Spectroscopy, Infrared Spectroscopy And Spectroscopy

The purpose of this experiment is to determine the maximum absorbance of fast green, and the chlorophylls, also in the case of fast green create a concentration curve to determine an unknown substance. Each test will use the spectrophotometer.

- The C-H bonds in this structure are shown at 1444 and 1368cm-1. These two bands indicate the two different types of C-H bends that occur on the molecule. One is that of the alkene and the other is that of the several alkanes on the molecule.

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

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

IR and 1H NMR analysis was also performed for the starting ketone, 4-tert¬-butylcyclohaxanone. The IR spectra, RM-07-Oi, contained peaks at 2947 cm-1, 2868 cm-1, 1720 cm-1, and 1365 cm-1. These peaks represent a carbon-hydrogen stretch, a sp3 hybridized carbon, a ketone and a ring carbon-carbon stretch, which are all present in 4-tert¬-butylcyclohaxanone. This confirms that the initial ketone used was 4-tert¬-butylcyclohaxanone. The 1H NMR also proves this. Peaks found at 2.34 ppm, 2.07 ppm, 1.45 ppm and 0.87 ppm show that only four chemical shifts occur in this compound. There are not four types of hydrogens in this molecule, but some chemical environments are similar resulting in only four

Coefficients show the correct proportions of atoms and molecules in a chemical reaction. They are normal sized numbers placed at the beginning of the chemical formulas in a chemical reaction during the process of balancing. They tell how many of an entire chemical formula is in a reaction.

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.

The signals labeled as “A” at 2.01ppm should be a singlet since it is next the ester group and has no neighbor. Two overlapping signals labeled as “D&E” have 3 and 5 neighbors, respectively. The signals are multiplet overlapping each other between 1.6 to 1.30 ppm. A doublet was seen in the spectrum at 1.18ppm labeled as “C” since the protons are on the carbon that attached to the tertiary carbon, which locates next to the ester group labeled as “B” and has one proton. Lastly, a triplet labeled “F” was seen at 0.90ppm, which is the most shielded since it locates the furthest from the electronegative ester group.

The H-NMR of the unknown alcohol was provided for analysis in order to determine the identity. Based on the peaks and their respective locations, the unknown was determined to be isopentanol. The peak at 0 ppm was the TMS. From this, the first doublet peak represented the six hydrogens located on the two terminal carbons, the signal was split as a result of the single hydrogen from the adjacent carbon. This peaks falls within the typical range for a primary alkyl hydrogen, 0.7-1.3 ppm. The next peak is a quartet that has a

The author of the article that appeared in a business magazine suggest that the numerous complaints of dizziness and nausea made by consumers of Promofoods tuna are not related to any chemicals that are found in the canned tuna and the same did not pose a health risk. However, the conclusion made by the author is based on numerous assumptions without any proof to support them. Before the author can make such a conclusion he needs to provide cogent unbiased proofs to bolster his claim.

-If possible, I suggest the authors to label the 6bʺ and 6aʺ protons of compound 1 in figure 1.

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.

Chemistry is the study of matter and energy and the interaction between them. Chemistry sometimes referred to as the central science because it connects other sciences to each other, such as biology, astronomy, medicine, physics, geology and environmental science. Chemistry is everywhere in the world around us; It 's in the food we eat, water we drink, medicine we take, clothes we wear, air, even the body is made of chemicals. Chemical reactions occur when we breathe, eat, run, or lay down. We are all chemists as we use chemistry every day and perform chemical reactions without knowing it or thinking about it.

Chemistry has been called the science of what things are. Its intent is the exploration of the nature of the materials that fabricate our physical environment, why they hold the different properties that depict them, how their atomic structure may be fathomed, and how they may be manipulated and changed.

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.