13. Structure Determination: Nuclear Magnetic Resonance Spectroscopy
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Transcript 13. Structure Determination: Nuclear Magnetic Resonance Spectroscopy
13. Structure Determination:
Nuclear Magnetic Resonance
Spectroscopy
Why This Chapter?
Mass Spec: Molecular size & formula
IR:
Functional Groups
NMR:
Map of Carbons with Hydrogens
NMR is the most valuable
Used to determine relative location of atoms
Maps carbon-hydrogen framework of molecules
Depends on very strong magnetic fields
More advanced NMR techniques are used in
biological chemistry to study protein structure and
folding
2
13.1 Nuclear Magnetic
Resonance Spectroscopy
Nuclei w/odd # of protons or odd # of
neutrons (~ 1H, 13C, 14N, 2H) spin so act like
tiny magnets randomly oriented.
Parallel: ~ lower E
external magnet
Antiparallel: ~ higher E
Internal magnetic fields align parallel to
or against an aligned external magnet
Parallel orientation is lower in energy making this spin state more populated
3
Nuclear Magnetic Resonance
Radio energy of exactly correct frequency (resonance) causes
the parallel nuclei to flip to anti-parallel state
Energy needed to flip a spinning parallel nucleus is related to its molecular
environment (proportional to field strength, B)
4
Nuclear Magnetic Resonance
Radio energy of exactly correct frequency (resonance) causes
the parallel nuclei to flip to anti-parallel state
Energy needed to flip a spinning parallel nucleus is related to its molecular
environment (proportional to field strength, B)
If nucleus is protected (shielded) from the magnet it takes less E to flip.
If nucleus is exposed to the magnet then takes more E to flip.
5
Example:
I takes 8.0x10-5 kJ/mol to spin-flip a proton at 200 MHz.
Calculate the Energy required to spin-flip a proton (1H) in a
spectrometer operating at 300 MHz.
8.0x10-5 kJ/mol
200 MHz
=
X kJ/mol
300 MHz
X = 1.2x10-4 kJ/mol
6
13.2 The Nature of NMR
•The sample is dissolved in a solvent that does not have a signal itself
Magnet (Applied Field)
Radiofrequency energy is transmitted
and absorption is detected
7
The Nature of NMR Absorptions
NMR of methyl acetate has 2 equivalent kinds of H’s so shows 2 peaks
Electrons in neighboring bonds shield or expose nuclei from magnetic
field
1H
H’s on C next to electron withdrawing C=O
H’s on C next to electron withdrawing O
Intensity of 1H NMR peak is proportional to # of equivalent H’s
8
NMR of methyl acetate has 3 kinds of C’s so shows 3 peaks
Electrons in neighboring bonds shield nuclei from magnetic field
13C
C next door to electron withdrawing C=O
C next to electron withdrawing O
C of C=O
Intensity of 13C NMR peak is not related to # of equivalent C’s
9
At room temperature cyclohexane conformations are
interconverting so rapidly that axial and equatorial 1H’s
appear identical.
When cold cyclohexane conformations intercovert so
slowly that axial and equatorial 1H’s appear different.
10
Example:
How many signals would you expect each to
have in its 1H and 13C spectra?
1H
CH3
13C
CH3
C C
CH3
CH3
CH3
CH3
O
11
Solution:
How many signals would you expect each to
have in its 1H and 13C spectra?
