Transcript CHEM 2125 Unit 3 - Oklahoma City Community College

NMR = Nuclear Magnetic Resonance
NMR spectroscopy is the most powerful technique
available to organic chemists for determining molecular
structures.
Looks at nuclei with odd mass numbers or odd number
of protons:
1H, 13C, 15N, 19F, 31P
NMR spectroscopy that is looking at 1H nuclei is called
proton NMR or 1H-NMR. If you just say NMR, it will
generally be assumed you’re talking about proton NMR.
NMR looking at
13C
nuclei is called
13C-NMR.
Nuclei with odd mass numbers or number of protons
have nuclear spin states.
Spinning nuclei generate magnetic fields.
Spinning nuclei will line up with or against an external
magnetic field (B0).
Aligned with
the field
Aligned against
the field
Alignment with the field (lower energy) is called the 
spin state. Alignment against the field (higher energy)
is called the  spin state.
The difference in energy between  and : E.
In the absence of an external magnetic field,
orientations are random. Application of the external
field forces nuclei into the  or  spin states.

B0
E

no external field
A photon of the correct energy (E) can cause
flipping from the  spin state to the  spin state.
When flipping occurs, the energy of that photon is
absorbed.
When the combination of the external magnetic field
strength and photon energy produce flipping, the
nucleus is “in resonance” with the magnetic field and
that frequency of electromagnetic radiation.
E is proportional to the external magnetic field
strength.
h B
E =  
0
2
where B0 is measured in gauss and  is the
gyromagnetic ratio – unique for each kind of nucleus
h B
E =  
0
2
For a proton,  = 26,753 sec1 gauss1
Since E = h
h B
h =  
0
2
Factoring out h, gives
1 B
 =  
0
2
The frequency () and field strength (B0 ) are
directly proportional.
1 B
 =  
0
2
This equation tells us that for a frequency of 60 MHz
(radio frequency or RF), a magnet with a strength of
14,092 gauss is needed for resonance with a bare
proton (H nucleus).
A radio frequency of 300 MHz requires a magnet 5x
stronger (70,459 gauss) for resonance with a bare
proton.
But nuclei in molecules are not bare, isolated nuclei –
they are surrounded by clouds of electrons.
In the presence of an external magnetic field, an
electron cloud has an induced magnetic field that
opposes the external field.
This means it takes a stronger external field to get
resonance – a stronger field is needed because the
electron cloud effectively shields the nucleus.
The effective magnetic field experienced by the
nucleus is the strength of the external field minus the
shielding effect.
Beffective = Bexternal  Bshielding
The electron cloud density (and thus shielding) is
different for different nuclei due to different chemical
environments (magnetic environments).
more shielded protons
requiring higher field
strength to get resonance
at a specific 
H
..
H C O:
H
H
less shielded proton due to
electron withdrawing effect of
electronegative oxygen requiring
lower field strength to get
resonance at a specific 
The differences in shielding mean that the we can
detect the differences in the chemical environments of
the nuclei in an NMR spectrometer.
NMR spectrometer:
RF is held constant – common
frequencies are 60 MHz, 100
MHz, and 300 MHz.
Magnetic strength is varied
downfield
upfield
increasing magnetic field strength
Chemical shift
Chemical shift for protons = the difference between magnetic
field strength needed for resonance of a given proton and the
magnetic field strength needed for resonance of a proton in
tetramethylsilane (TMS).
CH3
H3C Si CH3
CH3
Due to the low electronegativity of the central silicon atom,
protons in TMS are more shielded than any other protons likely
to show up in organic compounds.
Proton signals in most organic compounds will therefore be shifted
downfield from TMS (i.e. require weaker fields for resonance).
The amount of downfield shifting from TMS is expressed in Hz
and is divided by the radio frequency of the instrument. (Even
though NMR spectrometers vary magnet strength, remember
that frequency is proportional to magnetic field strength.)
chemical shift =
shift downfield from TMS in Hz
instrument frequency in MHz
Since Hz units cancel, the chemical shift is unitless.
Since MHz are 1,000,000 times Hz, the amount of chemical
shift is called parts per million (ppm).
The chemical shift is symbolized by  (delta scale).
The advantage of expressing chemical shift data on the
delta scale is that it is easily allows for the comparison
of data from instruments that are operating at
different field strengths.
