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 $\text{C=C}$ bond forms a vicinal dibromide product. In the first step, the electron-rich $\text{C=C}$ bond attacks on one of the bromine atom, and the lone pair of the same bromine atom attacks back to another carbon of $\text{C=C}$ bond forming bromonium ion intermediate and simultaneously breaks $\text{Br}-\text{Br}$ 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 ${\text{S}}_{\text{N}}\text{2}$ 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 $\text{C}-\text{Br}$ 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 $\text{C=C}$ bonds, the addition takes place at $\text{1, 4}$ 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 $\text{C=C}$ bond forms a vicinal dibromide product. In the first step, the electron-rich $\text{C=C}$ bond attacks one of the bromine atom and the lone pair of the same bromine atom attacks back another carbon of the $\text{C=C}$ bond forming bromonium ion intermediate and simultaneously breaks the $\text{Br}-\text{Br}$ 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 ${\text{S}}_{\text{N}}\text{2}$ 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 $\text{C}-\text{Br}$ 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 $\text{C=C}$ bonds, the addition takes place at $\text{1, 4}$ position due to resonance by formation of a five-membered intermediate with positively charged bromine atom.