Lecture 24 Organic Chemistry 1
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1 CEM 232 Organic Chemistry I at Chicago Lecture 24 Organic Chemistry 1 Professor Duncan Wardrop April 6,
2 Which shorthand orbital diagram best represents the LUMO of a dienophile in a Diels-Alder reaction? 1. Step one: determine the number of p AOs in the - system. 2. Step two: draw the same number of unshaded molecular orbitals as p AOs. 3. Assign nodes to MOs. Lowest = 0. Each higher MO has one more node. 4. Shade MOs to reflect node assignment in step Determine # of electrons in - system. Place those electrons into MO according to und, Pauli & Aufbau principles. Self Test Question π 2 π 1 EWG A. B. C. D. E. University of Illinois at Chicago CEM 232, Spring 2010 Slide 2 2
3 Self Test Question University of Which shorthand orbital diagram best represents the LUMO of a dienophile in a Diels-Alder reaction? 1. Step one: determine the number of p AOs in the - system. 2. Step two: draw the same number of unshaded molecular orbitals as p AOs. 3. Assign nodes to MOs. Lowest = 0. Each higher MO has one more node. 4. Shade MOs to reflect node assignment in step Determine # of electrons in - system. Place those electrons into MO according to und, Pauli & Aufbau principles. π 2 π 1 EWG Illinois at Chicago CEM 232, Spring 2010 A. B. C. D. E. Slide 3 3
4 CEM 232 Organic Chemistry I at Chicago Chapter 11 Arenes & Aromaticity Sections You are responsible for sections
5 Classes of ydrocarbons ydrocarbons Aliphatic aleiphar Greek for fat sources were fats and oils only C & atoms framework for fxnl groups generally, non-reactive Aromatic aka: arenes many isolated from plants; benzene cyclic conjugated alkenes Alkanes Alkenes Alkynes only single bonds nomenclature: -ane suffix contain a double bond nomenclature: -ene suffix contain a triple bond nomenclature: -yne suffix at Chicago CEM 232, Spring 2010 Slide 5 5
6 Examples of Aromatic ydrocarbons Aromatic aka: arenes many isolated from plants; impurities were fragrant cyclic conjugated alkenes many are derivatives of benzene resonance: delocalized electrons adjacent double bonds planar (120º bond angles) polyaromatic hydrocarbons (PAs) contain two or more aromatic rings; highly carcinogenic and environmental toxin Arene: aromatic hydrocarbon Aromatic: (not smell) typically based on structure of benzene; we ll define aromaticity later C 3 C C C C C C benzene benzene toluene napthalene pyridine furan N O at Chicago CEM 232, Spring 2010 Slide 6 6
7 Kekule Formulation of Benzene Kekule proposed a cyclic structure for C66 with alternating single and double bonds. at Chicago CEM 232, Spring 2010 Slide 7 7
8 Kekule Formulation of Benzene Later Kekule revised his proposal by suggesting a rapid equilibrium between two equivalent structures. at Chicago CEM 232, Spring 2010 Slide 8 8
9 Kekule Formulation of Benzene owever, this suggested that isomers of the kind shown were possible. Yet, none were ever isolated. X Y X Y at Chicago CEM 232, Spring 2010 Slide 9 9
10 Curious Case of Benzene Furthermore, structural studies did support the Kekule model. Instead of alternating single and double bonds, all of the C-C bonds are the same length. Predicted Actual C C bond length: 150 pm C=C bond length: 134 pm All bonds = 140 pm at Chicago CEM 232, Spring 2010 Slide 10 10
11 Resonance Formulation of Benzene The structure of benzene is best represented, not as an equilibrium between two isomers, but as a resonance hybrid of two Lewis structures. Electrons are not localized in alternating single and double bonds, but are delocalized over all six ring carbon atoms. at Chicago CEM 232, Spring 2010 Slide 11 11
12 Resonance Formulation of Benzene A common representation for benzene resonance is a circle inscribed in a hexane ring (Robinson symbol); the circle represents the 6 delocalized π-electrons. Kekule Robinson Electrons are not localized in alternating single and double bonds, but are delocalized over all six ring carbon atoms. at Chicago CEM 232, Spring 2010 While common, this representation will not be used in this class. The Kekule structure is preferred since it allows us to more easily keep track of electrons and electron-flow in mechanism. Slide 12 12