The higher field strength instruments are used
because they give higher resolution and higher
sensitivity. The main drawbacks to the higher field
strength instruments are initial cost and very high
maintenance cost.
300 MHz NMR spectrum of methanol:
Ultimately, the number of
signals indicates the number
of different chemical
environments in the molecule.
Since the number of signals is related to how many different
chemical environments are present in a molecule, it is useful to be
able to recognize groups of chemically equivalent protons.
How many groups of chemically equivalent protons are present in
each of the following compounds. Hint: Think about symmetry.
CH3
NO2
CH3-CH2-CH2-Br
CH3
O
If different chemical shifts are so close that the signals cannot
be resolved, the protons are called accidentally equivalent.
Accidental equivalence can only be determined experimentally.
Diastereotopic protons are in different chemical environments
(i.e. are not chemically equivalent) and therefore give
different chemical shifts (may only resolve in 300 MHz
instrument).
Diastereotopic protons are identified by seeing if you get
diastereomers by replacing them.
Are these protons
chemically equivalent?
H
H
H
OH
CH2OH
CHO
CHO
CHO
Br
imagine replacing
each with Br or some
other atom
H
H
H
Br
OH
H
OH
CH2OH
CH2OH
What is the relationship between
these two structures?
Are these protons
chemically equivalent?
H
H
Cl
imaginary
replacement
Br
H
H
Cl
Cl
R enantiomer
Br
S enantiomer
Replacement did not produce diastereomers, therefore the
chemical environments are the same. Therefore both
protons will give the same chemical shift.
The protons in question are enantiotopic protons which are
chemically equivalent. NMR can not distinguish enantiomer.
How many groups of chemically equivalent protons are
present in the following compounds?
OH
H
CH2CH2CH3
H
H
Again, different protons have different expected
chemical shifts – depending upon their different
chemical environments.
X in CH3-X
F
OH
Cl
Br
I
EN
4.0
3.4
3.2
3.0
2.7

4.3
3.4
3.0
2.7
2.2
CH3-CH2-CH2-CH3
 = 0.9
1.3
1.3
CH3-CH2-CH2-CH2-Br
0.9
deshielding caused by nearby
Br =
0.9
1.3
1.7 3.4
0.0
0.0
0.4
2.5
Typical Values of Chemical Shifts (see Table 13-3)
Type of proton
alkane –CH3
Approximate 
0.9
Ph-H
Approximate 
7.2
alkane –CH2
1.3
Ph-CH3
2.3
alkane –CH
1.4
R-CHO
9-10
2.1
R-COOH
10-12
2.5
R-OH
variable, about 2-5
3-4
Ar-OH
variable, about 4-7
R-NH2
variable, about 1.5-4
R-CH2-NH2
2.6
Type of proton
O
-C-CH3
C
C H
R-CH2-X
(X = halogen, O)
C
C
5-6
H
C
C
1.7
CH3
Predict the expected chemical shifts of the following
groups of chemically equivalent protons.
CH3
CH3
Chemical shift values are approximately additive –
when you predict chemical shifts you need to take
into account additive effects.
CH2CH3
CH2CH3
O
Groups on a benzene ring can cause downfield or
upfield shifts depending upon whether the group
is electron withdrawing or electron donating.
NO2
OMe
OMe
O
Origins of chemical shifts
In some cases, electron withdrawing effects cause
deshielding (and electron donation causes more
shielding).
Ex.
NO2
CH3-CH2-CHO
CH3-CH2-
In other cases, shielding and deshielding are caused
by magnetic effects of nearby pi systems…
 = 7.2
=5-6
 = 2.5
Aldehyde protons are strongly deshielded by a
combination of electron withdrawing and magnetic
effects.
 = 9 - 10
Carboxylic acid protons are very strongly deshielded
due to being attached to an O which is attached to
an electron-withdrawing C=O.
( = 10 – 12)
What else can you say about the compound that gave the above
spectrum?
Peak area – related to number of protons in a given chemical
environment. Indicated by an “integral trace” on NMR spectrum.
