1
Chapter 4, The Coupling Constant
Chapter 4: The Coupling Constant
I. First- and Second-order Spectra
A. for 1st order v/ J > 10
1. AX 1st order, AB 2nd order
a) multiplicities based on n+1 rule for I = ½
(1)
(2nI + 1)
1
2
Chapter 4, The Coupling Constant
b) intensity follows binomial expansion – Pascal’s triangle
c) nuclei with same chemical shift do not split each other,
even when the coupling constant between them is nonzero
d) spacings between adjacent components of spin-spin
multiplet are equal to J
e) spin-spin mulitplets are centered on the resonance
frequency
2. 2nd Order effects:
a) deviation of intensity from binomial pattern
b) observation of spacings not corresponding to coupling
constants
c) appearance of additional peaks
3. Pople notation AX Vs AB
II. Chemical and Magnetic Equivalence
A. Also for 1st order, all chemically equivalent nuclei
be magnetically equivalent
must
1. Any pair of chemically equivalent nuclei must have the same coupling
constant to any other nucleus or they are magnetically nonequivalent
2. Chemically equivalent if they can be interchanged by a
symmetry operation
3. A spin system consists of sets of nuclei that interact
(spin couple) among each other but do not interact
with any nuclei outside the spin system
a) spin systems are "insulated" from one another
b) A set of nuclei consists of chemical shift
equivalent nuclei
B. symmetry operations
2
3
Chapter 4, The Coupling Constant
1. rotation about a symmetry axis (Cn)
3
2. inversion at a center of symmetry (i)
3. reflection at a plane of symmetry (sigma)
4. higher orders of rotation about an axis followed
by reflection in a plane normal to this axis
5. symmetry element (axis, center or plane) must be
symmetry element for the entire molecule
(Sn)
a
C. Types of nuclei
1. homotopic - interchangeable through an axis of
symmetry
a) homotopic - interchanged by rotational symmetry
b) chemical shift equivalent in any environment
c) 1,1 difluoroethene, 180o, n-fold axis of symmetry
d) difluoromethane, 180o
2. enantiotopic - related by other symmetry elements such
plane (if there is not also a rotational symmetry)
as a
a) chemical shift equivalent only in achiral medium
b) bromo-chloromethane
c) replacement of 1 H -> enantiomers
d) homotopic produces identical molecules
e) enantiotopic nuclei have identical chemical
properties except in chiral environments
(solvents, enzymes)
f) chemically equivalent - same letter A2X2
g) enantiotopic or homotopic H need not be on the same C
atom
4
Chapter 4, The Coupling Constant
(1) the 2 vinyl H on cyclopropene are
homotopic, A2X2
4
(2) the 2 vinyl protons on 3-methylcyclopropene
are enantiotopic, AX2
3. diastereotopic - noninterchangeable geminal protons
a) not chemically equivalent in any environment
4. heterotopic - noninterchangeable protons on different
carbon atoms
a) not chemically equivalent in any environment
D. Chemically equivalent are magnetically equivalent if they
are
equally coupled to any other given nucleus in the
molecule outside
the set
1. difluoromethane, cyclopropene
2. if not equally coupled - magnetically nonequivalent
a) to determine whether chemical shift equivalent
nuclei
are magnetic equivalent, it must be
determined
whether they are coupled equally to
each nucleus
(probe nucleus) in every other set
in the spin system
(1)
examining geometrical relationships
(2) are bond angles & bond distances from each
nucleus in relation to the probe nucleus
identical
(3) 2 chemical shift equivalent nuclei are
magnetic equivalent if they are symmetrically
disposed with respect to each nucleus (probe) in
any other set in the spin system
(4) a test for magnetic equivalence is valid
only when the 2 nuclei are chemical shift
equivalent
(5)
1,1-difluoroethene AA'XX', JAX & JAX'
5
Chapter 4, The Coupling Constant
5
3. Any spin system containing magnetically nonequivalent nuclei is 2nd
order by definition.
a) remains 2nd order at highest accessible fields because due
to symmetry considerations only.
