Oxazolone TEMPLATE
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TEMPLATE: Oxazolone Synthesis
Please make a copy before editing.
Name: Olivia Mertes
Section: TA First Name: Sepi, Moha
TA email: abbais@iastate.edu
Combination: 38-14-28
Pre-lab Questions
Day 1: Hippuric Acid Synthesis
1.
(4 pts) The first synthesis for this experiment is shown below. Examine the product of the reaction. Given the most common types of reactions for acyl chlorides, what type of reaction is this? (e.g. nucleophilic substitution, addition, elimination, etc.)
This type of reaction is a nucleophilic substitution since the lone pairs in the nitrogen will attack the carbonyl and the chlorine will be kaicked off.
2.
(4 pts) Consider the structure of glycine at various pHs as shown below. Explain why the addition of NaOH is necessary to facilitate the reaction between benzoyl chloride and glycine.
The addition of NaOH is necessary to facilitate the reaction between benzoyl chloride and glycine since we want glycine to act as the nucleophile instead of the base in the reaction. The lone pair on the nitrogen
needs to be there so that it can attack the carbonyl of the benzoyl chloride. It the reaction did not have NaOH the amine group would be protonated and the lone pairs on notrogen wouldn’t be there and thereofr wouldn’t attack the carbonyl aand a whole different product would form.
3.
(16 pts) Draw the complete mechanism for the reaction between benzoyl chloride and glycine. 4.
(20 pts) Fill in the following tables with amounts to use.
Amounts for Synthesis of Hippuric Acid
Reactant
Reactant
Base
Acid
Product
benzoyl chloride
glycine
NaOH (aq)
conc HCl (aq)
hippuric acid
Moles
2.5 mmol
2.5 mmol
3x excess Enough to neutralize NaOH
2.5 mmol
MW
140.57 g/mol
75.07 g/mol
NA
NA
179.17 g/mol
Concentration
NA
NA
6 M
12 M NA
Amount 0.35g
0.19g
1.25mL
0.625mL
0.448g
Benzoyl chloride: 0.0025mol x 140.57g/mol = 0.35g
Glycine: 0.0025mol x 75.07g/mol = 0.19g
NaOH: 0.0075mol / 6M = 1.25mL
HCl: 0.0075mol / 12M = 0.625mL
Hippuric Acid: 0.0025mol x 179.17g/mol = 0.448g
Day 2: Oxazolone Sythesis
Mechanism
The reaction of hippuric acid and an aromatic aldehyde to produce a fluorescent oxazolone has multiple steps. These questions will walk you through the steps.
1.
(4 pts) The first step is simply a proton transfer. Fill in the missing curved arrows and draw product A to complete the step.
2.
(4 pts) The second step is a nucleophilic attach on the anhydride. Draw product A and fill in the missing arrows to complete the step.
3.
(4 pts) The third step is loss of a leaving group. Draw product B to complete the step.
4.
(4 pts) Compound B tautomerizes as shown below. Draw product B and curved arrows to show tautomerization.
5.
(8 pts) Next step is a nucleophilic attack of the double bond on the carbonyl carbon of the aldehyde. This is followed by a proton transfer step to remove the charge on the oxygen. Draw the missing product C and fill in missing curved arrows to complete these two steps.
6.
(4 pts) Notice that the product of the last step is a beta hydroxy ketone (circled in red below). Remember that a beta hydroxy ketone is also the product of an aldol addition. The next step in the oxazolone reaction is like an aldol condensation: Base removes a proton, a double bond is formed, and a hydroxide leaves. Using this information, draw product D and fill in missing curved arrows.
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7.
(8 pts) The next step involves intramolecular nucleophilic attack to form a five membered ring. Draw the missing product E. The curved arrows are drawn for you this time.
8.
(4 pts) The final step involves loss of a leaving group to form the final product, an oxazolone. Draw the missing product E and fill in missing curved arrows.
Amounts
9.