1H
CH3
CH3
1
CH3
C C
CH3
CH3
2 CH3
C C
CH3
CH3
CH3
O
13C
3
C C
CH3
CH3
H H
H
H
CH3
CH3
O
H
H
H H
CH3
CH3
CH3
5
C
O
C
C
C
CH3
C CH3
C
12
13.3 Chemical Shifts
Shift = relative energy of resonance
Downfield = deshielded
(more exposed to
magnet)
Upfield = shielded
(more protected
from magnet)
tetramethylsilane [TMS]
Reference point
CH3
H3C Si
CH3
CH3
13
13.3 Chemical Shifts
Other signals measured in ppm relative to TMS
tetramethylsilane [TMS]
Reference point
CH3
H3C Si
CH3
CH3
14
Measuring Chemical Shift
Numeric value of chemical shift: difference between
strength of magnetic field at which the observed
nucleus resonates and field strength for resonance of
a reference
Difference is very small but can be accurately
measured
Taken as a ratio to the total field and multiplied by
106 so the shift is in parts per million (ppm)
Absorptions normally occur downfield of TMS, to the
left on the chart
Calibrated on relative scale in delta () scale
Independent of instrument’s field strength
15
13.4 Signal Averaging & FT-NMR
Carbon-13: (only carbon
isotope with a nuclear spin)
Natural abundance =1.1%
of C’s so sample is very
dilute in this isotope
Sample measured using
repeated accumulation of data
and averaging of signals,
incorporating pulse and the
operation of Fourier transform
(FT-NMR)
All signals are obtained
simultaneously using a broad
pulse of energy and
resonance recorded
Frequent repeated pulses give
many sets of data that are
averaged to eliminate noise
Fourier-transform of averaged
pulsed data gives spectrum
Single run
Average of 200 runs
16
13.5 13C NMR Spectroscopy
Each signal shows different types of environments of carbon
13C resonances are 0 to 220 ppm downfield from TMS
C’s shift downfield (deshield) if next to electron-withdrawing
Like O, N, X (halogens)
sp2 C ~
sp3 C signal ~
110 to 220
0 to 90
C(=O) at low field,
160 to 220
17
13C
NMR Example: 2-butanone
18
13C
NMR Example: p-bromoacetopheone
19
Learning Check:
Assign resonances in the given 13C spectrum of methyl propanoate
1
CH3
O
2
3
4
O C CH2 CH3
20
Solution:
Assign resonances in the given 13C spectrum of methyl propanoate
1
CH3
O
2
3
4
O C CH2 CH3
21
13.6 DEPT 13C NMR
DEPT (distortionless enhancement by polarization transfer)
Normal spectrum shows all C’s then:
Obtain spectrum of all C’s except quaternary
(broad band decoupled)
Change pulses to obtain separate information for
CH2, CH
Subtraction reveals each type
22
8
DEPT
13C
NMR
7
CH3
Normal spectrum shows all C’s:
(Difficult to Assign so many C’s)
5
CH3
C
5
6
C
H
HH
C
4
H
3
OH
C
C
H H
2
CH3
1
2
6
DEPT-90: shows only CH’s
Quaternary C’s don’t show
(Can now narrow our assignments)
23
8
DEPT
13C
NMR
7
CH3
Normal spectrum shows all C’s:
(Difficult to Assign so many C’s)
5
CH3
C
5
6
C
H
HH
C
4
H
3
C
C
H H
2
OH
2
CH3
1
7,8
1
6
DEPT-135:
Positive =shows CH’s and CH3’s
Negative =shows CH2’s
(Can narrow assignments even further)
24
8
DEPT
13C
NMR
7
CH3
Normal spectrum shows all C’s:
(Difficult to Assign so many C’s)
5
CH3
C
5
6
C
H
HH
C
4
H
3
C
C
H H
2
OH
2
CH3
1
4
7,8
2
1
6
DEPT-135:
Positive =shows CH’s and CH3’s
Negative =shows CH2’s
(Can narrow assignments even further)
25
13.7 Uses of13C NMR: Example
Evidence for product of elimination of 1-chloro-methyl cyclohexane
Cl
CH3
CH2
CH3
KOH
ethanol
or
Expect 5 different C’s;
Expect 7 different C’s;
3 sp3 resonances 20-50
2 sp2 resonances 100-150
5 sp3 resonances 20-50
2 sp2 resonances 100-150
CH3
26
13.8 1H NMR & Proton Equivalence
Proton NMR is much more sensitive than
13C
and the
active nucleus (1H) is nearly 100 % of the natural
abundance
Shows how many kinds of nonequivalent hydrogens
are in a compound
Theoretical equivalence can be predicted by seeing if
replacing each H with “X” gives the same or different
outcome
Equivalent H’s have the same signal while
nonequivalent are different
There are degrees of nonequivalence
27
Nonequivalent H’s
If replacement of each H with “X” gives a different constitutional
isomer then the H’s are in constitutionally heterotopic
environments and will have different chemical shifts
– they are nonequivalent under all circumstances
28
Equivalent H’s
Two H’s that are in identical environments (homotopic) have the
same NMR signal
Test by replacing each with X if they give the identical result, they