(1)
o- & p-dichlorobenzene AA’XX’
(2)
butyrolactone AA’XX’
(3)
p-nitrotoluene
(4)
rotamers of 2-chloroethanol
4. Magnetically equivalent must be chemically equivalent
have same chemical shift (be isochronous)
&
a) Ha & Hb on phenyl ethane
(1)
chemically equivalent – enantiotopic
(2)
magnetically equivalent on average
(3)
A2X3
b) Ha & Hb on 1-bromo-1-chloro-2-phenylethane
(1)
each rotamer separate & distinct
(2) in none, are Ha & Hb related by symmetry
element
(3) therefore Ha & Hb have different chemical
shifts & coupling constants
(4)
ABX
(5) diastereotopic - replacement of Ha by D
gives a diastereomer
(6) -CH2- diastereotopic when there is already a
chiral center elsewhere in the molecule
(7) frequently, diastereotopic Hs are
degenerate
6
c)
Chapter 4, The Coupling Constant
alpha-thujene - double resonance for the 2 methyls
d)
1,1-diethoxyethane
(1)
replace Ha on ethoxy by D, get 2 chiral
centers
(2)
therefore even though no chiral C, Ha
& Hb diastereotopic
e)
6
2 Hs (at C3) on cis-1,2-dichlorocyclopropane
diastereotopic
f)
axial & equitorial Hs on single C in ring-frozen cyclohexane
are diastereotopic
g)
Other Examples
(1)
p-fluoronitrobenzene
(a)
AA'BB'X but para couplings essentially zero
(b)
therefore treat as 2 equivalent ABX systems
(2)
meta dichlorobenzene
(a)
AB2C
(3)
ortho dichlorobenzene
(a)
AA'BB'
(4)
1,1-difluoroethylene
(a)
AA'XX'
7
Chapter 4, The Coupling Constant
III.
7
Signs & mechanisms: One Bond Couplings
A.
spin-spin coupling arises because info about nuclear spin
is transferred from nucleus to nucleus by the electron cloud"
B.
Fermi contact mechanism
1. electron in X-Y where both X & Y are magnetic, spends
amount of time at same point in space as
finite
nucleus X
a) electron & nucleus in "contact"
b) if nucleus Iz = +1/2, then electron is -1/2
(1)
therefore polarize the electron
c) electron with -1/2 polarizes second electron with
+1/2
d) this polarizes Y to -1/2
(1) therefore Y has tendency to -1/2 when X
+1/2
e) not averaged to 0 since through the bond
(1)
molecular tumbling not effect
8
Chapter 4, The Coupling Constant
2. When preferred arrangement of 2 coupled spins is
opposite, coupling constant J has a + sign by
convention
8
3. negative coupling occurs when preferred arrangement is
parallel
4. H-C-H
a) 1H – electron
b) bond
c) Hund's rule - an electron in another similar
orbital about Y preferentially has the same
spin
d) bond
e) electron - 1H
5. Dirac vector model - little theoretical basis & many
exceptions but useful mnemonic device
a) coupling over odd # of bonds expected to be +
b) coupling over even # of bonds expected to be –
6. High resolution NMR spectra not dependent on absolute
coupling but relative sign can be important
a) ABX JAB, JAX, JBX
b) different spectra if JAX & JBX have same or
different signs
7. Convention
a) superscript - # of bonds between coupling nuclei
on left
b) subscript - any other descriptive info
(1)
2J
H-C-H
or 2J(H-C-H) geminal
sign of
9
Chapter 4, The Coupling Constant
(2)
3J
H-C-C-H
vicinal
(3)
beyond 3 bonds - long range coupling
9
10
C.
Chapter 4, The Coupling Constant
C coupling dominated by 13C-1H 1 bond coupling
13
1. because p orbital has node at the nucleus. only
electrons in s orbital contribute to Fermi coupling
mechanism
a) 1H - all in s
2. Linear relationship between s character & coupling
constant
a) C13 dominated by J coupling when decoupler turned off
b) methane 125
1
J(13C-1H)
c) ethylene 157
d) benzene 159
e) acetylene 249
f) %s(C-H) = 0.2 J(13C - 1H) linear relationship
g) zero intercept - no coupling with no s character
3. measuring s character via J-s relationship
a) cyclopropane is sp2
b) 4-15 is 29% or sp2.4 for tricyclopropane
c) cubane is 32% or sp2
d) quadricyclane 1-17 is 36% s or sp1.8
e) not work well with polar molecules
4. coupling constant between x-y
a) J(x-y) proportional to gammax * gammay
(1)
product of magnetogyric ratios
10
11
Chapter 4, The Coupling Constant
5. one bond coupling between 2 carbons - INADEQUATE
D.
