2Reduction of Vanillin to Vanillyl Alcohol Organic Synthesis andInfrared Identification
3How are organic reactions planned and conducted? ? 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?
4Synthesis starting material product Purpose:To conduct organic reaction, isolate product & characterize product using infrared spectroscopyConcepts:Synthesis starting material producttheoretical yield percent yield reductionorganic functional groupscharacteristic infrared absorptionsTechniques:handling micro-scale quantitiesquantitative transfer of liquids and solidsinfrared spectroscopy analyzing infrared spectracrystallization vacuum filtrationThe checkmarks note concepts and techniques with which they should already be familiar from the alum exercise.
5principal flavoring agent in vanilla beans Vanillinprincipal flavoring agent in vanilla beanscured, unripe fruit of aplant in the orchid family.Vanillin is our starting material for synthesis of the related compound:Grew initially only in central america because the particular variety of bee that pollinates the orchid is found only there. Now use manual pollination.Vanillyl alcohol
6NOMENCLATURE – FUNCTIONAL GROUPS aldehyde162methyl oxymethoxy354This introduces the concept of drawing organic structures that suppress the display of H atoms directly bound to carbons (and violates it with the aldehyde and methyl substituents). Must go slowly. Benzene with all atoms to skeleton representation. Some of the students have not yet encountered this notation and are seeing the groups and group names for the first time.benzene3-methoxy4-hydroxybenzaldehyde(Vanillin)
7NOMENCLATURE – FUNCTIONAL GROUPS 3-methoxy4-hydroxybenzyl alcohol(Vanillyl Alcohol)
8PRINCIPAL CONSTITUENTS OF THE RIPE VANILLA BEAN A chromatogramA chromatogram and the (related) primary constituents of pure vanilla extract. An analogous paper chromatogram is simulated with spots moving from right to left. Right left analogy to paper would have the arrow going in the opposite direction.Rf1) 4-Hydroxybenzyl alcohol3) vanillyl alcohol6) 4-hydroxybenzaldehyde9) vanillin10) coumaric acid
9Our ObjectiveN.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.
10In organic chemistry, reduction often means addition of a hydrogen molecule to 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+The reference is to the identification of an unknown substance exercise. The animation shows a hydrogen molecule. Our original definition survives since we are adding electrons along with associated protons.Hydrogen can be added to organic compounds in many ways. As hydrogen gas - or using inorganic hydrides.
11Different ways of adding hydrogen give different results depending on type of multiple bonds in reactant.We seek a way to add twohydrogen atoms (i.e., reduce)to the C==O bondCO≈without reducingbonds in benzene ring.CCRemind them that the ring bonds in benzene are all identical 1½ order bonds.A reagent which accomplishes this is:
12Na+ Sodium Borohydride - H Hydride Ion H- H B H H We introduce the hydride ion and develop the Lewis structure of the Borohydride ion. By now, the tetrahedron should be a familiar structure to the student.H
13In organic chemistry, reduction often refers to the removal of atoms of hydrogen across a multiple bondTrueFalse
14In organic chemistry, reduction often refers to the removal of atoms of hydrogen across a multiple bond.additionH - H+H - H+B = False
15infrared spectroscopy Structures of starting and product molecules differ in a way that makesinfrared spectroscopyan 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).Those absorptions were due to transitions between ELECTRONIC energy levels(UV) 350 nm – 700 nm (Red)We talk about wavelengths here to connect with the visible-UV spectroscopy they did earlier.Absorption of light in infrared region isprimarily due toVIBRATION of molecules.(1 μm = )1,000 nm – 100,000 nm(Infra-red)
16CHARACTERISTIC WAVELENGTHS Study of many thousands of substances shows that SPECIFIC MOLECULAR FRAGMENTS absorb light at well-defined,CHARACTERISTIC WAVELENGTHS = cThis and the next slide attempt to explain that IR spectra are displayed with a decreasing frequency axis. O has a greater mass than C, but the C=O bond strength is 177 vs 146. (nm) ~ ~ ~ ~ 3400or, since 1 m = 1000 nm (µm) ~ ~ ~ ~ 3.4Wavelength is convenient measure of light in visible region
