Chemistry 341
Spectroscopy of Organic
Compounds
Modern Spectroscopic Methods

Revolutionized the study of Organic Chemistry

Determine the exact structure of small to
medium size molecules in a few minutes.

Nuclear Magnetic Resonance (NMR) and
Infrared Spectroscopy (IR) are particularly
powerful techniques which we will focus on.
Interaction of Light and Matter
The Physical Basis of Spectroscopy

Quantum properties of light (photons)

Quantum properties of matter (quantized
energy states).

Photons of light act as our “quantum
probes” at the molecular level giving us
back precise information about the energy
levels within molecules.
The Electromagnetic Spectrum

Continuous

Covers a wide range of wavelengths of
“light” from radio waves to gamma rays.

Wavelengths (l) range from more than ten
meters to less than 10-12 meter.
The Electromagnetic Spectrum
Relationship Between Wavelength,
Frequency and Energy

Speed of light (c) is the same for all wavelengths.

Frequency (n), the number of wavelengths per second, is
inversely proportional to wavelength:
n = c/l

Energy of a photon is directly proportional to frequency
and inversely proportional to wavelength:
E = hn = hc/l
(where h = Plank’s constant)
Wavelength/Spectroscopy
Relationships
Spectral Region Photon Energy
UV-Visible
Molecular
Energy
Changes
~ 100 kcal/mole Electronic
Infrared (IR)
~ 10 kcal/mole
Bond vibrations
Radio
< 0.1 kcal/mol
Nuclear Spin
states in a
magnetic field
Spin of Atomic Nuclei

Spin 1/2 atoms:
mass number is odd.
examples: 1H and 13C.

Spin 1 atoms: mass number is even.
examples: 2H and 14N.

Spin 0 atoms: mass number is even.
examples: 12C and 16O.
Magnetic Properties of the Proton
Related to Spin
Energy States of Protons in a Magnetic
Field
D E = l absorbed light
Applied
M agnetic
Field
H ext
Nuclear Magnetic Resonance
(NMR)

Nuclear – spin ½ nuclei (e.g. protons) behave as
tiny bar magnets.

Magnetic – a strong magnetic field causes a
small energy difference between + ½
and – ½ spin states.

Resonance – photons of radio waves can match
the exact energy difference between the + ½
and – ½ spin states resulting in absorption of
photons as the protons change spin states.
The NMR Experiment

The sample, dissolved in a suitable NMR solvent
(e.g. CDCl3 or CCl4) is placed in the strong
magnetic field of the NMR.

The sample is bombarded with a series of radio
frequency (Rf) pulses and absorption of the
radio waves is monitored.

The data is collected and manipulated on a
computer to obtain an NMR spectrum.
The NMR Spectrometer
The NMR Spectrometer
The NMR Spectrum




The vertical axis shows the intensity of Rf
absorption.
The horizontal axis shows relative energy at
which the absorption occurs (in units of parts per
million = ppm)
Tetramethylsilane (TMS) in included as a
standard zero point reference (0.00 ppm)
The area under any peak corresponds to the
number of hydrogens represented by that peak.
The NMR Spectrum
Chemical Shift (d)

The chemical shift (d) in units of ppm is defined as:
d = distance from TMS (in hz)
radio frequency (in Mhz)

A standard notation is used to summarize NMR spectral
data. For example p-xylene:
d 2.3 (6H, singlet)
d 7.0 (4H, singlet)

Hydrogens in identical chemical environments
(equivalent hydrogens) have identical chemical shifts.
Shielding – The Reason for
Chemical Shift Differences

Circulation of electrons within molecular
orbitals results in local magnetic fields that
oppose the applied magnetic field.

The greater this “shielding” effect, the
greater the applied field needed to achieve
resonance, and the further to the right
(“upfield”) the NMR signal.
Structure Effects on Shielding

Electron donating groups increase the
electron density around nearby hydrogen
atoms resulting in increased shielding,
shifting peaks to the right.

Electron withdrawing groups decrease the
electron density around nearby hydrogen
atoms resulting in decreased shielding,
(deshielding) shifting peaks to the left.
Structure Effects on Shielding
The Deshielding effect of an electronegative
substituent can be seen in the NMR
spectrum of 1-Bromobutane
Br – CH2-CH2-CH2-CH3
d (ppm):
3.4 1.8 1.5 0.9
No. of H’s:
2
2
2
3
Some Specific Structural Effects on
NMR Chemical Shift
Type of Hydrogen
d (ppm
Alkyl (C – H)
0.8 – 1.7
Alkyl Halide (RCH2X)
3-4
Alkene (R2C=CH2)
4-6
Aromatic (e.g. benzene) 6 - 8
Carboxylic Acid
(RCOOH)
10 - 12
Spin-Spin Splitting

Non-equivalent hydrogens will almost always
have different chemical shifts.

