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TRANSCRIPT
Introduction to Spectroscopy for Organic Compounds
Dr. Theeraphan Machan
Program of Natural Products Chemistry
School of Science
Mae Fah Luang University
What is spectroscopy?
2
Electromagnetic radiation
3
Electromagnetic componentsElectric component
magnetic component
nl
= c
El
= hc
4
time distance
1 cycle/sec = 1 Hz 1 cycle/cm = 1 cm-11 complete cycle
hv
absorption
matter transmitting
emission scattering
5
Types of Spectroscopy
Ultraviolet (UV) spectroscopy uses electron
transitions to determine bonding patterns.
Infrared (IR) spectroscopy measures the bond
vibration frequencies in a molecule and is used to
determine the functional groups.
Nuclear magnetic resonance (NMR)
spectroscopy detects signals from hydrogen
atoms and can be used to distinguish isomers.
Mass spectrometry (MS) fragments the
molecules and measures the masses.
6
7
Ultraviolet and visible spectroscopy
Theeraphan Machan, Ph.D
Program of Natural Products Chemistry
School of Science
Mae Fah Luang University
8
Ultraviolet and visible radiation
The nature of electronic excitations
When continuous radiation passes
through a transparent material, a portion
of the radiation may be absorbed.
As a result of energy absorption, atom or
molecules pass from a state of low energy
(ground state) to a state of higherenergy (excited state)
9
The excitation process.
10
Atomic Orbitals, Molecular Orbitals, and Nonbonding orbitals
The way to construct such molecular orbitals
(MOs) is to combine the atomic orbitals
(AOs) of the atoms that make up the molecules.
Linear Combination of Atomic Orbitals
(LCAO)
11
Sigma bonds,
12
13
Pi bonds,
14
15
Electronic energy levels
16
Electronic transitions
17
*
*
nE
n * **n
(> 270 nm)< 105 kcal (> 185 nm)
< 150 kcal
(> 165 nm)< 170 kcal
(< 165 nm)> 170 kcal
*
antibonding
bonding
18
Principles of absorption spectroscopy
19
Vibrational sublevels
E E1
E2
sublevels
sublevels
Rotational sublevels
UV-Vis spectrum
20
Wavelength (nm)
Abso
rbance
200 400 600
lmax
21
* Transitions
An electron in a bonding s orbital is excitedto the corresponding antibonding orbital.The energy required is large.
For example, methane (which has only C-Hbonds, and can only undergo *transitions) shows an absorbance maximumat 125 nm.
Absorption maxima due to *transitions are not seen in typical UV-VISspectra (200 - 700 nm)
22
* Transitions in an alkane
23
Saturated compounds containing atoms withlone pairs (non-bonding electrons) arecapable of n * transitions.
These transitions usually need less energythan * transitions.
They can be initiated by light whosewavelength is in the range 150 - 250 nm.
The number of organic functional groupswith n * peaks in the UV region is small,such as C-N, C-O, C-S, and C-X.
n * Transitions
n * Transitions in an amine
24
25
n * and * Transitions
Most absorption spectroscopy of organic
compounds is based on transitions of n or
electrons to the * excited state.
These transitions fall in an experimentally
convenient region of the spectrum (200 -
700 nm).
These transitions need an unsaturated
group in the molecule to provide the
electrons such as alkene (170 nm), Alkyne
(175 nm) and carbonyl (188 and 280 to
290).
n * and * Transition in carbonyl
26
* Transition in alkene
27
28
Auxochromes
Substituents that increase the intensity of the absorption,
and possibly the wavelength are call auxochromes.
CH3, OH, OR, NH2, and X (halogens).
Auxochromes may have any of four kinds of effects on the
absorption of chromophores.
◦ Bathochromic shift (red shift) – a shift to lower
energy or longer wavelength.
◦ Hypsochromic shift (blue shift) – a shift to high
energy or shorter wavelength.
◦ Hyperchromic effect – an increase in intensity.
◦ Hypochromic effect – a decrease in intensity.
29
Effects of auxochromes to chromophores
30
ความยาวคลื่น (nm)
Absorbance
blue red
Hyperchromic effect
Hypochromic effect
Hypsochromic shift Bathochromic shift
The effect of conjugation on alkenes
31
32
A comparison of the * energy gap in a series of polyenes of increasing chain length.
