CHEM 430 * NMR Spectroscopy

Report
Introduction to Spectroscopy
Spectroscopy is the study of the interaction of matter with the
electromagnetic (EM) spectrum
1.
EM radiation displays the properties of both particles and waves
2.
This “packet” of wave and particle properties is called a photon
The term “photon” is implied to mean a small, massless particle that
contains a small wave-packet of EM radiation/light
3.
The energy E component of a photon is proportional to the frequency
n
E = hn
The constant of proportionality is Plank’s constant, h
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Introduction to Spectroscopy
4.
Because the speed of light (c ) is constant, the frequency (n)
(number of cycles of the wave per second) can complete in the
same time, must be inversely proportional to how long the
oscillation is, or wavelength (l):
n=
c
___
l
D E = hn =
hc
___
l
5.
Amplitude describes the wave height, or strength of the oscillation
6.
Because the atomic particles in matter also exhibit wave and
particle properties (though opposite in how much) EM radiation
can interact with matter in two ways:
•
Collision – particle-to-particle – energy is lost as heat
•
Coupling – the wave property of the radiation matches
and movement
the wave property of the particle and “couple” to the
next higher quantum mechanical energy level
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Introduction to Spectroscopy
Remember atoms and molecules are quantum mechanical particles
9.
Where a photon is a wave with some particle character, matter is made of particles
with some wave character – wave/particle duality
10.
As a result of this, the energy of these particles can only exist at discrete energies
– we say these energy levels are quantized
11.
It is easy to understand if we visualize the
“wave” property of matter as an oscillating
string in a box—only certain “energy levels”
can exist as the string is bound at both ends:
CHEM 430 – NMR Spectroscopy
Energy
8.
4
The Spectroscopic Process
2. Absorption: Molecule takes on
the quantum energy of a photon
that matches the energy of a
transition and becomes excited
excited state
Energy
hn
5. Detection: Relaxation as
photons are reemitted.
Spectrometers differ whether
they measure actual emission or
absorbance
rest state
hn
hn
hn
1. Irradiation: Molecule is
bombarded with photons of
various frequencies over
CHEMdesired
430 – NMR Spectroscopy
the range
rest state
5
5
Types of Spectroscopy
g-rays
(X-ray cryst.)
X-rays
Frequency, n (Hz)
Wavelength, l
Energy (kcal/mol)
~1017
~0.01 nm
> 300
electronic
excitation
(p to p*)
molecular molecular
vibration rotation
Visible
nuclear
excitation
(PET)
core
electron
excitation
UV
~1015
10 nm
300-30
IR
Microwave
Nuclear Magnetic
Resonance
NMR & MRI
Radio
~1013
~1010
~105
1000 nm
0.01 cm
100 m
300-30
~10-4
~10-6
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6
NMR Spectroscopy
• NMR spectroscopy has emerged as the ultimate spectroscopic
method for organic structural analysis
• Currently, the development of novel NMR methods is in its
“golden age” with some of the 2-D methods entering their
maturation period as routine spectroscopic methods
• A typical NMR sample consists of 1-10 mg of sample, with which
a full analysis of 1H, 13C, DEPT, COSY, HMBC, HSQC and NOESY
could be done in a few hours on a high-field instrument
• Important spin-offs of NMR spectroscopy include a host of
medical and security imaging equipment
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Brief History of NMR
• First NMR spectrum of H2O, 1946:
Bloch, F.; Hansen, W. W.; Packard,
M. Phys. Rev. 1946, 70 474-85.
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Brief History of NMR
• First observation of chemical shift
1H
spectrum of ethanol – 1951 vs. 2011
Arnold, J.T., S.S. Dharmatti, and M.E.
Packard, J. Chem. Phys., 1951, 19, 507.
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Brief History of NMR
•
Fourier transform NMR by Ernst - 1966
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Brief History of NMR
•
2D NMR – 1975 Jeener and Ernst
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Brief History of NMR (MRI)
• First magnetic resonance image – 1973
Lauterbur and Mansfield
2011
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Brief History of NMR
• First 3-D spectrum of small protein - 1985 Wüthrich
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Brief History of NMR
Nobel Prizes for NMR
•1944 Physics Rabi (Columbia)
•1952 Physics Bloch (Stanford), Purcell (Harvard)
•1991 Chemistry Ernst (ETH)
•2002 Chemistry Wüthrich (ETH)
•2003 Medicine Lauterbur (University of Illinois in Urbana ) and
Mansfield (University of Nottingham)
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Basis of NMR Spectroscopy
2.1 Magnetic Properties of Nuclei
• The sub-atomic particles within atomic nuclei possess a spin
quantum number just like electrons
• Just as when using Hund’s rules to fill atomic orbitals with
electrons, nucleons must each have a unique set of quantum
numbers
• The spin quantum number of a nucleus is a physical constant, I
• For each nucleus, the total number of spin states allowed is given
by the equation:
2I + 1
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Basis of NMR Spectroscopy
Spin Quantum Numbers of Common Nuclei
Element
1H
2H
12C
13C
14N
16O
17O
19F
31P
35Cl
Nuclear Spin
Quantum Number
½
1
0
½
1
0
5/2
½
½
3/2
# of spin states
2
3
0
2
3
0
6
2
2
4
• Observe that for atoms with no net nuclear spin, there are zero
allowed spin states
• Nuclear Magnetic Resonance can only occur where there are
allowed spin states
• Note that two nuclei, prevalent in organic compounds have
allowed nuclear spin states – 1H and 13C, while two others do not
12C and 16O
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Basis of NMR Spectroscopy
2.1 Magnetic Properties of Nuclei
• There are three types of nuclei:
• No spin: I = O
•
12C, 16O
• Cannot be observed by NMR
• Spinning sphere: I = ½
•
1H, 13C, 15N, 19F, 29Si, 31P)
• Easiest to observe by NMR
• Spinning ellipsoid I = 1, 3/2, 2…
•
2H, 11B, 14N, 17O, 33S, 35Cl
• Difficult to observe by NMR
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Basis of NMR Spectroscopy
2.1 Magnetic Properties of Nuclei
• A nucleus contains protons, which each bear a +1 charge
• If the nucleus has a net nuclear spin, and an odd number of
protons, the rotation of the nucleus will generate a magnetic field
along the axis of rotation
m
I = +½
H
H
I = -½
m
• A hydrogen atom with its lone proton making up the nucleus, can
have two possible spin states—degenerate in energy
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Basis of NMR Spectroscopy
2.1 Magnetic Properties of Nuclei
• The magnitude of m varies from atom to atom :
m = għI
• ħ is Planck’s constant divided by 2p
• g is the characteristic gyromagnetic ratio of the nucleus
• The larger g is, the greater the magnetic moment
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Nuclear Magnetic Resonance
In the absence
of stimulus all
nuclear spin
sates are
degenerate
When a large
magnetic field
B0 is applied the
two spin states
become nondegenerate
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As B0 increases,
the larger DE
becomes
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Basis of NMR Spectroscopy
2.1 Magnetic Properties of Nuclei
• The large external magnetic field is defined as B0 in units of Tesla, T
• The axis of B0 is defined as the z-direction
• Splitting of spins into quantized groups is called the Zeeman effect
DE
direction
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0 +z Spectroscopy
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Basis of NMR Spectroscopy
2.1 Magnetic Properties of Nuclei
• The force of B0 causes m to move in a circular motion about the zdirection – precession
• B0 field in z-direction operates on the x component of m to create a
force in the y-direction (F = m X B0)
• This occurs with an angular frequency w0 known as the Larmor
frequency (rad s-1)
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Basis of NMR Spectroscopy
2.1 Magnetic Properties of Nuclei
• A quantum mechanical particle can absorb a photon of energy
equal to DE and become promoted to a higher state –
spectroscopic process
• As B0 increases so does w0 (B0  w); the constant of
proportionality is g:
w0 = g B0
• By equating w0 with Planck’s relationship:
DE = hn0 = ħw0 = g B0
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Basis of NMR Spectroscopy
2.1 Magnetic Properties of Nuclei
• What does this mean for the NMR experiment (observing DE)?
•
Solving for the frequency of EM radiation we are observing:
• For a bare hydrogen nucleus (H+), g = 267.53 (106 rad/T·s)
• In a B0 of 1.41 Tesla:
DE = 60 MHz
DE corresponds to the highly weak radio region of the EM spectrum:
l > 5 meters and energies of < 0.02 cal·mol-1
•
This causes technical challenges to observing NMR
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Basis of NMR Spectroscopy
2.1 Magnetic Properties of Nuclei
Boltzmann distribution – more problems with NMR observation
•DE at 60 MHz (DE = hn) is 2.39 x 10-5 kJ mol-1 (tiny) – thermal energy at room
temperature (298 oK) is sufficient to populate both energy levels
•DE is small, so rapid exchange is occurring between the two populations, but
there is always a net excess of protons in the lower energy state
•From the Boltzman’s Law we can calculate the population of each energy
state:
Nupper/Nlower = e-DE/kT = e-hn/kT
@ 298 oK the ratio is 1,000,000 / 1,000,009 !
There is an excess population of 9 nuclei in the lower energy state!
