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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0:54 minutes

Problem 11a

Textbook Question

Show how the following compounds could be prepared from 2-methylpropane:
a. 2-bromo-2-methylpropane

Verified step by step guidance

Start with 2-methylpropane (isobutane), which has the structure (CH₃)₂CHCH₃. This compound is an alkane and can undergo free radical halogenation to introduce a halogen atom.

Perform a free radical bromination reaction using bromine (Br₂) in the presence of ultraviolet (UV) light or heat. This reaction will selectively brominate the tertiary carbon atom due to its higher stability compared to primary or secondary carbons.

The bromination will yield 2-bromo-2-methylpropane, which has the structure (CH₃)₃CBr. This occurs because the tertiary carbon forms the most stable radical intermediate during the reaction.

Isolate and purify the product, 2-bromo-2-methylpropane, using standard organic chemistry techniques such as distillation or recrystallization, depending on the physical properties of the product.

Verify the structure of the product using spectroscopic methods such as NMR (nuclear magnetic resonance) or IR (infrared spectroscopy) to confirm the presence of the bromine atom on the tertiary carbon.

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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 alkanes react with halogens (like Br2) in the presence of heat or light to form alkyl halides. This process involves the formation of free radicals, which are highly reactive species that can abstract hydrogen atoms from the alkane, leading to the substitution of hydrogen with a halogen atom. Understanding this mechanism is crucial for synthesizing compounds like 2-bromo-2-methylpropane from 2-methylpropane.

Rearrangement Reactions

Rearrangement reactions involve the structural reorganization of a molecule to form a more stable or reactive isomer. In the case of preparing 2-bromo-2-methylpropane, the formation of a more stable tertiary carbocation from a secondary one can occur during the halogenation process. Recognizing how carbocation stability influences product formation is essential for predicting the outcome of reactions involving branched alkanes.

Nucleophilic Substitution

Nucleophilic substitution is a fundamental reaction type in organic chemistry where a nucleophile replaces a leaving group in a molecule. In the context of preparing 2-bromo-2-methylpropane, this concept is relevant if considering the conversion of an alcohol derived from 2-methylpropane into the corresponding bromo compound. Understanding the conditions that favor either SN1 or SN2 mechanisms is important for predicting the efficiency and selectivity of the substitution reaction.

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Related Practice

Textbook Question

Using the bond-dissociation energies in Table 5.6,

(a) predict whether or not an iodine radical would be selective for forming a single radical propane.

Textbook Question

What is the major product of the reaction in Problem 7 when the alkane reacts with Cl₂ instead of with Br₂? Disregard stereoisomers.

Textbook Question

Radical addition to alkenes is not effective for the synthesis of iodo- and chloroalkanes. Using your knowledge of the mechanism of this reaction, along with bond dissociation energies, explain why the radical additions of HI and HCl are not effective. (Assume ∆H = 65 kcal/ mol for the C–C π bond.)

Textbook Question

The benzene ring alters the reactivity of a neighboring group in the benzylic position much as a double bond alters the reactivity of groups in the allylic position.

Benzylic cations, anions, and radicals are all more stable than simple alkyl intermediates.

b. Toluene reacts with bromine in the presence of light to give benzyl bromide. Propose a mechanism for this reaction.

Textbook Question

Draw the products of the following reactions, including all stereoisomers:

Textbook Question

When a student attempted a bromination to produce compound A, they generated compound B instead. Rationalize the formation of B using the arrow-pushing formalism.

Multiple Choice

Predict the major, organic product of the following reaction.

Multiple Choice

Which step of a radical reaction is responsible for the formation of the bulk of the product?

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