Lecture 6: September 7, 2018

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1 CM 223 Organic Chemistry I Prof. Chad Landrie Lecture 6: September 7, 2018 Ch. 4: Nomenclature of Cylcoalkanes and their Physical and Chemical Properties ( ) Conformational Isomers of Cycloalkanes ( )

2 i>clicker Question Which set of molecules are conformational isomers? A 3 C C 3 C 3 B 3 C 3 C 3 C C 3 3 C C3 C 3 C 3 C Cl Cl D O O C 3 Br O O Br C 3 Slide 2 Organic Chemistry 1 (CM 223) Lecture 6: September 6

3 Conformational Analysis of Ethane Torsional strain is the increase in potential energy when torsion angles (dihedral angles) are other than 60º (gauche) Staggered conformations are more stable (lower potential energy) than eclipsed. Why? Slide 3 Organic Chemistry 1 (CM 223) Lecture 6: September 6 3

4 Conformational Analysis of Ethane Two hypotheses have been posited that explain the preference for staggered conformations. 1. Steric Repulsions: Electrons in vicinal (adjacent) bonds destabilize (raise the potential energy) in eclipsed conformations due to repulsion; they are closer 2. yperconjugation: Electrons in vicinal (adjacent) bonds are delocalized by overlap between bonding and anti-bonding molecular orbitals Slide 4 Organic Chemistry 1 (CM 223) Lecture 6: September 6 4

5 1. Steric Repulsion maximum electronelectron repulsion eclipsed staggered widely accepted explanation until 2001 electron-electron repulsion is greatest in eclipsed conformation Slide 5 Organic Chemistry 1 (CM 223) Lecture 6: September 6 5

6 2. yperconjugation: Role in Conformations yperconjugation is the donation (transfer) of electrons from a filled orbital to an empty orbital; orbitals must overlap to allow transfer. Eclipsed Staggered σ & σ* weak overlap = no hyperconjugation = no electron delocalization = no additional stabilization σ & σ* strong overlap = hyperconjugation (σ σ*) = delocalized electrons = lower energy (more stable) Slide 6 Organic Chemistry 1 (CM 223) Lecture 6: September 6 6

7 Constructing Molecular Orbitals (2) 1. Determine the valence orbitals (basis set) on each atom involved in the bond. 2. Combine the the basis set of orbitals subtractively or additively to obtain the MOs. 3. Arrange basis set and MOs on a MO-diagram and add electrons to each orbital. Slide 7 Organic Chemistry 1 (CM 223) Lecture 6: September 6 7

8 Constructing Molecular Orbitals (2) 1. Determine the valence orbitals (basis set) on each atom involved in the bond. 2. Combine the the basis set of orbitals subtractively or additively to obtain the MOs. 3. Arrange basis set and MOs on a MO-diagram and add electrons to each orbital. Basis Set (1s) (1s) Slide 8 Organic Chemistry 1 (CM 223) Lecture 6: September 6 8

9 Constructing Molecular Orbitals (2) 1. Determine the valence orbitals (basis set) on each atom involved in the bond. 2. Combine the the basis set of orbitals subtractively or additively to obtain the MOs. 3. Arrange basis set and MOs on a MO-diagram and add electrons to each orbital. Basis Set Combination Molecular Orbitals subtractive combination (destructive interference) antibonding sigma MO (σ*) (1s) (1s) additive combination (constructive interference) bonding sigma MO (σ) Slide 9 Organic Chemistry 1 (CM 223) Lecture 6: September 6 9

10 Energy Constructing Molecular Orbitals (2) 1. Determine the valence orbitals (basis set) on each atom involved in the bond. 2. Combine the the basis set of orbitals subtractively or additively to obtain the MOs. 3. Arrange basis set and MOs on a MO-diagram and add electrons to each orbital. Molecular Orbitals antibonding sigma MO (σ*) (1s) stabilization energy (bond energy) (1s) bonding sigma MO (σ) 2018, 2013, Dr. Chad L. Landrie Slide 10 Organic Chemistry 1 (CM 223) Lecture 6: September 6

11 1. Determine the valence orbitals (basis set) on each atom involved in the bond. 2. Combine the the basis set of orbitals subtractively or additively to obtain the MOs. 3. Arrange basis set and MOs on a MO-diagram and add electrons to each orbital. Basis Set Constructing Molecular Orbitals (C-) C C (2sp 3 ) (1s) Slide 11 Organic Chemistry 1 (CM 223) Lecture 6: September 6

