Lecture16x hetcor

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Unformatted text preview: CHEM / BCMB 4190/6190/8189 Introductory NMR Lecture 16 supplement 1 Lets consider the two-spin AX system (13CHCl3) with A=1H = sensitive nuclei and X=13C = insensitive nuclei A) At equilibrium: N4 = N N3 = N + C N2 = N + H N1 = N + C + H N4 X1 A1 N3 A2 N2 N2 - N4 N1 - N3 = H N3 - N4 N1 - N2 = C H=4* C X2 For 13C spectrum: X1 transition: N3 – N4 = C X2 transition: N1 – N2 = C X1 X2 2 N1 A) After a selective 180° pulse exciting the A2 transition: The populations of N1 and N3 are inverted: before 180 A2 pulse N4 = N3 = N2 = N1 = N N+ C N+ H N+ C+ H N N+ C+ H N+ H N+ C After 180° A2 pulse: X1 transition: N3 – N4 = C + H = 5 C X2 transition: N1 – N2 = C - H = -3 C M HC z N4 X1 after 180 A2 pulse M HC z A1 N3 A2 N2 N1 180°x X1 X2 X2 M HC 180°x y y on A2 M HC x x z M CH z M CH M CH 180°x 90°x y y on A2 x x on 13C M CH After selective inversion of the A1 or A2 transition, the signal amplification factors for the spectra of X are given by: 1 + A / X and 1 - A / X 3 B) How about if we “selectively” apply a 180° pulse to both the A1 and A2 transitions: this is essentially, therefore, just a 180 pulse on 1H The populations of N1 and N3, and N2 and N4 are inverted: before 180° pulse N4 = N N3 = N + C N2 = N + H N1 = N + C + H N4 180°x X1 A1 after 180° pulse N+ H N+ C+ H N N+ C N3 A2 N2 X1 transition: N3 – N4 = C X2 transition: N1 – N2 = C X2 X1 180°x N1 X2 As you would expect, a non-selective 180° 1H pulse does not change the sensitivity of the 13C spectrum: 1...
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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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