HOMO
LUMO
HOMO
LUMO
33
34
Infrared spectroscopy
Theeraphan Machan, Ph.D
Program of Natural Products Chemistry
School of Science
Mae Fah Luang University
35
The IR Region
Just below red in the visible region.
Wavelengths usually 2.5-25 mm.
More common units are wavenumbers, or
cm-1, the reciprocal of the wavelength in
centimeters.
Wave numbers are proportional to
frequency and energy.
36
Molecular Vibrations
Covalent bonds vibrate at only certain
allowable frequencies.
37
Stretching Frequencies
Frequency decreases with increasing atomic weight.
Frequency increases with increasing bond
energy.
38
Vibrational Modes
Nonlinear molecule with n atoms usually has
3n-6 fundamental vibrational modes.
39
Fingerprint of Molecule
Whole-molecule vibrations and bending
vibrations are also quantitized.
No two molecules will give exactly the same
IR spectrum (except enantiomers).
Simple stretching: 1600-3500 cm-1.
Complex vibrations: 600-1400 cm-1, called
the “fingerprint region.”
=>
40
IR-Active and Inactive
A polar bond is usually IR-active.
A nonpolar bond in a symmetrical molecule
will absorb weakly or not at all.
=>
41
An Infrared Spectrometer
42
FT-IR Spectrometer
Uses an interferometer.
Has better sensitivity.
Less energy is needed from source.
Completes a scan in 1-2 seconds.
Takes several scans and averages them.
Has a laser beam that keeps the
instrument accurately calibrated.
43
Carbon-Carbon Bond Stretching
Stronger bonds absorb at higher frequencies:◦ C-C 1200 cm-1
◦ C=C 1660 cm-1
◦ CC 2200 cm-1 (weak or absent if internal)
Conjugation lowers the frequency:◦ isolated C=C 1640-1680 cm-1
◦ conjugated C=C 1620-1640 cm-1
◦ aromatic C=C approx. 1600 cm-1
=>
44
Carbon-Hydrogen Stretching
Bonds with more s character absorb at a
higher frequency.
◦ sp3 C-H, just below 3000 cm-1 (to the right)
◦ sp2 C-H, just above 3000 cm-1 (to the left)
◦ sp C-H, at 3300 cm-1
=>
45
An Alkane IR Spectrum
=>
46
An Alkene IR Spectrum
=>
47
An Alkyne IR Spectrum
=>
48
O-H and N-H Stretching
Both of these occur around 3300 cm-1,
but they look different.
◦ Alcohol O-H, broad with rounded tip.
◦ Secondary amine (R2NH), broad with one
sharp spike.
◦ Primary amine (RNH2), broad with two sharp
spikes.
◦ No signal for a tertiary amine (R3N) =>
49
An Alcohol IR Spectrum
=>
50
An Amine IR Spectrum
=>
51
Carbonyl Stretching
The C=O bond of simple ketones,
aldehydes, and carboxylic acids absorb
around 1710 cm-1.
Usually, it’s the strongest IR signal.
Carboxylic acids will have O-H also.
Aldehydes have two C-H signals around
2700 and 2800 cm-1.
=>
52
A Ketone IR Spectrum
=>
53
An Aldehyde IR Spectrum
=>
54
O-H Stretch of a Carboxylic Acid
This O-H absorbs broadly, 2500-3500 cm-1, due
to strong hydrogen bonding.
55
Variations in C=O Absorption
Conjugation of C=O with C=C lowers the
stretching frequency to ~1680 cm-1.
The C=O group of an amide absorbs at an
even lower frequency, 1640-1680 cm-1.
The C=O of an ester absorbs at a higher
frequency, ~1730-1740 cm-1.
Carbonyl groups in small rings (5 C’s or
less) absorb at an even higher frequency.
56
An Amide IR Spectrum
57
Carbon – Nitrogen stretching
C - N absorbs around 1200 cm-1.
C = N absorbs around 1660 cm-1 and is
much stronger than the C = C absorption in
the same region.
C N absorbs strongly just above 2200 cm-1.
