Download Nuclear Magnetic Resonance (NMR) The most common tool for an

Nuclear Magnetic Resonance (NMR)
The most common tool for an organic chemist to determine structure
Any nucleus with either an odd atomic number or odd mass has a “nuclear spin”
A charge species that is spinning creates a current loop
A current loop creates magnetic field lines
B
+
I
+
+
Spinning
direction
Spinning nucleus
creates a current loop
+
proton
Current loop creates
magnetic field lines
Nuclear Magnetic Resonance
This magnetic field is analogous to a bar magnet
B
+
B
N
S
Thus a spinning nucleus behaves similar to tiny bar magnets
Nuclear Magnetic Resonance
In solution there are many nuclear spins occurring,
hence many tiny bar magnets are present
+
Each spinning direction is random
(there is no energy difference between spinning in any one direction)
and hence the magnetic moments will cancel
Nuclear Magnetic Resonance
Consider the bar magnet analogy
While each individual bar magnet will have a random orientation, in the presence of a larger magnetic field the bar magnet will align itself with the field
N
B0
Field is aligned
S
The same result will occur if we place a spinning nucleus in the field of a large magnetic field
Nuclear Magnetic Resonance
For any S = ½ nucleus, obtain two quantized spin directions in large external magnetic field (B0)
B0
N
S
α-spin state, align with field
β-spin state, align against field
In the presence of an external magnetic field the spinning nucleus can align in one of two ways (with or against the field) when S = ½
Most nuclei of interest for organic chemists have a S = ½ state (1H, 13C, 19F)
Nuclear Magnetic Resonance
The nuclear spins are therefore no longer random
β-spin state
ΔE
N
S
B0
N
N
N
N
S
S
S
S
α-spin state
Orientation when B0 = 0
B0 increasing
There will be more spins in the α-state than β-state, the difference in energy is dependent upon the external magnet strength
Nuclear Magnetic Resonance
The difference in energy is dependent upon two terms
-external magnetic field and gyromagnetic ratio
Energy difference between α- and β-state spins ΔE = γ (h/2π) B0
γ = gyromagnetic ratio
This term is dependent upon which nucleus is spinning,
It is constant for a given nucleus
1H
for example has γ = 26,753 sec-1 gauss-1
Resonance
A key remaining point is the nucleus in the α-state can be changed to the β-state
-need the exact amount of energy that separates the two states
We know this energy difference (ΔE)
We can supply this energy by absorption of the appropriate wavelength of light
ΔE = hν
With infrared spectroscopy we supplied light in the infrared region
to cause a bond vibration
In NMR we supply energy in the radio frequency region to cause the transition
of nuclei from the α-state to the β-state
Nuclear Magnetic Resonance
This absorption of energy will cause the spin state to change
β-spin state
B0
N
N
N
N
S
S
S
S
α-spin state
ΔE = hν
β-spin state
B0
N
N
N
S
S
S
α-spin state
In presence of external magnetic field the spin states are quantized with more spins in the α-state than the β-state With exact combination of external magnetic field (B0) and energy of photons (ΔE, which happens to be in the radio frequency for NMR) the nuclear spins can flip from the
α-spin to β-spin state, thus they are in “resonance”
Called nuclear magnetic resonance (NMR)
Nuclear Magnetic Resonance
The NMR machine can detect this absorption of energy for a sample
Once the field strength and the radiofrequency are in “resonance” the nucleus will absorb the energy of the radio waves
If this was all a NMR machine did it would be practically useless for an organic chemist
-we could detect whether a compound had hydrogens
But that would be the only piece of information
Almost all organic compounds have hydrogens so how does a NMR machine
distinguish between different compounds?
Shielding
Need to remember the structure of a compound
(consider only an isolated C-H bond)
B0
C
H
Bnet = B0 - Belectron
To reach the nucleus the magnetic field must past through the electron cloud surrounding the nucleus
The electrons surrounding the nucleus are charged species that can rotate in the presence of the external magnetic field
What this means is that the external magnetic field (B0) is effectively reduced by the time it reaches the nucleus (B0 minus the field of the electron cloud)
Shielding
This process is called shielding
The electrons are shielding the nucleus from the full external magnetic field
-the field at the nucleus is thus always less than the actual B0
The critical point for a NMR experiment is that each hydrogen nucleus will be shielded by different amounts
The amount of shielding is a direct result of the electron density around the nucleus
The greater the electron density, the more shielded the hydrogen nucleus
Therefore a NMR experiment infers the amount of electron density around each hydrogen in the compound!!!
