1. Organic Chemistry Chapter 9 17:40:49 1

2. The Competition among S N 2, S N 1, E2, and E1 Reactions Competition can exist between reaction mechanisms. This can lead to the formation of different products. an S N 2 example… 17:40:49 2

3. Competition an S N 1 example… an E2 example… 17:40:493

4. Competition an E1 example… 17:40:494

5. Reasons for the Competition The products of all four mechanisms can be different. The stereochemistry of each unimolecular reaction is different from that of the corresponding bimolecular reaction. 17:40:49 5

6. Kinetic Control or Thermodynamic Control Before you decide how to predict the major product of any competition, you must know whether the competition takes place under kinetic control or thermodynamic control. 17:40:49 6

7. Rate-Determining Steps Revisited Because substitution and elimination reactions generally take place under kinetic control, predicting the outcome of an S N 2/S N 1/E2/E1 competition means we have to know how to predict the relative rates of the competing reactions. The rate-determining step of a reaction dictates the rate of the overall reaction. 17:40:49 7

8. S N 2 Rate Determining Step The S N 2 mechanism consists of just a single step so that single step must be rate determining. 17:40:49 8

9. S N 1 Rate Determining Step The S N 1 reaction takes place in two steps. The first step is rate determining because formation of the carbocation is much slower than formation of the final product. 17:40:49 9

10. A Way of Viewing the S N 2 versus S N 1 Rate Determining Steps The rate of an S N 2 reaction is highly sensitive to factors that affect the nucleophile’s ability to attack the substrate and to factors that affect the ability of the leaving group to depart. The rate of an S N 1 reaction is highly sensitive only to factors that help the leaving group depart. The S N 1 reaction rate is independent of the nucleophile. 17:40:49 10

11. E2 Rate Determining Step Like an S N 2 reaction, an E2 reaction consists of a single step that must be rate determining. 17:40:49 11

12. E1 Rate Determining Step As in an S N 1 reaction, the rate-determining step of an E1 reaction is the first of its two steps. Just as with the S N 1 reaction rate, the E1 reaction rate is independent of the attacking species—in this case, the base. 17:40:49 12

13. A Way of Viewing the E2 versus E1 Rate Determining Steps In the E2 reaction, the substrate is deprotonated at the same time the leaving group leaves. In an E1 reaction, the base does not enter the picture until the second step, even though it is in solution throughout the reaction. 17:40:49 13

14. Factor 1: Strength of the Attacking Species Because the S N 2, S N 1, E2, and E1 reactions are sensitive to the attacking species in different ways, the identity of the attacking species can play a major role in the outcome of the S N 2/S N 1/E2/E1 competition. 17:40:49 14

15. Factors affecting Reaction Rates 1.Streangth of the attacking species. 2. Concentration of the attacking species. 3. Leaving group ability 4. Type of carbon bonded to the leaving group 5. Solvent effects 6. Heat 17:40:49 15

16. The Nucleophile Strength in S N 2 and S N 1 Reactions The rate of the S N 2 reaction depends very heavily upon the identity of the nucleophile. For example, when DMF is the solvent, Cl⁻ undergoes the S N 2 reaction about twice as fast as Br⁻, and the S N 2 reaction involving the cyanide anion (NC⁻) is about 250 times faster than that involving Br⁻. 17:40:49 16

17. 17:40:49 17

18. The Nucleophile Strength in S N 2 and S N 1 Reactions continued… These rate differences reflect differences in nucleophile strength, or nucleophilicity. 17:40:49 18

19. Hammond Postulate Differences in rate reflect differences in the sizes of the energy barriers. The smaller the energy barrier, the faster the reaction. In turn, the size of the energy barrier is determined by the energy of the transition state relative to that of the reactants. This problem can be greatly simplified by applying the Hammond postulate. 17:40:49 19

