Products A and B formed in the following set of reactions are [figure]
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Answer
(4) [figure]
Solution
[figure]
A: H+/H2O ⇒ Markovnikov hydration through the 3° carbocation ⇒ OH on the ring C carrying CH3
∵ 3° carbocation is more stable than 2° ⇒ A = 1-methylcyclohexan-1-ol
[figure]
B: hydroboration–oxidation ⇒ anti-Markovnikov (syn) addition ⇒ OH on the less substituted C
∵ boron adds to the less hindered carbon ⇒ B = 2-methylcyclohexan-1-ol
Answer: (4)