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NMR Spectroscopy
Chemical Shifts
δ =
(Frequency shift from Me 4 Si in Hz)
(Spectrometer frequency, MHz)
• e^ -
Be
Bo^ B = Bo^ - B^ e H (^) A νo = γB/2π
(magnetic field at nucleus)
(Larmor precession frequency of H (^) A)
Chemical shifts have their origin in the circulation of electrons induced by the magnetic field, which reduces the actual field at the nucleus. Thus a higher magnetic field has to be applied to achieve resonance. Different types of protons in a molecule are surrounded by different electron densities, and thus each one sees a slightly different magnetic field.
Reich Chem 345 Univ. Wisconsin, Madison
The Larmour precession frequency νo depends on the magnetic field strength. Thus at a magnet strength of 1. Tesla protons resonate at a frequency of 60 MHz, at 2.35 Tesla at 100 MHz, and so on. Although Hz are the fundamental energy unit of NMR spectroscopy, the use of Hz has the disadvantage that the position of a peak is dependent on the magnetic field strength. This point is illustrated by the spectra of 2-methyl-2-butanol shown below at several different field strengths, plotted at a constant Hz scale.
60 MHz
100 MHz
220 MHz
For this reason, the distance between the reference signal (Me 4 Si) and the position of a specific peak in the spectrum (the chemical shift) is not usually reported in Hz, but rather in dimensionless units of δ, which is the same on all spectrometers.
HO C
CH 3
CH 3
CH 2 CH 3
b a
c
c
d
a b
c
d
Me 4 Si
Effect of Spectrometer Magnetic Field Strength
δ ppm
H R
O (^) H
H (^) X H
X=O,Cl,Br X H
X=N,S
Alkanes
H
H
1 H Chemical Shifts
The ranges above provide an estimate of the chemical shift for simple molecules, but don't help very much when there are multiple substituents. A simple scheme can be used to estimate chemical shifts of protons on sp 3 carbons. Use the base shift for methyl groups. CH 2 groups, and CH groups, and add to these the increments for each^ α substituent:
OC(=O)R 3.
OR 2.
Br 2. Cl 2. Aryl 1. C(=O)R 1. C=C
Ph
Cl
Base shift CH: 1.5 OH α Ph: 1. α OH: 2.
Base shift CH 2 : 1. α Cl: 2.
Calculated:
Calculated:
Observed: 4.
Observed: 3.
Bo increases
νo decreases
Upfield Shielded
Bo decreases
νo increases
Downfield Deshielded
High frequency Low frequency
H
O
H
H
Base Shift Increment
CH 3 0.
CH 2 1.
CH 1.
8 7.9^ 7.8^ 7.7^ 7.6^ 7.5^ 7.4^ 7.3^ 7.2^ 7.1^ 7 6.9^ 6.8^ 6.7^ 6.6^ 6.5^ 6.4^ 6.3^ 6.2^ 6.1^ 6
HO
H
H
H
10 9.9^ 9.8^ 9.7^ 9.6^ 9.5^ 9.4^ 9.3^ 9.2^ 9.1^ 9 8.9^ 8.8^ 8.7^ 8.6^ 8.5^ 8.4^ 8.3^ 8.2^ 8.1^ 8
12 11.9^ 11.8^ 11.7^ 11.6^ 11.5^ 11.4^ 11.3^ 11.2^ 11.1^ 11 10.9^ 10.8^ 10.7^ 10.6^ 10.5^ 10.4^ 10.3^ 10.2^ 10.1^ 10
CCl 3 CCl 2 H
H
H Ph
NMe
H OMe
O
H NMe 2
O
EtO 2 C
H H
CO 2 Et
O H
Ph
H
H
H
OMe
CH 3 H
HO 2 C H
H H (^) H
Me
O
H
AcO
H
H
H
N
H H
H
N
H
H H
N H
H
OMe
O
H
H^ N^ H
NO 2
NO 2
O 2 N Me H
O H Ph
O
O H
O
OMe
O
H
H
O
O
H
OMe
O
t (^) Bu H CH 3 CO 2 H
S
14 13.9^ 13.8^ 13.7^ 13.6^ 13.5^ 13.4^ 13.3^ 13.2^ 13.1^ 13 12.9^ 12.8^ 12.7^ 12.6^ 12.5^ 12.4^ 12.3^ 12.2^ 12.1^ 12
O
H
Me
H O
S
H
Se
H
O
O (^) O H
H CCl 3
O H
O
H
i Pr 3 Si H
O
Hg H
Representative Proton Chemical Shift Values ( δ 6 to 12)
If given the molecular formula (C 9 H 10 O), there are 10H in molecule
Total area: 26.5 + 11.8 + 16.2 = 54.5 mm Thus 5.5 mm per H
26.5 / 5.5 = 4.86 i.e. 5H
11.8 / 5.5 = 2.16 i.e. 2H 16.2 / 5.5 = 2.97 i.e. 3H
NMR is unique among common spectroscopic methods in that signal intensities are directly proportional to the number of nuclei causing the signal (provided certain conditions are met). In other words, all absorption coefficients for a given nucleus are identical. This is why proton NMR spectra are routinely integrated, whereas IR and UV spectra are not. A typical integrated spectrum is shown below, together with an analysis.
