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

Haloform Reaction

Organic Chemistry

24. Enolate Chemistry: Reactions at the Alpha-Carbon

Haloform Reaction

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11:21 minutes

Problem 12

Textbook Question

Propose a mechanism for the reaction of cyclohexyl methyl ketone with excess bromine in the presence of sodium hydroxide.

Verified step by step guidance

Identify the reaction type: This is a classic example of the haloform reaction, where a methyl ketone reacts with excess bromine in the presence of a base (sodium hydroxide) to produce a carboxylate ion and a haloform (in this case, bromoform).

Step 1: Enolate formation. The base (OH⁻) abstracts an α-hydrogen from the cyclohexyl methyl ketone, forming an enolate ion. The enolate is stabilized by resonance between the oxygen and the α-carbon.

Step 2: Bromination of the enolate. The enolate ion reacts with bromine (Br₂), leading to the substitution of one α-hydrogen with a bromine atom. This step repeats two more times, as excess bromine is present, until all three α-hydrogens on the methyl group are replaced by bromine atoms, forming a tribromo intermediate.

Step 3: Base-induced cleavage. The hydroxide ion attacks the carbonyl carbon of the tribromo intermediate, leading to the formation of a tetrahedral intermediate. This intermediate collapses, expelling the tribromomethane (CHBr₃, bromoform) and forming a carboxylate ion.

Step 4: Final products. The reaction concludes with the formation of cyclohexanecarboxylate ion (the conjugate base of cyclohexanecarboxylic acid) and bromoform (CHBr₃).

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Key Concepts

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

Electrophilic Bromination

Electrophilic bromination is a reaction where bromine acts as an electrophile, attacking nucleophilic sites in organic molecules. In the case of cyclohexyl methyl ketone, the carbonyl group can enhance the electrophilicity of adjacent carbon atoms, making them more susceptible to bromination. Understanding this mechanism is crucial for predicting the products formed during the reaction.

Nucleophilic Addition to Carbonyls

Nucleophilic addition to carbonyls involves the attack of a nucleophile on the electrophilic carbon of a carbonyl group, leading to the formation of an alkoxide intermediate. In the presence of sodium hydroxide, hydroxide ions can act as nucleophiles, facilitating the addition to the carbonyl of cyclohexyl methyl ketone. This step is essential for understanding how the reaction proceeds in the presence of a base.

Mechanism of Base-Catalyzed Reactions

Base-catalyzed reactions often involve the deprotonation of acidic protons, which can stabilize intermediates or enhance nucleophilicity. In this reaction, sodium hydroxide not only acts as a base but also helps in the formation of more reactive species, such as enolates, which can further react with bromine. Grasping this concept is vital for predicting the overall reaction pathway and the final products.

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Related Practice

Textbook Question

Predict the products of the following reactions.

(b) 1-phenylethanol + excess I₂ in base

Textbook Question

Which compounds will give positive iodoform tests?

(a) 1-phenylethanol

(b) pentan-2-one

(d) pentan-3-one

(e) acetone

(f) isopropyl alcohol

Textbook Question

Predict the products of the following reactions.

(a) cyclopentyl methyl ketone + excess Cl₂ + excess NaOH

(b) 1-cyclopentylethanol + excess I₂ + excess NaOH

Textbook Question

Show how you would accomplish the following conversions in good yields. You may use any necessary reagents.

(c)

Multiple Choice

Provide the major product for the following reaction.

Textbook Question

In the presence of excess base and excess halogen, a methyl ketone is converted to a carboxylate ion. The reaction is known as the haloform reaction because one of the products is haloform (chloroform, bromoform, or iodoform). Before spectroscopy became a routine analytical tool, the haloform reaction served as a test for methyl ketones: the formation of iodoform, a bright yellow compound, signaled that a methyl ketone was present. Why do only methyl ketones form a haloform?

Textbook Question

Predict the products of the following reactions.

(b) 1-phenylethanol + excess I₂ in base

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