(16 pts) Fill in the following tables with amounts to use.
Amounts for Synthesis of Oxazolone
Reactant
Reactant
Base
Product
4-(dimethylamino)benzaldehyde
Hippuric acid
anhydrous sodium acetate
Moles
1.4 mmol
1.4 mmol
1.4 mmol 1.4 mmol
MW
149.19 g/mol
179.17 g/mol
82.03 g/mol
292.3 g/mol
Amount 0.2089g
0.251g
0.115g
0.409g
Reactant: (4-dimethyl): 1.4mmol = 0.0014mol x 149.19g/mol = 0.2089g
Reactant (Hippuric): 0.0014mol x 179.17g/mol = 0.251g
Base: 0.0014mol x 82.03g/mol = 0.115g
Product: 0.0014mol x 292.3g/mol = 0.409g
In-lab Notes
Week 1:
o
Mass of benzoyl chloride: 0.3mL
o
Mass of glycine: 0.185g
o
Mass of watch glass: 22.641g
o
Mass of watch glass + product: 22.702g
o
Mass of product: 0.061g
o
Melting point: 185-189 degrees C
Week 2:
o
Mass of aldehyde: 0.16mL
o
Mass of anhydrous sodium acetate: 0.118g
o
Mass of hippuric acid: 0.253g
o
Mass of acetic anhydride: 0.7 mL
o
Mass of watch glass: 24.185g
o
Mass of watch glass + product: 24.306g
o
Mass of product: 0.121g
o
Melting point: 140-142 degrees C
Analysis and Application Questions
Do not copy and paste structures or mechanisms from any source.
Answers must be legible or they will not be graded.
Note that there are no application questions and point structure is different due to this being a 2-
day experiment.
Part 1: Hippuric Acid Synthesis
Data and Observations
1.
(5 pts) Calculate your percent yield. Show your work for full credit.
2.
(1 pts) Melting point range of crude and pure product.
a.
Our melting point range was 185-189 degrees
3.
Copy of FTIR spectra of crude and pure product annotated with:
Crude Product:
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Pure Product:
a.
(6 pts) Structure (3 pts ea)
b.
(8 pts) Peaks labeled according to functional groups in product. (4 pts ea)
4.
Copy of Proton NMR spectra of product annotated with:
3 . 6
3 . 8
4 . 0
4 . 2
4 . 4
4 . 6
4 . 8
5 . 0
5 . 2
5 . 4
5 . 6
5 . 8
6 . 0
6 . 2
6 . 4
6 . 6
6 . 8
7 . 0
7 . 2
7 . 4
7 . 6
7 . 8
8 . 0
8 . 2
8 . 4
8 . 6
8 . 8
9 . 0
9 . 2
f 1 ( p p m
)
- 5 0 0 0
0
5 0 0 0
1 0 0 0 0
1 5 0 0 0
2 0 0 0 0
2 5 0 0 0
3 0 0 0 0
3 5 0 0 0
4 0 0 0 0
4 5 0 0 0
5 0 0 0 0
5 5 0 0 0
6 0 0 0 0
6 5 0 0 0
7 0 0 0 0
A
p r 1 2 2 0 2 3 _
3 3 2 L S e c 5 . 3 3 . f i d
O
x a z o
l o n e p 1
2 . 1 0
2 . 1 6
1 . 0 3
2 . 0 0
1 . 0 0
3 . 8 5
3 . 9 2
3 . 9 4
6 . 1 5
7 . 3 4
7 . 3 6
7 . 3 7
7 . 3 8
7 . 3 8
7 . 3 8
7 . 4 2
7 . 4 3
7 . 4 4
7 . 4 6
7 . 4 7
7 . 4 7
7 . 4 8
7 . 4 9
7 . 4 9
7 . 5 0
7 . 5 1
7 . 5 3
7 . 5 4
7 . 5 4
7 . 5 4
7 . 5 5
7 . 5 6
7 . 5 6
7 . 5 7
7 . 5 8
7 . 5 8
7 . 6 9
7 . 7 0
7 . 7 1
7 . 7 1
7 . 7 2
7 . 8 6
7 . 8 7
7 . 8 7
7 . 8 8
7 . 8 8
7 . 8 9
7 . 8 9
7 . 9 4
7 . 9 4
7 . 9 6
8 . 8 2
8 . 8 3
8 . 8 5
a.