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Basis of NMR Spectroscopy
2.1 Magnetic Properties of Nuclei
• As the applied B0 increases, exchange becomes more difficult and the excess
increases:
Field, B0
T
Frequency
MHz
Excess nuclei
1.41
60
9
1.88
80
12
7.05
300
48
9.40
400
64
14.1
600
96
6. In each case, it is these few nuclei that allow us to observe NMR
7. When radio radiation is applied to a sample both transitions upward and
downward are stimulated – if too much radiation is applied both states
completely equilibrate – a state called saturation – no observed NMR signal
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Basis of NMR Spectroscopy
2.2 Commonly Studied Nuclides
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Basis of NMR Spectroscopy
2.2 Commonly Studied Nuclides
•
Remember that the greater DE the easier it is to detect NMR active
nuclei and have greater S/N ratios:
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Basis of NMR Spectroscopy
2.2 Commonly Studied Nuclides
1. Spin
•
•
In general spin ½ nuclei are the easiest to observe
Quadrapolar (I > ½) nuclei are more difficult to observe
•
Unique mechanism for relaxation gives very short relaxation time
•
Heisenberg uncertainty principle dictates:
DE Dt ~ ħ
•
As relaxation times become very short, the uncertainty in
energies becomes large and peaks broaden greatly
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Basis of NMR Spectroscopy
2.1 Magnetic Properties of Nuclei
2. Natural Abundance
• Remember from Boltzman’s Law we have only a small excess of nuclei
we can observe: 60 MHz, 1.41 T: 1,000,000 / 1,000,009 for 1H
•1H is 99.985% of natural hydrogen - only 9 excess nuclei
•Consider the excess population when the nuclei we are observing is 13C –
1.11% of natural carbon
•Spin couplings between low-abundant nuclei are also hampered:
•
•
Chance of two 1H-1H on adjacent Cs: 99.985 x 99.985 = 99.97%
Chance of two 13C-13C adjacent to one another: 1.11 x 1.11 = 0.1%
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Basis of NMR Spectroscopy
2.2 Commonly Studied Nuclides
3. Natural Sensitivity - g
• Remember - DE = hn0 = ħw0 = g B0
NMR signal a function of only B0 and g
•
The larger DE the greater the excess population for observation
(Boltzmann distribution)
•
1H
is best followed by 19F in routine observations
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Basis of NMR Spectroscopy
2.2 Commonly Studied Nuclides
4. Receptivity
• Mathematical product of abundance and g
•
Good quick measure of amenable a nuclei is for observation
•
Your text uses 13C as a guidepost rather than 1H
•
Quick survey:
•
•
•
1H
is 5680 times easier to observe than 13C
15N is 2.2% as easy to observe as 13C
19F is 4730 times easier to obersve than 13C; 31P is 2 times easier
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Basis of NMR Spectroscopy
2.3 The Chemical Shift
• Observation of the NMR phenomenon would be of little use if all
protons resonated at the same frequency
• In organic compounds protons are not bare nuclei, they are
surrounded by an s-orbital of containing an e- shared with an e- in
a hybridized orbital of another atom to form a covalent bond
• In the presence of an external magnetic field, an induced
circulation of electrons opposite to that of a proton is observed
since the two are of opposite charges
• This induced circulation generates a magnetic field in opposition to
the applied magnetic field – a local diamagnetic current
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Basis of NMR Spectroscopy
2.3 The Chemical Shift
• Since the magnetic field “felt” by the proton within this electron cloud is
lowered, the resonance condition frequency is also lowered
• This effect of lowering the energy of transition by a cloud of electrons is
called diamagnetic shielding or shielding - represented as s
• The opposite effect – if electron density is removed from the vicinity of
the proton is called deshielding
• The actual field around the nucleus becomes B0(1 – s); substitution in the
energy equation gives:
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Basis of NMR Spectroscopy
2.3 The Chemical Shift
• Since the magnetic field “felt” by the proton within this electron cloud is
lowered, the resonance condition frequency is also lowered
• This effect of lowering the energy of transition by a cloud of electrons is
called diamagnetic shielding or shielding
• The opposite effect – if electron density is removed from the vicinity of
the proton is called deshielding
DE
direction
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0 +z Spectroscopy
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NMR Spectroscopy Introduction
2-6
Basis
of NMR Spectroscopy
• In acetic acid, the –CH3 protons
are in an e- rich environment
relative to the –OH proton.
• Shielding
of the electrons
opposes B0 and therefore DE is
lower than that observed for
the –OH proton. Here DE is
large as the full effect of B0 is
felt
DE –CH3
DE –OH
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direction
0 +z Spectroscopy
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DE
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B0
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Basis of NMR Spectroscopy
• The effect of electrons on a 1.41 T magnetic field is negligible, but
measurable
• Compare the resonance frequencies for the protons in fluoromethane vs.
chloromethane
CH3F
CH3Cl
• The stronger inductive w/d of electrons by fluorine reduces the resonance
frequency by 72 Hz (not MHz) compared to an operating frequency of the
instrument at 60 MHz @ 1.41 T – barely 1 part per million (ppm)
• Using units of 60000072 vs. 60000000 is clunky at best
• There needs to be a reference “proton” by which these “chemical shifts”
can be related - the best candidate would be a completely deshielded
proton (H+) which does not exist in the solution phase
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Basis of NMR Spectroscopy
• NMR spectroscopists chose the other end of the spectrum- a proton that
was more shielded than any other known proton (at the time) – those in
tetramethylsilane (TMS)
• The 12 chemically identical protons in TMS were used as the standard zero
for an NMR spectrum
• The resonance frequency of any proton to be studied (since all were less
shielded) would be at parts per million of the operating frequency of the
instrument greater than this zero
• This allowed NMR instruments of varying field (and thus operating
frequency) strengths to use the same scale
• Here’s how:
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Basis of NMR Spectroscopy
• In an applied field of 1.41 T, the resonance frequency for a typical proton is
60 MHz, at 2.35 T it is at 100 MHz – a ratio of 5/3
• Thus, for a given proton, the shift in Hz from the TMS standard should be 5/3
greater in the 100 MHz instrument compared to the 60 MHz
• Since these are simple ratios, we can simply factor out the effect of field
strength by defining d, or chemical shift to be
d =
(shift from TMS in Hz)
(spectrometer frequency in MHz)
…or ppm of the instruments operating frequency
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Basis of NMR Spectroscopy
90 Mhz spectrum
90 Hz
300 MHz spectrum
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300 Hz
41
Basis of NMR Spectroscopy
• In an applied field of 1.41 T, the resonance frequency for a typical proton is
60 MHz, at 2.35 T it is at 100 MHz – a ratio of 5/3
• Thus, for a given proton, the shift in Hz from the TMS standard should be 5/3
greater in the 100 MHz instrument compared to the 60 MHz
• Since these are simple ratios, we can simply factor out the effect of field
strength by defining d, or chemical shift to be
d =
(shift from TMS in Hz)
(spectrometer frequency in MHz)
…or ppm of the instruments operating frequency
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Basis of NMR Spectroscopy
• A detailed study of chemical shifts is the basis of Chapter 3
CH3
H
Si
O
H3C
C
H
C
CH
CH3 3
O
C
H
O
H
S
H
O
R
H
O
O
H
O
C
C
H
10
R
R
H
C
C
Ph
15
C
9
8
7
downfield
deshielded
higher DE
6
5
d (ppm)
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R = H or alkyl
EWG
3
2
1
0.0
upfield
shielded
lower DE
43
Spectrometer Design
Basis of NMR Spectroscopy
Continuous-Wave (CW) Instrument
An NMR spectrometer needs to perform several functions:
• Generate a high (>1 Tesla) magnetic field to split the energy levels of the
spin states enough to:
– Create an excess nuclei population large enough to observe
– Make the radio n that correspond to the transition be
observable
• Ensure that the field is homogeneous (shimming)
• Vary either the applied field or the radiofrequency (RF) to observe
different nuclei at their various energies of transition
• Receive the faint signal of the relaxation of the excited nuclei to their
ground state
• Process the signal into a usable spectrum vs. a reference
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Basis of NMR Spectroscopy
Continuous-Wave (CW) Instrument:
RF Detector
RF (60 MHz)
oscillator
Permanent
Magnet
Variable magnetic field – 1.41 T ± few millionths of T
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Basis of NMR Spectroscopy
How it works (CW NMR):
• The sample is placed in a 5 mm solution cell or tube (experimental aspects
we will cover shortly) in the center of a large permanent or electromagnet
• A RF oscillator coil at 90° to the sample generates a radio signal at the
operating frequency of the instrument (60 MHz for a 1.41 T field)
• The overall magnetic field is varied by a small electromagnet capping the
poles of the larger field magnet
• Remember: DE = n = (g/2p) B0, so variations of either magnetic field or
frequency will cover the observed spectral width if the other is held
constant
• As with older dispersive IR instruments, the sweep of magnetic fields is
simultaneous with the movement of the chart paper
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Basis of NMR Spectroscopy
How it works (CW NMR):
• As a particular proton population comes into resonance, a second receiver
coil at 90° to the transmitter coil will pick up the change in orientation of
nuclear spin
• This is recorded by the chart as a voltage response, proportional to the
size of the proton population that generated the resonance
• One artifact of CW instruments is that the relaxation of the protons is
slower than the movement (sweep) of the chart paper
• This causes the ringing effect – a decreasing oscillation of the signal after
the spectrometer has moved past a given resonance
• CW instruments operate by bringing each individual population of protons
into resonance individually.
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Basis of NMR Spectroscopy
Limitations- CW NMR:
• Since the spectrum is collected once, the sample must possess enough
protons to give a suitable excess population that can be observed – need a
concentrated sample
• Due to the limitations of the relatively low magnetic field (CW instruments
top out at 60-90 MHz) the coupling constants for JHH are relatively large
compared to the spectral width – so only simple molecules can be
observed and their structures elucidated
• For nuclei of lower magnetogyric ratios, g, or natural abundance (13C most
specifically) the ratio of radio noise to signal is high
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Basis of NMR Spectroscopy
2.4 Excitation and Relaxation – FT-NMR
• Let’s revisit the NMR experiment we considered earlier
• Recall that at even a high applied B0 only a small excess of nuclei are in
the lower spin state (+ ½ ) as per the Boltzman distribution
Ex: @ 7.04 T an excess of 50 spins per million
• It is important to also note that 298 K imparts enough energy to the
system such that all spins are interchanging rapidly:
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Basis of NMR Spectroscopy
2.4 Excitation and Relaxation – FT-NMR
• So when we are discussing an NMR sample we have quadrillions of
protons creating a net magnetization M in the z direction (B0)
• The xy components are distributed randomly, and we can think of
the rest state as M = Mz
z
z
M
y
x
y
Bo
x
• M precesses at the Larmor frequency. We are using the rotating
frame of reference to make the visualization easier.
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Basis of NMR Spectroscopy
2.4 Excitation and Relaxation – FT-NMR
• Important note:
• Each individual spin can only exist in either quantized state (±
½) separated by DE
• The bulk magnetization however can exist at a continuum of states
z
M
y
Bo
x
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Basis of NMR Spectroscopy
2.4 Excitation and Relaxation – FT-NMR
• If a RF (w0 = w) is applied that creates a magnetic oscillation along
the x-axis a torque is applied to M rotating it towards y
z
z
Mo
y
B1
B1
y
(or off-resonance)
x
Mxy
x
wo
wo
• With the application of energy, a small amount of the excess
population has flipped spins. Mxy < M0
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Basis of NMR Spectroscopy
2.4 Excitation and Relaxation – FT-NMR
• If RF energy continues to be applied the excess of + ½ nuclei
disappears – sample at saturation – no NMR signal can be
detected
• Natural mechanisms return an excess population to equilibrium
• Any process that returns the z-magnetization to equilibrium with
an excess of + ½ spins is called spin-lattice or longitudinal
relaxation
• It is usually a first-order process and with a time constant of T1
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Basis of NMR Spectroscopy
2.4 Excitation and Relaxation – FT-NMR
• Other magnetic nuclei in the NMR sample tumble, generating a
field from its motion
• If the field is at w, excess spin energy can pass to this motional
energy as – ½ nuclei become + ½ nuclei
• For this process to be effective the nuclei need to be spatially
proximate to the tumbling molecule.
