Chemistry 125: Lecture 60
March 23, 2011
NMR Spectroscopy
This
Chemical Shift and
Diamagnetic Anisotropy,
Spin-Spin Coupling
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see final page of this
file
Components of
Effective Magnetic Field.
Bmolecular (diamagnetic)
Applied Field
Bapplied
Beffective
Molecular Field:
Net electron orbiting - “Chemical Shift”
(Range ~12 ppm for 1H, ~ 200 ppm for 13C)
Nearby magnetic nuclei - “Spin-Spin Splitting”
(In solution JHH 0-30 Hz ; JCH 0-250 Hz)
The Chemical Shift:
Electron Orbiting and
Diamagnetic Anisotropy
high
electron
density
Chemical Shift and Shielding
Note: Electron orbiting to give B is driven by B; so B  B.
d+
Bmolecular (diamagnetic)
R-OH
(depends on conc, T)
O
RC
O
RC
OH
Bapplied
Beffective
11
H
H
10
9
deshielded
downfield
low e- density
high chemical shift
high frequency
8
7
CH3C
C C
H
6
5
d (ppm)
C-H
d-
TMS
X = O, Hal, N
O
RC
CH Alkyl
X CH
R-H
4
3
2
1
! ???
shielded
upfield
high e- density
low chemical shift
low frequency
0

Bapplied
Suppose the
studied nucleus is fixed
relative to the other
nucleus
by bond(s).
PPM
Suppose molecule
in fluid undergoes
rotational averaging.
1/r3
Electrons
Orbiting
Other
Nuclei
net from
average
around
over
sphere
circle
ZERO!
Ignore
Diamagnetism
electrons
other
from on
Orbiting
atoms!
Electrons
NOT
Bapplied
Electrons
Orbiting
Other
Nuclei
reinforces
Bapplied
suppose less orbiting
for this molecular
orientation
Diamagnetic
“Anisotropy”
net from
average
over
sphere
ZERO!
Unless orbiting
depends on
molecular
orientation
Diamagnetic Anisotropy
Benzene “Ring Current”
B0 can only drive circulation about a
path to which it is perpendicular.
B0
Net deshielding
of aromatic
protons;
shifted downfield
If the ring rotates so
that it is no longer perpendicular
to B0,
the ring current stops.
Aromaticity: PMR Chemical Shift Criterion
14  electrons
(43) + 2
?
TMS
DIAMAGNETIC
ANISOTROPY 10  electrons
TMS
(distorted – less overlap & ring current)
DIAMAGNETIC
ANISOTROPY!
HCCl3
9
8
7
8H
d -4.23
6
5
4
3
2
d (ppm)
1
0
2H
-1
-2
-3
-4
Boekelheide (1969)
Aromaticity: PMR Chemical Shift Criterion
“Anti-Aromatic” Dianion
Metallic K adds
-2
CH3 signals shift downfield
by 26 ppm despite addition
of “shielding” electrons.
2  electrons
14  electrons
to give 16
(43) + 2
(4n)
TMS
DIAMAGNETIC
ANISOTROPY
HCCl3
922
820
718
DIAMAGNETIC
THF ANISOTROPY!
solvent
Shrink Scale
616
514
412
d -4.23
310
28
d (ppm)
16
04
-12
-20
-3-2
-4-4
Boekelheide (1969)
Diamagnetic Anisotropy
Acetylene “Ring Current”
H
H
H
The H nuclei of benzene
lie beside the orbital path
when there is ring current.
(B0 at H reinforced;
signal shifts downfield).
H
H
H
The H nuclei of acetylene
lie above the orbiting path
when there is ring current.
(B0 at H diminshed;
signal shifts upfield).
Warning!
This handy picture of diamagnetic
anisotropy due to ring current
may well be nonsense!
(Prof. Wiberg showed it /
/to be nonsense for 13C.)
Spin-Spin Splitting
Chem 220
NMR Problem 1
O
(of 40)
H
C.2CH3
OCH
Four (22) sets of H
CH3C
Triplet
(1:2:1)
molecules that
differ in spins of
adjacent H nuclei
“Spin Isomers”
so similar in energy
that equilibrium
keeps them
equally abundant
8
7
6
5
4
3
d (ppm)
2
1
0
Chem 220
NMR Problem 1
(of 40)
Influence of CH2 on CH3 must be
O
the same as that of CH3 on CH2
H
CH3C
and independent of Bo
C.3H
OCH2CH
Triplet
(23)
Eight
sets of
molecules that
differ in spins of
adjacent H nuclei
1:
1 1
2:
1 2 1
3: 1 3 3 1
4: 1 4 6 4 1
(1:2:1)
H
J in Hz
Quartet
7.3
(1:3:3:1)
7.3
7.3
7.3
7.3
binomial
coefficients
8
7
6
5
vs. Chemical Shift in
4
3
d (ppm)
2
1
0
(Orbiting driven by Bo)
DMSO-d5
Subtle
Asymmetry
5.1 Hz
7.2 Hz
HO-CH2-CH3
Doublet of Quartets
7.2
Dd 0.018 ppm
× 400 MHz
J = 7.2 Hz
1.052
d
1.070
5.1
124 Hz
1:4:6:4:1
Quintet?
D is a weaker magnet than
H.
13CH
7.2be oriented 3 ways in Bo.3
D can
O 1.8 Hz
1.1% of C
CD3SCD2H
?H2O
1:2:3:2:1
Quintet
?
d (ppm)
?
What determines
the Strength of
Spin-Spin Splitting?
Isotropic JH-H
is mediated by
bonding electrons
(the anisotropic through-space part
is averaged to zero by tumbling)
In tumbling molecules, nuclear spins
communicate not through space, but
through paired electrons on the nuclei.
When the “up” electron of
this MO is on Nucleus A
J depends on the s-orbital
content of molecular orbitals.
HOMO-3
only its “down” electron is
available to be on Nucleus B
J = 1-3 Hz
J = 6-8 Hz
J = 0-1 Hz
Might overlap be greater for anti C-H bonds ??
2.38 Å
1.85 Å
J = 0-3 Hz
J = 6-12 Hz
J = 12-18 Hz
Not spatial proximity!
Through-space interaction of dipoles averages to zero on tumbling.
Which gives better overlap?
Examine the overlap
of the components.
+
Backside overlap
is counterintuitive.

good p-p
+
+
2 bad s-p
good s-s
Better
bad p -p
Overlap! good s-p ; good p -s



s-p > s-s or p-p
(See Lecture 12)

bad s-s

+
1.0
Overlap Integral
0.8
0.6
C
C
C
C
C
s-p
0.4
s-s
0.2
p-p
0.0
1.2
1.3
1.4
C
Overlap
1.5 Å
C
invisible
10No “handle” for rf if same chem shift
(see Frame 26 below)
H
132 Hz
11
Hz
H
gauche ~7 Hz
2-13 Hz, depends on conformation (overlap)
(approximate way to measure a rigid torsional angle!)
End of Lecture 60
March 23, 2011
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J. M. McBride, Chem 125. License: Creative Commons BY-NC-SA 3.0