NMR%20Spectroscopy

1
Chapter 13

• NMR Spectroscopy

2
Recall that electrons have two spin states
spin up (1/2) and spin down (-1/2). Similarly
nuclei have spin quantum states.
Nuclei of interest. By coincidence, each of
these has two states, ½ and ½ .
By the way, note that 14N has three states -1,
0, 1. They each differ by 1.
3
Spectroscopy involves using energy to excite a
system from one state (ground state) to another
of higher energy (excited state).
The nuclear spin quantum number determines how
many spin states there are
Nuclear spin quantum number Quantum numbers of spin states Number of spin states
0 0 1
1/2 - ½ , ½ 2
1 -1, 0, 1 3
3/2 -3/2, -1/2, 1/2, 3/2 4
4
Normally, nuclei in different spin states have
the same energy. Can not do spectroscopy. We
need to have a ground state and excited state. In
a magnetic field they have different energies.
Now we can do spectroscopy. We apply a
magnetic field and create a ground state and a
higher energy excited state (perhaps more than
one).
5
Apply a strong external field..
Both orientations have same energy if no magnetic
field
6
(No Transcript)
7
Figure 13.3, p.497
8
Figure 13.4, p.499
9
Example of nmr spectrum methyl acetate.
More shielded, nuclei experience lesser magnetic
field. Less energy to excite.

• Two kinds of hydrogens in methyl acetate two
  peaks. (Peak at zero is tetramethyl silane to
  standardize the instrument. )
• Chemical shift where on horizontal axis the
  signal from a nucleus occurs. Question What
  causes nuclei to appear with different chemical
  shift??
• Answer the sigma bonding electrons in a molecule
  will be set in motion to establish a magnetic
  field that opposes the external magnetic field.
  The nuclei are shielded.
• The shielded nuclei experience less of a magnetic
  field, closer energy states.
• The shielded nuclei require less energy to excite
  and their signal occurs to the right in the
  spectrum.

10
p. 481
11
More Shielding

• Doing nmr spectroscopy
• the magnetic field creates the energy difference
  between the spin states of the nucleus and
• Radio waves provide the energy needed to excite
  the nucleus from the lower energy state to the
  excited state.

• Simplifying
• The energy supplied by the radio waves has to
  match the energy gap created by the magnetic
  field.
• We can vary either the magnetic field or the
  frequency of the radio waves to match the
  exciting radiation energy with the energy needed
  to reach the excited state.

12
More Shielding
Since we control energy of excited state
(magnetic field) and the energy being supplied by
radiowaves two ways for an nmr spectrometer to
function
Hold the external magnetic field constant, vary
radio frequency. Less energy needed to excite the
nuclei when more shielded.
More Shielded, Less energy needed from radio waves
or
Hold excitation energy (radio waves) constant,
vary magnetic field. Stronger magnetic field
needed to overcome shielding.
More shielded, stronger magnetic field needed to
create the right energy difference.
Terminology based on this approach downfield
(lower ext field) on left upfield on right
13
Remember that methyl acetate only gave two peaks
in its spectrum. There were two sets of
equivalent hydrogens. Equivalent hydrogens

• Hydrogens are equivalent if
• They are truly equivalent by symmetry.
• -or-
• They are bonded to same atom and that carbon
  atom can rotate freely at room temperature to
  interchange the positions of the hydrogens making
  the equivalent to the spectrometer.

14
Equivalence by Symmetry
Figure 13.6, p.500
15
Equivalent by rotation
Note that if it were not for rotation the methyl
hydrogens would not be equivalent. Two are
gauche to the Cl and one is anti.
equivalent
16
Some molecules which have only one type of
hydrogen - only one signal
p.501
17
p. 484
18
Signal area proportional to the number of
hydogens producing the signal
Looking at the molecular structure Methyl
hydrogens tert butyl hydrogens 39 13 In
the spectrum we find two peaks 23 67 1
2.91 Conclude smaller peak due to methyl
hydrogens larger due to tert butyl hydrogens.
Figure 13.7, p.503
19
Now return to chemical shift and factors
affecting it. Look at two isomeric esters to get
some feeling for chemical shift. The
electronegative oxygens play the key role here.
Most electron density around the H atoms, most
shielded, upfield.
Less shielded, more deshielded, downfield
Most deshielded, furthest downfield. Sigma
electrons pulled away by oxygen.
p.504
20
Chemical shift table
Figure 13.8, p.505
21
Relationship of chemical shift to
electronegativity
Further left, downfield, Less shielded
Less electrons density around hydrogens as ascend
table.
22
For C-H bond as the hybridization of the carbon
changes sp3 to sp2 to sp the electronegativity of
the C increases and expect to deshield (move
left) the H peak.
sp3 sp sp2
Expect vinylic hydrogens to be deshielded due to
hybridization but acetylenic (recall acidity)
should be even more deshielded and they arent.
Some other factor is at work. Magnetic induction
of pi bonds.
23