• Most H-atoms are on the outside of molecules and have roughly
equal probability to relax by this process - #1Hs ~ signal
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Basis of NMR Spectroscopy
2.4 Excitation and Relaxation – FT-NMR
• C-atoms differ substantially in their contact with external
molecules – T1 relaxation usually by attached protons
• Therefore T1 relaxation is more efficient for a –13CH3 rather than a
quaternary carbon or 13C=O.
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Basis of NMR Spectroscopy
2.4 Excitation and Relaxation – FT-NMR
• T1 relaxation is responsible for generating the excess of + ½ nuclei
in the first place!
• When the sample is first placed in the B0 field, all spins are
degenerate
• Magnetization builds up as spins flip from the effect of interactions
with surrounding magnetized nuclei
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Basis of NMR Spectroscopy
2.4 Excitation and Relaxation – FT-NMR
•Any process that returns xy magnetization to its equilibrium position of
zero is called spin-spin or transverse relaxation
•It is a first order process with time constant T2
•By definition T2 < T1
•For T2 relaxation the phases of nuclear spins must become randomized
•The mechanism for this is when spin is transferred between nuclei of
opposite spin
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Basis of NMR Spectroscopy
2.4 Excitation and Relaxation – FT-NMR
•When one spin goes from +1/2 to -1/2 while the other goes from -1/2 to
+1/2 there is no net change in z magnetization
•However switches in spin cause dephasing; as the process continues xy
magnetization disappears
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Basis of NMR Spectroscopy
2.4 Excitation and Relaxation – FT-NMR
•Detection – as xy magnetization dephases the vectors begin to rotate
about z in the rotating frame of reference
•A second coil 90o to the transmitter coil detects the decay in the xy
z
y
Mxy
wo
x
Receiver coil (x)  NMR signal
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Basis of NMR Spectroscopy
2.4 Excitation and Relaxation – FT-NMR
•The NMR process:
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Basis of NMR Spectroscopy
2.4 Excitation and Relaxation – FT-NMR
•Introduction to Pulse Sequences
•Shorthand:
90x
Mixing time
Detection
Pulse
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Pulsed Fourier Transform (FT) Instrument
First, what is a Fourier Transform?
Fourier transforms interconvert mathematical functions in the frequency domain to the
time domain:

f(n) = ∫ f(t) e-int dt
-

f(t) = ½ p ∫ f(n) eint dt
-
For purposes of this discussion, we will black box the actual calculations and derivations
of these functions, but we need to understand what they do
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Spectrometer Design
Pulsed Fourier Transform (FT) Instrument
If we feed two simple oscillating equations into a FT, here are the results:
f (t) = cos (nt)
FT
-n
+n
f (t) = sin (nt)
FT
+n
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-n
63
Spectrometer Design
NMR Spectroscopy
Pulsed Fourier Transform (FT) Instrument
• In the FT instrument, all proton populations are excited simultaneously by a short,
intense burst of RF energy
• Due to a variation of the Heisenberg Uncertainty Principle, even if the RF generator
is set at 90 MHz, if the duration of the pulse is short, the radio waves do not have
time to establish a solid fundamental frequency
• This can be illustrated by the following cartoon, showing the combination of a short
pulse being added to a step function:
on
+
off
on
off
=
off
off
tp = pulse duration
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Spectrometer Design
Pulsed Fourier Transform (FT) Instrument
4. If this short pulse is converted into the frequency domain by a FT:
on
FT
off
off
tp
n
Observe that we have a continuum of frequency content centered at the
operating frequency of the instrument
We will talk more about the effects of pulse time and width when we
discuss advanced 1-D and 2-D NMR
CHEM 430 – NMR Spectroscopy
65
Spectrometer Design
Pulsed Fourier Transform (FT) Instrument
5. For now, if a sample containing one unique population of hydrogens was excited
over tp by a pulse, it would then “relax” back to its original spin state
6. As each nuclei relaxes it will emit RF radiation of a given frequency; since different
nuclei will relax at different rates, the signal decays over time
7. This emission is recorded by the spectrometer as a free-induction decay or FID
off
off
CHEM 430 – NMR Spectroscopy
66
Spectrometer Design
Pulsed Fourier Transform (FT) Instrument
8. The actual frequency of the FID is the interference signal of the relaxing protons
superimposed with the frequency of the RF source
9. Conversion of this decay signal by FT back into the frequency domain gives us the
actual n of resonance for the proton being observed
Pulse n
Proton signal
FT
n
time
10.Again to due Heisenburg and other factors, NMR signals are not single lines, but a
Lorentzian shaped continuum of lines centered at the n of the signal
CHEM 430 – NMR Spectroscopy
67
Spectrometer Design
Pulsed Fourier Transform (FT) Instrument
Advantages:
Since all nuclei are excited and observed simultaneously, the pulse can be repeated
after each relaxation period (for 1H, about 10 seconds) and the resulting signals
added together
Because we are observing weak radiofrequency signals in a sea of RF noise for dilute
samples (or those observed once as in CW NMR) noise becomes an issue
If several to hundreds of FIDs are added together, signals will tend to constructively
add together and become more pronounced; since noise is random, it will tend to
destructively add and become less pronounced
Signal to noise ratio improves as a function of the square root of the scans (FIDs)
performed: S/N = f (n)
CHEM 430 – NMR Spectroscopy
68
Basis of NMR Spectroscopy
2.4 Excitation and Relaxation – FT-NMR
•Back to Pulse Sequences
•Standard 1-D observation of 1H or other nuclei
90x
Detection
Delay
Pulse
n = scans
•Delay must be longer than T1 (and T2 < T1)
•Scans are repeated until S/N ratio is high enough
CHEM 430 – NMR Spectroscopy
69
Basis of NMR Spectroscopy
2.4 Excitation and Relaxation – FT-NMR
•How can you know what T1 is?
•We cannot measure it directly – in the z-axis
•Pulse sequence is used called inversion recovery:
180y (or x)
90y
tD
CHEM 430 – NMR Spectroscopy
70
NMR Spectroscopy
Basis of NMR Spectroscopy
2-6
2.4 Excitation and Relaxation – FT-NMR
tD = 0
z
z
x
y
tD > 0
x FT
y
z
z
x
y
tD >> 0
x FT
y
z
z
x
y
x FT
y
CHEM 430 – NMR Spectroscopy
71
NMR Spectroscopy
2-6
Basis of NMR Spectroscopy
Intensity, (I)
• If we plot the intensity versus time we get the following curve:
time
I(t) = I * ( 1 - 2 * e - t / T1 )
• It is an exponential with a time constant equal to the T1 relaxation
time.
• Most 1Hs decay in less than 10 seconds – no need to run routinely
CHEM 430 – NMR Spectroscopy
72
NMR Spectroscopy
Basis of NMR Spectroscopy
2-6
2.4 Excitation and Relaxation – FT-NMR
•How can you know what T2 is?
•We cannot measure it directly like T1
•Pulse sequence is used called spin-echo sequence
90y
180y (or x)
tD
CHEM 430 – NMR Spectroscopy
tD
73
NMR Spectroscopy
Basis of NMR Spectroscopy
2-6
2.4 Excitation and Relaxation – FT-NMR
•What it does:
z
y
y
tD
x 
x
x
y
y
dephasing
y
tD
x
x
180y (or x)
refocusing
CHEM 430 – NMR Spectroscopy
74
NMR Spectroscopy
2-6
Basis of NMR Spectroscopy
2.4 Excitation and Relaxation – FT-NMR
• If we acquire the FID right after the spin-echo sequence, the intensity of
the signal after FT will only be affected by T2 relaxation and not by
dephasing due to B0 imperfections.
• Upon repetition for different tD values, we plot the intensity versus 2 * tD
and get a graph similar to the one we got for inversion recovery, but in this
case the decay rate will be equal to T2.
CHEM 430 – NMR Spectroscopy
75
The Coupling Constant
• Consider the spectrum of ethyl alcohol:
• Why does each resonance “split” into smaller peaks?
HO
CH3
C
H2
CHEM 430 – NMR Spectroscopy
76
The Coupling Constant
• The magnetic effects of nuclei in close proximity to those being observed
have an effect on the local magnetic field, and therefore DE
• Two examples of close proximity are geminal and vicinal protons
(homonuclear) and protons attached to 13C (heteronuclear):
• On any one of the 108 of these molecules in a typical NMR sample, there is
an equal statistical probability each proton is either spin + ½ or – ½
CHEM 430 – NMR Spectroscopy
77
NMR Spectroscopy
2-6
The Coupling Constant
• This creates two different magnetic environments for a proton being
observed – one where its neighbor is +½ the other where its neighbor is –
½
• If there is more than one proton on an adjacent carbon – all the statistical
probabilities exist that each one is either + ½ or – ½ in spin
• The summation of these effects over all of the observed nuclei in the
sample is observed as the spin-spin splitting of resonances
CHEM 430 – NMR Spectroscopy
78
The Coupling Constant
• Consider 1-chloro-4-nitrobenzene
If we are observing the resonance for the HA proton,
in half the molecules in the sample HX would be –½
in the other half the spin would be +½
We would see the resonance of HA “split” into two
different resonances (energy states):
CHEM 430 – NMR Spectroscopy
79
The Coupling Constant
• Consider 1-chloro-4-nitrobenzene
Likewise, if we are observing the resonance for the
HX proton, in half the molecules in the sample HA
would be –½ in the other half the spin would be +½
We would see the resonance of HX “split” into two
different resonances (energy states):
CHEM 430 – NMR Spectroscopy
80
The Coupling Constant
• Consider 1-chloro-4-nitrobenzene
The observed 1H NMR spectrum shows two doublets,
one for HA the other for HX
HA and HX are described as a spin-system
CHEM 430 – NMR Spectroscopy
81
NMR Spectroscopy
The Coupling Constant
2-6
• The influence of neighboring spins on peak multiplicity is called spin-spin
coupling, indirect coupling or J-coupling
• The difference between the component peaks of a resonance is a measure
of how strong the interaction is between adjacent nuclei
• This difference is called the coupling constant, J, measured in Hz
# of bonds
separating nuclei
3J
H-cis Nuclei that are coupled
and stereochemistry where
appropriate
CHEM 430 – NMR Spectroscopy
82
NMR Spectroscopy Introduction
2-6
The Coupling Constant
60 MHz – propyl bromide
• Since J is a measure of
J-values are =
interaction between two
nuclei, it must be the same for
both nuclei
J-values are =
• Similarly, J is independent of
the applied B0 and will have
the same value regardless of
the field strength of the NMR
300 MHz – propyl bromide
CHEM 430 – NMR Spectroscopy
83
NMR Spectroscopy Introduction
2-6
The Coupling Constant
60 MHz – propyl bromide
• Since J is a measure of
J-values are =
interaction between two
nuclei, it must be the same for
both nuclei
J-values are =
• Similarly, J is independent of
the applied B0 and will have
the same value regardless of
the field strength of the NMR
300 MHz – propyl bromide
CHEM 430 – NMR Spectroscopy
84
NMR Spectroscopy
2-6
The Coupling Constant
• Recall, we are observing the frequency (E = hn) where a proton goes into
resonance
Any change in B0 will cause a change in
energy at which the resonance condition will
occur for a proton of a given chemical shift
CHEM 430 – NMR Spectroscopy
85
NMR Spectroscopy
2-6
The Coupling Constant
• In solution we are not looking at a single molecule but about 108
• On some molecules the proton being observed may be next to another
proton of spin + 1/2 :
CHEM 430 – NMR Spectroscopy
86
NMR Spectroscopy
2-6
The Coupling Constant
• On some molecules the proton being observed may be next to another
proton of spin – 1/2 :
CHEM 430 – NMR Spectroscopy
87
NMR Spectroscopy Introduction
2-6
The Coupling Constant
• Mechanism for coupling: usually oversimplified in early studies of
NMR - If molecules can rotate in solution any spatial effect of one
protons magnetic field on another is averaged to zero
• The most common mechanism involves the interaction of electrons
along the bonding path between the nuclei.
