### NMR SPectroscopy

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NMR SPECTROSCOPY
CHEM 212
Introduction to Spectroscopy
2
Spectroscopy is the study of the interaction of matter with the
electromagnetic spectrum
1.
Electromagnetic 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
Introduction to Spectroscopy
3
5.
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
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 and
movement
•
Coupling – the wave property of the radiation matches the wave
property of the particle and “couple” to the next higher quantum
mechanical energy level
Introduction to Spectroscopy
4
9.
10.
11.
Remember atoms and molecules are quantum mechanical particles
Where a photon is a wave with some particle character, matter is made of particles
with some wave character – wave/particle duality
As a result of this, the energy of these particles can only exist at discrete energies
– we say these energy levels are quantized
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:
Energy
8.
The Spectroscopic Process
5
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: Photons that are
reemitted and detected by the
spectrometer correspond to
quantum mechanical energy
levels of the molecule
rest state
hn
hn
hn
bombarded with photons of
various frequencies over
the range desired
rest state
5
Types of Spectroscopy
6
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
~1013
~1010
~105
1000 nm
0.01 cm
100 m
300-30
~10-4
~10-6
6
Basis of NMR Spectroscopy
7
Nuclear Spin States

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 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
Basis of NMR Spectroscopy
8
6.
Observe that for atoms with no net nuclear spin, there are zero
allowed spin states
7.
Nuclear Magnetic Resonance can only occur where there are allowed
spin states
8.
Note that two nuclei, prevalent in organic compounds have allowed
nuclear spin states – 1H and 13C, while two others do not 12C and 16O
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
Basis of NMR Spectroscopy
9
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
m
I = +½
H
H
I = -½
m


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
Creating Non-Degenerate Nuclei
10
In the absence
of stimulus all
nuclear spin
sates are
degenerate
When a large
magnetic field
B0 is applied
the two spin
states become
non-degenerate
As B0 increases,
the larger DE
becomes
Nuclear Magnetic Resonance
11
When a nuclei
of spin +½
encounters a
photon where
n = E/h, the two
“couple”
The nuclei “flips”
its spin state
from +½ to
–½ and is now
opposed to B0
The nuclei
“relaxes” with
the re-emission
of a photon and
returns to the
+ ½ spin state
Nuclear Magnetic Resonance
12




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
Origin of the Chemical Shift
13
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
The Proton
14
1
( H)
NMR Spectrum
The
1H
NMR Spectrum
15



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
The
1H
NMR Spectrum
16



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
The
1H
NMR Spectrum
17

We need to consider four aspects of a 1H spectrum:
a.
Number of signals
b.
Position of signals
c.
Intensity of signals.
d.
Spin-spin splitting of signals
The Number of Signals
18



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
The Number of Signals
19


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
H
CH 3
The Number of Signals
20

Examples
Important: To determine equivalent protons in cycloalkanes and alkenes, always draw
all bonds to show specific stereochemistry:
The Number of Signals
21

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.
The Number of Signals
22

Proton equivalency in cycloalkanes can be determined similarly:
The Number of Signals
23

Enantiotopic Protons – when substitution of two H atoms by Z
forms enantiomers:
a.
b.
The two H atoms are equivalent and give the same NMR signal
These two atoms are called enantiotopic
The Number of Signals
24

Diastereotopic Protons - when substitution of two H atoms by
Z forms diastereomers
a.
b.
The two H atoms are not equivalent and give two NMR signals
These two atoms are called diastereotopic
Chemical Shift – Position of Signals
25

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
Chemical Shift – Position of Signals
26
•
•
•
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
Upfield, shielded 
Chemical Shift – Position of Signals
27

There are three principle effects that contribute to
local diamagnetic shielding:
a.
b.
c.
Electronegativity
Hybridization
Proton acidity/exchange
Chemical Shift – Position of Signals
28

Electronegative groups comprise most organic functionalities:
-F
-Cl
-Br
-I
-OH
-OR
-NH2
-NHR
-NR2
-NH3+
-C=O
-NO2
-NO
-SO3H
-PO3H2
-SH
-Ph
-C=C
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
Chemical Shift – Position of Signals
29

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
Chemical Shift – Position of Signals
30
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
Chemical Shift – Position of Signals
31
Hybridization
3
2
 Increasing s-character (sp  sp  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)
Chemical Shift – Position of Signals
32

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
Chemical Shift – Position of Signals
33

Magnetic Anisotropy – Aromatic Protons
a.
b.
c.
In a magnetic field, the six p electrons in benzene circulate
around 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.
Chemical Shift – Position of Signals
34

Similarly this effect operates in alkenes:
Chemical Shift – Position of Signals
35

In alkynes there are two perpendicular sets of p-electrons—the molecule
orients with the field lengthwise—opposing B0 shielding the terminal H atom
Chemical Shift – Position of Signals
36
Chemical Shift – Position of Signals
37
Intensity of Signals—Integration
38





