24.10 Carboxylation of Phenols

1. O OCCH3 COH O 24.10Carboxylation of Phenols • Aspirin and the Kolbe-Schmitt Reaction

2. O O O OH OCCH3 CH3COCCH3 COH COH O O Aspirin is prepared from salicylic acid • how is salicylic acid prepared? H2SO4

3. OH ONa CONa O Preparation of Salicylic Acid CO2 • called the Kolbe-Schmitt reaction • acidification converts the sodium salt shownabove to salicylic acid 125°C, 100 atm

4. •• •• – O H O •• •• •• •• – C O •• •• O •• •• What Drives the Reaction? • acid-base considerations provide an explanation:stronger base on left; weaker base on right + CO2 stronger base: pKa of conjugate acid = 10 weaker base: pKa of conjugate acid = 3

5. OH ONa CONa O Preparation of Salicylic Acid CO2 • how does carbon-carbon bond form? • recall electron delocalization in phenoxide ion • negative charge shared by oxygen and by the ring carbons that are ortho and para to oxygen 125°C, 100 atm

6. – •• •• O O •• •• •• H H – H H •• H H H H H H •• •• O O •• •• H H – H H •• •• H H – H H H H

7. O – •• O •• – •• •• O •• •• C H O •• •• Mechanism of ortho Carboxylation •• •• •• C O •• O •• •• H

8. •• – O •• •• C O •• •• •• O H •• •• – C O •• •• O •• •• Mechanism of ortho Carboxylation •• O •• – •• •• C O O •• •• •• O •• •• H H

9. – •• O •• •• •• •• O H O H •• •• •• – – •• C O C O •• •• •• •• O •• •• O •• •• Why ortho?Why not para?

10. – •• O •• •• •• •• O H O H •• •• •• – – •• C O C O •• •• •• •• O •• •• O •• •• Why ortho?Why not para? stronger base: pKa of conjugate acid = 4.5 weaker base: pKa of conjugate acid = 3

11. O H – O C O Intramolecular Hydrogen Bondingin Salicylate Ion Hydrogen bonding between carboxylate and hydroxylgroup stabilizes salicylate ion. Salicylate is less basic than para isomer and predominates under conditionsof thermodynamic control.

12. 24.11Preparation of Aryl Ethers

13. ONa OR Typical Preparation is by Williamson Synthesis SN2 + + NaX RX

14. ONa OR + X RONa Typical Preparation is by Williamson Synthesis SN2 + + NaX RX but the other combination fails because aryl halides are normally unreactivetoward nucleophilic substitution

15. OCH3 ONa Example acetone + CH3I heat (95%)

16. H2C CHCH2Br OH OCH2CH CH2 Example + K2CO3 acetone, heat (86%)

17. F OCH3 NO2 NO2 Aryl Ethers from Aryl Halides • nucleophilic aromatic substitution is effective with nitro-substituted (ortho and/or para) aryl halides CH3OH + + KOCH3 KF 25°C (93%)

18. 24.12Cleavage of Aryl Ethersby Hydrogen Halides

19. Ar – •• •• •• •• + Br O H Br H •• •• •• •• •• •• R Cleavage of AlkylAryl Ethers Ar + O + R

20. Ar – •• •• •• •• + Br O H Br H •• •• •• •• •• •• R Ar •• •• Br R O H •• •• •• Cleavage of AlkylAryl Ethers Ar • An alkyl halide is formed; never an aryl halide! + O + R +

21. OCH3 OH OH OH Example HBr + CH3Br heat (85-87%) (57-72%)

22. 24.13Claisen Rearrangementof Allyl Aryl Ethers

23. OCH2CH CH2 200°C OH CH2CH CH2 Allyl Aryl Ethers Rearrange on Heating • allyl group migrates to ortho position (73%)

24. OCH2CH CH2 O O H Mechanism rewrite as keto-to-enolisomerization OH

25. O O H Sigmatropic Rearrangement Claisen rearrangement is an example of a sigmatropic rearrangement. A  bond migratesfrom one end of a conjugated  electron systemto the other. this  bond breaks “conjugated  electron system” is the allyl group this  bond forms

26. 24.14Oxidation of Phenols:Quinones

27. OH O Na2Cr2O7, H2SO4 H2O OH O Quinones • The most common examples of phenol oxidationsare the oxidations of 1,2- and 1,4-benzenediolsto give quinones. (76-81%)

28. OH OH CH3 CH3 Quinones • The most common examples of phenol oxidationsare the oxidations of 1,2- and 1,4-benzenediolsto give quinones. O O Ag2O diethyl ether (68%)

29. O OH OH O Some quinones are dyes Alizarin(red pigment)

30. O CH3 CH3O CH3O n O Some quinones are important biomolecules Ubiquinone (Coenzyme Q) n = 6-10 involved in biological electron transport

31. O CH3 CH3 CH3 CH3 CH3 CH3 O Some quinones are important biomolecules Vitamin K (blood-clotting factor)

32. Section 24.15Spectroscopic Analysis of Phenols

33. Infrared Spectroscopy • infrared spectra of phenols combine features of alcohols and aromatic compounds • O—H stretch analogous to alcohols near 3600 cm-1 • C—O stretch at 1200-1250 cm-1

34. CH3 OH 3500 3000 2500 2000 1500 1000 500 Figure 24.3: Infrared Spectrum of p-Cresol C—H C—O O—H Wave number, cm-1

35. H H CH3 HO H H 1H NMR Hydroxyl proton of OH group lies between alcoholsand carboxylic acids; range is ca.  4-12 ppm(depends on concentration). For p-cresol the OHproton appears at  5.1 ppm (Figure 24.4).

36. H H HO CH3 H H 10.0 9.0 8.0 7.0 6.0 5.0 4.0 3.0 2.0 1.0 0 Chemical shift (, ppm)

37. OH 155.1 112..3 116.1 139.8 129.4 CH3 21.3 121.7 13C NMR Oxygen of hydroxyl group deshields carbonto which it is directly attached. The most shielded carbons of the ring are those thatare ortho and para to the oxygen.

38. OH max 204 nm 256 nm max 210 nm 270 nm max 235 nm 287 nm UV-VIS Oxygen substitution on ring shifts max to longerwavelength; effect is greater in phenoxide ion. – O

39. •+ OH •• Mass Spectrometry Prominent peak for molecular ion. Most intense peak in phenol is for molecular ion. m/z 94

40. End of Chapter 24