• Electrons, like protons, act like spinning particles and have a
magnetic moment. The X proton ( HX) influences, or polarizes, the
spins of its surrounding electrons, making the electron spins favor
one Iz state very slightly.
CHEM 430
88
NMR Spectroscopy Introduction
2-6
The Coupling Constant
• Thus, a proton of spin +½ polarizes the electron to –½. The electron in
turn polarizes the other electron of the bond, and so on, finally
reaching the resonating A proton (HA).
• Because J normally represents an interaction through bonds, it is a
useful parameter for drawing conclusions about molecular bonding,
such as bond order and stereochemistry.
CHEM 430
89
The Coupling Constant
• Observe what effect this has on an isolated ethyl group:
• The two methylene Ha protons have three neighbors, Hb, on the adjacent
methyl carbon
• Each one of these hydrogens can be + ½ or – ½ , and since we are not
looking at one molecule, but billions, we will observe all combinations
CHEM 430 – NMR Spectroscopy
90
The Coupling Constant
•
The first possibility is that all three Hb protons have a + ½ spin; in this case the
three protons combine to generate three small magnetic fields that aid B0 and
deshield the protons – pushing part of the resonance for Ha slightly downfield (the
magnetic field of a proton is tiny compared to B0)
All 3 Hb
protons + ½
+++
Resonance, n, in absence of coupling
CHEM 430 – NMR Spectroscopy
91
The Coupling Constant
•
The second possibility is that two Hb protons have a + ½ spin and the third a - ½ ;
in this case the two protons combine to enhance B0 and the other against it, a net
deshielding; there are 3 different combinations that generate this state
or
3 combinations of
+½, +½ , -½
+++
or
-++
+-+
++-
Resonance, n, in absence of coupling
CHEM 430 – NMR Spectroscopy
92
The Coupling Constant
•
The third possibility is that two Hb protons have a –½ spin and the third +½; here,
the two protons combine to reduce B0 and the other enforce it, a net shielding
effect; there are 3 different combinations that generate this state
or
3 combinations of
+½, -½ , -½
+++
-++
+-+
++-
or
--+
-++--
Resonance, n, in absence of coupling
CHEM 430 – NMR Spectroscopy
93
The Coupling Constant
•
The last possibility is that all three Hb protons have a – ½ spin; in this case the three
protons combine to oppose B0, a net shielding effect; there is one combination
that generates this state:
All 3 Hb
protons -½
+++
-++
+-+
++-
--+
-++--
---
Resonance, n, in absence of coupling
CHEM 430 – NMR Spectroscopy
94
The Coupling Constant
•
The result is instead of one resonance (peak) for Ha, the peak is “split” into four, a
quartet, with the constituent peaks having a ratio of 1:3:3:1 centered at the d (n)
for the resonance :
-CHa2- is ‘split’
into a quartet by
three adj. Hbs
+++
-++
+-+
++-
--+
-++--
---
Resonance, n, in absence of coupling
CHEM 430 – NMR Spectroscopy
95
NMR Spectroscopy
•
2-6
The Coupling Constant
Similarly, the Hb protons having two protons, on the adjacent carbon each
producing a magnetic field, cause the Hb resonance to be split into a triplet
-CHb3- is ‘split’
into a triplet by two
adj. Has
-+
++ +- -Resonance, n, in absence of coupling
CHEM 430 – NMR Spectroscopy
96
NMR Spectroscopy
The Coupling Constant
2-6
• Rather than having to do this exercise for every situation, it is quickly
recognized that a given family of equivalent protons (in the absence of
other spin-coupling) will have its resonance split into a multiplet
containing n+1 peaks, where n is the number of hydrogens on carbons
adjacent to the carbon bearing the proton giving the resonance – this is
the n + 1 rule
# of Hs on
adj. C’s
Multiplet
# of
peaks
0
singlet
1
1
1
doublet
2
1 1
2
triplet
3
1 2 1
3
quartet
4
1 3 3 1
4
quintet
5
1 4 6 4 1
5
sextet
6
1 5 10 10 5 1
6
septet
7
1 6 15 20 15 6 1
The relative ratios of the peaks are a mathematical
progression given by Pascal’s triangle:
CHEM 430 – NMR Spectroscopy
97
NMR Spectroscopy
The Coupling Constant
2-6
• Common patterns:
tert-butyl - singlet
methyl - singlet
ethyl – quartet - triplet
n-propyl – triplet - quintet - triplet
iso-propyl – septet - doublet
CHEM 430 – NMR Spectroscopy
98
NMR Spectroscopy
2-6
The Coupling Constant
CHEM 430 – NMR Spectroscopy
99
NMR Spectroscopy
2-6
The Coupling Constant
CHEM 430 – NMR Spectroscopy
100
NMR Spectroscopy
2-6
The Coupling Constant
• Heteronuclear coupling between 1H and 13C are not apparent in 1H spectra
(13C low abundance - 1.1%), In 99 of 100 cases, 1H are attached to
nonmagnetic 12C atoms.
• In the 13C spectrum, carbon nuclei are coupled to 1H directly (99%
abundant) attached to the carbon.
• Thus, the 13C resonance of a methyl carbon is split into a quartet, that of a
methylene carbon CH2 into a triplet, and that of a methine carbon CH into a
doublet. A quaternary carbon is not split by one bond coupling.
CHEM 430
101
NMR Spectroscopy
2-6
The Coupling Constant
• Shown here (top) the 13C spectrum
of 3- hydroxybutyric acid which
contains a carbon resonance with
each type of multiplicity.
undecoupled
• From right to left are seen a quartet
CH3 , a triplet CH2 , a doublet CH ,
and a singlet CO2H (C=O)
• Instrumental procedures, called
decoupling, are available by which
spin– spin split-tings may be
removed. These methods, discussed
in Section 5- 3, involve irradiating
one
CHEM 430
decoupled
102
NMR Spectroscopy
2-6
The Coupling Constant
• Instrumental procedures, called
decoupling, are available by which
spin– spin splittings are removed for
clarity.
undecoupled
• These methods involve irradiating
one nucleus with an additional field
B2 while observing another nucleus
resonating in the B1 field.
•
13C
spectra are usually run decoupled
as other spectral techniques are
used to establish 1H-13C connectivity
more rapidly and with more clarity
CHEM 430
decoupled
103
The NMR Spectrum - 1H
NMR Spectroscopy
Spin-spin splitting – 1H NMR
The next level of complexity (which we will cover in detail in Chapter 4) is when protons on
adjacent carbons exert different J’s than one another.
Consider the ethylene fragment:
The influence of the geminal-relationship
is over the shortest distance
The magnetic influence of the transrelationship is over the longest distance
The cis-relationship, is over an
intermediate distance
CHEM 430 – NMR Spectroscopy
104
The NMR Spectrum - 1H
NMR Spectroscopy
Spin-spin splitting – 1H NMR
For ethylene we would then observe three chemically distinct resonances with spin-spin
splitting exerted by the other two protons:
J couplings:
2J
gem
= 0 – 1 Hz
The observed
multiplet for Ha
is a “doublet of
doublets”
3J
AC
3J
trans
= 11- 18 Hz
3J
AB
3J
cis
3J
AB
= 6 - 15 Hz
CHEM 430 – NMR Spectroscopy
105
The NMR Spectrum - 1H
NMR Spectroscopy
Spin-spin splitting – 1H NMR
Similar behavior is observed with aromatic rings; since the ring structure is fairly rigid and
electronic effects are conducted over a longer distance, J – couplings are observed across
the ring system:
3J
ortho
4J
meta
5J
para
In low-field 1H NMR the signal for this proton
would be split into a doublet by the proton
ortho to it.