11
15
N
1. very small gamma & negative
a) smaller coupling constant than 13C
b) one-bond couplings between 1H & 15N have – sign
2. examples
a) ammonium ion -73 Hz for sp3
b) phenyl2C=NH2+ -93 for sp2
c) CH3 C#NH+ -130 sp
3. lone pair of N add contribution to contact term which
is +. Therefore consideration of s character
meaningless
a) phenyl2C=NH -51 but sp2
4. similar effects in P coupling
a) gamma(31P) is +
b) Me2P-PMe2
(1)
1J(31P-31P)
= -180Hz
(2)
results from lone pair contribution
(3)
Me2P(=S)-P(=S)Me2 +19 (no lone pair)
12
Chapter 4, The Coupling Constant
5. When neither nucleus is H, coupling constant determined
by product of s characters of orbitals
12
IV. Factors that influence J: Preliminary Considerations
A. Factors influencing the sign of J
1. magnitude of J tends to decrease as the number of
intervening bonds increase
2. the coupling constant between two nuclei is positive if
the spins of the coupled nuclei prefer to be paired
3. in a single bond, the electrons are paired
a) therefore the two nuclei prefer to be paired
B. Factors influencing the magnitude of J
1. the magnitude of J is directly proportional to the
product of the magnetogyric ratios of the two
coupled nuclei
a) all other things being equal
V. One-bond coupling constants
A. 1JCH
1. all H use s atomic orbital to overlap for molecular
2. C uses different hybrid orbitals
a) spn, s character = 1/(n + 1)
b) 1JCH is directly proportional to the fraction of scharacter
c) s orbital has more overlap with nucleus
d) 1JCH = (500 Hz)(fs)
VI. Geminal Couplings
orbital
13
Chapter 4, The Coupling Constant
A. Macomber
1. two-bond (geminal)
2. much smaller magnitude than one-bond coupling constants
3. often (but not always) negative signs
4. sensitive to magnetogyric ratios, s-character
5. also bond angle
6. range of -9 to -15 Hz as in H-C-H
7. in cyclic compounds, as ring gets smaller, value of 2J
gets more positive
B. H-C-H
1. often chemically equivalent & magnetically equivalent
a) therefore measure by substituting with D
2. Pople & Bothner-By - molecular orbital approach
a) 2JH-C-H
(1)
in CH4 -12.4 Hz
(2)
in ethylene +2.3 Hz
b) measured using deuterated samples
c) inductive withdrawing group
(1)
sigma withdrawal
(2)
2J
(3)
if inductive donating, J less +
more +
d) pi withdrawal (hyperconjugation) 2J less +
(1)
pi donor J more +
13
14
Chapter 4, The Coupling Constant
3. H-C-H on sp2 C small
a) often can't observe for exomethylene (C=CH2)
(1)
or terminal vinyl (CH=CH2)
b) H2C=O 42 Hz (sign is unknown)
(1)
sigma withdrawal - O - 2J more +
(2)
pi donor, lone pair on O, 2J more +
c) H2C=N-tBu
(1)
17, sign unknown
(2)
lower electronegativity of N
(3)
less effective overlap
d) allenes =C=CR2
(1) more distant double bond can accept
electrons
(a) pi withdrawal
(2)
more negative J
-9 Hz
4. H-C-H on sp3 C compared to -12.4 Hz
a) sigma withdrawal - 2J more positive
(1)
methanol -10.8 Hz
(2)
methyl iodide -14.1
(3)
methylene bromide -5.5
14
15
Chapter 4, The Coupling Constant
b) sigma donation tetramethyl silane -14.1
c) in open chain, free rotation around O-C inhibits
pi overlap, 4-19
d) 4-21 has optimal overlap, 1.5, an increase of 10
Hz
e) 4-21 adjacent double bond withdraw pi electrons,
therefore 2J less positive (-21.5)
f) 4-22 -22.3
g) C#C has no such angular dependence, very large
overlap
(1)
acetonitrile 2J = -16.9
(2)
CH2(CN)2 2J = -20.4
15
16
Chapter 4, The Coupling Constant
C. Geminal coupling constants between different atoms
1. H - CH2 - CH3 -4.8
2. H - CCl2 - CHCl2 +1.2
3. H - CH2 - C= -4 to –7
4. H - (C=) - C
+20 to 50
5. 4-23 & 4-24
6. H - C - 14N dependent on presence or absence of lone
pair
a) 4-25, 4-26, 4-27
7.