(height of trace is proportional to peak area) (peaks with bigger
areas are usually taller)
a
b
c
a
3.0
d
6 spaces
total
d
a
b
1.5
c
1.0
0.5
Ratio of protons - a:b:c:d = 3.0 : 1.5 : 1.0 : 0.5
To get whole numbers multiply by 2 to get: 6 : 3 : 2 : 1
a
b
c
a
3.0
d
6 spaces
total
d
a
b
1.5
c
1.0
0.5
A ratio of 6 : 3 : 2 : 1 could represent a compound with a total
of 12 H’s (6 + 3 + 2 + 1 = 12) or…
Any compound with the same ratio and a multiple of 12 H’s
Ex. 12 : 6 : 4 : 2
ratio for a total of 24 H’s
a
b
c
a
3.0
d
6 spaces
total
d
a
b
1.5
c
1.0
0.5
Or if total number of H’s is known to start with, figure out how
many spaces = 1 H (You have a molecular formula.)
6 spaces = 12 H’s
means 0.5 spaces = 1 H
The integral with 3 spaces then represents 6 H’s giving that
signal.
a
b
c
a
3.0
d
6 spaces
total
d
a
b
1.5
c
1.0
0.5
Integral traces are not always this neat and well-defined.
Sometimes deciding where traces start and end is tricky – may
require some trial and error to get ratios that make sense.
Insert NMR with integration curve and have class determine ratio
on a handout.
Spin-Spin Splitting
Signals of protons often do not show up as single peaks – their
signal can be split by magnetically coupled protons.
Because a proton we’re interested in may be near other
protons, the magnetic fields of those nearby protons can
reinforce (add to) or oppose (subtract from) the external
magnetic field – shifting a signal downfield or upfield.
Downfield shift due
to one adjacent
proton adding to
field strength
Upfield shift due to
one adjacent proton
subtracting from
field strength
Expected signal
location
The more magnetically coupled protons there are, the more
complex splitting gets because all adjacent protons could be
spinning in the same direction, all adding or all opposing the
external field or some could be spinning in opposite directions
giving mixed effects. Ultimately, this splitting will tell you the
number of protons on adjacent carbons.
Notice that the
peak area ratios
reflect the
number of
combinations
that cause each
peak
The splitting pattern ultimately follows an N + 1 Rule:
N equivalent coupled protons split a signal into N + 1 peaks.
Relative peak
areas
N
Number of peaks
(multiplicity)
0
1 (singlet)
1
1
2 (doublet)
1 : 1
2
3 (triplet)
1 : 2 : 1
3
4 (quartet)
1 : 3 : 3 : 1
4
5 (quintet)
1 : 4 : 6 : 4 : 1
5
6 (sextet)
1 : 5 : 10 : 10 : 5 : 1
6
7 (septet)
Most splitting is caused by adjacent protons.
Ph-CH2-CH3
These protons should split each other’s signals – splitting is reciprocal.
Some splitting is caused by protons on the same carbon (only if
diastereotopic).
These protons will split each
other’s signal
H
CH2CH2CH3
H
H
Splitting by protons separated by a total of 4 or more bonds is
normally not observed (but there are exceptions)
H
H
1
CCC
2
3
4
Too far away to be
magnetically coupled
Characteristic splitting patterns to know:
Peaks representing
magnetically
coupled protons
often “lean” toward
one another.
• Pattern for isopropyl group
• Pattern for p-disubstituted benzene
(1 withdrawing group, 1 donating group)
Coupling constants (J) – how much proton signals split each
other. The magnitude of the J value is also reciprocal.
Areas of the NMR Spectra can be expanded to show greater
detail. Below it is easier to measure the value of thecoupling
constant in the expanded view.
Assign the proton signals in the following spectrum of
4,4-dimethylcyclohe-2-en-1-one.
Complex splitting
When a signal is split by two or more protons that have
different coupling constants, the pattern of the peaks no longer
follows an N + 1 rule.
Ha
Hc
Hb
Protons a and b are trans with coupling
constant (Jab) = 17 Hz
Protons a and c are cis with Jac = 11 Hz
You can draw a splitting tree
to show the expected pattern
of peaks for the signal of
proton a – starting with the
biggest split.
H
a
H
H
c
b
Protons a and b are trans with
coupling constant (Jab) = 17 Hz
Protons b and c are geminal
with Jbc = 1.4 Hz
Label each set of peaks in the NMR spectrum with the letter of
the corresponding protons in the following structure:
a
O
H3 C
CH C
H3 C
b
a
OH
c
Label the groups of protons in the structure below with the letter
corresponding to the peak in the NMR spectrum associated with
those protons:
d
O
O
e
a
b
c
Determine the structure for a compound with a formula of
C9H10O2 if an IR spectrum shows a strong peak at 1705 cm1.