• Diamagnetic shielding
• Hydrogen on axis and shielded effectively.
• Hydrogen experiences reduced magnetic field.
• Less energy needed to excite.
• Peak moves upfield to the right.

Figure 13.9, p.507
24

• Diamagnetic shielding
• Hydrogen off axis and external field increased.
• Hydrogen experiences increased magnetic field.
• More energy needed to excite.
• Peak moves downfield to the left.

Figure 13.10, p.507
25
In benzene the H atoms are on the outside and the
induced magnetic field augments the external
field.
Figure 13.11, p.508
26
Spin Spin Splitting

• If a hydrogen has n equivalent neighboring
  hydrogens the signal of the hydrogen is split
  into (n 1) peaks.
• The spin-spin splitting hydrogens must be
  separated by either two or three bonds to observe
  the splitting. More intervening bonds will
  usually prevent splitting.

27
Example
Expect the signal for this hydrogen to be split
into seven by the six equivalent neighbors.
Expect the peak for the methyl hydrogens to be
split into two peaks by the single neighbor.
Overall small peak split into seven (downfield
due to the Cl). larger peak (six times larger)
split into two (further upfield).
28
Attempt to anticipate the splitting patterns in
each molecule.
p. 491
29
p. 491
30
Spin-spin splitting. Coupling constant, J.
The actual distance, J, between the peaks is the
same within the quartet and the doublet.
Split into a group of 4
Split into a group of 2
31
In preparation for discussion of origin of
Spin-Spin recall earlier slide
More Shielding due to electrons at nucleus being
excited.
Due to shielding, less of the magnetic field
experienced by nucleus, Lower energy needed to
excite. Peaks on right are upfield.
Reduced shielding, more of the magnetic field
experienced, higher energy of excitation. Peaks
are downfield.
32
Origin of spin-spin splitting
In the presence of a external magnetic field each
nuclear spin must be aligned with or against the
external field. Approximately 50 aligned each
way. Non-equivalent hydrogen nuclei separated by
two or three bonds can spin spin split each
other. What does that mean? Consider excitation
of a hydrogen H1. Energy separation of ground and
excited states depends on total magnetic field
experienced by H1. Now consider a neighbor
hydrogen H2 (passive, not being excited) which
can increase or decrease the magnetic field
experienced by H1.
About 50 of the neighboring hydrogens will
augment the applied magnetic field and about 50
will decrement it. Get two peaks, a double
The original single peak of H1 has been split
into two peaks by the effect of the neighbor H2.
The energy difference is J
Energy
H1, being excited
Here H2 augments external field, peak moved
downfield.
Here H2 decreases external field, peak moved
upfield.
33
Same as gap here.
Coupling constant, J, in Hz
3-pentanone
The left side of molecule unaffected by right
side.
Peak identification
Figure 13.14, p.511
34
Magnitude of Coupling Constant, J
The magnitude of the coupling constant, J, can
vary from 0 to about 20 Hz. This represents an
energy gap (E hn) due to the interaction of the
nuclei within the molecule. It does not depend
on the strength of the external field. J is
related to the dihedral angle between bonds. J
largest for 0 (eclipsed) or 180 (anti), smallest
for 90, intermediate for gauche.
35
gauche
anti
vinyl systems
Table 13.4, p.511
36
Spin-Spin Splitting
Now look at some simple examples. Examine the
size of the peaks in the splitting.
Hb is augmenting external field causing a larger
energy gap. Hb decrementing external field
causing a smaller energy gap.
Ha is being excited. Hb is causing spin-spin
splitting by slightly increasing or decreasing
the magnetic field experienced by Ha.
37
Two neighboring atoms assist external field.
More energy needed to excite. Peak is downfield.
Again Ha is flipping, resonating. The two Hb are
causing spin-spin splitting by slightly changing
the magnetic field experienced by Ha.
One neighbor assists, one hinders. No effect.
Both neighbors oppose. Less energy needed to
excite, upfield.
Recall that for the two Hb atoms the two states
(helping and hindering the external field) are
almost equally likely. This give us the 1 2
1 ratio.
Figure 13.15b, p.512
38
Three neighboring Hbs causing splitting when Ha
is excited.
All Hb augment Two augment, one decrement. One
augment, two decrement. All decrement.
Ha being excited.
Three equivalent Hb causing spin spin splitting.
Figure 13.15c, p.512
39
Intensities of peaks in a multiplet corresponds
to Pascals triangle.
40
Naturally if there are two non-equivalent nuclei
they split each other.
Figure 13.17, p.513
41
Three nonequivalent nuclei. Ha and Hb split
each other. Also Hb and Hc split each
other. Technique use a tree diagram and consider
splittings sequentially.
Figure 13.19, p.513
42
More complicated system
Figure 13.20, p.514
43

