(a) Interpretation: The stereochemistry of the product formed in the given reaction according to the proposed mechanism is to be predicted. Concept introduction: The electrophilic addition of bromine molecule across the C=C bond forms a vicinal dibromide product. In the first step, the electron-rich C=C bond attacks on one of the bromine atom, and the lone pair of the same bromine atom attacks back to another carbon of C=C bond forming bromonium ion intermediate and simultaneously breaks Br − Br bond. The bromonium ion is the three-membered ring with one positively charged bromine atom. The bromide ion produced in the first step acts as a nucleophile, and the positively charged bromine of bromonium ion intermediate is the leaving group. Thus, the reactions proceed through S N 2 where the nucleophilic bromide ion attacks on one of the carbon of the three-membered intermediate of bromonium ion from the opposite side of the positively charged bromine atom which results in breaking of one C − Br bond to form the vicinal dibromide product. As the bromide ion approached from the opposite side of the positively charged bromine atom, the stereochemistry of both bromine atoms with respect to each other in the product is trans. In case of a conjugated system of C=C bonds, the addition takes place at 1, 4 position due to resonance by the formation of a five-membered intermediate with positively charged bromine atom.

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Answer 1

Question

Chapter 12, Problem 12.38P

Interpretation Introduction

(a)

Interpretation:

The stereochemistry of the product formed in the given reaction according to the proposed mechanism is to be predicted.

Concept introduction:

The electrophilic addition of bromine molecule across the C=C bond forms a vicinal dibromide product. In the first step, the electron-rich C=C bond attacks on one of the bromine atom, and the lone pair of the same bromine atom attacks back to another carbon of C=C bond forming bromonium ion intermediate and simultaneously breaks Br−Br bond. The bromonium ion is the three-membered ring with one positively charged bromine atom. The bromide ion produced in the first step acts as a nucleophile, and the positively charged bromine of bromonium ion intermediate is the leaving group. Thus, the reactions proceed through SN2 where the nucleophilic bromide ion attacks on one of the carbon of the three-membered intermediate of bromonium ion from the opposite side of the positively charged bromine atom which results in breaking of one C−Br bond to form the vicinal dibromide product. As the bromide ion approached from the opposite side of the positively charged bromine atom, the stereochemistry of both bromine atoms with respect to each other in the product is trans. In case of a conjugated system of C=C bonds, the addition takes place at 1, 4 position due to resonance by the formation of a five-membered intermediate with positively charged bromine atom.

Interpretation Introduction

(b)

Interpretation:

It is to be determined whether the observations in the proposed mechanism support or discredit the formation of cis and trans products.

Concept introduction:

The electrophilic addition of bromine molecule across the C=C bond forms a vicinal dibromide product. In the first step, the electron-rich C=C bond attacks one of the bromine atom and the lone pair of the same bromine atom attacks back another carbon of the C=C bond forming bromonium ion intermediate and simultaneously breaks the Br−Br bond. The bromonium ion is the three-membered ring with one positively charged bromine atom. The bromide ion produced in the first step acts as a nucleophile, and the positively charged bromine of bromonium ion intermediate is the leaving group. Thus, the reactions proceed through SN2 where the nucleophilic bromide ion attacks one of the carbons of the three-membered intermediate of bromonium ion from the opposite side of positively charged bromine atom which results in breaking of one C−Br bond to form a vicinal dibromide product. As the bromide ion approaches from the opposite side of the positively charged bromine atom, the stereochemistry of both bromine atoms with respect to each other in the product is trans. In case of a conjugated system of C=C bonds, the addition takes place at 1, 4 position due to resonance by formation of a five-membered intermediate with positively charged bromine atom.

