10/30/2015
H
O
H
C H
? QUESTIONS ?
How are organic reactions planned and executed?
O CH3
O
What characterizes reduction in organic reactions?
H
What reagents can be used to conduct
hydrogenations?
Synthesis of
Vanillyl Alcohol
What is the basis for the absorption of IR
radiation by molecules?
Organic Synthesis and
Infrared Identification
How is IR spectroscopy used to ascertain the
molecular structure of a substance?
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Organization of this Pre-lab Lecture
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
group vibration frequencies
characteristic infrared absorptions
organic nomenclature
Techniques:
handling semi micro-scale quantities of reagents
quantitative transfer of liquids and solids
crystallization
vacuum filtration
melting point determination
infrared spectroscopy analyzing infrared spectra
Apparatus:
vacuum filtration apparatus (filter flask, Buchner funnel)
melting point apparatus
Fourier Transform IR Spectrometer
2
Nomenclature – Functional Groups
Comments on the Procedure
Vanilla Beans
Verification of the Product – Melting Point
Synthesis - Principles
Verification of the Product – Infra-red
Spectrum
Our Synthetic Objective
Molecular Vibrations
Reduction in Organic Reactions
Functional Group Vibrations
The Synthetic Procedure
The IR of Vanillin
The Reaction / Stoichiometry
IR of the Product – What to Expect
Calculations – Limiting Reagent
Procedure for IR Spectrum
Calculations – Percent Yield
Interpretation of the Spectrum
Isolating the Product – Vacuum Filtration
Transmission vs FTIR/HATR
Why We Add HCl
4
3
NOMENCLATURE – FUNCTIONAL GROUPS
H
H
H
C
C
H
C
C
C
C
C
H
O
H
C
H
Vanillin
O
aldehyde
H
H
H
C
C
C
6
C
C
H
H
2
3
4
C
H
1
5
H
NOMENCLATURE – FUNCTIONAL GROUPS
O
H
methoxy
C
H
C
C
C
O
O
H
H
O
C
C
H
O
H
C
H
C
H
H
H
Vanillyl Alcohol
3-methoxy 4-hydroxy benzyl alcohol
H
3-methoxy 4-hydroxy benzaldehyde
H
benzene
C
H
5
N.B. In synthetic exercises, you are expected to know the
formulas and structures of the reactants and products!
E.g., Vanillyl Alcohol, etc.
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MAJOR CONSTITUENTS OF
THE RIPE VANILLA BEAN
SYNTHESIS
Conducting one or more chemical REACTIONS using
STARTING MATERIALS to produce a desired
PRODUCT
What makes a synthesis EFFECTIVE ?
H
C
O
If it produces the desired product:
• from reasonable STARTING MATERIALS
• under practical CONDITIONS
e.g., accessible temperature & pressure
• at a reasonable RATE
(CATALYSTS may be helpful)
• with relatively high YIELD
(Few by-products)
CH2OH
OCH3
OCH 3
OH
OH
vanillyl alcohol
vanillin
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8
OUR OBJECTIVE
• in a chemical environment that permits
practical SEPARATION OF PRODUCT
from excess starting materials and byproducts
To synthesize:
• in an ENVIRONMENTALLY RESPONSIBLE way
- GREEN CHEMISTRY
• in a CO$T-EFFECTIVE manner
For substances that may be consumed by humans or
animals, there is an additional concern:
+ H2
SYNTHESIS must not involve
TOXIC AGENTS – either chemical
or biological - that are
NOT EASILY REMOVED
How do you decide what starting materials to use?
What readily available substance has a related structure?
What reactions would accomplish the change in structure?
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10
REDUCTION
In organic chemistry, reduction often means addition of a
hydrogen molecule to a multiple (e.g., double) bond.
H-H +
H
H
C C
H
H
ethylene
C C H
H
H
H
ethane
C O
H
H
formaldehyde
+ 3 H2
Ni
H
C O
C=O
H
+ 3 H2
Pd
methyl alcohol
Hydrogen can be added to organic compounds in many ways.
As hydrogen gas, H2:
REDUCTION USING H2
H
H
H
H
H-H +
Vanillyl Alcohol
Vanillin
11
CH4 + H2O
H2 will reduce the C≈C bonds in the benzene ring
in vanillin and the C=O bond to CH4.
