Product A and B formed in the following set of reactions are:

Step 1: Identify the starting material

Methylenecyclohexane = a cyclohexane ring with an exocyclic $\ce{=CH2}$ double bond.

Step 2: Reaction with H+/H2O (acid-catalysed hydration = Markovnikov)

• $\ce{H+}$ protonates the exocyclic $\ce{=CH2}$ (less substituted end) → generates a tertiary carbocation on the ring carbon.
• Water attacks the tertiary carbocation → 1-methylcyclohexanol (tertiary alcohol).
• A = 1-methylcyclohexanol ✓

Step 3: Reaction with BH3, then H2O2/NaOH (hydroboration-oxidation = anti-Markovnikov)

• BH3 adds to the less hindered terminal $\ce{=CH2}$ carbon (boron goes to the less substituted end per steric control).
• After oxidation ($\ce{H2O2/NaOH}$), -OH replaces the boron at the terminal carbon.
• Product: cyclohexylmethanol (1-(hydroxymethyl)cyclohexane), a primary alcohol.
• B = cyclohexanemethanol (cyclohexylmethanol) ✓

Step 4: Key contrasting principle

Both reactions work on the same substrate but give completely different products:

• H+/H2O = Markovnikov → OH on more substituted C → tertiary alcohol.
• BH3/H2O2 = anti-Markovnikov → OH on less substituted C → primary alcohol.

Key Points to Remember:

• Exocyclic double bond in methylenecyclohexane: more substituted carbon is the ring carbon.
• Markovnikov addition: OH at tertiary (ring) carbon → 1-methylcyclohexanol.
• Anti-Markovnikov (BH3): OH at exo-methylene (CH2) → cyclohexylmethanol (primary).