교과서 질문
Explain why an amide ion cannot be used to form a carbanion from an alkane in a reaction that favors products.
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Ch. 7 - The Reactions of Alkynes • An Introduction to Multistep Synthesis
Bruice8th EditionOrganic ChemistryISBN: 9780135213711
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TOC.tableOfContents
- Ch. 1 - Remembering General Chemistry: Electronic Structure and Bonding31
- 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
모든 교과서
Bruice 8th EditionCh. 7 - The Reactions of Alkynes • An Introduction to Multistep Synthesis문제 21
Bruice 8th EditionCh. 7 - The Reactions of Alkynes • An Introduction to Multistep Synthesis문제 217장, 문제 21
Which of the following bases can remove a proton from a terminal alkyne in a reaction that favors products?
검증된 단계별 안내1
Step 1: Understand the acidity of terminal alkynes. Terminal alkynes have a hydrogen atom attached to the sp-hybridized carbon. This hydrogen is relatively acidic compared to hydrogens attached to sp2 or sp3 carbons due to the high s-character of the sp orbital, which stabilizes the negative charge on the conjugate base.
Step 2: Recall the concept of acid-base reactions. For a base to remove a proton from a terminal alkyne in a reaction that favors products, the base must be strong enough to deprotonate the alkyne. The reaction will favor products if the conjugate acid of the base is weaker (less acidic) than the terminal alkyne.
Step 3: Compare the pKa values. Terminal alkynes typically have a pKa of around 25. Bases with conjugate acids having pKa values higher than 25 are strong enough to deprotonate terminal alkynes. For example, CH3O− (methoxide) has a conjugate acid (CH3OH) with a pKa of ~16, making it too weak to favor the reaction. NH3 (ammonia) has a conjugate acid (NH4+) with a pKa of ~9, also too weak.
Step 4: Evaluate the bases provided. CH3CH2− (ethyl anion) has a conjugate acid (ethane) with a pKa of ~50, making it strong enough to deprotonate a terminal alkyne. HC≡C− (acetylide ion) has a conjugate acid (acetylene) with a pKa of ~25, which is comparable to the terminal alkyne, so it can also deprotonate the alkyne. F− (fluoride ion) has a conjugate acid (HF) with a pKa of ~3, making it too weak.
Step 5: Conclude which bases can deprotonate the terminal alkyne. Based on the analysis, CH3CH2− and HC≡C− are strong enough bases to remove a proton from a terminal alkyne in a reaction that favors products.
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Acidity of Terminal Alkynes
Terminal alkynes, such as HC≡CH, are weak acids due to the presence of a hydrogen atom bonded to a sp-hybridized carbon. The acidity is attributed to the high s-character of the sp orbital, which holds the electrons closer to the nucleus, stabilizing the negative charge on the conjugate base (alkynide ion) formed after deprotonation.
추천 영상:
02:43
Anti-Markovnikov addition of alcohols to terminal alkynes yields aldehydes
Basicity of Compounds
The ability of a compound to act as a base is determined by its capacity to accept protons (H+). Strong bases, such as CH3O− (methoxide) and CH3CH2− (ethyl), are more effective at deprotonating terminal alkynes compared to weaker bases like NH3 (ammonia) or F− (fluoride), which are less likely to favorably react in this context.
추천 영상:
01:40Defining meso compounds.
Equilibrium and Reaction Favorability
In organic reactions, the position of equilibrium determines whether products or reactants are favored. A reaction that favors products will have a stronger base that can effectively remove a proton from the terminal alkyne, leading to the formation of a more stable alkynide ion. Understanding the relative strengths of acids and bases is crucial for predicting the outcome of such reactions.
추천 영상:
04:00Determining Acid/Base Equilibrium
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