1.1 Zwitterionic Intermediate Lab Report

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

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.

Day 1. Michael was coming home for vacation from college. When he got home he found out that

An NMR spectrum was also taken for our product sample which was close to the ideal but was a little off. Ideally there should be five peaks for the five chemical shifts from the five hydrogen groups on the product structure. The first peak should occur in the mid six ppm and can be described as a quartet. The second peak should occur at a ppm in the high 6s and should be another quartet. The third peak should be a triplet occurring at a low 7ppm. The fourth peak should be close by to the third, being another triplet. Finally, the fifth peak should be a doublet occurring at a mid 7ppm. The ppms for the ideal and the actual differ greatly in the fact that they span from approximately 5.3-6.25 instead of the ideal 6.5-7.5. It is fact that chemical shifts caused by benzene rings occur in the 6 to 8 ppm range, thus the nmr we received as a result can be somewhat doubted.. Ideally the first and second peaks should be a result of the hydrogens furthest away from the rings and the

2. (5 pts) List and explain the names and affiliations of the various characters/stakeholders in this story – I’m looking for us to use the story to map out the complexities that are generally associated with solving public health puzzles – the stakeholders you list and explain here should apply to many of the cases we consider going forward.

The 13C NMR Spectrum reveals the chemical environments of the C-H bonds. In this experiment, the carbons were assigned a number, from C1 to C4. C4 appeared the most downfield because it was the most deshielded because it is double bonded directly to an O and thus the O pulls the most on electron density. C3 is the most upfield because it is not bonded directly to any electron withdrawing groups or electronegative atom. C1 is the second most deshielded because this carbon is involved in a double bond which draws electron density. C2 appears second most upfield because this carbon is single bonded to an oxygen, but it is not as electron pulling as C4 or C1 that is double

Through this experiment, I used IR and NMR to identify several unknown compounds. In order to identify my pure unknown compound, IR-12, I first looked for any medium to strong peaks on my IR spectra. The peaks that were useful in identifying my unknown were: a C-O bond of an ester at 1043.8, a C-O bond of an ester at 1243.6, a stretching C=O bond at 1743.2, a stretching alkyl C-H bond at 2860.4, and a stretching alkyl C-H bond at 2958.7. When figuring out which IR unknown was my compound, I first looked to see if my IR spectra showed an alcohol or an amine. Since my IR spectra didn’t show an alcohol or an amine, I was able narrow down my choices of possible compounds to twelve. Next, I looked for an aromatic ring in my IR spectra and since my IR didn’t show an aromatic ring, I was able to narrow down my choices to six possible compounds. Then I looked for any nitriles in my IR and since my IR didn’t contain any nitriles, I was able to narrow down my choice to five possible compounds. Then, I looked to see if my IR contained a ketone, an aldehyde, or an ester. My IR spectra didn’t contain a ketone or an aldehyde, however, it did contain an ester, which is how I was able to identify my unknown, IR-12.

There are four main regions of IR absorptions: region 4000 – 3000 cm-1 corresponds to N-H, C-H and O-H stretching, region 2250- 2100 cm-1 is triple-bond stretching , region 2000- 1500 cm-1 is double bonds and the region below 1500 cm-1 is the fingerprint region where a variety of single bonds are absorbed.3 The chromic acid test is a test for oxidizability and gives a positive result for primary and secondary alcohols as well as aldehydes2. A positive result in the chromic acid test is indicated by a color change and the formation of a precipitate. Tertiary alcohols give negative results for the chromic acid test since there must be a hydrogen present on the alcoholic carbon for oxidation to occur. The 2,4 DNP test, tests for a carbonyl and is therefore a dependable test for aldehydes and ketones. Finally, 13C NMR spectroscopy is a test to determine the structure of a compound. 13C NMR detects the 13C isotope of carbon. Each carbon has a different chemical shift. A carbon’s chemical shift is affected by the electronegativity of nearby atoms. Carbons that are bonded to highly electronegative atoms resonant downfield because the electronegative atom pulls electrons away from the nearby carbons and cause those carbons to resonant downfield1 (John McMurry, 2008). A general trend is that sp3-hybridized carbons absorb from 0 to 90 ppm, sp2-hybridized carbons resonant between 110

