1. Lone Pair acting as Base. Note the change in formal charges. As reactant oxygen had complete ownership of lone pair. In product it is shared. Oxygen more positive by 1. Similarly, B has gained half of a bonding pair; more negative by 1.

2. An example: pi electrons as bases Bronsted Lowry Acid Bronsted Lowry Base The carbocations are conjugate acids of the alkenes. For the moment, just note that there are two possible carbocations formed.

3. Sigma bonding electrons as bases. Much more unusual!! Super acid A very, very electronegative F!! A very positive S!! The OH becomes very acidic because that would put a negative charge adjacent to the S.

4. Trends for Relative Acid Strengths Totally ionized in aqueous solution. Aqueous Solution Totally unionized in aqueous solution

5. Example Ethanol, EtOH, is a weaker acid than phenol, PhOH. It follows that ethoxide, EtO -, is a stronger base than phenolate, PhO -. For reaction PhOH + EtO - PhO - + EtOH where does equilibrium lie? pK a = 9.95 Stronger acid H 2 O + PhOH H 3 O + + PhO - K a = [H 3 O + ][PhO - ]/[PhOH] = 10 -9.95 H 2 O + EtOH H 3 O + + EtO - K a = [H 3 O + ][EtO - ]/[EtOH] = 10 -15.9 pK a = 15.9 Weaker acid Stronger base Weaker base. Query: What makes for strong (or weak) acids? K = 10 -9.95 / 10 -15.9 = 10 6.0

6. What affects acidity? 1. Electronegativity of the atom holding the negative charge. Increasing electronegativity of atom bearing negative charge. Increasing stability of anion. Increasing acidity. Increasing basicity of anion. 2. Size of the atom bearing the negative charge in the anion. Increasing size of atom holding negative charge. Increasing stability of anion. Increasing acidity. Increasing basicity of anion.

7. What affects acidity? - 2 3. Resonance stabilization, usually of the anion. Increasing resonance stabilization. Increased anion stability. Acidity Increasing basicity of the anion. No resonance structures!! Note that phenol itself enjoys resonance but charges are generated, costing energy, making the resonance less important. The more important resonance in the anion shifts the equilibrium to the right making phenol more acidic.

8. An example: competitive Bases & Resonance Two different bases or two sites in the same molecule may compete to be protonated (be the base). Acetic acid can be protonated at two sites. Which conjugate acid is favored? The more stable one! Which is that? Recall resonance provides additional stability by moving pi or non-bonding electrons. Pi bonding electrons converted to non-bonding. Non-bonding electrons converted to pi bonding. No valid resonance structures for this cation.

9. An example: competitive Bases & Resonance All atoms obey octet rule! The carbon is electron deficient – 6 electrons, not 8. Lesser importance Comments on the importance of the resonance structures.

10. What affects acidity? - 3 4. Inductive and Electrostatic Stabilization. Due to electronegativity of F small positive charges build up on C resulting in stabilization of the anion. Increasing anion stability. Acidity. Increasing anion basicity.  Effect drops off with distance. EtOH pK a = 15.9

11. What affects acidity? - 4 5. Hybridization of the atom bearing the charge. H-A  H + + A: -. sp 3 sp 2 sp More s character, more stability, more “electronegative”, H-A more acidic, A: - less basic. Increasing Acidity of HAIncreasing Basicity of A - Note. The NH 2 - is more basic than the RCC - ion. Know this order.

12. Example of hybridization Effect.

13. What affects acidity? - 5 6. Stabilization of ions by solvents (solvation). Solvation provides stabilization. Crowding inhibiting solvation Solvation, stability of anion, acidity pK a = 15.9 1718 (CH 3 ) 3 CO -, crowded Comparison of alcohol acidities.