On a high field instrument one finds this
3J
4
5
ortho as well as a Jmeta and a J para from
the effect of the protons meta and para to it
Typically:
3J
ortho = 7-10 Hz
4J
meta = 1-3 Hz
5J
para = 0-1 Hz
CHEM 430 – NMR Spectroscopy
106
The NMR Spectrum - 1H
NMR Spectroscopy
Spin-spin splitting – 1H NMR
For our initial treatment of 1H NMR the alkenyl, aromatic and the following J values should
be learned:
3J
3J
= 6-8
= 8-14
3J
a,e = 0-7
3J
e,e = 0-5
3J
trans
= 11-18
3J
= 6-15
cis
a,a
3J
= 8-11
3J
= 5-7
3J
= 4-8
3J = 6-12
cis
trans
3J
3J
allyl
= 4-10
ortho =
7-10 Hz
meta = 1-3 Hz
5J
para = 0-1 Hz
4J
3J
= 4-8
3J = 6-12
cis
CHEM 430 – NMR Spectroscopy
trans
107
The NMR Spectrum - 1H
A typical 1H NMR is recorded from -2 to 15 d (ppm); what is typically reported is the
region from 0 to 10 d
Remember, if a proton is shielded (e- circulation reduces “felt” magnetic field) DE for the
transition is lowered and the signal is near the high field or upfield region of the
spectrum (right)
If the proton is deshielded (e- circulation doesn’t reduce the “felt” magnetic field) DE
for the transition is raised and the signal is near the low field or downfield region of
the spectrum (left)
high DE
deshielded 1H sees full B0
downfield
10
or ppm
CHEM 430 –dNMR
Spectroscopy
low DE
shielded 1H reduces B0
upfield
0
108
The NMR Spectrum - 1H
The number of signals observed will be equal to the number of unique populations of
chemically equivalent protons
To determine if two protons are chemically equivalent, substitute “X” for that each
respective hydrogen in the compound and compare the structures
If the two structures are fully superimposible (identical) the two hydrogens are
chemically equivalent; if the two structures are different the two hydrogens were not
chemically equivalent
CH3
H
A simple example: p-xylene
CH3
X
H
CH3
Same structure
H
CH3
CHEM 430 – NMR Spectroscopy
CH3
X
H
CH3
109
The NMR Spectrum - 1H
The position (v) of each resonance is dependant on the electronic environment around
the proton – chemical shift as a result of local diamagnetic shielding
There are three principle effects that contribute to local diamagnetic shielding:
1) Electronegativity
2) Hybridization
3) Proton acidity/exchange
CHEM 430 – NMR Spectroscopy
110
The NMR Spectrum - 1H
Local Diamagnetic Shielding - Electronegativity
1. Electronegative groups comprise most organic functionalities:
-F
-Cl -Br -I
-OH -OR -NH2
-NHR -NR2 -NH3+
-C=O
-PO3H2 -SH -Ph -C=C
-NO2-NO -SO3H
and most others
In all cases, the inductive w/d of electrons of these groups decreases the electron
density in the C-H covalent bond – proton is deshielded – higher DE of transition
CHEM 430 – NMR Spectroscopy
111
The NMR Spectrum - 1H
Local Diamagnetic Shielding - Electronegativity
2. Protons bound to carbons bearing electron withdrawing groups are deshielded based
on the magnitude of the withdrawing effect – Pauling electronegativity:
CH3F
CH3O-
CH3Cl
CH3Br
CH3I
CH4
(CH3)4Si
Pauling
Electronegativity
4.0
3.5
3.1
2.8
2.5
2.1
1.8
d of H
4.26
3.40
3.05
2.68
2.16
0.23
0.0
CHEM 430 – NMR Spectroscopy
112
The NMR Spectrum - 1H
Local Diamagnetic Shielding - Electronegativity
3. The magnitude of the withdrawing effect is cumulative:
CH3Cl
CH2Cl2
CHCl3
3.05
5.30
7.27
d of H
4. The magnitude of the withdrawing effect is reduced by distance, as the inductive
model suggests
d of H
-CH2Br
-CH2CH2Br
-CH2CH2CH2Br
3.30
1.69
1.25
CHEM 430 – NMR Spectroscopy
113
The NMR Spectrum - 1H
NMR Spectroscopy
Local Diamagnetic Shielding - Hybridization
1. The hybridization of the carbon the proton is bound exerts a strong electronic effect
2. The greater the s-character, the more tightly bound the electrons are to carbon, raising
its effective electronegativity (sp = 50% s, sp2, 33% s and sp3 25% s)
Type of H
Name of H
Chemical Shift, d
R-CH3, R2CH2, R3CH
alkyl
0.8-1.7
C=C-CH3
allyl
1.6-2.6
CC-H
Acetylenic
2.0-3.0
C=C-H
Vinylic
4.6-5.7
Ar-H
aromatic
6.5-8.5
O=C-H
aldehydic
9.5-10.1
CHEM 430 – NMR Spectroscopy
Something odd
is happening
here, as we will
discuss
114
The NMR Spectrum - 1H
NMR Spectroscopy
Local Diamagnetic Shielding - Proton Acidity/Exchange
1. If an organic molecule that possesses hydrogen atoms of low pKA are dissolved in a
deuterated solvent that also has a low pKA, the “visible” protons will exchange with
“deuterium” from solvent and become “invisible” to the NMR spectrometer
OH
OD
D2O
Such studies are useful, if it is desired to see which H-atoms on an organic are acidic!
CHEM 430 – NMR Spectroscopy
115
The NMR Spectrum - 1H
NMR Spectroscopy
Local Diamagnetic Shielding - Proton Acidity/Exchange
2. Due to H-bonding effects, the resonance for certain functional groups (esp. –OH and –
NH2) can change drastically dependent on concentration and the extent of the H-
bonding
3. Just as in IR spectroscopy, peaks corresponding to these resonances are broad and
often undefined – observing a continuum of bond strengths/electron densities about
the observed proton
4. The correlation tables for the position of such protons tend to be broad and unreliable:
•
•
•
•
•
•
Acid –OH
Phenol –OH
Alcohol –OH
Amine –NH2
Amide –NH2
Enol CH=CH-OH
10.5-12.0 d
4.0-12.0 d
0.5-5.0 d
0.5-5.0 d
5.0-8.0 d
>15 d
CHEM 430 – NMR Spectroscopy
116
NMR Spectroscopy
Some observed 1H resonances can not be fully explained by local diamagnetic shielding
effects
Magnetic Anisotropy – literally “magnetic dissimilarity”
For example, by our hybridization model, a proton bound to an sp2 C should be observed
at lower d than a proton bound to an sp C
Type of H
Name of H
Chemical Shift, d
R-CH3, R2CH2, R3CH
alkyl
0.8-1.7
C=C-CH3
allyl
1.6-2.6
CC-H
Acetylenic
2.0-3.0
C=C-H
Vinylic
4.6-5.7
Ar-H
aromatic
6.5-8.5
O=C-H
aldehydic
9.5-10.1
CHEM 430 – NMR Spectroscopy
117
The NMR Spectrum - 1H
NMR Spectroscopy
Magnetic Anisotropy –
1. This effect is primary due to the fact that there is an additional effect of circulating
electrons, observed in p-systems
2. In benzene, the 6-p-orbitals overlap to allow full circulation of electrons; as these
electrons circulate in the applied magnetic field they oppose the applied magnetic field
at the center – just like the circulation of electrons in the 1-s orbital about hydrogen –
at the middle!:
B0
CHEM 430 – NMR Spectroscopy
118
The NMR Spectrum - 1H
NMR Spectroscopy
Magnetic Anisotropy –
3. On the periphery of the ring, the effect is opposite – the magnetic effect reinforces the
applied B0, and DE becomes greater – deshielding effect
CHEM 430 – NMR Spectroscopy
119
The NMR Spectrum - 1H
NMR Spectroscopy
Magnetic Anisotropy –
4. This theory can easily be tested by the observation of large aromatic systems that
possess protons inside the ring (now a shielding effect):
-1.8 d
8.9 d
Or over a ring system:
H
H
2.0 d
H2
C
CH2
CHEM 430 – NMR Spectroscopy
-1.0 d
120
The NMR Spectrum - 1H
NMR Spectroscopy
Magnetic Anisotropy –
5. In alkynes, a similar situation (to the central protons in large aromatic systems) arises
where the terminal proton is in the region of maximum shielding
CHEM 430 – NMR Spectroscopy
121
The NMR Spectrum - 1H
NMR Spectroscopy
General Correlation Chart – 1H NMR
Due to the three effects on local diamagnetic shielding, in conjunction with the effect of
magnetic anisotropy 1H NMR chemical shifts are variable
• Avoid using hard and fast rules (tables of numbers)
• Instead, start from the general correlation table and deduce structural features
based on the effects just discussed
• After a structural inference has been made, then use the more specific correlation
tables to confirm the analysis
CHEM 430 – NMR Spectroscopy
122
The NMR Spectrum - 1H
NMR Spectroscopy
General Correlation Chart – 1H NMR
CH3
Here are the general regions for 1H chemical shifts:
H
Si
O
H3C
C
H
C
CH
CH3 3
O
C
H
O
H
S
H
R
O
H
O
O
H
O
H
C
C
Ph
15
10
9
8
7
downfield
deshielded
higher DE
6
R
R
H
C
C
C
5
4
d (ppm)
CHEM 430 – NMR Spectroscopy
R = H or alkyl
EWG
3
2
1
0.0
upfield
shielded
lower DE
123
The NMR Spectrum - 1H
NMR Spectroscopy
Spin-spin splitting – 1H NMR
1. The magnetic effects of nuclei in close proximity to those being observed have an
effect on the local magnetic field, and therefore DE
2. Specifically, when proton is close enough to another proton, typically by being on an
adjacent carbon (vicinal), it can “feel” the magnetic effects generated by that proton
3. On any one of the 108 of these molecules in a typical NMR sample, there is an equal
statistical probability that the adjacent (vicinal) proton is either in the + ½ or – ½ spin
state
4. If there is more than one proton on an adjacent carbon – all the statistical probabilities
exist that each one is either + ½ or – ½ in spin
5. The summation of these effects over all of the observed nuclei in the sample is
observed as the spin-spin splitting of resonances
CHEM 430 – NMR Spectroscopy
124
Intensity of Signals—Integration
•
The area under an NMR signal is proportional to the number of absorbing
protons
•
An NMR spectrometer automatically integrates the area under the peaks, and
prints out a stepped curve (integral) on the spectrum
•
The height of each step is proportional to the area under the peak, which in
turn is proportional to the number of absorbing protons
•
Modern NMR spectrometers automatically calculate and plot the value of
each integral in arbitrary units
•
The ratio of integrals to one another gives the ratio of absorbing protons in a
spectrum; note that this gives a ratio, and not the absolute number, of
absorbing protons
CHEM 430 – NMR Spectroscopy
125
The NMR Spectrum - 1H
NMR Spectroscopy
Integration – 1H NMR
1. Like instrumental chromatography, in NMR spectroscopy, the area under a peak (or
multiplet) is proportional to the number of protons in the sample that generated that
particular resonance
2. The NMR spectrometer typically will print this information on the spectrum as an
integral line (stepped line on the spectrum below)
3. The height of the integral is proportional to that proton population; by comparing the
ratios of the integrals on an NMR spectrum you can determine the number of protons
as a least common multiple of these ratios
CHEM 430 – NMR Spectroscopy
126
The NMR Spectrum - 1H
NMR Spectroscopy
Integration – 1H NMR
