CHM 227: Organic Chemistry Laboratory
Iodination of Vanillin and Subsequent Suzuki-Miyaura Coupling Lab Report
Names: Ali Chedid and Derek Carter
Answers must be TYPED
Draw structures using ChemDraw and copy the figure onto the report. Properly label and
format any reactions, mechanism, table and figures using ACS style formatting guidelines
(refer ‘how to record data and results’ presentation slides on Canvas)
Session 01 – Iodination of Vanillin
1. Materials and Methods. Develop a Materials and Method Section for the electrophilic
aromatic iodination of vanillin you performed in the lab. Follow Materials and Method
Section Guide on Canvas.
Material:
All the reagents were provided by the University of Michigan-Dearborn. Reagents used:
4-hydroxy-3-methoxybenzaldehyde (121-33-5), potassium iodide (7681-11-0),
Potassium peroxymonosulfate (70693-62-8), and oxidane (9028-76-6). Equipment used was
reaction reflux apparatus, vacuum filtration apparatus, gas chromatography mass spectrometry
machine, and Fourier-transform infrared spectroscopy machine.
Method:
First 1.0638g of oxone was dissolved with 4mL of DI oxidane inside of a reaction flask. In a
25mL round bottom flask 0.5015g of 4-hydroxy-3-methoxybenzaldehyde was dissolved in 4mL
of DI oxidane. To the round bottom flask a stir bar was added. The round bottom flask was then
placed inside the reflux apparatus and set to spin and heat was applied to it. Into the separatory
flask attached to the reflux condenser via a Claisen adapter the solution with
4-hydroxy-3-methoxybenzaldehyde was added. Slowly over ~5 minutes, the solution in the
separatory flask was added. Once all of the solution was added to the round bottom flask the
separatory flask was removed from the Claisen adapter and the adapter was sealed with a glass
stopper and the reaction was allowed to reflux for ~1 hour. Once the reaction finished refluxing,
10mL of DI oxidane was used to wash the walls of the condenser and Claisen adapter into the
round bottom flask. Once cooled the mixture was tested for excess oxidizing agent using starch
paper. Once a negative test was read vacuum filtration was performed. The product was left to
dry for ~5 minutes. Once dried the final mass of the crude product was recorded and was
analyzed by gas chromatography mass spectrometry and Fourier-transform infrared
spectroscopy.
Data and Results
2. A) Amount of product isolated. Show all the calculation work –
Mass of crude product = (Mass of watch glass + filter paper + crude product) – (Mass of
empty watch glass + filter paper)
Mass of crude product = 23.168g – 22.241g = 0.927g
B) Physical appearance of the product –
Brown/Sand-like
Calculations
3. A) Calculate the theoretical yield of the product (use the exact amounts you measured/used
in the lab). Record the overall reaction and show all the calculation steps.
Figure 01: Reaction scheme for Iodination of Vanillin
Determination of Masses
● Mass of Vanillin
Mass = 0.515g
● Mass of Oxone
Mass = 1.0638g
● Mass of Potassium Iodide
Mass = 0.554g
Vanillin = .5015 g x (1 mol / 152.15 g ) = .00329 mol
Oxone = 1.0638 g x (1 mol / 152.2 g ) = .0070 mol
Potassium Iodide = .554g x (1 mol / 166.0028 g ) = .00333 mol
Determining Limiting Reagant:
# moles of product from Vanillin = .00329 mol x ( 1 mol product / 1 mol Vanillin) = .00329 mol
# moles of product from Oxone = .0070 mol x ( 1 mol product / 1 mol Oxone) = .0070 mol
# moles of product from Potassium Iodide= .00333 mol x ( 1 mol product / 1 mol Potassium
Iodide) = .00333 mol
Theoretical Yield = .00329 mol x ( 1 mol product / 1 mol LR) x 277.9 = .914 g
B) Calculate the % yield of the product.
Mass of empty watch glass + filter paper: 22.241 g
Mass of watch glass + filter paper + product: 23.163 g
● Amount of product isolated – 23.163 g - 22.241 g = .922 g
● Show all the calculation steps.