20. Hammond Postulate continued… The Hammond postulate can be used to provide insight into the structure and energy of a transition state by considering the reactants that immediately precede it and the products that immediately follow it. 17:40:49 20

21. Hammond Postulate continued… 17:40:4921

22. The Hammond Postulate Applied to S N 2 Reactions Cl⁻ is less stable (smaller in size) than Br⁻ An S N 2 reaction with Cl⁻ as the nucleophile is more exothermic than with Br⁻ as the nucleophile The S N 2 reaction is faster with Cl⁻ as the nucleophile. 17:40:4922

23. Rate of an S N 1 Reaction 17:40:49 23

24. Nucleophile Strength and S N 2 versus S N 1 Strong nucleophiles tend to favor S N 2 over S N 1. Weak nucleophiles tend to favor S N 1 over S N 2. 17:40:4924

25. Generating Carbon Nucleophiles Nucleophilic carbon atoms are often important in synthesis, especially in the formation of carbon–carbon bonds. In uncharged molecules, carbon atoms are typically non- nucleophilic for two reasons: (1) They do not possess a lone pair of electrons, and (2) they are rarely considered electron-rich because they are generally bonded to atoms with electronegativities comparable to or greater than their own. 17:40:49 25

26. Generating Carbon Nucleophiles continued… A carbon atom is quite nucleophilic when it bears a formal negative charge—that is, when it is a carbanion. Carbanions possess a lone pair of electrons that can be used to form a bond. 17:40:49 26

27. Generating Carbon Nucleophiles continued… The simplest way to generate carbon nucleophiles is to deprotonate the uncharged carbon. Alkanes have pKa values around 50, so they are such weak acids that deprotonation is unfeasible. The pKa values of alkynes are lower than corresponding alkanes. 17:40:49 27

28. The Base Strength in E2 and E1 Reactions The relative rates of E2 reactions depend on the identity of the base—more specifically, on the strength of the base. 17:40:49 28

29. The Base Strength in E2 and E1 Reactions continued… 17:40:4929

30. Rate Dependence with E2 and E1 Reactions The base does not participate in an E1 reaction until the leaving group has left, at which point it removes the proton on the substrate. Because the base does not help the leaving group to leave, the specific identity of the base has little effect on the reaction rate. 17:40:49 30

31. Base Strength and Competition Reactions 17:40:4931

32. 17:40:4932

33. Strong, Bulky Bases The tert-butoxide anion, (CH 3 ) 3 CO⁻, should be both a strong nucleophile (because it has a full negative charge) and a strong base (because it is stronger than HO⁻). Thus, it should favor both S N 2 and E2 reactions. The tert-butoxide anion usually favors just E2 products because (CH 3 ) 3 CO⁻ is a much weaker nucleophile (due to its steric bulk) than we would expect based on charge stability. 17:40:49 33

34. Steric Hindrance in S N 2 Reactions Steric hindrance prevents the O atom from forming a bond to a C atom to displace a leaving group, which slows the reaction. The basicity of the anion remains relatively unaffected by steric hindrance because protons are very small and are usually well exposed. 17:40:49 34

35. Structures of Some Strong, Bulky Bases 17:40:4935

36. Rank the following alkoxide nucleophiles in order of increasing SN2 reaction 17:40:4936

37. Factor 2: Concentration of the Attacking Species The dependence of each reaction upon the concentration of the substrate and the attacking species is summarized in its respective empirical rate law. 17:40:49 37

38. Concentration Effects and S N 2/S N 1/E2/E1 Reactions Note that the S N 2 and E2 reaction rates depend on the concentration of the attacking species, [Att⁻], whereas the S N 1 and E1 reactions do not. S N 2 and E2 reactions proceed faster with higher concentration of the attacking species because the role of the attacking species is to force off the leaving group. 17:40:49 38

