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

11. Radical Reactions

Free Radical Halogenation

Organic Chemistry

11. Radical Reactions

Free Radical Halogenation

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14:05 minutes

Problem 22b(1,2)

Textbook Question

For each reaction, show which stereoisomers are obtained
1. NBS/∆/peroxide
2. Br₂/CH₂Cl₂

Verified step by step guidance

Step 1: Understand the context of the reactions. Both reactions involve bromination, but they proceed through different mechanisms. NBS (N-Bromosuccinimide) with heat and peroxide typically leads to allylic bromination, while Br2 in CH2Cl2 results in electrophilic addition to alkenes.

Step 2: For the NBS/∆/peroxide reaction, identify the allylic position in the molecule. The allylic position is the carbon atom adjacent to a double bond. This reaction proceeds via a radical mechanism, where the bromine radical selectively reacts at the allylic position, leading to the formation of allylic bromides.

Step 3: Consider stereoisomer formation in the NBS reaction. Since the reaction involves a radical intermediate, the product may exhibit stereoisomerism depending on the geometry of the double bond and the position of the bromine atom. Analyze the molecule to determine if E/Z isomers or other stereoisomers are possible.

Step 4: For the Br2/CH2Cl2 reaction, understand that this is an electrophilic addition reaction. Bromine adds across the double bond in a syn or anti fashion, depending on the reaction conditions. This typically leads to the formation of vicinal dibromides (two bromine atoms on adjacent carbons).

Step 5: Evaluate stereoisomer formation in the Br2/CH2Cl2 reaction. The addition of bromine can lead to stereoisomers (e.g., enantiomers or diastereomers) depending on the geometry of the starting alkene and the mechanism of bromine addition. Draw the possible stereoisomers based on the molecular structure.

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Key Concepts

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

Stereoisomerism

Stereoisomerism refers to the phenomenon where compounds have the same molecular formula and connectivity of atoms but differ in the spatial arrangement of those atoms. This can lead to different physical and chemical properties. The two main types of stereoisomers are enantiomers, which are non-superimposable mirror images, and diastereomers, which are not mirror images of each other.

Radical Bromination

Radical bromination, often initiated by N-bromosuccinimide (NBS) in the presence of heat or light, involves the formation of bromine radicals that abstract hydrogen atoms from alkenes or alkanes. This process can lead to the formation of allylic bromides and can generate different stereoisomers depending on the position of the bromine addition and the nature of the starting material.

Electrophilic Addition

Electrophilic addition is a reaction mechanism where an electrophile reacts with a nucleophile, typically involving alkenes or alkynes. In the case of bromination with Br2 in a non-polar solvent like CH2Cl2, the alkene undergoes addition across the double bond, resulting in vicinal dibromides. The stereochemistry of the addition can lead to the formation of different stereoisomers, including meso compounds and enantiomers.

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Textbook Question

The following compounds have the same molecular formula as benzene. How many monobrominated products could each form?

1. HC≡CC≡CCH₂CH₃

2. CH₂=CHC≡CCH=CH₂

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.

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.

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