Textbook Question
We have studied a number of pericyclic reactions previously. Draw the mechanism of the steps shown. The section number where this material was first studied is given for your review.
(a)
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Ch. 22 - Conjugated Systems II: Pericyclic 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
Chapter 21, Problem 3a
Draw the molecular orbital picture for (a) ethene. Label the HOMO and LUMO of each.
Verified step by step guidance1
Step 1: Ethene (C2H4) consists of a double bond between two carbon atoms. The double bond is made up of one sigma (σ) bond and one pi (π) bond. The sigma bond is formed by the overlap of sp² hybrid orbitals, while the pi bond is formed by the sideways overlap of unhybridized p orbitals.
Step 2: To draw the molecular orbital picture for ethene, start by representing the bonding and antibonding molecular orbitals for the π system. The two p orbitals from the carbon atoms combine to form two molecular orbitals: a bonding π orbital (lower energy) and an antibonding π* orbital (higher energy).
Step 3: Label the Highest Occupied Molecular Orbital (HOMO) and Lowest Unoccupied Molecular Orbital (LUMO). In ethene, the bonding π orbital is the HOMO because it contains the electrons from the π bond. The antibonding π* orbital is the LUMO because it is the next available orbital that does not contain electrons.
Step 4: Represent the π bonding orbital as a region of electron density above and below the plane of the molecule, showing constructive overlap of the p orbitals. Represent the π* antibonding orbital with nodes (regions of zero electron density) between the carbon atoms, indicating destructive overlap.
Step 5: Ensure the molecular orbital diagram clearly shows the energy levels of the π and π* orbitals, with the π orbital lower in energy than the π* orbital. Label the HOMO and LUMO explicitly in the diagram to complete the molecular orbital picture for ethene.
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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 ethene (C2H4), the π bonds are created from the overlap of p orbitals, resulting in bonding and antibonding molecular orbitals. Understanding how these orbitals are formed and their energy levels is crucial for analyzing the electronic structure of the molecule.
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Review of Molecular Orbitals
HOMO and LUMO
The Highest Occupied Molecular Orbital (HOMO) is the molecular orbital that contains the highest energy electrons, while the Lowest Unoccupied Molecular Orbital (LUMO) is the lowest energy orbital that is unoccupied. In ethene, the HOMO is typically a π orbital, and the LUMO is the corresponding π* antibonding orbital. Identifying these orbitals is essential for predicting reactivity and electronic transitions.
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Ethene Structure
Ethene, or ethylene, is a simple alkene with a double bond between two carbon atoms. This double bond consists of one σ bond and one π bond, which influences the molecule's geometry and reactivity. Understanding the structure of ethene is fundamental for drawing its molecular orbital diagram and labeling the HOMO and LUMO accurately.
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Related Practice
Textbook Question
Draw two important resonance structures involving the C―O π bond for the molecule shown. To which carbons would you expect a nucleophile to add?
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Textbook Question
Draw two important resonance structures involving the lone pair on oxygen for the molecule shown. Which carbons are most likely to act as nucleophiles?
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Textbook Question
What frequency of light would be required to excite an electron if the HOMO–LUMO energy gap was 33.9 kcal/mol (142 kJ/mol)?
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Textbook Question
The SN2 reaction is the concerted, backside displacement of a good leaving group by a nucleophile. Why do nucleophiles attack from the back in SN2 reactions?
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