Table of contents

1. A Review of General Chemistry5h 5m
2. Molecular Representations1h 14m
3. Acids and Bases2h 46m
4. Alkanes and Cycloalkanes4h 17m
5. Chirality3h 39m
6. Thermodynamics and Kinetics1h 22m
7. Substitution Reactions1h 48m
8. Elimination Reactions2h 30m
9. Alkenes and Alkynes2h 9m
10. Addition Reactions3h 19m
11. Radical Reactions1h 58m
12. Alcohols, Ethers, Epoxides and Thiols2h 42m
13. Alcohols and Carbonyl Compounds2h 17m
14. Synthetic Techniques1h 26m
15. Analytical Techniques:IR, NMR, Mass Spect7h 3m
16. Conjugated Systems6h 13m
17. Ultraviolet Spectroscopy51m
18. Aromaticity2h 34m
19. Reactions of Aromatics: EAS and Beyond5h 1m
20. Phenols55m
21. Aldehydes and Ketones: Nucleophilic Addition4h 56m
22. Carboxylic Acid Derivatives: NAS2h 51m
23. The Chemistry of Thioesters, Phophate Ester and Phosphate Anhydrides1h 10m
24. Enolate Chemistry: Reactions at the Alpha-Carbon1h 53m
25. Condensation Chemistry2h 9m
26. Amines1h 43m
27. Heterocycles2h 0m
28. Carbohydrates5h 53m
29. Amino Acids4h 20m
30. Peptides and Proteins2h 42m
31. Catalysis in Organic Reactions1h 30m
32. Lipids 2h 50m
33. The Organic Chemistry of Metabolic Pathways2h 52m
34. Nucleic Acids1h 32m
35. Transition Metals6h 14m
36. Synthetic Polymers1h 49m

10. Addition Reactions

Epoxidation

Organic Chemistry

10. Addition Reactions

Epoxidation

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4:12 minutes

Problem 18b

Textbook Question

When producing a chiral molecule, epoxide formation still results in a mixture of enantiomers, despite its stereospecificity.
(b) How is it that a reaction can be stereospecific while still producing two enantiomers?

Verified step by step guidance

Step 1: Analyze the reaction provided in the image. The starting material is an alkene (with a phenyl group attached), and the reagent is m-chloroperbenzoic acid (mCPBA), which is commonly used for epoxidation reactions. The products are two enantiomers of an epoxide.

Step 2: Understand the concept of stereospecificity. A stereospecific reaction is one in which the stereochemistry of the reactant determines the stereochemistry of the product. In this case, the alkene undergoes epoxidation, and the stereochemistry of the double bond influences the formation of the epoxide.

Step 3: Explain why two enantiomers are formed. The alkene is planar, meaning it has no inherent chirality. When the mCPBA approaches the double bond, it can attack from either the top face or the bottom face of the planar alkene. This leads to the formation of two enantiomers, as the epoxide ring can form with opposite configurations depending on the face of attack.

Step 4: Clarify the stereospecificity aspect. Despite the formation of two enantiomers, the reaction is still stereospecific because the stereochemistry of the starting alkene dictates the relative stereochemistry of the products. For example, if the alkene had substituents that created a cis or trans configuration, the epoxide products would reflect that configuration.

Step 5: Summarize the key point. A reaction can be stereospecific while producing enantiomers because the stereospecificity refers to the relationship between the reactant and product stereochemistry, not the exclusivity of a single stereoisomer. The planar nature of the alkene allows for attack from both faces, resulting in a racemic mixture of enantiomers.

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Key Concepts

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

Chirality

Chirality refers to the geometric property of a molecule that makes it non-superimposable on its mirror image, much like left and right hands. Chiral molecules typically have at least one carbon atom bonded to four different substituents, leading to two distinct configurations known as enantiomers. Understanding chirality is essential for grasping how certain reactions can yield different stereoisomers.

Stereospecificity

Stereospecificity is a characteristic of certain chemical reactions where the reactants lead to a specific stereoisomer of the product. In stereospecific reactions, the mechanism dictates the spatial arrangement of atoms, resulting in a preferred configuration. However, even stereospecific reactions can produce a mixture of enantiomers if the starting material is not entirely chiral or if the reaction conditions allow for different pathways.

Epoxide Formation

Epoxide formation involves the creation of a three-membered cyclic ether, which can occur through various mechanisms, often involving the reaction of alkenes with peracids. While the formation of epoxides can be stereospecific, the presence of a prochiral center or the reaction conditions can lead to the generation of both enantiomers. This illustrates how a stereospecific reaction can still yield a racemic mixture when starting materials or conditions allow for such outcomes.

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Related Practice

Textbook Question

Predict the product(s) that would result when the alkenes are allowed to react under the following conditions: (v) mCPBA;

(h)

Textbook Question

Predict the product(s) that would result when the alkenes are allowed to react under the following conditions: (vii) 1. mCPBA 2. H₂SO₄, H₂O

(h)

Textbook Question

Predict the product(s) that would result when the alkenes are allowed to react under the following conditions: (v) mCPBA;

(k)

Textbook Question

Predict the product(s) when each of the following are reacted with mCPBA, making sure to indicate the relative stereochemical outcome. Indicate any racemic mixtures by drawing both enantiomers.

(a)

Textbook Question

Rank the reactivity of the following alkenes with mCPBA ( 1 = most reactive , 5 least reactive ).

Textbook Question

One way to think about concerted reactions is to imagine them as being stepwise reactions where, besides the slowest step, all others have infinitesimally small activation energies. Considering the hypothetical stepwise mechanism and actual concerted mechanism of epoxide formation, show what a reaction coordinate diagram might look like for each possibility.

(a) Stepwise, hypothetical mechanism:

Textbook Question

(b) Concerted , actual mechanism (butterfly transition state:

Textbook Question

How does the carbonyl in mCPBA weaken the O―O σ bond (i.e., make a better leaving group)?

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