Reduction of Vanillin to Vanillyl Alchol

Report
1
O
C
H
H
O
H
C H
[ H2 ]
O CH3
O
H
O CH3
O
H
Reduction of Vanillin to
Vanillyl Alcohol
Organic Synthesis and
Infrared Identification
2
? 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?
3
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
product 
reduction
Techniques:
handling micro-scale quantities
quantitative transfer of liquids and solids
infrared spectroscopy analyzing infrared spectra
crystallization 
vacuum filtration 
4
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:
Vanillyl alcohol
5
NOMENCLATURE – FUNCTIONAL GROUPS
aldehyde
1
6
2
3
5
methyl oxy
methoxy
4
benzene
3-methoxy 4-hydroxy benzaldehyde
(Vanillin)
6
NOMENCLATURE – FUNCTIONAL GROUPS
H
O C
H
H
3-methoxy 4-hydroxy benzyl alcohol
(Vanillyl Alcohol)
7
PRINCIPAL CONSTITUENTS
OF THE RIPE VANILLA BEAN
CH2OH
A chromatogram
OCH3
OH
H
C
O
Rf
OCH 3
OH
1)
3)
6)
9)
10)
4-Hydroxybenzyl alcohol
vanillyl alcohol
4-hydroxybenzaldehyde
vanillin
coumaric acid
8
Our Objective
O
C
H O
H
H
C
H
OCH3
OH
OCH3
OH
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.
9
In
chemistry,
reduction
oftenas
means
addition of
Weorganic
have earlier
described
reduction
the addition
-)
a
to a multiple
ofhydrogen
electronsmolecule
to a molecule
(e.g., I2 (e.g.,
+ 2e- double)
 2 Ibond.
H
H-H +
H
C C
H
H
ethylene
H
H
C C H
H
H
H
ethane
H
H
H
C O
H-H +
H
formaldehyde
H
C O
H
methyl alcohol
Hydrogen can be added to organic compounds in many
ways. As hydrogen gas - or using inorganic hydrides.
10
Different ways of adding hydrogen give different
results depending on type of multiple bonds in reactant.
O
H
C
C
C
H
C
H
H
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
H
C
O
H
O
H
A reagent which accomplishes this is:
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Sodium Borohydride
H
+
Na
H
 B 

Hydride Ion H-
H

H
12
In organic chemistry, reduction often refers to
the removal of atoms of hydrogen across a
multiple bond
A. True
B. False
0%
A.
0%
B.
13
In organic chemistry, reduction often
refers to the addition
removal of atoms of
H
H
bond.
H hydrogen
H across a multiple
H-H +
C C
C C
H
H
ethylene
H
H
H
H
ethane
H
H
H
C O
H-H +
H
H
formaldehyde
C O
H
methyl alcohol
B = False
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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).
Absorbance vs Wavelength
0.7
0.6
0.5
0.4
0.3
0.2
0.1
Absorption of light in infrared region is
primarily due to
VIBRATION of molecules.
(1 μm = ) 1,000 nm – 100,000 nm (Infra-red)
15
0
0
74
0
0
0
Wavelength (nm)
71
68
65
0
0
0
0
0
0
0
62
59
56
53
50
47
44
41
0
0
38
Absorbance
0.8
0
(UV) 350 nm – 700 nm (Red)
0.9
35
Those absorptions were due to transitions
between ELECTRONIC energy levels
1
Study of many thousands of substances shows
that SPECIFIC MOLECULAR FRAGMENTS
absorb light at well-defined,
CHARACTERISTIC WAVELENGTHS

=c
C C
 (nm) ~ 9100
 (µm)
C C
C O
C H
~ 6100
~ 5800
~ 3400
or, since 1 m = 1000 nm
~ 9.1
~ 6.1
~ 5.8
~ 3.4
16
Wavelength is convenient measure of light in visible region
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
1 cm
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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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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The C=C bond absorbs IR radiation of
6.1 m wavelength. To what wavenumber does
that correspond?
A.
B.
C.
D.
1600 cm-1
0.16 sec-1
6100 mm
1600 cm
0%
A.
0%
B.
0%
C.
0%
D.
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The C=C bond absorbs IR radiation of
6.1 m wavelength. To what wavenumber does
that correspond?
A
1600 cm-1
B
0.16 sec
X -1
C
6100 mm
X
D
1600 cm
X
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 = 6.1 μm = 6.1 X 10-6 m
wavenumber = 1 /  = 1 / 6.1 X 10-6
= 1.6 X
105
m-1
X
1m
100 cm
= 1.6 X 103 cm-1
A = 1600 cm-1
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GROUP VIBRATIONS
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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)
C — H (aromatic)
C — H (alkane)
C == O (aldehyde)
O — H (phenol)
O — H (alcohol)
1600, 1500
H
3030 – 3050
2850 – 2960
1680 – 1750
3200
H
C
H C
H
C
C
C
O
O
H
H
H
H
C O C H
C
H
3400 – 3650
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Vanillin IR Spectrum: 500 cm-1 – 4000 cm-1
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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
% Transmittance
So, absorption peaks point DOWNWARD
Wavelength
etc.
25
Vanillin IR Spectrum: 1500 cm-1 – 4000 cm-1
O-H
C-H3
-H
4000
cm-1
3000
cm-1
HC=O
2000
cm-1
CC
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1500 cm-1
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
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
All but one of these peaks should show up in spectrum of
product, vanillyl alcohol.
27
The infrared spectrum of the product,
vanillyl alcohol, will absorb near 1700 cm-1
due to the vibration of a C=O double
bond.
A. True
B. False
0%
0%
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A.
B.
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.
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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.
30
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)
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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
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comparing their spectra.
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
Transmission
Spectroscopy:
Reflection
Spectroscopy:
I0 ()
I0 (t)
It ()
Ir (t)
HATR: Horizontal Attenuated Total Reflectance
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IR Spectrometer
The ZnSe
sample
area
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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 ]
400 ± 40 mg
but exactly
2.5 ± 0.2 mL
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 ]
Must exercise care in transferring
such amounts between containers.
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STOICHIOMETRY
O
C
H O
H
4
H
C
H
4
OCH3
OH
+ BH4- + 4 H2O
OCH3
OH
+ H3BO3
+ OH-
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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
37
If 2.6 mmol of vanillin react with 2.1 mmol of NaBH4
which is the limiting reagent?
A.
B.
C.
vanillin
NaBH4
it depends on the molar
mass of vanillyl alcohol
0%
A.
0%
B.
0%
C. 38
If 2.6 mmol of vanillin react with 2.1 mmol of
NaBH4 what is the limiting reagent?
O
C
H
H O
2.6 mmol
H
C
H
4
4
OCH3
OCH3
OH
OH
+ 1 BH4- + 4 H2O
+ H3BO3 + OH-
2.1 mmol X 4 = 8.4 mmol
A
vanillin
39
As defined
earlier
Calculations
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 = 152)
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 154
= 0.405 g
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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.
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NEXT WEEK
Acid content of fruit juices and soft drinks
SUSB-010
Do Pre-Lab Exercise
42
43
Grading - Reminder
As of today, have completed:
3 Preliminary Exercises @ 55
2 Final Exercises @ 105
1 Quiz @ 50
6 Lectures @ 5
Total
ABC
D
/
/
/
/
B+
C+
D
F
0.90
0.80
0.70
0.60
X 475 =
X 475 =
X 475 =
X 475 =
165
210
50
30
475
428
380
333
285
Grading details are on the course web site.
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