4. For example observe the integration of the ethanol spectrum below:
HO
CH3
C
H2
-CH3
3.75 units high
-OH
1.25 units high
2.5 units high
CHEM 430 – NMR Spectroscopy
127
Intensity of Signals—Integration
CHEM 430 – NMR Spectroscopy
128
Intensity of Signals—Integration
CHEM 430 – NMR Spectroscopy
129
Nuclear Magnetic Resonance
When a nuclei of
spin +½
encounters a
photon where
n = E/h, the two
“couple”
The nuclei “tips”
its spin state and
is now opposed
to B0
CHEM 430 – NMR Spectroscopy
The nuclei
“relaxes” and
returns to the
+ ½ spin state
130
Basis of NMR Spectroscopy
2.1 Excellent Simulators for NMR Phenomenon
• 2-D compass needle analogy:
http://www.drcmr.dk/JavaCompass/
• 3-D Bloch equation simulator
http://www.drcmr.dk/BlochSimulator/
CHEM 430 – NMR Spectroscopy
131
General Theory
NMR Spectroscopy
Nuclear Spin States
• The sub-atomic particles within atomic nuclei possess a spin quantum number just
like electrons
• As with electrons, the nucleons are organized in energy levels
• Just as when using Hund’s rules to fill atomic orbitals with electrons, nucleons must
each have a unique set of quantum numbers
• The total spin quantum number of a nucleus is a physical constant, I
• For each nucleus, the total number of spin states allowed is given by the equation:
2I + 1
CHEM 430 – NMR Spectroscopy
132
General Theory
NMR Spectroscopy
Nuclear Spin States
Spin Quantum Numbers of Common Nuclei
Element
1H
2H
12C
13C
14N
16O
17O
19F
31P
35Cl
Nuclear Spin
Quantum Number
½
1
0
½
1
0
5/2
½
½
3/2
# of spin states
2
3
0
2
3
0
6
2
2
4
6. Observe that for atoms with no net nuclear spin, there are zero allowed spin states
7. All the spin states of a given nucleus are degenerate in energy
CHEM 430 – NMR Spectroscopy
133
General Theory
NMR Spectroscopy
Nuclear Magnetic Moments
• A nucleus contains protons, which each bear a +1 charge
• If the nucleus has a net nuclear spin, and an odd number of protons, the rotation of
the nucleus will generate a magnetic field along the axis of rotation
• Thus, a nucleus has a magnetic moment, m, generated by its charge and spin
• A hydrogen atom with its lone proton making up the nucleus, can have two possible
spin states, degenerate in energy
m
+½
H
m
H
CHEM 430 – NMR Spectroscopy
-½
134
General Theory
Nuclear Spin States
5. In the presence of an externally applied magnetic field, these two spin states are no
longer degenerate in energy
6. The spin opposed orientation is slightly higher in energy than the spin aligned
orientation
m
-½
H
B0 – externally applied magnetic field
DE
m
+½
H
CHEM 430 – NMR Spectroscopy
135
General Theory
Absorption of Energy
• The energy difference between the two non-degenerate spin states in the presence
of an applied magnetic field is quantized
• At low B0 is easy to surmise that the potential energy of the spin opposed state
would be low, and as B0 grows in strength, so would the potential energy
• Thus, with increasing strength of B0, DE between the two spin states also increases
-1/2
-1/2
+1/2
DE
+1/2
CHEM 430 –BNMR
Spectroscopy
0 – increasing
136
General Theory
NMR Spectroscopy
Absorption of Energy
4. From theory we have already discussed, we say that a quantum mechanical particle
can absorb a photon of energy equal to DE and become promoted to the higher
state
5. This energy is proportional to the frequency of the photon absorbed, and in the
case of nuclear spin, is a function of the magnetic field applied:
DE = hn = f (B0)
6. Every nucleus has a different ratio of m to angular momentum (each has a different
charge and mass) – this is referred to as the magnetogyric ratio, g
DE = hn = f (gB0)
Angular momentum is quantized in units of h/2p, thus:
DE = hn = g (h/2p)B
CHEM
0 430 – NMR Spectroscopy
137
General Theory
Absorption of Energy
7. Solving for the frequency of EM radiation we are observing:
DE = n = (g/2p) B0
8. For a bare hydrogen nucleus (H+), g = 267.53 (106 radians/T·sec)
9. In a field strength of 1 Tesla, DE = 42.5 MHz (for our discussion, at 1.41 T, DE = 60
MHz)
DE = hn = f (B0)
This energy difference corresponds to the highly weak radio frequency region of the
EM spectrum – with wavelengths of >5 meters equal to < 0.02 cal·mol-1
g-rays
X-rays
UV
IR
Microwave Radio
CHEM 430 – NMR Spectroscopy
138
General Theory
Mechanism of absorption – nuclear magnetic resonance
• What we are actually observing for DE is the precessional or Larmor frequency (w)
of the spinning nucleus – this is analogous to a spinning toy top precessing as a
result of the influence of the earth’s magnetic field:
w
B0
CHEM 430 – NMR Spectroscopy
139
General Theory
NMR Spectroscopy
Mechanism of absorption – Nuclear Magnetic Resonance
2. When a photon of n = 60 MHz encounters this spinning charged system (a bare
proton) the two can couple and change the spin state of the proton
This state is called nuclear
magnetic resonance, and the
nucleus is said to be in
resonance with the incoming
radio wave
w
DE
B0
n=w
w
CHEM 430 – NMR Spectroscopy
140
General Theory
NMR Spectroscopy
Mechanism of absorption – Nuclear Magnetic Resonance
3. The energy difference corresponding to 60 MHz (DE = hn) is 2.39 x 10-5 kJ mol-1
(tiny) – thermal energy at room temperature (298 oK) is sufficient to populate both
energy levels
4. The energy difference is small, so rapid exchange is occurring between the two
populations, but there is always a net excess of protons in the lower energy state
5. From the Boltzman distribution equation we can calculate the population of each
energy state:
Nupper/Nlower = e-DE/kT = e-hn/kT
@ 298 oK the ratio is 1,000,000 / 1,000,009 !
There is an excess population of 9 nuclei in the lower energy state!
CHEM 430 – NMR Spectroscopy
141
General Theory
Mechanism of absorption – Nuclear Magnetic Resonance
6. As the applied B0 increases, exchange becomes more difficult and the excess
increases:
Frequency (MHz)
Excess nuclei
60
9
80
12
100
16
200
32
300
48
600
96
7. In each case, it is these few nuclei that allow us to observe NMR
8. When radio radiation is applied to a sample both transitions upward and downward
are stimulated – if too much radiation is applied both states completely equilibrate
– a state called saturation – no NMR signal can be observed
CHEM 430 – NMR Spectroscopy
142
General Theory
NMR Spectroscopy
Chemical Shift
6. Spectroscopic observation of the NMR phenomenon would be of little use if all
protons resonated at the same frequency
7. The protons in organic compounds are not bare nuclei, they are surrounded by an s
-orbital of containing an electron shared with an electron in a hybridized orbital of
another atom to form a covalent bond
8. In the presence of an external magnetic field, an induced circulation of electrons
opposite to that of a proton is observed since the two are of opposite charges
9. This induced circulation generates a magnetic field in opposition to the applied
magnetic field – a local diamagnetic current
CHEM 430 – NMR Spectroscopy
143
Basis of NMR Spectroscopy
2.1 Magnetic Properties of Nuclei
• The magnitude of m from atom to atom varies:
m = għI
• ħ is Planck’s constant divided by 2p
• g is the characteristic gyromagnetic ratio of the nucleus
• The larger g is, the greater the magnetic moment
CHEM 430 – NMR Spectroscopy
144
Nuclear Magnetic Resonance
• For the 1H nucleus (proton) this resonance condition occurs at low
energy (lots of noise) unless a very large magnetic field is applied
• Early NMR spectrometers used a large permanent magnet with a
field of 1.4 Tesla—protons undergo resonance at 60 MHz (1 MHz =
106 Hz)
• Modern instruments use a large superconducting magnet—our
NMR operates at 9.4 T where proton resonance occurs at 400 MHz
• In short, higher field gives cleaner spectra and allows longer and
more detailed experiments to be performed
CHEM 430 – NMR Spectroscopy
145
Origin of the Chemical Shift
Electrons
surrounding the
nucleus are
opposite in charge
to the proton,
therefore they
generate an
opposing b0
Deshieding
Shielding
Factors which lower edensity allow the
nucleus to “see” more
of the B0 being
applied – resonance
occurs at higher
energy
Factors which raise
e- density reduce
the amount of B0
the nucleus “sees”
– resonance
condition occurs at
lower energy
CHEM 430 – NMR Spectroscopy
146
The Proton (1H) NMR Spectrum
CHEM 430 – NMR Spectroscopy
147
The 1H NMR Spectrum
 A reference compound is needed—one that is inert and does not
interfere with other resonances
 Chemists chose a compound with a large number of highly shielded
protons—tetramethylsilane (TMS)
 No matter what spectrometer is used the resonance for the protons on
this compound is set to d 0.00
CH3
Si
H3C
CH
CH3 3
CHEM 430 – NMR Spectroscopy
148
The 1H NMR Spectrum
 The chemical shift for a given proton is in frequency units (Hz)
 This value will change depending on the B0 of the particular spectrometer
 By reporting the NMR absorption as a fraction of the NMR operating
frequency, we get units, ppm, that are independent of the spectrometer
CHEM 430 – NMR Spectroscopy
149
The 1H NMR Spectrum
• We need to consider four aspects of a 1H spectrum:
a.
b.
c.
d.
Number of signals
Position of signals
Intensity of signals.
Spin-spin splitting of signals
CHEM 430 – NMR Spectroscopy
150
The Number of Signals
 The number of NMR signals equals the number of different types of
protons in a compound
 Protons in different environments give different NMR signals
 Equivalent protons give the same NMR signal
CHEM 430 – NMR Spectroscopy
151
The Number of Signals



To determine if two protons are chemically equivalent, substitute “X” for
that each respective hydrogen in the compound and compare the
structures
If the two structures are fully superimposible (identical) the two
hydrogens are chemically equivalent; if the two structures are different
the two hydrogens were not equivalent
CH 3
A simple example: p-xylene
H
CH 3
Z
H
CH 3
Same Compound
H
CH 3
CH 3
Z
CHEM 430 – NMR Spectroscopy
H
CH 3
152
The Number of Signals
• Examples
Important: To determine equivalent protons in cycloalkanes and alkenes, always draw
all bonds to show specific stereochemistry:
CHEM 430 – NMR Spectroscopy
153
The Number of Signals
 In comparing two H atoms on a ring or double bond, two protons are
equivalent only if they are cis or trans to the same groups.
