Ch.5 - Stereochemistry

Wade9th EditionOrganic ChemistryISBN: 9780135213728Not the one you use?Change textbook

Wade 9th Edition

Ch.5 - Stereochemistry

Problem 36a,b,c

Chapter 5, Problem 36a,b,c

Free-radical bromination of the following compound introduces bromine primarily at the benzylic position next to the aromatic ring. If the reaction stops at the monobromination stage, two stereoisomers result.

a. Propose a mechanism to show why free-radical halogenation occurs almost exclusively at the benzylic position.
b. Draw the two stereoisomers that result from monobromination at the benzylic position.
c. Assign R and S configurations to the asymmetric carbon atoms in the products.

Verified step by step guidance

Step 1: Understand the reaction conditions. Free-radical bromination occurs under the presence of bromine (Br₂) and light (hv). The benzylic position is highly reactive due to the stability of the benzylic radical formed during the reaction.

Step 2: Propose the mechanism for free-radical bromination. The reaction begins with the homolytic cleavage of Br₂ under light to form two bromine radicals. The benzylic hydrogen is abstracted by a bromine radical, forming a benzylic radical. This radical is stabilized by resonance with the aromatic ring, making the benzylic position the preferred site for bromination.

Step 3: The benzylic radical reacts with another bromine molecule (Br₂), leading to the formation of the monobrominated product at the benzylic position. This product has a new chiral center due to the introduction of bromine.

Step 4: Draw the two stereoisomers resulting from monobromination. Since the benzylic position is now a chiral center, two stereoisomers (R and S configurations) are formed. These stereoisomers differ in the spatial arrangement of the substituents around the chiral center.

Step 5: Assign R and S configurations to the products. Use the Cahn-Ingold-Prelog priority rules to determine the configuration of the chiral center in each stereoisomer. Assign priorities based on atomic number and analyze the arrangement of substituents to label each stereoisomer as R or S.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Free-Radical Halogenation

Free-radical halogenation is a reaction mechanism where halogens, such as bromine, react with alkanes or other hydrocarbons in the presence of heat or light. This process involves the formation of free radicals, which are highly reactive species with unpaired electrons. The reaction typically proceeds through three steps: initiation, propagation, and termination, leading to the substitution of hydrogen atoms with halogen atoms.

Benzylic Position

The benzylic position refers to the carbon atom directly attached to a benzene ring. This position is particularly reactive in free-radical reactions due to the stability of the resulting benzylic radical, which is resonance-stabilized by the aromatic system. As a result, bromination at the benzylic position is favored over other positions, as the formation of a stable radical intermediate lowers the activation energy of the reaction.

Stereoisomerism

Stereoisomerism occurs when molecules have the same molecular formula and connectivity but differ in the spatial arrangement of their atoms. In the context of the monobromination reaction, the introduction of bromine at the benzylic position creates a chiral center, leading to two possible stereoisomers. These isomers can be assigned R or S configurations based on the Cahn-Ingold-Prelog priority rules, which help determine the spatial orientation of the substituents around the chiral carbon.

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Related Practice

Textbook Question

3,4-Dimethylpent-1-ene has the formula CH₂=CH—CH(CH₃)—CH(CH₃)₂. When pure (R)-3,4-dimethylpent-1-ene is treated with hydrogen over a platinum catalyst, the product is (S)-2,3-dimethylpentane.

a. Draw the equation for this reaction. Show the stereochemistry of the reactant and the product.

b. Has the chiral center retained its configuration during this hydrogenation, or has it been inverted?

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Textbook Question

The specific rotation of (S)-2-iodobutane is +15.90°.

a. Draw the structure of (S)-2-iodobutane.

b. Predict the specific rotation of (R)-2-iodobutane.

c. Determine the percentage composition of a mixture of (R)- and (S)-2-iodobutane with a specific rotation of –7.95°.

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Textbook Question

For each structure,

1. draw all the stereoisomers.

2. label each structure as chiral or achiral.

3. give the relationships between the stereoisomers (enantiomers, diastereomers).

(a)

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Textbook Question

(+)-Tartaric acid has a specific rotation Of +12.0°. Calculate the specific rotation of a mixture of 68% (+)-tartaric acid and 32% (–)-tartaric acid.

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Textbook Question

c. The reactant is named (R), but the product is named (S). Does this name change imply a change in the spatial arrangement of the groups around the chiral center? So why does the name switch from (R) to (S)?

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Textbook Question

Free-radical bromination of the following compound introduces bromine primarily at the benzylic position next to the aromatic ring. If the reaction stops at the monobromination stage, two stereoisomers result.

d. What is the relationship between the two isomeric products?

e. Will these two products be produced in identical amounts? That is, will the product mixture be exactly 50:50?

f. Will these two stereoisomers have identical physical properties such as boiling point, melting point, solubility, etc.? Could they be separated (theoretically, at least) by distillation or recrystallization?

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