Protons in Resonance
The UTD Chemistry department has a 400 MHz NMR to measure samples
With no shielding, what is the magnetic field strength required to cause a proton to be in resonance at this radiofrequency?
ΔE = hν = γ (h/2π) B0
B0 = 2νπ / γ
400 MHz = 400 x 106 sec-1
B0 = 2(4 x 108 sec-1)(3.14) / 26,753 sec-1 gauss-1
B0 = 93,896 gauss for 400 MHz NMR
Shielding
But this is for an unshielded proton
The actual proton MUST experience a lower effective magnetic field
If the machine is set at 93,896 gauss the proton will not be in resonance and
therefore it will not absorb the energy to change spin states
Must supply a greater external magnetic field to cause the resonance for a shielded proton
The amount of extra magnetic field required is dependent upon the amount of shielding
MORE electron density around the proton, REQUIRES a larger external magnetic field
Shielding
How can we predict how much the magnetic field must be increased?
By knowing the amount of electron density around a proton
Therefore WHERE ARE THE ELECTRONS
If we know where the electrons are located, we can begin to predict how much the
external magnetic field needs to be increased to cause resonance
Chemical Shifts
To record the position of an absorption scientists use a chemical shift
A standard is chosen, for NMR the standard chosen was tetramethylsilane (TMS)
H3C CH3
Si
H3 C
CH3
Since silane is less electronegative than carbon, the methyl groups (and hence the hydrogens on the methyl groups) will have more electron density than a typical carbon compound
The chemical shift of TMS is defined as 0.00
The chemical shift for a different hydrogen is defined as follows:
Chemical shift (ppm or δ) = (shift downfield from TMS, Hz) / (total frequency, MHz)
Chemical Shifts
A NMR spectrum is recorded with increasing magnetic field towards the right
A more shielded hydrogen (more electron density around the hydrogen) would therefore be further to the right (called upfield)
A less shielded hydrogen (less electron density around the hydrogen) would therefore be further to the left (called downfield)
Since the δ is measured by shift downfield, a positive δ number means the hydrogen has less electron density (or less shielding) than TMS
An advantage of δ is the number is the same regardless of which instrument is used
Chemical Shifts
downfield
upfield
Chemical Shifts
Predicting chemical shifts is therefore the same as predicting electron density around nucleus
Consider methyl halides
CH3Cl
CH3Br
As the electronegativity of the halide increases there is less electron density around the hydrogens
CH3I
Chemical Shifts
H
H
Cl
δ of H’s 3.0 ppm
Br
δ of H’s 2.7 ppm
H
H
H
H
As amount of electron density around
hydrogens decrease, the peak is shifted
more downfield in a 1H NMR
H
H
I
H
δ of H’s 2.2 ppm
Chemical Shifts
The chemical shift is lowered the further removed from electronegative atoms
Saw same trend for pKa effect in acidity
δ1.87
H H
H3 C
Br
H H
δ1.03
δ3.39
Also notice that more than one peak is obtained for each type of hydrogen
Number of Signals
A NMR supplies more information than merely the amount of electron density around nuclei of interest (although that alone is a tremendously important piece of information)
In a 1H NMR, each electronically different hydrogen will have its own unique chemical shift
In the example above with 1-bromopropane there are three different types of hydrogens (one α to bromine, one β to bromine and one γ to bromine)
Each different hydrogen has a different chemical shift
α = 3.4 ppm, β = 1.9 ppm, γ = 1.0 ppm
Number of Signals
By counting number of different peaks we can determine the number of different hydrogens
O
H3 C
O
CH3
The compound for this 1H NMR therefore must have two different types of hydrogens
Area of Peaks
In a 1H NMR spectrum, the area of each peak indicates the relative number of hydrogens of that type
