Textbook Question
Choose the molecule in each pair you'd expect to have the higher boiling point. Explain your reasoning.
(a) eicosane ( C20H42) vs pentadecane
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Ch. 3 - Alkanes and Cycloalkanes: Properties and Conformational Analysis
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. 3 - Alkanes and Cycloalkanes: Properties and Conformational AnalysisProblem 7
Mullins 1st EditionCh. 3 - Alkanes and Cycloalkanes: Properties and Conformational AnalysisProblem 7Chapter 2, Problem 7
Draw the molecular orbital picture for propane. Your picture should clearly show the shape and hybridization of the carbons. Label all σ bonds.
Verified step by step guidance1
Step 1: Begin by identifying the molecular structure of propane (C3H8). Propane consists of three carbon atoms bonded in a chain, with each carbon atom bonded to sufficient hydrogen atoms to satisfy the octet rule.
Step 2: Determine the hybridization of each carbon atom. In propane, all carbon atoms are sp³ hybridized because they form four sigma (σ) bonds each (single bonds only). This hybridization results in tetrahedral geometry around each carbon atom.
Step 3: Draw the molecular orbital picture for propane. Represent each carbon atom with its sp³ hybrid orbitals arranged in a tetrahedral geometry. Show the overlap of sp³ orbitals between adjacent carbon atoms to form C-C σ bonds, and the overlap of sp³ orbitals with hydrogen's 1s orbitals to form C-H σ bonds.
Step 4: Label all σ bonds in the diagram. There are two C-C σ bonds formed by the overlap of sp³ orbitals from adjacent carbon atoms, and eight C-H σ bonds formed by the overlap of sp³ orbitals from carbon atoms with 1s orbitals from hydrogen atoms.
Step 5: Ensure the diagram clearly shows the spatial arrangement of the orbitals and bonds. Highlight the tetrahedral geometry around each carbon atom and the linear arrangement of the three carbon atoms in the propane molecule.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Molecular Orbitals
Molecular orbitals (MOs) are formed by the combination of atomic orbitals when atoms bond together. In propane, the MOs help visualize how electrons are distributed among the carbon and hydrogen atoms. Understanding MOs is crucial for predicting the shape and stability of the molecule, as well as the types of bonds formed.
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Review of Molecular Orbitals
Hybridization
Hybridization is the process by which atomic orbitals mix to form new, equivalent hybrid orbitals that can accommodate bonding. In propane, the carbon atoms undergo sp³ hybridization, resulting in four equivalent sp³ hybrid orbitals that form σ bonds with hydrogen and other carbon atoms, leading to a tetrahedral geometry.
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Sigma (σ) Bonds
Sigma bonds are the strongest type of covalent bond formed by the head-on overlap of atomic orbitals. In propane, each carbon forms σ bonds with three hydrogen atoms and one other carbon atom. These bonds are essential for the molecule's structure and stability, and they can be identified in the molecular orbital diagram.
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Related Practice
Textbook Question
How many hydrogens would you expect a 24-carbon compound from each of the following molecular classes to have?
(a) Alkane
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Textbook Question
Choose the molecule in each pair you'd expect to have the higher boiling point. Explain your reasoning.
(b)
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Textbook Question
How many hydrogens would you expect a 24-carbon compound from each of the following molecular classes to have?
(c) Alkyne
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Textbook Question
How many hydrogens would you expect a 24-carbon compound from each of the following molecular classes to have?
(b) Alkene
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Textbook Question
Rank the following molecules from least water soluble to most water soluble. Explain your reasoning.
(a) H3C ― OH
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