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Bromination can occur in a 1,4 fashion across conjugated double bonds, as shown here for cyclohexa-1,3-diene. One mechanism that has been proposed involves a five-membered ring, bromonium ion intermediate, as shown below. (a) According to this mechanism, what should the stereochemistry be for the products-namely, all cis, all trans, or a mixture of both? (b) Observations from experiment show that both cis and trans products are formed. Does this support or discredit the proposed mechanism shown here? Br Br2 Cyclohexa-1,3-diene Br :Br: Br: Br:

When considering the photochemical cis-trans isomerization of trans-stilbene, an electrocyclization is observed. The product of this electrocyclization is usually oxidized to a condensed aromatic product (by oxygen or addition of other oxidizing reagents such as iodine). Discuss the mechanism of electrocyclic ring formation and ring opening non- verbally by showing orbital images, indicate the orientation of the cyclization / opening, and shoe how this reaction is crucial in the synthesis of helicenes.

The alkene shown undergoes bromination. (a) Draw the product(s) of bromination of this compound, including all expected stereoisomers (if any). Use wedge‑and‑dash bonds to designate the stereochemistry at any chirality centers, and make sure to draw an explicit hydrogen if a chirality center has one. (b) Characterize the starting alkene as having the E or Z configuration. (c) characterize the product(s).

Answer 2

Question

Chapter 12, Problem 12.38P

Interpretation Introduction

(a)

Interpretation:

The stereochemistry of the product formed in the given reaction according to the proposed mechanism is to be predicted.

Concept introduction:

The electrophilic addition of bromine molecule across the C=C bond forms a vicinal dibromide product. In the first step, the electron-rich C=C bond attacks on one of the bromine atom, and the lone pair of the same bromine atom attacks back to another carbon of C=C bond forming bromonium ion intermediate and simultaneously breaks Br−Br bond. The bromonium ion is the three-membered ring with one positively charged bromine atom. The bromide ion produced in the first step acts as a nucleophile, and the positively charged bromine of bromonium ion intermediate is the leaving group. Thus, the reactions proceed through SN2 where the nucleophilic bromide ion attacks on one of the carbon of the three-membered intermediate of bromonium ion from the opposite side of the positively charged bromine atom which results in breaking of one C−Br bond to form the vicinal dibromide product. As the bromide ion approached from the opposite side of the positively charged bromine atom, the stereochemistry of both bromine atoms with respect to each other in the product is trans. In case of a conjugated system of C=C bonds, the addition takes place at 1, 4 position due to resonance by the formation of a five-membered intermediate with positively charged bromine atom.

Interpretation Introduction

(b)

Interpretation:

It is to be determined whether the observations in the proposed mechanism support or discredit the formation of cis and trans products.

Concept introduction:

The electrophilic addition of bromine molecule across the C=C bond forms a vicinal dibromide product. In the first step, the electron-rich C=C bond attacks one of the bromine atom and the lone pair of the same bromine atom attacks back another carbon of the C=C bond forming bromonium ion intermediate and simultaneously breaks the Br−Br bond. The bromonium ion is the three-membered ring with one positively charged bromine atom. The bromide ion produced in the first step acts as a nucleophile, and the positively charged bromine of bromonium ion intermediate is the leaving group. Thus, the reactions proceed through SN2 where the nucleophilic bromide ion attacks one of the carbons of the three-membered intermediate of bromonium ion from the opposite side of positively charged bromine atom which results in breaking of one C−Br bond to form a vicinal dibromide product. As the bromide ion approaches from the opposite side of the positively charged bromine atom, the stereochemistry of both bromine atoms with respect to each other in the product is trans. In case of a conjugated system of C=C bonds, the addition takes place at 1, 4 position due to resonance by formation of a five-membered intermediate with positively charged bromine atom.

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Answer 3

Question

Chapter 12, Problem 12.38P

Interpretation Introduction

(a)

Interpretation:

The stereochemistry of the product formed in the given reaction according to the proposed mechanism is to be predicted.

Concept introduction:

The electrophilic addition of bromine molecule across the C=C bond forms a vicinal dibromide product. In the first step, the electron-rich C=C bond attacks on one of the bromine atom, and the lone pair of the same bromine atom attacks back to another carbon of C=C bond forming bromonium ion intermediate and simultaneously breaks Br−Br bond. The bromonium ion is the three-membered ring with one positively charged bromine atom. The bromide ion produced in the first step acts as a nucleophile, and the positively charged bromine of bromonium ion intermediate is the leaving group. Thus, the reactions proceed through SN2 where the nucleophilic bromide ion attacks on one of the carbon of the three-membered intermediate of bromonium ion from the opposite side of the positively charged bromine atom which results in breaking of one C−Br bond to form the vicinal dibromide product. As the bromide ion approached from the opposite side of the positively charged bromine atom, the stereochemistry of both bromine atoms with respect to each other in the product is trans. In case of a conjugated system of C=C bonds, the addition takes place at 1, 4 position due to resonance by the formation of a five-membered intermediate with positively charged bromine atom.

