Textbook Question
Suggest a reagent and a reactant that could be combined to make each of the following alcohols.
(d)
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Ch. 13 - Alcohols, Ethers and Related Compounds: Substitution and Elimination
Mullins1st EditionOrganic Chemistry: A Learner Centered ApproachISBN: 9780137566471
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Table of contents
- Ch. 2 - General Chemistry Translated: Finding the Electrons114
- Ch. 3 - Alkanes and Cycloalkanes: Properties and Conformational Analysis109
- Ch. 4 - Acids and Bases: Electron Flow144
- Ch. 5 - Chemical Reaction Analysis: Thermodynamics and Kinetics118
- Ch. 6 - Stereoisomerism: Arrangement of Atoms in Space112
- Ch. 7 - Principles of Green (Organic) Chemistry3
- Ch. 8 - Alkenes I: Properties and Electrophilic Additions158
- Ch. 9 - Alkenes II: Oxidation and Reduction150
- Ch. 10 - Alkynes: Electrophilic Addition and Redox Reactions101
- Ch. 11 - Properties and Synthesis of Alkyl Halides: Radical Reactions108
- Ch. 12 - Substitution and Elimination: Reactions of Haloalkanes172
- Ch. 13 - Alcohols, Ethers and Related Compounds: Substitution and Elimination239
- Ch. 14 - Structural Identification I: Infrared Spectroscopy and Mass Spectrometry54
- Ch. 15 - Structural Identification II: Nuclear Magnetic Resonance Spectroscopy93
- Ch. 16 - Metals in Organic Chemistry48
- Ch. 17 - Carbonyl Addition Reactions: Aldehydes and Ketones45
- Ch. 18 - Nucleophilic Acyl Substitution I: Carboxylic Acids18
- Ch. 19 - Nucleophilic Acyl Substitution II: Carboxylic Acid Derivatives39
- Ch. 20 - Enolates: Carbonyl Addition and Substitution13
- Ch. 21 - Conjugated Systems I: Stability and Addition Reactions56
- Ch. 22 - Conjugated Systems II: Pericyclic Reactions54
- Ch. 23 - Benzene I: Aromatic Stability and Substitution Reactions64
- Ch. 24 - Benzene II: Reactions Influenced by the Aromatic Ring62
- Ch. 25 - Amines: Structure, Reactions, and Synthesis27
- Ch. 26 - Amino Acids, Proteins, and Peptide Synthesis9
- Ch. 27 - Carbohydrates, Nucleic Acids, and Lipids54
Mullins1st EditionOrganic Chemistry: A Learner Centered ApproachISBN: 9780137566471Not the one you use?Change textbook
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Mullins 1st EditionCh. 13 - Alcohols, Ethers and Related Compounds: Substitution and EliminationProblem 18
Mullins 1st EditionCh. 13 - Alcohols, Ethers and Related Compounds: Substitution and EliminationProblem 18Chapter 12, Problem 18
When a secondary haloalkane is treated with sodium ethoxide in ethanol, we predict alkene formation over ether formation. How did we make this determination?
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Identify the type of reaction: Sodium ethoxide (NaOCH₂CH₃) in ethanol is a strong base and a good nucleophile. This suggests that the reaction could proceed via either an elimination (E2) or substitution (SN2) mechanism.
Analyze the substrate: A secondary haloalkane is sterically hindered, which makes it less favorable for a nucleophile to attack the carbon bearing the leaving group (SN2 mechanism). This steric hindrance favors elimination (E2) over substitution.
Consider the reaction conditions: Sodium ethoxide is a strong base, and ethanol is a polar protic solvent. These conditions favor the E2 elimination mechanism, as strong bases promote the removal of a β-hydrogen to form an alkene.
Predict the major product: In an E2 reaction, the β-hydrogen is abstracted by the base, and the leaving group departs simultaneously, leading to the formation of a double bond (alkene). Ether formation would require an SN2 mechanism, which is less likely under these conditions.
Conclude the reasoning: The combination of a strong base, polar protic solvent, and steric hindrance of the secondary haloalkane makes alkene formation via E2 elimination the predominant pathway over ether formation.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Nucleophilic Substitution vs. Elimination
In organic chemistry, reactions involving haloalkanes can proceed via nucleophilic substitution (SN) or elimination (E) mechanisms. Secondary haloalkanes can undergo both SN2 and E2 reactions. The choice between these pathways depends on factors such as the strength of the nucleophile, the solvent, and the structure of the substrate.
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00:38
Intro to Substitution/Elimination Problems
Role of Sodium Ethoxide
Sodium ethoxide (NaOEt) is a strong base and a good nucleophile. In the presence of a secondary haloalkane, it can promote elimination reactions, leading to alkene formation. The basicity of sodium ethoxide favors the E2 mechanism, especially in a protic solvent like ethanol, which can stabilize the transition state of the elimination pathway.
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03:19Sodium Alkynide Alkylation
Solvent Effects on Reaction Pathways
The choice of solvent significantly influences the reaction mechanism. Ethanol, being a polar protic solvent, can stabilize ions and transition states. In this case, it favors elimination over substitution because the solvent can stabilize the alkene product and the leaving group, thus promoting the formation of alkenes from secondary haloalkanes.
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02:26General format of reactions and how to interpret solvents.
Related Practice
Textbook Question
The intended SN2 displacement of the 1° chloride by acetylide is unsuccessful for the molecule below. Why?
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Textbook Question
Which of the following bases would favorably deprotonate a hydroxyl group?
(a)
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Textbook Question
In contrast to the results of Assessment 13.18, when a secondary haloalkane is treated with sodium ethanethiolate, we predict formation of a thioether. How is this rationalized?
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Textbook Question
Rationalize the difference in pKₐ values for the two hydroxyl groups.
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Textbook Question
Suggest an arrow-pushing mechanism for the following reaction. Is the hydroxyl group acting as a base, acid, Lewis base, or Lewis acid?