12 Constructing Molecular Orbitals (C-) 1. Determine the valence orbitals (basis set) on each atom involved in the bond. 2. Combine the the basis set of orbitals subtractively or additively to obtain the MOs. 3. Arrange basis set and MOs on a MO-diagram and add electrons to each orbital. Basis Set Combination Molecular Orbitals subtractive combination (destructive interference) C antibonding sigma MO (σ*) C C C (2sp 3 ) (1s) additive combination (constructive interference) C bonding sigma MO (σ) Slide 12 Organic Chemistry 1 (CM 223) Lecture 6: September 6

13 Energy Constructing Molecular Orbitals (C-) 1. Determine the valence orbitals (basis set) on each atom involved in the bond. 2. Combine the the basis set of orbitals subtractively or additively to obtain the MOs. 3. Arrange basis set and MOs on a MO-diagram and add electrons to each orbital. C Molecular Orbitals C antibonding sigma MO (σ*) C stabilization energy (bond energy) C C bonding sigma MO (σ) Slide 13 Organic Chemistry 1 (CM 223) Lecture 6: September 6

14 Energy Constructing Molecular Orbitals (C-) 1. Determine the valence orbitals (basis set) on each atom involved in the bond. 2. Combine the the basis set of orbitals subtractively or additively to obtain the MOs. 3. Arrange basis set and MOs on a MO-diagram and add electrons to each orbital. C Note: 1. The C orbital is a little lower in energy since C is more electronegative and can better stabilize electron density. C stabilization energy (bond energy) C 2. Stabilization energy (bond energy) is less the more different in energy the basis set orbitals are. Slide 14 Organic Chemistry 1 (CM 223) Lecture 6: September 6

15 2. yperconjugation: Role in Conformations yperconjugation is the donation (transfer) of electrons from a filled orbital to an empty orbital; orbitals must overlap to allow transfer. Eclipsed Staggered σ & σ* weak overlap = no hyperconjugation = no electron delocalization = no additional stabilization σ & σ* strong overlap = hyperconjugation (σ σ*) = delocalized electrons = lower energy (more stable) Slide 15 Organic Chemistry 1 (CM 223) Lecture 6: September 6

16 2. yperconjugation: Role in Conformations yperconjugation is the donation (transfer) of electrons from a filled orbital to an empty orbital; orbitals must overlap to allow transfer. Eclipsed Staggered σ & σ* weak overlap = no hyperconjugation = no electron delocalization = no additional stabilization σ & σ* strong overlap = hyperconjugation (σ σ*) = delocalized electrons = lower energy (more stable) Slide 16 Organic Chemistry 1 (CM 223) Lecture 6: September 6

orbital to an empty orbital; orbitals must overlap to allow

Eclipsed Staggered σc σ*c σc σ*c σ & σ* weak overlap = no

additional stabilization σ & σ* strong overlap =

17 2. yperconjugation: Role in Conformations yperconjugation is the donation (transfer) of electrons from a filled orbital to an empty orbital; orbitals must overlap to allow transfer. Eclipsed Staggered σc σ*c σc σ*c σ & σ* weak overlap = no hyperconjugation = no electron delocalization = no additional stabilization σ & σ* strong overlap = hyperconjugation (σ σ*) = delocalized electrons = lower energy (more stable) Slide 17 Organic Chemistry 1 (CM 223) Lecture 6: September 6

18 i>clicker Question A Which picture below accurately depicts the C C antibonding sigma bond in ethylene (C24)? p p π-bond sp 2 sp 2 C C C C Basis Set: C C σ-bond B C C Combination: C C C C C D E C C C C C C Slide 18 Organic Chemistry 1 (CM 223) Lecture 6: September 6

19 Conformational Analysis of Ethane Torsional strain is the increase in potential energy when torsion angles (dihedral angles) are other than 60º (gauche) Staggered conformations are more stable (lower potential energy) than eclipsed. Why? Slide 19 Organic Chemistry 1 (CM 223) Lecture 6: September 6

20 Conformational Analysis of Butane Slide 20 Organic Chemistry 1 (CM 223) Lecture 6: September 6

Steric Strain in Staggered Conformations of Butane Gauche Anti The gauche

The gauche conformation is destabilized by van der Waals strain (also called

Waals strain = destabilization that results from atoms being too close together;

