1. Alkene Reactions

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

3. 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.

4. 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.

5. 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.

6. 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

7. 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.

8. 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.

9. 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.

10. 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.

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

12. 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

13. Electrophilic Addition We now address regioselectivity….

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

15. Mechanism Step 2 Step 1

16. 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.

17. 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.

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

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

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

21. 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. -- ++ -- --

22. 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

23. 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.

24. 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!

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

26. 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.

27. 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

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

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

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

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

32. 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 ).

33. Addition of Br 2 and Cl 2

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

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

36. Step 2

37. 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.

38. 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.

39. 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.

40. 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.

41. 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. =

42. 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.

43. 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.

44. 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.

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

46. 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!!

47. 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?

48. 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.

49. 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!!!

50. 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.

51. 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.

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

53. Mechanism 1 2 3 4

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

55. Borane, a digression Isoelectronic with a carbocation

56. 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…

57. Cont’d

58. Oxidation and Reduction Reactions

59. 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.

60. 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 ()

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

62. 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

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

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

65. 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.

66. 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.

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

68. Mechanism - 2

69. Mechanism - 3

70. Mechanism - 4

71. Hydrogenation No regioselectivity Syn addition

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

73. 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

74. 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.

75. 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.

76. 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

77. 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?

78. 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)

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

80. 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!!