EBK ORGANIC CHEMISTRY-PRINT COMPANION (
4th Edition
ISBN: 9781119776741
Author: Klein
Publisher: WILEY CONS
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Chapter 4.2, Problem 3ATS
Interpretation Introduction
Interpretation:
The bond line structure and isomers of hexane should be drawn. The parent chain for each isomer should be identified.
Concept Introduction:
Compounds consisting of carbon and hydrogen are known as hydrocarbons. Hydrocarbons are classified as a saturated hydrocarbon and unsaturated hydrocarbon. Saturated hydrocarbons are those hydrocarbons in which a carbon-carbon single bond is present as carbon is linked with four atoms.
Saturated hydrocarbon is known as
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The unsaturation number or degree of unsaturation (U) can be used to determine the number of rings and multiple bonds in a compound from its molecular formula. Given a structure, you can determine the number of hydrogens without having to count them explicitly. Consider three compounds and their degree of unsaturation.
(a) A compound A has the molecular formula C7H13ClN2OC7H13ClN2O. How many rings and/or π bonds does it contain?
Explain how covalent bonds are formed in each of the following compounds in terms of
orbital hybridisation and overlap of orbitals
(i) Ethene, C2H4
(ii) Ethyne, C2H2
These molecules are called constitutional (or structural) isomers because they share the same chemical formula but have different patterns of atomic connectivity. Draw as many constitutional isomers as you can for C5H8 in bond-line notation. It may be helpful to draw a Lewis Structure for one of the isomers first.
Chapter 4 Solutions
EBK ORGANIC CHEMISTRY-PRINT COMPANION (
Ch. 4.2 - Prob. 1LTSCh. 4.2 - Prob. 1PTSCh. 4.2 - Prob. 2PTSCh. 4.2 - Prob. 3ATSCh. 4.2 - Prob. 2LTSCh. 4.2 - Prob. 4PTSCh. 4.2 - Prob. 5ATSCh. 4.2 - Prob. 3LTSCh. 4.2 - Prob. 6PTSCh. 4.2 - Prob. 7ATS
Ch. 4.2 - Prob. 4LTSCh. 4.2 - Prob. 8PTSCh. 4.2 - Prob. 9PTSCh. 4.2 - Prob. 10ATSCh. 4.2 - Prob. 5LTSCh. 4.3 - Prob. 6LTSCh. 4.3 - Prob. 14PTSCh. 4.6 - Prob. 7LTSCh. 4.6 - Prob. 16PTSCh. 4.8 - Prob. 8LTSCh. 4.12 - Prob. 11LTSCh. 4.12 - Prob. 25PTSCh. 4.14 - Prob. 33CCCh. 4.14 - Prob. 34CCCh. 4.14 - Prob. 35CCCh. 4 - Prob. 36PPCh. 4 - Prob. 51PPCh. 4 - Prob. 52PPCh. 4 - Prob. 53PPCh. 4 - Prob. 54PPCh. 4 - Prob. 55PPCh. 4 - Prob. 56PPCh. 4 - Prob. 57PPCh. 4 - Prob. 58PPCh. 4 - Prob. 59PPCh. 4 - Prob. 69ACPCh. 4 - Prob. 73IPCh. 4 - Prob. 76IP
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- 2. Account for the bond angle differences between (i) H-C-H (109.5°) in methane and H-S-H (90°); H-C-H (109.5°) and H-O-H (107.5°) in water The H-S-H has two lone pairs; The H-O-H has two lone pairs The H-S-H has no hybridization at p-orbitals; The H-O-H has two lone pairs The H-S-H has two lone pairs; The H-O-H has no hybridization at p-orbitals The H-S-H has no hybridization at p-orbitals; The H-O-H has no hybridization at p- orbitalsarrow_forwardSketch the ultraviolet absorption spectrum of 1,3-Butadiene, matching peaks to electronic transitions. show that four isolated p orbital on carbon atoms combine to form pi orbitalarrow_forwardDraw the structure of the two tertiary (3°) amines with molecular formula C6H15N that contain a group with 3 carbon atoms in the largest group. You do not have to consider stereochemistry. You do not have to explicitly draw H atoms.arrow_forward
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- Which of the following arrangements of p atomic orbitals gives the highest energy p-bonding molecular orbital of 1,3- butadiene?arrow_forwardIf we extend the formation of the four pi-molecular orbitals to forming pi-molecular orbitals by the combination of the six un-hybridized p-orbitals in 1,3,5-hexatriene? Among those pi- molecular orbitals for 1,3,5-hexatriene, how many electrons are in the lowest-enery pi- molecular orbital? 1,3,5-hexatrienearrow_forwardUsing Hydrogenation Data to Determine the Number of Rings and π Bonds in a Molecule How many rings and π bonds are contained in a compound of molecular formula C8H12 that is hydrogenated to a compound of molecular formula C8H14?arrow_forward
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