2. Reduction of Vanillin to Vanillyl Alcohol
Organic Synthesis and
Infrared Identification

3. How 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?

4. Synthesis starting material product
Purpose:
To conduct organic reaction, isolate product & characterize product using infrared spectroscopy
Concepts:
Synthesis starting material product
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
The checkmarks note concepts and techniques with which they should already be familiar from the alum exercise.  

5. principal flavoring agent in vanilla beans
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:
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

6. NOMENCLATURE – FUNCTIONAL GROUPS
aldehyde 1 6 2
methyl oxy
methoxy 3 5 4
This 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.
benzene
3-methoxy
4-hydroxy
benzaldehyde
(Vanillin)

7. NOMENCLATURE – FUNCTIONAL GROUPS
3-methoxy
4-hydroxy
benzyl alcohol
(Vanillyl Alcohol)

8. PRINCIPAL CONSTITUENTS OF THE RIPE VANILLA BEAN
A chromatogram
A 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. Rf
1) 4-Hydroxybenzyl alcohol
3) vanillyl alcohol
6) 4-hydroxybenzaldehyde
9) vanillin
10) coumaric acid

9. 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.

10. In 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.

11. 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 C O ≈
without reducing
bonds in benzene ring. C C
Remind them that the ring bonds in benzene are all identical 1½ order bonds.
A reagent which accomplishes this is:

12. Na+ 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

13. In organic chemistry, reduction often refers to the removal of atoms of hydrogen across a multiple bond
True
False

14. In organic chemistry, reduction often refers to the removal of atoms of hydrogen across a multiple bond.
addition
H - H +
H - H +
B = False

15. infrared spectroscopy
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).
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 is
primarily due to
VIBRATION of molecules.
(1 μm = )
1,000 nm – 100,000 nm
(Infra-red)

16. CHARACTERISTIC WAVELENGTHS
Study of many thousands of substances shows that SPECIFIC MOLECULAR FRAGMENTS absorb light at well-defined, 
CHARACTERISTIC WAVELENGTHS
  = c
This 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) ~ ~ ~ ~ 3400
or, since 1 m = 1000 nm
 (µm) ~ ~ ~ ~ 3.4
Wavelength is convenient measure of light in visible region

17. 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
This slide has to be done slowly. It is their first exposure to wavenumbers.
1 cm
Infrared 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

18. CHARACTERISTIC WAVELENGTHS
or FREQUENCIES (wavenumber) in cm-1
This 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 ~ 3.4
f (cm-1) ~1100 ~ ~ ~2900

19. The 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-1
0.16 sec-1
6100 mm
1600 cm

20. The C=C bond absorbs IR radiation of
6.1 m wavelength. To what wavenumber does that correspond?
A cm-1
B sec-1
C mm
D 1600 cm X X
We 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-1
A = cm-1

22. GROUP VIBRATIONS
A 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 fragments
Here are some that are related to
today’s exercise.
“Aromatic” means benzene or benzene-like
C  C (aromatic) 1600, 1500
C — H (aromatic) – 3050
C — H (alkane) – 2960
C == O (aldehyde) – 1750
The 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

24. Vanillin 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.

25. So, absorption peaks point DOWNWARD
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
So, absorption peaks point DOWNWARD
This 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.
% Transmittance
etc.
Wavelength

26. Vanillin 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-H
C-H3
-H
HC=O
CC
4000 cm-1
3000 cm-1
2000 cm-1
1500 cm-1

27. Explanation 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 stretch
C−H3 C−H stretch in the methoxy (O-CH3)group
HC=O C=O stretch in the aldehyde group
CC two peaks due to the ring CC stretch
All 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.

28. The infrared spectrum of the product, vanillyl alcohol, will absorb near 1700 cm-1 due to the vibration of a C=O double bond.
True
False

29. 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.
B = False
The C=O double bond is the one which is reduced in the reaction.

30. What 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, 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.
This completes the expectations for the product IR. Some students will understandably not appreciate this reasoning yet.

31. 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)
The sample must BE DRY!!!!!!!!!!

32. 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 comparing their spectra.
A review of the IR analysis and a reminder that they must also look at the rest of the product spectrum.

33. A 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 Red
Transmission Spectroscopy:
Reflection
Spectroscopy:
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

34. IR 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

35. Must exercise care in transferring
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 ]
80 mg of sodium borohydride (NaBH4)- [ 2.1 mmol ]
Less than 10 mL of 2.5 M HCl - [ 25 mmol ]
Must exercise care in transferring
such amounts between containers.
400 ± 40 mg
but exactly
2.5 ± 0.2 mL
80 ± 8 mg
but exactly

36. 4 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-

37. Calculations
100 X Actual yield
Pct yield =
Theoretical yield
Theoretical yield = maximum yield that could be produced from actual amount of limiting reagent.
400 mg / 152
= 2.6 mmol
E.g., g vanillin
(MM = 152)
80 mg / 38
= 2.1 mmol
E.g., g NaBH4
(MM = 38)

38. If 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?
vanillin
NaBH4
it depends on the molar
mass of vanillyl alcohol

39. If 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 mmol 4 4
+ 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 mmol
A vanillin

40. Calculations Limiting Reagent
As defined earlier
Calculations
100 X Actual yield
Pct yield =
Theoretical yield
Limiting
Reagent
Theoretical yield = maximum yield that could be produced from actual amount of limiting reagent.
400 mg / 152
= 2.63 mmol
E.g., g vanillin
(MM = 152)
2.63 X 154
= g
Could make 2.63 mmol vanillyl alcohol
This should be a review from similar calculations in the alum exercise.
If you actually recover g
100 X 0.349
% Yield = = 86.2%
0.405

41. PROCEDURE – Special Notes Dry sample for melting point and IR.
Pay close attention to directions:
Add NaBH4 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.
For
30 min
For
10 min
Excess acid will decompose the product!
Dry sample for melting point and IR.

42. Acid content of fruit juices and soft drinks
NEXT WEEK
Acid content of fruit juices and soft drinks
SUSB-010
Do Pre-Lab Exercise

44. Grading details are on the course web site.
Grading - Reminder
As of today, have completed:
3 Preliminary
2 Final
Total
A- / B X 475 = 428
B- / C X 475 = 380
C / D X 475 = 333
D / F X 475 = 285
Grading details are on the course web site.