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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
Infrared Spectroscopy
Chapter 12
Energy
Table 12.1, p.472
Final Exam Schedule, Thursday, May 22, 10:30 AM
Fang, MD10A
Kunjappu, MD10B
Kunjappu, MD10C
Metlitsky, MD10D
1127N
Zamadar
2143N
320A
Infrared spectroscopy causes
molecules to vibrate
A non-linear molecule having n atoms may have many different vibrations. Each
atom can move in three directions: 3n. Need to subtract 3 for translational motion
and 3 for rotations
# vibrations = 3 n – 6
(n = number of atoms in non-linear molecule)
Infrared radiation does not cause all possible vibrations to vibrate.
For a vibration to be caused by infrared radiation (infrared active) requires that the
vibration causes a change in the dipole moment of the molecule. (Does the
moving of the atoms in the vibration causes the dipole to change. Yes: should
appear in spectrum. No: should not appear.)
Consider
C=C bond
stretch…
H
H
F
H
H
H
F
H
ethylene
1,1 difluoro ethylene
What about 1,2 difluoro ethylene?
Different bonds have different resistances to stretching, different frequencies of vibration
Table 12.4, p.478
Typical Infra-red spectrum.
wavelength
Frequency, measured in “reciprocal centimeters”, the number
of waves in 1 cm distance.
Energy.
Figure 12.2, p.475
C-H
C=O
“fingerprint region”, complex
vibrations of the entire molecule.
Vibrations characteristic of
individual groups.
Figure 12.2, p.475
BDE of C-H
414
464
556
472
Table 12.5, p.480
BDE and CC stretch
376
727
966
Table 12.5, p.480
Alkane bands
Figure 12.4, p.480
Recognition of Groups: Alkenes (cyclohexene).
Compare these two
C-H stretches
Sometimes weak
if symmetric
Recognition of Groups: Alkynes (oct-1-yne)
This is a terminal alkyne and we expect to see
1. Alkyne C-H
2. Alkyne triple bond stretch (asymmetric)
Recognition of Groups: Arenes. (methylbenzene, toluene)
Out-of-plane bend;
strong
Recognition of Groups: Alcohols
The O-H stretch depends on whether there is hydrogen bonding present
Compare –O-H vs -O-H….O Hydrogen bonding makes it easier to move the H
with H bonding as it is being pulled in both directions; lower frequency
Recognition of Groups: Alcohols
Recognition of Groups: Ethers
No O-H bond stretch present but have C-O in
same area as for alcohol.
C-O stretch in assymetric ethers
sp3
CH3
O
sp2
Recognition of Groups: Amines
Easiest to recognize is N-H bond stretch: 3300 – 3500 cm-1. Same area as
alcohols. Note tertiary amines, NR3, do not have hydrogen bonding.
Hydrogen bonding can shift to lower frequency
Esters
One C=O stretch and two C-O stretches.
Recognition of Groups: Carbonyl
C=O stretch can be recognized reliably in area of 1630 – 1820 cm-1
•Aldehydes will also have C(O)-H stretch
•Esters will also have C-O stretch
•carboxylic acid will have O-H stretch
•Amide will frequently have N-H stretch
•Ketones have nothing extra
What to check for in an IR
spectrum
C-H vibrations about 3000 cm-1 can detect vinyl and terminal alkyne
hydrogens.
O-H vibrations about 3500 cm-1
C=O vibrations about 1630 – 1820 cm-1
C-O vibrations about 1000-1250 cm-1