Hemiacetals and Acetals, carbonyls and alcohols

1. Hemiacetals and Acetals, carbonyls and alcohols Addition reaction. (Unstable in Acid; Unstable in base) (Unstable in Acid; Stable in base) Substitution reaction

2. Acetals as Protecting Groups Br-Mg E Synthetic Problem, do a retrosynthetic analysis Target molecule N Form this bond by reacting a nucleophile with an electrophile. Choose Nucleophile and Electrophile centers. Grignard would react with this carbonyl. The nucleophile could take the form of an organolithium or a Grignard reagent. The electrophile would be a carbonyl. Do you see the problem with the approach??

3. Use Protecting Group for the carbonyl… Acetals are stable (unreactive) in neutral and basic solutions. Create acetal as protecting group. protect Now create Grignard and then react Grignard with the aldehyde to create desired bond. react Remove protecting group. deprotect Same overall steps as when we used silyl ethers: protect, react, deprotect.

4. Tetrahydropyranyl ethers (acetals) as protecting groups for alcohols. Recall that the key step in forming the acetal was creating the carbocation as shown… There are other ways to create carbocations…… Recall that we can create carbocations in several ways: 1. As shown above by a group leaving. 2. By addition of H+ to a C=C double bond as shown next. This resonance stabilized carbocation then reacts with an alcohol molecule to yield the acetal. An acid This cation can now react with an alcohol to yield an acetal. The alcohol becomes part of an acetal and is protected.

5. Sample Problem Provide a mechanism for the following conversion First examination: have acid present and will probably protonate Forming an acetal. Keep those mechanistic steps in mind. Ok, what to protonate? Several oxygens and the double bond. Protonation of an alcohol can set-up a better leaving group. Protonation of a carbonyl can create a better electrophile. We do not have a carbonyl but can get a similar species as before.

6. Strongly electrophilic center, now can do addition to the C=O The protonation of the C=C Now do addition, join the molecules Product Now must open 5 membered ring here. Need to set-up leaving group.

7. Leaving group leaves…. Followed by new ring closure. Done. Wow!

8. Sulfur Analogs Consider formation of acetal Sulfur Analog

9. The aldehyde hydrogen has been made acidic Why acidic? Sulfur, like phosphorus, has 3d orbitals capable of accepting electrons: violating octet rule.

10. Recall early steps from the Wittig reaction discussed earlier This hydrogen is acidic. Why acidic? The P is positive and can accept charge from the negative carbon into the 3d orbitals

11. Some Synthetic Applications

12. Umpolung – reversed polarity What we have done in these synthetic schemes is to reverse the polarity of the carbonyl group; change it from an electrophile into a nucleophile. electrophile nucleophilic Can you think of two other examples of Umpolung we have seen?

13. Nitrogen Nucleophiles

14. Mechanism of Schiff Base formation Attack of nucleophile on the carbonyl Followed by transfer of proton from weak acid to strong base. Protonation of –OH to establish leaving group. Leaving group departs, double bond forms.

15. Hydrazine derivatives

16. Note which nitrogen is nucleophilic Nucleophilic nitrogen Favored by resonance Less steric hinderance

17. Reductive Amination Pattern: R2C=O + H2N-R’   R2CH-NH-R’

18. Enamines Recall primary amines react with carbonyl compounds to give Schiff bases (imines), RN=CR2. Primary amine But secondary amines react to give enamines See if you can write the mechanism for the reaction. Secondary Amine

19. Acidity of a Hydrogens a hydrogens are weakly acidic Weaker acid than alcohols but stronger than terminal alkynes. Learn this table….

20. Keto-Enol Tautomerism (Note: we saw tautomerism before in the hydration of alkynes.) Fundamental process Mechanism in base: Negative carbon, a carbanion, basic, nucleophilic carbon. Additional resonance form, stabilizing anion, reducing basicity and nucleophilicity. Protonation to yield enol form.

21. Details… Base strength Alkoxides will not cause appreciable ionization of simple carbonyl compounds to enolate. Strong bases (KH or NaNH2) will cause complete ionization to enolate. Double activation (1,3 dicarbonyl compounds) will be much more acidic. For some 1,3 dicarbonyl compounds the enol form may be more stable than the keto form.

22. More details… Nucleophilic carbon nucleophilicity Some examples:

23. Some reactions related to acidity of a hydrogens Racemization Exchange

24. Oxidation: Aldehyde  Carboxylic Recall from the discussion of alcohols. Milder oxidizing reagents can also be used Tollens Reagent test for aldehydes

25. “Drastic Oxidation” of Ketones Obtain four different products in this case.

26. Reductions: two electron NaBH4 Or LiAlH4

27. Reductions: Four Electron Clemmenson Wolf-Kishner

28. Mechanism of Wolf-Kishner, C=O  CH2 Recall reaction of primary amine and carbonyl to give Schiff base. Here is the formation of the Schiff base. We expect this to happen. These hydrogens are weakly acidic, just as the hydrogens a to a carbonyl are acidic. Weakly acidic hydrogen removed. Resonance occurs. Same as keto/enol tautomerism. Protonation (like forming the enol) Perform an elimination reaction to form N2.

