Transcript CHEM 430 * NMR Spectroscopy

NMR - The
Coupling Constant
4-1
FIRST ORDER SPECTRA
For a spectrum to be 1st order, the Dn between the chemical shifts of any given
pair of nuclei must be much larger than the value of the coupling constant J
between them:
Dn / J > 10
1st order spectra exhibit the following characteristics:
• Spin– spin multiplets are centered on the resonance frequency.
• Spacings between adjacent components of a spin–spin multiplet = J.
• Multiplicities that result from coupling exactly reflect the n + 1 rule for I = ½
• The intensities of spin–spin multiplets correspond to the coefficients of the
binomial expansion given by Pascal’s triangle for spin- ½ nuclei
• Nuclei with the same chemical shift do not split each other, even when the
coupling constant between them is nonzero.
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NMR - The
Coupling Constant
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FIRST ORDER SPECTRA
When the chemical shift difference is less than about 10 times J, 2nd order
effects appear in the spectrum, including deviations in intensities from the
binomial pattern and other exceptions from the preceding characteristics.
Pople notation:
•Nuclei that have a 1st order relationship are represented by letters that are
far apart in the alphabet (AX)
•Nuclei that are close in chemical shift and may exhibit a second- order
relationship are represented by adjacent letters (AB)
•Nuclei in the middle of AX are represented as M
Higher field spectrometers – > 300 MHz increase Dn (i.e. 1 ppm represents a
greater number of Hz) and minimize 2nd order effects
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NMR - The
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FIRST ORDER SPECTRA
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AB Spin System - 1st to 2nd order
Dn/J = 0.4
Dn/J = 1
Dn/J = 4
Dn/J = 15
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NMR - The
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FIRST ORDER SPECTRA
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AB2 Spin System – 1st to 2nd order
Dn/J = 0.4
Dn/J = 1
Dn/J = 4
Dn/J = 15
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NMR - The
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CHEMICAL AND MAGNETIC EQUIVALENCE
In addition to meeting the requirement that Dn / J > 10 , 1st order spectra must
pass a symmetry test.
Any two chemically equivalent nuclei must have the same coupling constant to
every other nucleus.
Nuclear pairs that fail this test are said to be magnetically nonequivalent, and
their spectral appearance is 2nd order
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NMR - The
Coupling Constant
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CHEMICAL AND MAGNETIC EQUIVALENCE
Symmetry
•Nuclei are chemically equivalent if they can be interchanged by a symmetry
operation of the molecule.
•Thus the two protons in 1,1- difluoroethene or in difluoromethane may be
interchanged by a 180° rotation. Nuclei that are interchangeable by rotational
symmetry are said to be homotopic.
•Rotation about C—C single bonds is so rapid that the chemist rarely considers
the fact that the three methyl protons in CH3CH2Br are not symmetrically
equivalent (dynamic effect)
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NMR - The
Coupling Constant
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CHEMICAL AND MAGNETIC EQUIVALENCE
Symmetry
• Nuclei related by a plane of symmetry are called enantiotopic, provided
there is no rotational axis of symmetry.
• For example, the protons in BrClCH2 are chemically equivalent and
enantiotopic because they are related by the plane of symmetry containing
C, Br, and Cl.
• If the molecule is placed in a chiral environment, (using a solvent composed
of an optically active material or by placing the molecule in the active site of
an enzyme) as represented by a small hand placed to one side of BrClCH2 the
protons are no longer equivalent because the hand is a chiral object.
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NMR - The
Coupling Constant
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CHEMICAL AND MAGNETIC EQUIVALENCE
Symmetry
• Because the plane of symmetry is lost in a chiral environment, the nuclei are
not enantiotopic and have become chemically nonequivalent ( no symmetry
operation can interchange them).
• The term enantiotopic was coined because replacement of one proton of the
pair by another atom or group, such as deuterium, produces the enantiomer
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NMR - The
Coupling Constant
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CHEMICAL AND MAGNETIC EQUIVALENCE
Symmetry
•Enantiotopic or homotopic protons need not be on the same carbon. For
example the alkenic protons in cyclopropene are homotopic, but those in 3methylcyclopropene are enantiotopic.
•Chemically equivalent nuclei (homotopic or enantiotopic) are represented by
the same letter in the spectral shorthand of Pople.
•Cyclopropene is A2X2, as is difluoromethane, since the two fluorine atoms have
spins of ½
•The ring protons of 3-methylcyclopropene constitute an AX2 group.
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NMR - The
Coupling Constant
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CHEMICAL AND MAGNETIC EQUIVALENCE
Symmetry
• To be magnetically equivalent, two nuclei must be chemically equivalent and
have the same coupling constant to every other nucleus.
• This test is more stringent than that for chemical equivalence, because it is
necessary to go beyond considering just the overall symmetry of the
molecule.
• For example, in CH2F2 each of the two hydrogens has the same coupling to a
specific fluorine atom because both hydrogens have the same spatial
relationship to that fluorine. Consequently, the protons are chemically and
magnetically equivalent - A2X2.
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CHEMICAL AND MAGNETIC EQUIVALENCE
Symmetry
Examples that once seemed simple:
p-nitrotoluene
• On first inspection, like CH2F2, it has a plane of symmetry
NO2
Jab≠Ja’b
Ha
H a'
Hb
H b'
CH3
• However, on careful inspection the coupling of what would be two
chemically identical nuclei (Ha and Ha’) is different to Hb so these are
magnetically non-equivalent
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CHEMICAL AND MAGNETIC EQUIVALENCE
Symmetry
Examples that once seemed simple:
1,1-difluroethene
• Like the previous example, the molecule has a plane of
symmetry
JaF≠JbF
•
Here, the fluorines are spin-active (±1/2), so each
hydrogen is coupled differently to Fa so these are
magnetically non-equivalent
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CHEMICAL AND MAGNETIC EQUIVALENCE
Symmetry
Why is this important?
