1. E2 is most favorable (lowest activation energy) when H and Lv are anti and coplanar Stereochemistry of E2 A B D E E DA B

2. Examples of E2 Stereochemistry cis Major product. Zaitsev product trans Only product Anti-Zaitsev Explain both regioselectivity and relative rates of reaction. But Faster reaction Slower reaction

3. In order for the H and the Cl to be anti, both must be in axial positions First the cis isomer. Reactive Conformation; H and Cl are anti to each other Iso-propyl groups is in more stable equatorial position. Dominant conformation is reactive conformation. Principles to be used in analysis Stereochemical requirement: anti conformation for departing groups. This means that both must be axial. Dominant conformation: ring flipping between two chair conformations, dominant conformation will be with iso propyl equatorial.

4. In the more stable chair of the trans isomer, there is no H anti and coplanar with Lv, but there is one in the less stable chair Now the trans Unreactive conformation Reactive but only with the H on C 6 Most of the compound exists in the unreactive conformation. Slow reaction. Anti Zaitsev

5. Example, Predict Product Problem!: Fischer projection diagram represents an eclipsed structure. Task: convert to a staggered structure wherein H and Br are anti and predict product. We will convert to a Newman and see what we get… H & Br not anti yet! Now anti and we can see where the pi bond will be.

6. Alternative Approach: CAR The H and Br will be leaving: just indicate by disks. Meso or Racemic?? Anti Geometry CRCR A Relationship works in both directions. Should get cis isomer. Note: As we have said before it may take some work to characterize a compound as “racemic” or “meso”. This may be recognized as one of the enantiomers of the racemic mixture. A C R

7. E1 or E2 (Carbocation)

8. ionization 1 o, 2 o, 3 o polar solvents, weak nucleophiles, weak bases 1 o strong, bulky bases 2 o strong bases 3 o strong bases 1 o good nucleophiles, aprotic solvents 2 o good nucleophiles but also poor bases, aprotic solvents 3 0 SN2SN2 SN1SN1 E2 E1 Rearrange ? 1 o 2 o heat, more hindered 3 o heat, more hindered 1 o 2 o lower hinderance, better nucleophile than base 3 o lower hinderance, better nucleophile than base

9. Recall Halohydrins and Epoxides Creation of Nucleophile Internal S N 2 reaction with inversion Creation of good leaving group. Attack by poor nucleophile

10. Neighboring Group Effect Mustard gases –contain either S-C-C-X or N-C-C-X –what is unusual about the mustard gases is that they undergo hydrolysis rapidly in water, a very poor nucleophile Bis(2-chloroethyl)sulfide (a sulfur mustard gas) Bis(2-chloroethyl)methylamine (a nitrogen mustard gas)

11. –the reason is neighboring group participation by the adjacent heteroatom –proton transfer to “solvent” completes the reaction Good nucleophile.

12. 5. Provide a clear, unambiguous mechanism to explain the following stereochemical results. Complete structures of intermediates, if any, should be shown. Use curved arrow notation consistently. Expect sulfur to attack the C-Cl, displacing the Cl and forming a three membered ring. Like this… But we have to be careful with stereochemistry Here is the crux of the matter: how can the non-reacting carbon change its configuration??? Further it does not always change but only if configuration of the reacting carbon changes!! We got a mixture of enantiomers, a racemic mixture. Something strange is happening!! From an old quiz

13. We have to put the molecule in the correct conformation. = Reactive conformation reached by 180 rotation around C-C bond And then the ring is opened by attack of water S and Cl are eclipsed, not anti. But let’s pause for a moment. Our reactant was optically active with two chiral carbons. Recall the problem: If reaction occurs only at the C bearing the Cl the other should remain chiral! Hmmmm? But now notice that the intermediate sulfonium ion is achiral. It has a mirror plane of symmetry. Only optically inactive products will result.