The alkyne C C signal is much weaker and
is just below 2200 cm-1 .
58
A Nitrile IR Spectrum
=>
59
Summary of IR Absorptions
60
Strengths and Limitations
IR alone cannot determine a structure.
Some signals may be ambiguous.
The functional group is usually indicated.
The absence of a signal is definite proof that
the functional group is absent.
Correspondence with a known sample’s IR
spectrum confirms the identity of the
compound. =>
61
Nuclear Magnetic Resonance Spectroscopy
Theeraphan Machan, Ph.D
Program of Natural Products Chemistry
School of Science
Mae Fah Luang University
62
Introduction
NMR is the most powerful tool available for
organic structure determination.
It is used to study a wide variety of nuclei:
◦ 1H
◦ 13C
◦ 15N
◦ 19F
◦ 31P
63
Nuclear Spin
A nucleus with an odd atomic number or an
odd mass number has a nuclear spin.
The spinning charged nucleus generates a
magnetic field.
64
External Magnetic Field
When placed in an external field, spinning protons act like bar magnets.
Spin +1/2Aligned
Spin -1/2Opposed
65
The spin state energy separation as a function of the
strength of the applied magnetic field B0
66
Two Energy States
The magnetic fields of the spinning nuclei will aligneither with the external field, or against the field.
A photon with the right amount of energy can be
absorbed and cause the spinning proton to flip.
67
E and Magnet Strength
Energy difference is proportional to the
magnetic field strength.
E = hn = h B0
2
Gyromagnetic ratio, , is a constant for each
nucleus (26,753 s-1gauss-1 for H).
In a 14,092 gauss field, a 60 MHz photon is
required to flip a proton.
Low energy, radio frequency.
Nuclear magnetic resonance process ; absorption occurs when n =
B0B0
+1/2 -1/2
69
The NMR Spectrometer
=>
70
The NMR Spectrum
=>
TMS
0 ppm
71
Magnetic Shielding
If all protons absorbed the same amount of
energy in a given magnetic field, not much
information could be obtained.
But protons are surrounded by electrons
that shield them from the external field.
Circulating electrons create an induced
magnetic field that opposes the external
magnetic field.
72
Shielded Protons
Magnetic field strength must be increased for a shielded proton to flip at the same frequency.
73
Protons in a Molecule
Depending on their chemical environment,
protons in a molecule are shielded by
different amounts.
Electronic structure of ethanol
74
75
NMR Signals
The number of signals shows how many
different kinds of protons are present.
The location of the signals shows how
shielded or deshielded the proton is.
The intensity of the signal shows the
number of protons of that type.
Signal splitting shows the number of
protons on adjacent atoms. =>
76
Number of Signals
77
Tetramethylsilane
TMS is added to the sample.
Since silicon is less electronegative than carbon, TMS protons are highly shielded. Signal defined as zero.
Organic protons absorb downfield (to the left) of the TMS signal.
Si
CH3
CH3
CH3
H3C
How many signal number are in the given compound?
78
2 3
How many signal number are in the given compound?
79
80
81
82
•Equivalent = proton magnetic environments
identical in every way
•Nonequivalent = proton magnetic environments
not identical in one or more ways
•Easier to test for nonequivalency than for
equivalency
•Models:
Build two copies; label protons in question
Superpose protons in question
If rest of molecule superposable then protons in
question are equivalent
How to test for proton equivalency?
83
Equivalent proton;
enantiomers
Non-equivalent proton;
diastereomers
84
Chemical Shift ()
Measured in parts per million (ppm.).
Ratio of shift downfield from TMS (Hz) to
total spectrometer frequency (Hz).
Same value for 60, 100, or 300 MHz
machine.
Called the delta scale.
Chapter 13 85
Delta Scale
86
Location of Signals
More electronegative atoms
deshield more and give larger
shift values.
Effect decreases with
distance.
Additional electronegative
atoms cause increase in
chemical shift.
87
Typical Values
88
89
Aromatic Protons, 7-8
90
Vinyl Protons, 5-6
91
Acetylenic Protons, 2.5
92
Aldehyde Proton, 9-10
Electronegative
oxygen atom
93
O-H and N-H Signals
Chemical shift depends on concentration.
