Identify the product A and product B in the following set of reactions: CH3-CH=CH2 H2O, H+ Major product A CH3-CH=CH2…

Step 1: Acid-catalysed hydration (Markovnikov) → A

In acid-catalysed hydration of propene:

• $\ce{H+}$ adds to C1 (terminal, less substituted) forming the more stable secondary carbocation at C2.
• $\ce{H2O}$ attacks C2.
• A = propan-2-ol ($\ce{CH3CH(OH)CH3}$, isopropanol) — Markovnikov product ✓

Step 2: Hydroboration-oxidation (Anti-Markovnikov) → B

In hydroboration:

• $\ce{BH3}$ adds B to the terminal C1 (less substituted, less hindered).
• $\ce{H2O2/OH-}$ oxidizes: -B is replaced by -OH at C1.
• B = propan-1-ol ($\ce{CH3CH2CH2OH}$, n-propanol) — Anti-Markovnikov product ✓

Key conceptual insight:

These two methods convert the same alkene to two different alcohol isomers:

• H+/H2O → secondary alcohol (Markovnikov)
• BH3/H2O2 → primary alcohol (anti-Markovnikov)

Key Points to Remember:

• Markovnikov hydration: -OH to more substituted C.
• Hydroboration-oxidation: -OH to less substituted C (anti-Markovnikov, syn addition).
• Both reactions are additions across the double bond but follow opposite regioselectivity.