Question

Q and R in the reaction sequences are respectively:

• A. Hg(OAc)₂, NaBH₄/OH⁻; B₂H₆, H₂O₂/OH⁻
• B. B₂H₆, H₂O₂/OH⁻; H⁺/H₂O
• C. Hg(OAc)₂, NaBH₄/OH⁻; H⁺/H₂O
• D. B₂H₆, H₂O₂/OH⁻; Hg(OAc)₂, NaBH₄/OH⁻

Hint:

Hydroboration-oxidation follows anti-Markovnikov addition, while oxymercuration-demercuration follows Markovnikov addition.

Answer 1

The Correct Option is D

The question asks to identify the reagents Q and R that convert an alkene into an alcohol. There are two common methods for hydration of alkenes: oxymercuration-demercuration and hydroboration-oxidation. Let's analyze the reaction sequences.

Reaction Sequence 1: 

The first step, converting the alkene to an alcohol, is achieved via **oxymercuration-demercuration**. This reaction involves two steps:

1. **Oxymercuration:** The alkene reacts with mercury(II) acetate \(Hg(OAc)_2\) in water. This step involves the electrophilic addition of \(Hg(OAc)^+\) to the double bond, followed by the addition of water to the more substituted carbon.

2. **Demercuration:** The resulting mercurinium ion is then reduced with sodium borohydride \(NaBH_4\) in the presence of a base (\(OH^-\)). This replaces the mercury with a hydrogen atom.

Oxymercuration-demercuration follows Markovnikov's rule, meaning the alcohol group (-OH) adds to the more substituted carbon of the alkene.

Therefore, reagent Q is \(Hg(OAc)_2, NaBH_4/OH^-\)

Reaction Sequence 2:

The second step converts the alkene to an alcohol via **hydroboration-oxidation**. This reaction also involves two steps:

1. **Hydroboration:** The alkene reacts with diborane \(B_2H_6\) (or BH3 THF complex). Boron adds to the double bond in a syn addition manner, with boron adding preferentially to the less substituted carbon due to steric reasons.

2. **Oxidation:** The resulting trialkylborane is then oxidized with hydrogen peroxide \(H_2O_2\) in the presence of a base (\(OH^-\)). This replaces the boron with a hydroxyl group (-OH).

Hydroboration-oxidation is an anti-Markovnikov addition of water. The alcohol group (-OH) adds to the less substituted carbon of the alkene, and it is a syn addition.

Therefore, reagent R is \(B_2H_6, H_2O_2/OH^-\)

Conclusion:

Q and R are respectively: \(Hg(OAc)_2, NaBH_4/OH^-\); \(B_2H_6, H_2O_2/OH^-\)

Answer 2

The reaction sequence involves:

1. Hydroboration-Oxidation: B₂H₆, H₂O₂/OH⁻ adds water across the double bond with anti-Markovnikov regioselectivity.
2. Oxymercuration-Demercuration: Hg(OAc)₂, NaBH₄/OH⁻ adds water across the double bond with Markovnikov regioselectivity.