• Any spin system that contains nuclei that are chemically equivalent but
magnetically nonequivalent is, by definition, 2nd order.
• Moreover, raising the magnetic field cannot alter basic structural
relationships between nuclei, so that the spectrum remains second order at
the highest accessible fields.
• Nuclei that do not have the same chemical shift (anisochronous) also are
magnetically nonequivalent because they resonate at different resonance
frequencies (chemical shift criterion).
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NMR - The
Coupling Constant
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CHEMICAL AND MAGNETIC EQUIVALENCE
Symmetry
•Isochronous nuclei that are magnetically non-equivalent by having unequal
couplings to another nucleus are said to fail the coupling constant criterion.
•Nuclei that are chemically equivalent but not magnetically equivalent are given
the same letter in the Pople notation, but one is denoted by a prime
•Thus 1,1-difluoroethene is a AA’XX’ system
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Coupling Constant
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CHEMICAL AND MAGNETIC EQUIVALENCE
Symmetry
•Spectra for these systems are complex – 1H spectrum of 1,1-difluoroethene:
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NMR - The
Coupling Constant
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CHEMICAL AND MAGNETIC EQUIVALENCE
Symmetry
•Another example – 1H spectrum of 1,2-dichlorobenzene:
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Coupling Constant
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CHEMICAL AND MAGNETIC EQUIVALENCE
Symmetry
• Even open- chain systems such as ClCH2CH2OH contain magnetically
nonequivalent spin systems:
A 1st order spectrum would have
comprised two 1: 2: 1 triplets (not
case here) and instead of three
peaks in each resonance, there are
four (look to sides of center
resonance)
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NMR - The
Coupling Constant
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CHEMICAL AND MAGNETIC EQUIVALENCE
Symmetry
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NMR - The
Coupling Constant
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SIGNS AND MECHANISMS
Fermi Contact Interaction
•Spin–spin coupling arises because information about nuclear spin is
transferred from nucleus to nucleus via the electrons.
•Both nuclei and electrons are magnetic dipoles, whose mutual interactions
normally are described by the point–dipole approximation (as used by
McConnell to describe diamagnetic anisotropy)
•Fermi found that this approximation breaks down when dipoles are very close
(comparable to the radius of a proton).
•Under these circumstances, when the nucleus and electron in essence are in
contact, their interaction is described by a new mechanism, the Fermi contact
term.
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NMR - The
Coupling Constant
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SIGNS AND MECHANISMS
Fermi Contact Interaction
•The energy of the interaction is proportional to the gyromagnetic ratios of the
nucleus and of the electron, the scalar (dot) product of their spins (I for a
nucleus, S for an electron), and the probability that the electron is at the
nucleus (the square of the electronic wave function evaluated with zero
distance from the nucleus): .
EFC  – gn geI · Sy2(0)
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SIGNS AND MECHANISMS
Fermi Contact Interaction
• Because the nuclear and electronic gyromagnetic ratios have opposite signs ,
the more stable arrangement is when the nucleus and the electron are
antiparallel (spins paired):
Nuclear spin
electron spin
Nuclear spin
electron spin
Energy
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Coupling Constant
SIGNS AND MECHANISMS
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Fermi Contact Interaction
• Here, a single bond (two electrons) joins two spin-active nuclei: 13C-1H
• The bonding electrons will tend to avoid one another, if one is near the
13C nucleus (in this example) the other will be near the 1H nucleus
• By the Pauli principle, these electrons must be opposite in spin
• The Fermi model then predicts that the most stable condition between
the two nuclei must be one in which they too are opposite in spin:
13C
spin
1H
spin
electrons opposite in spin
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NMR - The
Coupling Constant
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SIGNS AND MECHANISMS
Fermi Contact Interaction
•When one spin slightly polarizes another spin oppositely the coupling constant
J between the spins has a positive sign by convention .
•A negative coupling occurs when spins polarize each other in the same
(parallel) direction.
•Qualitative models indicate that coupling over two bonds, as in H—C—H, is
negative, while coupling over three bonds, as in H—C—C—H is positive.
•There are numerous exceptions to this qualitative model, but it is useful in
understanding that J has sign as well as magnitude.
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NMR - The
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SIGNS AND MECHANISMS
Fermi Contact Interaction
•There are many variations of the subscripts and superscripts associated with J
constants
•In general, the superscript numeral to the left of J is the number of intervening
bonds through which the coupling is taking place
•
•
•
2J
is a coupling constant operating through two bonds - geminal
3J is a coupling constant operating through three bonds – vicinal
⩾4J is a coupling constant operating through ⩾ four bonds – long range
•Subscripts to the right of J can be used to show the type of coupling, such as
HH for homonuclear between protons or HC for heteronuclear between a
carbon and proton
•Often, this subscript will be used to “order” the various J-constants within a
complex multiplet: J1, J2, J3, etc. or JAB, JBC, JAC
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NMR - The
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SIGNS AND MECHANISMS
Fermi Contact Interaction
• High resolution NMR spectra normally are not dependent on the absolute
sign of coupling constants.
• Simultaneous reversal of the sign of every coupling constant in a spin system
results in an identical spectrum.
• Many spectra, however, depend on the relative signs of component
couplings.
• For example, the general ABX spectrum is determined in part by three
couplings, JAB, JAX, and JBX
• Different spectra can be obtained when JAX and JBX have the same sign from
when they have opposite signs even when the magnitudes are the same.
