Introduction to NMR Theory in 8 Easy Steps
The nuclei of some atoms have a purely quantum mechanical property called spin (a.k.a.
are ‘spin-active’) and have a nonzero spin quantum number S. If a nucleus does not have
spin then S=0. Sometimes one isotope of an element is spin active, while another is not.
For example, the 13C nucleus has S=1/2 but the 12C nucleus has S=0.
Nuclei with S=1/2, such as 1H, 13C, 15N, 31P, 19F, are most often studied in biomolecular
NMR, but it is possible to take NMR spectra of nuclei that have S>1/2.
Nuclei with S=1/2 can be in one of two spin states, ms=1/2 or ms =-1/2. If the nucleus
does not experience a magnetic field, then the energy of the nucleus is the same no matter
what the value of ms is. When a magnetic field is present, these two states have different
energies as shown below.
Em  (ms ) (h /2 )B0
Energy
s
ms=-1/2
ms=1/2,-1/2

ms=1/2
Field Off
Field On (B0)
Notice the factors that influence the energy levels: the strength of the magnetic field (B0
which is usually in Tesla) and the gyromagnetic ratio (), which is also known as the
magnetogyric ratio. The constant  is specific to each nucleus:
atom
1
H
13
C
15
N
31
P
 (107 rad T-1 s-1)
26.752
6.7283
-2.712
10.841
A nucleus can absorb energy corresponding to the difference in energy levels. We apply
this energy in the form of oscillating magnetic fields. When B0=7.0 Tesla, NMR
transitions range from 30 MHz for 15N to 300 MHz for 1H.
NMR was initially practiced using a method called “continuous wave” or CW-NMR.
This was a reasonable approach, but was slow and had poor sensitivity. Modern NMR is
performed using short pulses of radio frequency magnetization, often termed pulsed
Fourier-Transform NMR or FT-NMR. New alternative sampling technologies in NMR
are still emerging, but FT-NMR is here to stay and will not disappear like CW NMR!
1. equilibrium magnetization in presence of external B0
2. applied oscillating magnetic field
3. splitting the oscillating field
4. the rotating frame
5. the right hand rule
6. detecting the signal
7. processing the signal
8. the return to equilibrium
B0
-1/2
E
=
B0
=
+1/2
M0
E   (h / 2 )B0
First, the number of spins in
the ms=1/2 and ms=-1/2 states
is governed by the Boltzman
equation and these levels are
not equally populated in the
presence of B0. A slight
excess of spins precess in
alignment with B0 and the
sample develops a net
magnetic moment M0. This is
the equilibrium state.

Second, we impose an oscillating
magnetic field on the sample. This is
accomplished by passing an oscillating
current through a coil that creates an
oscillating magnetic field inside the coil.
The coil is wrapped around the sample.
The coil is placed so that the oscillating
field is perpendicular to B0 and therefore
also M0. By convention, the field
oscillates between –2B1 and +2B1. This
will help us out soon.
B0
M0
coil
x
2B1sin(t)
y
Third, a single
oscillating
magnetic field
of magnitude
2B1 is formally
equivalent to
two counterrotating fields,
each of
magnitude B1.
One oscillating field of
magnitude 2B1
Two counter-rotating fields
of magnitude B1 each.
Fourth, choose the one rotating field that moves in the same direction as the Larmor
precession, and jump onto it! Imagine sitting on the rim of a spinning record. To you,
the record looks still since you are moving with it. Now “sit” on one of the rotating
components: it looks
static, and the other
M0
M0
component appears
to be moving twice
as fast. From now
on we ignore the
B1
B1
other component.
This is the rotating
frame.
LAB Frame
Rotating Frame
Fifth, the B1 field exerts a force on M0, (torque) and causes M0 to change orientation.
Point the fingers of your right hand along the applied field B1. Curl your fingers towards
M0. Your thumb points in the direction
z
of motion for M0. In fact, M0 will
rotate around in the x-z plane until B1 is
M0
turned off! (B0 is not drawn in the
rotating frame for reasons that can be
skipped here).
A 90 degree pulse : if we turn off the B1
field exactly when M0 has rotated by
90o and is aligned with the x axis, then
we have applied a 90 degree pulse.
A 180 degree pulse : iturn off the B1
field at exactly the time when M0 has
revolved to point along the –z axis.
y
B1
x
B1 along y rotates M0 in the x-z plane
Sixth, following a 90o pulse, the B1 field is off and the sample only sees the static field
B0. The right hand rule again tells us that M0 will rotate around and around in the x-y
plane. This will occur at the Larmor frequency plus the chemical shift.
z
B0
... and induces an oscillating current in the
coil that is digitized by a computer
V
tim
M
x
M0 rotates in the x-y plane
Seventh, the time-domain oscillation occurs at the NMR frequency (the base Larmor
frequency plus the chemical shift). It is not easy to interpret the data in this form! To see
the data as a histogram of frequencies present and their relative proportions, perform a
fast-Fourier transform (FFT). A method called ‘quadrature detection’ is used to
determine if each NMR signal appears at + or –.
analog signal is digitized
FFT, Quadrature

To illustrate the importance of chemical
shift, recall the example of ethanol, which
exhibits three signal regions for: methyl,
methylene, and hydroxyl hydrogen nuclei.
The signal at 0 ppm is due to tetramethylsilane (TMS) whose proton
frequency is defined to be 0 ppm. The
signal at about 4.7 ppm could be the –OH
signal but could also be residual water in the sample.
The convention for reporting NMR frequencies is the ppm scale, and is based on the
frequency of a bare nucleus. If a bare proton nucleus resonates at exactly 500 MHz, then
1 ppm = 500 Hz on the proton scale. Frequencies are reported as a shift relative to a
defined 0 point. A nucleus gives the same shift in ppm units no matter what field is used.
(e.g. the methyl triple is at 1.1 ppm at any field strength)
The ethanol spectrum shows split resonances. This effect is also called the scalar or J
coupling and is a through-bond effect. Splitting of NMR signals tells a lot about the
local bonded environment of a
molecule. It is common to prepare
proteins isotopically enriched in 15N,
coupled spectrum
and/or 13C. In these cases the 13C-1H
or 1H-15N J couplings can arise and a
great deal of effort is devoted to
collapsing such heteronuclear
splittings using a method called
decoupling. We will not describe the
decoupled spectrum
specifics of how to carry out
decoupling; but the result should be a
A cartoon of the collapse of J-doublets upon
collapse of multiplets to single lines.
decoupling. Doublets are common in 1H(X)
situations such as 1H-15N.
Eighth, a sample of spins will return to the initial equilibrium state, given enough time.
This is relaxation. The relaxation of NMR signals for macromolecules is very sensitive to
the shape and size of the molecule and to all of the possible motions of the molecule.
There are two fundamental ways that we view relaxation:
(i) the disappearance of M0 in the x-y plane (R2 relaxation) and
(ii) the reappearance of M0 along z (R1 relaxation)
A key feature of macromolecules is that R2 is faster than R1, as shown below. The
symbol R represents the rate of relaxation.
Longitudinal (R1) Relaxation
y
x
Transverse (R2) Relaxation
y
x
time
M0