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ARE THERE OTHER WAYS TO ADD H2?
SODIUM BOROHYDRIDE – NaBH4
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 C==O
O bond
≈
C
without reducing C
bonds in benzene ring.
H

Na+
H
Hydride Ion H-
 B 

H

H
A reagent which accomplishes this is:
13
BH4- is isoelectronic with CH4 and NH4+
SYNTHETIC PROCEDURE
STOICHIOMETRY OF THE REACTION
You will be handling small quantities of materials.
400 mg of vanillin (C8H8O3) - [ 2.6 mmol ]
2.5 mL of 1.0 M NaOH -
O
400 ± 40 mg
but exactly
The reagents
2.5 ± 0.2 mL
[ 2.5 mmol ]
80 ± 8 mg
but exactly
C
H O
H
4
H
C
H
4
OCH3
80 mg of sodium borohydride (NaBH4)- [ 2.1 mmol ]
After reaction is complete,
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OCH3
OH
+ BH4- + 4 H2O
OH
+ H3BO3
+ OH-
Less than 10 mL of 2.5 M HCl - [ 25 mmol ]
Must exercise care in transferring
such small amounts between containers.
15
16
CALCULATIONS – LIMITING REAGENT
CALCULATIONS – PERCENT YIELD
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., 0.4120 g vanillin
(MM = 152)
412 mg / 152
= 2.71 mmol
E.g., 0.0825 g NaBH4
(MM = 38)
82.5 mg / 38
= 2.2 mmol
Theoretical yield = maximum yield that could be produced
from actual amount of limiting reagent.
Limiting
Reagent
E.g., 0.4120 g vanillin
(MM = 152)
Could make 2.71 mmol vanillyl alcohol
But, stoichiometry is 1 NaBH4 ↔ 4 Vanillin
so, we could reduce 4 X 2.2 mmol = 8.8 mmol Vanillin
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412 mg / 152
= 2.71 mmol
2.71 X 154
= 417 mg
If you actually recover 349 mg
100 X 349
% Yield = ---------------- = 83.7%
417
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ISOLATING THE PRODUCT
WHY WE ADD HCl
The vanillyl alcohol product is isolated from the
reaction mixture by crystallization from water and
vacuum filtration (See web supplement.)
What is in the reaction vessel after the reaction?
We assume the most insoluble component is the
product and that its solubility is lowest at low
temperatures (ice bath).
In addition,
• we dissolved the vanillin in NaOH and
• OH- was also produced in the reaction
Vanillin is the limiting reagent so some NaBH4 remains
unreacted.
But, H3BO3 can react with NaOH to form Na3BO3
Vanillyl alcohol decomposes in either
strongly acid OR basic solutions.
HCl is added to decompose BH4- & make solution neutral.
BH4- (and OH-) react with HCl (aq)
http://www.ic.sunysb.edu/Class/che133/techniques/suctfilt/suctfilt.html
BH4- + 3H2O + H+ → H3BO3 + 4 H2 (g)
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PROCEDURE – SOME NOTES
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VERIFYING SYNTHETIC SUCCESS – 1
THE MELTING POINT
Pay close attention to directions:
• Add NaBH4 slowly to cold solution
Melting points are a simple way to obtain evidence in
support of the identity and purity of a substance.
(See the web supplement).
• Let reaction mixture stand at room temperature
They are determined using the following apparatus:
• Chill Solution to 0oC
For 30 min
• Chill with ice for recommended period
For 10 min
• Add HCl, adjusting pH to acid litmus test slowly.
Be sure that entire solution is just acidic (pink),
but not excessively.
You must dry a small amount (½ spatulaful) of
the sample for melting point and IR.
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http://www.ic.sunysb.edu/Class/che133/techniques/meltpts/meltpts.html
MOLECULAR VIBRATIONS
VERIFYING SYNTHETIC SUCCESS - IR
The structures of the starting and product molecules in this
exercise (and many syntheses) 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 a virtual lab on absorption
spectroscopy of food dyes (ultraviolet and visible).
0.9
0.8
where, 1/μ = 1/m1 + 1/m2
k is a the strength of
the spring (bond)
0.7
0.6
0.5
Some Typical Infra-red Frequencies
0.4
0.3
0.2
0.1
Wavelength (nm)
Absorption of light in infrared region is
primarily due to VIBRATION of molecules.