39. Concentration Effects and S N 2/S N 1/E2/E1 Reactions continued… In both the S N 1 and E1 mechanisms the attacking species does not participate in the reaction until the leaving group has departed. Therefore, a higher concentration of the attacking species simply means more attacking species waiting for the leaving group to come off, but the reaction rate does not change. Therefore, a high concentration of the attacking species tends to favor S N 2 and E2 reactions A low concentration tends to favor S N 1 and E1. 17:40:49 39

40. Factor 3: Leaving Group Ability A leaving group must leave in the rate-determining step of an S N 2, S N 1, E2, or E1 reaction. The identity of the leaving group has an effect on the rate of each reaction. 17:40:49 40

41. Factor 3: Leaving Group Ability continued… 17:40:4941

42. Leaving Group Ability, Charge Stability, and Base Strength 17:40:4942

43. Leaving Group Ability, Charge Stability, and Base Strength continued… Relative leaving group abilities are governed largely by the stabilities of the leaving groups in the form in which they come off the substrate. The stability as a leaving group is reflected in its base strength. 17:40:49 43

44. Sulfonate Anions as Leaving Groups The tosylate anion (TsO⁻), the mesylate anion (MsO⁻), and the triflate anion (TfO⁻) are among the best leaving groups. They are weak bases, as they are similar in structure to the conjugate base of H 2 SO 4. 17:40:49 44

45. Leaving Group Ability and S N 2/S N 1/E2/E1 Reactions S N 1 reactions are more sensitive to leaving group ability than S N 2 reactions are. Excellent leaving groups favor S N 1 and E1 reactions over corresponding S N 2 and E2 reactions. 17:40:49 45

46. Converting a Bad Leaving Group into a Good Leaving Group Frequently, a desired nucleophilic substitution or elimination reaction is unfeasible because the leaving group is unsuitable. The HO⁻ leaving group, for example, is a very poor leaving group, so the following will not lead to a reaction under neutral conditions. 17:40:49 46

47. Converting a Bad Leaving Group into a Good Leaving Group continued… Under acidic conditions a reaction does take place. The acidic conditions facilitate the substitution reaction because the O atom on the OH group is weakly basic. 17:40:49 47

48. Water Leaving Group Mechanism The leaving group leaves as H 2 O, not HO⁻. Being uncharged, H 2 O is much more stable than HO⁻, and is an excellent leaving group. 17:40:49 48

49. Factors affecting Reaction Rates 1.Streangth of the attacking species. 2. Concentration of the attacking species. 3. Leaving group ability 4. Type of carbon bonded to the leaving group 5. Solvent effects 6. Heat 17:40:49 49

50. Factor 4: Type of Carbon Bonded to the Leaving Group An important characteristic of the carbon atom bonded to the leaving group is its hybridization. This is partly because sp 2 - and sp-hybridized carbons form stronger σ bonds than sp 3 hybridized carbons. A stronger bond makes it more difficult for the leaving group to leave. 17:40:49 50

51. S N 2 Reactions and the Hybridization of the Carbon S N 2 reactions are further hindered by electrostatic repulsion when the leaving group is attached to an sp 2 - or sp-hybridized carbon. 17:40:49 51

52. S N 1 and E1 Reactions and the Hybridization of the Carbon S N 1 and E1 reactions are further hindered by charge stability in the carbocation intermediate. Because sp- and sp 2 -hybridized carbon atoms possess greater s character than an sp 3 - hybridized carbon does, these atoms have a higher effective electronegativity. 17:40:49 52

53. The Number of Alkyl Groups on the Carbon Bonded to the Leaving Group Nucleophilic substitution and elimination reactions can be strongly influenced by the number of alkyl groups bonded to the carbon atom with the leaving group. A carbon atom bonded to three alkyl groups is called a tertiary (3°) carbon; if it is bonded to two alkyl groups, then it is called a secondary (2°) carbon; and if it is bonded to one alkyl group, then it is called a primary (1°) carbon. If the carbon is bonded only to hydrogen atoms, then it is called a methyl carbon and the substrate takes the form CH 3 -L. 17:40:49 53

