Reduction of Vanillin to 10/24/2014
Transcription
Reduction of Vanillin to 10/24/2014
10/24/2014 O C H H C H O H ? QUESTIONS ? How are organic reactions planned and conducted? [ H2 ] O CH3 O CH3 O O H What reagents can be used to conduct hydrogenations? H Reduction of Vanillin to Vanillyl Alcohol What is the basis for the absorption of IR radiation by molecules? Organic Synthesis and How is IR spectroscopy used to ascertain the structure of a substance? Infra-red Identification 1 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 2 Vanillin • principal flavoring agent in vanilla beans • cured, unripe fruit of a plant in the orchid family. product reduction Techniques: handling micro-scale quantities quantitative transfer of liquids and solids infrared spectroscopy analyzing infrared spectra crystallization vacuum filtration Vanillin is our starting material for the synthesis of the related compound: Vanillyl alcohol 3 NOMENCLATURE – FUNCTIONAL GROUPS aldehyde NOMENCLATURE – FUNCTIONAL GROUPS Standard Representation H O 1 6 H C H 2 benzene 3 5 hydroxy 4 4 methoxy 3-methoxy 4-hydroxy Benz aldehyde (Vanillin) 3-methoxy 4-hydroxy benzyl alcohol 5 (Vanillyl Alcohol) 6 1 10/24/2014 Hydrogen can be added to organic compounds in many ways. As hydrogen gas - or using inorganic hydrides. Our Objective H O H C H OCH3 O H OCH3 OH OH H N.B. In synthetic exercises, you are expected to know the names, formulas and structures of the reactants and products! C C H C C C O H C C H O H C H H We seek a way to add two hydrogen atoms (i.e., reduce) to the C C==O O bond without reducing C C bonds in benzene ring. ≈ A reagent which accomplishes this is: 7 Sodium Borohydride 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) Study of many thousands of substances shows that SPECIFIC MOLECULAR FRAGMENTS absorb light at well-defined, =c (nm) ~ 9100 ~ 6100 ~ 5800 ~ 3400 or, since 1 m = 1000 nm (µm) ~ 9.1 ~ 6.1 ~ 5.8 Absorbance 0.1 68 0 65 0 62 0 590 56 0 53 0 50 0 47 0 44 0 0 (1 μm = ) 1,000 nm – 100,000 nm (= 100μm ) ~ 3.4 11 Wavelength is convenient measure of light in visible region 10 But 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. = c = 50 μm -1 = 50 μm C H 0.2 (Infra-red) 9 C O 0.4 0.3 Absorption of light in infrared region is primarily due to VIBRATION of molecules. BH4- + 4 H2O → H3BO3 (s)+ 4 H2 (g)+ OH- C C 0.6 0.5 Wavelength (nm) Soluble in water, but reacts slowly to liberate hydrogen. C C 0.8 0.7 35 0 A white crystalline solid which can be a source of hydrogen in reactions. Absorbance vs Wavelength 1 0.9 41 0 Absorptions in visible and UV are due to transitions between ELECTRONIC energy levels. (UV) 350 nm – 700 nm (Red) 380 Na+ CHARACTERISTIC WAVELENGTHS 8 740 C Different ways of adding hydrogen give different results depending on type of multiple bonds in reactant. 71 0 O H f = 1/ = 200 cm = 3 X 1010/50 X 10-4 = 6 X 1013 sec-1 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 12 2 10/24/2014 GROUP VIBRATIONS 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 13 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) 1600, 1500 C — H (aromatic) 3030 – 3050 C — H (alkane) 2850 – 2960 H C == O (aldehyde) 1680 – 1750 H C O — H (phenol‡) O — H (alcohol) ‡ H 3200 C C C C O 14 Vanillin IR Spectrum: 500 cm-1 – 4000 cm-1 O H H H C O C H C H 3400 – 3650 O-H groups attached directly to benzene ring 15 Note that like visible spectra, IR spectra are displayed as intensity vs increasing wavelength 16 Vanillin IR Spectrum: 1500 cm-1 – 4000 cm-1 BUT as Percent Transmittance (instead of absorbance), and indicating the (decreasing) wavenumber scale instead of wavelength (100% transmittance = 0 absorbance) % Transmittance So, absorption peaks point DOWNWARD O-H -H Wavelength etc. 17 4000 cm-1 3000 cm-1 C-H3 HC=O 2000 cm-1 CC 18 1500 cm-1 3 10/24/2014 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 So, the product spectrum should show the absence of the C=O absorption near 1700 cm-1. What other difference should there be? 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 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: All but one of these peaks should show up in spectrum of product, vanillyl alcohol. H2C O—H between 3400 and 3600 cm-1 in the alcohol O-H region. 19 20 INTERPRETATION OF IR SPECTRA Use infrared spectrum to verify the presence or absence of functional groups Procedure for IR Spectrum of Vanillyl Alcohol When sample is DRY, Reaction replaces a -HC=O group by a –H2C-O-H. • obtain the spectrum of a small sample using the FTIR Spectrometer. Follow the posted instructions So, starting material will show: absorption by -HC=O absence of absorptions by –H2C-O-H •Analyze the spectrum to identify the peaks due to the product (and, if any, due to the starting material) Product should show: absorption by –H2C-O-H absence of absorption by –HC=O 21 Should also be able to identify absorptions of other functional groups common to vanillin and vanillyl alcohol by 22 comparing their spectra. IR Spectrometer IR IRSpectrometer Spectrometer The ATR Press The ZnSe sample area Only a very small, but dry, sample is needed. 23 24 4 10/24/2014 SYNTHETIC PROCEDURE STOICHIOMETRY Will be handling small quantities of materials. 400 ± 40 mg but exactly 400 mg of vanillin (C8H8O3) - [ 2.6 mmol ] 2.5 mL of 1.0 M NaOH - 2.5 ± 0.2 mL [ 2.5 mmol ] 80 mg of sodium borohydride (NaBH4)- [ 2.1 mmol ] O H O H 4 H C H 4 OCH3 80 ± 8 mg but exactly Less than 10 mL of 2.5 M HCl - [ 25 mmol ] C OCH3 OH OH + BH4- + 4 H2O + H3BO3 + OH- Must exercise care in transferring such amounts between containers. 25 26 Calculations Calculations 100 X Actual yield Pct yield = ----------------------Theoretical yield 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 Theoretical yield = maximum yield that could be produced Limiting from actual amount of limiting reagent. Reagent E.g., 0.400 g vanillin (MM = 152) 400 mg / 152 = 2.63 mmol Could make 2.63 mmol vanillyl alcohol 4 X 2.1 > 2.6 So, limiting reagent is vanillin As defined earlier 27 If you actually recover 0.349 g 100 X 0.349 % Yield = ---------------- = 86.2% 0.405 2.63 X 154 = 0.405 g 28 PROCEDURE – Special Notes Pay close attention to directions: • Add NaBH4 slowly to cold solution • Let reaction mixture stand at room temperature For 30 min • 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. You should need much less than the suggested 10 mL of HCl Dry small amount of sample for melting point and IR. RFS 10/21/14 29 5