(3 pts) Structure
b.
(7 pts) Prediction table showing shift, integrations and splitting
c.
(5 pts) Each peak labeled with associated proton(s) in structure
Analysis
5.
Explain how your data give evidence that you synthesized the product that you expected and that
it is pure.
a.
(3 pts) Compare melting point with what was expected. Comment on possible contaminants if relevant. Was the melting point improved with recrystallization? Explain.
i.
Our expected melting point was 190 degrees C and we had a melting point of 185-189 degrees C for our pure product. Pur product does not show any contaminants or deviation from what we expected with our observed melting point.
b.
(5 pts) Compare FTIR spectrum with what was expected. Comment on possible contaminants if relevant. Was the FTIR spectrum improved with recrystallization? Explain.
i.
We expected the FTIR spectrums to have four relevent peaks. These peaks were
N-H which is expected to show at around 3250cm^-1, O-H stretch which is supposed to appear around 3000cm^-1, a C-H stretch which is supposed to appear around 2750cm^-1 and a C=O stretch that is supposed to appear around 1750cm^-1. Our pure spectra shows all founr relevent peaks. The N-H stretch appears at around 3300cm^-1, O-H appears at about 3000cm^-1, C-H appears right under 3000cm^-1, C=O appears right about 1700cm^-1. Yes, the FTIR was improved with recrystallization as the peaks in the pure product were a lot atringer and easier to identify. This also indicated our product had little contaminates and our product is pure.
c.
(5 pts) Compare 1
H-NMR spectrum with what was expected. Comment on possible contaminants if relevant.
i.
We predicted the final product of Hippuric acid to have 5 relevent peaks. Proton A around the aromatic were expected to have a shift between 7-8ppm, integration of 3H and multiplet splitting. Our spectra shows a peak at about 7.5ppm with 3H integration and multiplet splitting. Proton B, connected to nitrgoen was predicted to habe a shift between 8-9ppm with 1H integration and triplet splitting. Our spectra shows a peak at about 8.8ppm that has 1H integration and triplet splitting. Proton C, which is the proton off of the carbon between the nitrogen and carbonyl was expected to have a shift between 3-4ppm
2H integration, and doublet splitting. Our spectra shows a peak at about 4.0ppm with 2H integration and doublet splitting. Proton D which is connected to the oxygen was predicted to have a shift between 12-13ppm with 1H integration and singlet splitting. Our spectra does not show a peak anywhere near 12ppm. Proton E whish are next to A on the aromatic ring were expected to have a shift between 7-8ppm with 2H integration and doublet splitting. Our spectra shows a peak at about 7.9ppm with 2H integration and doublet splitting. Our spectra does not show any contaminants, but it also seems to be missing a peak which indicates that our synthesis had some issues and our product is probably not as pure as it could have been.
Part 2: Oxazolone Sythesis
Data and Observations
6.
(5 pts) Calculate your percent yield. Show your work for full credit.
7.
(1 pts) Melting point range
a.
140-142 degrees C
8.
(4 pts) Photo of TLC plate with R
values calculated.
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9.
(1 pts) Photo of fluorescence properties of product.
10.
Include an annotated FTIR spectrum of your product annotated with:
a.
(3 pts) Structure
b.
(4 pts) Label each peak with the corresponding functional group on the structure.
11.
