Slide 1

Slide 2 1 Slide 3 [ H 2 ] Reduction of Vanillin to Vanillyl Alcohol Organic Synthesis and Infrared Identification 2 Slide 4 How are organic reactions planned and conducted? ? QUESTIONS ? How is IR spectroscopy used to ascertain the structure of a substance? What reagents can be used to conduct hydrogenations? What is the basis for the absorption of IR radiation by molecules? 3 Slide 5 Purpose: To conduct organic reaction, isolate product & characterize product using infrared spectroscopy Concepts: Synthesis starting materialproduct 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 4 Slide 6 principal flavoring agent in vanilla beans Vanillin is our starting material for synthesis of the related compound: cured, unripe fruit of a plant in the orchid family. Vanillyl alcohol Vanillin 5 Slide 7 NOMENCLATURE FUNCTIONAL GROUPS benzaldehyde4-hydroxy3-methoxy (Vanillin) benzene methyl oxy 2 1 4 3 methoxy aldehyde 6 5 6 Slide 8 NOMENCLATURE FUNCTIONAL GROUPS benzyl alcohol4-hydroxy3-methoxy (Vanillyl Alcohol) 7 Slide 9 A chromatogram 1) 4-Hydroxybenzyl alcohol 3) vanillyl alcohol 6) 4-hydroxybenzaldehyde 9) vanillin 10) coumaric acid PRINCIPAL CONSTITUENTS OF THE RIPE VANILLA BEAN RfRf 8 Slide 10 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. 9 Slide 11 In organic chemistry, reduction often means addition of a hydrogen molecule to a multiple (e.g., double) bond. hydrides. Hydrogen can be added to organic compounds in many ways. As hydrogen gas - or using inorganic hydrides. H - H + + We have earlier described reduction as the addition of electrons to a molecule (e.g., I 2 + 2e - 2 I - ) 10 Slide 12 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 A reagent which accomplishes this is: CC CO without reducing bonds in benzene ring. 11 Slide 13 B H H H H - Na + Sodium Borohydride Hydride Ion H - 12 Slide 14 removal In organic chemistry, reduction often refers to the removal of atoms of hydrogen across a multiple bond A.True B.False 13 Slide 15 H - H + B = False removal In organic chemistry, reduction often refers to the removal of atoms of hydrogen across a multiple bond. addition + 14 Slide 16 Structures of starting and product molecules differ in a way that makes infrared spectroscopy identity and a rough indication of its purity 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). Absorption of light in infrared region is primarily due to VIBRATION VIBRATION of molecules. Those absorptions were due to transitions between ELECTRONIC energy levels (UV) 350 nm 700 nm (Red) 1,000 nm 100,000 nm (1 m = ) (Infra-red) 15 Slide 17 Study of many thousands of substances shows that SPECIFIC MOLECULAR FRAGMENTS absorb light at well-defined, (nm) ~ 9100 ~ 6100 ~ 5800 ~ 3400 CHARACTERISTIC WAVELENGTHS or, since 1 m = 1000 nm (m) ~ 9.1 ~ 6.1 ~ 5.8 ~ 3.4 Wavelength is convenient measure of light in visible region = c 16 Slide 18 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 IR range becomes 100 cm -1 10,000 cm -1 Infrared photon wavelengths are in the approximate range 1 m - 100 m (or 1 X 10 -4 cm 1 X 10 -2 cm) 1 cm = 0.2 cm f = 1/ = 5 cm -1 = 3 X 10 10 / 0.2 2 X 10 11 sec -1 = c The rational unit for this form of frequency is cm -1. 17 Slide 19 (m) ~ 9.1 ~ 6.1 ~ 5.8~ 3.4 f (cm -1 ) ~1100 ~1650 ~1720 ~2900 CHARACTERISTIC WAVELENGTHS or FREQUENCIES (wavenumber) in cm -1 18 Slide 20 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 -1 C. 6100 mm D. 1600 cm 19 Slide 21 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 -1 C 6100 mm D1600 cm X X X 20 Slide 22 wavenumber = 1 / = 6.1 m = 6.1 X 10 -6 m = 1 / 6.1 X 10 -6 A = 1600 cm -1 1 m 100 cm X = 1.6 X 10 5 m -1 = 1.6 X 10 3 cm -1 21 Slide 23 GROUP VIBRATIONS 22 Slide 24 Table 2 of SUPL-005 shows the absorption frequencies in cm -1 of some molecular fragments Here are some that are related to todays exercise. C C (aromatic) 1600, 1500 C H (alkane) 2850 2960 C H (aromatic) 3030 3050 C == O (aldehyde) 1680 1750 O H (alcohol) 3400 3650 O H (phenol) 3200 Aromatic means benzene or benzene-like 23 Slide 25 Vanillin IR Spectrum: Vanillin IR Spectrum: 500 cm -1 4000 cm -1 24 Slide 26 BUT as Percent Transmittance (instead of absorbance), and indicating the (decreasing) wavenumber scale instead of wavelength So, absorption peaks point DOWNWARD % Transmittance