Chemistry 2213a Fall 2012 Western University
Topic 8. Structure Elucidation: IR, C-NMR, H-NMR 13 1
A. Electromagnetic Radiation
Energy is transmitted through space in the form of electromagnetic radiation,
which are oscillating waves of electric and magnetic fields, and this term used
to describe all the component waves found in a continuous spectrum known as
the electromagnetic spectrum. Structure Elucidation 2
Regardless of wavelength, all electromagnetic radiation propagates through
space at the speed of light (c). It is characterized its wavelength ( in metres)
and frequency (oscillations per second, in s ).−1
The shorter the , the higher the (more waves go by
per second if the wavelength is shorter).
B. Review of Infrared Spectroscopy
IR spectra are normally scanned from 4000 to 600 cm (wavenumbers, which
are reciprocal of in cm). Spectra can be divided into the functional group and
Aspirin Structure Elucidation 3
The absorption of IR light causes vibrational excitation. Different functional
groups absorb at different wavelengths, which in turn allows the identification
of functional groups.
Absorption appears as peaks of lowered transmittance. Correlation tables list
the characteristic absorptions, as well as their intensities, for various functional
Alkanes, alkenes, and alkynes Structure Elucidation 4
Sample spectra are shown below. Note that the alkene and alkyne signals
near 3100 and 3300 cm , respectively, are only present if an alkenyl or alkynyl
hydrogen is present. Structure Elucidation 5
Question: The IR shown is for which compound, H C=CH CH CH CH C2 or 2 2 2 2 3
HC≡CH CH CH CH CH
2 2 2 2 3 Structure Elucidation 6
Alcohols and ethers
Both contain a C–O
bond, but only alcohols
contain an O−H bond. Structure Elucidation 7
Compounds with benzene rings give distinctive IR spectra. Structure Elucidation 8
If an N–H bond is present, the region from 3100 – 3500 cm will show two
peaks in the case of 1°amines (two N–H bonds), but 2° amines will have just
one peak. Structure Elucidation 9
C. Nuclear Magnetic Resonance Spectroscopy
1. Energy absorption by nuclei
An electron has a spin, and spinning charges create magnetic fields (spin
quantum number, m = ±½s. A nucleus that has an odd mass number (total
number of protons and neutrons) also has a spin. All nuclei with an odd mass
number can have two spin states, +½ and −½ .
o Examples include H, C, F, P 19 31
For nuclei with the two possible spin states, they can be viewed as spinning
bar magnets. If such nuclei are placed in a magnetic field (B ), t0e two spin
states assume two different energy levels. Structure Elucidation 10
Once the energy separation has been achieved, the nuclei can the promoted
from the lower- to higher-energy state using radiofrequency radiation.
The absorption of radiation by nuclei in a magnetic field is called nuclear
magnetic resonance, NMR.
For organic chemists, the two most useful types are:
o C NMR (“carbon NMR”)
o H NMR (“proton NMR”)
For all types of NMR, the stronger the magnetic field (B ),0the greater the
energy separation between the two states. Structure Elucidation 11
Very strong magnetic fields are required to make the separation large enough
to be useful. For example, for C:
o At B 0 7.5 T (Tesla), E = 0.0072 cal/mol
o This energy separation corresponds to an electromagnetic frequency of 75
MHz and a wavelength of 1000 m (radiofrequency).
Most commercial NMR instruments consist of a fixed magnetic field, a
radiofrequency generator to “radiate” the nuclei, a receiver to detect the
absorption, and a recorder.
The sample is dissolved in
a solvent that does not
interfere with the magnetic
field. The solvent chosen
must be not contain nuclei
with spin, so no hydrogens!
A common solvent used is
CDCl .3 Structure Elucidation 12
2. C-NMR spectroscopy
The most abundant isotope of carbon is C (98.9%), which has no nuclear
spin. Yet, all substances containing carbon will contaC isotope (1.1%),
which does have spin and will give a signal in an NMR instrument.
The local magnetic fields of carbon atoms in a molecule are affected by all
neighbouring atoms. This makes the actual magnetic field around a specific
atom slightly different than the applied field.
Therefore, every different carbon environment in a molecule has a d13ferent
energy separation and therefore absorbs a different radiofrequency. Thus, C-
NMR spectra reveal the number of magnetically non-equivalent C atoms and
their identity (chemical shifts).
We can determine the O 1 signal O OH citric acid
number of diffe13nt C HO 4 signals
atoms (and thus C
signals) by looking at O HO OH
molecular symmetry. 3 signals
O Structure Elucidation 13
Instead of plotting signals using energy on the x-axis, they are plotted as
chemical shifts () in units of ppm.
Furthermore, absorption 13gnals are plotted
relative to the single C absorption signal of a tetramethylsilane (TMS)
reference compound, TMS. So, the signal at 0 Si
ppm, corresponds to that of TMS.
E13’s, and greater s-character in hybrid C orbitals, shift
C signals to the left of TMS. This is termed downfield. Structure Elucidation 14
Shown below is a correlation table for C NMR. Structure Elucidation 15
Example: The C-NMR spectrum
of 1-pentanol shows the five
signals, with the ones furthest
downfield closest to the OH.
Example: diethyl ether Structure Elucidation 16
Question: Predict the number of signals for the following compounds. Do the
spectra agree with their structure?
4-chlorophenol Structure Elucidation 17
Important note: signal heights on C-NMR spectra do not correspond with the
number of C atoms they represent (see 1-pentene). However, higher signals
usually result from greater numbers of equivalent nuclei (see 4-chlorophenol).
The analysis of IR and C-NMR spectra, together with info about units of
unsaturation, often gives enough information to elucidate molecular structure
(see assigned problems). Structure Elucidation 18
3. H-NMR spectroscopy
Like C, H atoms have spins of ±½ and are NMR-active. Non-equivalent
protons in a molecule will absorb a different energy of radiofrequency energy.
Absorptions are plotted as chemical shifts that typically range from 0 – 13 ppm.
For example, methyl acetate contains two H environments.
In methyl acetate, the peak at 3.7 ppm is due to the threeO H
protons of the methoxy group, while the 2.0 ppm peak is H
due to the three protons of the ac3l CH group. H O H
See textbook “How To 12.2” for more information on H
how to determine the number of proton signals. Structure Elucidation 19
Chemical shifts are determined by the type, hybridization, and electronegativity
of the atom to which the H is bonded. H chemical shifts give us the same
structural information for protons that C shifts give for carbon atoms. Structure Elucidation 20
However, H NMR has two extra features that make increase its usefulness for