15
N & 13C
a) 4-28, 4-29
8. H - C - F
a) 50 Hz for sp3 C - fluoroethane 47.5
b) 80 Hz for sp2 C - fluoroethylene 84.7
9. F - C - F 150 to 250 for sp3
a) less than 100 Hz for sp2 C
(1)
CH2=CF2 35.6
16
17
VII.
Chapter 4, The Coupling Constant
17
Three-Bond (Vicinal) Coupling Constants
A. dihedral angle (theta) - the angle between two planes
1. book as an example
2. vicinal (three-bond) coupling can tell us the most
the three-dimensional arrangement of the atoms
molecule
about
within a
a) angle between Ha - C - C plane & C - C - Hb
b) vary between 0 & 360 by rotation around the C – C
(1) certain arrangements more stable than
others
(2) each one has its own specific dihedral
angle
3. Karplus equation, modified by Bothner-By
a) 3J = A cos2(theta) + C for theta = 0-90o
b) 3J = A' cos2(theta) + C' for theta = 90-180o
c) J reaches maximum values at 0 & 180 degrees (J =
11 Hz)
(1)
largest when antiparallel (theta = 150-180o)
(2)
large when eclipsed (0-30o)
(3)
small when staggered (60-120o)
d) A & A' different, 2 maxima
(1)
C & C' often ignored
18
Chapter 4, The Coupling Constant
e) 3JHH = (7 - cos theta + 5 cos 2 theta)Hz
4. Shape of Karplus curve - ability to pass spin info
through bonds
a) theta = 0, 180 good overlap
(1) backside of sp3 or sp2 allow coupling
pathway
18
19
Chapter 4, The Coupling Constant
b) theta = 90o orbitals orthogonal, minimal overlap
5. in most molecules, constant & rapid (on nmr time scale)
interconversion of all possible conformations, so
dihedral angles vary among all possible values
19
the
a) since each conformation has a different value of
values of 3J represent a weighted average over
all conformations
b) also show dependence on temperature because the ratio of
the populations of the various conformations is a function of
temperature
c) minimum at 90 degrees (J = 2 Hz)
B. Examples
1. cyclohexane – chair
a) Jaa large, theta ~ 180, 8-12 Hz
b) Jae small, theta ~ 60, 1-5 Hz
c) Jee small, theta ~ 60, 0 - 4Hz
d) example 9-6 in Macomber, p 116
2. cyclopropane 4-31
a) theta = 0, 8.97
b) theta = 120, 5.58
3. 5 membered ring - Jc or Jt can be large
4. Vicinal coupling across double bonds shows a similar
variation with structure
a) double bond does not allow rotation
(1) therefore limits vicinal relationships to
either 0 (cis) or 180 (trans)
20
Chapter 4, The Coupling Constant
20
(a) because of extra intervening bond, there are more electrons
communicating spin information between the nuclei
(i) 3J is about 20% greater than in the case of
single bonds
three
(b) acrylonitrile
(i) Jc = 11.3
(ii) Jt = 18.2
(2)
3J
HH
in linear molecule acetylene
(a) only 9 Hz but
(i) the presence of the triple bond
(ii) s-character effects might have lead us to
a larger value
(3)
predict
benzene
(a) ortho H, theta = 0, 6-10 Hz
(b) meta H, 1-4 Hz
(c) para H, 0-1.5 Hz
(4)
1,4-dinitro-2,3,5,6-tetraacetoxycyclohexane
(a) 2.02 (s,6 H)
(b) 2.12 (s,6 H)
(c) 5.03 (d of d (J = 3.5, 11 Hz), 2 H)
(d) 5.89 (d of d (J = 3.5, 11 Hz), 2 H)
(e) 6.22 (t (J = 3.5 Hz), 2 H)
(f) d of d - axial H flanked by 1 axial & 1 equitorial
(g) triplet, J = 3.5, - equitorial H
(h) only 4-32 is consistent structure
H
21
Chapter 4, The Coupling Constant
C. Limitations of Karplus equation - H - C - C – H
21
1. dependent on:
a) bond length
b) H - C - C valence angle
(1)
both (a) & (b) can be controlled
c) electronegativity of substituents - most difficult
to handle
d) orientation of substituents
e) H - C - C - H dihedral angle
2. Approaches to electronegativity
a) derivations of mathematical dependence of J on
electron
b) empirical allowance by use of chemical shifts due
electron
c) eliminate by using ratio (R value) of 2 coupling
constants that correspond to the same dihedral
angle & have same multiplicative dependence on
substituent electronegativity
3. so far talking about homonuclear (1H) vicinal coupling
a) also found for heteronuclear vicinal coupling
constants as well
