EBK ORGANIC CHEMISTRY
6th Edition
ISBN: 8220103151757
Author: LOUDON
Publisher: MAC HIGHER
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Chapter 15, Problem 15.34P
Interpretation Introduction
(a)
Interpretation:
The
Concept introduction:
The
Interpretation Introduction
(b)
Interpretation:
The
Concept introduction:
The
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The pentadienyl radical, H2C“CH¬CH“CH¬CH2#, has its unpaired electron delocalized over three carbon atoms.(a) Use resonance forms to show which three carbon atoms bear the unpaired electron.(b) How many MOs are there in the molecular orbital picture of the pentadienyl radical?(c) How many nodes are there in the lowest-energy MO of the pentadienyl system? How many in the highest-energy MO?(d) Draw the MOs of the pentadienyl system in order of increasing energy. (continued)762 CHAPTER 15 Conjugated Systems, Orbital Symmetry, and Ultraviolet Spectroscopy(e) Show how many electrons are in each MO for the pentadienyl radical (ground state).(f) Show how your molecular orbital picture agrees with the resonance picture showing delocalization of the unpairedelectron onto three carbon atoms.(g) Remove the highest-energy electron from the pentadienyl radical to give the pentadienyl cation. Which carbon atomsshare the positive charge? Does this picture agree with the resonance picture?(h) Add an…
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?
The pentadienyl radical, H2C“CH¬CH“CH¬CH2#, has its unpaired electron delocalized over three carbon atoms.(a) Use resonance forms to show which three carbon atoms bear the unpaired electron.(b) How many MOs are there in the molecular orbital picture of the pentadienyl radical?
Chapter 15 Solutions
EBK ORGANIC CHEMISTRY
Ch. 15 - Prob. 15.1PCh. 15 - Prob. 15.2PCh. 15 - Prob. 15.3PCh. 15 - Prob. 15.4PCh. 15 - Prob. 15.5PCh. 15 - Prob. 15.6PCh. 15 - Prob. 15.7PCh. 15 - Prob. 15.8PCh. 15 - Prob. 15.9PCh. 15 - Prob. 15.10P
Ch. 15 - Prob. 15.11PCh. 15 - Prob. 15.12PCh. 15 - Prob. 15.13PCh. 15 - Prob. 15.14PCh. 15 - Prob. 15.15PCh. 15 - Prob. 15.16PCh. 15 - Prob. 15.17PCh. 15 - Prob. 15.18PCh. 15 - Prob. 15.19PCh. 15 - Prob. 15.20PCh. 15 - Prob. 15.21PCh. 15 - Prob. 15.22PCh. 15 - Prob. 15.23PCh. 15 - Prob. 15.24PCh. 15 - Prob. 15.25PCh. 15 - Prob. 15.26PCh. 15 - Prob. 15.27PCh. 15 - Prob. 15.28PCh. 15 - Prob. 15.29PCh. 15 - Prob. 15.30PCh. 15 - Prob. 15.31PCh. 15 - Prob. 15.32PCh. 15 - Prob. 15.33PCh. 15 - Prob. 15.34PCh. 15 - Prob. 15.35PCh. 15 - Prob. 15.36PCh. 15 - Prob. 15.37PCh. 15 - Prob. 15.38PCh. 15 - Prob. 15.39PCh. 15 - Prob. 15.40PCh. 15 - Prob. 15.41PCh. 15 - Prob. 15.42APCh. 15 - Prob. 15.43APCh. 15 - Prob. 15.44APCh. 15 - Prob. 15.45APCh. 15 - Prob. 15.46APCh. 15 - Prob. 15.47APCh. 15 - Prob. 15.48APCh. 15 - Prob. 15.49APCh. 15 - Prob. 15.50APCh. 15 - Prob. 15.51APCh. 15 - Prob. 15.52APCh. 15 - Prob. 15.53APCh. 15 - Prob. 15.54APCh. 15 - Prob. 15.55APCh. 15 - Prob. 15.56APCh. 15 - Prob. 15.57APCh. 15 - Prob. 15.58APCh. 15 - Prob. 15.59APCh. 15 - Prob. 15.60APCh. 15 - Prob. 15.61APCh. 15 - Prob. 15.62APCh. 15 - Prob. 15.63APCh. 15 - Prob. 15.64APCh. 15 - Prob. 15.65APCh. 15 - Prob. 15.66APCh. 15 - Prob. 15.67APCh. 15 - Prob. 15.68APCh. 15 - Prob. 15.69APCh. 15 - Prob. 15.70APCh. 15 - Prob. 15.71APCh. 15 - Prob. 15.72APCh. 15 - Prob. 15.73APCh. 15 - Prob. 15.74APCh. 15 - Prob. 15.75APCh. 15 - Prob. 15.76APCh. 15 - Prob. 15.77APCh. 15 - Prob. 15.78APCh. 15 - Prob. 15.79APCh. 15 - Prob. 15.80APCh. 15 - Prob. 15.81APCh. 15 - Prob. 15.82APCh. 15 - Prob. 15.83APCh. 15 - Prob. 15.84APCh. 15 - Prob. 15.85APCh. 15 - Prob. 15.86AP