14. Example Para nitrophenol is more acidic than phenol. Offer an explanation The lower lies further to the right. Why? Could be due to destabilization of the unionized form, A, or stabilization of the ionized form, B. A B

15. Examine the equilibrium for p-nitrophenol. How does the nitro group increase the acidity? Resonance structures A, B and C are comparable to those in the phenol itself and thus would not be expected to affect acidity. But note the + to – attraction here Structure D occurs only due to the nitro group. The stability it provides will slightly decrease acidity. Examine both sides of equilibrium. What does the nitro group do? First the unionized acid. Note carefully that in these resonance structures charge is created: + on the O and – in the ring or on an oxygen. This decreases the importance of the resonance.

16. Resonance structures A, B and C are comparable to those in the phenolate anion itself and thus would not be expected to affect acidity. But note the + to – attraction here Structure D occurs only due to the nitro group. It increases acidity. The greater amount of significant resonance in the anion accounts for the nitro increasing the acidity. Now look at the anion. What does the nitro group do? Remember we are interested to compare with the phenol phenolate equilibrium. In these resonance structures charge is not created. Thus these structures are important and increase acidity. They account for the acidity of all phenols.

17. Sample Problem

18. Alkenes 1

19. Shape Alkenes, 6 coplanar atoms. All atoms in same plane except for these hydrogens on sp 3 carbon.

20. Arene shapes Planar ring structure. 12 atoms coplanar. Phenyl group, C 6 H 5,, Ph Ph = C 6 H 5

21. Pi bonds Energy

22. Nomenclature Z / E generalization of cis / trans Use R, S priorities to compare substituents on same carbon. High priority on same side, Z. Opposite, E.

23. Cis / Trans in Cycloalkenes For small rings normally have cis double bonds.

24. trans cyclooctene

25. Terpenes and the isoprene Rule A terpene is composed of isoprene units joined head to tail (the isoprene rule). This molecule has additional cross links. Note that location of functional groups such as OH or double bonds is not addressed.

26. Vitamin A Four isprene units joined head to tail One cross link (non-head to tail) linkage.

27. Fatty Acids triglycerideAnimal fats and vegetable oils are both triesters of glycerol, hence the name triglyceride. –Hydrolysis of a triglyceride in aqueous base followed by acidification gives glycerol and three fatty acids. –Fatty acids with no C=C double bonds are called saturated fatty acid. –Those with one or more C=C double bonds are called unsaturated fatty acids.

28. Fatty Acids –The most common fatty acids have an even number of carbons, and between 12 and 20 carbons in an unbranched chain. –The C=C double bonds in almost all naturally occurring fatty acids have a cis configuration. –The greater degree of unsaturation, the lower the melting point. oils –Triglycerides rich in unsaturated fatty acids are generally liquid at room temperature and are called oils. fats –Triglycerides rich in saturated fatty acids are generally semisolids or solids at room temperature and are called fats.

29. Fatty Acids –the four most abundant fatty acids

30. Alkene Reactions

31. Pi bonds Plane of molecule Reactivity above and below the molecular plane!

32. Addition Reactions Important characteristics of addition reactions Orientation (Regioselectivity) If the doubly bonded carbons are not equivalent which one get the A and which gets the B. Stereochemistry: geometry of the addition. Syn addition: Both A and B come in from the same side of the alkene. Both from the top or both from the bottom. Anti Addition: A and B come in from opposite sides (anti addition). No preference.

33. Reaction Mechanisms Mechanism: a detailed, step-by-step description of how a reaction occurs. A reaction may consist of many sequential steps. Each step involves a transformation of the structure. For the step C + A-B  C-A + B Reactants Products Transition State Energy of Activation. Energy barrier. Three areas to be aware of.

34. Energy Changes in a Reaction Enthalpy changes,  H 0, for a reaction arises from changes in bonding in the molecule. –If weaker bonds are broken and stronger ones formed then  H 0 is negative and exothermic. –If stronger bonds are broken and weaker ones formed then  H 0 is positive and endothermic.