13 Based upon what you ve already learned about hydrogenation, rank the molecules below in order of increasing heat of hydrogenation ( hydrog = - ). If you ve already read this chapter, do not consider aromaticity when answering this question. Self Test Question Current Assumptions: 1. double bonds = hydrog 2. conjugation = hydrog a b c d A. a,b,c,d B. a,c,b,d C. d,c,b,a D. d,b,c,a E. d,c,a,b University of Illinois at Chicago CEM 232, Spring 2010 Slide 13 13
14 Resonance Energy of Benzene Observed hydrog is 152 kj/mol less than expected for cyclohexatriene Benzene is 152 kj/ mol more stable than expected for cyclohexatriene The extra 152 kj/ mol stability of benzene is called resonance energy at Chicago CEM 232, Spring 2010 Slide 14 14
15 Resonance Energy of Benzene When compared to (Z)-1,3,5-hexatriene, conjugation alone cannot account for the extra stability of benzene. Pt, 3 2 º = 208 kj/mol Pt, 3 2 at Chicago CEM 232, Spring 2010 º = 337 kj/mol initial observation: benzene posses a cyclic conjugation that imparts a significant amount of extra stability benzene and similar compounds with this extra stability as a result of cyclic conjugation are called aromatic; we ll refine this definition later Slide 15 15
16 Self Test Question Now, considering the extra stability imparted by aromaticity (crudely: cyclic conjugation), rank the molecules below in order of increasing heat of hydrogenation ( hydrog = - ). Revised Assumptions: 1. double bonds = hydrog 2. conjugation = hydrog 3. aromatic = more stable = hydrog a b c d A. a,b,c,d B. a,c,b,d C. a,d,c,b D. a,c,d,b E. a,b,d,c University of Illinois at Chicago CEM 232, Spring 2010 Slide 16 16
17 Resonance Energy of Benzene The extra stability of benzene compared to 1,3,5- hexatriene implies that the electrons in the molecular orbitals (MOs) of benzene are lower in energy than the MOs of 1,3,5-hexatriene. We must consider the energies of the MOs for both molecules to explain this mysterious extra stability inherent in benzene. at Chicago CEM 232, Spring 2010 Slide 17 17
18 Molecular Orbitals of Benzene 1. Step one: determine the number of p AOs in the -system. 2. Step two: draw the same number of unshaded molecular orbitals as p AOs. 3. Assign nodes to MOs. Lowest = 0. Each higher MO has one more node. 4. Shade MOs to reflect node assignment in step Determine # of electrons in -system. Place those electrons into MO according to und, Pauli & Aufbau principles. 0 nodes at Chicago CEM 232, Spring 2010 Slide 18 18
19 Fronteir Molecular Orbitals of Benzene 1. Step one: determine the number of p AOs in the -system. 2. Step two: draw the same number of unshaded molecular orbitals as p AOs. 3. Assign nodes to MOs. Lowest = 0. Each higher MO has one more node. 4. Shade MOs to reflect node assignment in step Determine # of electrons in -system. Place those electrons into MO according to und, Pauli & Aufbau principles. 1 nodes 1 nodes 0 nodes at Chicago CEM 232, Spring 2010 Slide 19 19
20 Fronteir Molecular Orbitals of Benzene degnerate orbitals = orbitals with the same number of nodes = orbitals with the same energy 1 nodes 1 nodes 0 nodes at Chicago CEM 232, Spring 2010 Slide 20 20
21 Fronteir Molecular Orbitals of Benzene antibonding MOs (π*) bonding MOs (π) 3 π-bonds = 6 π-electrons at Chicago CEM 232, Spring 2010 Slide 21 21
22 Fronteir Molecular Orbitals of Benzene antibonding MOs (π*) bonding MOs (π) benzene has a closed shell = all bonding MOs are full at Chicago CEM 232, Spring 2010 Slide 22 22
23 Bonding vs. Antibonding MOs Qualitative Observation net bonding = bonding interactions antibonding interactions bonding interaction: constructive interference between adjacent orbitals antibonding interactions: destructive interference between adjacent orbitals more bonding interactions = more stable = lower in energy 6 bonding - 0 antibonding 6 net bonding 4 bonding - 2 antibonding 2 net bonding 2 bonding - 0 antibonding 2 net bonding at Chicago CEM 232, Spring 2010 Slide 23 23