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NMR - The
Coupling Constant
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COUPLINGS OVER ONE BOND
• The one- bond coupling between 13C and 1H is readily measured from the 13C
spectrum when the decoupler is turned off
• Although it complicates routine 13C interpretation, this coupling can provide
useful information and illustrates several important principles.
• Because a p orbital has a node at the nucleus, only electrons in s orbitals can
contribute to the Fermi coupling mechanism (s orbitals have a maximum in
electron density at the nucleus).
• For protons, all electrons reside in the 1s orbital, but, for other nuclei, only
that proportion of the orbital that has s character can contribute to coupling.
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NMR - The
Coupling Constant
COUPLINGS OVER ONE BOND
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• When a proton is attached to an sp3 carbon atom (25% s character), 1JHC is
about half as large as that for a proton attached to an sp carbon atom ( 50%
s character).
• These numbers define a linear relationship between the %-s character of the
carbon orbital and the one- bond coupling:
%-s(C—H) = 0.2 J(13C—H)
1J
ethane = 125 Hz (sp3)
1J
2
HC ethene = 156 Hz (sp )
1J
HC ethyne = 249 Hz (sp)
HC
• The zero intercept of this equation indicates that there is no coupling when
the s character is zero, in agreement with the Fermi contact model.
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COUPLINGS OVER ONE BOND
• The 1JCH coupling ranges from about 100 to 320 Hz, and may be interpreted
in terms of the J– s relationship
• The coupling constant in cyclopropane (162 Hz) demonstrates that the
carbon orbital to hydrogen is approximately sp2 hybridized!
• Other examples include tricyclopentane (144 Hz, 29% s, sp2.4), cubane (160
Hz, 32% s, sp2) and quadricyclane (179 Hz, 36% s, sp1.8) .
• Although the J–s relationship works well for hydrocarbons not as applicable
to polar bonds.
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Coupling Constant
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COUPLINGS OVER ONE BOND
• For other nuclei variations in the effective nuclear charge and hybridization
effects, may alter the coupling constants.
• Just as the resonance frequency of a nucleus is proportional to its
gyromagnetic ratio , the coupling constant between two nuclei, as noted
above, is proportional to the product of both gyromagnetic ratios.
• Nuclei with very small gyromagnetic ratios, such as
correspondingly small couplings.
15N,
tend to have
• For 15N g has a negative sign, whereas 1H and 13C are positive. Therefore 1JNH
between 15N and hydrogen have a negative sign
• One bond couplings have been studied for other nuclei but are more
complex
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Coupling Constant
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COUPLINGS OVER ONE BOND
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NMR - The
Coupling Constant
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GEMINAL COUPLINGS
• By the Fermi model for geminal coupling (H—C—H) 2J is usually negative
• Geminal coupling (H–C–H) cay be measured directly from the spectrum
when the coupled nuclei are chemically nonequivalent, (the AB or AM part
of an ABX, AMX, ABX3… spectrum).
• If the relationship is 1st order (AM), the coupling may be measured by
inspection. In 2nd order cases (AB), the spectrum must be simulated
computationally, unless the two spins are isolated (a two- spin system)
• When nuclei are chemically equivalent but magnetically nonequivalent (as in
the AA' part of an AA'XX' spectrum) their coupling constant is accessible by
computational methods
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NMR - The
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GEMINAL COUPLINGS
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• Here, a intervening atom (usually a spin-inactive 12C) communicates spin
information between the two interacting nuclei
• The Fermi model then predicts that the most stable condition between the
these two geminal nuclei must be one in which they are parallel in spin:
12C
1H
spin
is spin inactive
12C
1H
spin
13H
spin
CHEM 430 – NMR Spectroscopy
is spin inactive
1H
spin
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GEMINAL COUPLINGS
IMPORTANT
•Splittings are not observed between coupled nuclei when they are
magnetically equivalent, but the coupling constant may be measured by
replacing one of the nuclei with deuterium.
•For example, in CH2Cl2- the geminal H– C– D coupling is seen as the spacing
between the components of the 1: 1: 1 triplet (deuterium has a spin of 1).
•Since coupling constants are proportional to the product of the gyro-magnetic
ratios of the coupled nuclei, J( HCH) may be calculated from J(HCD):
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GEMINAL COUPLINGS
• As the ∠H-C-H decreases, the amount of electronic interaction between the
two orbitals increases, the electronic spin correlations also increase and J
becomes larger (more negative)
H-C-H 109o
2J
HH = -12-18 Hz
In general:
40
2J
HH
H-C-H 118o
2J
HH = -4.3 Hz
H-C-H 120o
2J
HH = +0-3 Hz
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90 100
110
120
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GEMINAL COUPLINGS
• Variations in J also result from hybridization and ring size
• As ring size decreases, ∠ C-C-C decreases, along with p-character; the
resulting ∠ H-C-H increases, along with the corresponding s -character – J
becomes smaller
2J
HH
(Hz) = -2
-4
-9
-11
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-13
-9 to -15
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GEMINAL COUPLINGS
• Electron withdrawal by induction tends to make the coupling constant more
positive - for alkanes the negative coupling thus decreases in absolute value
(becoming less negative)
CH4 -12.4 to CH3OH -10.8 Hz
CH3I -9.2 to CH2Br2 -5.5 Hz
• Electron donation makes the coupling more negative
CH4, -12.4 to TMS -14.1 Hz
• Analogous substitution on sp2 carbon changes the coupling profoundly:
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GEMINAL COUPLINGS
• These effects of withdrawal or donation of electrons through the s-bonds
(induction) can be augmented or diminished by p-effects such as
hyperconjugation.