Integration of NMR Spectra - Number of Protons
The vertical displacement of the integral gives the relative number of protons. It is not possible to determine the absolute numbers without additional information (such as a molecular formula). Sometimes a numeric value will be given, or sometimes, as in the example above, you have to measure the distance with a ruler. In this example, if we add up all of the integrals, we get 54.5; dividing by the number of hydrogens in the molecular formula gives 5.5 mm per H. We can then directly estimate the number of protons corresponding to each multiplet by rounding to the nearest integer. It is generally possible to reliably distinguish signals with intensities of 1-8, but it becomes progressively harder to make a correct assignment as the number of protons in a multiplet increases beyond 8, because of the inherent inaccuracies in the method.
The two parts of aromatic proton integral at δ 7.5 - 8.0 can be separately measured as a 2:3 ratio of ortho to meta+para protons.
Reich Chem 345
O
26.5 mm
11.8 mm 16.2 mm
0
Hz 302010
C
H
H
(^2) J = 2-15 Hz
H
C
C
H
(^3) J = 2-20 Hz
C H
C
C H
(^4) J = 0-3 Hz
geminal vicinal long-range
Typical: -12 Hz Typical: 7 Hz
Coupling constants J vary widely in size, but the vicinal couplings in acyclic molecules that we are mostly going to be interested in are usually 7 Hz. The leading superscript ( 3 J ) indicates the number of bonds between the coupled nuclei.
There are also a few situations where coupling across 4 bonds are observed in NMR spectra. This is rarely seen across single bonds, but small couplings (typically 1-3 Hz) are seen when there are intervening double or triple bonds. H H
(^4) J = 2 to 3 Hz
H H
Allylic (^4) J = 0 to 3 Hz
H
H
Propargylic (^4) J = 2 to 4 Hz
Allenic (^4) J = 6 to 7 Hz
H H
One situation where the size of J provides important information is in the vicinal coupling across double bonds, where trans couplings are always substantially larger than cis couplings.
H
H
J = 14 - 18 Hz
H H
J = 8 - 12 Hz
Coupling Constants
Meta
C
H 4
O 8
2
300 MHz
(^1) H NMR Spectrum
Solv: CDCl
3
Source: Aldrich Spectral Viewer/Reich
O
OMe
ppm
2.
3.
3.
Methoxyacetone
C
H 5
O 8
4
300 MHz
(^1) H NMR Spectrum
Solv: CDCl
3
Source: Aldrich Spectra Viewer/Reich^10
ppm
2.
5.
MeO
O
OMe O
Dimethyl malonate
NMR Spectra with no Coupling
ReichChem 345
Problem R-18H C 2 H 4 Cl 2 300 MHz 1 H NMR spectrum in CDCl 3 Source: Aldrich Spectra Viewer/Reich
Problem R-18N C 2 H 3 Cl 3 300 MHz 1 H NMR spectrum in CDCl 3 Source: Aldrich Spectra Viewer/Reich
Simple Coupling Patterns
(^10 9 8 7 6) ppm 5 4 3 2 1 0
1.
2.
Cl
Cl
Cl
Cl
(^10 9 8 7 6) ppm 5 4 3 2 1 0
1.
2.
Cl
Reich Chem 345
Problem R-18G C 2 H 5 300 MHz 1 H NMR spectrum in CDCl 3 Source: Aldrich Spectra Viewer/Reich Br
1.
1.
Bromoethane
1,1-Dichloroethane
1,1,2-Trichloroethane
Simple Coupling Patterns
ppm
1.
5.
4.4 4.3 4.2 1.75^ 1.70^ 1.
Problem R-18E C 3 H 7 Br 300 MHz 1 H NMR spectrum in CDCl 3 Source: Aldrich Spectra Viewer/Reich (^) Br
Reich Chem 345
2-Bromopropane
Br
1-Bromopropane
Problem R-18F C 3 H 7 Br 300 MHz 1 H NMR spectrum in CDCl 3 Source: Aldrich Spectra Viewer/Reich
ppm
1.00 1.
1.
Problem R-18P C 3 H 3 ClO 2 300 MHz 1 H NMR spectrum in CDCl 3 Source: Aldrich Spectra Viewer/Reich
Size of Coupling Constants
Problem R-18Q C 3 H 3 ClO 2 300 MHz 1 H NMR spectrum in CDCl 3 Source: Aldrich Spectra Viewer/Reich
2258.52244.9 1883.91870.
2063.72055.6 1879.91871.
HO
O
Cl
HO
O Cl
J = 13.6 Hz
J = 8.1 Hz
Reich Chem 345
Vicinal coupling across double bonds shows a strong stereochemical dependence, with cis couplings (typically 10 Hz) always being less than trans couplings (typically 15 Hz).
