Transcript Chapter 13

Chapter 13
NMR Spectroscopy
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.
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
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).
Apply a strong external field…..
Both orientations have
same energy if no
magnetic field
Figure 13.4, p.499
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.
p. 481
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.
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
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.
Equivalence by Symmetry
Figure 13.6, p.500
Equivalent by rotation
H
H
H
Cl
H
H
equivalent
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.
Some molecules which have only one type of hydrogen - only one
signal
p.501
p. 484
Signal area: proportional to the number of hydogens producing the signal
Looking at the molecular structure
# Methyl hydrogens : # tert butyl hydrogens = 3:9 = 1:3
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
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.
Less shielded,
more deshielded,
downfield
Most electron density
around the H atoms,
most shielded, upfield.
Most deshielded, furthest
downfield. Sigma electrons
pulled away by oxygen.
p.504
Chemical shift table…
Figure 13.8, p.505
Less shielded
Further left,
downfield,
Relationship of chemical shift to electronegativity
Less electrons density around hydrogens as ascend table.
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 aren’t. Some other factor is at
work. Magnetic induction of pi bonds.
•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
In benzene the H atoms are on the outside and the induced magnetic field
augments the external field.
Figure 13.11, p.508
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.
Example
H
Expect the signal for
this hydrogen to be
split into seven by the
six equivalent
neighbors.
Cl
C
H3C
CH3
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).
Attempt to anticipate the splitting patterns in each molecule.
p. 491
p. 491
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
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”.
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.
Energy
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
The original
hydrogens will
single peak of
augment the
H1 has been
applied magnetic
split into two
field and about
peaks by the
50% will
effect of the
decrement it. Get
neighbor H2.
two peaks, a
The energy
double
difference is J
H1, being
excited
Here H2 augments external
field, peak moved downfield.
Here H2 decreases external
field, peak moved upfield.
3-pentanone
Coupling
constant, J, in
Hz
Same as gap
here.
The left side of molecule unaffected by right side.
Peak identification…
O
Figure 13.14, p.511
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.
gauche
anti
vinyl systems
Table 13.4, p.511
Spin-Spin Splitting
Now look at some simple
examples. Examine the
size of the peaks in the
splitting.
Ha is being excited.
Hb is causing spinspin splitting by
slightly increasing or
decreasing the
magnetic field
experienced by Ha.
Hb is augmenting external
field causing a larger
energy gap.
Hb decrementing external
field causing a smaller
energy gap.
Again Ha is flipping,
resonating. The two
Hb are causing spinspin splitting by
slightly changing the
magnetic field
experienced by Ha.
Two neighboring atoms
assist external field.
More energy needed to
excite. Peak is
“downfield”.
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
Three neighboring Hb’s causing
splitting when Ha is excited.
All Hb augment
Ha being
excited.
Two augment, one
decrement.
Three
equivalent Hb
causing spin
spin splitting.
One augment, two
decrement.
All decrement.
Figure 13.15c, p.512
Naturally if there are two non-equivalent nuclei they split each other.
Figure 13.17, p.513
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
More complicated system
Figure 13.20, p.514
Return to Vinyl Systems
Not equivalent (R1 is not
same as R2) because
there is no rotation about
the C=C bond.
Figure 13.21, p.514
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.
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….
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
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-C=C-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.
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-C=C-H) and middle (cis H-C=C-H) splittings.
The H being excited must have both a trans and
cis H. The H must be Hb.
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-C=C-H)
splittings. The H being excited must have both a
geminal and cis H. The H must be Hc.
As with pi bonds, cyclic structures also prevent rotation about bonds
Approximately
the same vinyl
system as
before.
No spin spin splitting of
these hydrogens.
Nothing close enough
Non equivalent geminal
hydrogens. Analyze this.
Note the “roof
effect”. For similar
hydrogens the
inner peaks can be
larger.
Figure 13.25, p.516
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.
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
2
1
1
1
First split the Hb
by Ha in ratio of
1:2:1.
2x2
Each component
is split by Hc in
ratio of 1:2:1.
Result for each final
peak is product of
probabilities
1x1
1x2
2x1
Examine
middle peak.
Let Jbc become
larger until it
equals Jab and
add overlapping
peaks together.
121242121
Peak heights shown when Jab does not equal Jbc.
1+4+1
Now adjacent
peak.
121242121
2+2
1
4
6
4
1
Fast Exchange
H
H
H
OH
H
H
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.
ethanol
The spectrometer sees an
“averaged hydrogen”. No
coupling and broad peak.
Return to Question of Equivalent hydrogens.
Stereotopicity – Equivalent or Not?
C(CH3)3
H2
C
C(CH3)3
H3C
H
CH3
Seem to be equivalent
until we look at most
stable conformation,
the most utilized
conformation.
H3C
H
Are these two hydrogens truly equivalent?
H
H
CH3
C(CH3)3
H
H
Seemingly equivalent hydrogens may be homotopic, enantiotopic,
diastereotopic.
H3C
H
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.
CH3
Homotopic
H
Achiral
H
replace one H with D
H
D
Achiral
The central hydrogens of propane are homotopic and have
identical chemical shifts under all conditions.
Enantiotopic
H
H
replace one H with D
Achiral
H
D
Chiral, have
two enantiomers.
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 R hydrogen
H D
This structure would be S
Pro S
hydrogen.
Diastereotopic
H D
H H
D H
H3C
H3C
H3C
replace H with D
Cl
Cl
H
H
Cl
H
produced diastereomers
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 R hydrogen
H D
H3C
Cl
H
This structure would be S
Pro S
hydrogen.
(Making this
a D causes
the
structure to
be S.)
Example of diastereotopic methyl groups.
H3C
c
a
H
d
OH
H3C
a'
a and a’
H
CH3
b
H
Diastereotopic methyl groups
(not equivalent), each split into
a doublet by Hc
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 H’s
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.
13C
NMR spectrum
4 peaks  4 types of carbons.
13C
chemical shift table
Hydrogen NMR: Analysis: Example 1
Fragments: (CH3)3C-, -CH2-, CH3-
-(C=O)-
O
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 nonequivalent 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.
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.
Possible structures
OH
O
HO
OCH3
O
O
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
Possible structures
OH
O
HO
OCH3
O
O
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.
Now look at chemical shifts
1. Formula told us
that there are two pi
bonds/rings in the
compound.
2. From chemical shift
conclude geminal CH2=CR2.
Thus one pi/ring left.
3. Conclude there are no single
C=CH- vinyl hydrogens. Have
CH2=C-R2.
This rules out a second pi bond
as it would have to be fully
substituted,
CH2=C(CH3)C(CH3)=C(CH3)2 , to
avoid additional vinyl hydrogens
which is C8H14.
In CH2=CR2 are there allylic
hydrogens: CH2=C(CH2-)2?
X
Do the R groups have allylic hydrogens, C=C-CH?
1. Four allylic hydrogens.
Unsplit. Equivalent!
2. Conclude CH2=C(CH2-)2
3. Subtract known structure from
formula of unknown…
C7H12
- CH2=C(CH2-)2
------------------------------------------
C3H6 left to identify
Remaining hydrogens
produced the 6H singlet.
Likely structure of this
fragment is –C(CH3)2-.
But note text book
identified the
compound as