are equivalent (homotopic)
29
Enantiotopic Distinctions
If H’s are in environments that are mirror images of each other, they
are enantiotopic
Replacement of each H with X produces a set of enantiomers
The H’s have the same NMR signal (in the absence of chiral
materials)
30
Diastereotopic Distinctions
In a chiral molecule, paired hydrogens can have different
environments and different shifts
Replacement of a pro-R hydrogen with X gives a different
diastereomer than replacement of the pro-S hydrogen
Diastereotopic hydrogens are distinct chemically and
spectrocopically
*
31
Learning Check:
Identify sets of H’s as
Unrelated (U), homotopic (H), enantiotopic (E), or diasterotopic (D)
32
Solution:
Identify sets of H’s as
Unrelated (U), homotopic (H), enantiotopic (E), or diasterotopic (D)
E
D
D
D
D
H
33
13.9 Chemical Shifts in 1H NMR
Proton signals range from 0 to 10
Electronegative atoms cause downfield shift
H’s on sp3 C
H’s on sp2 C
Lower field
Higher field
34
Shifts in 1H NMR
35
13.10 Integration of 1H NMR
Absorptions: Proton Counting
The relative intensity of a signal (integrated area) is proportional
to the number of protons causing the signal
For example in ethanol (CH3CH2OH), the signals have the
integrated ratio 3:2:1
For narrow peaks, the heights are the same as the areas and
can be measured with a ruler
3
1
36
13.11 Spin-Spin Splitting in 1H NMR
Peaks are often split into multiple peaks due to
interactions between nonequivalent protons on
adjacent carbons, called spin-spin splitting
The splitting is into one more peak than the number
of H’s on the adjacent carbon (“n+1 rule”)
The relative intensities are in proportion of a binomial
distribution and are due to interactions between
nuclear spins that can have two possible alignments
with respect to the magnetic field
The set of peaks is a multiplet
(2 = doublet, 3 = triplet, 4 = quartet)
37
Simple Spin-Spin Splitting
In bromoethane see 2 kinds of H’s
One at 3.42 and one at 1.68
Each signal split by neighbors
H H
H C C Br
H H
38
Simple Spin-Spin Splitting
An adjacent
CH2 gives a
ratio of 1:2:1
H H
An adjacent
CH3 group can
have four
different spin
alignments as
1:3:3:1
H C C Br
H H
J (coupling constant) = The separation of
peaks in a multiplet is a constant, in Hz
39
Rules for Spin-Spin Splitting
Equivalent
protons do not
split each other
The signal of a
proton with n
equivalent
neighboring H’s
is split into n + 1
peaks
Protons farther than 2 C’s
apart do not split each other
40
n+1
41
Spin-Spin Splitting Example:
Integration shows ~6:1 ratio
Shift ~1.7 (shielded)
6 H’s see 1 neighbor
(1+1=2 doublet)
Shift ~4.3 (deshielded)
1 H see’s 6 neighbors
(6+1=7 septuplet)
6
1
42
Spin-Spin Splitting Example:
Integration shows ratio
Singlet at 3.8
(deshielded)
3 H’s see 0
neighbors
Shift ~7.8 & 6.8
(deshielded)
2 H’s see 1 neighbor
2x (1+1=2 doublet)
•
a
2
b
b
b
Triplet ~1.2
(shielded)
3 H’s see 2
neighbors
(typical
CH3-CH2-c=o)
(typical CH3-CH2)
3
(typical CH3-O)
(typical para pattern)
a
Quartet ~2.8
(~deshielded)
2 H’s see 3
neighbors
2
3
2
a
43
Learning Check:
From the 1H NMR of C4H10O propose a structure.
44
Solution:
From the 1H NMR of C4H10O propose a structure.
CH3
CH2
O CH2
CH3
45
13.12 More Complex Spin-Spin
Splitting Patterns: trans-cinnamaldehyde
Spectra more complex if overlapping signals, multiple
nonequivalence
Doublet of
Doublets (dd)
~6.7
(~deshielded)
1 H see’s 2
different
neighbors
Doublet ~9.8
(deshielded)
1 H see 1
neighbor
•(typical aldehyde)
1
3
3
b
(typical
CH-CH-CH)
a c
1
a
1
2
1
b
2
c
a
b
46
47
13.12 More Complex Spin-Spin
Splitting Patterns: trans-cinnamaldehyde
Spectra more complex if overlapping signals, multiple
nonequivalence
Doublet ~9.8
(deshielded)
1 H see 1
neighbor
Doublet at 7.5
(deshielded)
1 H see’s 1
neighbor
a=d
2 H’s see 1
neighbor
(typical CH-CH)
(large J value
typical trans alkene H)
•(typical aldehyde)
1
3
3
a c
b
b = dd
2 H’s see 2
different
neighbors
(typical CH-CH-CH)
1
1
(typical
CH-CH-CH)
a
1
2
Doublet of
Doublets (dd)
~6.7
(~deshielded)
1 H see’s 2
different
neighbors
b
2
c
a
b
48
13.13 Uses of 1H NMR Spectroscopy
Determine the regiochemistry of hydroboration/oxidation of
methylenecyclohexane. Which structure gives this 1H NMR?