14. Two modes of attack by water. And… Again note the ring structure is achiral and that we must, of course, produce optically inactive product. Enantiomers, racemic mixture

15. Alcohols

16. Hydrogen Bonding Three ethanol molecules.

17. Hydrogen Bonding & boiling point Increases boiling point, higher temperature needed to separate the molecules. Hexane 69 deg. 1-pentanol 138 1,4-butanediol 230 Ethanol 78 deg Dimethyl ether 24

18. Earlier Discussion of Acidity Methanol Ethanol 2-Propanol 2-Methyl-2-propanol Increasing Hinderance of Solvation RO-H  RO – (solvated) + H + (solvated) Increasing Acidity of the alcohol Recall: H 2 O + Na  Na + + OH - + ½ H 2 (g) Alcohols behave similarly ROH + Na  Na + + OR - + ½ H 2 (g) Alkoxide, strong base, strong nucleophile (unless sterically hindered) Also: ROH + NaH  Na + + OR - + ½ H 2 (g) Increasing Basicity of Alkoxide Anion, the conjugate base Alkoxide ion, base Alkoxides can be produced in several ways…

19. -OH as a Leaving Group R-OH + H +  R-OH 2 + Protonation of the alcohol sets-up a good leaving group, water. Poor leaving group, hydroxide ion. Another way to turn the –OH into a leaving group…

20. Conversion to Alkyl Halide, HX + ROH  RX + H 2 O When a carbocation can be formed (Tertiary, Secondary alcohols) beware of rearangement. S N 1 Expect both configurations. When a carbocation cannot be formed. Methanol, primary. S N 2

21. But sometimes experiment does not agree with our ideas… Observed reaction The problem: Rearrangement of carbon skeleton which usually indicates carbocations. Reacting alcohol is primary; do not expect carbocation. Time to adjust our thinking a bit…. Not a primary carbocation

22. Other ways to convert: ROH  RX We have used acid to convert OH into a good leaving group There are other ways to accomplish the conversion to the halide. Leaving group. Next, a very useful alternative to halide…

23. An alternative to making the halide: ROH  ROTs p-toluenesulfonyl chloride Tosyl chloride TsCl Tosylate group, -OTs, good leaving group, including the oxygen. The configuration of the R group is unchanged. Preparation from alcohols.

24. Example Preparation of tosylate. Retention of configuration

25. Substitution on a tosylate The –OTs group is an excellent leaving group

26. Acid Catalyzed Dehydration of an Alcohol, discussed earlier as reverse of hydration Protonation, establishing of good leaving group. Elimination of water to yield carbocation in rate determining step. Expect tertiary faster than secondary. Rearrangements can occur. Elimination of H + from carbocation to yield alkene. Zaitsev Rule followed. Secondary and tertiary alcohols, carbocations

27. Primary alcohols Problem: primary carbocations are not observed. Need a modified, non-carbocation mechanism. Recall these concepts: 1.Nucleophilic substitution on tertiary halides invokes the carbocation but nucleophilic substitution on primary RX avoids the carbocation by requiring the nucleophile to become involved immediately. 2.The E2 reaction requires the strong base to become involved immediately. Note that secondary and tertiary protonated alcohols eliminate the water to yield a carbocation because the carbocation is relatively stable. The carbocation then undergoes a second step: removal of the H +. The primary carbocation is too unstable for our liking so we combine the departure of the water with the removal of the H +. What would the mechanism be???

28. Here is the mechanism for acid catalyzed dehydration of Primary alcohols 1. protonation 2. The carbocation is avoided by removing the H at the same time as H 2 O departs (like E2). As before, rearrangements can be done while avoiding the primary carbocation.

29. Principle of Microscopic Reversibility Same mechanism in either direction.

30. Pinacol Rearrangement: an example of stabilization of a carbocation by an adjacent lone pair. Overall:

31. Mechanism Reversible protonation. Elimination of water to yield tertiary carbocation. 1,2 rearrangement to yield resonance stabilized cation. Deprotonation. This is a protonated ketone!

32. Oxidation Primary alcohol RCH 2 OH RCH=O RCO 2 H Na 2 Cr 2 O 7 Na 2 Cr 2 O 7 (orange)  Cr 3+ (green) Actual reagent is H 2 CrO 4, chromic acid. Secondary R 2 CHOH R 2 C=O Tertiary R 3 COH NR KMnO 4 (basic) can also be used. MnO 2 is produced. The failure of an attempted oxidation (no color change) is evidence for a tertiary alcohol. Na 2 Cr 2 O 7

33. Example…

34. Oxidation using PCC Primary alcohol RCH 2 OH RCH=O PCC Secondary R 2 CHOH R 2 C=O Stops here, is not oxidized to carboxylic acid

35. Periodic Acid Oxidation

36. Mechanistic Notes Cyclic structure is formed during the reaction. Evidence of cyclic intermediate.

37. Sulfur Analogs, Thiols Preparation RI + HS -  RSH S N 2 reaction. Best for primary, ok secondary, not tertiary (E2 instead) Acidity H 2 SpK a = 7.0 RSH pK a = 8.5 Oxidation