Some of the aromatic protons were hard to distinguish due to their overlapping nature and resonance. At the same time, there was still enough evidence to verify that the unknown aldehyde was p-anisaldehyde using the number of signals, the chemical shifts, the proton integration, and the splitting patterns. The expected number of signals was roughly 5 peaks, however there were some impurities present in the NMR spectrum. The aromatic region contained the peaks of Hb and Hc with a chemical shift and an integration value of 7.570 and 6.931 ppm respectfully. With that, the protons Hd and He were very distinguishable from the rest of the spectrum due to their large J-value of 16.

I am Kristal Lopez in your Chemistry 120B– lab section 8 on Tuesdays and Thursdays from 11am to 1:50pm. I am writing to you because yesterday, Thursday September 22, in lab I forgot to get my lab notebook sheet signed and turn in the carbon copy. I was wondering if there was anything I could do about it? Thank you for your time and I hope you have a nice day.

Throughout this project, there were a few things that stood out to me, which created obstacles, making it difficult for my group to remediate our pond. One of the major things that affected how much Zn2+ was extracted from our pond was the precipitant we chose. Even though our precipitant, PO43- , would create the most precipitation with the Zn2+, my group and I did not realize that an excess amount of this precipitant would result in some of the Zn2+ going back into solution. Based on the results in our experiment, when 0.1771g of Na3PO4 was added, centrifuged, filtered, and placed in the atomic absorbance spectrometer (AAS), we were able to determine that 0.846 ppm of Zn2+ remained in the pond, which is way below the maximum contamination

The overall experiment wasn’t difficult but just needed to be preformed carefully. We needed accurate information to get accurate results and answers. Without properly collecting and testing the gas we would have never known the types of gases and liquids that were being released into the environment when the reaction took place. We would have also not been able to know the amount of products needed in order to make this experiment happen since finding a complete chemical formula would have been difficult without having an understanding of what “Zinc Carbonate” really meant. Because of the experiments and the well collected data, we were successfully able to conclude that the overall chemical formula for this experiment was 2ZnCO3 *3Zn(OH)2

The observed increased zinc resistance in these laboratory-evolved strains is affirmative to the finding that prior exposure to toxic zinc concentrations could render a selection advantage in the presence of even more toxic concentrations as observed with the 24mM zinc (18). Besides, prior exposure to stress has been shown not only to offer a selection advantage to a once-experienced stress factor but to other stressors as well (62), as several mechanisms are upregulated altering the general metabolism of an organism with possible mutations ensuing. This was observed with the laboratory-evolved strains as they also displayed increased resistance when exposed to high cadmium concentrations (18). Several studies have also reported an association

The results from the NMR of 1-propanol showed 3 different prominent peaks with the peak at 2.2 cm-1 being the acetone. Because 1-bromopropane has three non-equivalent hydrogens it was found to represent this set of NMR data. The other product, 2-bromopropane only had 2 different types of hydrogens and would have only had 2 peaks. Further analysis of the structure of 1-bromopropane showed that the hydrogens closest the bromine group were an indication of peak A in the graph. Because of the electronegativity of the bromine, this peak was located further downfield. There were 2 neighboring hydrogens so using the n+1 rule gave the 3 peaks. Going down peak B showed the next carbon which had 5 neighboring hydrogens thus giving 6 peaks. Finally, the carbon furthest away from the bromine was found at peak C. It had 2 neighboring hydrogens and provided 3 peaks.

The purpose of this experiment was to identify one ketone with Thin Layer Chromatography and one using NMR spectrometry. We will do this by making 2, 4 a DNPH derivative and checking the melting points.