Return to Vinyl Systems
Not equivalent (R1 is not same as R2) because
there is no rotation about the CC bond.
Figure 13.21, p.514
44
Example of alkenyl system
We will perform analysis of the vinyl system and
ignore the ethyl group.
Three different kinds of H in the vinyl group.
We can anticipate the magnitude of the coupling
constants.
45
Analysis
Each of these patterns is different from the
others.
JAB 11-18 Hz, BIG JAC 0 5 Hz, SMALL JBC
5 10 Hz, MIDDLE
Now examine the left most signal.
46
Ha being excited. Both Hb and Hc are coupled and
causing splitting.
Hb causes splitting into two peaks (big
splitting, JAB)
Hc causes further splitting into a total of four
peaks (smallest splitting, JAC)
JAB 11-18 Hz JAC 0 - 5 JBC 5 - 10
47
Analysis in greater depth based on knowing the
relative magnitude of the splitting constants.
Aim is to associate each signal with a particular
vinyl hydrogen.
Each of these patterns is different from the
others.
JAB 11-18 Hz, BIG JAC 0 5 Hz, SMALL JBC
5 10 Hz, MIDDLE
Look at it this way...
This signal appears to have big (caused by trans
H-CC-H) and small (caused by geminal HHC)
splittings. The H being excited must have both a
trans and geminal H. The H must be Ha.
48
Analysis in greater depth - 2.
Each of these patterns is different from the
others.
JAB 11-18 Hz, BIG JAC 0 5 Hz, SMALL JBC
5 10 Hz, MIDDLE
And the middle signal.
This signal appears to have big (caused by trans
H-CC-H) and middle (cis H-CC-H) splittings. The
H being excited must have both a trans and cis H.
The H must be Hb.
49
Analysis in greater depth - 3.
Each of these patterns is different from the
others.
JAB 11-18 Hz, BIG JAC 0 5 Hz, SMALL JBC
5 10 Hz, MIDDLE
And the right signal.
This signal appears to have small (caused by
geminal HHC) and middle (cis H-CC-H)
splittings. The H being excited must have both a
geminal and cis H. The H must be Hc.
50
As with pi bonds, cyclic structures also prevent
rotation about bonds
Approximately the same vinyl system as before.
Non equivalent geminal hydrogens. Analyze this.
No spin spin splitting of these hydrogens.
Nothing close enough
51
Note the roof effect. For similar hydrogens
the inner peaks can be larger.
Figure 13.25, p.516
52
Coincidental Overlap Non-equivalent nuclei have
same coupling constant.
Ha will be a triplet (two Hb) Likewise for Hc.
We analyze Hb.
A triplet of triplets
Here Ha and Hc have same coupling with Hb (Jab
Jbc), ,, coincidental overlap splits to 5, four
equivalent neighbors.
53
Analyze what happens as Jab becomes equal to
Jbc. First get peak heights when Jab does not
equal Jbc.
Recall heights in a triplet are 1 2 1
2
1
1
2
1
1
First split the Hb by Ha in ratio of 121.
2 x 2
Each component is split by Hc in ratio of 121.
Result for each final peak is product of
probabilities
1 x 1
1 x 2
2 x 1
54
Examine middle peak. Let Jbc become larger until
it equals Jab and add overlapping peaks together.
1 2 1 2 4 2 1 2 1
141
Peak heights shown when Jab does not equal Jbc.
55
Now adjacent peak.
1 2 1 2 4 2 1 2 1
22
1 4 6 4 1
56
Fast Exchange
Expect coupling between these hydrogens. Three
bond separation. There is no coupling observed
especially in acid or base. Reason exchange of
weakly acidic hydrogen with solvent. The
spectrometer sees an averaged hydrogen. No
coupling and broad peak.
ethanol
57
Return to Question of Equivalent hydrogens.
Stereotopicity Equivalent or Not?
Seem to be equivalent until we look at most
stable conformation, the most utilized
conformation.
Are these two hydrogens truly equivalent?
Seemingly equivalent hydrogens may be homotopic,
enantiotopic, diastereotopic. How to tell
replace one of the hydrogens with a D. If
produce an achiral molecule then hydrogens are
homotopic, if enantiomers then hydrogens are
enantiotopic, if diastereomers then
diastereotopic.
We look at each of these cases.
58
Homotopic
The central hydrogens of propane are homotopic
and have identical chemical shifts under all
conditions.
59
Enantiotopic
The hydrogens are enantiotopic and equivalent in
the NMR unless the molecule is placed in a chiral
environment such as a chiral solvent..
The hydrogens are designated as Pro R or Pro S
Pro S hydrogen.
Pro R hydrogen
This structure would be S
60
Diastereotopic
If diastereormers are produced from the
substitution then the hydrogens are not
equivalent in the NMR. Diastereotopic hydrogens.
The hydrogens are designated as Pro R or Pro S
Pro S hydrogen. (Making this a D causes the
structure to be S.)
Pro R hydrogen
This structure would be S
61
Example of diastereotopic methyl groups.
a and a
Diastereotopic methyl groups (not equivalent),
each split into a doublet by Hc
62
13C NMR

