Reduction of Vanillin to Vanillyl Alcohol

2. [ H2 ] Reduction of Vanillin to Vanillyl Alcohol Organic Synthesis and Infrared Identification

3. ? QUESTIONS ? How are organic reactions planned and conducted? What reagents can be used to conduct hydrogenations? What is the basis for the absorption of IR radiation by molecules? How is IR spectroscopy used to ascertain the structure of a substance?

4. Purpose: To conduct organic reaction, isolate product & characterize product using infrared spectroscopy Concepts: Synthesis starting material product theoretical yield percent yield reduction organic functional groups characteristic infrared absorptions      Techniques: handling micro-scale quantities quantitative transfer of liquids and solids infrared spectroscopy analyzing infrared spectra crystallization vacuum filtration  

5. Vanillin • principal flavoring agent in vanilla beans • cured, unripe fruit of a • plant in the orchid family. Vanillin is our starting materialfor synthesis of the related compound: Vanillyl alcohol

6. NOMENCLATURE – FUNCTIONAL GROUPS aldehyde 1 6 2 methyl oxy methoxy 3 5 4 benzene 3-methoxy 4-hydroxy benzaldehyde (Vanillin)

7. NOMENCLATURE – FUNCTIONAL GROUPS 3-methoxy 4-hydroxy benzyl alcohol (Vanillyl Alcohol)

8. PRINCIPAL CONSTITUENTS OF THE RIPE VANILLA BEAN A chromatogram Rf 1) 4-Hydroxybenzyl alcohol 3)vanillyl alcohol 6) 4-hydroxybenzaldehyde 9)vanillin 10) coumaric acid

9. Our Objective N.B. In synthetic exercises, you are expected to know the formulas and structures of the reactants and products! E.g., Alum, Vanillin, Vanillyl Alcohol, etc.

10. In organic chemistry, reduction often means addition of a hydrogen moleculeto a multiple (e.g., double) bond. We have earlier described reduction as the addition of electrons to a molecule (e.g., I2 + 2e- 2 I-) H - H + H - H + Hydrogen can be added to organic compounds in many ways. As hydrogen gas - or using inorganic hydrides.

11. Different ways of adding hydrogen give different results depending on type of multiple bonds in reactant. We seek a way to add two hydrogen atoms (i.e., reduce) to the C==O bond C O ≈ C C without reducing bonds in benzene ring. A reagent which accomplishes this is:

12. Sodium Borohydride - H Hydride Ion H-   Na+   H B H     H

13. In organic chemistry, reductionoften refers to the removalof atoms of hydrogen across a multiple bond • True • False

14. In organic chemistry, reductionoften refers to the removalof atoms of hydrogen across a multiple bond. addition H - H + H - H + B= False

15. Structures of starting and product molecules differ in a way that makes infrared spectroscopy an appropriate analytical tool to establish identity of the product (and a rough indication of its purity) Have previously done absorption spectroscopy of food dyes (ultravioletand visible). Those absorptions were due to transitions between ELECTRONIC energy levels (UV) 350 nm – 700 nm (Red) Absorption of light in infrared region is primarily due to VIBRATIONof molecules. (1 μm = ) 1,000 nm – 100,000 nm (Infra-red)

16. Study of many thousands of substances shows that SPECIFIC MOLECULAR FRAGMENTS absorb light at well-defined,  CHARACTERISTICWAVELENGTHS  = c  (nm) ~ 9100 ~ 6100 ~ 5800 ~ 3400 or, since 1 m = 1000 nm  (µm) ~ 9.1 ~ 6.1 ~ 5.8 ~ 3.4 Wavelength is convenient measure of light in visible region

17. Convention in infrared spectra is to report frequency, f, in terms of number of oscillations in 1 cm ( # / cm ) I.e., f (cm-1)= 1 /  =  / c or wavenumber instead of the wavelength,  The rational unit for this form of frequency is cm-1. •  = 0.2 cm • f = 1/ = 5 cm-1 • = 3 X 1010 / 0.2  2 X 1011 sec-1  = 0.2 cm  = c 1 cm Infrared photon wavelengths are in the approximate range 1 m - 100 m (or 1 X 10-4cm– 1 X 10-2cm) The IR range becomes 100 cm-1 – 10,000 cm-1

18. CHARACTERISTICWAVELENGTHS or FREQUENCIES (wavenumber) in cm-1  (µm) ~ 9.1 ~ 6.1 ~ 5.8 ~ 3.4 f (cm-1) ~1100 ~1650 ~1720 ~2900

19. The C=C bond absorbs IR radiation of 6.1 m wavelength. To what wavenumberdoes that correspond? • 1600 cm-1 • 0.16 sec-1 • 6100 mm • 1600 cm

20. The C=C bond absorbs IR radiation of 6.1 m wavelength. To what wavenumberdoes that correspond? A1600 cm-1 B 0.16 sec-1 C 6100 mm D 1600 cm X X X

21.  = 6.1 μm = 6.1 X 10-6 m wavenumber = 1 / = 1 / 6.1 X 10-6 X 1 m 100 cm = 1.6 X 105 m-1 = 1.6 X 103 cm-1 A= 1600 cm-1

22. GROUP VIBRATIONS

23. Table 2 of SUPL-005 shows the absorption frequencies in cm-1 of some molecular fragments Here are some that are related to today’s exercise. “Aromatic” means benzene or benzene-like C  C (aromatic) 1600, 1500 C — H (aromatic) 3030 – 3050 C — H (alkane) 2850 – 2960 C == O (aldehyde) 1680 – 1750 O — H (phenol) 3200 O — H (alcohol) 3400 – 3650

