Markovnikov-type or allylic product, not from hydroboration-oxidation. - Option (c) shows an epoxide, which is not the product of hydroboration-oxidation. - Option (d) shows OH on the more substituted carbon (Markovnikov product), which would result from acid-catalyzed hydration, not hydroboration-oxidation. Therefore, the correct answer is B.