54. The Number of Alkyl Groups on the Carbon Bonded to the Leaving Group continued… 17:40:4954

55. Leaving Group on Primary, secondary, Tertiary carbons 17:40:4955

56. Rate of S N 2 and Steric Hindrance With each additional alkyl group bonded to the carbon, steric hindrance of the nucleophile increases, which slows the reaction 17:40:49 56

57. Rate of S N 1 and E1 versus Carbocation Stability 17:40:49 57

58. Which substrate C or D under goes an E1 mechanism faster 17:40:49 58

59. E2 Reactions 17:40:49 59

60. Which substrate X, Y or Z will undergo SN2 reaction faster? 17:40:49 60

61. Summary of the Influence of the Number of Alkyl Groups on the Carbon 17:40:4961

62. Leaving Groups Bonded to an Allyl or Benzyl Group Exceptions arise when the leaving is attached to the allyl (CH 2 =CH-CH 2 -L) or benzyl (C 6 H 5 -CH 2 -L) group. The allyl cation or the benzyl cation that is formed is stabilized by resonance. 17:40:49 62

63. Formation and stability of the Allyl Cation and Benzyl Cation 17:40:49 63

64. Factors affecting Reaction Rates 1.Streangth of the attacking species. 2. Concentration of the attacking species. 3. Leaving group ability 4. Type of carbon bonded to the leaving group 5. Solvent effects 6. Heat 17:40:49 64

65. Factor 5: Solvent Effects There are two types of solvent in which S N 2, S N 1, E2, and E1 reactions can take place: polar protic solvents and polar aprotic solvents. 17:40:49 65

66. Factor 5: Solvent Effects. 17:40:49 66

67. Protic Solvents, Aprotic Solvents, and the S N 2/S N 1/E2/E1 Competition The choice of solvent can have a significant influence on the outcome of nucleophilic substitution and elimination reactions. 17:40:49 67 Polar Aprotic Solvents => Favor SN2 an E2reactions Polar Protic Solvents => Favor SN1 and E1 Reactions.

68. Solvation With protic solvents, strong ion–dipole interactions weaken the attacking species. With aprotic solvents, ion–dipole interactions are much weaker because the positive end of the net dipole is typically buried inside the solvent molecule. 17:40:49 68

69. Solvation. 17:40:49 69

70. Solvation and Relative Nucleophile Strength Some nucleophilicities are reversed in protic versus aprotic solvents. 17:40:49 70

71. Factor 6: Heat When substitution and elimination reactions are both favored under a specific set of conditions, it is often possible to influence the outcome by changing the temperature under which the reactions take place. 17:40:49 71 Increaded Temperature=> Favor Elimination

72. Elimination Favored by Heat 17:40:4972

73. Factor 6: Heat continued… This temperature effect is due to entropy. ∆S ° rxn is more positive for an elimination reaction than for a substitution reaction. 17:40:49 73

74. Predicting the Outcome of Nucleophilic Substitution and Elimination Reactions The leaving group, Cl⁻, is at least as stable as F⁻, so proceed to Step 2. 17:40:49 74

75. Predicting the Outcome of Nucleophilic Substitution and Elimination Reactions continued… The leaving group in (S)-2-chloropentane is attached to a 2° carbon, so all four reactions must be considered. – If it were a 3° carbon, don’t consider S N 2. – If it were a 1° or methyl carbon, don’t consider S N 1 or E1 unless the resulting carbocation is resonance stabilized. 17:40:49 75

76. Predicting the Outcome of Nucleophilic Substitution and Elimination Reactions continued… 17:40:4976

77. Predicting the Outcome of Nucleophilic Substitution and Elimination Reactions continued… 17:40:4977

78. Predicting the Outcome of Nucleophilic Substitution and Elimination Reactions continued… In this case, heat is not factored in because S N 2 is the clear winner. 17:40:49 78

