C8953
NMR structural analysis - seminar
Vector model of NMR experiments + 13C APT
Martin Novák
323460@mail.muni.cz
March 16, 2016
Sketch the estimate of 13
C spectrum of attached
hypothetical molecule.
Analysis of simple pulse sequences using vector
model
simple model based on rotation of the vector of bulk magnetization in the plane
perpendicular to the vector of magnetic field, direction is determined by the
"right-hand rule"
NMR signal is detectable only as coherent magnetization oscillating in xy plane
the free precession ω (due to the B0) of magnetization vector is eliminated by
introducing rotating frame ω0 ⇒ magnetic field of excitation pulses (B1) is
motionless and the individual resonance frequencies differs in so called offset
Ωi = ωi − ω0
applicability of vector model is rather limited to simple single-quantum
experiments without transfer of polarisation
x
y
z
B0
Bpulse
T1 relaxation
Apply following sequence (inversion recovery) to isolated spin characterized by a)
T1 = τ/2 and b) T1 = 5τ. Draw semi-quantitatively resulting spectrum.
τ
180x 90y-
T1 relaxation
Apply following sequence (inversion recovery) to isolated spin characterized by a)
T1 = τ/2 and b) T1 = 5τ. Draw semi-quantitatively resulting spectrum.
τ
180x 90y-
x
y
z
x
y
z
x
y
z
x
y
z
x
y
z
x
y
z
+0.73M0
-0.65M0
1-1 sequence
Draw the evolution of macroscopic magnetization through the sequence:
90(y) - τ - 90(y) - aq
Consider the evolution of an isolated spin due to the chemical shift.
1. How does the result differ for the following offsets: Ωτ = 0, π/2, π.
2. Draw lineshapes of resulting signal assuming the a) y+ b) x+ corresponds to zero
phase of receiver.
τ
90y 90y
1-1 sequence
Draw the evolution of macroscopic magnetization through the sequence:
90(y) - τ - 90(y) - aq
Consider the evolution of an isolated spin due to the chemical shift.
1. How does the result differ for the following offsets: Ωτ = 0, π/2, π.
2. Draw lineshapes of resulting signal assuming the a) y+ b) x+ corresponds to zero
phase of receiver.
τ
90y 90y
x
y
z
x
y
z
x
y
z
0
π/2
π
x
y
z
0
π/2
π
ω ω
y+ x+
Heteronuclear spin echo
By using vector diagrams determine the result of attached pulse sequence.
1. First ignore 180 pulse in hydrogen channel. Explain the role of CPD block.
2. Lets consider the complete sequence and isolated spin systems a) 13C-1H and b)
13C-1H2.
τ=1/2J
90y 180x
13
C
τ=1/2J
180x
1
H
CPD
Heteronuclear spin echo
By using vector diagrams determine the result of attached pulse sequence.
1. First ignore 180 pulse in hydrogen channel. Explain the role of CPD block.
2. Lets consider the complete sequence and isolated spin systems a) 13C-1H and b)
13C-1H2.
τ=1/2J
90y 180x
13
C
τ=1/2J
1
H
CPD
x
y
z
x
y
z
x
y
z
CH...J=JHC
CH2...J=2*JHC
CHα
CHβ
y
+Ωτ
x
y
z
y
+Mx
+Mx
+πJτ
-πJτ
CHαCHβ
CHβ
CHα
+Ωτ
CH2αα CH2αα
CH2αα
CH2ββ CH2ββ
CH2ββ
Heteronuclear spin echo
By using vector diagrams determine the result of attached pulse sequence.
1. First ignore 180 pulse in hydrogen channel. Explain the role of CPD block.
2. Lets consider the complete sequence and isolated spin systems a) 13C-1H and b)
13C-1H2.
τ=1/2J
90y 180x
13
C
τ=1/2J
1
H
CPD
x
y
z
CH...J=JHC
CH2...J=2*JHC
y
x
y
z
y
180x
y
y
-Mx
+Mx
CH2ββCH2αα
CH2αα
CH2αα
CH2αα CH2ββ
CH2ββ
CH2ββ
CHβ CHβ
CHβ
CHβ
CHα
CHα
CHα
CHα
APT - Attached Proton Test
based on heteronuclear spin echo
t1 = 1/1JCH
13
C signals are differentiated according to the number of directly
bound 1
H
Cq, CH2 positive
CH, CH3 negative
Evolution of signal governed by the value of 1
JCH =⇒ reflected by
the intensity of APT signal
90x 180y
180
H-1
C-13 acq
dec
t1
-1
-0.5
0
0.5
1
0 0.5 1 1.5 2
Relativeintensity
t1 (1/1
JCH)
Cq
CH
CH2
CH3
13
C APT Cinnamic acid
13
C APT Cinnamic acid
9 DMF
7
1
2,6 3,5
4 8
Notes:
C9, C1 positive
quaternary
C7 deshielded by -M
effect of carboxyl
group + in
neighbourhood of
aromatic system
equivalent C2/6,
C3/5 in aromatic
region, para C4 less
sensitive
13
C APT of Nicotine
13
C APT of Nicotine
11
10
9
127
3
2,6
54
Notes:
C2, C6 CH negative
connected to N
C3 quaternary, C4
more deshielded
C7 tertiary carbon,
in neighbourhood of
aromatic system
and N
C9 secondary, close
to N; C12 primary
attached to N
C11 connected to
tertiary carbon
13
C APT 4
Cq Cq CH2
13
C APT 4
Cq Cq
2 6 4
8
13
12
5 11
15
14
1'
CH2
Next topic
2D spectroscopy