21 Steric Strain in Staggered Conformations of Butane Gauche Anti The gauche conformation of butane is 3 kj/mol less stable than the anti. The gauche conformation is destabilized by van der Waals strain (also called steric strain); repulsive van der Waals force between methyl groups van der Waals strain = destabilization that results from atoms being too close together; nuclear-nuclear & electron-electron repulsions dominate Slide 21 Organic Chemistry 1 (CM 223) Lecture 6: September 6

22 Van der Waals Strain (Steric Strain) conformation of butane with two methyl group eclipsed is the least stable (highest in energy) destabilized by both torsional strain (eclipsed vicinal bonds) and van der Waals strain (atoms close together) Slide 22 Organic Chemistry 1 (CM 223) Lecture 6: September 6

23 Conformations of igher Alkanes anti arrangements of C-C-C-C units all vicinal (adjacent) bonds = gauche or anti minimize torsional strain; minimize steric strain described as zig-zag backbone Slide 23 Organic Chemistry 1 (CM 223) Lecture 6: September 6

24 i>clicker Question Rank the following conformations of butane in order of increasing potential energy. b c a c 1 C3-C3 gauche 1 C3-C3 eclipsed 2 - eclipsed 2 C3- eclipsed 1 - eclipsed b d lowest NRG conformation C3-C3 anti d A. d < a < b < c B. a < d < c < b C. d < b < a < c D. b < c < a < d E. d < a < c < b a Slide 24 Organic Chemistry 1 (CM 223) Lecture 6: September 6

25 Conformational Analysis total energy cost = sum of steric strain and torsional strain 1 kcal = kj Adjacent Groups Relationship strain energy gauche 0 anti no strain components 1 - eclipsed 2 -C3 eclipsed 2 - eclipsed 1 C3-C3 eclipsed C3 gauche 0 C3 C3 gauche 0.9 kcal/mol individual components of total energy cost for each conformation 1(1.0) = 1.0 2(1.4) = 2.8 tot = 3.8 kcal/mol 2(1.0) = 2.0 1(3.1) = 3.1 tot = 5.1 kcal/mol eclipsed 1.0 kcal/mol C3 eclipsed 1.4 kcal/mol C3 C3 eclipsed 3.1 kcal/mol Slide 25 Organic Chemistry 1 (CM 223) Lecture 6: September 6

26 Conformational Analysis total energy cost = sum of steric strain and torsional strain 1 kcal = kj Adjacent Groups Relationship strain energy gauche 0 anti no strain components 1 C3-C3 gauche 1 C3-C3 gauche C3 gauche 0 C3 C3 gauche 0.9 kcal/mol individual components of total energy cost for each conformation 1(0.9) = 0.9 tot = 0.9 kcal/mol 1(0.9) = 0.9 tot = 0.9 kcal/mol eclipsed 1.0 kcal/mol C3 eclipsed 1.4 kcal/mol C3 C3 eclipsed 3.1 kcal/mol Slide 26 Organic Chemistry 1 (CM 223) Lecture 6: September 6

27 i>clicker Question 3 C Determine the total energy cost for the least stable conformer below. a. b. c. d. C 3 C 3 3(1.4) = 3.1 tot = 3.1 kcal/mol 3 C C 3 C 3 3 C 3 C3- eclipsed 2 C3-C3 gauche 1 C3-C3 gauche 2(0.9) = 1.8 tot = 1.8 kcal/mol C 3 C 3 1(0.9) = 0.9 tot = 0.9 kcal/mol 3 CC 3 3 C 1 C3-C3 eclipsed 1 C3- eclipsed 1 - eclipsed 1(3.1) = 3.1 1(1.4) = 1.4 1(1.0) =1.0 tot = 5.5 kcal/mol Adjacent Groups Relationship strain energy gauche 0 C3 gauche 0 C3 C3 gauche 0.9 kcal/mol eclipsed 1.0 kcal/mol C3 eclipsed 1.4 kcal/mol C3 C3 eclipsed 3.1 kcal/mol A. 3.1 B. 4.5 C. 3.6 D. 0.9 E. 5.5 Slide 27 Organic Chemistry 1 (CM 223) Lecture 6: September 6

28 Strain energy: additional potential energy associated with unfavorable interactions in different conformations 1. Torsional strain: increase in energy associated with eclipsed electron pairs (typically in covalent bonds). 3 C C 3 staggered conformation all groups gauche lowest energy (most stable) Summary of Strain Energy 3 C C 3 eclipsed conformation all groups eclipsed (θ=0º) highest energy (least stable) Slide 28 Organic Chemistry 1 (CM 223) Lecture 6: September 6

Summary of Strain Energy Strain energy: additional potential energy associated with unfavorable interactions in different conformations 2.

