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7/29/2019 Proton NMR.pdf
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NMR SPECTROSCOPY
Handout 2
Felipe Garcia
fgarcia@ntu.edu.sg
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Proton NMR
The scale of the chemical shift1H NMR is smaller than that in
13C NMR. Proton NMR spectra get
complicated really quickly due to the coupling. However,1H NMR is incredibly useful since a lot of
information can be derived from the spectra.
Most organic compounds resonate between 0 and about 14 ppm. As we saw for13
C NMR, the
chemical shift depends on the local magnetic field a nucleus experiences. All the explanations and
principles applied to 13NMR are also applicable to 1H NMR:
For example if we have a look at the proton NMR of ethylbenzene
As we can see there are signals at very distinctive areas. The chemical shifts tent to be difficult to
predict and interpret. However, we can make certain generalizations about where certain groups and
resonances will appear in the spectrum.
The magnetic field experienced by a proton is influenced by various structural factors. Since electrons
shield the nuclei from the external magnetic field, the less electron density around an atom, the less
shielded it is, and hence it resonates a higher frequencies with a larger chemical shift.
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However some shifts seem counterintuitive, such as the chemical shifts of protons on a benzene rings
are found at higher chemical shifts than protons on a double bond, but the the ones attached to a triple
bond are found at much lower chemical shift. This is the way the electron move in a system.
Electronegative or electropositive substituents affect the chemical shifts, the presence ofelectropositive atoms (donating) or electron withdrawing atoms affect the local electrondensity of the nucleus of study
X C H X C HLi -14 -1.94 NH2 26.9 2.47SiMe3 0 0 OH 50.2 3.39H -2.3 0.23 F 75.2 4.27Me 8.4 0.86 SMe 19.3 2.09Et 15.4 0.91 Cl 24.9 3.06
Hydrogen is more electropositive than carbon , with the result of every replacement of hydrogen by an
alkyl group causes a downfield shift in the resonance of that carbon and any remaining hydrogens on
it.
CH4 MeCH3 Me2CH2 Me3CH Me4C
H 0.23ppm 0.86 1.33 1.68C -2.3ppm 8.4 15.9 25.0 27.7
For example, in the case of H3COOCH2CH3
CH3 protons (A) not deshielded. But next to a CH2 group..slightlyShielded..upfield
CH3 protons (B) 3 bonds away from Oxygens..less deshielded than C..downfieldfrom A
CH2 protons (C) next to a electronegative O, just two bonds awayhighly deshielded -downfield
Aromaticity and (magnetic anisotropy): There are some types of nuclei whose chemicalshifts are not easily explained by simple considerations of the electronegativity of the groups
attached. The anomalous shift is due to the presence of an unsaturated system (one with -
electrons) in the vicinity of the nuclei in question. All groups in a molecule that have -
electrons generate an anisotropic field. In electromagnetic terminology an isotropic field is
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one of either uniform density or spherically symmetric distribution; and anisotropic field is
not isotropic (not uniform). All the groups in a molecule that have -electrons generate
secondary anisotropic fields.
For example, in the previously mentioned
ethyl benzene the aromatic protons are
found between 6.0-9.5
For example the protons on a benzene ring typically have larger magnetic fields than protons
on a C=C double bond, but protons on a triple bond have much smaller shifts. This is due to
the way the electrons in the system of the aromatic ring move in a magnetic field.
When the benzene ring is placed in a magnetic field the electrons in the -system of the
benzene ring move in such a way as to generate a magnetic field that opposes the field applied
at the centre of the ring. This field is said to be due to what is called ring current. Thus a
proton attached to a benzene ring in influenced by three magnetic fields:
i) the strong field applied by the electromagnets of the NMR spectrometer
ii) a weak magnetic field by the shielding by the valence electrons around the proton
iii) a weak magnetic field due to the anisotropy generated by the ring-system -electrons
This field is opposed to the applied field in the centre of the ring, but is aligned with the
applied field for protons on the outside of the ring. Is this anisotropy effect ring current -
that gives the benzene protons a chemical shift higher than expected.
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CH2H2C
H2C CH2
H2C
ringcurrent6electrons
= 1.0ppm
= 2.0ppm
H
H
H
H
H
H
H
HH
H
H
H
H
H
H
H
= 1.8ppm
= 8.9ppmringcurrent18electrons
bonds (magnetic anisotropy).Similar ring currents are set up by the -electronsof an alkyne; but is this case the proton
attached to an alkyne is in the region where the magnetic field is decreased slightly.
The same effect discussed for benzene - aryl protons are found between 6.0-9.5- also happens
for vinylic protons-deshielded to a lesser extend than aryl protons.