(1 μm = ) 1,000 nm – 100,000 nm (Infra-red)
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740
71
0
68
0
65
0
62
0
590
56
0
53
0
50
0
47
0
44
0
41
0
380
0
35
0
Absorbance
The frequency of molecular vibrations depends on the mass
of the atoms involved and the strength of the bonds
between them. For two atoms, the relationship is:
To involve numbers in a convenient range, it is customary
to report f/c where c = velocity of light in cm/sec.
Since f = c, this just 1/, which we measure in cm-1.
Absorbance vs Wavelength
1
Those absorptions were due to transitions
between ELECTRONIC energy levels
(UV) 350 nm – 700 nm (Red)
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Molecule
f/c
(cm-1)
H-H
4161
H-D
3632
H-F
3961
H-I
Molecule
f/c
(cm-1)
Molecule
f/c
(cm-1)
H-Cl
2885
N-O
2220
H-Br
2558
C-O
2143
2230
Cl-Cl
557
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WHAT ABOUT POLYATOMIC MOLECULES?
GROUP VIBRATIONS
Study of many thousands of substances shows that
SPECIFIC MOLECULAR FRAGMENTS absorb light at
well-defined, characteristic frequencies, e.g.,
H2O
3835 cm-1
3939 cm-1
1648 cm-1
Symmetric Stretch
Bend
Asymmetric Stretch
f (cm-1)
CH4
3025 cm-1
1367 cm-1
3157 cm-1
C — H (aromatic)
3030 – 3050
C — H (alkane)
2850 – 2960
C == O (aldehyde)
1680 – 1750
H
H
C
H C
C
C
C
O
O — H (phenol)
O — H (alcohol)
C O
~1100
~1650
~1720
C H
~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.
1600, 1500
C C
1367 cm-1
http://www2.ess.ucla.edu/~schauble/molecular_vibrations.htm
C  C (aromatic)
C C
3200
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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
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Note that like visible spectra, IR spectra are displayed as
intensity vs increasing wavelength
http://sdbs.db.aist.go.jp/sdbs/cgi-bin/cre_index.cgi?lang=eng
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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
% Transmittance
So, absorption peaks point DOWNWARD
O-H
-H
Wavelength
etc.
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4000 cm-1
3000 cm-1
C-H3
HC=O
2000 cm-1
CC
30
1500 cm-1
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H
H cm
O -1
C
VANILLIN
IR SPECTRUM: 1500 cm-1 – 4000
H
C
H
H
C
H C
O
C
C
O
C
H
H
H
H
C O C H
C
H C
C
C
O
H
C
H
H
H
H
H
C H
C
O
CC
C
H
IR OF THE PRODUCT
H
H
H
H
C O C H
C
H C
H
C
C
C
C
C
O
H
O
H
H
O
H
H
C H
H
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).
ring hydrogen stretch
C−H stretch in the methoxy (O-CH3)group
C−H3
C
O
So, the product spectrum should show the absence of the
C=O absorption near 1700 cm-1.
-O−H stretch due to the OH group on the ring
HC=O
H C
C
O
H
The vanillin spectrum shows 6 major absorption peaks
between 1500 and 4000 cm-1.
−H
C
C
C=O stretch in the aldehyde group
From the table we see that we expect it at:
two peaks due to the ring CC stretch
All but one of these peaks should show up in the spectrum
of the product, vanillyl alcohol.
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H2C O—H between 3400 and 3600 cm-1
in the alcohol O-H region.
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IR Spectrometer
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)
You may wish to examine the IR spectrum
of pure vanillyl alcohol before coming to lab.
You can download it from:
The ZnSe
sample
area
http://sdbs.db.aist.go.jp/sdbs/cgi-bin/cre_index.cgi?lang=eng
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INTERPRETATION OF IR SPECTRA
Use the infrared spectrum to verify the presence or
absence of functional groups
Reaction replaces a -HC=O group by a –H2C-O-H.
So, while starting material will show:
absorption by -HC=O
no absorption by –H2C-O-H
Product should show:
absorption by –H2C-O-H
no absorption by –HC=O
a new O-H absorption
Should also be able to identify absorptions of other
functional groups common to vanillin and vanillyl alcohol by
35
comparing their spectra.
34
TRANSMISSION vs FTIR/HATR SPECTROSCOPY
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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