Textbook Question
Some of the following examples can show geometric isomerism, and some cannot. For the ones that can, draw all the geometric isomers, and assign complete names using the E-Z system.
(i)
Ch. 7 - Structure and Synthesis of Alkenes; Elimination
Wade9th EditionOrganic ChemistryISBN: 9780135213728
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Table of contents
- Ch.1 - Structure and Bonding107
- Ch. 2 - Acids and Bases; Functional Groups115
- Ch.3 - Structure and Stereochemistry of Alkanes100
- Ch.4 - The Study of Chemical Reactions99
- Ch.5 - Stereochemistry93
- Ch.6 - Alkyl Halides; Nucleophilic Substitution118
- Ch. 7 - Structure and Synthesis of Alkenes; Elimination158
- Ch.8 - Reactions of Alkenes171
- Ch.9 - Alkynes91
- Ch.10 - Structure and Synthesis of Alcohols143
- Ch.11 - Reactions of Alcohols104
- Ch. 12 - Infrared Spectroscopy and Mass Spectrometry22
- Ch. 13 - Nuclear Magnetic Resonance Spectroscopy45
- Ch. 14 - Ethers, Epoxides, and Thioethers72
- Ch. 15 - Conjugated Systems, Orbital Symmetry, and Ultraviolet Spectroscopy89
- Ch. 16 - Aromatic Compounds74
- Ch. 17 - Reactions of Aromatic Compounds126
- Ch. 18 - Ketones and Aldehydes193
- Ch. 19 - Amines129
- Ch. 20 - Carboxylic Acids77
- Ch. 21 - Carboxylic Acid Derivatives141
- Ch. 22 - Condensations and Alpha Substitutions of Carbonyl Compounds175
- Ch. 23 - Carbohydrates and Nucleic Acids93
- Ch. 24 - Amino Acids, Peptides, and Proteins54
Chapter 7, Problem 9a
How many stereogenic double bonds are in octa-1,3,5-triene? How many stereocenters are there? Draw and name the four stereoisomers of octa-1,3,5-triene.
Verified step by step guidance1
Step 1: Identify the structure of octa-1,3,5-triene. This compound is an eight-carbon chain with three double bonds located at positions 1, 3, and 5. Draw the structure to visualize the molecule.
Step 2: Determine the stereogenic double bonds. A stereogenic double bond is a double bond that can exhibit cis-trans (E/Z) isomerism. For this to occur, each carbon in the double bond must have two different substituents. Analyze the double bonds at positions 1, 3, and 5 to see if they meet this criterion.
Step 3: Count the stereocenters. A stereocenter is a carbon atom bonded to four different groups. Examine the structure of octa-1,3,5-triene to identify any such carbons. Note that stereogenic double bonds are not stereocenters, as they involve sp2-hybridized carbons.
Step 4: Draw the four stereoisomers of octa-1,3,5-triene. Since there are stereogenic double bonds, assign E/Z configurations to each double bond. Combine these configurations to generate all possible stereoisomers (e.g., E,E,E; E,E,Z; E,Z,Z; Z,Z,Z).
Step 5: Name each stereoisomer using the E/Z nomenclature. For each isomer, specify the configuration of the double bonds in the name (e.g., (1E,3E,5E)-octa-1,3,5-triene). Ensure the names are consistent with IUPAC rules.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Stereogenic Double Bonds
Stereogenic double bonds are carbon-carbon double bonds that can create geometric isomerism due to the restricted rotation around the double bond. In compounds with multiple double bonds, such as octa-1,3,5-triene, each double bond can potentially contribute to the overall stereochemistry of the molecule, leading to different spatial arrangements of substituents.
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Single bonds, double bonds, and triple bonds.
Stereocenters
Stereocenters are atoms in a molecule, typically carbon, that have four different substituents attached, leading to non-superimposable mirror images, or enantiomers. Identifying stereocenters is crucial for determining the number of possible stereoisomers, as each stereocenter can exist in two configurations (R or S), significantly increasing the complexity of the molecule's stereochemistry.
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Stereoisomers
Stereoisomers are compounds that have the same molecular formula and connectivity of atoms but differ in the spatial arrangement of their atoms. In the case of octa-1,3,5-triene, the presence of stereogenic double bonds and stereocenters allows for the formation of multiple stereoisomers, which can be categorized into cis and trans forms, as well as different configurations at the stereocenters.
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Related Practice
Textbook Question
Draw the six stereoisomers of octa-2,4,6-triene. Explain why there are only six stereoisomers, rather than the eight we might expect for a compound with three stereogenic double bonds.
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Textbook Question
Teflon-coated frying pans routinely endure temperatures that would cause polyethylene or polypropylene to oxidize and decompose. Decomposition of polyethylene is initiated by free-radical abstraction of a hydrogen atom by O2. Bond-dissociation energies of C—H bonds are about 400 kJ/mol, and C—F bonds are about 460 kJ/mol. The BDE of the H—OO bond is about 192 kJ/mol, and the F—OO bond is about 63 kJ/mol. Show why Teflon (Figure 7-5) is much more resistant to oxidation than polyethylene is.
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Textbook Question
For each pair of compounds, predict the one with a higher boiling point. Which compounds have zero dipole moments?
a. cis-1,2-dichloroethene or cis-1,2-dibromoethene
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Textbook Question
Some of the following examples can show geometric isomerism, and some cannot. For the ones that can, draw all the geometric isomers, and assign complete names using the E-Z system.
e. 3-ethyl-5-methyloct-3-ene
f. 3,7-dichloroocta-2,5-diene
Textbook Question
Some of the following examples can show geometric isomerism, and some cannot. For the ones that can, draw all the geometric isomers, and assign complete names using the E-Z system.
(g)
(h)