b) Karplus works for H - C - X – H
(1)
X = O,N,S,Se,Te,Si
c) Also for H-C-C-C, H-C-C-F, C-C-C-C
(1)
new A
to
22
Chapter 4, The Coupling Constant
d) F-C-C-F & H-C-C-P not follow
(1)
22
possibly due to lone pairs
VIII. Long-Range Coupling Constants
A. one or more multiple bonds between the coupling nuclei
1. electron pairs in multiple bonds behave differently
those in single bonds
a) geometry, hybridization, s-character
b) delocalization through the other moleuclar
orbitals in a molecule
(1)
resonance
(2)
conjugation
2. H-C-C=CH - allylic coupling
a) examples
b) C-Ha orthogonal to pi
(1)
no overlap
(2)
sigma-pi coupling small (< 1 Hz)
c) in acyclic - average over all arrangements
d) rings freeze bonds
3. homoallylic coupling HC-C=C-CH
a) 5J ~ +2
4. allene & propyne
a) C#C places no steric limitations
(1) therefore observe over 5 to 7 bonds in
polyalkynes
from
23
Chapter 4, The Coupling Constant
5. conjugation stereochemically more complicated
a) 1,3-butadiene
(1)
2 4-bond couplings -8.6, -.83
(2)
3 5-bond couplings .6, 1.3, .69
b) aromatic
(1)
4J
m
1.37
(2)
5J
p
.69
(3)
3J
o
7.54
23
24
Chapter 4, The Coupling Constant
c) also benzylic couplings
24
B. The molecule is rigid & the 2 nuclei possess a so-called W
spatial relationship
1. this overlap "short-circuits" or sidesteps the other
intervening C-C bonds
a) resembles a three-bond coupling
b) usually involves relatively small coupling
constants
2. examples
3. Usually through the bond but if within van der Waals
contact as in
IX. Measurement of the Coupling Constant
A. Spectral analysis - derivation of chemical shift vi of each
nucleus & coupling constant Jij
1. if vi & Jij known, spectrum precisely reproduced by
computer calculation
2. 1st order spectra - distance between pairs of peaks
a) other nuclei (than H) large chemical shift range
b) other nuclei often low natural abundance - no
coupling to each other
c) therefore other nuclei have 1st order
d) B may show 2nd order behavior
3. proton NMR often has 2nd order behavior
a) but can often remove with high magnetic field
(1)
400 MHz
two
25
Chapter 4, The Coupling Constant
b) magnetic nonequivalence independent of field
25
4. 2nd order 2 spin system (AB) has 4 lines
a) J obtained directly from doublet spacing
b) no specific peak positions correspond to chemical shift
c) The A peaks have unequal intensities
(1)
therefore va is weighted average
(2)
delta(vab) = (4C2 - J2)1/2
5. ratio of intensities of inner & outer peaks =
(1 + J/2C)/(1 - J/2C)
inner/outer =
6. 3 spin system analysis possible only in a few instances
a) AX2
b) AMX
c) but AB2 can contain up to 9 peaks
(1)
4 from flip of A
(2)
4 from flip of B
(3)
1 from simultaneous flips
(4)
When 8 peaks observed:
(a) chemical shift of A = v3
(b) chemical shift of B = vb = 1/2(v5 + v7)
(c) JAB = 1/3(v1 - v4 + v6 - v8
(d) if less than 8, confusion - best to use
analysis
computer
26
Chapter 4, The Coupling Constant
7. ABX, ABC, AA'XX' & AA'BB' often require computer
analysis
26
a) trial & error guessing of chemical shift &
coupling
b) shifts varied 1st until widths of observed &
calculated spectra match
c) then coupling constants or sums & differences
varied
d) iterative methods to refine
X. 2nd Order Effects
A. 3 major classes of 2nd order effects that can lead to
misinterpretations
1. spacings do not correspond to coupling constants even
the spectrum contains an apparently 1st order
when
pattern
2. deceptive simplicity) certain unequal couplings appear
to be
equal simplicity) so that there are fewer
lines than 1st order
considerations would predict
3. virtual coupling) zero or nearly zero couplings appear
to give
rise to significant splittings, so that
there are more lines
than 1st order considerations
would predict
B. Peak spacings do not correspond to J
1. Hs is t of t
a) 12.4 Hz Hs to Hax
b) 4.0 Hz Hs to Heq
c) but coupling constants of 14.1 & 2.3 required in
computer generated