Interpretation Introduction

(b)

Interpretation:

It is to be determined whether the observations in the proposed mechanism support or discredit the formation of cis and trans products.

Concept introduction:

The electrophilic addition of bromine molecule across the C=C bond forms a vicinal dibromide product. In the first step, the electron-rich C=C bond attacks one of the bromine atom and the lone pair of the same bromine atom attacks back another carbon of the C=C bond forming bromonium ion intermediate and simultaneously breaks the Br−Br bond. The bromonium ion is the three-membered ring with one positively charged bromine atom. The bromide ion produced in the first step acts as a nucleophile, and the positively charged bromine of bromonium ion intermediate is the leaving group. Thus, the reactions proceed through SN2 where the nucleophilic bromide ion attacks one of the carbons of the three-membered intermediate of bromonium ion from the opposite side of positively charged bromine atom which results in breaking of one C−Br bond to form a vicinal dibromide product. As the bromide ion approaches from the opposite side of the positively charged bromine atom, the stereochemistry of both bromine atoms with respect to each other in the product is trans. In case of a conjugated system of C=C bonds, the addition takes place at 1, 4 position due to resonance by formation of a five-membered intermediate with positively charged bromine atom.

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Answer 4

Question

Chapter 12, Problem 12.38P

Interpretation Introduction

(a)

Interpretation:

The stereochemistry of the product formed in the given reaction according to the proposed mechanism is to be predicted.

Concept introduction:

The electrophilic addition of bromine molecule across the C=C bond forms a vicinal dibromide product. In the first step, the electron-rich C=C bond attacks on one of the bromine atom, and the lone pair of the same bromine atom attacks back to another carbon of C=C bond forming bromonium ion intermediate and simultaneously breaks Br−Br bond. The bromonium ion is the three-membered ring with one positively charged bromine atom. The bromide ion produced in the first step acts as a nucleophile, and the positively charged bromine of bromonium ion intermediate is the leaving group. Thus, the reactions proceed through SN2 where the nucleophilic bromide ion attacks on one of the carbon of the three-membered intermediate of bromonium ion from the opposite side of the positively charged bromine atom which results in breaking of one C−Br bond to form the vicinal dibromide product. As the bromide ion approached from the opposite side of the positively charged bromine atom, the stereochemistry of both bromine atoms with respect to each other in the product is trans. In case of a conjugated system of C=C bonds, the addition takes place at 1, 4 position due to resonance by the formation of a five-membered intermediate with positively charged bromine atom.

Interpretation Introduction

(b)

Interpretation:

It is to be determined whether the observations in the proposed mechanism support or discredit the formation of cis and trans products.

Concept introduction:

The electrophilic addition of bromine molecule across the C=C bond forms a vicinal dibromide product. In the first step, the electron-rich C=C bond attacks one of the bromine atom and the lone pair of the same bromine atom attacks back another carbon of the C=C bond forming bromonium ion intermediate and simultaneously breaks the Br−Br bond. The bromonium ion is the three-membered ring with one positively charged bromine atom. The bromide ion produced in the first step acts as a nucleophile, and the positively charged bromine of bromonium ion intermediate is the leaving group. Thus, the reactions proceed through SN2 where the nucleophilic bromide ion attacks one of the carbons of the three-membered intermediate of bromonium ion from the opposite side of positively charged bromine atom which results in breaking of one C−Br bond to form a vicinal dibromide product. As the bromide ion approaches from the opposite side of the positively charged bromine atom, the stereochemistry of both bromine atoms with respect to each other in the product is trans. In case of a conjugated system of C=C bonds, the addition takes place at 1, 4 position due to resonance by formation of a five-membered intermediate with positively charged bromine atom.

