bcmb8190_problemSet1-answers

bcmb8190_problemSet1-answers - The single unpaired proton...

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*BCMB/CHEM 8190* *ANSWERS TO PROBLEM SET 1* 1) Receptivities given in the table are reduced by abundances (except 3H). Divide by abundance to get receptivity for equal number of nuclei. The dependence on gamma can be seen by equating the ratio of the gammas raised to some power, n, to the receptivity of an equal number of nuclei and finding n. For 13C, ( γ C/ γ H)^n = 1.6e-2 or n*ln( γ C/ γ H) = ln(1.6e-2) or n = ln(1.6e-2)/ln(.25) or n = 3.0. Receptivity also is proportional to 1/T. Therefore at 4 versus 300K the relative receptivity is 75. 2) Li-7 has 3 protons and 4 neutrons. The neutrons will be paired and will not contribute to the spin properties. The first 2 protons go into the 1s 1/2 level and are paired - they don't contribute.
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Unformatted text preview: The single unpaired proton goes into a 1p 3/2 level. Here the proton spin has added to the ordital spin. Hence gamma has the same sign as that of the proton - positive S=3/2 N-15 has 7 protons, 8 neutrons. Again all neutrons are paired so we only consider protons. The single unpaired proton ends up in the 1p level. Here the proton spin has subtracted from the orbital part, so the moment is opposite in sign from the proton, Hence, gamma is negative, S=1/2. In both cases predictions agree with experiment. 3) The difference in precession frequency is (2.6752x10^8 x 11.7 x (60-55.2) x 10^-6)/2pi, or 2391Hz. The methyl resonance is 4.8 ppm upfiled of the OH resonance....
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