9H’s
δ = 1.0
O
3H’s
2H’s
δ = 2.1
δ = 2.3
Integrating the signals will indicate
the area under each curve, which is directly related to number
of hydrogens causing signal
Spin-Spin Splitting
We can observe a 1H NMR with additional signals
There are three types of hydrogens in this molecule, but some of the signals display more than one peak
Spin-Spin Splitting
In addition to the electron density surrounding the hydrogen causing shielding, each hydrogen acts like its own magnetic field
If one magnet is close to another, it will feel the effect of that magnetic field
This occurs in a 1H NMR a hydrogen will feel the effect of neighboring hydrogens
H
H
Each hydrogen is similar
to a tiny bar magnet
N
S
N
S
The bar magnets could
thus be aligned in same
direction
N
S
Or in opposite directions
These two scenarios lead to a different field strength required to get hydrogen in resonance
In the field of one additional hydrogen therefore we would observe two signals
Spin-Spin Splitting
As the number of neighboring hydrogens increases,
the number of signals increases
Can predict the splitting by knowing the number of magnetically equivalent hydrogens causing the splitting
Pascal’s triangle
# of hydrogens causing splitting
Peak area
Splitting pattern
0
1
singlet
1
2
3
4
1
1
1
1
doublet
1
2
1
3
3
1
4
6
4
triplet
quartet
1
pentet
N+1 Rule
We can use Pascal’s triangle to determine the number of signals expected (and the relative intensity of each signal), but a quick guide is to remember the observed splitting is equal to N+1
If you have N number of magnetically equivalent hydrogens causing the splitting
then you have N+1 peaks in the spectrum
A hydrogen does not cause splitting with itself, but only with neighboring hydrogens
Must be within a short distance to allow the small magnetic field of one hydrogen to affect the magnetic field around another
For alkanes normally observe splitting only for hydrogens attached to adjacent carbons
Nuclear Magnetic Resonance
Knowing these “rules” we can interpret a spectrum
2H’s
3 adjacent H’s
Therefore quartet
O
H3 C
O
CH3
3H’s
2 adjacent H’s
Therefore triplet
3H’s
no adjacent H’s
Therefore singlet
Distance of Splitting
In normal organic compounds, splitting is only observed with hydrogens attached to adjacent atoms
O
singlet
hextet
triplet
too far
removed
triplet
Coupling Constants
The splitting caused by adjacent hydrogens (or any magnetically active species)
is called J coupling and it is a constant value
A hydrogen, which splits a neighboring hydrogen by a J value, will also be split itself by the same J value (reciprocal splitting)
The magnitude of this J coupling can also be characteristic of the type of hydrogen
Most alkane hydrogens have a J constant of ~7 Hz
Coupling Constants
Other characteristic J couplings
H
H
H
Cis alkene
6-12 Hz
Trans > Cis
H
Ortho
6-8 Hz
H
H
H
Trans alkene
12-18 Hz
H
Meta
1-3 Hz
H
Only observe if
geminal hydrogens
are different
H
H
Gem alkene
0-3 Hz
Para
0-1 Hz
H
Most convenient to distinguish E from Z alkenes
Can observe
coupling farther
than adjacent
carbons with
extended
conjugation
Complex Splitting
Most of the cases we have observed so far are relatively quite simple In many examples, though, the splitting is caused by more than one magnetically equivalent adjacent hydrogen
doublet
H
O
H
H
doublet of
triplet
The β hydrogens of a conjugated carbonyl
are more downfield due to resonance
forms placing positive charge at this site
doublet of
doublet
O
O
H
H
Stereochemical Nonequivalence
Remember that all magnetically equivalent hydrogens will have the same chemical shift
If the hydrogens are diastereotopic, however, they will have different chemical shifts
To determine if hydrogens are diastereotopic, replace hydrogens in question with a different