Percent Yield = (.922 g / .914 g) x 100% = 100.8 %
Discussion
4. Explain the following based on the iodination experiment you performed in the lab.
A. Why did you keep Vanillin and potassium iodide separated from Oxone until you
complete the experiment set up? Why didn’t you mix all the three reagents at once?
We keep Vanillin and potassium iodide separated from oxone until we complete the reaction
setup for specific reasons. The main one would be to control when the reaction should happen.
This helps control the timing of the reaction as well as the reaction condition. If we were to add
it at once we would most likely have other side reactions taking place and our desired product
would not be as expected. Keeping the oxone separated provides that stability until we start
that reaction.
B. Why it is necessary to have a reflux condition during the reaction.
It is critical to have the reflux conditions in the reaction for a variety of reasons. The main
reason would be to prevent any loss of reactants in the mixture. If we did not have the reflux
condtiions, it is possible that some of our reactants may evaporate. The effects of that can
reduce our percent yield and in addition to that, the reaction could be incomplete since that
reactant would have been used in the synthesis. So we can say that while it's in the process of
refluxing, we are extending the reaction in order for the reaction itself to be under completion.
Lastly, doing the reflux can help with the reactants interactions during the chemical reaction.
While mixing and heating it up for a long period of time, this allows the reactant to become
more soluble thus making way for the interaction to occur.
C. Did you see any distinct observation once the reaction starts or when Oxone is added? If
yes, what was the observation? and what causes that observation?
Once the oxone was added and we began to observe the reaction taking place, we could
see a neon green color developing in the round bottom flask. We already know that
Iodine can produce a variety of colors when reacted with the right compound. To
produce the neon color we believe the oxone, being a strong oxidizing agent, was able to
react with the iodine to produce that vibrant color. Along with that, we also believe
factors such as the concentration and temperature also play a role in producing that
color too.
D. How did you check for any remaining oxidizing reagents at the end of the reaction? How
did you quench the excess oxidizing reagent?
We checked for any remaining oxidizing reagents using the starch-iodide paper. When
applying a droplet, the paper acts as an indicator. If we get a dark blue color, that means we
still have some oxidizing reagents, if we see no color change in the paper, then we know that
we do not have any remaining oxidizing reagents. If we do end up having oxidizing reagents,
we quench the excess by adding sodium bisulfate or sodium thiosulfate until the negative
test is observed.
E. What are the advantages of this method/reaction over the traditional methods of
iodination? Explain with suitable examples from literature.
Suzuki coupling reactions are generally speaking much more environmentally friendly than
traditional methods of iodination. The reaction conditions are generally much more mild than a
typical iodination reaction. Both the catalyst and solvent can also be reused which helps it to
align more closely with the principles of green chemistry. An example of a traditional method
stated in the article would be electrophilic aromatic substitution (EAS). The reading mentions
that in EAS, the nitration of methybenzoate, used as a traditional experiment, uses nitric and
sulfuric acids which are strong mineral acid mixtures. These chemicals are not environmentally
friendly which makes this a disadvantage compared to the greeen chemistry we have learned.
Product confirmation and Spectroscopic analysis
5. GCMS analysis:
a) Attach the GCMS spectra and label all suitable peaks on the mass spectrum
Figure 02: GCMS spectrum for Iodination of Vanillin
b) Discuss the purity of your product based on GC spectrum
Judging by the gas chromatography portion of the spectrum it tells us a great deal about the
purity of our sample. Since we only see one peak in the gc spectrum it indicates that there is
only the presence of a single compound. If there were multiple peaks present, it would indicate
impurities/unwanted products.
c) How could you use the mass spectrum to identify the reaction completion or success of
your reaction? Explain.
Due to the molecular ion peak of 277.9 we can say with a high level of certainty that our desired
product was produced. The molecular ion peak is by far the strongest peak present indicating
that this was the most abundant compound present. Also the molecular weight of 5-iodo
vanillin (4-hydroxy-3iodo-5-methoxybenzaldehyde) is 278 g/mol. This aligns very closely with
that of the molecular ion peak. We can see a few smaller peaks to the left of the molecular ion
peak (these are likely mostly fragments). The peaks to the right of the parent peak must be
other compounds (as their molecular weight exceeds that of our product and cannot be
fragments). However the peaks of the compounds to the right of the molecular ion peak are too
small to indicate a significant presence.