29 Summary of Strain Energy Strain energy: additional potential energy associated with unfavorable interactions in different conformations 2. Steric strain (van der Waals strain): increase in energy associated adjacent groups being close together (i.e. overlapping van der Waals radii); most pronouced for large groups (i.e. two eclipsed -C3) no steric strain between adjacent hydrogen atoms slight steric strain between adjacent & C3 groups (~1.4 kcal/mol) large steric strain between adjacent, eclipsed C3 groups (~3.1 kcal/mol) Slide 29 Organic Chemistry 1 (CM 223) Lecture 6: September 6

30 Summary of Strain Energy Strain energy: additional potential energy associated with unfavorable interactions in different conformations 3. Angle strain (Baeyer strain): increase in energy associated with deviation of natural bond angles predicted by VSEPR (e.g., in cycloalkanes). natural bond angle for sp 3 carbon = 109.5º actual bond angle in cyclopropane = 60º Slide 30 Organic Chemistry 1 (CM 223) Lecture 6: September 6

31 CM 223 Organic Chemistry I Prof. Chad Landrie Nomenclature and Properties of Cylcoalkanes Sections:

32 IUPAC: Monosubstituted Cycloalkanes C 1 2 C C C 2 2 propylcyclobutane Steps: 1. Count the number of carbons in the ring. Precede the parent name with cyclo. 2. Identify straight chain, common or branched substituent groups. 3. Name the compound according to the figure below. Conventions: If the number of carbons in the substituent is greater, name the ring as the substituent as a cycloaklyl substituent (e.g. cyclobutyl) If the ring is monosubstituted, no locant is neccessary; substituent locant is assumed to be 1. C 2 C 2 C 3 substituent cyclo parent ring name Slide 33 Organic Chemistry 1 (CM 223) Lecture 6: September 6

33 IUPAC: Disubstituted Cycloalkanes 1,1-diethyl-4-(3-methylbutyl)cyclononane Steps: 1. Count the number of carbons in the ring. Precede the parent name with cyclo. 2. Identify straight chain, common or branched substituent groups. 3. Name the compound according to the figure below. Conventions: If the number of carbons in the substituent is greater, name the ring as the substituent as a cycloaklyl substituent (e.g. cyclobutyl). Follow all previous steps and conventions for naming substituents. List substituents in alphabetical order. Ignore replicating prefixes. Follow first point of difference rule. locant substituent 1st alphabetically locant substituent 2nd alphabetically 3 C 2 C 3 C 2 C 2 C 2 C C C C C 2 C C 2 C 3 C2 C 2 C C 2 C 3 cyclo parent ring name Slide 34 Organic Chemistry 1 (CM 223) Lecture 6: September 6

34 i>clicker Question What is the IUPAC name for the molecule below? A. 1-ethyl-3-pentylcyclohexane B. 3-ethyl-1-pentylcyclohexane C. 1-ethyl-3-(3-methylbutyl)cyclohexane D. 1-butyl-3-ethylcycloheptane E. 1-(2-methylbutyl)-3-ethylcyclohexane Slide 35 Organic Chemistry 1 (CM 223) Lecture 6: September 6

35 i>clicker Question Which structure below is 1-(2-methylbutyl)-4-(1-methylethyl)cyclooctane? A B C D E Slide 36 Organic Chemistry 1 (CM 223) Lecture 6: September 6

36 i>clicker Question Which structure below is 1-(2-methylbutyl)-4-(1-methylethyl)cyclooctane? When two numbering schemes give the same locants, choose the direction that gives the substituent first in the alphabet the lowest locant value Slide 37 Organic Chemistry 1 (CM 223) Lecture 6: September 6