• Lone pairs of electrons can donate electron density and make 2J more
positive, whereas the orbitals of double or triple bonds can withdraw
electrons and make 2J less positive (or more negative).
• The above-mentioned large increase in the geminal coupling of imines or
formaldehyde compared with ethene results from reinforcement of the
effects of s withdrawal and p donation
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GEMINAL COUPLINGS
• These effect of p-withdrawal occurs for carbonyl, nitrile, and aromatic
groups as in the values for acetone (14.9 Hz), acetonitrile (– 16.9 Hz), and
dicyanomethane (– 20.4 Hz).
• The effect is some-what reduced by free rotation in open-chain systems, so
that particularly large effects are created by constraints of rings:
• p-donation by lone pairs makes J more positive. This effect also explains the
difference in the geminal couplings of three-membered rings: cyclopropane
and oxirane
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GEMINAL COUPLINGS
• Remember splittings are not observed for magnetically equivalent nuclei
(like the 4 hydrogens on CH4) ; the 2J values in this table are used as
reference values and generated by observing the 2JHD for the deuterated
analog of these compounds (top page 92 in text).
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GEMINAL COUPLINGS
• Geminal couplings between protons and other nuclei also have been studied.
• The H–C–13C coupling responds to substituents in much the same way as
does the H–C–H coupling; values are smaller, due to the smaller g of 13C.
• Unlike the H–C–H case, the H–C–13C geminal coupling pathway can include a
double or triple bond; such couplings can be useful to determine
stereochemistry:
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GEMINAL COUPLINGS
• The 2JHCN between hydrogen and 15N strongly depends on the presence and
orientation of the nitrogen lone pair.
•
2J
is a useful structural diagnostic for syn–anti isomerism in imines,
oximes, and related compounds as the H–C–15N coupling in imines is larger
and negative when the proton is cis to the lone pair but smaller and positive
for a proton trans to the lone pair:
HCN
• The cis relationship between the nitrogen lone pair and hydrogen also is
found in heterocycles such as pyridine 2JHCN -10.8.
• In saturated amines with rapid bond rotation values typically are quite small
and negative (CH3NH2 , -1.0).
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NMR - The
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•
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GEMINAL COUPLINGS
2J
between 15N and 13C follow a similar pattern and also can be used for
structural and stereochemical assignments.
• The carbon on the same side as the lone pair (syn) in imines again has a
larger, negative coupling (- 11.6 Hz). The anti-isomer has a 2JCCN of 1.0 Hz.
• Likewise, the two indicated carbons in quinoline have couplings
differentiated by their geometry - as one is syn and the other anti to the
nitrogen lone pair.
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•
2J
HCP
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GEMINAL COUPLINGS
between 31P and hydrogen also have been exploited stereochemically.
• The maximum positive value of 2JHCP is observed when the H—C bond and
the phosphorus lone pair are eclipsed (syn), and the maximum negative
value when they are orthogonal or anti.
• The situation is similar to that for couplings between hydrogen and 15N, but
signs are reversed as a result of the opposite signs of the gyromagnetic
ratios of 15N and 31P.
• The coupling also is structurally dependent, as it is larger for P(III) than for
P(V): 27 Hz for (CH3)3P and 13.4 Hz for (CH3)3P=O.
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GEMINAL COUPLINGS
• Geminal H– C– F couplings are usually close to for an sp3 carbon (47.5 Hz for
CH3CH2F) and for an sp2 carbon (84.7 Hz for CH2=CHF).
• Geminal F–C–F couplings are quite large for saturated carbons (240 Hz for
1,1-difluorocyclohexane), but less than 100 Hz for unsaturated carbons
(35.6 Hz for CH2=CF2).
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VICINAL COUPLINGS
• Coupling constants between protons over three bonds have provided the
most important early stereochemical application of NMR spectroscopy vicinal coupling
• As with geminal coupling, the overall lowest energy spin state is one where
the 1H nuclei and electron spins are paired (12C is spin inactive)
• Observe the two possible spin interactions:
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VICINAL COUPLINGS
4-6
• Observe that the orbitals must overlap for this communication to take place
– weaker J-constants
• To communicate spin information, one additional “flip” must take place, and
the J-values are usually positive
• The magnitude of the interaction, it can readily be observed, is greatest
when the orbitals are at angles of 0o and 180o to one another:
0o dihedral angle
180o dihedral angle
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VICINAL COUPLINGS
• In 1961, Karplus derived a mathematical relationship between 3JHCCH and
dihedral ∠ H‒C‒C‒H .
• The cos2 relationship results from strong coupling when orbitals are
parallel. They can overlap at the syn-periplanar or anti-periplanar
geometries.
• When orbitals are staggered or orthogonal , coupling is weak.
• A and C are empirically determined constants; C and C’ usually are
neglected, as they are thought to be less than 0.3 Hz while A and A’ imply
that J is different at the syn and the anti-maximum
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NMR - The
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VICINAL COUPLINGS
4-6
• Unfortunately, these multiplicative constants vary from system to system in
the range 8–14 Hz and quantitative applications cannot be transferred easily
from one structure to another.
• In general:
12
8
3J
(Hz)
6
4
2
0
45
90
ao
135
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180
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VICINAL COUPLINGS
• In chair cyclohexane Jaa is large as faa is close to 180°
• Jee (0– 5 Hz) and Jae (1– 6 Hz) are small as fee and fae are close to 60°
• When cyclohexane rings are flipping between two chair forms, Jaa is
averaged with Jee to give a Jtrans in the range 4– 9 Hz, and Jae is averaged with
Jea to give a smaller Jcis, still in the range 1– 6 Hz.