35. Gibbs Free Energy Gibbs Free Energy controls the position of equilibrium for a reaction. It takes into account enthalpy, H, and entropy, S, changes. An increase in H during a reaction favors reactants. A decrease favors products. An increase in entropy (eg., more molecules being formed) during a reaction favors products. A decrease favors reactants.  G 0 : if positive equilibrium favors reactants (endergonic), if negative favors products (exergonic).  G 0 =  H 0 – T  S 0

36. How do enthalpy and entropy changes work out. Favor reactants Favor Products. Low temperatures favors reactants but at higher temperature favors products. Low temperatures favors products but at higher temperature favors reactants.

37. Multi-Step Reactions Step 1: A + B  Intermediate Step 2: Intermediate  C + D Step 1: endergonic, high energy of activation. Slow process Step 2: exergonic, small energy of activation. Fast Process. Step 1 is the “slow step”, the rate determining step.

38. Characteristics of two step Reaction 1.The Intermediate has some stability. It resides in a valley. 2.The concentration of an intermediate is usually quite low. The Energies of Activation for reaction of the Intermediate are low. 3.There is a transition state for each step. A transition state is not a stable structure. 4.The reaction coordinate can be traversed in either direction: A+B  C+D or C+D  A+B.

39. Hammond Postulate The transition state for a step is close to the high energy end of the curve. For an endothermic step the transition state resembles the product of the step more than the reactants. For an exothermic step the transition state resembles the reactants more than the products. Reaction coordinate.

40. Example Endothermic Transition state resembles the (higher energy) products. Almost broken. Almost formed. Almost formed radical. Only a small amount of radical character remains.

41. Electrophilic Additions –Hydrohalogenation using HCl, HBr, HI –Hydration using H 2 O in the presence of H 2 SO 4 –Halogenation using Cl 2, Br 2 –Halohydrination using HOCl, HOBr –Oxymercuration using Hg(OAc) 2, H 2 O followed by reduction

42. Electrophilic Addition We now address regioselectivity….

43. Regioselectivity (Orientation) The incoming hydrogen attaches to the carbon with the greater number of hydrogens. This is regioselectivity. It is called Markovnikov orientation.

44. Mechanism Step 2 Step 1

45. Now examine Step 1 Closely Rate Determining Step. The rate at which the carbocation is formed controls the rate of the overall reaction. The energy of activation for this process is critical. Electron rich, pi system. Showed this reaction earlier as an acid/base reaction. Alkene was the base. New term: the alkene is a nucleophile, wanting to react with a positive species. Acidic molecule, easily ionized. We had portrayed the HBr earlier as a Bronsted- Lowry acid. New term: the HBr is an electrophile, wanting to react with an electron rich molecule (nucleophile). The carbocation intermediate is very reactive. It does not obey the octet rule (electron deficient) and is usually present only in low concentration.

46. Carbocations Electron deficient. Does not obey octet rule. Lewis acid, can receive electrons. Electrophile. sp 2 hybridized. p orbital is empty and can receive electrons. Flat, planar. Can react on either side of the plane. Very reactive and present only in very low concentration.

47. Step 2 of the Mechanism Mirror objects :Br -

48. Regioselectivity (Orientation) Secondary carbocation Primary carbocation Secondary carbocation more more stable and more easily formed. Or

49. Carbocation Stabilities Order of increasing stability: Methyl < Primary < Secondary < Tertiary Order of increasing ease of formation: Methyl < Primary < Secondary < Tertiary Increasing Ease of Formation

50. Factors Affecting Carbocation Stability - Inductive 1.Inductive Effect. Electron redistribution due to differences in electronegativities of substituents. Electron releasing, alkyl groups, -CH 3, stabilize the carbocation making it easier to form. Electron withdrawing groups, such as -CF 3, destabilize the carbocation making it harder to form. -- ++ -- --

51. Factors Affecting Carbocation Stability - Hyperconjugation 2. Hyperconjugation. Unlike normal resonance or conjugation hyperconjugation involves  bonds. Hyperconjugation spreads the positive charge onto the adjacent alkyl group

52. Hyperconjugation Continued Drifting of electrons from the filled C-H bond into the empty p orbital of the carbocation. Result resembles a pi bond. Another description of the effect.