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Answer 5

Chapter 12, Problem 12.38P

Interpretation Introduction

(a)

Interpretation:

The stereochemistry of the product formed in the given reaction according to the proposed mechanism is to be predicted.

Concept introduction:

The electrophilic addition of bromine molecule across the C=C bond forms a vicinal dibromide product. In the first step, the electron-rich C=C bond attacks on one of the bromine atom, and the lone pair of the same bromine atom attacks back to another carbon of C=C bond forming bromonium ion intermediate and simultaneously breaks Br−Br bond. The bromonium ion is the three-membered ring with one positively charged bromine atom. The bromide ion produced in the first step acts as a nucleophile, and the positively charged bromine of bromonium ion intermediate is the leaving group. Thus, the reactions proceed through SN2 where the nucleophilic bromide ion attacks on one of the carbon of the three-membered intermediate of bromonium ion from the opposite side of the positively charged bromine atom which results in breaking of one C−Br bond to form the vicinal dibromide product. As the bromide ion approached from the opposite side of the positively charged bromine atom, the stereochemistry of both bromine atoms with respect to each other in the product is trans. In case of a conjugated system of C=C bonds, the addition takes place at 1, 4 position due to resonance by the formation of a five-membered intermediate with positively charged bromine atom.

Interpretation Introduction

(b)

Interpretation:

It is to be determined whether the observations in the proposed mechanism support or discredit the formation of cis and trans products.

Concept introduction:

The electrophilic addition of bromine molecule across the C=C bond forms a vicinal dibromide product. In the first step, the electron-rich C=C bond attacks one of the bromine atom and the lone pair of the same bromine atom attacks back another carbon of the C=C bond forming bromonium ion intermediate and simultaneously breaks the Br−Br bond. The bromonium ion is the three-membered ring with one positively charged bromine atom. The bromide ion produced in the first step acts as a nucleophile, and the positively charged bromine of bromonium ion intermediate is the leaving group. Thus, the reactions proceed through SN2 where the nucleophilic bromide ion attacks one of the carbons of the three-membered intermediate of bromonium ion from the opposite side of positively charged bromine atom which results in breaking of one C−Br bond to form a vicinal dibromide product. As the bromide ion approaches from the opposite side of the positively charged bromine atom, the stereochemistry of both bromine atoms with respect to each other in the product is trans. In case of a conjugated system of C=C bonds, the addition takes place at 1, 4 position due to resonance by formation of a five-membered intermediate with positively charged bromine atom.

Answer 6

Chapter 12, Problem 12.38P

Interpretation Introduction

(a)

Interpretation:

The stereochemistry of the product formed in the given reaction according to the proposed mechanism is to be predicted.

Concept introduction:

The electrophilic addition of bromine molecule across the C=C bond forms a vicinal dibromide product. In the first step, the electron-rich C=C bond attacks on one of the bromine atom, and the lone pair of the same bromine atom attacks back to another carbon of C=C bond forming bromonium ion intermediate and simultaneously breaks Br−Br bond. The bromonium ion is the three-membered ring with one positively charged bromine atom. The bromide ion produced in the first step acts as a nucleophile, and the positively charged bromine of bromonium ion intermediate is the leaving group. Thus, the reactions proceed through SN2 where the nucleophilic bromide ion attacks on one of the carbon of the three-membered intermediate of bromonium ion from the opposite side of the positively charged bromine atom which results in breaking of one C−Br bond to form the vicinal dibromide product. As the bromide ion approached from the opposite side of the positively charged bromine atom, the stereochemistry of both bromine atoms with respect to each other in the product is trans. In case of a conjugated system of C=C bonds, the addition takes place at 1, 4 position due to resonance by the formation of a five-membered intermediate with positively charged bromine atom.

Answer 7

Chapter 12, Problem 12.38P

Interpretation Introduction

(a)

Interpretation:

The stereochemistry of the product formed in the given reaction according to the proposed mechanism is to be predicted.