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- Sketch 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_forwardWhy does carbon monoxide have a greater bond-dissociation energy than molecular nitrogenarrow_forward(a)The concept of hybridization was first proposed by Linus Pauling in 1931, that explain about the concept of mixing orbitals to form new hybrid orbital suitable for the pairing of electrons to form chemical bonds. Draw the kekule structure of a compound that contains only carbon and hydrogen atoms that has: (1) One sp hybridize carbon, two sp? hybridized carbons and two sp hybridized carbon (ii) Two sp' hybridized carbon and two sp hybridized carbons (ii) Two sp? hybidized carbons and two sp hybydized carbonarrow_forward
- (a) Draw a diagram showing the hybrid atomic orbitals on carbon in ethene (ethylene, H2C=CH2) and the way they overlap to form the C=C bond. Clearly label the orbitals in your diagram. (b) Draw an energy level diagram for the formation of the orbitals in part (a), making sure to include both bonding and anti-bonding orbitals.arrow_forwardTwo useful organic compounds that contain Cl atoms are vinyl chloride (CH2 = CHCI) and chloroethane (CH;CH;CI). Vinyl chloride is the starting material used to prepare poly (vinyl chloride), a plastic used in insulation, pipes, and bottles. Chloroethane (ethyl chloride) is a local anesthetic. Why is the C- Cl bond of vinyl chloride stronger than the C– CIl bond in chloroethane.arrow_forward(c) Calculate the n-electron energy of the pi electrons a pentadiene radicalarrow_forward
- what is the complete lewis structure of acetic anhydride such that all atoms have zero formal charge?arrow_forwardExplain the Carbon-Carbon σ Bond Length in Buta- 1,3-diene ?arrow_forwardDescribe two significant differences between the infrared spectra of ethyl alcohol(CH3CH2OH) and ethylene (CH=CH).arrow_forward
- Specify whether the two structures are resonance contributors to the same resonance hybrid. Be sure to explain your reasoning. If yes, be sure to specify which is preferred and why. H2C=NH2arrow_forwardDHA is a fatty acid derived from fish oil and an abundant fatty acid in vertebrate brains. Hydrogenation of DHA forms docosanoic acid [CH3(CH2)20CO2H], and ozonolysis forms CH3CH2CHO, CH2(CHO)2 (five equivalents), and HCOCH2CH2CO2H. What is the structure of DHA if all double bonds have the Z configuration?arrow_forward4. Octatetraene, C8H10, is an 8-membered linear carbon chain with alternating single and double bonds. As a result, each carbon is left with an unhybridized p-orbital containing one electron. This problem will take you through steps to construct and fill the molecular orbital diagram for octatetraene based on the way these different p-orbitals can come together. a) The various molecular orbitals formed from octatetraene’s 8 unhybridized p-orbitals are shown on the next page. Draw the nodal planes for each molecular orbital. Hint: A nodal plane is a line through the molecule that denotes the sites of destructive interference. An example is shown below with ethylene. b) Based on the nodal planes, arrange the energy levels corresponding to the 8 molecular orbitals on the energy level diagram (label them using the numbering provided on the left side of the molecular orbitals) on the next page, in order of lowest to highest energy. c) Populate your molecular…arrow_forward
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