When non-equivalent hydrogens are on adjacent
carbon atoms spin-spin splitting will occur due to
the hydrogens on one carbon feeling the
magnetic field from hydrogens on the adjacent
carbon.

The magnitude of the splitting between two
hydrogens (measured in Hz) is the coupling
constant, J.
Spin-Spin Splitting
Origin of the Doublet
Spin-Spin Splitting
Origin of the Triplet
Spin-Spin Splitting
Origin of the Quartet
Pascal’s Triangle
Pattern
Singlet
Doublet
Triplet
Quartet
Quintet
Relative Peak Height
1
1:1
1:2:1
1:3:3:1
1 :4 : 6 : 4 : 1
The n + 1 Rule
If Ha is a set of equivalent hydrogens and Hx is an adjacent
set of equivalent hydrogens which are not equivalent to
Ha:

The NMR signal of Ha will be split into n+1 peaks by Hx.
(where n = # of hydrogens in the Hx set.)

The NMR signal of Hx will be split into n+1 peaks by Ha.
(where n = # of hydrogens in the Ha set.)
1H
NMR Spectrum of Bromoethane
1H
NMR Spectrum of
1-Nitropropane
Unknown A
(Figure 9.20 Solomons 7th ed.)

Formula: C3H7I

IHD = 0

No IR data provided
 1HNMR
d: 1.90 (d, 6H), 4.33 (sept., 1H)
Unknown B
(Figure 9.20 Solomons 7th ed.)

Formula C2H4Cl2

IHD = 0

No IR data given
 1HNMR
d: 2.03 (d, 3H), 4.32 (quartet, 1H)
Unknown C
(Figure 9.20 Solomons 7th ed.)

Formula: C3H6Cl2

IHD = 0

No IR data given
 1HNMR
d: 2.20 (pent., 2H), 3.62 (t, 4H)
Exceptions to the n+1 Rule

The n+1 rule does not apply when a set of
equivalent H’s is split by two or more other
non-equivalent sets with different coupling
constants.

The n+1 rule does not apply to second
order spectra in which the chemical shift
difference between two sets of H’s is not
much larger than the coupling constant.
NMR: Some Specific Functional
Group Characteristics

O-H and N-H will often show broad peaks with
no resolved splitting, and the chemical shift can
vary greatly.

Aldehyde C-H is strongly deshielded.
(d = 9-10 ppm) and coupling to alkyl H’s on
adjacent carbon is small.

Carboxylic Acid O-H is very strongly deshielded.
(d = 10-12 ppm)
NMR: Some Specific Functional
Group Characteristics

Ortho splitting on aromatic rings is often
resolved, but meta and para splitting is
not.

Cis and trans H’s on alkenes usually show
strong coupling, but geminal H’s on
alkenes show little or no resolved
coupling.
Infrared Spectroscopy

Energy of photons in the IR region corresponds
to differences in vibrational energy levels within
molecules.

Vibrational energy levels are dependent on bond
types and bond strengths.

IR is highly useful to determine if certain types of
bonds (functional groups) are present in the
molecule.
IR Spectrum of Ethanol
IR Correlation Table
Key Functional Groups by Region
of the IR Spectrum
IR Spectrum of Benzaldehyde
IR Spectrum of Cyclohexanone
IR Spectrum of Propanoic Acid
Unknown A
(Figure 14.27 Solomons 7th ed.)

Formula = C9H12

IHD = 4

IR shows no medium or strong bands above
1650 cm-1 except C-H stretching bands around
3,000 cm-1
 1HNMR
d: 1.26 (d, 6H), 2.90 (sept., 1H),
7.1-7.5 (m, 5H)
Unknown B
(Figure 14.27 Solomons 7th ed.)

Formula = C8H11N

IHD = 4

IR shows two medium peaks between 3300 and
3500 cm-1 . No other medium or strong bands
above 1650 cm-1 except C-H stretching bands
around 3,000 cm-1
 1HNMR
d: 1.4 (d, 3H), 1.7 (s, br, 2H),
4.1(quart., 1H), 7.2-7.4 (m, 5H)
Unknown C
(Figure 14.27 Solomons 7th ed.)

Formula = C9 H10

IHD = 5

IR shows no medium or strong bands above
1650 cm-1 except C-H stretching bands around
3,000 cm-1
 1H
NMR d: 2.05 (pent., 2H),
2.90 (trip., 4H), 7.1-7.3 (m, 4H)
Unknown H
(Figure 9.48 Solomons 7th ed.)

Formula = C3H4Br2

IHD = 1

No IR data given

1HNMR d: 4.20 (2H), 5.63 (1H), 6.03 (1H)
Unknown Y
(Figure 14.34 Solomons 7th ed.)

Formula = C9H12O

IHD = 4

IR shows a strong, broad, absorbance centered
at 3400 cm-1
 1HNMR
d: 0.85 (t, 3H), 1.75 (m, 2H),
4.38 (s, br, 1H), 4.52 (t, 1H),
7.2-7.4 (m, 5H)