Why does hot KMnO 4 produce different products compared to cold KMnO 4 ?

Hot KMnO 4 produces different products compared to cold KMnO 4 due to the increased reactivity at higher temperatures. Cold KMnO 4 typically adds vicinal diols (syndials) to the double bond, resulting in the formation of alcohols. In contrast, hot KMnO 4 causes oxidative cleavage, breaking the double bond completely and forming carbonyl-containing products such as ketones, carboxylic acids, and carbon dioxide, depending on the structure of the original alkene.

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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

Oxidative Cleavage

10. Addition Reactions

Oxidative Cleavage: Videos & Practice Problems

Video Lessons Practice Worksheet

Topic summary

Strong oxidative cleavage of double bonds occurs with hot KMnO₄, resulting in the formation of ketones, carboxylic acids, and carbon dioxide. Internal alkenes yield ketones, while terminal alkenes produce carboxylic acids due to the presence of hydrogen. A one-carbon fragment fully oxidizes to carbon dioxide. Understanding the geometry of the molecule is crucial; cutting a ring produces a straight chain, while multiple cuts yield distinct fragments. This reaction emphasizes the addition of oxygen, illustrating the concept of oxidation in organic chemistry.

When you react a double bond with potassium permanganate (KMnO₄), your first thought may be to make syn-diols, as seen in previous videos. However, looking at the temperature is extremely important. In heat, KMnO₄ will actually cleave the double bond, rather than add diols. Let’s check it out.

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General properties of strong oxidative cleavage.

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General properties of strong oxidative cleavage. Video Summary

Strong oxidative cleavage is a significant reaction involving double bonds, particularly when treated with hot potassium permanganate (KMnO₄). This reaction is crucial to understand because the behavior of KMnO₄ varies with temperature. At lower temperatures, KMnO₄ performs a syn-dihydroxylation, adding vicinal diols to the double bond. However, when heated, KMnO₄ undergoes strong oxidative cleavage, effectively breaking the double bond and producing distinct products.

During strong oxidative cleavage, the reaction can yield three primary types of products: ketones, carboxylic acids, and carbon dioxide. The type of product formed depends on the structure of the alkene being cleaved. Internal alkenes, which have substituents (R groups) on both sides of the double bond, will produce ketones. In contrast, terminal alkenes, which have at least one hydrogen atom attached to the carbon of the double bond, will yield carboxylic acids. If the cleavage results in a one-carbon fragment, that fragment will be fully oxidized to carbon dioxide, which is the most oxidized form of carbon.

To summarize the outcomes based on the structure of the alkene: internal double bonds (0 hydrogens) lead to ketones, terminal double bonds (1 hydrogen) produce carboxylic acids, and fragments with 2 hydrogens result in carbon dioxide. This oxidation process can be viewed as adding oxygen atoms to the resulting products, emphasizing the transformation of the carbon structure during the reaction.

Additionally, when considering cyclic compounds, the cleavage of a ring structure will yield a straight-chain product if cut once, while two cuts will produce multiple fragments. This analogy can be likened to cutting a rubber band: a single cut transforms it into a straight line, while two cuts create separate pieces.

Understanding the geometry and hydrogen count of the alkene is essential for predicting the products of strong oxidative cleavage. By analyzing the number of hydrogens attached to the double bond, one can determine the resulting fragments and their respective chemical identities.

Note: I added an extra carbon when rotating the structure to a bondline conformation in the video below:)

Problem

Predict the product of the following reaction.

Product structure with two hydroxyl groups on a cyclohexane ring.

Product structure with an aldehyde and a ketone on a carbon chain.

Product structure with two hydroxyl groups and a ketone on a carbon chain.

Product structure with two carbonyl groups and one hydroxyl group.

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Here’s what students ask on this topic:

What is oxidative cleavage in organic chemistry?

Oxidative cleavage in organic chemistry refers to the reaction where a double bond in an alkene is broken, resulting in the formation of two separate carbonyl-containing fragments. This reaction typically involves strong oxidizing agents like hot potassium permanganate (KMnO₄). The products can include ketones, carboxylic acids, and carbon dioxide, depending on the structure of the original alkene. Internal alkenes yield ketones, terminal alkenes produce carboxylic acids, and a one-carbon fragment fully oxidizes to carbon dioxide.

What are the products of oxidative cleavage with hot KMnO₄?

When oxidative cleavage is performed with hot KMnO₄, the products depend on the type of alkene. Internal alkenes (those with R groups on both sides of the double bond) yield ketones. Terminal alkenes (those with at least one hydrogen attached to the double-bonded carbon) produce carboxylic acids. If the cleavage results in a one-carbon fragment, it is fully oxidized to carbon dioxide (CO₂).

How does the geometry of a molecule affect the products of oxidative cleavage?

The geometry of a molecule significantly affects the products of oxidative cleavage. For example, if you start with a ring structure and make a single cut, the ring will open up into a straight chain. Multiple cuts in a ring or chain can yield distinct fragments. The position of the double bonds and the presence of hydrogen atoms on the carbons involved in the double bond determine whether the products will be ketones, carboxylic acids, or carbon dioxide.

Why does hot KMnO₄ produce different products compared to cold KMnO₄?

Hot KMnO₄ produces different products compared to cold KMnO₄ due to the increased reactivity at higher temperatures. Cold KMnO₄ typically adds vicinal diols (syndials) to the double bond, resulting in the formation of alcohols. In contrast, hot KMnO₄ causes oxidative cleavage, breaking the double bond completely and forming carbonyl-containing products such as ketones, carboxylic acids, and carbon dioxide, depending on the structure of the original alkene.

What is the role of hydrogen atoms in determining the products of oxidative cleavage?

The presence of hydrogen atoms on the carbons involved in the double bond is crucial in determining the products of oxidative cleavage. Internal alkenes, which have no hydrogen atoms on the double-bonded carbons, yield ketones. Terminal alkenes, which have at least one hydrogen atom, produce carboxylic acids. If the cleavage results in a one-carbon fragment with two hydrogen atoms, it is fully oxidized to carbon dioxide (CO₂).