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ExamsJEE AdvancedChemistry

In which of the following reactions is a secondary (2°) alcohol obtained as the major product when a cyclohexene derivative undergoes the given reaction?

  1. Acid-catalyzed hydration: cyclohexene + H2O / H+
  2. Oxymercuration-demercuration: (i) Hg(OAc)2, H2O then (ii) NaBH4 / OH-
  3. Hydroboration-oxidation: (i) B2H6, THF then (ii) H2O2 / OH-
  4. None of these

Correct answer: None of these

Solution

In unsubstituted cyclohexene, the two alkene carbons are both secondary (each attached to two CH2 groups in the ring). Regardless of regioselectivity: (A) Acid-catalyzed hydration follows Markovnikov's rule but since the ring is symmetric, both C1 and C2 are equivalent => 2° alcohol. (B) Oxymercuration-demercuration is Markovnikov-selective => 2° alcohol from the more substituted carbon. But with symmetric cyclohexene, again 2° alcohol. (C) Hydroboration-oxidation is anti-Markovnikov but in symmetric cyclohexene still gives 2° alcohol. So actually all three give 2° alcohol for cyclohexene. If the question has a substituted cyclohexene where some carbons become tertiary, then specific regioselectivity matters. Without that information, the standard answer for this type of JEE question (with symmetric cyclohexene) is that all give 2° alcohol, making 'None of these' correct if the question asks which EXCLUSIVELY gives 2° as major while others don't. However, if one of the reactions gives a 3° alcohol due to rearrangement (acid-catalyzed), and we need the one that avoids rearrangement, then hydroboration-oxidation (anti-Markovnikov, no rearrangement) reliably gives the less-substituted 2° alcohol. Given the options, the expected answer in JEE context where all give 2° alcohol from cyclohexene is 'None of these' is WRONG; all three give 2° alcohols. The answer is likely that option D (None) is incorrect and the question refers to a specific substituted cyclohexene not provided — marking as best answerable: all of A, B, C give 2° alcohols for simple cyclohexene.

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