17Convention 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 wavenumberinstead of the wavelength, The rational unit for this form of frequency is cm-1. = 0.2 cmf = 1/ = 5 cm-1= 3 X 1010 / 0.2 2 X 1011 sec-1 = 0.2 cm = cThis slide has to be done slowly. It is their first exposure to wavenumbers.1 cmInfrared photon wavelengths are in the approximate range 1 m m (or 1 X 10-4 cm – 1 X 10-2 cm)The IR range becomes 100 cm-1 – 10,000 cm-1
18CHARACTERISTIC WAVELENGTHS or FREQUENCIES (wavenumber) in cm-1This and the next slide are an attempt to explain why IR spectra are, and will be, displayed with both a wavelength and a frequency axis (which increase in opposite directions) (µm) ~ ~ 6.1 ~ 5.8 ~ 3.4f (cm-1) ~1100 ~ ~ ~2900
19The C=C bond absorbs IR radiation of 6. 1 m wavelength The C=C bond absorbs IR radiation of 6.1 m wavelength. To what wavenumber does that correspond?1600 cm-10.16 sec-16100 mm1600 cm
20The C=C bond absorbs IR radiation of 6.1 m wavelength. To what wavenumber does that correspond?A cm-1B sec-1C mmD 1600 cmXXWe can eliminate these three answers based on the (incorrect) units alone, regardless of the values.X
21 = 6.1 μm = 6.1 X 10-6 m wavenumber = 1 / = 1 / 6.1 X 10-6 100 cm= 1.6 X 103 cm-1A = cm-1
22GROUP VIBRATIONSA given group of atoms can vibrate in a number of different ways. (3n – 6 for a non linear molecule). These are called normal modes.
23“Aromatic” means benzene or benzene-like Table 2 of SUPL-005 shows the absorption frequencies in cm-1 of some molecular fragmentsHere are some that are related totoday’s exercise.“Aromatic” means benzene or benzene-likeC C (aromatic) 1600, 1500C — H (aromatic) – 3050C — H (alkane) – 2960C == O (aldehyde) – 1750The atoms involved in each vibration type are in red. Must do this slowly. In the last picture, the molecule changes to the product. The information on this slide is posted in the laboratory (without the structure). The words aromatic, alkane and phenol need some elaboration.O — H (phenol)O — H (alcohol) – 3650
24Vanillin IR Spectrum: 500 cm-1 – 4000 cm-1 Problem 3 of the pre-lab requires them to go to the library database web site and look up the spectrum of vanillin. Because the resolution is best, mention that they should copy the spectrum labeled as Nujol Mull! Since it is just an example, no need to elaborate on the spectrum here. The full spectrum (450 – 4000 cm-1) is shown. A later slide focuses on the region between 1500 and 4000 cm-1.
25So, absorption peaks point DOWNWARD Note that like visible spectra, IR spectra are displayed as intensity vs increasing wavelengthBUTas Percent Transmittance (instead of absorbance), and indicating the (decreasing) wavenumber scale instead of wavelengthSo, absorption peaks point DOWNWARDThis slide is critical. In their earlier experience with spectra, the students displayed the absorbance vs wavelength. They need to understand that convention in IR is %T.% Transmittanceetc.Wavelength
26Vanillin IR Spectrum: 1500 cm-1 – 4000 cm-1 Do this slowly. It is what they are supposed to do with the vanillyl alcohol spectrum in lab.O-HC-H3-HHC=OCC4000 cm-13000 cm-12000 cm-11500 cm-1
27Explanation 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−H ring hydrogen stretchC−H3 C−H stretch in the methoxy (O-CH3)groupHC=O C=O stretch in the aldehyde groupCC two peaks due to the ring CC stretchAll but one of these peaks should show up in spectrum of product, vanillyl alcohol.This begins with a review of the previous slide and makes a transition to the spectrum of the product.
28The infrared spectrum of the product, vanillyl alcohol, will absorb near 1700 cm-1 due to the vibration of a C=O double bond.TrueFalse
29The 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.B = FalseThe C=O double bond is the one which is reduced in the reaction.
30What other difference should there be? 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, theO—H absorption due to the OH group on the ring (phenolAt ~3200 cm-1).From the table we see that we expect it at:H2C O—H between 3400 and 3600 cm-1in the alcohol O-H region.This completes the expectations for the product IR. Some students will understandably not appreciate this reasoning yet.
31Procedure for IR Spectrum of Vanillyl Alcohol When sample is DRY,obtain the spectrum of a small sample using the FTIR Spectrometer. Follow the posted instructionsAnalyze the spectrum to identify the peaks due to the product (and, if any, due to the starting material)The sample must BE DRY!!!!!!!!!!