24 Self Test Question What is the net bonding in the MO below? 2 bonding - 4 antibonding 2 antibonding A. 4 bonding B. 2 bonding C. 2 antibonding University of Illinois at Chicago CEM 232, Spring 2010 D. 4 antibonding E. 6 antibonding Slide 24 24
25 Bonding vs. Antibonding MOs Qualitative Observation 6 antibonding 2 antibonding 2 antibonding antibonding MOs (π*) bonding MOs (π) 2 bonding 2 bonding As the energy of the π MOs increases, the number of nodes increases and the net number of bonding interactions decreases. 6 bonding at Chicago CEM 232, Spring 2010 Total bonding interactions for filled molecular orbitals = 10 Slide 25 25
26 Aromaticity Revealed Q: Why is benzene more stable than 1,3,5-hexatriene? A: Benzene contains filled MOs with more net bonding. net bonding interactions for filled MOs = 10 net bonding interactions for filled MOs = 9 1 bonding 2 bonding 2 bonding 3 bonding 5 bonding 6 bonding at Chicago CEM 232, Spring 2010 Only the first three MOs for each are shown. Slide 26 26
27 a 1 bonding Self Test Question Unlike benzene, cyclobutadiene is not stabilized by resonance. Using the MO diagrams below, determine the total number of bonding interactions in all filled orbitals for each molecule. List them in the order a,b. b 0 bonding 0 bonding A. 4,4 B. 4,3 C. 2,3 D. 1,1 E. 0,0 3 bonding 4 bonding University of Illinois at Chicago CEM 232, Spring 2010 Slide 27 27
28 Nomenclature: Monosubstituted Benzene Benzene is considered the parent and comes last in the name. at Chicago CEM 232, Spring 2010 Slide 28 28
29 Frequently Encountered Monosubstituted Benzene Derivatives at Chicago CEM 232, Spring 2010 Slide 29 29
30 Nomenclature: Polysubstituted Benzene 1. Benzene is considered as the parent and comes last in the name. 2. List substituents in alphabetical order in the name. 3. Number ring in the direction that gives lowest locant at first point of difference. Do not base numbering on CIP rules!!! Cl 1 4 F Cl F Br Br 3 2-bromo-1-chloro-4-fluorobenzene locants: 1,2,4 First point of difference locants: 1,3,4 3-bromo-4-chloro-1-fluorobenzene X at Chicago CEM 232, Spring 2010 Slide 30 30
31 Nomenclature: Polysubstituted Benzene 1. Benzene is considered as the parent and comes last in the name. 2. List substituents in alphabetical order in the name. 3. Number ring in the direction that gives lowest locant at first point of difference. Do not base numbering on CIP rules!!! at Chicago CEM 232, Spring 2010 Slide 31 31
32 Nomenclature: Polysubstituted Benzene 4. Common names of benzene derivatives may be used as the parent name. 5. In this case, the substituent that defines the derivative is at C-1. OC 3 C 3 N 2 anisole toluene aniline at Chicago CEM 232, Spring 2010 Slide 32 32
33 Nomenclature: Disubstituted Benzene 6. ortho (o), meta (m) and para (p) may be used to indicated 1,2-; 1,3; and 1,4 substituted benzene, respectively. at Chicago CEM 232, Spring 2010 Slide 33 33
34 Nomenclature: Benzene 7. If the substitution is symmetrical (and ortho, meta & para are not used) the substituents are numbered in alphabetical order. C 3 C 3 C 3 2 N OC 3 3 C O 1-amino-3-methoxybenzene (m-aminomethoxybenzene) 1-tert-butyl-2-ethoxybenzene (o-tert-butylethoxybenzene) at Chicago CEM 232, Spring 2010 Slide 34 34
35 Nomenclature: Benzene as a Substituent 6. When benzene is not the highest priority group, it is named phenyl (substituent name). O C 3 3 C C 3 (S)-1-phenyl-1-propanol (E)-2-methyl-1-phenyl-1-butene at Chicago CEM 232, Spring 2010 Slide 35 35
36 Self Test Question Which IUPAC name is correct? A Cl NO 2 p-chloronitrobenzene C C 3 OC 3 Br 5-bromo-2-methyl-1-methoxybenzene B D Br N 2 C 3 3,5-diethyl-1-methylbenzene C 3 4-bromo-2-methylaniline University of Illinois at Chicago CEM 232, Spring 2010 Slide 36 36
37 CEM 232 Organic Chemistry I at Chicago Next Lecture... Chapter 11: Sections Quiz Next Week... Synthesis Problem Chapter 11 37
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