79. Regioselectivity in Elimination Reactions: Zaitsev’s Rule A substrate can possess two or more distinct hydrogen atoms that can be removed in an elimination reaction, leading to two or more possible alkene products. This elimination reaction exhibits regioselectivity. 17:40:49 79

80. Another Example of Zaitsev’s Rule 17:40:4980

81. An Exception to Zaitsev’s Rule An E2 reaction can exhibit anti-Zaitsev regiochemistry with a strong, bulky base. 17:40:4981

82. Intermolecular Reactions versus Intramolecular Cyclizations A chemical reaction between two separate species is an intermolecular (between molecule) reaction. Reactions typically require two separate functional groups. Those functional groups need not be on separate molecules. Instead, they can be part of the same molecule, attached at different places on the molecule’s backbone. In that case, the reaction is said to be intramolecular (within the same molecule). 17:40:49 82

83. Example of an Intramolecular Reaction 17:40:4983

84. Example of an Intermolecular Reaction 17:40:4984

85. Reversible Reactions For products from competing reactions to be in equilibrium, there must be a way that those products can interconvert. This can happen with reversible competing reactions, which take place readily in both the forward and reverse directions. 17:40:49 85

86. If competing reactions are irreversible, in which case they do not take place readily in the reverse direction, then equilibrium is not established between the products from the respective reactions. This is illustrated using irreversible reaction arrows to connect reactants to products. Irreversible Reactions 17:40:49 86

87. Reversibility and Kinetic versus Thermodynamic Control Reversible reactions tend to take place under thermodynamic control. Irreversible reactions tend to take place under kinetic control. 17:40:49 87

88. Free Energy Diagram to Determine Whether a Reaction is Reversible or Irreversible Whether a reaction is reversible or irreversible can be determined by carefully examining its free energy diagram. In the free energy diagram of an irreversible reaction, the products are much lower in energy than the reactants, making ∆G° rxn substantially negative. 17:40:49 88

89. Free Energy Diagram of an Irreversible Reaction ∆G° ‡ reverse is much larger than ∆G ° ‡ forward, so the reaction in the reverse direction is much slower than the reaction in the forward direction, thus making the reaction virtually irreversible. 17:40:4989

90. A Reversible S N 2 Reaction 17:40:49 90

91. A Reversible S N 2 Reaction continued… The products are higher in energy than the reactants, making ∆G° rxn somewhat positive. This is primarily because Br⁻ is less stable than I⁻. Consequently, ∆G° ‡ reverse is smaller than ∆G° ‡ forward, making the reaction faster in the reverse direction than in the forward direction under standard conditions. 17:40:49 91

92. Summary and Conclusions S N 2, S N 1, E2, and E1 reactions compete with one another under kinetic control. The fastest reaction yields the major product. S N 2 reaction rates are highly sensitive to nucleophilicity, whereas S N 1 reactions are not. E2 reaction rates are highly sensitive to the strength of the attacking base, whereas E1 reactions are not. Nucleophilicity can be weakened significantly by bulky alkyl groups surrounding the nucleophilic site. Base strength, however, is not significantly affected. 17:40:49 92

93. Summary and Conclusions continued… If a nucleophile is strong, then high concentration favors S N 2 and low concentration favors S N 1. All four reaction rates are affected by leaving group ability. Leaving group ability is determined largely by charge stability. Nucleophilic substitution and elimination reactions generally do not occur when leaving groups are on sp 2 - or sp- hybridized carbon atoms. 17:40:49 93

94. Summary and Conclusions continued… Aprotic solvents favor S N 2 and E2; protic solvents favor S N 1 and E1. Heat favors elimination over substitution due to the greater entropy in the elimination products. When multiple elimination products are possible, the major product is usually the one that is the most stable, in accordance with Zaitsev’s rule. Intramolecular cyclization reactions are favored over their corresponding intermolecular reactions when a five- or six- membered ring can be formed. 17:40:49 94