32INTERPRETATION OF IR SPECTRA Use infrared spectrum to verify the presence or absence of functional groupsReaction replaces a -HC=O group by a –H2C-O-H.So, starting material will show:absorption by -HC=Oabsence of absorptions by –H2C-O-HProduct should show:absorption by –H2C-O-Habsence of absorption by –HC=OShould also be able to identify absorptions of other functional groups common to vanillin and vanillyl alcohol by comparing their spectra.A review of the IR analysis and a reminder that they must also look at the rest of the product spectrum.
33A Brief Description of FTIR and HATR Transmission Spectroscopy: 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 RedTransmission Spectroscopy:ReflectionSpectroscopy:A very brief explanation of FTIR contrasting it with normal transmission spectroscopy. A web page explains this a little bit further.I0 ()It ()I0 (t)Ir (t)HATR: Horizontal Attenuated Total Reflectance
34IR Spectrometer The ZnSe sample area To give them an idea of the size of the sample required. It should dispel the notion that they need to have weighed the product before taking the IR spectrum. A couple of mg is all they need.The ZnSe sample area
35Must exercise care in transferring SYNTHETIC PROCEDUREWill 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 ]80 mg of sodium borohydride (NaBH4)- [ 2.1 mmol ]Less than 10 mL of 2.5 M HCl - [ 25 mmol ]Must exercise care in transferringsuch amounts between containers.400 ± 40 mgbut exactly2.5 ± 0.2 mL80 ± 8 mgbut exactly
364 4 + BH4- + 4 H2O + H3BO3 + OH- STOICHIOMETRY The stoichiometry is shown in an acidic medium. In the initially basic solution, the product is a sodium borate.Since 1 H2 per mole is added to the reactant in this reaction, students should realize that the molar mass of the product = molar mass of vanillin On the next quiz, they will be expected to understand and remember that. You can tell them that on a quiz, they would not be given the molar masses of both.+ BH4-+ 4 H2O+ H3BO3+ OH-
37Calculations100 X Actual yieldPct yield =Theoretical yieldTheoretical yield = maximum yield that could be produced from actual amount of limiting reagent.400 mg / 152= 2.6 mmolE.g., g vanillin(MM = 152)80 mg / 38= 2.1 mmolE.g., g NaBH4(MM = 38)
38If 2. 6 mmol of vanillin react with 2 If 2.6 mmol of vanillin react with 2.1 mmol of NaBH4 which is the limiting reagent?vanillinNaBH4it depends on the molarmass of vanillyl alcohol
39If 2. 6 mmol of vanillin react with 2 If 2.6 mmol of vanillin react with 2.1 mmol of NaBH4 what is the limiting reagent?2.6 mmol44+ 1 BH4-+ 4 H2O+ H3BO3+ OH-Note that 1 mole of BH4- can reduce (add H2) 4 moles of vanillin. Four hydrogens must be coming from somewhere else (must be H2O). An indication that the detailed mechanism of the borohydride reduction reaction must be quite complex.2.1 mmol X 4 = 8.4 mmolA vanillin
40Calculations Limiting Reagent As defined earlierCalculations100 X Actual yieldPct yield =Theoretical yieldLimitingReagentTheoretical yield = maximum yield that could be produced from actual amount of limiting reagent.400 mg / 152= 2.63 mmolE.g., g vanillin(MM = 152)2.63 X 154= gCould make 2.63 mmol vanillyl alcoholThis should be a review from similar calculations in the alum exercise.If you actually recover g100 X 0.349% Yield = = 86.2%0.405
41PROCEDURE – Special Notes Dry sample for melting point and IR. Pay close attention to directions:Add NaBH4 slowly to cold solutionLet reaction mixture stand at room temperatureChill with ice for recommended periodAdjust pH to acid litmus test slowly. Be sure that entire solution is acidic, but not excessively.For30 minFor10 minExcess acid will decompose the product!Dry sample for melting point and IR.
42Acid content of fruit juices and soft drinks NEXT WEEKAcid content of fruit juices and soft drinksSUSB-010Do Pre-Lab Exercise
44Grading details are on the course web site. Grading - ReminderAs of today, have completed:3 Preliminary2 FinalTotalA- / B X 475 = 428B- / C X 475 = 380C / D X 475 = 333D / F X 475 = 285Grading details are on the course web site.