Proton NMR spectra of your product annotated with:
2 . 5
3 . 0
3 . 5
4 . 0
4 . 5
5 . 0
5 . 5
6 . 0
6 . 5
7 . 0
7 . 5
8 . 0
8 . 5
f 1 ( p p m
)
- 2 0 0 0 0
0
2 0 0 0 0
4 0 0 0 0
6 0 0 0 0
8 0 0 0 0
1 0 0 0 0 0
1 2 0 0 0 0
1 4 0 0 0 0
1 6 0 0 0 0
1 8 0 0 0 0
2 0 0 0 0 0
2 2 0 0 0 0
2 4 0 0 0 0
2 6 0 0 0 0
2 8 0 0 0 0
A
p r 1 9 2 0 2 3 _
3 3 2 L S e c 5 . 3 3 . f i d
O
x a z o
l o n e p t . 2
3 . 0 0
1 . 9 3
2 . 1 1
1 . 9 8
0 . 9 9
1 . 9 7
1 . 9 2
2 . 4 6
2 . 4 7
7 . 2 7
7 . 2 8
7 . 3 1
7 . 3 3
7 . 3 4 C
D
C l 3
7 . 5 4
7 . 5 4
7 . 5 5
7 . 5 6
7 . 5 8
7 . 5 9
7 . 6 1
7 . 6 2
7 . 6 2
7 . 6 4
7 . 6 4
7 . 6 5
7 . 6 5
7 . 6 6
8 . 1 3
8 . 1 5
8 . 2 0
8 . 2 0
8 . 2 1
8 . 2 2
8 . 2 3
a.
(3 pts) Structure
b.
(7 pts) prediction table showing shift, integrations and splitting
c.
(5 pts) each peak labeled with associated proton(s) in structure
Analysis
12.
Explain how your data give evidence that you synthesized the product that you expected and that
it is pure.
a.
(2 pts) Compare melting point with what was expected. Comment on possible
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contaminants if relevant. a.
Our expected melting point of our final product was around 149-151 degrees C. Our product had a melting point range of 140-142 degrees C. Based on the melting point, our product shows that is was decently pure as it was in the reasonable range of acceptable melting point of 10 degrees.
b.
(4 pts) Did the TLC plate give evidence of successful synthesis? Explain
a.
Our TLC plate does give evidence of a pretty successful synthesis. Our product only has one spot showing that it had no contaminants and it had a higher Rf value than 2 out of 3 of our starting materials, one matching our product Rf value which indicates evidence of a successful synthesis, but there could be some contaminants in it with the matching Rf value of one of the starting materials.
c.
(3 pts) Did the color and fluorescence properties give evidence of a successful synthesis of oxazolone? Explain.
a.
Yes, the color and flourescence properties do give evidence of a successful synthesis since we know oxazilone has flourescence properties due to its amino acid light sensitivity. Our product showed a neon yellow color and flourescent properties under the UV light which indicates a succeddful synthesis and creation
of oxazolone.
d.
(5 pts) Compare FTIR spectrum with what was expected. Comment on possible contaminants if relevant. a.
The FTIR spectra is what we expected. We expected three rlelvent peaks of our product. The first peak was a C-H stretch around 2500cm^-1, a C=O stretch at about 1750cm^-1, and a C=C stretch at right about 1500cm^-1. Our soectra shows these three relevent peaks, C-H right above 2500cm^-1, C=O right above 1500cm^-1 ans C=C right below 1500cm^-1. There do not appear to be any contaminants in our spectra.
e.
(5 pts) Compare 1
H-NMR spectrum with what was expected. Comment on possible contaminants if relevant.
a.
