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exam2-2009 - Name Exam 2 CHEM/BCMB 4190/6190/8189(119...

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Name _________________________ Exam 2, 2009: CHEM/BCMB 4190/6190/8189 1 Exam 2: CHEM/BCMB 4190/6190/8189 (119 points) Tuesday, 6 October, 2009 1 ). In the example (right), the effect of a 90 ° ( ± /2) pulse applied along the “ x ” axis (90 ° x ) is shown for a bulk magnetization vector ( M 0 ) at equilibrium (on the ‘ z ’ axis). For ‘b’, ‘c’, ‘d’ and ‘e’ below, show the effects of the indicated pulses by drawing the missing (originating or resulting) vectors on the coordinate axes. For ‘a’ and ‘f’, fill in the blank with the correct pulse (angle and axis along which it is applied) that will promote the indicated movement of the bulk magnetization vector. (12 points) a. d. b. e. c. f. 90 ° x M 0 y z x y z x 90 ° - x y z x y z x 90 ° - y y z x y z x 315° y y z x y z x y z x y z x 270 ° - x , 180 y y z x y z x ± /4 y y z x y z x 180 ° y this vector is in the x-z plane this vector is in the x-z plane
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Name _________________________ Exam 2, 2009: CHEM/BCMB 4190/6190/8189 2 2 ). In the [PF 6 ] - ion, the fluorides are octahedrally arranged and they are chemically equivalent. a. Please explain completely what is meant by chemical equivalence. ( 4 points ) In NMR, nuclei are chemically equivalent if they have the same chemical shift and if this does not occur because of coincidence. A symmetry operation that relates the nuclei or effective symmetry imposed by rotation (for instance, the hydrogens of a methy group) is required. b. Why would an octahedral arrangement would result in chemical equivalence. ( 4 points ) The octahedral arrangement imposes symmetry on the fluorides such that each is chemically equivalent to the others. c. Sketch the simple, one-dimensional 31 P and 19 F NMR spectra of the [PF 6 ] - ion. You will have to justify/explain all aspects of your spectra . Assume a magnetic field strength ( B 0 ) of 400 MHz (9.4 T), a 31 P chemical shift of -140 ppm, a 19 F chemical shift of -72 ppm, and a 31 P- 19 F coupling constant of 800 Hz. ( 8 points ) Both 19 F and 31 P are spin ± nuclei, and are the essentially the only isotopes present at natural isotopic abundance. The six 19 F nuclei are symmetrically disposed about the 31 P nucleus, and are chemically equivalent. Thus, the single signal from the 19 F nuclei will be split into a doublet (relative intensities 1:1) by the directly-bonded 31 P nucleus, and the signal from the 31 P nucleus will be split into a septet (relative intensities 1:6:15:20:15:6:1) by the six equivalent 19 F nuclei. The Larmor frequencies are directly proportional to the gyromagnetic ratios ( ± L = | ² /(2 ³ )| B 0 ), and are 25.1815*400/26.7519=376.519 MHz and 10.8394*400/26.7519=162.073 MHz for 19 F and 31 P, respectively. An 800 Hz coupling constant at 376.519 MHz is 2.125 ppm, and at 162.073 MHz is 4.936 ppm. d. Compare the 31 P spectrum of the [PF 6 ] - ion acquired with the pulse sequence shown (right) to the spectrum described above in part ‘b’. Assume that an equal number of scans were acquired for each spectrum and that the delay is properly set. You will have to clearly define the changes to the spectrum and why they occur. ( 4 points ) This pulse sequence results in broadband decoupling of the 19 F nuclei from the 31 P nuclei.
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