1 O C H H O H C H [ H2 ] O CH3 O H O CH3 O H Reduction of Vanillin to Vanillyl Alcohol Organic Synthesis and Infrared Identification 2 ? 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? 3 Purpose: To conduct organic reaction, isolate product & characterize product using infrared spectroscopy Concepts: Synthesis starting material theoretical yield percent yield organic functional groups characteristic infrared absorptions product reduction Techniques: handling micro-scale quantities quantitative transfer of liquids and solids infrared spectroscopy analyzing infrared spectra crystallization vacuum filtration 4 Vanillin • principal flavoring agent in vanilla beans • cured, unripe fruit of a plant in the orchid family. Vanillin is our starting material for synthesis of the related compound: Vanillyl alcohol 5 NOMENCLATURE – FUNCTIONAL GROUPS aldehyde 1 6 2 3 5 methyl oxy methoxy 4 benzene 3-methoxy 4-hydroxy benzaldehyde (Vanillin) 6 NOMENCLATURE – FUNCTIONAL GROUPS H O C H H 3-methoxy 4-hydroxy benzyl alcohol (Vanillyl Alcohol) 7 PRINCIPAL CONSTITUENTS OF THE RIPE VANILLA BEAN CH2OH A chromatogram OCH3 OH H C O Rf OCH 3 OH 1) 3) 6) 9) 10) 4-Hydroxybenzyl alcohol vanillyl alcohol 4-hydroxybenzaldehyde vanillin coumaric acid 8 Our Objective O C H O H H C H OCH3 OH OCH3 OH 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. 9 In chemistry, reduction oftenas means addition of Weorganic have earlier described reduction the addition -) a to a multiple ofhydrogen electronsmolecule to a molecule (e.g., I2 (e.g., + 2e- double) 2 Ibond. H H-H + H C C H H ethylene H H C C H H H H ethane H H H C O H-H + H formaldehyde H C O H methyl alcohol Hydrogen can be added to organic compounds in many ways. As hydrogen gas - or using inorganic hydrides. 10 Different ways of adding hydrogen give different results depending on type of multiple bonds in reactant. O H C C C H C H H We seek a way to add two hydrogen atoms (i.e., reduce) to the C C==O O bond C without reducing C bonds in benzene ring. ≈ H C C C H C O H O H A reagent which accomplishes this is: 11 Sodium Borohydride H + Na H B Hydride Ion H- H H 12 In organic chemistry, reduction often refers to the removal of atoms of hydrogen across a multiple bond A. True B. False 0% A. 0% B. 13 In organic chemistry, reduction often refers to the addition removal of atoms of H H bond. H hydrogen H across a multiple H-H + C C C C H H ethylene H H H H ethane H H H C O H-H + H H formaldehyde C O H methyl alcohol B = False 14 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 (ultraviolet and visible). Absorbance vs Wavelength 0.7 0.6 0.5 0.4 0.3 0.2 0.1 Absorption of light in infrared region is primarily due to VIBRATION of molecules. (1 μm = ) 1,000 nm – 100,000 nm (Infra-red) 15 0 0 74 0 0 0 Wavelength (nm) 71 68 65 0 0 0 0 0 0 0 62 59 56 53 50 47 44 41 0 0 38 Absorbance 0.8 0 (UV) 350 nm – 700 nm (Red) 0.9 35 Those absorptions were due to transitions between ELECTRONIC energy levels 1 Study of many thousands of substances shows that SPECIFIC MOLECULAR FRAGMENTS absorb light at well-defined, CHARACTERISTIC WAVELENGTHS =c C C (nm) ~ 9100 (µm) C C C O C H ~ 6100 ~ 5800 ~ 3400 or, since 1 m = 1000 nm ~ 9.1 ~ 6.1 ~ 5.8 ~ 3.4 16 Wavelength is convenient measure of light in visible region 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 10-4 cm – 1 X 10-2 cm) The IR range becomes 100 cm-1 – 10,000 cm-1 17 CHARACTERISTIC WAVELENGTHS or FREQUENCIES (wavenumber) in cm-1 C C C C C O C H (µm) ~ 9.1 ~ 6.1 ~ 5.8 ~ 3.4 f (cm-1) ~1100 ~1650 ~1720 ~2900 18 The C=C bond absorbs IR radiation of 6.1 m wavelength. To what wavenumber does that correspond? A. B. C. D. 1600 cm-1 0.16 sec-1 6100 mm 1600 cm 0% A. 0% B. 0% C. 0% D. 19 The C=C bond absorbs IR radiation of 6.1 m wavelength. To what wavenumber does that correspond? A 1600 cm-1 B 0.16 sec X -1 C 6100 mm X D 1600 cm X 20 = 6.1 μm = 6.1 X 10-6 m wavenumber = 1 / = 1 / 6.1 X 10-6 = 1.6 X 105 m-1 X 1m 100 cm = 1.6 X 103 cm-1 A = 1600 cm-1 21 GROUP VIBRATIONS 22 Table 2 of SUPL-005 shows the absorption frequencies in cm-1 of some molecular fragments “Aromatic” means Here are some that are related to benzene or benzene-like today’s exercise. C C (aromatic) C — H (aromatic) C — H (alkane) C == O (aldehyde) O — H (phenol) O — H (alcohol) 1600, 1500 H 