Textbook Question
What are products of the following reactions?
e.
Ch. 7 - The Reactions of Alkynes • An Introduction to Multistep Synthesis
Bruice8th EditionOrganic ChemistryISBN: 9780135213711
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Table of contents
- Ch. 1 - Remembering General Chemistry: Electronic Structure and Bonding (Part 1)31
- Ch. 1 - Remembering General Chemistry: Electronic Structure and Bonding (Part 2)65
- Ch. 2 - Acids and Bases: Central to Understanding Organic Chemistry84
- Ch. 3 - An Introduction to Organic Compounds:Nomenclature, Physical Properties, and Structure183
- Ch. 4 - Isomers: The Arrangement of Atoms in Space121
- Ch. 5 - Alkenes: Structure, Nomenclature, and an Introduction to Reactivity • Thermodynamics and Kinetics83
- Ch. 6 - The Reactions of Alkenes • The Stereochemistry of Addition Reactions175
- Ch. 7 - The Reactions of Alkynes • An Introduction to Multistep Synthesis119
- Ch. 8 - Delocalized Electrons: Their Effect on Stability, pKa, and the Products of a Reaction • Aromaticity and Electronic Effects: An Introduction to the Reactions of Benzene148
- Ch. 9 - Substitution and Elimination Reactions of Alkyl Halides212
- Ch. 10 - Reactions of Alcohols, Ethers, Epoxides, Amines, and Sulfur-Containing Compounds131
- Ch. 11 - Organometallic Compounds60
- Ch. 12 - Radicals90
- Ch. 13 - Mass Spectrometry; Infrared Spectroscopy; UV/Vis Spectroscopy62
- Ch. 14 - NMR Spectroscopy95
- Ch. 15 - Reactions of Carboxylic Acids and Carboxylic Acid Derivatives123
- Ch. 16 - Reactions of Aldehydes and Ketones • More Reactions of Carboxylic Acid Derivatives141
- Ch. 17 - Reactions at the Alpha-Carbon101
- Ch. 18 - Reactions of Benzene and Substituted Benzenes144
- Ch. 19 - More About Amines • Reactions of Heterocyclic Compounds49
- Ch. 20 - The Organic Chemistry of Carbohydrates80
- Ch. 21 - Amino Acids, Peptides, and Proteins57
- Ch. 22 - Catalysis in Organic Reactions and in Enzymatic Reactions35
- Ch. 23 - The Organic Chemistry of the Coenzymes—Compounds Derived from Vitamins1
- Ch. 28 - Pericyclic Reactions58
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Bruice 8th EditionCh. 7 - The Reactions of Alkynes • An Introduction to Multistep SynthesisProblem 22
Bruice 8th EditionCh. 7 - The Reactions of Alkynes • An Introduction to Multistep SynthesisProblem 22Chapter 8, Problem 22
Explain why an amide ion cannot be used to form a carbanion from an alkane in a reaction that favors products.
Verified step by step guidance1
Step 1: Begin by understanding the nature of the amide ion (NH₂⁻). The amide ion is a strong base, meaning it has a high tendency to accept protons (H⁺) from other molecules. This property is crucial in determining its reactivity in deprotonation reactions.
Step 2: Consider the acidity of alkanes. Alkanes are hydrocarbons with single bonds between carbon atoms, and their hydrogen atoms are bonded to sp³-hybridized carbons. The C-H bond in alkanes is nonpolar and has a very high bond dissociation energy, making the hydrogen atoms extremely weakly acidic.
Step 3: Compare the basicity of the amide ion to the acidity of the alkane. For a reaction to favor the formation of products, the base must be strong enough to deprotonate the acid. However, the amide ion is not strong enough to overcome the very low acidity of the alkane's hydrogen atoms. This means the equilibrium of the reaction would strongly favor the reactants rather than the products.
Step 4: Analyze the stability of the carbanion that would form if deprotonation occurred. Carbanions formed from alkanes are highly unstable because the negative charge is localized on an sp³-hybridized carbon atom, which is not electronegative enough to stabilize the charge. This instability further discourages the formation of a carbanion from an alkane.
Step 5: Conclude that the combination of the weak acidity of alkanes and the instability of the resulting carbanion makes it thermodynamically unfavorable for an amide ion to deprotonate an alkane in a reaction that favors products.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Amide Ion
An amide ion (NH2-) is a strong base and a nucleophile, typically used in organic reactions to deprotonate compounds or to react with electrophiles. However, its strong basicity can lead to side reactions, making it less effective in forming stable carbanions from alkanes, which are generally unreactive due to their saturated nature.
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Amide Nomenclature
Carbanion Stability
Carbanions are negatively charged carbon species that are highly reactive and unstable, especially when derived from alkanes. The stability of a carbanion is influenced by the surrounding groups; for alkanes, the lack of electron-withdrawing groups makes it difficult to stabilize the negative charge, thus making the formation of carbanions unfavorable in reactions that favor products.
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Reaction Favorability
In organic chemistry, the favorability of a reaction is determined by the Gibbs free energy change, which considers both enthalpy and entropy. Reactions that lead to more stable products or lower energy states are favored. Since the formation of a carbanion from an alkane using an amide ion does not lead to a more stable product, the reaction is not favored under standard conditions.
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Related Practice
Textbook Question
Which carbocation is more stable?
a.
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Textbook Question
Any base whose conjugate acid has a pKa greater than ______ can remove a proton from a terminal alkyne to form an acetylide ion (in a reaction that favors products).
2
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Textbook Question
What are products of the following reactions?
f.
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Textbook Question
Which of the following bases can remove a proton from a terminal alkyne in a reaction that favors products?
Textbook Question
Which carbocation is more stable?
b.
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