• In conformationally locked systems (no ring flip) the effect can be used to
assign stereochemistry
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VICINAL COUPLINGS
• Further examples:
3J
aa = 10-14 Hz
a=
180o
3J
ee = 4-5 Hz
a = 60o
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3J
ae
= 4-5 Hz
a = 60o
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VICINAL COUPLINGS
• For alkenes 3Jtrans (f = 180o) is always larger than 3Jcis (f = 0o)
3J
trans
= 11-18 Hz
3J
cis =
a = 180o
•
3J
6-15 Hz
a = 0o
> 3Jcis > 2Jgem allows assignment of the vinyl system (AMX, ABX or ABC
spectrum) trivial
trans
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VICINAL COUPLINGS
• For cyclic alkenes internal bond angles may affect 3Jcis
120o angle in H-C-C
bond reduces overlap
3J
trans
= 11-18 Hz
3J
cis =
a = 180o
3J
cis =
0-2
2-4
5-7
6-15 Hz
a = 0o
8-11
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6-15 Hz
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VICINAL COUPLINGS
• Despite the potentially general application of the Karplus equation to
dihedral angle problems, there are quantitative limitations.
• The 3J H–C–C–H depends on the C–C bond length or bond order, the H–C–C
valence angle, the electronegativity and orientation of substituents on the
carbon atoms in addition to the H–C–C–H dihedral angles.
• A properly controlled calibration series of molecules must be rigid (monoconformational) and have unvarying bond lengths and valence angles. Three
approaches have been developed to take the only remaining factor,
substituent electronegativity, into account:
Derive the mathematical dependence of 3J on electronegativity.
2. Empirical allowance by the use of chemical shifts that depend on
electronegativity in a similar fashion as 3J.
3. Eliminate the problem through the use of the ratio (the R value) of two 3J
coupling constants that respond to the same or related dihedral angles and that
have the same multiplicative dependence on substituent electronegativity,
which divides out in R.
1.
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VICINAL COUPLINGS
• These more sophisticated versions of the Karplus method have been used
quite successfully to obtain reliable quantitative results.
• The existence of factors other than the dihedral angle results in ranges of
vicinal coupling constants at constant even in structurally analogous
systems.
• Saturated hydro-carbon chains (H–C–C H) exhibit vicinal couplings in the
range 3– 9 Hz, depending on substituent electronegativity and rotamer
mixes and 8.90 Hz for . Higher substituent electronegativity always lowers
the vicinal coupling constant. In small rings, the variation is almost entirely
the result of substituent electronegativity, with cis ranges of 7– 13 Hz and
trans ranges of 4– 10 Hz in cyclopropanes.
CHEM 430 – NMR Spectroscopy
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NMR - The
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4-6
VICINAL COUPLINGS
• These more sophisticated versions of the Karplus method have been used
quite successfully to obtain reliable quantitative results.
• The existence of factors other than the dihedral angle results in ranges of
vicinal coupling constants at constant even in structurally analogous
systems.
• Saturated hydro-carbon chains (H–C–C–H) exhibit vicinal couplings in the
range 3–9 Hz, depending on substituent electronegativity and rotamer
mixes.
• Higher substituent electronegativity always lowers the vicinal coupling
constant. In small rings, the variation is almost entirely the result of
substituent electronegativity, with cis ranges of 7– 13 Hz and trans ranges of
4– 10 Hz in cyclopropanes.
CHEM 430 – NMR Spectroscopy
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NMR - The
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4-6
VICINAL COUPLINGS
• In small rings, the variation is almost entirely the result of substituent
electronegativity, with cis ranges of 7– 13 Hz and trans ranges of 4– 10 Hz in
cyclopropanes.
• Coupling constants in oxiranes ( epoxides) are smaller because of the effect
of the electronegative oxygen atom.
•
3J
is proportional to the overall bond order, as in benzene and in
naphthalene. The ortho-coupling in benzene derivatives varies over the
relatively small range of 6.7– 8.5 Hz, depending on the resonance and polar
effects of the substituents.
• The presence of heteroatoms in the ring expands the range at the lower end
down to 2 Hz, because of the effects of electronegativity ( pyridines) and of
smaller rings (furans, pyrroles).
CHEM 430 – NMR Spectroscopy
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4-6
VICINAL COUPLINGS
CHEM 430 – NMR Spectroscopy
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NMR - The
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4-1
FIRST ORDER SPECTRA
LONG- RANGE COUPLINGS Coupling between protons over more than three
bonds is said to be long range. Sometimes coupling between 13C and protons
over more than one bond also is called long range, but the term is
inappropriate for 2J( CCH) and 3J( CCCH). Long- range cou-pling constants
between protons normally are less than 1 Hz and frequently are unob-servably
small. In at least two structural circumstances, however, such couplings commonly become significant. Overlap. Interactions of bonds with electrons of
double and triple bonds and aromatic rings along the coupling pathway often
increase the magnitude of the coupling constant. One such case is the four bond
allylic coupling, , with a range of about to and typical values close to . Larger
values are ob-served when the saturated C ¬ Ha bond is parallel to the p
orbitals ( 4- 31). This s– p + 1 - 3 Hz - 1 Hz HC— C“ CH S– P C ¬ H( s) p
CHEM 430 – NMR Spectroscopy
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4-7
LONG RANGE COUPLINGS
• As can be deduced from the reduced J values for vicinal coupling and the
Karplus relationship, the greater the number of intervening bonds the less
opportunity for orbital overlap over long range (> 3 bond)
• Long- range coupling constants between protons normally are less than 1 Hz
and frequently are unobservably small.
• In at least two structural circumstances, however, such couplings commonly
become significant.