6. NMR Analysis:
Consider the following 1H-NMR spectrum.
a) Tabulate the NMR peaks, chemical shift, splitting pattern, and integration values of the
product
Table 01: Predicted NMR peak assignment data for product
NMR Peak
Chemical Shift
Splitting Pattern
Integration
Ha
~9.9
Singlet
1H
Hb
~7.8
Singlet
1H
Hc
~9.7
Singlet
1H
Hd
~3.8
Singlet
3H
He
~7.2
Singlet
1H
b) Draw the structure of the product and identify the sets of hydrogens with the
corresponding signals in the 1H NMR (use Ha, Hb, Hc, etc notation)
7. Discussion:
How could you use the 1H-NMR data to confirm the reaction completion or success of your
reaction? How can you identify if there is still unreacted starting material (Vanillin) present
in the final compound? (Hint: which distinct peak(s) or region on NMR you should look for)
It is apparent that the reaction was completed successfully by looking at several factors.
Firstly, the number of hydrogens aligns with our desired product (5 unique peaks). The tallest
(most integrated) peak of our spectrum aligns perfectly with what we would expect, as our final
product should have exactly 1 methyl group present. We would expect to see 2 aromatic peaks
between 7 and 8 ppm because there are only 2 spots on the benzene ring of our product that
are free. And the 2 peaks we see close to 10 ppm make sense since we have 2 hydrogen atoms
close to electronegative groups (oxygen). The splitting patterns (seeing as how they are all
singlet peaks) indicates that all of our hydrogens do not have any neighboring hydrogens
(otherwise we would see splitting beyond singlets). And finally the chemical shift of the
hydrogens aligns quite nicely with what we would expect. By simply looking at the number of
hydrogens present in the spectrum we can tell the difference between our starting material and
our final product. If there were still unreacted starting material (vanillin) present in the final
product then we would see an additional hydrogen peak present. This hydrogen should’ve been
replaced by electrophilic aromatic substitution. Where the hydrogen used to be in vanillin, an
Iodine should have replaced it. We would see an extra hydrogen around ~7 ppm indicating that
it was within the aromatic region if vanillin were still present.
Session 02 – Suzuki-Miyaura Coupling of 5-iodovanillin.
Data and Results
8. A) Amount of product isolated. Show all the calculation work –
Mass of empty round bottom flask: 57.490 g
Mass of round bottom flask + product: 57.734 g
Amount of isolated product: 57.734 g - 57.490 g = .244 g
B) Physical appearance of the product –
The product was brown with a grain-like sand texture.
Calculations
9. A) Calculate the theoretical yield of the product (use the exact amounts you measured/used
in the lab). Record the overall reaction and show all the calculation steps.
Figure 03: Reaction scheme for Suzuki-Miyaura Coupling of 5-iodovanillin
Determination of Masses:
Mass of 5-iodovanillin
Mass = 0.258 g
Mass of 4-methylphenylboronic acid
Mass = .1607 g
Mass of Potassium Carbonate
Mass = 0.4008 g
5-iodovanillin = .258 g x (1 mol / 278.04 g ) = .000928 mol
4-methylphenylboronic acid = .1607 g x (1 mol / 151.96 g ) = .0011 mol
Potassium Carbonate = .4008 g x (1 mol / 138.205 g ) = .0029 mol
Determining Limiting Reagent:
# moles of product from 5-iodovanillin = .000928 mol x ( 1 mol product / 1 mol
5-iodovanillin) = .000928 mol
# moles of product from 4-methylphenylboronic acid = .0011 mol x ( 1 mol product / 1 mol
4-methylphenylboronic acid) = .0011 mol
# moles of product from Potassium Carbonate = .0029 mol x ( 1 mol product / 1 mol
Potassium Carbonate) = .0029 mol
Theoretical Yield = .000928 mol x (1 mol product / 1 mol LR) x 242.1 = .225 g
B) Calculate the % yield of the product.