Determine the structure for a compound with a formula of
C9H10O2 if an IR spectrum shows a strong peak at 1735 cm1.
Determine the structure for a compound with a chemical
formula of C5H11Cl given the following NMR spectrum:
Determine the structure for a compound with the chemical
formula C4H10O given the below NMR spectrum:
Exchangeable protons:
Some protons are exchangeable – they come off the atom to
which they are attached and can be replaced with a proton from
another molecule or the solvent.
R-COOH
R-OH
R-NH2
R-CONH2
The proton of an alcohols rapidly exchanges in water or weakly
acidic solutions giving a single peak – no splitting.
Splitting is observed in ultra-pure alcohols (but not always).
Moderate exchange gives a broadened peak. Protons on N often
show broadened peaks.
You generally will not see spin-spin splitting from protons that
are attached to O or N.
Deuterium Exchange:
Peaks that are do to OH or NH protons can be
identified by exchange with deuterium.
Deuterium is not observed in the 1H NMR spectra.
Shaking a sample containing an exchangeable proton
with D2O causes the exchangeable proton to be
replaced by deuterium. And as a result the signal
for the exchangeable proton disappears from the
spectrum.
R-OH + D2O  R-OD + H-O-D
R-NH2 + 2 D2O  R-ND2 + 2 D-O-H
13C-NMR
Signal is weaker because:
a) gyromagnetic ratio is only about 1/4th that of a proton, and
b) only about 1% of C’s in a compound are
13C.
To obtain a 13 C NMR spectrum hundreds of spectra have to be
averaged to get a good signal to noise ratio.
FT-NMR allows hundreds of spectra to be taken in a hour or
two – computer converts free induction decay signal (transients)
to an NMR spectrum.
(Lower gyromagnetic ratio means radiofrequencies used are
lower – 75.6 MHz and 15.1 MHz vs. 300 MHz and 60 MHz)
Chemical shifts in
13C-NMR
TMS is still used as the reference standard.
Chemical shifts are 15-20 times larger than for a proton on
the corresponding C.
Ex:
CH3-OH
for CH3-OH
H has a  of 3-4 ppm
C has a  of 55-85 ppm
CHO
H has 9-10
CHO
C has 180-200
This chart will be given on the test…
…but multiplying proton shift values by 15-20
probably just as useful.
The acquisition of 13C NMR is more complex than 1H NMR. Each of
the protons on a carbon are spin-spin coupled to that carbon, which
results in splitting of the carbon signal. Splitting by adjacent 13C’s
is not observed because of the low natural abundance of 13C.
Splitting of the carbon signal by the protons attached to the
carbon can be used to identify the type of carbon that is
responsible for a particular peak.
However, proton splitting does complicate the 13C spectrum. To
eliminate the proton coupling, 13C NMR are generally run decoupled.
Spin decoupling of the protons involves irradiation of the sample
with broadband rf noise at the proton resonance frequency. This
causes rapid proton flipping which averages out their magnetic
effects.
An added benefit of running a proton decoupled spectrum is that
the energy put into the proton from the decoupler is transferred to
the carbons causing a change in the distribution of the alpha and
beta spin states. This ultimately result in an enhancement of the
carbon signal. (less time to get the spectrum)
Two decoupling protocols:
Proton-spin decoupling – removes splitting effect of protons
completely. All peaks are singlets (most common).
Off-resonance decoupling – noise removes splitting effect of all
but the protons directly attached to the C. Number of peaks
follows an N + 1 where N = number of protons directly on the
C – not adjacent as in 1H-NMR.
(q)
(t)
(s)
(q)
Interpreting
13C-NMR
1. Number of peaks = number of kinds of C’s.
2. Chemical shifts indicate functional groups containing the C’s.
3. Splitting pattern in off-resonance decoupled spectra indicate
how many H’s are attached to the C’s.
4. Peak areas – only indicate numbers of C’s in some types of
experiments.
CH3-CH2-CH2-CH2-CH2-Cl
O
CH3-C-CH2-CH2-O-CH2-CH3
C9H10O2
Given the following three spectra, determine the structure of the
compound.
Assign peaks in the
final structure.
Given the following three spectra, determine the structure of the
compound.
Assign peaks in the final structure.