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Unformatted text preview: CHEM / BCMB 4190/6190/8189 Introductory NMR Lecture 16 supplement 1 Lets consider the twospin 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 nonselective 180° 1H pulse does not change the
sensitivity of the 13C spectrum:
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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.
 Fall '08
 Staff
 Equilibrium

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