53. Factors Affecting Carbocation Stability - Resonance Utilizing an adjacent pi system. Positive charge delocalized through resonance. Another very important example. Positive charge delocalized into the benzene ring. Increased stability of carbocation. Note: the allylic carbocation can react at either end! The benzylic carbocation will react only at the benzylic position even though delocalization occurs!

54. Another Factor Affecting Carbocation Stability – Resonance Utilizing an adjacent lone pair. Look carefully. This is the conjugate acid of formaldehyde, CH 2 =O.

55. Production of Chiral Centers. Goal is to see all the possibilities. The H will attach here. Regioselectivity Analysis: the positive charge will go here and be stabilized by resonance with the phenyl group. Enantiomeric carbocations. What has been made? Two pairs of enantiomers. React alkene with HBr. Note that the ends of the double bond are different.

56. Production of Chiral Centers - 2 Racemic Mixture 1Racemic Mixture 2 The product mixture consists of four stereoisomers, two pairs of enantiomers The product is optically inactive. Distillation of the product mixture yields two fractions (different boiling points). Each fraction is optically inactive. Rule: optically inactive reactants yield optically inactive products (either achiral or racemic). diastereomers

57. Acid Catalyzed Hydration of Alkenes What is the orientation???Markovnikov

58. Mechanism Step 1 Step 2 Step 3 Note the electronic structure of the oxonium ion.

59. Carbocation Rearrangements Expected product is not the major product; rearrangement of carbon skeleton occurred. The methyl group moved. Rearranged.

60. Also, in the hydration reaction. The H moved.

61. Mechanism including the “1,2 shift” Step 1, formation of carbocation Step 2, the 1,2 shift of the methyl group with its pair of electrons. Step 3, the nucleophile reacts with the carbocation Reason for Shift: Converting a less stable carbocation (2 0 ) to a more stable carbocation (3 0 ).

62. Addition of Br 2 and Cl 2

63. Stereochemistry Anti Addition (halogens enter on opposite sides); Stereoselective Syn addition (on same side) does not occur for this reaction.

64. Mechanism, Step 1 Step 1, formation of cyclic bromonium ion.

65. Step 2

66. Detailed Stereochemistry, addition of Br 2 enantiomers Alternatively, the bromine could have come in from the bottom! enantiomers S,S R,R Only two compounds (R,R and S,S) formed in equal amounts. Racemic mixture. Bromide ion attacked the carbon on the right. But can also attack the left-side carbon.

67. Number of products formed. enantiomers S,S R,R We have formed only two products even though there are two chiral carbons present. We know that there is a total of four stereoisomers. Half of them are eliminated because the addition is anti. Syn (both on same side) addition does not occur.

68. Attack of the Bromide Ion Starts as R Becomes S The carbon was originally R with the Br on the top-side. It became S when the Br was removed and a Br attached to the bottom. In order to preserve a tetrahedral carbon these two substituents must move upwards. Inversion.

69. Progress of Attack Things to watch for: Approach of the red Br anion from the bottom. Breaking of the C-Br bond. Inversion of the C on the left; Retention of the C on the right.

70. Using Fischer Projections Not a valid Fischer projection since top vertical bond is coming forward. Convert to Fischer by doing 180 deg rotation of top carbon. =

71. There are many variations on the addition of X 2 to an alkene. Each one involves anti addition. Br - I - + The iodide can attach to either of the two carbons. I - Instead of iodide ion as nucleophile can use alcohols to yield ethers, water to yield alcohols, or amines.

72. Regioselectivity If Br 2 is added to propene there is no regioselectivity issue. If Br 2 is added in the presence of excess alternative nucleophile, such as CH 3 OH, regioselectivity may become important.