Hydrogen bonding in concentrated solutions
deshield the protons, so signal is around 3.5
for N-H and 4.5 for O-H.
Proton exchanges between the molecules
broaden the peak.
94
Carboxylic Acid Proton, 10+
Chapter 13 95
Number of Signals
Equivalent hydrogens have the same chemical
shift.
96
Intensity of Signals
The area under each peak is proportional to the
number of protons.
Shown by integral trace (integration line).
97
How Many Hydrogens?
When the molecular formula is known, each integral rise can be assigned to a particular number of hydrogens.
98
Spin-Spin Splitting
Nonequivalent protons on adjacent carbons
have magnetic fields that may align with or
oppose the external field.
This magnetic coupling causes the proton
to absorb slightly downfield when the
external field is reinforced and slightly
upfield when the external field is opposed.
All possibilities exist, so signal is split.
99
1,1,2-Tribromoethane
Nonequivalent protons on adjacent carbons.
100
Doublet: 1 Adjacent Proton
101
Triplet: 2 Adjacent Protons
102
The N + 1 Rule
If a signal is split by N equivalent protons,
it is split into N + 1 peaks.
103
Range of Magnetic Coupling
Equivalent protons do not split each other.
Protons bonded to the same carbon will split each other only if they are not equivalent.
Protons on adjacent carbons normally will couple.
Protons separated by four or more bonds will not couple.
104
Splitting for Ethyl Groups
105
Splitting for Isopropyl Groups
106
Coupling Constants ; J
Distance between the peaks of multiplet
Measured in Hz
Not dependent on strength of the external
field
Multiplets with the same coupling constants
may come from adjacent groups of protons
that split each other.
107
Coupling Constants
108
Complex Splitting
Signals may be split by adjacent protons, different from each other, with different coupling constants.
Example: Ha of styrene which is split by an adjacent Hb trans to its (J = 17 Hz) and an adjacent Hc cis to its (J = 11 Hz).
C C
H
H
Ha
b
c
109
Splitting Tree C C
H
H
Ha
b
c
110
Spectrum for Styrene
111
Stereochemical Nonequivalence
Usually, two protons on the same C are
equivalent and do not split each other.
If the replacement of each of the protons of a
-CH2 group with an imaginary “Z” gives
stereoisomers, then the protons are non-
equivalent and will split each other.
112
Some Nonequivalent Protons
C C
H
H
Ha
b
cOH
H
H
H
a
b
c
d
CH3
H Cl
H H
Cl
a b
113
Time Dependence
Molecules are tumbling relative to the
magnetic field, so NMR is an averaged
spectrum of all the orientations.
Axial and equatorial protons on cyclohexane
interconvert so rapidly that they give a single
signal.
Proton transfers for OH and NH may occur
so quickly that the proton is not split by
adjacent protons in the molecule.
114
Hydroxyl Proton
Ultrapure samples of ethanol show splitting.
Ethanol with a small amount of acidic or basic
impurities will not show splitting.
115
N-H Proton
Moderate rate of exchange.
Peak may be broad.
116
Identifying the O-H or N-H Peak
Chemical shift will depend on concentration and solvent.
To verify that a particular peak is due to O-H or N-H, shake the sample with D2O
Deuterium will exchange with the O-H or N-H protons.
On a second NMR spectrum the peak will be absent, or much less intense.
117
Carbon-13 NMR
12C has no magnetic spin.
13C has a magnetic spin, but is only 1% of the
carbon in a sample.
The gyromagnetic ratio of 13C is one-fourth
of that of 1H.
Signals are weak, getting lost in noise.
Hundreds of spectra are taken, averaged.
118
Fourier Transform NMR
Nuclei in a magnetic field are given a radio-
frequency pulse close to their resonance
frequency.
The nuclei absorb energy and precess
(spin) like little tops.
A complex signal is produced, then decays
as the nuclei lose energy.
Free induction decay is converted to
spectrum.
Chapter 13 119
Hydrogen and Carbon Chemical
Shifts
120
Combined 13C and 1H Spectra
121
Differences in 13C Technique
Resonance frequency is ~ one-fourth, 15.1
MHz instead of 60 MHz.
Peak areas are not proportional to number
of carbons.