29. Haloform Reaction, overall The last step which produces the haloform, HCX3 only occurs if there is an a methyl group, a methyl directly attached to the carbonyl. a methyl If done with iodine then the formation of iodoform, HCI3, a bright yellow precipitate, is a test for an a methyl group (iodoform test).

30. Steps of Haloform Reaction The first reaction: • All three H’s replaced by • This must happen • stepwise, like this: Pause for a sec: We have had three mechanistic discussions of how elemental halogen, X2, reacts with a hydrocarbon to yield a new C-X bond. Do you recall them? Radical Reaction: R. + X-X  R-X + X. (initiation required) Addition to double bond: C=C + X-X  + Br- (alkene acts as nucleophile, ions) Nucleophilic enolate anion:

31. Mechanism of Haloform Reaction-1 Using the last of the three possibilities One H has been replaced by halogen. Repeat twice again to yield Where are we? The halogens have been introduced. First reaction completed. But now we need a substitution reaction. We have to replace the CBr3 group with OH.

32. Mechanism of Haloform - 2 This is a substitution step; OH- replaces the CX3 and then ionizes to become the carboxylate anion. Here’s how: Attack of hydroxide nucleophile. Formation of tetrahedral intermediate. Anticipate the attack… Reform the carbonyl double bond. CX3- is ejected. The halogens stabilize the negative carbon. Neutralization.

33. Cannizaro Reaction Overall: Restriction: no a hydrogens in the aldehydes. a hydrogens No a hydrogens Why the restriction? The a hydrogens are acidic leading to ionization.

34. Mechanism • What can happen? Reactants are the aldehyde and concentrated hydroxide. • Hydroxide ion can act both as • Base, but remember we have no acidic hydrogens (no a hydrogens). • Nucleophile, attacking carbonyl group. Attack of nucleophilic HO- Acid-base Re-establish C=O and eject H- which is immediately received by second RCHO

35. Experimental Evidence These are the hydrogens introduced by the reaction. They originate in the aldeyde and do not come from the aqueous hydroxide solution.

36. Kinetic vs Thermodynamic Contol of a Reaction Examine Addition of HBr to 1,3 butadiene

37. Mechanism of reaction. Allylic resonance But which is the dominant product?

38. Nature of the product mixture depends on the temperature. Product mixture at -80 deg 80% 20% Product mixture at + 40 deg 20% 80% Goal of discussion: how can temperature control the product mixture?

39. When two or more products may be formed in a reaction A  X or A  B Thermodynamic Control: Most stable product dominates Kinetic Control: Product formed fastest dominates Thermodynamic control assumes the establishing of equilibrium conditions and the most stable product dominates. Kinetic Control assumes that equilibrium is not established. Once product is made it no longer changes. Equilibrium is more rapidly established at high temperature. Thermodynamic control should prevail at high temperature where equilibrium is established. Kinetic Control may prevail at low temperature where reverse reactions are very slow.

40. Nature of the product mixture depends on the temperature. Product mixture at -80 deg 80% 20% Product mixture at + 40 deg 20% 80% More stable product Thermodynamic Control Kinetic Control Product formed most quickly, lowest Ea

41. Formation of the allylic carbocation. Can react to yield 1,2 product or 1,4 product.

42. Most of the carbocation reacts to give the 1,2 product because of the smaller Ea leading to the 1,2 product. This is true at all temperatures. At low temperatures the reverse reactions do not occur and the product mixture is determined by the rates of forward reactions. No equilibrium.

43. Most of the carbocation reacts to give the 1,2 product because of the smaller Ea leading to the 1,2 product. This is true at all temperatures. At higher temperatures the reverse reactions occur leading from the 1,2 or 1,4 product to the carbocation. Note that the 1,2 product is more easily converted back to the carbocation than is the 1,4. Now the 1,4 product is dominant.

44. Diels Alder Reaction/Symmetry Controlled Reactions Quick Review of formation of chemical bond. Electron donor Electron acceptor Note the overlap of the hybrid (donor) and the s orbital which allows bond formation. For this arrangement there is no overlap. No donation of electrons; no bond formation.

45. Diels Alder Reaction of butadiene and ethylene to yield cyclohexene. We will analyze in terms of the pi electrons of the two systems interacting. The pi electrons from the highest occupied pi orbital of one molecule will donate into an lowest energy pi empty of the other. Works in both directions: A donates into B, B donates into A. B HOMO donates into A LUMO Note the overlap leading to bond formation LUMO acceptor LUMO acceptor A HOMO donates into B LUMO HOMO donor HOMO donor Note the overlap leading to bond formation B A

46. Try it in another reaction: ethylene + ethylene  cyclobutane LUMO LUMO Equal bonding and antibonding interaction, no overlap, no bond formation, no reaction HOMO HOMO

47. Reaction Problem

48. Synthesis problem

49. Mechanism Problem Give the mechanism for the following reaction. Show all important resonance structures. Use curved arrow notation.