73. Regioselectivity - 2 Consider, again, the cyclic bromonium ion and the resonance structures. Weaker bond More positive charge Stronger bond Expect the nucleophile to attack here. Remember inversion occurs.

74. Regioselectivity, Bromonium Ion –Bridged bromonium ion from propene.

75. Example Regioselectivity, addition of Cl and OH Cl, from the electrophile Cl 2, goes here OH, the nucleophile, goes here Stereochemistry: anti addition Note: non-reacting fragment unchanged Put in Fisher Projections. Be sure you can do this!!

76. Bromination of a substituted cyclohexene Consider the following bromination. Expect to form two bromonium ions, one on top and the other on bottom. Expect the rings can be opened by attack on either carbon atom as before. But NO, only one stereoisomer is formed. WHY?

77. Addition to substituted cyclohexene The tert butyl group locks the conformation as shown. The cyclic bromonium ion can form on either the top or bottom of the ring. How can the bromide ion come in? Review earlier slide showing that the bromide ion attacks directly on the side opposite to the ring.

78. Progress of Attack Things to watch for: Approach of the red Br anion from the bottom. Breaking of the C-Br bond. Inversion of the C on the left; Retention of the C on the right. Notice that the two bromines are maintained anti to each other!!!

79. Addition to substituted cyclohexene Observe Ring is locked as shown. No ring flipping. Attack as shown in red by incoming Br ion will put both Br into equatorial positions, not anti. This stereoisomer is not observed. The bromines have not been kept anti to each other but have become gauche as displacement proceeds. Be sure to allow for the inversion motion at the carbon attacked by the bromide ion.

80. Addition to substituted cyclohexene Attack as shown in green by the incoming Br will result in both Br being axial and anti to each other This is the observed diastereomer. We have kept the bromines anti to each other.

81. Oxymercuration-Reduction Regioselective: Markovnikov Orientation Occurs without 1,2 rearrangement, contrast the following No rearrangement Alkene  Alcohol

82. Mechanism 1 2 3 4

83. Hydroboration-Oxidation Alkene  Alcohol Anti-Markovnikov orientation Syn addition

84. Borane, a digression Isoelectronic with a carbocation

85. Mechanism Syn stereochemistry, anti- Markovnikov orientation now established. Two reasons why anti-Markovnikov: 1.Less crowded transition state for B to approach the terminal carbon. 2.A small positive charge is placed on the more highly substituted carbon. Just call the circled group R. Eventually have BR 3. Next…

86. Cont’d

87. Oxidation and Reduction Reactions

88. We think in terms of Half Reactions Write reactants and products of each half reaction. Cr 2 O 7 2- 2 Cr 3+ Balance oxygen by adding water + 7 H 2 O In acid balance H by adding H + 14 H + + Balance charge by adding electrons 6 e - + Inorganic half reaction… If reaction is in base: first balance as above for acid and then add OH - to both sides to neutralize H +. Cancel extra H 2 O. Will be oxidized. Will be reduced.

89. Cont’d Now the organic half reaction… Balance oxygen by adding water In acid balance H by adding H + Balance charge by adding electrons CH 3 CH 2 OHCH 3 CO 2 HH 2 O ++ 4 H + + 4 e - Combine half reactions so as to cancel electrons… CH 3 CH 2 OHCH 3 CO 2 HH 2 O ++ 4 H + + 4 e - Cr 2 O 7 2- 2 Cr 3+ + 7 H 2 O14 H + +6 e - + 3 x () 16 H + + 2 Cr 2 O 7 2- + 3 CH 3 CH 2 OH 4 Cr 3+ + 3 CH 3 CO 2 H + 11 H 2 O 2 x ()

90. Formation of glycols with Syn Addition Osmium tetroxide Syn addition also KMnO 4

91. Anti glycols Using a peracid, RCO 3 H, to form an epoxide which is opened by aq. acid. Peracid: for example, perbenzoic acid The protonated epoxide is analagous to the cyclic bromonium ion. epoxide

