exam2-2011 - Name _ Exam 2: CHEM/BCMB 4190/6190/8189 (114...

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Name _________________________ Exam 2, 2011: CHEM/BCMB 4190/6190/8189 1 Exam 2: CHEM/BCMB 4190/6190/8189 (114 points) Tuesday, 4 October, 2011 1 ). During a 13 C NMR experiment, often broadband decoupling of 1 H nuclei is implemented (see diagram, right). a. Please list and explain the advantages and disadvantages of performing broadband 1 H decoupling, as diagrammed above, during a simple 13 C experiment. ( 6 points ) Broadband 1 H decoupling in 13 C experiments produces a significant increase in signal- to-noise due both to collapse of multiplets into singlets, and due to the nuclear Overhauser effect (NOE). The removal of splittings also decreases signal overlap often times allowing the number of carbon signals to be unambiguously determined. The disadvantages are nominal. One possible disadvantage is that the spectra no longer include information about carbon type (i.e., methyl group, methene, methine), and the integrals of the signals no longer are reliable indicators of the number of nuclei giving rise to the signal, because the sensitivity enhancement by the NOE is not the same for all carbon types. 90 ° x 13 C 1 H d 1
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Name _________________________ Exam 2, 2011: CHEM/BCMB 4190/6190/8189 2 2 ). When the cyclohexanone shown (right) is dissolved in methanol, the three-bond scalar coupling constant between H A and H B ( 3 J HAHB ) is 11 Hz. However, when it is dissolved in benzene, this coupling is 3 Hz. Please explain in detail why the couplings are different and the origins of the difference. For credit, your answer must include the word “Karplus” used in a contextually correct way. ( 6 points ) These observations suggest that the conformation of the molecule in methanol is different than the conformation in benzene. For a three bond 1 H- 1 H coupling, the Karplus relationship suggests that the dihedral angle between H A and H B is near 0 ° or 180 for the molecule in methanol, but that this angle is perhaps 60 in benzene. This cyclohexanone most likely exists in one of two stable chair conformations, in which case the hydrogens H A and H B can both be axial (dihedral angle ~180 ), both equatorial (dihedral angle ~60 ), or one axial and the other equatorial (dihedral angle also ~180 ). Structure ‘a’ (below) shows the most likely conformation in methanol (dihedral angle ~180 , 3 J HAHB ~11 Hz), whereas structure ‘b’ (below) shows the most likely conformation in benzene (dihedral angle ~0 , 3 J HAHB ~3 Hz). Incidentally, whereas the hydroxyl group is hydrogen bonded to the solvent in methanol, and the isopropyl group is in the more stable equatorial position, in benzene, where hydroxyl group hydrogen bonding with solvent is not possible, intramolecular hydrogen bonding of the hydroxyl with the carbonyl oxygen stabilizes this conformation greatly and is the reason this conformation predominates in benzene.
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Name _________________________ Exam 2, 2011: CHEM/BCMB 4190/6190/8189 3 3 ). A 1 H pulse applied at the carrier frequency excites a bandwidth equal to + 1/ Δ τ p from the carrier frequency, where Δ p is the pulse width. One can estimate that at + 1/(2 Δ p ) half maximal excitation will occur, meaning that a 90 ° pulse applied on resonance (i.e. the carrier frequency) will produce only a 45
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This note was uploaded on 11/13/2011 for the course CHEM 4190 taught by Professor Staff during the Fall '08 term at University of Georgia Athens.

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exam2-2011 - Name _ Exam 2: CHEM/BCMB 4190/6190/8189 (114...

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