CHEM 430 – NMR Spectroscopy
154
The Number of Signals
 Proton equivalency in cycloalkanes can be determined similarly:
CHEM 430 – NMR Spectroscopy
155
The Number of Signals
 Enantiotopic Protons – when substitution of two H atoms by Z forms
enantiomers:
a. The two H atoms are equivalent and give the same NMR signal
b. These two atoms are called enantiotopic
CHEM 430 – NMR Spectroscopy
156
The Number of Signals
 Diastereotopic Protons - when substitution of two H atoms by Z forms
diastereomers
a. The two H atoms are not equivalent and give two NMR signals
b. These two atoms are called diastereotopic
CHEM 430 – NMR Spectroscopy
157
Chemical Shift – Position of Signals
• Remember:
Electrons
surrounding the
nucleus are
opposite in charge
to the proton,
therefore they
generate an
opposing b0
Deshieding
Shielding
Factors which lower
e- density allow the
nucleus to “see”
more of the B0 being
applied – resonance
occurs at higher
energy
Factors which raise
e- density reduce the
amount of B0 the
nucleus “sees” –
resonance condition
occurs at lower
energy
CHEM 430 – NMR Spectroscopy
158
Chemical Shift – Position of Signals
• The less shielded the nucleus becomes, the more of the applied magnetic
field (B0) it feels
• This deshielded nucleus experiences a higher magnetic field strength, to it
needs a higher frequency to achieve resonance
• Higher frequency is to the left in an NMR spectrum, toward higher
chemical shift—so deshielding shifts an absorption downfield
 Downfield, deshielded
CHEM 430 – NMR Spectroscopy
Upfield, shielded 
159
Chemical Shift – Position of Signals
• There are three principle effects that contribute to local
diamagnetic shielding:
a. Electronegativity
b. Hybridization
c. Proton acidity/exchange
CHEM 430 – NMR Spectroscopy
160
Chemical Shift – Position of Signals
 Electronegative groups comprise most organic functionalities:
-F
-Cl -Br -I
-NHR -NR2 -NH3+
-PO3H2 -SH -Ph -C=C
-OH -OR -NH2
-C=O
-NO2-NO -SO3H
and most others
In all cases, the inductive WD of electrons of these groups
decreases the electron density in the C-H covalent bond –
proton is deshielded – signal more downfield of TMS
CHEM 430 – NMR Spectroscopy
161
Chemical Shift – Position of Signals
 Protons bound to carbons bearing electron withdrawing groups are
deshielded based on the magnitude of the withdrawing effect – Pauling
electronegativity:
CH3F
CH3O-
CH3Cl
CH3Br
CH3I
CH4
(CH3)4Si
Pauling
Electronegativity
4.0
3.5
3.1
2.8
2.5
2.1
1.8
d of H
4.26
3.40
3.05
2.68
2.16
0.23
0.0
CHEM 430 – NMR Spectroscopy
162
Chemical Shift – Position of Signals
3. The magnitude of the deshielding effect is cumulative:
CH3Cl
CH2Cl2
CHCl3
3.05
5.30
7.27
d of H
As more chlorines are added d becomes larger
3. The magnitude of the deshielding effect is reduced by distance, as the
inductive model suggests
d of H
-CH2Br
-CH2CH2Br
-CH2CH2CH2Br
3.30
1.69
1.25
CHEM 430 – NMR Spectroscopy
163
Chemical Shift – Position of Signals
Hybridization
 Increasing s-character (sp3  sp2  sp) pulls e- density closer to nucleus
effectively raising electronegativity of the carbon the H atoms are bound
to – a deshielding effect
 We would assume that H atoms on sp carbons should be well downfield
(high d) and those on sp3 carbons should be upfield (low d)
CHEM 430 – NMR Spectroscopy
164
Chemical Shift – Position of Signals
• What we observe is slightly different:
Type of H
Carbon
hybridization
Name of H
Chemical Shift, d
R-CH3, R2CH2, R3CH
sp3
alkyl
0.8-1.7
C=C-CH3
sp3
allyl
1.6-2.6
CC-H
sp
acetylenic
2.0-3.0
C=C-H
sp2
vinylic
4.6-5.7
Ar-H
sp2
aromatic
6.5-8.5
O=C-H
sp2
aldehydic
9.5-10.1
Chemists refer to this observation as magnetic anisotropy
CHEM 430 – NMR Spectroscopy
165
Chemical Shift – Position of Signals
 Magnetic Anisotropy – Aromatic Protons
a. In a magnetic field, the six p electrons in benzene circulate around
b.
c.
the ring creating a ring current.
The magnetic field induced by these moving electrons reinforces the
applied magnetic field in the vicinity of the protons.
The protons thus feel a stronger magnetic field and a higher
frequency is needed for resonance. Thus they are deshielded and
absorb downfield.
CHEM 430 – NMR Spectroscopy
166
Chemical Shift – Position of Signals
• Similarly this effect operates in alkenes:
CHEM 430 – NMR Spectroscopy
167
Chemical Shift – Position of Signals
• In alkynes there are two perpendicular sets of p-electrons—the molecule
orients with the field lengthwise—opposing B0 shielding the terminal H atom
CHEM 430 – NMR Spectroscopy
168
Chemical Shift – Position of Signals
CHEM 430 – NMR Spectroscopy
169
Chemical Shift – Position of Signals
CHEM 430 – NMR Spectroscopy
170
Intensity of Signals—Integration
 The area under an NMR signal is proportional to the number of
absorbing protons
 An NMR spectrometer automatically integrates the area under the
peaks, and prints out a stepped curve (integral) on the spectrum
 The height of each step is proportional to the area under the peak,
which in turn is proportional to the number of absorbing protons
 Modern NMR spectrometers automatically calculate and plot the value
of each integral in arbitrary units
 The ratio of integrals to one another gives the ratio of absorbing
protons in a spectrum; note that this gives a ratio, and not the absolute
number, of absorbing protons
CHEM 430 – NMR Spectroscopy
171
Intensity of Signals—Integration
CHEM 430 – NMR Spectroscopy
172
Intensity of Signals—Integration
CHEM 430 – NMR Spectroscopy
173
Spin-Spin Splitting
• Consider the spectrum of ethyl alcohol:
• Why does each resonance “split” into smaller peaks?
HO
CH3
C
H2
CHEM 430 – NMR Spectroscopy
174
Spin-Spin Splitting
 The magnetic effects of nuclei in close proximity to those being
observed have an effect on the local magnetic field, and therefore DE
 Specifically, when proton is close enough to another proton, typically
by being on an adjacent carbon (vicinal), it can “feel” the magnetic
effects generated by that proton
 On any one of the 108 of these molecules in a typical NMR sample,
there is an equal statistical probability that the adjacent (vicinal)
proton is either in the + ½ or – ½ spin state
 If there is more than one proton on an adjacent carbon – all the
statistical probabilities exist that each one is either + ½ or – ½ in spin
 The summation of these effects over all of the observed nuclei in the
sample is observed as the spin-spin splitting of resonances
CHEM 430 – NMR Spectroscopy
175
Spin-Spin Splitting
• Recall, we are observing the frequency (E = hn) where a proton goes
into resonance
Any change in B0 will cause a change in
energy at which the resonance condition will
occur for a proton of a given chemical shift
CHEM 430 – NMR Spectroscopy
176
 In solution we are not looking at a single molecule but about 108
 On some molecules the proton being observed may be next to another
proton of spin + 1/2 :
CHEM 430 – NMR Spectroscopy
177
Spin-Spin Splitting
• On some molecules the proton being observed may be next to
another proton of spin – 1/2 :
CHEM 430 – NMR Spectroscopy
178
Spin-Spin Splitting
 Observe what effect this has on an isolated ethyl group:
 The two methylene Ha protons have three neighbors, Hb, on the
adjacent methyl carbon
 Each one of these hydrogens can be + ½ or – ½ , and since we are not
looking at one molecule, but billions, we will observe all combinations
CHEM 430 – NMR Spectroscopy
179
Spin-Spin Splitting
 The first possibility is that all three Hb protons have a + ½ spin; in this
case the three protons combine to generate three small magnetic fields
that aid B0 and deshield the protons – pushing the resonance for Ha
slightly downfield (the magnetic field of a proton is tiny compared to B0)
All 3 Hb
protons + ½
CHEM 430 – NMR Spectroscopy
resonance for Ha in absence of spin-spin splitting
180
Spin-Spin Splitting
 The second possibility is that two Hb protons have a + ½ spin and the third
a - ½ ; in this case the two protons combine to enhance B0 and the other
against it, a net deshielding;
there are 3 different combinations that generate this state
or
or
2 Hb
protons + ½
CHEM 430 – NMR Spectroscopy
resonance for Ha in absence of spin-spin splitting
181
Spin-Spin Splitting
 The third possibility is that two Hb protons have a –½ spin and the third
+½; here, the two protons combine to reduce B0 and the other enforce
it, a net shielding effect;
there are 3 different combinations that generate this state
or
or
2 Hb
protons - ½
CHEM 430 – NMR Spectroscopy
resonance for Ha in absence of spin-spin splitting
182
Spin-Spin Splitting
 The last possibility is that all three Hb protons have a – ½ spin; in this case
the three protons combine to oppose B0, a net shielding effect;
there is one combination that generates this state
All 3 Hb
protons - ½
CHEM 430 – NMR Spectroscopy
resonance for Ha in absence of spin-spin splitting
183
Spin-Spin Splitting
 The result is instead of one resonance (peak) for Ha, the peak is “split”
into four, a quartet, with the constituent peaks having a ratio of 1:3:3:1
centered at the d (n) for the resonance
CHEM 430 – NMR Spectroscopy
resonance for Ha in absence of spin-spin splitting
184
Spin-Spin Splitting
 Similarly, the Hb protons having two protons, on the adjacent carbon each
producing a magnetic field, cause the Hb resonance to be split into a
triplet
CHEM 430 – NMR Spectroscopy
resonance for Ha in absence of spin-spin splitting
185
Spin-Spin Splitting
 Rather than having to do this exercise for every situation, it is quickly
recognized that a given family of equivalent protons (in the absence of other
spin-coupling) will have its resonance split into a multiplet containing n+1
peaks, where n is the number of hydrogens on carbons adjacent to the carbon
bearing the proton giving the resonance – this is the n + 1 rule
# of Hs
on adj.