49
50
Which of the following nuclei does not show
magnetic behavior?
20%
1.
1H
2.
2H
3.
12C
4.
13C
5.
17O
1
20%
2
20%
20%
3
4
20%
5
Which of the following is true of 13C NMR
spectra?
1.
2.
3.
4.
5.
The number of carbon atoms
in a molecule can be
ascertained.
The number of hydrogen
atoms in a molecule can be
ascertained.
Certain functional groups can
be deduced from the
locations of the peaks.
Both the number of carbon
atoms and the number of
hydrogen atoms in a
molecule can be ascertained.
All of these.
20%
1
20%
2
20%
20%
3
4
20%
5
Chemically equivalent nuclei always show a
single absorption.
50%
50%
True
2. False
1.
1
2
How many signals will appear in the 13C NMR
spectrum of the following molecule?
20%
1.
2.
3.
4.
5.
20%
20%
20%
3
4
20%
1
2
3
4
5
1
2
5
How many signals will appear in the 13C NMR
spectrum of the following molecule?
OCH3
1.
2.
3.
4.
5.
20%
20%
20%
20%
3
4
20%
3
4
5
6
7
1
2
5
Which of the following molecules best fits the
following 13C NMR data?
13C
NMR data: 20, 22, 32, 44, 67 ppm
25%
1.
CH3
HO
25%
25%
2.
HO
CH3
H3C
H3C
3.
4.
HO
HO
H3C
25%
CH3
H3C
CH3
1
2
3
4
What is the relationship between Ha and Hb in
the following compound?
Cl
20%
20%
20%
20%
3
4
20%
Ha
Hb
1.
2.
3.
4.
5.
chemically unrelated
homotopic
enantiotopic
diastereotopic
none of these
1
2
5
What is the relationship between Ha and Hb in
the following compound?
20%
OH
20%
20%
20%
3
4
20%
Hc
Ha
1.
2.
3.
4.
5.
Hb
chemically unrelated
homotopic
enantiotopic
diastereotopic
none of these
1
2
5
Order the following protons from lowest to
highest chemical shift value.
20%
O
Hb
Hc
20%
20%
20%
3
4
20%
Hd
Ha
1.
2.
3.
4.
5.
Ha < Hc < Hb < Hd
Ha < Hc < Hd < Hb
Hc < Ha < Hd < Hb
Hc < Ha < Hb < Hd
Hc < Hd < Ha < Hb
1
2
5
1H
NMR will allow one to distinguish between
the following two molecules:
H Br
Br
H
50%
50%
True
2. False
1.
1
2
The painkiller Demerol has the structure shown below. How
many peaks would you expect to see in the 13C NMR spectrum of
this substance?
20%
CH3
N
20%
20%
20%
3
4
20%
O
O
1.
2.
3.
4.
5.
10
11
12
14
15
1
2
5
The spectrum shown could represent the
molecule in the illustration.
50%
True
2. False
50%
1.
1
2
Which of the following molecules best fits the
following NMR spectrum?
20%
1.
3.
20%
20%
20%
3
4
20%
2.
4.
5.
1
2
5
Which of the following molecules best fits the
following NMR spectrum?
20%
1.
20%
20%
20%
3
4
20%
2.
OCH3
OCH3
3.
O
4.
O
5.
O
1
2
5
Which of the following molecules best fits the
following NMR spectrum?
20%
1.
O
2.
OH
3.
OH
4.
5.
HO
HO
HO
HO
20%
20%
20%
3
4
20%
O
OH
OH
1
2
5
Which of the following molecules best fits the
following NMR spectrum?
20%
1.
20%
20%
3
4
20%
2.
OH
Cl
3.
20%
5.
4.
NH2
NH2
1
2
5
Which of the following molecules best fits the
following NMR spectrum?
20%
1.
20%
20%
20%
3
4
20%
2.
O
O
OCH3
3.
5.
4.
O
O
O
O
OH
OH
1
2
5