For our product we expected four major peaks in the NMR spectra. The first proton A the methyl group was expected to have a shift of 2-3ppm, integration of 3H and singlet splitting. Our graph shows a peak at about 2.5ppm with a 3H integration and singlet splitting. Proton B which is around the aromatic right connected to the methyl group was expected to have a shirt of 7-8ppm with 4H integration and multiplet splitting. Our spectra shows a peak at about 7.4ppm with 3H integration and multiplet splitting. Proton C which is connected to the alkene was expected to have a shihft between 4-5ppm with 1H integration and singlet splitting. Our spectra does not show a peak for proton C. Proton D which is connected to the second aromatic ring was expected to show between 7-8ppm
with 5H integration and multiplet splitting. Our soectra shows a peak at about 7.6ppm with 3H integration and multiplet splitting. Our spectra shows many contaminants are present. There are peaks that are not accounted for in the product, and there are peaks we expected to be there that aren’t. Since we are missing a whole peak from our product we can conclude that our product was not
synthesized correctly, and there are contaminants in our product.
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- The synthesis of an azo dye is shown below. Please answer the following questions. NH2 *N2 1) Na2CO3 2) NANO2, HCI B N-N A ŠOH Step 1 Step 2 1 2. In step 2, Why compound 2 selectively adds to the para position of aniline? (multiple answers) Oa) amino group is an electron-withdrawing group ob) amino group is an electron-donating group Oc) amino group is an ortho-para directing group Od) para positions is more sterically hinderedarrow_forwardPrelab Questions: 1. Write the balanced equation for the bromination of 4-methylbenzoic acid. 2. Complete the following table for the bromination with the values from the procedure given: density, g/mL XXX XXX MW, mp/bp, °C mp, 179-181 mp, 175-180 (dec) mp, 105 bp, 132 equiv mmol compound 4-methylbenzoic acid NBS Benzoyl peroxide chlorobenzene g/mol mL XXX XXX XXX XXX XXX XXX XXX XXX 3. What is the theoretical yield for this reaction?arrow_forwardI am not sure how to do number 10arrow_forward
- Pkease give me the letter to fill the blank !arrow_forwardReset Help N-Acetylserotonin to melatonin Serotonin to N-acetylserotonin 5-Hydroxytryptophan to serotonin Acetylation Methylation Decarboxylation Submit Request Answer P Pearson Copyright © 2021 Pearson Education Inc. All rights reserved. Terms of Use | Privacy Policy | Permissions Contact Us / Claearrow_forwardAcids: Reaction of HCI with CH3CH2OH The pK, of CH3CH₂OH is 15.9 and the pK, of HCl is -7. Part A Draw the product that is formed when the compound shown below is treated with an excess of HCI. Draw only the organic ion. Interactive 3D display modearrow_forward
- I I I Draw the missing organic structures in the following multistep synthesis. Show all structures at physiological pH (pH = 7.4). Ignore any inorganic byproducts formed. H3N, (BOC)₂0 Et3N O Select to Draw O Q DCC + + H3N₁ ···||| O CH3OH, H+ Select to Draw Iarrow_forward16 eBook Print References 10 attempts left Click the "draw structure" button to launch the drawing utility. The following nucleophilic substitution reaction has been reported in the chemical literature. Draw the product by analogy to what you know about nucleophilic substitution in simple systems. Na + Check my work Br draw structure ...arrow_forwardDraw the missing organic structures in the following multistep synthesis. Show all structures at physiological pH (pH 7.4). Ignore any inorganic byproducts formed. = HaN, ° 回ǒ Θ HaN. 00 (BOC)20 EtзN CH3OH, H+ Select to Draw + Select to Draw DCC Select to Drawarrow_forward
- Draw the structure of the major organic product(s) of the reaction. ممله + H30* • You do not have to consider stereochemistry. • All carboxyl and amino groups should be drawn in the neutral form. • Draw one structure per sketcher. Add additional sketchers using the drop-down menu in the bottom right corner. Separate multiple products using the + sign from the drop-down menu.arrow_forwardI need help with predicting the ester.arrow_forwardThe following chemical equation is for an elimination reaction, where the importan H H- H- H catalyst A Explanation Use the information provided to predict the missing organic product and draw its stru- skeletal ("line") style for the structure as the rest of the equation. ? + H-H Check 27 Click and drag to start drawing a structure.. útv 2023 McGrarrow_forward
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