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
Intensity of Signals—Integration
39
Intensity of Signals—Integration
40
Spin-Spin Splitting
41


Consider the spectrum of ethyl alcohol:
Why does each resonance “split” into smaller
peaks?
HO
CH3
C
H2
Spin-Spin Splitting
42





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
Spin-Spin Splitting
43

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
44


In solution we are not looking at a single molecule but
On some molecules the proton being observed may be
next to another proton of spin + 1/2 :
Spin-Spin Splitting
45

On some molecules the proton being observed may
be next to another proton of spin – 1/2 :
Spin-Spin Splitting
46



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
Spin-Spin Splitting
47

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
resonance for Ha in absence of spin-spin splitting
Spin-Spin Splitting
48

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
2 Hb
protons + ½
resonance for Ha in absence of spin-spin splitting
or
Spin-Spin Splitting
49

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
2 Hb
protons - ½
resonance for Ha in absence of spin-spin splitting
or
Spin-Spin Splitting
50

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 for Ha in absence of spin-spin splitting
Spin-Spin Splitting
51

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
resonance for Ha in absence of spin-spin splitting
Spin-Spin Splitting
52

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
resonance for Ha in absence of spin-spin splitting
Spin-Spin Splitting
53

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
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:
1H
NMR—Spin-Spin Splitting
54

Common patterns:
tert-butyl - singlet
methyl - singlet
ethyl – quartet - triplet
n-propyl – triplet - quintet - triplet
iso-propyl – septet - doublet
1H
55
NMR—Spin-Spin Splitting
1H
NMR—Spin-Spin Splitting
56

Another Example:
Br
Br
Br
=
Br
C
Br
Ha
C
Br
Hb Hb
1H
NMR—Spin-Spin Splitting
57

Another Example:
1H
NMR—Spin-Spin Splitting
58
Three general rules describe the splitting patterns commonly
seen in the 1H NMR spectra of organic compounds:
Equivalent protons do not split each other’s signals
1.
A set of n nonequivalent protons splits the signal of a nearby
proton into n + 1 peaks
2.
Splitting is observed for nonequivalent protons on the same
3.
If Ha and Hb are not equivalent, splitting is observed when:
1H
NMR—Spin-Spin Splitting
59

Magnetic influence falls off dramatically with distance

The n + 1 rule only works in the following situations:
H H
Aliphatic compounds that have free
H H
G
H
Ha
H
Hb
Hc
Aromatic compounds where each proton is
held in position relative to one another
1H
NMR—Spin-Spin Splitting
60



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-
-CH3
1H
NMR—Spin-Spin Splitting
61


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
1H
NMR—Spin-Spin Splitting
62

For this substituted ethylene we see the following
spectrum:
The observed multiplet
2J
gem
= 0 – 1 Hz
for Ha is a “doublet of
doublets”
3J
AC
3J
trans
= 11- 18 Hz
3J
AB
3J
cis
= 6 - 15 Hz
3J
AB
1H
NMR—Spin-Spin Splitting
63


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
1H
NMR—Spin-Spin Splitting
64

3J
3J
Some common J-values
= 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
trans
3J
= 4-8
3J = 6-12
cis
trans
3J
ortho = 7-10 Hz
4J
meta = 1-3 Hz
5J
para = 0-1 Hz
1H
NMR—Spin-Spin Splitting
65

We can now tell stereoisomers apart through 1H NMR:
1H
NMR—Spin-Spin Splitting
66

A combined example:
1H
NMR—Spin-Spin Splitting
67
•
•
Under usual conditions, an OH proton does not split the NMR signal of
Protons on electronegative atoms rapidly exchange between molecules in the
presence of trace amounts of acid or base (usually with NH and OH protons)
Structure Determination
68
Structure Determination
69
Structure Determination
70
Structure Determination
71
13C
NMR
72
•
•
•
•
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
13C
73
NMR
13C
NMR
74
•
•
•
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.
13C
NMR
75
•
•
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 (0220 ppm).
The chemical shifts of carbon atoms in 13C NMR depend on the same
effects as the chemical shifts of protons in 1H NMR.
13C
76
NMR
13C
77
NMR
Shoolery Tables
78


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
H
X C H
H
methyl
H
X C Y
H
methylene
H
X C Z
Y
methine
Shoolery Values for Methylene
79
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
-N3
1.97
-OH
2.56
-NO2
3.36
-OR
2.36
-SR or H
1.64
-OPh
2.94
-OSO2R
3.13
Shoolery Values for Methine
80
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
-Ph
0.99
-OPh
1.79
Shoolery Tables
81


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
Shoolery Tables
82

Work for aromatics as well (.pdf posted)
Running an NMR Experiment
83


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!
Running an NMR Experiment
84




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)
Running an NMR Experiment
85

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
resonances are so weak to begin with
13C,
as the
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
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