3030 – 3050 2850 – 2960 1680 – 1750 3200 H C H C H C C C O O H H H H C O C H C H 3400 – 3650 23 Vanillin IR Spectrum: 500 cm-1 – 4000 cm-1 24 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 % Transmittance So, absorption peaks point DOWNWARD Wavelength etc. 25 Vanillin IR Spectrum: 1500 cm-1 – 4000 cm-1 O-H C-H3 -H 4000 cm-1 3000 cm-1 HC=O 2000 cm-1 CC 26 1500 cm-1 Explanation of Spectrum Notation We examine vanillin spectrum between 1500 and 4000 cm-1. There are 6 major peaks in this region. O−H −H C−H3 HC=O CC stretch due to the OH group on the ring ring hydrogen stretch C−H stretch in the methoxy (O-CH3)group C=O stretch in the aldehyde group two peaks due to the ring CC stretch All but one of these peaks should show up in spectrum of product, vanillyl alcohol. 27 The infrared spectrum of the product, vanillyl alcohol, will absorb near 1700 cm-1 due to the vibration of a C=O double bond. A. True B. False 0% 0% 28 A. B. 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. 29 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 absorption due to O−H in 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—H between 3400 and 3600 cm-1 in the alcohol O-H region. 30 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 •Analyze the spectrum to identify the peaks due to the product (and, if any, due to the starting material) 31 INTERPRETATION OF IR SPECTRA Use infrared spectrum to verify the presence or absence of functional groups Reaction replaces a -HC=O group by a –H2C-O-H. So, starting material will show: absorption by -HC=O absence of absorptions by –H2C-O-H Product should 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 32 comparing their spectra. A Brief Description of FTIR and HATR UV-visible spectrometer: FTIR: Scans individual wavelength – measures %T at that wavelength - proceeds to next wavelength, etc. 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 () I0 (t) It () Ir (t) HATR: Horizontal Attenuated Total Reflectance 33 IR Spectrometer The ZnSe sample area 34 SYNTHETIC PROCEDURE Will be handling small quantities of materials. 400 mg of vanillin (C8H8O3) - [ 2.6 mmol ] 2.5 mL of 1.0 M NaOH - [ 2.5 mmol ] 400 ± 40 mg but exactly 2.5 ± 0.2 mL 80 mg of sodium borohydride (NaBH4)- [ 2.1 mmol ] 80 ± 8 mg but exactly Less than 10 mL of 2.5 M HCl - [ 25 mmol ] Must exercise care in transferring such amounts between containers. 35 STOICHIOMETRY O C H O H 4 H C H 4 OCH3 OH + BH4- + 4 H2O OCH3 OH + H3BO3 + OH- 36 Calculations 100 X Actual yield Pct yield = ----------------------Theoretical yield Theoretical yield = maximum yield that could be produced from actual amount of limiting reagent. E.g., 0.400 g vanillin (MM = 152) 400 mg / 152 = 2.6 mmol E.g., 0.080 g NaBH4 (MM = 38) 80 mg / 38 = 2.1 mmol 37 If 2.6 mmol of vanillin react with 2.1 mmol of NaBH4 which is the limiting reagent? A. B. C. vanillin NaBH4 it depends on the molar mass of vanillyl alcohol 0% A. 0% B. 0% C. 38 If 2.6 mmol of vanillin react with 2.1 mmol of NaBH4 what is the limiting reagent? O C H H O 2.6 mmol H C H 4 4 OCH3 OCH3 OH OH + 1 BH4- + 4 H2O + H3BO3 + OH- 2.1 mmol X 4 = 8.4 mmol A vanillin 39 As defined earlier Calculations 100 X Actual yield Limiting Pct yield = ----------------------Reagent Theoretical yield Theoretical yield = maximum yield that could be produced from actual amount of limiting reagent. E.g., 0.400 g vanillin (MM = 152) 400 mg / 152 = 2.63 mmol Could make 2.63 mmol vanillyl alcohol If you actually recover 0.349 g 100 X 0.349 % Yield = ---------------- = 86.2% 0.405 2.63 X 154 = 0.405 g 40 PROCEDURE – Special Notes Pay close attention to directions: • Add NaBH4 slowly to cold solution For 30 min • Let reaction mixture stand at room temperature • Chill with ice for recommended period For 10 min • Adjust pH to acid litmus test slowly. Be sure that entire solution is acidic, but not excessively. Dry sample for melting point and IR. 41 NEXT WEEK Acid content of fruit juices and soft drinks SUSB-010 Do Pre-Lab Exercise 42 43 Grading - Reminder As of today, have completed: 3 Preliminary Exercises @ 55 2 Final Exercises @ 105 1 Quiz @ 50 6 Lectures @ 5 Total ABC D / / / / B+ C+ D F 0.90 0.80 0.70 0.60 X 475 = X 475 = X 475 = X 475 = 165 210 50 30 475 428 380 333 285 Grading details are on the course web site. 44