Reduction of Vanillin to Vanillyl Alcohol 11/13/2014 Organic Synthesis and
Transcription
Reduction of Vanillin to Vanillyl Alcohol 11/13/2014 Organic Synthesis and
11/13/2014 Reduction of Vanillin to Vanillyl Alcohol Organic Synthesis and Infrared Identification Last Update: 11/13/2014 1:32 PM ? 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? 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 NOMENCLATURE – FUNCTIONAL GROUPS aldehyde product reduction 1 6 2 3 5 Techniques: handling micro-scale quantities quantitative transfer of liquids and solids infrared spectroscopy analyzing infrared spectra crystallization vacuum filtration methoxy 4 3-methoxy 4-hydroxy benzaldehyde (Vanillin) Our Objective NOMENCLATURE – FUNCTIONAL GROUPS H O H C O H C H O H H C H OCH3 OH 3-methoxy 4-hydroxy benzyl alcohol OCH3 OH N.B. you are expected to know the formulas and structures of the reactants and products! E.g., Vanillin, Vanillyl Alcohol, etc. (Vanillyl Alcohol) 1 11/13/2014 In organic is chemistry, means addition of Reduction generally reduction defined asoften the addition of - double) a hydrogen to(e.g., a multiple (e.g., electrons to molecule a molecule I2 + 2e 2 I-). bond. In organic chemistry, it often involves addition of H2 across a multiple bond, e.g., H H H H ethylene H H H-H + C C H-H + H C C H H H ethane H H H C O H H Different ways of adding hydrogen give different results depending on type of multiple bonds in reactant. C O H formaldehyde methyl alcohol O H C C C C H Sodium Borohydride, NaBH4 Study of many thousands of substances shows that SPECIFIC MOLECULAR FRAGMENTS absorb light at well-defined, CHARACTERISTIC WAVELENGTHS Absorption of light in infrared region is primarily due to VIBRATION of molecules. (1 μm = ) 1,000 nm – 100,000 nm (Infra-red) 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 C (nm) ~ 9100 Infrared photon wavelengths are in the approximate range 1 m - 100 m (or 1 X 10-4 cm – 1 X 10-2 cm) C C C O C H ~ 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 CHARACTERISTIC WAVELENGTHS or FREQUENCIES (wavenmber) in cm-1 C C 1 cm =c Have previously studied absorption spectroscopy of food dyes (ultraviolet and visible). Those absorptions were due to transitions between ELECTRONIC energy levels (UV) 350 nm – 700 nm (Red) ≈ H C H C C H O C H O H A reagent which accomplishes this is: Hydrogen can be added to organic compounds in many ways. As hydrogen gas, - or using inorganic hydrides. 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) 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 O C H (µm) ~ 9.1 ~ 6.1 ~ 5.8 ~ 3.4 f (cm-1) ~1100 ~1650 ~1720 ~2900 The IR range becomes 100 cm-1 – 10,000 cm-1 2 11/13/2014 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. 1600, 1500 C — H (aromatic) 3030 – 3050 C — H (alkane) 2850 – 2960 C == O (aldehyde) 1680 – 1750 O — H (phenol) O — H (alcohol) H H H C C C H C C O 3200 O BUT as Percent Transmittance (instead of absorbance), and indicating the (decreasing) wavenumber scale instead of wavelength So, absorption peaks point DOWNWARD H H H C H C O C H H 3400 – 3650 % Transmittance C C (aromatic) Note that like visible spectra, IR spectra are displayed as intensity vs increasing wavelength Wavelength Vanillin IR Spectrum: 1500 cm-1 – 4000 cm-1 etc. 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 All but one of these peaks should show up in spectrum of product, vanillyl alcohol. O-H -H 4000 cm-1 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 3000 cm-1 C-H3 HC=O 2000 cm-1 CC 1500 cm-1 Procedure for IR Spectrum of Vanillyl Alcohol 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). 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) 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. 3 11/13/2014 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 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 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 comparing their spectra. Transmission Spectroscopy: Reflection Spectroscopy: I0 () I0 (t) It () Ir (t) HATR: Horizontal Attenuated Total Reflectance 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 ] 80 ± 8 mg but exactly Less than 10 mL of 2.5 M HCl - [ 25 mmol ] O C H O H 4 H C H 4 OCH3 OCH3 OH OH + BH4- + 4 H2O + H3BO3 + OH- Must exercise care in transferring such amounts between containers. Calculations 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 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 = 154) 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 38 = 0.405 g 4 11/13/2014 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. 5