Textbook Question
Draw the Lewis structure of the methoxide ion ( CH3O-). Draw the Lewis structure of a proton (H+) . To which atom of methoxide would you expect a proton to add?
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Ch. 4 - Acids and Bases: Electron Flow
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
Chapter 3, Problem 2
Which is more stable, free hydrogen atoms or the diatomic H2 molecule? Why do you know this to be true?
Verified step by step guidance1
Understand the question: The problem is asking about the relative stability of free hydrogen atoms (H) versus the diatomic hydrogen molecule (H₂). Stability in this context is related to energy—lower energy corresponds to greater stability.
Recall the concept of bond formation: When two hydrogen atoms combine to form an H₂ molecule, they form a covalent bond by sharing their electrons. Bond formation releases energy, which means the system moves to a lower energy state.
Apply the principle of energy minimization: The H₂ molecule is more stable than free hydrogen atoms because the energy released during bond formation makes the molecule lower in energy compared to the separate atoms.
Consider experimental evidence: The bond dissociation energy of H₂ (the energy required to break the H-H bond) is a positive value, indicating that energy must be supplied to break the bond and separate the molecule into free hydrogen atoms. This further supports that H₂ is more stable.
Conclude based on thermodynamics: The diatomic H₂ molecule is more stable than free hydrogen atoms because the formation of the H-H bond lowers the system's energy, and systems naturally favor lower-energy, more stable configurations.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Bond Formation and Stability
The stability of a molecule is often determined by the presence of chemical bonds. In the case of diatomic hydrogen (H₂), two hydrogen atoms share a pair of electrons, forming a covalent bond. This bond lowers the energy of the system compared to free hydrogen atoms, which exist as individual, unbonded entities. Thus, the formation of H₂ results in a more stable configuration.
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Energy Considerations
Chemical stability is closely related to energy levels. Free hydrogen atoms have higher potential energy due to their unbonded state. When they form H₂, energy is released, indicating that the H₂ molecule is at a lower energy state. This release of energy during bond formation is a key factor in determining the stability of molecules.
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Thermodynamics and Reaction Favorability
Thermodynamics helps explain why certain reactions occur spontaneously. The formation of H₂ from free hydrogen atoms is favored because it leads to a decrease in Gibbs free energy, making the process thermodynamically favorable. This principle underlines the preference for stable molecular forms over individual atoms in chemical systems.
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