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# bcmb8190_problemSet4-answers - 1 Ix 2 Iy 1 Iy 2 Iz 1 Iz 2...

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BCMB/CHEM 8190 ANSWERS TO PROBLEM SET 4 1)

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2) The separation between outer lines will give the coupling constant for an AB system (this is not true for all splittings in more complex systems – ie ABC or ABX). Here 5.32-5.35 = 0.03ppm. At 500 MHz this is 500X0.03 = 15Hz. The intensities would be expected to be lower on the two outer lines – perhaps 0.6, 1.4, 1.4, 0.6. 3) μ 1 •μ 2 is proportional to Ix
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Unformatted text preview: 1 Ix 2 + Iy 1 Iy 2 + Iz 1 Iz 2 . Ix = (1/ √ 2)(I + + I-) Iy = (i/ √ 2)(I +- I-). Hence Ix 1 Ix 2 = ½ (I +1 I +2 + I-1 I-2 + I +1 I-2 + I-1 I +2 ) and Iy 1 Iy 2 = ½ (-I +1 I +2- I-1 I-2 + I +1 I-2 + I-1 I +2 ). ( μ 1 • r)( μ 2 • r) will contain in addition to terms like the above, terms proportional to Ix 1 Iy 2 etc. These will result in finite I +1 I +2 and I-1 I-2 contributions....
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bcmb8190_problemSet4-answers - 1 Ix 2 Iy 1 Iy 2 Iz 1 Iz 2...

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