37 Summary of Strain Energy Strain energy: additional potential energy associated with unfavorable interactions in different conformations 3. Angle strain (Baeyer strain): increase in energy associated with deviation of natural bond angles predicted by VSEPR (e.g., in cycloalkanes). natural bond angle for sp 3 carbon = 109.5º actual bond angle in cyclopropane = 60º Slide 38 Organic Chemistry 1 (CM 223) Lecture 6: September 6

38 i>clicker Question 3 x 180º 5 Geometry: What are the angles in a regular pentagon? = 108º A. 110º B º C. 60º D. 120º E. 108º Slide 39 Organic Chemistry 1 (CM 223) Lecture 6: September 6

39 i>clicker Question 4 x 180º 6 Geometry: What are the angles in a regular hexagon? = 120º A. 90º B. 120º C. 144º D. 150º E. 30º Slide 40 Organic Chemistry 1 (CM 223) Lecture 6: September 6

40 eats of Combustion of Cycloalkanes Angle strain (Baeyer strain) is the increase in energy associated with bond angles that deviate from tetrahedral (109.5º) in sp3 hybridized carbon atoms. Cycloalkane Shape Geometric Angles Difference from 109.5º eat of Combustion (- ) per C2 Group cyclopropane 60º 49.5º 167 kcal/mol cyclobutane 90º 19.5º 163 kcal/mol cyclopentane 108º 1.5º 157 kcal/mol cyclohexane 120º 10.5º 156 kcal/mol Slide 41 Organic Chemistry 1 (CM 223) Lecture 6: September 6

41 Cyclopropane VB Combination MO Picture Electrostatic Potential Map only planar cycloalkane bent C-C bonds: sp 3 orbitals unable to overlap along internuclear axis; weaker C- C σ-bonds angle strain: 60º is a large deviation from 109.5º torsional strain: all bonds are eclipsed Slide 42 Organic Chemistry 1 (CM 223) Lecture 6: September 6

42 Conformations of Cyclobutane Nonplanar puckered Conformation torsional strain reduced in puckered conformation puckered = fewer eclipsed relationships less angle strain than cyclopropane Slide 43 Organic Chemistry 1 (CM 223) Lecture 6: September 6

43 Conformations of Cyclopentane planar conformation least stable; all bonds eclipsed some torsional strain relieved in envelope and half-chair envelope & half-chair have similar energies; interconvert rapidly Slide 44 Organic Chemistry 1 (CM 223) Lecture 6: September 6

i>clicker Question Which of the following

relationships ighest Potential NRG: 6 eclipsed

44 i>clicker Question Which of the following conformations of cyclohexane would you expect to have the highest potential energy? A C B Lowest Potential NRG: all gauche & anti relationships ighest Potential NRG: 6 eclipsed relationships Slide 45 Organic Chemistry 1 (CM 223) Lecture 6: September 6

45 CM 223 Organic Chemistry I Prof. Chad Landrie Exam One: Tuesday, September 19

46 CM 223 Organic Chemistry I Prof. Chad Landrie Experiment 5A: Extraction Determining a Partition Coefficient

47 Determining the Partition Coefficient Add 2O Add C2Cl2 remove C2Cl2 mix thoroughly Evaporate C2Cl2 measure mass Slide 48 Organic Chemistry 1 (CM 223) Lecture 6: September 6

48 Determining the Partition Coefficient Add 2O Add C2Cl2 remove C2Cl2 mix thoroughly Requirements for an Extracting Solvent 1. Does not react irreversibly with the solute. 2. Both solvents are immiscible (form two layers). 3. Selectively removes desired component. 4. Easly separated from the solute. (e.g., cyrstallization, precipitation/filtration, distillation/evaporation) Evaporate C2Cl2 measure mass Slide 49 Organic Chemistry 1 (CM 223) Lecture 6: September 6

49 Extraction Process Add 2O Add C2Cl2 remove C2Cl2 Add C2Cl2 remove C2Cl2 mix thoroughly mix thoroughly Evaporate C2Cl2 measure mass Slide 50 Organic Chemistry 1 (CM 223) Lecture 6: September 6

Pre-Lab Question #2 A student is extracting a

caffeine in 2 ml of water (C1) and extracts

caffeine will be in the dichloromethane extract?