Carbon atoms with more hydrogens absorb
more strongly.
122
Spin-Spin Splitting
It is unlikely that a 13C would be adjacent to another 13C, so splitting by carbon is negligible.
13C will magnetically couple with attached protons and adjacent protons.
These complex splitting patterns are difficult to interpret.
123
Proton Spin Decoupling
To simplify the spectrum, protons are continuously irradiated with “noise,” so they are rapidly flipping.
The carbon nuclei see an average of all the possible proton spin states.
Thus, each different kind of carbon gives a single, unsplit peak.
124
Off-Resonance Decoupling
13C nuclei are split only by the protons
attached directly to them.
The N + 1 rule applies: a carbon with N
number of protons gives a signal with
N + 1 peaks.
125
Interpreting 13C NMR
The number of different signals indicates the
number of different kinds of carbon.
The location (chemical shift) indicates the
type of functional group.
The peak area indicates the numbers of
carbons (if integrated).
The splitting pattern of off-resonance
decoupled spectrum indicates the number of
protons attached to the carbon.
126
Two 13C NMR Spectra
127
MRI
Magnetic resonance imaging, noninvasive
“Nuclear” is omitted because of public’s fear that it would be radioactive.
Only protons in one plane can be in resonance at one time.
Computer puts together “slices” to get 3D.
Tumors readily detected.
128
Mass Spectrometry
Theeraphan Machan, Ph.D
Program of Natural Products Chemistry
School of Science
Mae Fah Luang University
129
Mass Spectrometry
Molecular weight can be obtained from a very small sample.
It does not involve the absorption or emission of light.
A beam of high-energy electrons breaks the molecule apart.
The masses of the fragments and their relative abundance reveal information about the structure of the molecule.
130
Electron Impact Ionization
A high-energy electron can dislodge an
electron from a bond, creating a radical
cation (a positive ion with an unpaired e-).
e- + H C
H
H
C
H
H
H
H C
H
H
C
H
H
H
H C
H
H
C
H
H
+ H
H C
H
H
C
H
H
H
+
131
Separation of Ions
Only the cations are deflected by the
magnetic field.
Amount of deflection depends on m/z.
The detector signal is proportional to the
number of ions hitting it.
By varying the magnetic field, ions of all
masses are collected and counted.
132
Mass Spectrometer
133
The Mass Spectrum
Masses are graphed or tabulated according to
their relative abundance.
134
The GC-MS
A mixture of compounds is separated
by gas chromatography, then identified
by mass spectrometry.
135
High Resolution MS
Masses measured to 1 part in 20,000.
A molecule with mass of 44 could be C3H8,
C2H4O, CO2, or CN2H4.
If a more exact mass is 44.029, pick the
correct structure from the table:
C3H8 C2H4O CO2 CN2H4
44.06260 44.02620 43.98983 44.03740
136
Molecules with Heteroatoms
Isotopes: present in their usual abundance.
Hydrocarbons contain 1.1% C-13, so there
will be a small M+1 peak.
If Br is present, M+2 is equal to M+.
If Cl is present, M+2 is one-third of M+.
If iodine is present, peak at 127, large gap.
If N is present, M+ will be an odd number.
If S is present, M+2 will be 4% of M+. =>
137
Isotopic Abundance
81Br
138
Mass Spectrum with Sulfur
139
Mass Spectrum with Chlorine
140
Mass Spectrum with Bromine
141
Mass Spectra of Alkanes
More stable carbocations will be more
abundant.
142
Mass Spectra of Alkenes
Resonance-stabilized cations favored.
143
Mass Spectra of Alcohols
Alcohols usually lose a water molecule.
M+ may not be visible.
References
1. Pavia, D.L.; Lampma, G.M.; Kriz, G.S.; and
Vyvyan, J.R., Introduction to spectroscopy,
2009, 4th ed, USA, Brooks/Cole, 2009.
2. Clayden, J.; Greeves, N.; Warren, S.; and
Wothers, P., Organic Chemistry, 1st ed.,UK
Oxford University Press, 2001.
3. Field, L.D.; Sternhell, S.; and Kalman, J.R.,
Organic Structures from Spectra, 3rd ed., UK,
John Wiley & Sons, 2002.
144