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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7:48 minutes

Problem 69b

Textbook Question

Predict the products of the following reactions.
(b) 1-phenylethanol + excess I₂ in base

Verified step by step guidance

Step 1: Recognize the reactants and reaction conditions. The reactants are 1-phenylethanol (Ph-CH(OH)-CH3) and excess iodine (I2) in a basic medium (OH⁻). This is a classic example of the haloform reaction.

Step 2: Understand the mechanism of the haloform reaction. In the presence of excess iodine and base, the methyl group attached to the alcohol is oxidized to a ketone (acetophenone, Ph-CO-CH3). The methyl group is then further converted into a trihalomethyl group (CCl3 or CI3) and subsequently cleaved to form a haloform (CHI3, iodoform).

Step 3: Analyze the intermediate steps. The alcohol group (-OH) is first oxidized to a ketone (acetophenone). Then, the methyl group adjacent to the carbonyl undergoes successive halogenation to form a trihalomethyl group (-CI3).

Step 4: Predict the cleavage products. The trihalomethyl group is cleaved in the presence of base, yielding iodoform (CHI3) and a carboxylate ion (Ph-COO⁻).

Step 5: Summarize the final products. The reaction produces iodoform (CHI3), which is a yellow precipitate, and sodium benzoate (Ph-COO⁻Na⁺) as the carboxylate salt in the basic medium.

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Key Concepts

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

Nucleophilic Substitution

Nucleophilic substitution is a fundamental reaction mechanism in organic chemistry where a nucleophile attacks an electrophile, resulting in the replacement of a leaving group. In the context of 1-phenylethanol reacting with iodine in a basic medium, the hydroxyl group (-OH) can be converted into a better leaving group, facilitating the substitution reaction with iodine.

Oxidation-Reduction Reactions

Oxidation-reduction (redox) reactions involve the transfer of electrons between species, leading to changes in oxidation states. In the reaction of 1-phenylethanol with iodine, the alcohol can undergo oxidation to form a carbonyl compound, while iodine is reduced. Understanding these changes is crucial for predicting the products of the reaction.

Iodination of Alcohols

Iodination of alcohols is a specific reaction where iodine reacts with alcohols in the presence of a base, often leading to the formation of iodoalkanes. In this case, the excess iodine reacts with the hydroxyl group of 1-phenylethanol, resulting in the formation of an iodo derivative. Recognizing the conditions and mechanisms of iodination is essential for predicting the final products.

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Related Practice

Textbook Question

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

(c)

Textbook Question

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

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

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