• Allylic and homoallylic coupling
• W-coupling
• In cases where a rigid structural feature preserves these overlaps, however,
long range couplings are observed – especially where C-H s-bonds interact
with adjacent p-systems
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4-7
LONG RANGE COUPLINGS
• Allylic systems are the simplest example of a 4J coupling
• Here, if the allyl C-Ha bond is orthogonal to the p system,4J = 0 Hz; if this bond
is parallel to the vinyl C-Ha bond, 4J = 3 Hz
• In acyclic systems, the dihedral angle is averaged over both favorable and
unfavorable arrangements, so an average 4J is found, as in 2- methylacryloin
• Ring constraints can freeze bonds into the favorable arrangement, as in
indene or in an exactly parallel arrangement as in allene (right)
CHEM 430 – NMR Spectroscopy
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LONG RANGE COUPLINGS
• When this type of coupling is extended over five bonds, it is referred to as
homoallylic coupling
• Examples include the meta- and para- protons to the observed proton on an
aromatic ring and acetylenic systems:
5J
= 0-1
4J
1-3
CHEM 430 – NMR Spectroscopy
5J
0-1 Hz
62
LONG RANGE COUPLINGS
• Rigid aliphatic ring systems exhibit a specialized case of long range coupling
– W-coupling – 4JW
• The more heavily strained the ring system, the less “flexing” can occur, and
the ability to transmit spin information is preserved
4J
= 0-1
4J
=3
CHEM 430 – NMR Spectroscopy
4J
= 7 Hz
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4-7
LONG RANGE COUPLINGS
• Although coupling information always is passed via electron-mediated
pathways, in some cases part of the through- bond pathway may be skipped,
and effect known as through-space coupling
• Two nuclei that are within van der Waals contact in space can interchange
spin information if at least one of the nuclei possesses lone pair electrons found most commonly, but not exclusively, in H– F and F– F pairs.
• The six- bond H--F coupling is negligible on the left (2.84 Å) but is 8.3 Hz on
the right ( 1.44 Å) ( the sum of the H and F van der Waals radii is 2.55 Å).
• This is likely is important in the geminal F– C– F coupling, which is unusually
large: 2JFCF for sp3 CF2 (~200 Hz, 109.5o) compared to sp2 CF2 (~50 Hz, 120o)
CHEM 430 – NMR Spectroscopy
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4-8
SPECTRAL ANALYSIS
• Analysis of coupling constants in first-order spectra
• The general scheme for the interpretation and analysis of 1st order
multiplets:
• Some helpful constraints:
• For every signal split into a multiplet, the component J-value(s) must
match some other multiplet in the spectrum
• The distance (Hz) between the two outermost peaks in a multiplet is
equal to the sum of each of the coupling constants
• The smallest J-value is typically given by the difference between the first
and second peaks in the multiplet
• First order multiplets are symmetrically distributed about the center
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SPECTRAL ANALYSIS
4-8
Analysis of coupling constants in first-order spectra
Method:
1. Your book suggests letting the spectrometer do the work, however this
method still requires ‘spectral common sense’
2. Our method is adapted from:
Hoye, T. R.; Hanson, P. R. Vyvyan, J. R. J. Org. Chem. 1994, 59, 4096-4103
and basically builds the same tree analysis from the branches on in
3. This paper is a well loved and regarded classic by 17 years of grateful
graduate students
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SPECTRAL ANALYSIS
Let’s start with a simple spectrum:
Crotonic acid (trans-2-butenoic acid)
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SPECTRAL ANALYSIS
Assuming we analyze the spectrum as we have before, we should get a
structure similar (or isomeric) to the actual one:
Simple spectrum:
13C NMR:
4 unique carbons (too few for aromatic)
1 C=O; 2-alkenyl, 1 alkyl
1H
NMR:
2 multiplets in the alkene region
1 multiplet in the alkyl region
In CDCl3, the acidic proton appears at d 12.4 (acid)
Integrals are 1(a):1:1:3
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SPECTRAL ANALYSIS
Without much effort, and any more detailed analysis of chemical shifts,
several possibilities arise:
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NMR - The
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4-8
SPECTRAL ANALYSIS
From analysis of J-values and knowledge of which protons are coupled by the
various J-values can finish the analysis:
Inspection of the three multiplets shows the following Hz values:
Ha
2150.34
2143.45
2136.39
2134.75
2129.50
2127.86
2120.97
2114.08
Hb
1767.89
1766.25
1764.61
1762.96
1752.46
1750.82
1749.02
1747.38
CHEM 430 – NMR Spectroscopy
Hc
580.98
579.18
573.93
572.29
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NMR - The
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SPECTRAL ANALYSIS
4-8
Start with the simplest multiplet:
Hc is an apparent doublet of doublets (dd)
Step 1:
• The distance between the first two lines always represents the
smallest J value
• If the ratio of these two lines (integral) is 1:1, this J is unique; if
it is 1:2, 1:3, etc. there are two or more identical smallest J s
• Label this Jsmall
580.98 – 579.18 = 1.80 Hz
CHEM 430 – NMR Spectroscopy
Hc
580.98
579.18
573.93
572.29
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SPECTRAL ANALYSIS
4-8
Step 2: (most difficult step for complex multiplets)
• Find the full set of pairs within the multiplet that are separated
by Jsmall
• Each pair will have a reflected partner through the center of the
multiplet
• For pairs where one of the lines has a relative intensity >1, that
line will be part of more than one pair
580.98 – 579.18 = 1.80 Hz
579.18 – 573.93 = 5.25 Hz
573.93 – 572.29 = 1.64 Hz
While these two
may not seem
equal, they must
be matched if 1st
order
Hc
580.98
579.18
573.93
572.29
580.98 – 579.18 = 1.80 Hz
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NMR - The
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SPECTRAL ANALYSIS
4-8
Step 3:
• Find the centers of each of the pairs generated in step 2
• These will collectively represent a new pattern (as if the Jsmall
was selectively decoupled)
• As with step 1, the spacing between the first two lines of this
multiplet represent the next smallest J
• Label this J as med-small, etc. as necessary
580.98
580.08
579.18
580.08 - 573.11 = 6.97
573.93
Hc
580.98
579.18
573.93
572.29
573.11
572.29
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SPECTRAL ANALYSIS
Step 4:
• Find the midpoint(s) of this new pair(s), and repeat step 3
Step 5:
• Repeat as necessary until all J-values have been found
• Remember, it must be internally consistent and all the J values must
add up to the difference between the outer peaks of the multiplet.
