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Inorg Chem Lab

by: Aurore Armstrong

Inorg Chem Lab CHEM 4100

Aurore Armstrong
GPA 3.55


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This 3 page Class Notes was uploaded by Aurore Armstrong on Wednesday October 28, 2015. The Class Notes belongs to CHEM 4100 at University of Wyoming taught by Staff in Fall. Since its upload, it has received 8 views. For similar materials see /class/230374/chem-4100-university-of-wyoming in Chemistry at University of Wyoming.


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Date Created: 10/28/15
Nuclear Maunetic quot NMR Many nuclei not all possess a property called spin angular momentum I Examples 1H 130 P These nuclei have either 2 or 2 spin at a given time In the absence of an applied magnetic field the two levels are degenerate In an applied magnetic field the two spin states are not identical in energy The spin of the nuclei can align or oppose with the applied field Therefore the two spin states become different in energy Depending on the strength of the applied field the two spin states are separated in energy by different amount and the energy gap fall in the radio frequency range AE on Hl hv AE the energy difference between the spin states H the strength ofthe applied magnetic field I the nuclear spin quantum number 2 for 1H 130 P and 1 for 2H 12 spin state opposed to H Both 12 and 12 spin states are degenerate in the absence of an applied magnetic field This energy gap falls in the radio frequency region 12 spin state aligned with H Applied Magnetic Field H Hydrogen and phosphorous occur in nature almost exclusively gt99 as 1H and P respectively and therefore are very sensitive to detection by NMR Whereas nuclei such as carbon do not possess spin angular momentum in their naturally occurring isotope Only 11 of natural carbon possesses spin angular momentum Therefore observing a carbon NMR spectrum requires a significantly large data acquisition Liquid or solution samples can be prepared for NMR measurement Usually Pyrex tubes of 5 mm outer diameter are used for the measurement When a Pyrex tube containing a sample is placed in between the magnets in an NMR instrument the solution at the center of the tube fees slightly smaller magnetic field than the sample close to the walls of the sample tube This would result in a broad signal in the spectrum Therefore liquid and solution samples are spun 20 rotations per second An NMR spectrum is a plot of frequencyfield versus observed intensity 10 9 8 7 6 5 4 3 2 1 0 ppm High Frequency Low Frequency Downfield Up field 1 ppm 400 Hz Chemical Shift Most organic compounds contain protons that are not chemically similar Let us consider ethylbromide CHsCHzBr The methyl and methylene protons are chemically dissimilar Therefore the two different kinds of protons resonate at two different field strengths and absorb the applied radio frequency light at different field strengths For the magnetic field to be felt by the nuclei of the methylene protons first the magnetic field must penetrate the electron cloud of the protons The methyl protons have a denser electron cloud than that of the methylene protons because the electronegative bromide pulls some of the electron cloud around the methylene protons away from the protons No such effect is possible with the methyl protons Therefore the methyl protons are more protected or more shielded by their electron clouds in comparison to the methylene protons which are less shielded or deshieded by the bromide Therefore the methyl and methylene protons absorb the applied radio frequency at different fields Therefore two signals or peaks are observed for ethylbromide Integration The peak observed at low frequency upfield is due to the methyl protons and the other is due to the methylene protons The intensities of the two peaks are not identical they are at the ratio of 32 The intensity ratio corresponds to the number of each type of protons S inspin coupling The two signals are split into multiplets in the spectrum One of them is a triplet and the other is a quartet The splitting of the signals is caused by interaction between the spins of neighboring atoms Let us first consider the three methyl CH3 protons All the three are magnetically and chemically identical and therefore the applied magnetic field cannot distinguish them at all Therefore the observed splitting is not due to the multiplicity of protons Actually the protons interact with the magnetic field caused by the neighboring CH2 protons This interaction causes the splitting The triplet can be explained as follows The two CH2 protons can orient in four different ways 2 2 1 2 z 0 z 2 0 z z 1 There are three net spin states namely 1 0 and 1 The state with 0 spin occurs twice Therefore the intensity pattern of the triplet state is 1 21 Now let us consider the CH2 protons The two are magnetically and chemically equivalent They interact with the neighboring three methyl protons The interaction causes splitting of their signal The three CH3 protons can orient in eight different ways as follows 2 2 2 1 12 2 2 2 2 2 2 2 2 2 2 z z 2 z 2 z z 2 There are four net spin states namely 1 2 z 2 and 1 2 States 2 and 2 occur three times Therefore the intensity pattern ofthe quartet is 1331 Expt 7 Metalloenzymes containing metals such as vanadium manganese and copper are involved in various oxidation reactions in nature Taking these enzymes as models chemists have developed a number of coordination complexes These complexes are capable of acting as catalysts in oxidation reactions Eg i epoxidation of olefins ii oxidation of benzylic methylene groups iii anilines to nitrobenzenes iv oxidation of sufides to sulfoxides A special behavior of the enzymatic oxidation is enantioselectivity The Mn3 complexes a series of sterically hindered ligands prepared by Jacobsen mimic the enantioselective oxidation properties of the enzymes In this experiment we will synthesize two metal complexes of a Jacobsen ligand We will use the Schiff base formed from RR 12 cyclohexanediamine and 35 di tert butylsalicylaldyhyde as the ligand The chiral diamine is obtained from the commercially available mixture of isomers H H NH2 NH3 H3N water 2 2 dtartaric acid gt dtartz dtartz AcOH NHz NH3 H3N H H mixture RR SS less soluble freer soluble The aldehyde is synthesized from the commercially available 24 di tert butylphenol OH tBu I tBu 1 C6H12N4 90 C 2 Aq sulfuric acid 33 heat at 100 C tBu tBu The Schiff base ligand is synthesized by the treatment of the RR diammonium tartrate with the aldehyde in dimethylformamide tBu O OH tBu H t NH Bu OH 2 2 DMF r gt d tart dtartaric acid 5 NH H20 OH tBu tBu tBu RRLigand HZL


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