etc. Wavelength Note that like visible spectra, IR spectra are displayed as intensity vs increasing wavelength 25 Slide 27 O-H -H C-H 3 HC=OCCCC Vanillin Vanillin IR Spectrum: 1500 cm -1 4000 cm -1 4000 cm -1 3000 cm -1 2000 cm -1 1500 cm -1 26 Slide 28 Explanation of Spectrum Notation We examine vanillin spectrum between 1500 and 4000 cm -1. There are 6 major peaks in this region. OHstretch due to the OH group on the ring Hring hydrogen stretch CH 3 CH stretch in the methoxy (O-CH 3 )group HC=OC=O stretch in the aldehyde group C Ctwo peaks due to the ring C C stretch All but one of these peaks should show up in spectrum of product, vanillyl alcohol. 27 Slide 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. A.True B.False 28 Slide 30 B = False The C=O double bond is the one which is reduced in the reaction. The infrared spectrum of the product, vanillyl alcohol, will absorb near 1700 cm -1 due to the vibration of a C=O double bond. 29 Slide 31 So, the product spectrum should show the absence of the C=O absorption near 1700 cm -1. What other difference should there be? From the table we see that we expect it at: H 2 C OH between 3400 and 3600 cm -1 in the alcohol O-H region. There should be a new absorption due to OH in alcohol group. That absorption is near, but distinct from, the OH absorption due to the OH group on the ring (phenol At ~3200 cm -1 ). 30 Slide 32 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 Slide 33 INTERPRETATION OF IR SPECTRA Use infrared spectrum to verify the presence or absence of functional groups Reaction replaces a -HC=O group by a H 2 C-O-H. starting material So, starting material will show: absorption by -HC=O absence of absorptions by H 2 C-O-H Product Product should show: absorption by H 2 C-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. 32 Slide 34 A Brief Description of FTIR and HATR 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. Transmission Spectroscopy: I 0 ( )I t ( ) I 0 (t)I r (t) ReflectionSpectroscopy: FTIRFTIR FTIR: Fourier Transform Infra Red HATRHATR HATR: Horizontal Attenuated Total Reflectance 33 Slide 35 IR Spectrometer The ZnSe sample area 34 Slide 36 Will be handling small quantities of materials. 2.6 mmol 400 mg of vanillin (C 8 H 8 O 3 ) - [ 2.6 mmol ] 2.5 mmol 2.5 mL of 1.0 M NaOH - [ 2.5 mmol ] 2.1 mmol 80 mg of sodium borohydride (NaBH 4 )- [ 2.1 mmol ] Less than 25 mmol Less than 10 mL of 2.5 M HCl - [ 25 mmol ] Must exercise care in transferring such amounts between containers. SYNTHETIC PROCEDURE 400 40 mg but exactly 2.5 0.2 mL 80 8 mg but exactly 35 Slide 37 STOICHIOMETRY + BH 4 - + 4 H 2 O+ H 3 BO 3 + OH - 44 36 Slide 38 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 NaBH 4 (MM = 38) 80 mg / 38 = 2.1 mmol 37 Slide 39 If 2.6 mmol of vanillin react with 2.1 mmol of NaBH 4 which is the limiting reagent? A. vanillin B. NaBH 4 C. it depends on the molar mass of vanillyl alcohol 38 Slide 40 A vanillin + 1 BH 4 - + 4 H 2 O+ H 3 BO 3 + OH - 4 4 If 2.6 mmol of vanillin react with 2.1 mmol of NaBH 4 what is the limiting reagent? 2.1 mmol X 4 = 8.4 mmol 2.6 mmol 39 Slide 41 Calculations 100 X Actual yield Pct yield = ----------------------- Theoretical yield 100 X 0.349 % Yield = ---------------- = 86.2% 0.405 Theoretical yield = maximum yield that could be produced from actual amount of limiting reagent. E.g., 0.400 g vanillin Could make 2.63 mmol vanillyl alcohol If you actually recover 0.349 g Limiting Reagent (MM = 152) 400 mg / 152 = 2.63 mmol 2.63 X 154 = 0.405 g 40 As defined earlier Slide 42 Pay close attention to directions: Add NaBH 4 slowly to cold solution Let reaction mixture stand at room temperature Chill with ice for recommended period Adjust pH to acid litmus test slowly. Be sure that entire solution is acidic, but not excessively. PROCEDURE Special Notes Dry sample for melting point and IR. For 30 min For 10 min 41 Slide 43 NEXT WEEK Acid content of fruit juices and soft drinks SUSB-010 Do Pre-Lab Exercise 42 Slide 44 43 Slide 45 Grading - Reminder As of today, have completed: 3 Preliminary Exercises @ 55165 2 Final Exercises @ 105210 1 Quiz @ 50 50 6 Lectures @ 5 30 Total475 A- / B+0.90 X 475 = 428 B- / C+0.80 X 475 = 380 C / D0.70 X 475 = 333 D / F0.60 X 475 =285 Grading details are on the course web site. 44