C’s
Multiplet
# of
peaks
0
singlet
1
1
1
doublet
2
1 1
2
triplet
3
1 2 1
3
quartet
4
1 3 3 1
4
quintet
5
1 4 6 4 1
5
sextet
6
1 5 10 10 5 1
6
septet
7 CHEM 430 – NMR Spectroscopy
The relative ratios of the peaks are a
mathematical progression given by Pascal’s
triangle:
1 6 15 20 15 6 1
186
1H
NMR—Spin-Spin Splitting
• Common patterns:
tert-butyl - singlet
methyl - singlet
ethyl – quartet - triplet
n-propyl – triplet - quintet - triplet
iso-propyl – septet - doublet
CHEM 430 – NMR Spectroscopy
187
1H
NMR—Spin-Spin Splitting
CHEM 430 – NMR Spectroscopy
188
1H
NMR—Spin-Spin Splitting
 Another Example:
Br
Br
Br
=
Br
C
Br
Ha
C
Br
Hb Hb
CHEM 430 – NMR Spectroscopy
189
1H
NMR—Spin-Spin Splitting
 Another Example:
CHEM 430 – NMR Spectroscopy
190
1H
NMR—Spin-Spin Splitting
Three general rules describe the splitting patterns commonly
seen in the 1H NMR spectra of organic compounds:
1.Equivalent protons do not split each other’s signals
2.A set of n nonequivalent protons splits the signal of a
nearby proton into n + 1 peaks
3.Splitting is observed for nonequivalent protons on the same
carbon or adjacent carbons
If Ha and Hb are not equivalent, splitting is observed when:
CHEM 430 – NMR Spectroscopy
191
1H
NMR—Spin-Spin Splitting
• Magnetic influence falls off dramatically with distance
• The n + 1 rule only works in the following situations:
H H
Aliphatic compounds that have free
rotation about each bond
H H
G
H
Ha
H
Hb
Hc
Aromatic compounds where each proton is
held in position relative to one another
CHEM 430 – NMR Spectroscopy
192
1H
NMR—Spin-Spin Splitting
 The amount of influence exerted by a proton on an adjacent
carbon is observed as the difference (in Hz) between component
peaks within the multiplet it generates. This influence is
quantified as the coupling constant, J
 Two sets of protons that split one another are said to be “coupled”
 J for two sets of protons that are coupled are equivalent—
therefore on complex spectra we can tell what is next to what
This J
Is equal to this J
-CH2-
CHEM 430 – NMR Spectroscopy
-CH3
193
1H
NMR—Spin-Spin Splitting
 The next level of complexity (which at this level, is only introduced) is
when protons on adjacent carbons exert different J’s than one another.
 Consider the ethylene fragment:
The influence of the geminal-relationship is
over the shortest distance
The magnetic influence of the transrelationship is over the longest distance
The cis-relationship, is over an
intermediate distance
CHEM 430 – NMR Spectroscopy
194
1H
NMR—Spin-Spin Splitting
• For this substituted ethylene we see the following spectrum:
2J
gem
= 0 – 1 Hz
The observed multiplet
for Ha is a “doublet of
doublets”
3J
AC
3J
trans
= 11- 18 Hz
3J
AB
3J
cis
3J
AB
= 6 - 15 Hz
CHEM 430 – NMR Spectroscopy
195
1H
NMR—Spin-Spin Splitting
 In general, when two sets of adjacent protons are different from each
other (n protons on one adjacent carbon and m protons on the other),
the number of peaks in an NMR signal = (n + 1)(M + 1)
 In general the value of J falls off with distance; J values have been
tabulated for virtually all alkene, aromatic and aliphatic ring systems
CHEM 430 – NMR Spectroscopy
196
1H
NMR—Spin-Spin Splitting
• Some common J-values
3J
3J
= 6-8
= 8-14
3J
a,e = 0-7
3J
e,e = 0-5
3J
trans
= 11-18
3J
= 6-15
cis
a,a
3J
allyl
= 4-10
3J
= 8-11
3J
= 5-7
3J
= 4-8
3J = 6-12
cis
CHEM 430 – NMR Spectroscopy
trans
3J
= 4-8
3J = 6-12
cis
trans
3J
ortho = 7-10 Hz
4J
meta = 1-3 Hz
5J
para = 0-1 Hz
197
1H
NMR—Spin-Spin Splitting
• We can now tell stereoisomers apart through 1H NMR:
CHEM 430 – NMR Spectroscopy
198
1H
NMR—Spin-Spin Splitting
• A combined example:
CHEM 430 – NMR Spectroscopy
199
1H
NMR—Spin-Spin Splitting
• Under usual conditions, an OH proton does not split the NMR signal
of adjacent protons
• Protons on electronegative atoms rapidly exchange between
molecules in the presence of trace amounts of acid or base (usually
with NH and OH protons)
CHEM 430 – NMR Spectroscopy
200
Structure Determination
CHEM 430 – NMR Spectroscopy
201
Structure Determination
CHEM 430 – NMR Spectroscopy
202
Structure Determination
CHEM 430 – NMR Spectroscopy
203
Structure Determination
CHEM 430 – NMR Spectroscopy
204
13C
NMR
• The lack of splitting in a 13C spectrum is a consequence of the low
natural abundance of 13C
• Recall that splitting occurs when two NMR active nuclei—like two
protons—are close to each other. Because of the low natural
abundance of 13C nuclei (1.1%), the chance of two 13C nuclei being
bonded to each other is very small (0.01%), and so no carboncarbon splitting is observed
• A 13C NMR signal can also be split by nearby protons. This 1H-13C
splitting is usually eliminated from the spectrum by using an
instrumental technique that decouples the proton-carbon
interactions, so that every peak in a 13C NMR spectrum appears as a
singlet
• The two features of a 13C NMR spectrum that provide the most
structural information are the number of signals observed and the
chemical shifts of those signals
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205
13C
NMR
CHEM 430 – NMR Spectroscopy
206
13C
NMR
• The number of signals in a 13C spectrum gives the number of different
types of carbon atoms in a molecule.
• Because 13C NMR signals are not split, the number of signals equals the
number of lines in the 13C spectrum.
• In contrast to the 1H NMR situation, peak intensity is not proportional to
the number of absorbing carbons, so 13C NMR signals are not integrated.
CHEM 430 – NMR Spectroscopy
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13C
NMR
• In contrast to the small range of chemical shifts in 1H NMR (1-10
ppm usually), 13C NMR absorptions occur over a much broader range
(0-220 ppm).
• The chemical shifts of carbon atoms in 13C NMR depend on the same
effects as the chemical shifts of protons in 1H NMR.
CHEM 430 – NMR Spectroscopy
208
13C
NMR
CHEM 430 – NMR Spectroscopy
209
13C
NMR
CHEM 430 – NMR Spectroscopy
210
Shoolery Tables
• After years of collective observation of 1H and 13C NMR it is possible to
predict chemical shift to a fair precision using Shoolery Tables
• These tables use a base value for 1H and 13C chemical shift to which are
added adjustment increments for each group on the carbon atom
H
X C H
H
methyl
H
X C Y
H
methylene
CHEM 430 – NMR Spectroscopy
H
X C Z
Y
methine
211
Shoolery Values for Methylene
X or Y
Substituent
Constant
X or Y
Substituent
Constant
-H
0.34
-OC(=O)OR
3.01
-CH3
0.68
-OC(=O)Ph
3.27
-C—C
1.32
-C(=O)R
1.50
-CC-
1.44
-C(=O)Ph
1.90
-Ph
1.83
-C(=O)OR
1.46
-CF2-
1.12
-C(=O)NR2 or
H2
1.47
-CF3
1.14
-CN
1.59
-F
3.30
-NR2 or H2
1.57
-Cl
2.53
-NHPh
2.04
-Br
2.33
-NHC(=O)R
2.27
-I
2.19
-OH
2.56
CHEM 430 – NMR Spectroscopy
-N
3
1.97
-NO2
3.36
212
Shoolery Values for Methine
X ,Y or Z
Substituent
Constant
X, Y or Z
Substituent
Constant
-F
1.59
-OC(=O)OR
0.47
-Cl
1.56
-C(=O)R
0.47
-Br
1.53
-C(=O)Ph
1.22
-NO2
1.84
-CN
0.66
-NR2 or H2
0.64
-C(=O)NH2
0.60
-NH3+
1.34
-SR or H
0.61
NHC(=O)R
1.80
-OSO2R
0.94
-OH
1.14
-CC-
0.79
-OR
1.14
-C=C
0.46
-C(=O)OR
2.07
-OPh
1.79
-Ph
CHEM 430 – NMR Spectroscopy
0.99
213
Shoolery Tables
• For methyl—use methylene formula and table
using the –H value
• For methylene—use a base value of 0.23 and
add the two substituent constants for X and Y
In 92% of cases experimental is within 0.2 ppm
• For methine—use a base value of 2.50 and add
the three substituent constants for X, Y and Z
Error similar to methylene
CHEM 430 – NMR Spectroscopy
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Shoolery Tables
• Work for aromatics as well (.pdf posted)
CHEM 430 – NMR Spectroscopy
215
Running an NMR Experiment
• Sample sizes for a typical high-field NMR (300-600 MHz):
• 1-10 mg for 1H NMR
• 10-50 mg for 13C NMR
• Solution phase NMR experiments are much simpler to run; solid-
phase NMR requires considerable effort
• Sample is dissolved in ~1 mL of a solvent that has no 1H hydrogens
• Otherwise the spectrum would be 99.5% of solvent, 0.5% sample!
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216
Running an NMR Experiment
• Deuterated solvents are employed—all 1H atoms replaced with 2H
which resonates at a different frequency
• Most common: CDCl3 and D2O
• Employed if necessary: CD2Cl2, DMSO-d6, toluene-d8, benzene-d6,
CD3OD, acetone-d6
• Sample is contained in a high-tolerance
thin glass tube (5 mm)
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Running an NMR Experiment
• IMPORTANT—no deuterated solvent is 100% deuterated, there is
always residual 1H material, and this will show up on the spectrum
• CHCl3 in CDCl3 is a singlet at d 7.27
• HOD in D2O is a broad singlet at d 4.8
• No attempt is made to make solvents for 13C NMR free of 13C, as the
resonances are so weak to begin with
•
NMR using CDCl3 shows a unique 1:1:1 triplet at d 77.00 (+1, 0, 1
spin states of deuterium coupled with 13C)
13C
CHEM 430 – NMR Spectroscopy
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