Textbook Question
Long, polarized bonds are also reducible in the same way that the C―C π bond of an alkyne is. Show a mechanism by which a C―Br bond might be reduced by sodium metal.
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Ch. 10 - Alkynes: Electrophilic Addition and Redox Reactions
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. 10 - Alkynes: Electrophilic Addition and Redox ReactionsProblem 27a
Mullins 1st EditionCh. 10 - Alkynes: Electrophilic Addition and Redox ReactionsProblem 27aChapter 9, Problem 27a
The following deprotonation step occurs during the trans reduction of an alkyne. Calculate Keq for this reaction.
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Identify the acids and bases in the reaction. The alkyne (CH≡CH) is the acid being deprotonated, and ammonia (NH₃) is the base. The products are the acetylide anion (C≡C⁻) and the ammonium ion (NH₂⁻).
Determine the pKa values of the acid and conjugate acid. The pKa of CH≡CH is approximately 25, and the pKa of NH₃ is approximately 38.
Use the relationship between pKa values and equilibrium constants (Keq). The formula is: Keq = 10^(pKa(conjugate acid) - pKa(acid)).
Substitute the pKa values into the formula: Keq = 10^(38 - 25).
Interpret the result qualitatively. Since the pKa of NH₃ is much higher than that of CH≡CH, the equilibrium strongly favors the products (C≡C⁻ and NH₂⁻).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Deprotonation
Deprotonation is the removal of a proton (H+) from a molecule, resulting in the formation of a conjugate base. In organic chemistry, this process is crucial for understanding acid-base reactions, particularly in the context of alkynes, which can act as weak acids due to the acidity of their terminal hydrogen. The ability to deprotonate an alkyne is essential for subsequent reactions, such as reductions.
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Equilibrium Constant (K_eq)
The equilibrium constant (K_eq) is a numerical value that expresses the ratio of the concentrations of products to reactants at equilibrium for a given chemical reaction. It provides insight into the extent of a reaction and whether it favors the formation of products or reactants. Calculating K_eq involves using the concentrations of the species involved in the reaction at equilibrium, which is vital for understanding the thermodynamics of the deprotonation step.
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02:19The relationship between equilibrium constant and pKa.
Trans Reduction of Alkynes
The trans reduction of alkynes refers to the conversion of alkynes into trans-alkenes through the addition of hydrogen (H2) in the presence of a catalyst, typically a metal such as palladium or platinum. This reaction is significant in organic synthesis as it allows for the selective formation of trans isomers, which have distinct physical and chemical properties compared to their cis counterparts. Understanding this reduction process is essential for predicting the outcome of the deprotonation step in the reaction.
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Related Practice
Textbook Question
Suggest alkynes that might be used to make the following trans-alkenes.
(a)
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Textbook Question
Suggest alkynes that might be used to make the following trans-alkenes.
(c)
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Textbook Question
When sodium generates electrons in the presence of ammonia, these electrons persist in solution, giving the blue color. However, electrons do not persist when sodium is added to water. Why?
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Textbook Question
Suggest alkynes that might be used to make the following trans-alkenes.
(b)
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Textbook Question
Sodium amide, the base we use to deprotonate terminal alkynes, is synthesized by reducing ammonia via a mechanism similar to the reduction of alkynes in Figure 10.21. Suggest a mechanism for this reaction.
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