• 13C has spin states similar to H.
• Natural occurrence is 1.1 making 13C-13C spin
  spin splitting very rare.
• H atoms can spin-spin split a 13C peak. (13CH4
  would yield a quintet). This would yield
  complicated spectra.
• H splitting eliminated by irradiating with an
  additional frequency chosen to rapidly flip
  (decouple) the Hs averaging their magnetic field
  to zero.
• A decoupled spectrum consists of a single peak
  for each kind of carbon present.
• The magnitude of the peak is not important.

63
13C NMR spectrum
4 peaks ? 4 types of carbons.
64
13C chemical shift table
65
Hydrogen NMR Analysis Example 1
-(CO)-
Fragments (CH3)3C-, -CH2-, CH3-

1. Molecular formula given. Conclude One pi bond or
   ring.

2. Number of hydrogens given for each peak,
integration curve not needed. Verify that they
add to 14!
3. Three kinds of hydrogens. No spin-spin
splitting. Conclude Do not have non-equivalent H
on adjacent carbons.
4. The 9 equivalent hydrogens likely to be tert
butyl group (no spin-spin splitting). The 3
equivalent hydrogens likely to be methyl group.
The two hydrogens a CH2.
5. Have accounted for all atoms but one C and one
O. Conclude Carbonyl group!
6. Absence of splitting between CH2 and CH3.
Conclude they are not adjacent.
66
Example 2, C3H6O
1. Molecular formula ? One pi bond or ring
2. Four different kinds of hydrogen 1,1,1,3
(probably have a methyl group).
3. Components of the 1H signals are about equal
height, not triplets or quartets
4. Consider possible structures.
67
Possible structures
68
Chemical shift table Observed peaks were 2.5
3.1
ethers
Observed peaks were 2.5 3.1. Ether!
vinylic
Figure 13.8, p.505
69
Possible structures
70
NMR example
What can we tell by preliminary inspection.
Formula tells us two pi bonds/rings
Three kinds of hydrogens with no spin/spin
splitting.
71
Now look at chemical shifts
2. From chemical shift conclude geminal CH2CR2.
Thus one pi/ring left.
1. Formula told us that there are two pi
bonds/rings in the compound.
X
3. Conclude there are no single CCH- vinyl
hydrogens. Have CH2C-R2. This rules out a
second pi bond as it would have to be fully
substituted, CH2C(CH3)C(CH3)C(CH3)2 , to avoid
additional vinyl hydrogens which is C8H14.
In CH2CR2 are there allylic hydrogens
CH2C(CH2-)2?
72
Do the R groups have allylic hydrogens, CC-CH?

1. Four allylic hydrogens. Unsplit. Equivalent!
2. Conclude CH2C(CH2-)2

• Subtract known structure from formula of unknown
• C7H12
• - CH2C(CH2-)2
• ------------------------------------------
• C3H6 left to identify

But note text book identified the compound as
Remaining hydrogens produced the 6H
singlet. Likely structure of this fragment is
C(CH3)2-.