Textbook Question
Predict the products of the following reactions.
(a) cyclopentyl methyl ketone + excess Cl2 + excess NaOH
(b) 1-cyclopentylethanol + excess I2 + excess NaOH
(c) propiophenone + excess Br2 + excess NaOH
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Ch. 22 - Condensations and Alpha Substitutions of Carbonyl Compounds
Wade9th EditionOrganic ChemistryISBN: 9780135213728
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Table of contents
- Ch.1 - Structure and Bonding107
- Ch. 2 - Acids and Bases; Functional Groups115
- Ch.3 - Structure and Stereochemistry of Alkanes100
- Ch.4 - The Study of Chemical Reactions99
- Ch.5 - Stereochemistry93
- Ch.6 - Alkyl Halides; Nucleophilic Substitution118
- Ch. 7 - Structure and Synthesis of Alkenes; Elimination158
- Ch.8 - Reactions of Alkenes171
- Ch.9 - Alkynes91
- Ch.10 - Structure and Synthesis of Alcohols143
- Ch.11 - Reactions of Alcohols104
- Ch. 12 - Infrared Spectroscopy and Mass Spectrometry22
- Ch. 13 - Nuclear Magnetic Resonance Spectroscopy45
- Ch. 14 - Ethers, Epoxides, and Thioethers72
- Ch. 15 - Conjugated Systems, Orbital Symmetry, and Ultraviolet Spectroscopy89
- Ch. 16 - Aromatic Compounds74
- Ch. 17 - Reactions of Aromatic Compounds126
- Ch. 18 - Ketones and Aldehydes193
- Ch. 19 - Amines129
- Ch. 20 - Carboxylic Acids77
- Ch. 21 - Carboxylic Acid Derivatives141
- Ch. 22 - Condensations and Alpha Substitutions of Carbonyl Compounds175
- Ch. 23 - Carbohydrates and Nucleic Acids93
- Ch. 24 - Amino Acids, Peptides, and Proteins54
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Wade 9th EditionCh. 22 - Condensations and Alpha Substitutions of Carbonyl CompoundsProblem 15
Wade 9th EditionCh. 22 - Condensations and Alpha Substitutions of Carbonyl CompoundsProblem 15Chapter 22, Problem 15
Propose a mechanism for the acid-catalyzed bromination of pentan-3-one.
Verified step by step guidance1
Step 1: Begin by identifying the functional group in pentan-3-one. It is a ketone, which contains a carbonyl group (C=O). The carbonyl group is reactive due to the polarization of the C=O bond, making the carbon atom electrophilic.
Step 2: In the presence of an acid catalyst, the carbonyl oxygen atom is protonated by the acid (H⁺). This increases the electrophilicity of the carbonyl carbon, making it more susceptible to nucleophilic attack. Represent this step as:
Step 3: The protonated ketone undergoes enolization. This involves the migration of a proton from an adjacent alpha-carbon to the oxygen atom, forming an enol intermediate. The enol form is characterized by a hydroxyl group (-OH) attached to a double-bonded carbon. Represent this as:
Step 4: Bromine (Br₂) reacts with the enol intermediate. The double bond in the enol acts as a nucleophile and attacks the bromine molecule, leading to the formation of a bromonium ion intermediate. This step is followed by the loss of a proton to regenerate the ketone structure with a bromine atom attached to the alpha-carbon. Represent this as:
Step 5: Finally, the acid catalyst is regenerated, completing the catalytic cycle. The product of the reaction is 3-bromo-pentan-3-one, where bromine has been added to the alpha-carbon adjacent to the carbonyl group. This step ensures the reaction is catalytic in acid.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Acid-Catalyzed Reactions
Acid-catalyzed reactions involve the use of an acid to increase the rate of a chemical reaction. In organic chemistry, acids can protonate substrates, making them more electrophilic and susceptible to nucleophilic attack. Understanding how acids interact with organic molecules is crucial for proposing mechanisms, as they often dictate the pathway and intermediates formed during the reaction.
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Acid Catalyzed
Bromination Mechanism
Bromination is the process of adding bromine to an organic compound, which can occur through various mechanisms, including electrophilic addition. In the case of ketones like pentan-3-one, the mechanism typically involves the formation of a bromonium ion or a similar intermediate, where bromine acts as an electrophile. Recognizing the steps involved in bromination helps in accurately proposing the reaction mechanism.
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Keto-Enol Tautomerism
Keto-enol tautomerism is a chemical equilibrium between a keto form (a carbonyl compound) and its corresponding enol form (an alcohol with a double bond). In the context of pentan-3-one, the enol form can be more reactive towards electrophiles like bromine. Understanding this tautomerism is essential for predicting which form will participate in the bromination reaction and how the mechanism will unfold.
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Related Practice
Textbook Question
Which compounds will give positive iodoform tests?
(a) 1-phenylethanol
(b) pentan-2-one
(c) pentan-2-ol
(d) pentan-3-one
(e) acetone
(f) isopropyl alcohol
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Textbook Question
Propose a mechanism for the reaction of cyclohexyl methyl ketone with excess bromine in the presence of sodium hydroxide.
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Textbook Question
Propose a mechanism for the aldol condensation of cyclohexanone. Do you expect the equilibrium to favor the reactant or the product?
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Textbook Question
Show the products of the reactions of these carboxylic acids with PBr3/Br2 before and after hydrolysis.
(a) pentanoic acid
(b) phenylacetic acid
(c) succinic acid
(d) oxalic acid
Textbook Question
Acid-catalyzed halogenation is synthetically useful for converting ketones to α,β-unsaturated ketones, which are useful in Michael reactions (Section 22-18). Propose a method for converting cyclohexanone to cyclohex-2-en-1-one, an important synthetic starting material.
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