Vinylic, R2C=CH2 4.6 - 5.0ppm
Vinylic, R2C=CRH 5.2 - 5.7ppm
In general the anisotropy could be represented as follows :
(-) (-)
(+)
(+)
(-) (-)
(+)
(+)
C C C C(-) (-)
(+)
(+)
C O HH
(-)
(-)
(+) (+)
The -bonded electrons of a -bonds create a magnetic field around them that causes a
downfield shift of the signals from trigonal (sp2) and diagonal (sp) carbons and of the signals
from the protons attached to them. All the protons falling into the conical areas are shielded
(lowed chemical shifht), the ones falling outside the conical areas are deshielded (higherchemical shift).
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A few examples of chemical shifts for carbon and proton are given in the table below,
C H C HCH3H -.23 0.23 CH3CHO 31.2 2.20CH3CH=CH2 22.4 1.71 CH3COCH3 28.1 2.09CH3CCH 5.9 1.80 CH3CN 1.30 1.98CH2=CH2 123.3 5.25 CH3CHO 199.7 9.80CH3CCH 66.9 1.80 CH3COCH3 206.0CH3CCCH3 79.2 CH3CN 117.1
Conjugation. A alkyl group is donating and has not effect on the framework. Whereas amethoxy group is donating and inductively withdrawing. The effect on the chemical shift is
similar, with the MeO group more powerful than the Me group
OMe OMe
inductiveeffect-withdrawingconjugativeeffect
-donating
H Me OMe
H Me OMe
H
H
H
H
H
H
H
H
H
H3.85
HH5.73 H6.38
H4.88
C123 C133.9 C152
C115.4 C84.4
Both groups induce similar effects but the influence of the -overlap is greater that the inductive
effect in the -framework.
C C
H
H
H
X
X Electron nature C C H HH Reference compound 123.3 123.3 5.28 5.28
Me Weak and donor 115.4 133.9 4.88 5.73
OMe donor and -aceptor 84.4 152.7 3.85 6.38
Cl -aceptor, weak donor 117.2 125.9 5.02 5.94
CH=CH2 Single conjugation 130.3 136.9 5.06 6.27
SiMe3 aceptor, donor 129.6 138.7 5.87 6.12
COMe aceptor, aceptor 129.1 138.3 6.40 5.85
Exchange and hydrogen bonding.The shifts for protons connected to oxygen or nitrogen are very variable and can come out almost
anywhere. Some appropriate shifts are shown below,
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Despite of being so variable the NH or OH signals are very easy to spot. They are much broader
than the resonances for protons attached to carbons. The more hydrogen bonding there is the more
the proton is deshielded. Another reason why these signals are easy to identify is that on shaking
with a little deuterium oxide, D2O, the signals from O-H and N-H disappear.
Why would a signal disappear?
Consider the alcohol case for example: R-OH + D2O R-OD + HOD
During the hydrogen bonding, the alcohol and heavy water can "exchange" -H and -D amongst
each other, so the alcohol becomes R-OD. Although D is NMR active, it's signals are of different
energy and are not seen in the H-NMR, hence the signal due to the -OH disappears. (Note that
the HOD will appear...).
Spin-spin coupling:
In the case of protons NMR the coupling follows the same rules as seen before, but in this case there
is coupling with other proton. We have already seen 13C-1H coupling what happens when two nuclei
with I=1/2 are coupled. Much the same is true for proton-proton coupling, except that we are now
observing longer range coupling principally two bond (2J) and three bond coupling (
3J)
HA
C
HB
A
C C
B
Consider the 1H-NMR of ethanol; CH3CH2OH,
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The multiplicity of a multiplet (whether it is a triplet or doublet and so on..) is given by the number of
equivalent protons in neighboring atoms plus one, i.e. the n + 1 rule (where n is the number of
neighboring atoms)
Hb
Hb
Cl
Ha
Cl
Cl
Hb
Hb
Cl
Ha
Cl
Cl
two neighbours
give a triplet
(n+1 = 3)(Area= 1)
one neighbours
give a doublet
(n+1 = 2)
Hb
Hb
Cl
Ha
Ha
Ha
Hb
Hb
Cl
Ha
Ha
Ha
two neighbours
give a triplet
(n+1 = 3)(Area=3)
Three neighbourgs
give a quartet
(n+1 = 3)(Area=2)
Ha
Hb
CN
Hc
The relative intensities of the lines in a coupling pattern is given by a binomial expansion or more
conveniently by Pascal's triangle as already discussed for13
C NMR
As seen before eequivalent nuclei do not interact with each other. The three methyl protons in
ethanol cause splitting of the neighboring methylene protons but they do not cause splitting among
themselves. The relative rations follow what we saw previously using Pascals triangle.
