Chapter 8.SE, Problem 48AP

Predict the products of the following reactions. Don’t worry about the size of the molecule; concentrate on the functional groups.

Interpretation Introduction

a)

Interpretation:

To predict the product formed when cholesterol reacts with bromine.

Concept introduction:

The reaction of halogens to alkenes occurs with anti stereochemistry, that is, the two halogen atoms come from opposite faces of the double bond- one from top face and other from bottom face. In the first step the addition of halogen to the double bond in the alkenes results in the formation of a cyclic halonium ion with the simultaneous elimination of a halide ion. The large halonium ion shields one side of the molecule. Hence the attack of the halide ion occurs from the opposite, unshielded side to yield the trans product.

To predict:

The product formed when cholesterol reacts with bromine.

Answer to Problem 48AP

The product formed when cholesterol reacts with bromine.

Explanation of Solution

In the first step, the addition of bromine to the double bond in cholesterol results in the formation of a cyclic bromonium ion with the elimination of a bromide ion. In the second step the bromide ion attacks the bromonium ion from the opposite, unshielded side to yield a dibromo derivative.

Conclusion

The product formed when cholesterol reacts with bromine.

Interpretation Introduction

b)

Interpretation:

To predict the product formed when cholesterol reacts with HBr.

Concept introduction:

The addition of hydrogen halides to alkenes takes place through the formation of a carbocation intermediate by the attack of the π electrons in the double bond on the positively polarized hydrogen of the hydrogen halide. The halogen is eliminated as a halide ion. In the second step, the halide ion reacts with the carbocation to yield the product.

To predict:

The product formed when cholesterol reacts with HBr.

Answer to Problem 48AP

The product formed when cholesterol reacts with HBr is

Explanation of Solution

The addition of hydrogen bromide to cholesterol takes place through the formation of a carbocation intermediate by the attack of the π electrons in the double bond on the positively polarized hydrogen of the hydrogen bromide. The bromine is eliminated as bromide ion which reacts in the second step with the carbocation to yield the product.

Conclusion

The product formed when cholesterol reacts with HBr is

Interpretation Introduction

c)

Interpretation:

To predict the product formed when cholesterol reacts first with OsO₄ and then with NaHSO₃.

Concept introduction:

Hydroxylation of double bonds can be carried out directly by treating the alkene with osmium tetroxide, O_sO₄, in the presence of N-phenylmorpholine N-oxide. The reaction occurs with syn stereochemistry through the formation of a cyclic intermediate, called osmate, formed by the addition of O_sO₄ to the alkene in a single step. The cyclic osmate is then cleaved to give the cis-1, 2-diol by treatment with NaHSO₃.

To predict:

The product formed when cholesterol reacts with first with O_sO₄ and then with NaHSO₃.

Answer to Problem 48AP

The product formed when cholesterol reacts with first with O_sO₄ and then with NaHSO₃ is

Explanation of Solution

The addition of O_sO₄ to the double bond in cholesterol takes place in a single step to give an osmate. The cyclic osmate when cleaved by treatment with NaHSO₃ gives the cis-diol as the product.

Conclusion

The product formed when cholesterol reacts with first with O_sO₄ and then with NaHSO₃ is

Interpretation Introduction

d)

Interpretation:

To predict the product formed when cholesterol reacts first with BH₃, THF and then with H₂O₂, OH^-.

Concept introduction:

The hydroboration-oxidation reaction takes place with syn stereochemistry and results in a non-Markovnikov addition of water to the double bond in the alkene. The resulting product has the hydroxyl group on the less highly substituted carbon.

To predict:

The product formed when cholesterol reacts with first with BH₃, THF and then with H₂O₂, OH^-.

Answer to Problem 48AP

The product formed when cholesterol reacts with first with BH₃, THF and then with H₂O₂, OH^- is

Explanation of Solution

During hydroboration, the addition of B-H to the double bond takes place with a syn stereochemistry. The boron atom being bigger than hydrogen gets attached to the less highly substituted carbon atom (less steric crowding). During the oxidation step, the boron is replaced by an –OH group with the same stereochemistry.

Conclusion

The product formed when cholesterol reacts with first with BH₃, THF and then with H₂O₂, OH^- is

Interpretation Introduction

e)

Interpretation:

To predict the product formed when cholesterol reacts with CH₂I₂ in the presence of Zn (Cu).

Concept introduction:

The reaction given is an example of Simmons-Smith reaction. When CH₂I₂ is treated with Zn/Cu couple, iodomethylzinc iodide, ICH₂ZnI, is formed. This ICH₂ZnI transfers a CH₂ group to the double bond in alkene to form a cyclopropane ring in the product.

To predict:

The product formed when cholesterol reacts with CH₂I₂ in the presence of Zn (Cu).

Answer to Problem 48AP

The product formed when cholesterol reacts with CH₂I₂ in the presence of Zn (Cu) is

Explanation of Solution

When CH₂I₂ is treated with Zn/Cu couple, iodomethylzinc iodide, ICH₂ZnI, is formed. This ICH₂ZnI transfers a CH₂ group to the double bond in cholesterol to form a cyclopropane ring in the product.

Conclusion

The product formed when cholesterol reacts with CH₂I₂ in the presence of Zn (Cu) is