We check:
The two J s we determined are 1.80 (1.64) and 6.97
(1.80+1.64)/2 + 6.97 = 8.69
The difference between the outermost peaks of the multiplet:
580.98 – 572.29 = 8.69
CHEM 430 – NMR Spectroscopy
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NMR - The
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SPECTRAL ANALYSIS
4-8
This effectively generates the tree-diagram discussed in the text:
n
Jmed = 6.97 Hz
This proton is coupled to two
other protons, with coupling
constants of 1.72 and 6.97;
(dd, 6.97, 1.72)
Jsmall = 1.72 Hz
CHEM 430 – NMR Spectroscopy
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NMR - The
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SPECTRAL ANALYSIS
4-8
Repeat the analysis with the next most complex multiplet, look for possible Jvalues of 1.72 and 6.97. Analyzing the Hb proton; apparent doublet of quartets
(dq):
Hb
1767.89
Step 1&2
Step 3&4
1.64 Hz
1768.71
1766.25
1764.61
1762.96
1752.46
1750.82
1.64 Hz
1.65 Hz
1.64 Hz 1769.53
1767.07
1765.43
1765.43
1.64 Hz
1.64 Hz 1767.89
15.51 Hz
1.64 Hz
1.80 Hz
1749.02
1747.38
Step 5
1.64 Hz
symmetrical
Check:
1749.92
1.64 + 1.64 + 1.64 + 15.51 = 20.43
1767.89 – 1747.38 = 20.51
CHEM 430 – NMR Spectroscopy
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NMR - The
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SPECTRAL ANALYSIS
4-8
The tree diagram for Hb:
n
Jlarge = 15.51 Hz
Jsmall = 1.64 Hz
Jsmall = 1.64 Hz
We see three generational
iterations of the 1.64 similar
to ~1.7 as found for Hc
We also see a new constant
of 15.51
Jsmall = 1.64 Hz
CHEM 430 – NMR Spectroscopy
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NMR - The
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SPECTRAL ANALYSIS
4-8
Lastly, repeat the process for Ha; looking for 1.64, 15.51 or 6.97:
Analyzing the Ha proton; apparent doublet of quartets (dq):
Ha
2150.34
Step 1&2
Step 3&4
6.89 Hz
2146.91
2143.45
2136.39
1.64 Hz
2134.75
2129.50
1.64 Hz
2127.86
7.06 Hz
6.89 Hz
7.07 Hz 2143.29
2139.84
2132.95
2139.85
6.89 Hz
6.89 Hz 2136.40
15.43 Hz
6.89 Hz
6.89 Hz
2120.97
2114.08
Step 5
6.89 Hz
symmetrical
Check:
2124.42
6.89 + 6.89 + 6.89 + 15.43 = 36.10
2150.34 – 2114.08 = 36.26
CHEM 430 – NMR Spectroscopy
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NMR - The
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SPECTRAL ANALYSIS
4-8
The tree diagram for Ha:
n
Jmed = 15.43 Hz
We
see
three
generational
iterations of the 6.89 similar to Hc
Jsmall = 6.89 Hz
We also see a new constant of a
similar constant to 15.51, at 15.43
Jsmall = 6.89 Hz
Jsmall = 6.89 Hz
CHEM 430 – NMR Spectroscopy
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NMR - The
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4-8
SPECTRAL ANALYSIS
From analysis of J-values we can finish the analysis:
Ha has coupling
constants of
15.51 and 6.89
Hb has coupling
constants of
1.64 and 15.51
Hc has coupling
constants of
1.72 and 6.97
Conclusion: Ha is coupled to Hb and Hc
Hb is coupled to Ha and Hc
Hc is coupled to Hb and Ha
Wow.