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So for H-NMR a proton with zero neighbours, n = 0, appears as a single line, a proton with one
neighbour, n =1 as two lines of equal intensity, a proton with two neighbours, n = 2, as three lines of
intensities 1 : 2 : 1, etc.
For example, (CH3)2CHCl
There are two different environments: environment a (composed by one Hproton)environment b (composed by six Hprotons)
(CH3)2CHCl: the environment composed by H couples to six protons H in environment bgiving rise to a septet.
(CH3)2CHCl: the environment composed by H couples to one proton H in environment agiving rise to a doublet
(CH3CH2)2CO
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There are two different environments: environment a (composed by three Hprotons)environment b (composed by two Hprotons)
(CH3CH2)2CO: the environment composed by Hprotons couples to two protons H inenvironment b giving rise to a triplet.
(CH3CH2)2CO: the environment composed by H protons couples to one three H inenvironment a giving rise to a quartet.
More complicated coupling patters
More complex splitting patterns can be seen when a nuclei couples with more than when Hb isadjacent to nonequivalent Ha on one side and Hc on the other, the resulting coupling gives rise to a
doublet of doublets.
In the proton nmr of acrylonitrile, all the hydrogens are different - Ha is on the same side as (orcis to)
rhe cyanide group, whereas Hb is opposite (or trans). The orange integrals indicate that there are three
different environments (Ha, Hb and Hc)
Ha
Hb
CN
Hc
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But unlike in previous examples, Hc couples to Ha and Hb (which are different, one cis and one trans
to Hc). Therefore two different coupling constants will be observed Ha-Hc and Hb-Hc. Thefore rather
then a triplet, the signal will split into a doubletdue to Ha-Hc and consequently into anotherdoublet
Hb-Hc; this leads into the formation of a doublets of doublets.
If we construct a tree diagram for the doublet for doublets the resulting signal patter will be
independent of the order in which we consider the coupling constants
Another example will be ethyl propenoate, an unsymmetrical terminal alkene, the three vinylic
hydrogens are nonequivalent due to the restricted rotation about the double bond.
a tree diagram for the complex coupling of the three vinylic hydrogens in ethyl propenoate
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Identifying multiplets - coupling constants
Multiplets can be easily distinguished from closely spaced chemical shift signals. They will have the
same separation between the signals - same coupling constant
To calculate J we can use the simple formula
Operating Frequency (MHz) x Splitting (ppm) = Coupling Constant (Hz)
For a quartet like the one on the left you there are 3
splitting values (x-y), (y-z) and (z-g) and final value is
taken from the average , final J value expressed usually
expressed with just one decimal place
Splitting is a scalar quantity so there is no sign
Remember - Coupled protons have same J values
The coupling between two atoms
The coupling between two nuclei appears on the spectrum as the separation between two lines and is
measured in Hz. The coupling constant between two atoms coupled to each other is the same.
Whether is the same spectrum or a different one. Remember the example for fluoroacetic acid
xx..xxxxxxxxppppmm
yy..yyyyyyyyppppmmzz..zzzzzzzzppppmm
gg..ggggggggppppmm
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Another example where we can observe the splitting between adjacent proton environments is
CH3CH2Br
We can observe how under every peak there is a number, this number indicates what is called the
integral of the peak
Equivalent hydrogens:
have the same chemical environment a molecule with 1 set of equivalent hydrogens gives 1 NMR
signal
a molecule with 2 or more sets of equivalent hydrogens gives a different NMR signal for each set
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Integration of signals
The NMR spectrum not only distinguished how many different types of protons in a molecule but also
reveals how many of each type are composing each environment. In NMR the area under the peak is
proportional to the number of hydrogens generating that peak.
For example, in the case of 2-butanone
The ration showed are 1.99 : 3.04 : 2.96 that approximates 2:3:3.
If we increase the number of carbon atoms and we move from bromo-ethane to bromo propane, the
spectra will look like the following:
-----------------------------------
Further reading:
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Spectroscopic methods in organic chemistry, 5th edition, Dudley H. Williams and Ian Fleling,
McGraw-Hill (1995)
Structure determination of organic compounds, 3rd
edition, E. Pretsch, P. buhlmann, C. Affolter and
M. Badertscher, Springer-Verlag (2001).
Structural Metods in Inorganic chemistry, 2nd
edition, E. A. V. Ebdsworth, David W. H. Rankin and
Stephen Cradocl, CRC Press (1991).
Spectrometric Identification of Organic Compounds, 5th edition, R. M. Silverstein, G. C. Bassler and
T. C. Morrill, Wiley (1991).
Spectrometric identification of organic compounds, 7th
edition, Robert M. Silverstein, Francis X.
Webster, David J. Kiemle, Chapter 3
Essential NMR for scientists and engineers Blumich, Bernhard.
Modern NMR spectroscopy. Oxford U.P
Recommended