Alkene Reactions

1. Alkene Reactions

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

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 Transition State Three areas to be aware of. Products Reactants Energy of Activation. Energy barrier.

5. Energy Changes in a Reaction Enthalpy changes, DH0, for a reaction arises from changes in bonding in the molecule. If weaker bonds are broken and stronger ones formed then DH0 is negative and exothermic. If stronger bonds are broken and weaker ones formed then DH0 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. DG0: if positive equilibrium favors reactants (endergonic), if negative favors products (exergonic). DG0 = DH0 – TDS0

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

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

9. 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.

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

11. Electrophilic Additions Hydrohalogenation using HCl, HBr, HI Hydration using H2O in the presence of H2SO4 Halogenation using Cl2, Br2 Halohydrination using HOCl, HOBr Oxymercuration using Hg(OAc)2, H2O followed by reduction

12. Electrophilic Addition We now address regioselectivity….

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

14. Mechanism Step 1 Step 2

15. Now examine Step 1 Closely 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). 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. The carbocation intermediate is very reactive. It does not obey the octet rule (electron deficient) and is usually present only in low concentration.

16. Carbocations sp2 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. Electron deficient. Does not obey octet rule. Lewis acid, can receive electrons. Electrophile.

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

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

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

20. Factors Affecting Carbocation Stability - Inductive • Inductive Effect. Electron redistribution due to differences in electronegativities of substituents. • Electron releasing, alkyl groups, -CH3, stabilize the carbocation making it easier to form. • Electron withdrawing groups, such as -CF3, destabilize the carbocation making it harder to form. d- d+ d- d-

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

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

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

24. Another Factor Affecting Carbocation Stability – Resonance Utilizing an adjacent lone pair. Look carefully.This is the conjugate acid of formaldehyde, CH2=O.

25. Production of Chiral Centers. Goal is to see all the possibilities. React alkene with HBr. Note that the ends of the double bond are different. 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.

26. Production of Chiral Centers - 2 diastereomers Racemic Mixture 1 Racemic 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).

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

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

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

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

31. 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. Reason for Shift: Converting a less stable carbocation (20) to a more stable carbocation (30). Step 3, the nucleophile reacts with the carbocation

32. Addition of Br2 and Cl2

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

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

35. Step 2

36. Detailed Stereochemistry, addition of Br2 Bromide ion attacked the carbon on the right. S,S enantiomers But can also attack the left-side carbon. R,R R,R enantiomers Alternatively, the bromine could have come in from the bottom! Only two compounds (R,R and S,S) formed in equal amounts. Racemic mixture. S,S

37. Number of products formed. S,S enantiomers 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. R,R R,R enantiomers S,S

38. Attack of the Bromide Ion In order to preserve a tetrahedral carbon these two substituents must move upwards. Inversion. 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.

39. 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.

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

41. There are many variations on the addition of X2 to an alkene. Each one involves anti addition. Br - I - I - I - + The iodide can attach to either of the two carbons. Instead of iodide ion as nucleophile can use alcohols to yield ethers, water to yield alcohols, or amines.

42. Regioselectivity If Br2 is added to propene there is no regioselectivity issue. If Br2 is added in the presence of excess alternative nucleophile, such as CH3OH, regioselectivity may become important.

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

44. Regioselectivity, Bromonium Ion Bridged bromonium ion from propene.

45. Example Stereochemistry: anti addition Note: non-reacting fragment unchanged Regioselectivity, addition of Cl and OH Put in Fisher Projections. Be sure you can do this!! Cl, from the electrophile Cl2, goes here OH, the nucleophile, goes here

46. 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?

47. 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.

48. Progress of Attack Notice that the two bromines are maintained anti to each other!!! • 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.

49. Addition to substituted cyclohexene 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. Observe Ring is locked as shown. No ring flipping. Be sure to allow for the inversion motion at the carbon attacked by the bromide ion.

50. 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.