30 20 10 0 Hz
0
Hz 30 20 10
H
H
H
H
ppm
The chemical shifts of OH and NH protons vary over a wide range depending on details of sample concentration and substrate structure. The shifts are very strongly affected by hydrogen bonding, with strong downfield shifts of H-bonded groups compared to free OH or NH groups. Thus OH signals tend to move downfield at higher substrate concentration because of increased hydrogen bonding (see the spectra of ethanol below).
OH and NH Protons
There is a general tendency for the more acidic OH and NH protons to be shifted downfield. This effect is in part a consequence of the stronger H-bonding propensity of acidic protons, and in part an inherent chemical shift effect. Thus carboxylic amides and sulfonamides NH protons are shifted well downfield of related amines, and OH groups of phenols and carboxylic acids are downfield of alcohols.
Pure ethanol
10% EtOH in CCl (^4)
5% EtOH in CCl (^4)
0.5% EtOH in CCl (^4)
δ ppm
13 12 11 10 9 8 7 6 5 4 3 2 1 0
concentrated R-OH^ dilute
Ar-OH
R-CO 2 H
R-NH 2
R-NH 3 +
R-SO 3 H
Ar-NH 2
R-SH
R-C
O
R=CF 3 NH^2 R=CH
3
R-SO 2 NH 2
Chemical Shift Ranges of OH, NH and SH Protons:
Except for alcohols, the shifts are for dilute solutions in CDCl (^3)
Ar-SH
×
×
×
×
OH proton
νAB = 10 Hz
νAB = 20 Hz
νAB = 30 Hz
νAB = 40 Hz
νAB = 50 Hz
A B
A B
A B
A B
Hz
A B
νAB = 3 Hz
The "leaning" or "roof" effect
Why equivalent protons do not show coupling
A B
C
H A
When the two protons are well separated
in chemical shift, each one is a doublet due
to coupling with the neighboring proton
As the chemical shift becomes smaller, the
the two peaks closest to each other (the
inner peaks) become larger, and the outer
peaks become smaller
Eventually the outer peaks disappear, and
the inner peaks merge in to one - one sees
only a singlet. So it is not that protons with
the same shift don't couple, it is that the
peaks that would show us the coupling (the
outer peaks) have all disappeared.
C
H B
Coupling to Different Protons
So far, we have seen only spectra where all of the couplings to a proton are the same, so that simple multiplets like triplets, quartets, etc are formed. However, there are many circumstances where a proton may be coupled to two protons by different coupling constants, leading not to a triplet, but to a doublet of doublets. One common situation of this type occurs in aromatic compounds, where both ortho and meta couplings are large enough to see, but the ortho coupling (8 Hz) is much larger than the meta (2 Hz). The para coupling is usually too small to see. This is thus one of the important exceptions to the rule that protons separated by more than 3 bonds do not show coupling.
(^9 8 7 6) ppm 5 4 3 2 1 0
CH 3
Br
O 2 N
H^3
H^6 H^4
J ortho (coupling to H^3 ) J meta (coupling to H^6 )
H^4
H^6
H^3
t dd
J 1
J 2
J1 = J 2
Other situations where protons separated by more than 3 bonds show coupling also involve intervening π bonds (double or triple bonds). Such couplings are typically smaller than the 7 Hz often seen for 3-bond couplings. See if you can assign the signals in the spectrum below, and identify the couplings.
Problem R-27L C 5 H 8 O (^2) 250 MHz 1 H NMR spectrum in CDCl 3 Source: Adam Fiedler/Reich
(^8 7 6 5) ppm 4 3 2 1 0
1.00 1.
3. 3.
7.05 7.00ppm 6.95 6.
5.90 (^) ppm5.85 5.80 3.75ppm 3.70 1.90ppm 1.
0
Hz 30 20 10
CH 3 O
O
CH 3
H
H
0
Hz 30 20 10
Problem R-23D C 7 H 6 BrNO (^2) 300 MHz 1 H NMR spectrum in CDCl 3
ppm
2.
3.
2.
3.
NH 2
Cl
NO
Effect of Electron Donating and Withdrawing Substituents on NMR Chemical Shifts
Problem R-19C (C 6 H 7 N) 300 MHz 1 H NMR spectrum in CDCl 3 Source: Aldrich Spectra Viewer/Reich
C 6 H 5 Cl 300 MHz 1 H NMR spectrum in CDCl 3 Source: Aldrich Spectra Viewer/Reich
C 6 H 5 NO
300 MHz 1 H NMR spectrum in CDCl 3 Source: Aldrich Spectra Viewer/Reich
2.
3.
OMe
C 7 H 8 O
300 MHz 1 H NMR spectrum in CDCl 3 Source: Aldrich Spectra Viewer/Reich
ppm
ppm
ppm
Problem R-19B (C 7 H 7 NO 3 ) 300 MHz 1 H NMR spectrum in CDCl 3 Source: Aldrich Spectra Viewer/Reich
Isomeric Methoxynitrobenzenes
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