24. Vanillin IR Spectrum: 500 cm-1 – 4000 cm-1

25. Note that like visible spectra, IR spectra are displayed as intensity vs increasing wavelength BUT as Percent Transmittance(instead of absorbance), and indicating the (decreasing) wavenumber scale instead of wavelength So, absorption peaks point DOWNWARD %Transmittance etc. Wavelength

26. Vanillin IR Spectrum: 1500 cm-1 – 4000 cm-1 O-H C-H3 -H HC=O CC 4000 cm-1 3000 cm-1 2000 cm-1 1500 cm-1

27. Explanation of Spectrum Notation • We examine vanillin spectrum between 1500 and 4000 cm-1. • There are 6 major peaks in this region. • O−H stretch due to the OH group on the ring • −Hring hydrogen stretch • C−H3C−H stretch in the methoxy (O-CH3)group • HC=O C=O stretch in the aldehyde group • CC two peaks due to the ring CC stretch All but one of these peaks should show up in spectrum of product, vanillyl alcohol.

28. The infrared spectrum of the product, vanillyl alcohol, will absorb near 1700 cm-1 due to the vibration of a C=O double bond. • True • False

29. The infrared spectrum of the product, vanillyl alcohol, will absorb near 1700 cm-1 due to the vibration of a C=O double bond. B= False The C=O double bond is the one which is reduced in the reaction.

30. So, the product spectrum should show the absence of the C=O absorption near 1700 cm-1. What other difference should there be? There should be a new absorptiondue to O−Hin alcohol group. That absorption is near, but distinct from, the O—H absorption due to the OH group on the ring (phenol At ~3200 cm-1). From the table we see that we expect it at: H2C O—Hbetween 3400 and 3600 cm-1 in the alcohol O-H region.

31. Procedure for IR Spectrum of Vanillyl Alcohol • When sample is DRY, • obtain the spectrum of a small sample using the FTIR Spectrometer. Follow the posted instructions • Analyzethe spectrum to identify the peaks due to the product (and, if any, due to the starting material)

32. INTERPRETATION OF IR SPECTRA Use infrared spectrum to verify the presence or absence of functional groups Reaction replaces a -HC=Ogroup by a –H2C-O-H. So, starting materialwillshow: absorption by -HC=O absence of absorptions by –H2C-O-H Productshould show: absorption by –H2C-O-H absence of absorption by –HC=O Should also be able to identify absorptions of other functional groups common to vanillin and vanillyl alcohol by comparing their spectra.

33. A Brief Description of FTIRandHATR UV-visible spectrometer: Scans individual wavelength – measures %T at that wavelength - proceeds to next wavelength, etc. FTIR: Scans all wavelengths at once - measures total %T – changes source intensity profile at high rate and measures total %T as a function of time. FTIR: Fourier Transform Infra Red Transmission Spectroscopy: Reflection Spectroscopy: I0 () It () I0 (t) Ir(t) HATR: Horizontal Attenuated Total Reflectance

34. IR Spectrometer The ZnSe sample area

35. SYNTHETIC PROCEDURE • Will be handling smallquantities of materials. • 400 mg of vanillin (C8H8O3) - [ 2.6 mmol ] • 2.5 mL of 1.0 M NaOH - [ 2.5 mmol ] • 80 mg of sodium borohydride (NaBH4)- [ 2.1 mmol ] • Less than10 mL of 2.5 M HCl - [ 25 mmol ] • Must exercise care in transferring • such amounts between containers. 400 ± 40 mg but exactly 2.5 ± 0.2 mL 80 ± 8 mg but exactly

36. STOICHIOMETRY 4 4 + BH4- + 4 H2O + H3BO3 + OH-

37. Calculations 100 X Actual yield Pct yield = ----------------------- Theoretical yield Theoretical yield = maximum yield that could be produced from actual amount of limiting reagent. 400 mg / 152 = 2.6mmol E.g., 0.400 g vanillin (MM = 152) 80 mg / 38 = 2.1mmol E.g., 0.080g NaBH4 (MM = 38)

38. If 2.6 mmol of vanillin react with 2.1 mmol of NaBH4 which is the limiting reagent? • vanillin • NaBH4 • it depends on the molar mass of vanillyl alcohol

39. If 2.6 mmol of vanillin react with 2.1 mmol of NaBH4 what is the limiting reagent? 2.6 mmol 4 4 + 1 BH4- + 4 H2O + H3BO3 + OH- 2.1 mmol X 4 = 8.4 mmol A vanillin

40. As defined earlier Calculations 100 X Actual yield Pct yield = ----------------------- Theoretical yield Limiting Reagent Theoretical yield = maximum yield that could be produced from actual amount of limiting reagent. 400 mg / 152 = 2.63 mmol E.g., 0.400 g vanillin (MM = 152) 2.63 X 154 = 0.405 g Could make 2.63 mmol vanillyl alcohol If you actually recover 0.349 g 100 X 0.349 % Yield = ---------------- = 86.2% 0.405

41. PROCEDURE – Special Notes • Pay close attention to directions: • Add NaBH4slowlyto coldsolution • Let reaction mixture stand at room temperature • Chill with ice for recommended period • Adjust pH to acid litmus test slowly. Be sure that entiresolution is acidic, but not excessively. For 30 min For 10 min Dry sample for melting point and IR.

42. NEXT WEEK Acid content of fruit juices and soft drinks SUSB-010 Do Pre-Lab Exercise

44. Grading - Reminder As of today, have completed: 3 Preliminary Exercises @ 55 165 2 Final Exercises @ 105 210 1 Quiz @ 50 50 6 Lectures @ 5 30 Total 475 A- / B+ 0.90 X 475 = 428 B- / C+ 0.80 X 475 = 380 C / D 0.70 X 475 = 333 D / F 0.60 X 475 = 285 Grading details are on the course web site.