Concept introduction:

The electrophilic addition of bromine molecule across the C=C bond forms a vicinal dibromide product. In the first step, the electron-rich C=C bond attacks on one of the bromine atom, and the lone pair of the same bromine atom attacks back to another carbon of C=C bond forming bromonium ion intermediate and simultaneously breaks Br−Br bond. The bromonium ion is the three-membered ring with one positively charged bromine atom. The bromide ion produced in the first step acts as a nucleophile, and the positively charged bromine of bromonium ion intermediate is the leaving group. Thus, the reactions proceed through SN2 where the nucleophilic bromide ion attacks on one of the carbon of the three-membered intermediate of bromonium ion from the opposite side of the positively charged bromine atom which results in breaking of one C−Br bond to form the vicinal dibromide product. As the bromide ion approached from the opposite side of the positively charged bromine atom, the stereochemistry of both bromine atoms with respect to each other in the product is trans. In case of a conjugated system of C=C bonds, the addition takes place at 1, 4 position due to resonance by the formation of a five-membered intermediate with positively charged bromine atom.

Answer 8

Interpretation Introduction

(b)

Interpretation:

It is to be determined whether the observations in the proposed mechanism support or discredit the formation of cis and trans products.

Concept introduction:

The electrophilic addition of bromine molecule across the C=C bond forms a vicinal dibromide product. In the first step, the electron-rich C=C bond attacks one of the bromine atom and the lone pair of the same bromine atom attacks back another carbon of the C=C bond forming bromonium ion intermediate and simultaneously breaks the Br−Br bond. The bromonium ion is the three-membered ring with one positively charged bromine atom. The bromide ion produced in the first step acts as a nucleophile, and the positively charged bromine of bromonium ion intermediate is the leaving group. Thus, the reactions proceed through SN2 where the nucleophilic bromide ion attacks one of the carbons of the three-membered intermediate of bromonium ion from the opposite side of positively charged bromine atom which results in breaking of one C−Br bond to form a vicinal dibromide product. As the bromide ion approaches from the opposite side of the positively charged bromine atom, the stereochemistry of both bromine atoms with respect to each other in the product is trans. In case of a conjugated system of C=C bonds, the addition takes place at 1, 4 position due to resonance by formation of a five-membered intermediate with positively charged bromine atom.

Answer 9

Interpretation Introduction

(b)

Interpretation:

It is to be determined whether the observations in the proposed mechanism support or discredit the formation of cis and trans products.

Concept introduction:

The electrophilic addition of bromine molecule across the C=C bond forms a vicinal dibromide product. In the first step, the electron-rich C=C bond attacks one of the bromine atom and the lone pair of the same bromine atom attacks back another carbon of the C=C bond forming bromonium ion intermediate and simultaneously breaks the Br−Br bond. The bromonium ion is the three-membered ring with one positively charged bromine atom. The bromide ion produced in the first step acts as a nucleophile, and the positively charged bromine of bromonium ion intermediate is the leaving group. Thus, the reactions proceed through SN2 where the nucleophilic bromide ion attacks one of the carbons of the three-membered intermediate of bromonium ion from the opposite side of positively charged bromine atom which results in breaking of one C−Br bond to form a vicinal dibromide product. As the bromide ion approaches from the opposite side of the positively charged bromine atom, the stereochemistry of both bromine atoms with respect to each other in the product is trans. In case of a conjugated system of C=C bonds, the addition takes place at 1, 4 position due to resonance by formation of a five-membered intermediate with positively charged bromine atom.

Answer 10

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Answer 11

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Answer 12

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Answer 13

Ch. 12 - Prob. 12.1P Ch. 12 - Prob. 12.2P Ch. 12 - Prob. 12.3P Ch. 12 - Prob. 12.4P Ch. 12 - Prob. 12.5P Ch. 12 - Prob. 12.6P Ch. 12 - Prob. 12.7P Ch. 12 - Prob. 12.8P Ch. 12 - Prob. 12.9P Ch. 12 - Prob. 12.10P

Solution Summary: The author explains how the electrophilic addition of bromine molecule across the C=C bond forms a vicinal dibromide product.

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