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
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2
3
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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)
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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
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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.
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
CC
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 CC stretch
3000 cm-1
C-H3
HC=O
2000 cm-1
CC
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.
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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
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
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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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