As expected the solubility tests (Table [table: solubilityresults]) indicates that acidic compound is a strong acid compound. The infrared spectrum in Figure: [fig: IRs1], supports the presence of a carboxylic acid since there is a broad (O-H) stretch, between 2521 cm-1 and 2971 cm-1, and a (C=O) peak at 1673 cm-1. Furthermore, the 13C NMR ( Figure [fig: C13nmrS1]) shows an acidic (C=O) at 174 \delta and the proton NMR (Figure:[fig: PNMRs1]) has a (OH) singlet at 11 \delta. The functional group tests ( Table: [table: qualitative functionalgroup tests]) demonstrates that the unknown acid contains an aromatic functional group. The IR spectra shows a peak at 1600 cm-1, the typical benzene (C=C) zone and the 13C NMR also has 6 aromatic peaks between 126 \delta and 141 \delta. …show more content…
Because the multiplet at 8.02 \delta is more upfield, that benzene hydrogen must be near the carboxylic acid. The first substituent is the carboxylic acid and the second substituent is most likely a methyl group since there is a 22 \delta (CH3) peak in the 13C NMR and a 2.6 \delta (CH3) peak in the proton NMR. Based on this information there are three possible structures for the acidic substance, which are o-toluic, m- toluic, and p-toluic acid. P-toluic acid cannot be the structure since it only has 6 unique carbons, which contradicts the 8 unique signals in the 13C NMR, leaving o-toluic acid and m-toluic acid
Fragment 7 is shown in the structure. Electron density map suggested two bulk electro-rich group present in the fragment, which corresponds to the two ring structure on 7. The heterocyclic ring has high e-density and forms a hydrophobic interaction with Leu144 residue. The e-rich NH2 group on this ring
We can know from the unknown melting point table that this acid is o-Toluic because o-Toluic melting point is (103-105) °C. This was coming from NaOH extraction, which we added 10% HCl to make it acidic and recovered the acid. The crystalized form of NaOH was still wet when doing the measurement, so that make the temperature a little bit higher than theoretical melting point. Even that, it was left for a week to dry but the recovered acid was accumulated over each other, and can feel it was still wet when taking the acid in pipette to measure the melting
Moreover, the lack of chemical shift around 5.3 indicates the lack of double bonds. The NMR shows 7 peaks, which were labeled as a,b,c,d,e,f, and g respectively from low ppm to high ppm. Peak a and b appears to be similar to a methyl group with shifts between .9-1.4 ppm. Since peak b is more down field, it is assumed to be closer to the carbonyl. Since a carbonyl ketone have a chemical shift around 2.1, peak c, d, e are probably connected on one of the side chains.
3. The IR spectrum of the starting material shows a medium/strong C-O bond at around 1500cm-1, also the starting material shows a strong C-H bond at around 3000cm-1 and another medium C-H bond at 2865cm-1 indicating an aldehyde group whereas the product does not. The IR spectrum of the product shows a two weak broad O-H peaks at around
Which means that it did not have any neighboring hydrogens but it still had a hydrogen on the compound. The next peak was a multiplet around 6-7.1 ppm and had 5 similar hydrogens. According to the H NMR chemical shifts, 5 similar hydrogens and a ppm range from 6-8.0 ppm, this particular peak was an aromatic. More specifically a benzene with 5 similar hydrogens, and with another hydrogen with attached substituents. This aromatic also contains 4 degrees of saturation, satisfying the requirement that the compound needed to have DoU of 4. The next peak was a quartet around 3.4-3.5 with 2 hydrogens. The quartet meant that the compound had 3 neighboring hydrogens and that particular peak was a CH2. The last pear was a triplet around 1.2-1.4 ppm with 3 hydrogens on the compound. The triplet meant that compound that 2 neighboring hydrogens indicating that particular peak was a methyl group (CH3). This help determine that the CH2 and CH3 were bonded to one another creating an ethyl
In this graph there is a peak at 3248 cm-1 (C-O) and a peak at 1589 cm-1 (O-H) which confirms the presence of carboxylic acid.
The unknown acid range had expected values, but the range of the fluorene product was a little low, which could be due to impurities. The recrystallization of fluorene occurred too quickly, which is likely why the range is low. TLC tests could be performed on the products, the pure benzoic acids, and the pure fluorene to determine the compositions of the products. IR spectra of all of the substances could be taken to determine the structures of the products. There are no side products. NaCl is not a side product because the Na+ and Cl- ions dissociate in
Identifying this organic acid was an extensive task that involved several different experiments. Firstly, the melting point had to be determined. Since melting point can be determined to an almost exact degree, finding a close melting point of the specific unknown can accurately point to the identification of the acid. In this case the best melting point
This now left 2-Butanone and Butyraldehyde as our remaining two compounds. So we went to the “Spectral Database for Organic Compounds SDBS” and looked the IR peaks they had on there website for the two compounds. After comparing the IR peaks we solved for in lab using the Infrared Spectroscopy and the two compounds IR peaks it was made obvious that 2-Butanone was our compound because it shared similar IR
The massive water peak at 3383cm-1 was to be expected from the NMR. Evidence of the chloride ligand was visible in the low frequency fingerprint region along with several aromatic CH and CN stretching features around the 1500cm-1 mark. Ring breathing also evident at 1009cm-1. There was little point in obtaining an IR for the complex other than to confirm the presence of the water
Next, I noticed that my spectra showed a 2H, quintet, a CH2 at 1.6. Then, I noticed that my spectra showed a 3H, doublet, a CH3 at 1.3 and a 3H, triplet, a CH3, at 0.98. Lastly, I noticed that my spectra showed a 1H, sextet at 2.7 and I knew that it was a C-H bond because it gave me the correct amount of carbon’s and hydrogen’s. Since my spectra had a monosubstituted ring, I knew that that piece went on the end. I knew that my two methyl groups also went on the ends of the structure. Then for the C-H bond, I knew that that piece had to go in between to the monosubstituted ring and one of the methyl groups, which left the
For the identification of the product, IR, 13C NMR and 1H NMR spectra were examined, and the product was found to be butyl propionate. In the IR spectrum, RM-11-Bi, five key peaks are observed. These peaks are sp3 hybridized carbons at 2961 cm-1 and 2877 cm-1, an ester at 1737 cm-1, a
Next, I observed my Proton NMR which had only one peak. The multiplicity is six. Therefore, I have six hydrogens. Using this information, I was able to conclude that my unknown is
1. Obtain a sample of the mixture. The mixture you will separate contains three components: NaCl, NH4Cl, and SiO2. Their separation will be accomplished by heating the mixture to sub-lime the NH4Cl, extracting the NaCl with water, and drying the remaining SiO2.
In week one we performed a qualitative solubility test of our fats and oils, synthesized our soaps and detergents, and performed a solubility test and lathering test for the soaps and detergents. We wanted to test the solubility of our starting materials of the soap making process to understand the properties of the materials. In our initial solubility test of the starting materials, we found that most of the materials were insoluble. As you can see in Table 2.0, olive oil and vegetable oil were only soluble in toluene and the shortening and lard were only partially soluble in acetone. In order to understand the solubility of the soaps and detergents, after our synthesis and filtration, we performed a qualitative solubility test with each of