CHEM 430 – NMR Spectroscopy
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SPECTRAL ANALYSIS
Conclusion:
• On such a small system, the conclusion that all three of these proton families
are coupled to one another is trivial; on more complex systems, this is a
powerful tool to determine the position and relationship of protons on chains
or rings
• In this system, it helps us deduce which isomer we are observing by analysis of
the magnitude of the coupling constants that were generated
For each isomer, hypothesize as to what J’s would be observed:
Jtrans 1Hs = 11-18
Jallyl = 1-3
Jtrans = 15.5
Jcis 1Hs = 6-15 Hz
Jallyl = 1-3
CHEM 430 – NMR Spectroscopy
Jgeminal 1Hs = 1-3 Hz
No allylic coupling
81
NMR - The
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4-8
SPECTRAL ANALYSIS
Analysis of coupling constants in first-order spectra
Another Method:
1. From our adapted method:
Hoye, T. R.; Hanson, P. R. Vyvyan, J. R. J. Org. Chem. 1994, 59, 4096-4103
There is a second method for determining coupling constants – graphical
analysis
2. For first-order spectra, there are a finite number of combinations of an
observed 1H with a maximum number of interactions
3. Theoretically, the maximum 3J interactions would be nine:
4. In synthetically interesting systems (middle of chains, rings, etc.) it is
typically 2-4 couplings
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SPECTRAL ANALYSIS
Another Method:
5. The authors of the article suggest the method of visual pattern recognition
to save some of the tedium of the more analytical analysis we just covered
6. In the paper, they generated a series of tables that systematically cover the
most commonly encountered spin systems
7. For each table, an example compound is used to show where such a spin
pattern would be observed
8. It is suggested that you go through the paper, in conjunction with your text
(as a reference for representative J-values) to see how each of the tables
apply to the example the authors used
CHEM 430 – NMR Spectroscopy
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4-8
SPECTRAL ANALYSIS
Example:
∙ In each table, one coupling constant is
varied (Jz) while all others are held
constant
∙ As a consequence for any table you see a
symmetrical distribution of the multiplet
which converge as Jz approaches zero at
the bottom of each table
∙ 1Hs that are exemplified are highlighted
by a circle or square
∙ Note, as we have discussed the SJs is equal
to the distance between the outermost
peaks in a multiplet (Hz scale below each
graphical analysis)
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SPECTRAL ANALYSIS
Example:
∙ It is important to note, that as additional
coupling decreases, or certain couplings
become equal, the patterns simplify to
what we recognize from the n+1 rule
∙ Therefore, it should start to be clear that
the n+1 rule is a highly coincidental case
where all possible couplings (3J s) become
equal
∙ In other words, the n+1 rule is the
exception, and the first order relationships
we have just discussed are the rule
CHEM 430 – NMR Spectroscopy
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4-9
SECOND ORDER SPECTRA
• Second- order spectra are characterized by peak spacings that do not
correspond to coupling constants, by nonbinomial intensities, by chemical
shifts that are not at resonance midpoints, or by resonance multiplicities
that do not follow the n + 1 rule
• Even when the spectrum has the appearance of being first order, it may not
be.
• Lines can coincide in such a way that the spectrum assumes a simpler
appearance than seems consistent with the actual spectral parameters ( a
situation termed deceptive simplicity).
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SECOND ORDER SPECTRA
4-9
• For example, in the ABX spectrum, the X
nucleus is coupled to two nuclei ( A and
B) that are closely coupled (DnAB/J
<<<10) .
• Under these circumstances, the A and B
spin states are fully mixed, and X
responds as if the nuclei were
equivalent.
• Thus the ABX spectrum resembles an
A2X spectrum, as if JAX = JBX .
• In the ABX spectra at the right (a) DnAB
= 3.0 Hz, in (b) DnAB = 8.0 Hz
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4-9
• The AA'XX' spectrum often is observed as a deceptively simple pair of
triplets, resembling A2X2.
• In this case, it is the A and A' nuclei that are closely coupled ( and JAA' is
large).
• Such deceptive simplicity is not eliminated by raising the field because A and
A' are chemically equivalent.
• The chemist should beware of the pair of triplets that falsely suggests
magnetic equivalence ( A2X2) and equal couplings , when the molecular
structure suggests AA'XX'.
• Sometimes the couplings between A and X may be observed by lowering the
field to turn the AA'XX' spectrum into AA'BB' with a larger number of peaks
that may permit a complete analysis.
CHEM 430 – NMR Spectroscopy
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4-9
SECOND ORDER SPECTRA
• Another example of second-order complexity occurs in the ABX spectrum
(or, more generally, AxByXz) when A and B are very closely coupled, JAX is
large, and JBX is zero.
• With no coupling to B, the X spectrum should be a simple doublet from
coupling to A.
• Since A and B are closely coupled, however, the spin states of A and B are
mixed, and the X spectrum is perturbed by the B spins (the phenomenon has
been termed virtual coupling, which is something of a misnomer, since B is
not coupled to X).
CHEM 430 – NMR Spectroscopy
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4-9
SECOND ORDER SPECTRA
•
Methyl resonance
at 60 MHz
CHEM 430 – NMR Spectroscopy
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4-9
SECOND ORDER SPECTRA
• Sometimes proton spectra are second order even at 500 MHz or higher (
aside from the AA' case).
• Some institutions still have access only to iron core, 60 MHz spectrometers,
which produce largely second- order proton spectra.
• In each case, these spectra may be clarified somewhat by the use of
paramagnetic shift reagents. These molecules contain unpaired spins and
form Lewis acid– base complexes with dissolved substrates.
• The unpaired spin exerts a strong paramagnetic shielding effect (downfield)
on nuclei close to it. The effect drops off rapidly with distance, so that those
nuclei in the substrate that are closest to the site of acid–base binding are
affected more.
• Consequently, the shift to higher frequency varies through the substrate and
hence leads to greater separation of peaks.
CHEM 430 – NMR Spectroscopy
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SECOND ORDER SPECTRA
4-9
• Two common shift reagents contain lanthanides: tris( dipivalomethanato)
europium(III) · 2( tris(dipivalomethanato)europium(III)·2( pyridine) [called
Eu(dpm)3 without pyridine] and 1,1,1,2,2,3,3-heptafluoro-7,7-dimethyloctanedionatoeuropium( III) [or Eu( fod)3].
• Shift reagents are available with numerous rare earths as well as other elements -- Almost all organic functional groups that are Lewis bases have been
found to respond to these reagents.
• When the shift reagent is chiral, it can complex with enantiomers and
generate separate resonances from which enantiomeric ratios may be
obtained.
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4-1
FIRST ORDER SPECTRA
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4-1
FIRST ORDER SPECTRA
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4-1
FIRST ORDER SPECTRA
CHEM 430 – NMR Spectroscopy
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