92. An example Are these unique? Diastereomers, separable (in theory) by distillation, each optically active

93. Ozonolysis Reaction can be used to break larger molecule down into smaller parts for easy identification.

94. Ozonolysis Example For example, suppose an unknown compound had the formula C 8 H 12 and upon ozonolysis yielded only 3-oxobutanal. What is the structure of the unknown? The hydrogen deficiency is 18-12 = 6. 6/2 = 3 pi bonds or rings. The original compound has 8 carbons and the ozonolysis product has only 4 Conclude: Unknown  two 3-oxobutanal. Unknown C 8 H 12 ozonolysys Simply remove the new oxygens and join to make double bonds. But there is a second possibility.

95. Another Example Hydrogen Deficiency = 8. Four pi bonds/rings. Unknown has no oxygens. Ozonolysis product has four. Each double bond produces two carbonyl groups. Expect unknown to have 2 pi bonds and two rings. To construct unknown cross out the oxygens and then connect. But there are many ways the connections can be made. a b c d Look for a structure that obeys the isoprene rule.

96. Mechanism Consider the resonance structures of ozone. These two, charged at each end, are the useful ones to think about. Electrophile capability. Nucleophile capability.

97. Mechanism - 2

98. Mechanism - 3

99. Mechanism - 4

100. Hydrogenation No regioselectivity Syn addition

101. Heats of Hydrogenation Consider the cis vs trans heats of hydrogenation in more detail…

102. Heats of Hydrogenation - 2 The trans alkene has a lower heat of hydrogenation. Conclusion: Trans alkenes with lower heats of hydrogenation are more stable than cis. We saw same kind of reasoning when we talked about heats of combustion of isomeric alkanes to give CO 2 and H 2 O

103. Heats of Hydrogenation Increasing substitution Reduced heat of Hydrogenation By same reasoning higher degree of substitution provide lower heat of hydrogenation and are, therefore, more stable.

104. Acid Catalyzed Polymerization Principle: Reactive pi electrons (Lewis base) can react with Lewis acid. Recall Which now reacts with a Lewis base, such as halide ion to complete addition of HX yielding 2-halopropane Variation: there are other Lewis bases available. THE ALKENE. The new carbocation now reacts with a Lewis base such as halide ion to yield halide ion to yield 2-halo-4-methyl pentane (dimerization) but could react with another propene to yield higher polymers.

105. Examples of Synthetic Planning Give a synthesis of 2-hexanol from any alkene. Planning: Alkene is a hydrocarbon, thus we have to introduce the OH group How is OH group introduced (into an alkene): hydration What are hydration reactions and what are their characteristics: Mercuration/Reduction: Markovnikov Hydroboration/Oxidation: Anti-Markovnikov and syn addition

106. What alkene to use? Must involve C2 in double bond. Which reaction to use with which alkene? Markovnikov rule can be applied here. CH vs CH 2. Want Markovnikov! Use Mercuration/Reduction!!! Markovnkov Rule cannot be used here. Both are CH. Do not have control over regioselectivity. Do not use this alkene. For yourself : how would you make 1 hexanol, and 3-hexanol?

107. Another synthetic example… How would you prepare meso 2,3 dibromobutane from an alkene? Analysis: Alkene must be 2-butene. But wait that could be either cis or trans! We want meso. Have to worry about stereochemistry Know bromine addition to an alkene is anti addition (cyclic bromonium ion)

108. This worked! How about starting with the cis? This did not work, gave us the wrong stereochemistry!

109. Addition Reaction General Rule… Characterize Reactant as cis or trans, C or T Characterize Reaction as syn or anti, S or A Characterize Product as meso or racemic mixture, M or R RelationshipCharacteristics can be changed in pairs and C A R will remain true. Want meso instead?? Have to use trans. Two changed!!