Transcript No Slide Title
NMR excitation (detecting NMR)
• Last time we saw how an ensemble of spins (of a single type and
I
= 1/2) generates an average magnetization,
M o
, upon interaction with an external magnetic field,
B o
:
z z
M o
x x y y
B o B o
• We also saw that this magnetization is proportional to the population difference of spins in the low and high energy levels, and that it is precessing at a frequency w
o
, the Larmor frequency of the particular observed spin at a particular
B o
• So far, nothing happened. We need to do something to the system to observe any kind of signal. What we do is take it away from this condition and observe how it goes back to equilibrium. This means affecting the
populations
...
NMR excitation (continued)
• We need the system to absorb energy. The energy source is an oscillating electromagnetic radiation generated by an alternating current:
z
B 1 = C * cos (
w
o t) M o B 1
y
i
Transmitter coil (y)
x
B o
• How is that something that has a linear variation can be thought as circular field? A linear variation in
y
is the linear combination of two counter-rotating circular fields:
y y y
-
w
o +
w
o
x
=
x
+
x
For part of the period of oscillation:
= + = • We go through zero and then it repeats… + = + • Only the one vector that rotates at
+
w
o
(in the same direction of the precession of
M o
) interacts with the bulk magnetization
Now we throw
M
o
on the mix
• When the frequency of the alternating current is w
o
, the frequency of the right vector of
B 1
is w
o
and we achieve a
resonant condition
. The alternating magnetic field and
M o
interact, there is a torque generated on
M o
, and the system absorbs energy :
z z
B 1
w
o
y
M o B 1 off…
x
(or off-resonance)
y x
M xy
w
o
• Since the system absorbed energy, the equilibrium of the system was altered. We modified the populations of the
N
a and
N
b energy levels.
• Again, keep in mind that individual spins flipped up or down (a single
quanta
), but
M o
can have a continuous variation.
Return of
M
o
to equilibrium (and detection)
• In the absence of the external
B 1
, (equilibrium) by restoring the same
M xy N
a will try to go back to
M o / N
b distributiuon. We’ll see the physics that rule this phenomenon (
relaxation
) later.
•
M xy
returns to the
z
axis precessing on the
damm hard to draw…): plane (to
z z
M xy
x
w
o equilibrium...
M o
x y y
• The oscillation of
M xy
generates a fluctuating magnetic field which can be used to generate a current in a coil:
z x
M xy
w
o
y
Receiver coil (x)
NMR signal
Laboratory and Rotating frames
• The coordinate system that we used for the previous example (
laboratory frame
) is really pathetic. The whole system is spinning at w
o
, which makes any kind of analysis impossible.
• Again, an out-of-date example. It would be like trying to read the label of a long play spinning in a turn table… • The solution is to take a coordinate system that moves at w
o
.
This is like jumping on top of the long play to read the label.
What we effectively do is remove the effect of
B o
. If we take magnetization on the
plane:
z z x
B o
y
M xy Laboratory Frame
w
o
x
M xy
y
Rotating Frame
• In this coordinate system,
M xy
resonant condition (the w of
B 1
does not move if we are at the is exactly the frequency of the nuclei, w
o
). If we are slightly off-resonance, the movement of the vectors is still slow with respect to w
o
.
Chemical shifts
• If each type of nucleus has its characteristic w
o
magnetic field, why is NMR useful?
at a certain • Depending on the
chemical environment
we have variations on the magnetic field that the nuclei feels, even for the same type of nuclei. It affects the local magnetic field.
B eff = B o - B loc --- B eff = B o ( 1 -
s
)
• s is the
magnetic shielding
of the nucleus. Factors that affect it include neighboring atoms, aromatic groups, etc., etc.
The polarization of the bonds to the observed nuclei are also important.
• As a crude example, ethanol looks like this:
H O-C H 2 -C H 3
low field high field
w
o
The NMR scale (
d
, ppm)
• We can use the frequency scale as it is. The problem is that since
B loc
is a lot smaller than
B o
, the range is very small (hundreds of Hz) and the absolute value is very big (MHz).
• We use a relative scale, and refer all signals in the spectrum to the signal of a particular compound.
d
=
w
-
w
ref
w
ref ppm (parts per million)
• The good thing is that since it is a relative scale, the d in a 100 MHz magnet (2.35 T) is the same as that obtained for the same sample in a 600 MHz magnet (14.1 T).
•
Tetramethyl silane
(
TMS
) is used as reference because it is soluble in most organic solvents, is inert, volatile, and has 12 equivalent 1 Hs and 4 equivalent 13 Cs:
H 3 C CH 3 Si CH 3 CH 3
• Other references can be used, such as the residual solvent peak, dioxane for 13 C, or
TSP
in aqueous samples for 1 H.
Scales for different nuclei
• For protons, ~ 15 ppm: Acids Aldehydes Alcohols, protons a Aromatics Amides Olefins to ketones Aliphatic
15 10 7 5 2 0 TMS ppm
• For carbon, ~ 220 ppm: C=O in ketones Aromatics, conjugated alkenes Olefins Aliphatic CH 3 , CH 2 , CH
210 150
C=O of Acids, aldehydes, esters
100 80 50
Carbons adjacent to alcohols, ketones
0 TMS ppm
Chemical shift in the rotating frame
• We will consider only magnetization in the
start with a signal with an w
o
equal to the w of
B 1
plane. We . After some time passes, nothing changes…
y y
Time (t)
x x
• Now, if we are slightly off-resonance ( w w
o
0), the
M xy
vector will evolve with time. The angle will be proportional to the evolution time and w w
o
(that’s why we use radians…)
y y
Time (t)
x
w w
o
f
= (
w
-
w
o
) * t
f
x
Coupling Constants
• The energy levels of a nucleus will be affected by the spin state of nuclei nearby. The two nuclei that show this are said to be
coupled
to each other. This manifests in particular in cases were we have through bond connectivity:
1 H 13 C 1 H 1 H
three-bond one-bond • Energy diagrams. Each spin now has two energy ‘sub-levels’ depending on the state of the spin it is coupled to:
I
bb
S
J (Hz)
a b b a
S I
aa
I S
• The magnitude of the separation is called
coupling constant
(
J
) and has units of Hz.
• Coupling patterns are crucial to identify spin systems in a molecule and to the determination of its chemical structure.
Couplings in the rotating frame
• We will consider an ensemble of spins
I
spin
S
coupled to another that is exactly at the resonant condition ( w of
B 1
is w
o
), and again, only what goes on in the
plane.
• The situation is analogous to what happened with chemical shift. In this case, since there are two new energy levels for the spin, we get two counter-rotating vectors. Their evolution will depend on the magnitude of
J
, not w
o
:
y y
- J / 2
t ...
x x
t = 1 / J
+ J / 2
y
t = 2 / J
y x x
f
=
p
* t * J
Take home message…
a) Under the effect of chemical shifts, components of the total magnetization of a sample will move faster or slower than the reference frequency (i.e. the chemical reference,
TMS
).
b) Under the effect of scalar coupling, an ensemble of spins
I
will separate in two counter rotating vectors at speeds of + and -
J / 2
. We have a p involved due to the Hz to radians conversion.
c) The time we let the system evolve is extremely important.
By varying the evolution time we’ll be able to see different things related to the sample.