● Amount of product isolated –> .244 g (math shown above)
● Show all the calculation steps.
Percent Yield = (.244 g / .225 g) x 100 = 108.4 % yield
Product confirmation and Spectroscopic analysis
10. GCMS analysis:
c) Attach the GCMS spectra and label all suitable peaks on the mass spectrum.
d) Discuss the purity of your product based on the GC spectrum.
e) How could you use the mass spectrum to identify the reaction completion or success of
your reaction? Explain.
Figure 04: GCMS Spectrum Suzuki-Miyaura Coupling product
b) The purity of the compound is ~90% pure judging by the gc spectrum. The reason why
we believe this to be true is due to the presence of multiple peaks. The presence of multiple
peaks tells us that there are more than one compound present in our final product. Seeing as
how there are 4 separate peaks, this tells us that there are 4 different compounds present.
However, one peak is clearly more intense/prominent than the rest of them. And judging by the
mass spectrum we can conclude that this peak matches our parent peak (especially due to the
intensity of the parent peak).
c) We can use the mass spectrum to identify whether or not our reaction was successful by
evaluating the peaks on the spectrum. On the spectrum there will be a variety of peaks shown
and the furthest one to the right known as the molecular ion peak or parent peak is the
strongest one. This would represent the molecular weight of our final product. If we compare
that to the actual weight of the product you can see that they are relatively the same indicating
our reaction was a success. In addition to that we have the purity of the compound. The lack of
purity as shown in the spectrum above could indicate the slight change in the molecular weight,
meaning there were some impurities in the reaction.
11. NMR Analysis:
a) Predict the NMR for the expected product of Suzuki-Miyaura coupling. Use the following
scale and clearly show the peak position and peak multiplicity. (Use can hand draw the
peaks)
b) Draw the structure of the product and assign the correct sets of hydrogens with the
corresponding signals in the predicted 1H NMR above (use Ha, Hb, Hc, etc notation)
NMR Peak
Chemical Shift
Splitting Pattern
Integration
Ha
~2.3
Singlet
3H
Hb
~7.1
Doublet
2H
Hc
~7.3
Doublet
2H
Hd
~10.4
Singlet
1H
He
~8
Singlet
1H
Hf
~9.8
Singlet
1H
Hg
~7.25
Singlet
1H
Hh
~3.8
Singlet
3H
6-hydroxy-5-methoxy-4'-methyl-[1,1'-biphenyl]-3-carbaldehyde
c) Explain how the NMR spectrum gets different from that of 5-iodovanillin (NMR spectrum
in Q.6). What are the similar NMR regions or regions that do not change, and which
NMR regions get change for Suzuki-Miyaura coupling product. Justify your answer with
suitable peak information.
Comparing the NMR spectra of our starting reagent (5-iodo vanillin) to our final product we
should notice a few significant differences. First we should notice more hydrogen peaks to be
present in our final product compared to that of our starting reagent. Specifically an increase in
the number of aromatic carbons present due to the presence of a second aromatic ring in our
final product. We can also notice that our final product has 2 very intense peaks between 2-4
ppm (representing the presence of 2 distinct methyl groups). This is in contrast to the NMR
spectrum of the starting reagent. That spectrum would have only one intense peak (as there is
only one methyl group in 5-iodo vanillin). Another difference is the splitting patterns each
spectrum exhibits. In our starting reagent all of the peaks were singlets as none of the
hydrogens had neighboring hydrogens surrounding them. This isn’t true of our final product’s
NMR spectrum as (specifically in the aromatic region) there are multiple instances of hydrogens
neighboring one another. Despite these differences both spectra show similar peaks around 10
ppm for the hydrogens that are closest to electronegative groups (the carbonyl groups and and
hydroxyl hydrogens). But judging by these differences in spectra, we can easily conclude that a
significant difference has been observed between the starting reagent and our final product
6-hydroxy-5-methoxy-4'-methyl-[1,1'-biphenyl]-3-carbaldehyde.