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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Problem 58

Textbook Question

Tributyltin hydride (Bu₃SnH) is often used as a 'radical carrier' in radical reactions. Which bond would you expect to be weaker, Sn–H or C–H? How might this relate to radical stability? Explain your answer.

Verified step by step guidance

Identify the bonds in question: Sn–H in tributyltin hydride (Bu₃SnH) and C–H in a typical alkane. The image shows the structures of these compounds, highlighting the Sn–H and C–H bonds.

Understand bond strength: Bond strength is related to the bond dissociation energy (BDE). Generally, bonds involving heavier atoms like tin (Sn) tend to be weaker than those involving lighter atoms like carbon (C). This is due to the larger atomic size and lower electronegativity of tin compared to carbon.

Consider radical stability: In radical reactions, the stability of the radical formed after bond cleavage is crucial. A weaker bond is more likely to break, forming a radical. The Sn–H bond is weaker, making it more prone to homolytic cleavage, thus forming a tin radical.

Relate bond strength to radical stability: The stability of the radical formed is influenced by the atom's ability to stabilize the unpaired electron. Tin radicals are relatively stable due to the ability of tin to delocalize the unpaired electron over its larger atomic orbitals.

Conclude the relationship: The weaker Sn–H bond in Bu₃SnH facilitates the formation of radicals, making it an effective radical carrier. This is because the tin radical formed is stable, allowing it to participate in further radical reactions efficiently.

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Key Concepts

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

Bond Strength

Bond strength refers to the energy required to break a bond between two atoms. In the context of Sn–H versus C–H bonds, Sn–H bonds are generally weaker due to the larger atomic size and lower electronegativity of tin compared to carbon, making it easier to break and form radicals.

Radical Stability

Radical stability is influenced by the ability of a radical to delocalize its unpaired electron. Radicals formed from weaker bonds, like Sn–H, are often more stable because the unpaired electron can be better accommodated by the larger atomic orbitals of tin, reducing the energy of the radical species.

Radical Reactions

Radical reactions involve species with unpaired electrons and are initiated by breaking weak bonds. Tributyltin hydride (Bu₃SnH) acts as a radical carrier, facilitating the generation of radicals due to its weak Sn–H bond, which is crucial for initiating and propagating radical chain reactions.

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Related Practice

Textbook Question

For the following reaction, answer questions (a)–(d).

(d) Given your choice in (b), why is it the weakest bond?

Textbook Question

A halogenation intended to make compound A formed B instead.

(a) Suggest a mechanism for the intended formation of A.

Textbook Question

A halogenation intended to make compound A formed B instead.

Textbook Question

In the following reaction, which C―H bond would be most likely to react with a bromine radical?

Textbook Question

Haloalkanes can be reduced to alkanes using radical reactions involving Bu₃SnH and a catalytic amount of AIBN. Suggest a mechanism for this reaction. [Pay close attention to the bonds formed and broken in developing your mechanism.]

Textbook Question

What are the product(s) of each of the following reactions? Disregard stereoisomers.

Textbook Question

What are the product(s) of each of the following reactions? Disregard stereoisomers.

Textbook Question

What are the product(s) of each of the following reactions? Disregard stereoisomers.

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Tributyltin hydride (Bu3SnH) is often used as a 'radical carrier' in radical reactions. Which bond would you expect to be weaker, Sn–H or C–H? How might this relate to radical stability? Explain your answer.

Key Concepts

Bond Strength

Radical Stability

Radical Reactions

Watch next

Tributyltin hydride (Bu₃SnH) is often used as a 'radical carrier' in radical reactions. Which bond would you expect to be weaker, Sn–H or C–H? How might this relate to radical stability? Explain your answer.