JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Hydrocarbons

The alkene in this sequence is 1-methylcyclohex-1-ene, and the two reagent sets add water across the same double bond with opposite regiochemistry. Hydroboration–oxidation (${\text{B}}_2{\text{H}}_6$ then ${\text{H}}_2{\text{O}}_2/\text{NaOH}$) proceeds through a concerted four-centred transition state in which boron bonds to the less hindered, less substituted alkene carbon with syn addition of H and B (anti-Markovnikov placement); oxidative workup replaces C–B with C–OH with retention of configuration, giving trans-2-methylcyclohexan-1-ol. Acid-catalysed hydration (${\text{H}}^{+}/{\text{H}}_2\text{O}$) instead proceeds through the more stable tertiary carbocation formed on protonating the less substituted alkene carbon, so water is captured at the methyl-bearing carbon — Markovnikov addition — giving the tertiary alcohol 1-methylcyclohexan-1-ol. A is the Markovnikov product, 1-methylcyclohexan-1-ol, and B is the anti-Markovnikov hydroboration–oxidation product, trans-2-methylcyclohexan-1-ol.