1. health and fitness
2. disease
3. cholesterol

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)
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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
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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
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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
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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
CC
18
1500 cm-1
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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
CC
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 CC 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
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comparing their spectra.
IR Spectrometer
IR
IRSpectrometer
Spectrometer
The ATR
Press
The ZnSe
sample
area
Only a very small, but dry, sample is needed.
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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
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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
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If you actually recover 0.349 g
100 X 0.349
% Yield = ---------------- = 86.2%
0.405
2.63 X 154
= 0.405 g
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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
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