d) spectrum gives sum of coupling constants by
inspection (16.4 Hz)
27
Chapter 4, The Coupling Constant
e) commonly occurs when 1H coupled to 2 or more 1H
turn are closely coupled (small delta v/J)
27
that in
f) spin states of closely coupled nuclei are mixed
(1) therefore 1st nucleus cannot distinguish
them properly
28
Chapter 4, The Coupling Constant
C. Deceptive Simplicity
28
1. lines coincide
2. example: ABX showes up as doublet & triplet
a) coupling could be unequal but spectrum sensitive
to their sum
b) sometimes remove at higher or lower field
D. Virtual Coupling
1. example 1
a) methine & methylene 1H closely coupled
b) -CH should be doublet but more complicated
2. example 2
a) CH & CH2 no longer closely coupled
b) therefore -CH3 doublet
3. example 3
a) quartet & doublet
4. example 4
a) more complex because of W zigzag pathway
XI. Aids in Spectral Analysis
A. Higher Fields
1. v/J increased in higher fields
2. increased sensitivity
3. therefore decrease 2nd order effects caused by small
v/J but not by magnetic nonequivalences
only
29
Chapter 4, The Coupling Constant
a) AA'XX'
B. Lanthanide Shift Reagents
1. unusual shielding effects in molecules with which they
form acid-base complexes
2. pseudocontact mechanism - spin info passed through
space within the complex rather than through the
bonding electrons
3. presence of unpaired electron spins in same region of
space as resonating protons causes very large
shielding effects (hence the term pseudocontact)
4. both a distance & an angular dependence
5. europium shift reagents minimize 2nd order spectral
effects by increasing v/J
a) 4-50 Eu(dpm)3
b) 4-51 Eu(fod)3
(1)
better solubility properties
(2)
complexes with less basic sites
29
30
Chapter 4, The Coupling Constant
6. shift reagents with range of ligands & rare earths
7. useful info for alcohols, amines, ketones, esters,
sulfoxide, ether, thiol ether, nitrile, oximes,
aldehydes
30
lactones,
8. used with 1H, 19F, 13C
C. Multiple Resonance
D. Deuterium Substitution
1. 4-52
a) X=H, ABCXY spectrum
(1)
X too close to -CH2- for double irradiation
b) X=D, ABXY, irradiate D
2. 4-53
a) AA'BB' instead of A2A2'B2B2'
b) therefore determine vicinal couplings
E. Satellite Spectra
1. measure coupling between chemically equivalent on
different C using 13C
2. H-12C-13C-H~ same as H-12C-12C-H
3. large 13C-H coupling produces multiplets, called
satellites on each side of center band
a) separated by (1/2)J13C-H
b) separation of each sideband from centerband
serves as effective chemical shift difference
c) H-H coupling between H-12C & H-13C observed in the satellite
4. N15,Si29,Se77,Cd111,Cd113,Sn117,Sn119,Te125,Pt195,Hg199 have
31
Chapter 4, The Coupling Constant
spin 1/2 with useful satellite resonances
31
F. Multiple Pulse Sequences - used for better sensitivity,
examination
of multiple quantum relationships &
measurement of relaxation
times
G. 2-dimensional NMR
1. examine 2 frequency domains simultaneously
a) 2-dimensional plots
b) i.e. 1H frequency plotted vs 13C
c) or chemical shift vs coupling constant
H. Liquid Crystal Solvents
1. nematic liquid crystals
a) mesophase - intermediate between solids & liquids
b) rod-shaped molecules
c) 4-54
2. domain - rod-shaped molecules align along long axis
a) oriented by a magnetic field
(1)
long axis parallel to Bo
b) small organic molecules dissolved, also aligned
(1)
but can still move
32
Chapter 4, The Coupling Constant
3. NMR not broad
32
a) peaks relatively sharp
b) much larger range of frequencies than ordinary
phase (isotropic) spectra
4. spectrum determined by direct dipole-dipole couplings (D)
a) no longer tumbling
b) also can determine absolute signs of J
5. coupling constants observed between chemically
equivalent nuclei
a) isolated Me (MeCN) triplet resonance
6. D - mathematical relationship with internuclear
distances & angles between Bo & internuclear vectors
a) used to get hard structural data
liquid