50 Pre-Lab Question #2 A student is extracting a caffeine/water solution with dichloromethane. The K value is 4.6. If the student starts with a total of 40 mg of caffeine in 2 ml of water (C1) and extracts once with 6 ml of dichloromethane (C2) how much caffeine will be in the dichloromethane extract? Equations Needed Slide 51 Organic Chemistry 1 (CM 223) Lecture 6: September 6

ow much caffeine total in all three extracts of dichloromethane?

51 Pre-Lab Question #3 Another student also starts with 40 mg of caffeine in 2 ml of water and decides to extract 3 x 2 ml of dichloromethane. ow much caffeine will be in each extract? ow much caffeine total in all three extracts of dichloromethane? 1st Extraction 2nd Extraction 3rd Extraction Which is better: one extraction with 6 ml or three extractions with 2 ml each? Slide 52 Organic Chemistry 1 (CM 223) Lecture 6: September 6

52 FA = ( ) n Vo KVx + Vo FA = Several Smaller Extractions More Effecient Example one: 3 extractions x 5 ml each (15 ml total) from 20 ml of Vo. K = 2. FA = fraction of solute remaining in original solvent K = partition coefficient = ([Ax]/[Ao]) larger K = more efficient extracting solvent Vo = volume of original solvent Vx = volume of extracting solvent (per extraction) n = number of extractions Example two: 1 extractions x 15 ml each (15 ml total) from 20 ml of Vo. K = 2. ( ) 20 ( 20 FA = 2* * ) 1 FA = 0.30 FA = 0.40 **30% of solute remains in the original solvent.** **40% of solute remains in the original solvent.** Slide 53 Organic Chemistry 1 (CM 223) Lecture 6: September 6

53 CM 223 Organic Chemistry I Prof. Chad Landrie Experiment 5B: Extraction Part B: Extraction of Benzoic Acid

54 Base Extraction of 1:1 Benzoic Acid/Acetanilide 1. Neutralize/ Protonate with 3 M Cl O O 1. recrystallize from boiling water O O 2 O O O Na 2. Vacuum filter precipitated benzoic acid crude solid: not pure/clean depressed mp possibly discolored 2. vacuum filter pure solid: white crystals sharp mp narrow mp range 2 O O O Na O O Add 3 M NaO (aq) O separate layers N N O C2Cl2 is denser than water = bottom layer O C 2 Cl 2 C 2 Cl 2 C 2 Cl 2 N O evaporate C 2 Cl 2 N O C 3 crude solid: not pure/clean depressed mp possibly discolored 1. recrystallize from boiling water 2. vacuum filter N C 3 O pure solid: white crystals sharp mp narrow mp range Slide 55 Organic Chemistry 1 (CM 223) Lecture 6: September 6

55 Base Extraction of 1:1 Benzoic Acid/Acetanilide Some differences in our experiment: O N C 2 Cl 2 O O Add 3 M NaO (aq) C2Cl2 is denser than water = bottom layer 2 O O O N O C 2 Cl 2 O Na 1.We are using diethyl ether, (C3C2)2O, instead of C2Cl2. ow do the polarities of these two solvents compare? Do you expect diethyl ether to be the top layer or the bottom layer in our extraction? ow can you check? 2.We will not recrystallize either solid. We will only collect the product from the ether layer and measure its melting point to determine which unknown we have. Slide 56 Organic Chemistry 1 (CM 223) Lecture 6: September 6

56 Ionizable Functional Groups in Water O R O deprotonation O + O + R O 2 O pk a ~ 4 pk a = 15.7 carboxylic acid (conjugate acid) R O phenol (conjugate acid) carboxylate anion (conjugate base) deprotonation O + O R + pk a ~ 4 pk a = 15.7 phenoxide anion (conjugate base) 2 O *ydroxide and hydronium were chosen as general examples; other specific acids and bases may be used. The counterion on the left of the equilibria will depend upon the specific acid or base on the left. R N 2 amine (conjugate base) protonation + O 2 R N 3 + pk a ~ 1.7 pk a ~ 9.0 ammonium cation (conjugate acid) 2 O Slide 57 Organic Chemistry 1 (CM 223) Lecture 6: September 6

Organic Chemistry 1 Lecture 5

CEM 232 Organic Chemistry I Illinois at Chicago Organic Chemistry 1 Lecture 5 Instructor: Prof. Duncan Wardrop Time/Day: T & R, 12:30-1:45 p.m. January 26, 2010 1 Self Test Question Which of the following