Addition of Br 2 and Cl 2

1. Addition of Br2 and Cl2

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

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

4. Step 2

5. 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

6. 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

7. 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.

8. 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.

9. 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.

10. 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.

11. 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.

12. 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.

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

14. 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

15. 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?

16. 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.

17. 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.

18. 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.

19. 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.

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

21. Mechanism 1 2 3 4

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

23. Borane, a digression Isoelectronic with a carbocation

24. Mechanism Syn stereochemistry, anti-Markovnikov orientation now established. Just call the circled group R. Eventually have BR3. • Two reasons why anti-Markovnikov: • Less crowded transition state for B to approach the terminal carbon. • A small positive charge is placed on the more highly substituted carbon. Next…

25. Cont’d

26. Oxidation and Reduction Reactions

27. We think in terms of Half Reactions Write reactants and products of each half reaction. Will be reduced. Will be oxidized. Inorganic half reaction… 6 e - + 14 H+ + Cr2O7 2- 2 Cr 3+ + 7 H2O Balance oxygen by adding water In acid balance H by adding H + Balance charge by adding electrons If reaction is in base: first balance as above for acid and then add OH- to both sides to neutralize H +. Cancel extra H2O.

28. Cont’d Now the organic half reaction… H2O + CH3CH2OH CH3CO2H + 4 H+ + 4 e- Balance oxygen by adding water In acid balance H by adding H + Balance charge by adding electrons Combine half reactions so as to cancel electrons… 3 x ( ) H2O + CH3CH2OH CH3CO2H + 4 H+ + 4 e- 2 x ( ) 6 e - + 14 H+ + Cr2O7 2- 2 Cr 3+ + 7 H2O 16 H+ +2 Cr2O7 2- + 3 CH3CH2OH 4 Cr 3+ + 3 CH3CO2H + 11 H2O

29. Formation of glycols with SynAddition Osmium tetroxide Syn addition also KMnO4

30. Anti glycols Using a peracid, RCO3H, to form an epoxide which is opened by aq. acid. epoxide The protonated epoxide is analagous to the cyclic bromonium ion. Peracid: for example, perbenzoic acid

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

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

33. Ozonolysis Example For example, suppose an unknown compound had the formula C8H12 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 C8H12 ozonolysys Simply remove the new oxygens and join to make double bonds. But there is a second possibility.

34. Another Example d Hydrogen Deficiency = 8. Four pi bonds/rings. a c 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. b To construct unknown cross out the oxygens and then connect. But there are many ways the connections can be made. Look for a structure that obeys the isoprene rule.

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

36. Mechanism - 2

37. Mechanism - 3

38. Mechanism - 4

39. Hydrogenation No regioselectivity Syn addition

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

41. 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 CO2 and H2O

42. 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.

43. 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.

44. 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

45. 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 CH2. 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?

46. 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)

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

48. 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 Relationship Characteristics can be changed in pairs and C A R will remain true. Want meso instead?? Have to use trans. Two changed!!