Chem Experiment #6

Chem Experiment #6 - 1 H NMR spectrum that shows four...

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5223 November 5, 2009 Thursday pm Unknown Sample: C 6 H 7 N Elements of Unsaturation: 3 The unknown organic sample with a molecular formula of C 6 H 7 N showed four unique signals on the 13 C NMR spectrum which indicates that symmetry is present in the molecule. All of the four signals were downfield between 110 and 150 ppm. The degrees of unsaturation along with the placement of the four signals point toward the structure being a monosubstitute aromatic compound. All six carbons can be accounted for in the aromatic structure, indicated by the four downfield signals. A ring with nitrogen bonded within could not produce as few as four signals because of the greater amount of unique carbons. This is consistent with an amine bonded to an aromatic compound. The furthest downfield signal signifies the carbon bonded to the amine functional group since it is deshielded due to the nitrogen’s electronegativity. This structure is consistent with the
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Unformatted text preview: 1 H NMR spectrum that shows four signals. The duplet shown between 6.5 and 7.2 represents the hydrogen bonded to carbons in the aromatic, while the signal upfield constitutes the two hydrogen of the amine. ChemDraw Simulated 1 H NMR spectrum ChemDraw Simulated 13 C NMR spectrum The data shown in the above simulated ChemDraw spectrums matches the spectrum shown in the experimental results. The experimental 1 H NMR spectrum has minor deviations comprised of some small signals upfield of the amine group. These signals are most likely due to contamination either from the vile in which the solution was mixed or in the deuterochloroform. In summary, only Aniline is consistent with the duplet of hydrogen and the upfield hydrogen signal. Aniline possesses the correct amount of degrees of unsaturation and correct chemical shifts for the 13 C NMR signals....
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This note was uploaded on 11/13/2009 for the course CHEM 210 taught by Professor Slough during the Fall '08 term at Kalamazoo.

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