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

24. Enolate Chemistry: Reactions at the Alpha-Carbon

Tautomerization

Organic Chemistry

24. Enolate Chemistry: Reactions at the Alpha-Carbon

Tautomerization

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6:52 minutes

Problem 17

Textbook Question

Explain why a racemic mixture is formed when (R)-2-methylpentanal is dissolved in an acidic or basicsolution.

Verified step by step guidance

Identify the chiral center in (R)-2-methylpentanal, which is the carbon atom bonded to four different groups.

Understand that in acidic or basic solutions, the carbonyl group in 2-methylpentanal can undergo keto-enol tautomerism, forming an enol intermediate.

Recognize that the enol form is planar around the sp2 hybridized carbon, allowing for the possibility of re-protonation from either side of the plane.

Realize that re-protonation of the enol can lead to the formation of either the (R) or (S) enantiomer, as the planar intermediate does not favor one side over the other.

Conclude that the equal probability of forming either enantiomer results in a racemic mixture, which is a 1:1 mixture of the (R) and (S) enantiomers.

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Key Concepts

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

Racemic Mixture

A racemic mixture consists of equal amounts of two enantiomers, which are molecules that are mirror images of each other. In organic chemistry, racemic mixtures often arise during reactions involving chiral centers, where the reaction can lead to the formation of both enantiomers without preference. This results in a 50:50 mixture, which is optically inactive because the optical activities of the two enantiomers cancel each other out.

Chirality and Enantiomers

Chirality refers to the property of a molecule that makes it non-superimposable on its mirror image, typically due to the presence of a chiral center, such as a carbon atom bonded to four different substituents. Enantiomers are the two distinct forms of a chiral molecule, each capable of rotating plane-polarized light in opposite directions. The formation of a racemic mixture indicates that both enantiomers are produced during a reaction, often due to the lack of a chiral influence in the reaction environment.

Acidic and Basic Conditions

Acidic and basic conditions can significantly influence the reactivity and stability of organic compounds. In the case of (R)-2-methylpentanal, the presence of an acid or base can facilitate the formation of a carbocation or anionic intermediate, which can then lead to the attack by nucleophiles from either side of the chiral center. This non-selective attack results in the formation of both enantiomers, contributing to the racemic mixture observed in the solution.

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