group selectively at each position and consider the stereochemical relationship
H H
oxidize
HO H
H OH
These products are identical, hydrogens are called homotopic
H H
oxidize
HO H
H OH
These products are enantiomers, hydrogens are called enantiotopic
(identical NMR signals)
HO H
H OH
These products are diastereomers, hydrogens are called diastereotopic
(NMR signals are different)
H H
oxidize
Br
Br
Br
The magnitude of this difference is entirely dependent on the structure (can be quite small and not observable)
If the hydrogens are diastereotopic they will in addition split each other
Time Dependence
NMR takes a certain amount of time to record
Therefore anything with a lifetime shorter than the recording time will not be observed, in practice need a lifetime of ~1 sec to observe
This is true for any of the alkanes where bond rotation is much quicker than this time scale
and hence the observed spectrum is an average of all the bond rotations
If the process is slower than the NMR, however,
the NMR can distinguish the different conformations
Sometimes a NMR will be taken at different temperatures to slow down the process to observe by NMR
Some Characteristic Functional Groups
We have observed many functional groups (and will see more as course continues), NMR can
distinguish these groups due to the characteristic chemical shifts for these functional groups
Aromatic Compounds (e.g. Benzene)
View benzene as flat
Remember how the p orbitals are in conjugation
In magnetic field, an electron will rotate in a circular motion
The circulating electrons cause a ring
current which will generate magnetic
H
field lines, causing deshielding of the
aromatic hydrogens
B0
Due to this deshielding effect, aromatic hydrogens have chemical shifts in the 7-8 ppm range
Alkenes
Similar to benzene, hydrogens on alkenes are attached to sp2 hybridized carbons
The sp2 hybridization causes the electrons to be closer to the carbon, further from the hydrogen
The hydrogens are thus deshielded more than alkanes
Alkenes have chemical shifts in the 5-6 ppm range,
Less deshielded than aromatics but more than alkanes
H
H
Alkynes
In terminal alkynes, a hydrogen is attached to a sp hybridized carbon
H
H
Therefore the electrons are further removed from the hydrogen than either an alkane or
alkene due to the greater s character in the bond to carbon (sp hybridized electrons are closer to carbon than sp2 or sp3)
Would expect hydrogen to be further downfield than an alkene, but need to consider the
effect of the induced magnetic field caused by the electrons moving in the orbitals
Bi
B0
H
H
H B =B +B
0
i
H net
Induced magnetic field adds to
applied field, therefore less B0 is
required to get into resonance
δ = 5-6 ppm
Bnet = B0 - Bi
H
B0
Bi
H
Induced magnetic field opposes
applied field, therefore more B0 is
required to get into resonance
δ = low 2 ppm
Aldehyde
An aldehyde is similar to an alkene
H
H
O
O
H
R
R
H
H
H
In addition to the effect of the sp2 hybridized carbon, the aldehyde hydrogen is further
deshielded by the electron withdrawing effect of the oxygen of the carbonyl
O
H
Due to these effects,
an aldehyde hydrogen has a chemical shift of 9-10 ppm
Hydrogen-Bonded Protons
Protons that can hydrogen bond, e.g. alcohols, amines, carboxylic acids, can exist in many states on the NMR time scale
Each state can have a slightly different chemical shift
Therefore these hydrogens can often appear broader than the traditional sharp peaks for other functional groups
The exact chemical shift can also depend on the degree of hydrogen bonding
(which is affected by concentration, temperature, solvent, etc.)
Alcohols Amines Carboxylic acids
2-6 ppm
2-5 ppm
10-12 ppm
1H
NMR Chemical Shift Scale
Bmolecular
RO-H
TMS
(standard)
O
R
RCO2H
11
Bapplied
10
Beffective
Aromatic-H
RCHO
9
8
7
C=C-H
6
CH3
alkane
CH-O, CH-X, CH-N
5
4
3
2
1
0
ppm (!)
Due to empirically observed effects, chemists can predict the position for functional groups
Applied Magnetic Field: Homogeneous for 1H NMR – therefore spin sample Molecular Magnetic Field: two effects will change the magnetic field experienced by nucleus
1)  Density of electron density around nucleus “shields” nucleus
(range ~12 ppm for 1H NMR, ~200 ppm for 13C NMR)
2) Nearby magnetic nuclei (spin-spin splitting) The effective magnetic field at the nucleus is thus Bapplied-Bmolecular
13C
NMR
Any nucleus with an odd atomic number or odd mass will have a nuclear magnetic resonance
A proton is the most common example in organic chemistry so by default when someone
talks about NMR it usually means a 1H NMR
Carbon-12, the most abundant source of carbon, has an even atomic number and even mass so therefore this nuclei is NMR inactive
The isotope C-13, though, has an odd mass and is therefore NMR active
(also has S = ½ like 1H)
Another very useful nucleus to determine organic structures
Carbon-13
The natural abundance of C-13 isotope is ~1%
Therefore only 1% probability that a given carbon atom is NMR active
-therefore the signal-to-noise ratio for carbon NMR is less
Put another way – if studying 1-chloroethane only 1 in 100 molecules will be active at C1
In addition the gyromagnetic ratio, γ, for carbon is ~1/4 that of hydrogen
(remember that each nucleus has its own unique gyromagnetic ratio)
-this also causes a lower signal-to-noise for carbon relative to hydrogen NMR
13C
NMR Spectra
O
H
Spectrum shown as the most common “broad-band” decoupled or also called “proton” decoupled
Unique Features of 13C NMR Relative to 1H NMR
1)  The scale is much larger for 13C than 1H NMR
Notice spectrum is 0-200 ppm instead of 0-11 ppm, due to carbon being more shielded than hydrogen [8 electrons around carbon rather than 2 electrons around hydrogen]
2)  Relative placement of functional groups though is the same in the above spectrum
Which functional group was more deshielded in 1H NMR is still more deshielded in 13C NMR
-aldehyde carbon 194 ppm, sp2 carbons 158 and 133 ppm, sp3 carbons below 40 ppm
3)  Each carbon has its own unique chemical shift
big advantage with 13C NMR, can instantly tell number of distinct symmetrically different carbons
[the spectrum shown is a proton-decoupled 13C NMR, most common example]
4)  Do not observe 13C-13C splitting
Due to low probability of finding two adjacent carbons both 13C isotope with only 1%
probability for each
Splitting
Notice that carbon-carbon splitting is not detected in NMR
Can still detect 13C-1H splitting from hydrogens attached to carbon of interest
This is called off-resonance decoupled
Have same N+1 rule
Normally, though, a 13C NMR is proton spin (broad-band) decoupled
In this mode a sharp singlet is observed for each carbon
-far easier to interpret spectra
Carbon-13 NMR
Remember to look for symmetry
Only distinguish symmetrically different carbons
Peak Areas
Unlike 1H NMR we cannot integrate peaks in 13C NMR to detect relative abundance
In a typical 13C NMR, the peak areas are dependent upon how many hydrogens are attached
to carbon, not the relative number of carbons causing the signal
CH3 groups are the biggest followed by CH2, CH and quaternary carbons are the smallest
Position of Functional Groups in 13C NMR
C-O
Alkenes
Carbonyl
220
200
180
C-X
Alkynes
Aromatic
160
140
120
100
80
Alkanes
60
40
20
ppm (!)
The relative placement is similar to 1H NMR, but the scale is much larger
Observe all carbons
(obviously do not “see” carbons that have no hydrogens attached in 1H NMR)
0
Examples
O
Four signals: 1 carbonyl, 2 sp2 hybridized alkene carbons, one sp3 hybridized carbon
Examples
Br
Four signals: all four are sp2 hybridized carbons
DEPT
(distortionless enhancement with polarization transfer)
A programmed sequence of radio pulses can be used which allow the determination of
number of hydrogens attached to each carbon
In this technique, 4 different 13C spectra are obtained with different pulse sequences:
0˚: a normal broad-band decoupled 13C NMR
45˚: only carbons with hydrogen attached will appear (no quaternary)
90˚: only CH groups will appear
135˚: CH and CH3 groups are positive, CH2 groups are negative
H3C
CH3
MRI Imaging
MRI (magnetic resonance imaging) is in theory identical to our discussions
MRI = NMR
Consumers are hesitant with anything that has “nuclear”
in its descriptor so the name was changed
One main difference is that instead of spinning the sample to make the magnetic field
homogeneous throughout the sample in a NMR experiment (thus the chemical shift will be identical for all hydrogens with the same amount of shielding),
in MRI the field is inhomogeneous throughout the sample
This inhomogeneity allows one to measure either placement of protons in a body (where is the water) or also changes in movement of the protons (i.e. how mobile is water inside a tumor relative to a normal tissue)