CHE334—Identification of an Unknown Compound By NMR/IR/MS
Purpose
The object of this experiment is to determine the structure of an unknown compound using IR, 1H-NMR, 13C-NMR and
Mass spectroscopy.
Infrared Spectroscopy
For Infrared spectroscopy we will run samples using an attenuated total reflectance accessory for our spectrometer.
Attenuated total reflectance infrared (ATR/IR) spectra are obtained by placing a drop of a liquid sample or by pressing small
amount of a solid sample on an internal reflection element, e.g., zinc selenide (ZnSe) or germanium (Ge). IR radiation enters
the end of the internal reflective element and reflects down the length of the crystal. At each internal reflection, the IR
radiation actually penetrates a short distance (~1 µm) from the surface of the IRE into the sample (Figure 1). It is this unique
physical phenomenon that enables one to obtain infrared spectra of samples placed in contact with the IRE. We are going to
use IR spectroscopy for functional group determination; therefore, we will be most interested in the regionsfrom 4000 cm-1 to
1500 cm-1.
evanescent wave
Sample
IRE
Sample
IRE
IR radiation
Detector
Figure 1. Schematic Representation of Workings of ATR Accessory
Instructions for using the IR100 Spectrometer are found on the lab’s Canvas website (Under PagesàTechnique).
NMR
Your samples will be submitted for 1H-NMR (“proton NMR”) and 13C-NMR on our department’s AV400 spectrometer.
This is a 400 MHz spectrometer equipped with an “autosampler” which runs samples robotically. All that you will have to do
is prepare your sample. Your TA will then submit your samples for analysis. You will then either be provided with the data to
process using the software package M-Nova (the use of which will be demonstrated by your TA), or with the spectra
themselves.
See the Canvas website (Under PagesàTechnique) for instructions on how to prepare a routine NMR sample.
Mass Spectrometry
The information provided by mass spectroscopy is used in three ways when identifying unknown organic compounds:
determination of molecular weight, determination of molecular formula, and the determination of molecular structures. The
sample is introduced into the mass spectrometer through an inlet system. In our case, the sample is introduced through a
capillary gas chromatograph. The mass spectrum is then run on a single pure compound as it eludes from the gas
chromatograph. From the gas chromatograph, the sample enters a section of the mass spectrometer called the ionization
chamber. In the ionization chamber, a ray of very high-energy electrons bombards the sample. If these electrons collide with
a sample molecule they have enough energy to knock electrons out of the sample molecules. The result is a radical cation of
the original sample molecule known as the molecular ion (M+.).
1
M
+
e
M
+
2e
From the ionization chamber the ionized molecules are accelerated and focused into a beam of rapidly moving ions by a
means of charged plates. These ions are introduced into a chamber that is spherically surrounded by electrodes. The electric
field produced by these electrodes causes the ions to travel in circles inside this chamber. The electric field is slowly varied
to allow sequential release of ions of increasing mass to the detector (Figure 2). The plot of ion abundance versus mass-tocharge (m/z) ratio is known as the mass spectrum. Since most molecules only lose one electron the charge on each ion is
typically one. If the charge is one the mass-to-charge ratio is equal to the mass of the ion. In a typical mass spectrum, many
different m/z peaks will be observed because after being hit by a high-energy electron, molecules have a tendency to break
apart or fragment. A peak is observed in the mass spectrum for each fragment of the molecule that retained a positive charge.
Figure 2.
Schematic Diagram of a Quadrupole Ion Storage Trap Mass Spectrometer with Attached Gas Chromatograph
The most abundant ion formed in the mass spectrometer gives the tallest peak in the mass spectrum. This peak, called
the base peak, is often not the heaviest ion (the peak most to the right). The tallest peak corresponds to the most stable cation
and is often the result of the molecular ion fragmenting due to the large amount of energy it possesses after being hit by an
electron in the ionization chamber. Since molecules tend to fragment to form more stable cations, carefully studying the
fragmentation peaks can provide a large amount of structural information about the compound. The heaviest ion peak is
usually the molecular ion; however, at times the molecular ion is too unstable to make it from the ionization chamber to the
detector before completely fragmenting in which case no molecular ion is seen. The mass of the molecular ion is the m/e
value from the spectrum assuming only one electron was removed from the molecule. (It is extremely rare for a single small
molecule to lose more than one electron so this is a safe assumption.) The mass of the molecular ion is equivalent to the
molecular weight of the compound.
The mass spectra shows the molecular weight of individual ions; the intensity of the peak corresponding to the number
of those ions hitting the detector. If a particular molecule of a compound contains an atom of a heavier isotope the spectra
will show a peak at a larger m/z value. The intensity of this new peak will depend on how many molecules contain this
heavier isotope. Since almost all atoms exist in nature in multiple isotopic forms, in theory, the mass spectrum should show
peaks for molecules containing all these isotopes. In practice, the natural abundance of most atomic isotopes is too low for
the mass spectrometer to detect molecules containing these isotopes. However, certain isotopes have a high enough natural
abundance that molecules containing these isotopes are observable in the mass spectrum. When these peaks are observable
in the mass spectrum they provide very valuable information about the molecular formula of the compound. The masses of
these isotopes are usually one or two mass units above the mass of the most abundant isotope. Therefore, in addition to
looking for the molecular ion peak (M+), it is worth attempting to locate the M + 1 and M + 2 peaks. Comparing the
intensities of the M+ peak and the M + 2 peak provides information about the number of sulfurs, silicons and particularly
halogens in the molecule. The intensity of the M + 1 peak is very valuable in determining the number of carbons in the
molecule.
2
In this lab, mass spectra of the unknowns will be taken and provided for you.
Determining the Structure of an Unknown
Just like a detective you now have collected all the evidence you need to determine the structure of your unknown. You
just need to put all the clues together to solve the mystery. Like any detective investigation there is not a set linear way of
examining the evidence that leads to the solution in every case. Every investigation and its corresponding evidence are
unique. The approach presented here will get you close but in the end you still need to be the one to put all the pieces
together. Also like a detective, it is useful to have a notebook to record key information in as you collect it.
Functional Groups From the Infrared:
Assign major peaks between 4000 cm-1 and 1500 cm-1 in the Infrared spectra to their corresponding functional
groups. Pay particular attention for peaks corresponding to OH and C=O
Molecular Weight and Molecular Formula by Mass Spectroscopy:
Identify the cluster of peaks corresponding to the molecular ion (the highest mass peaks not the tallest). Determine
if the unknown compound contains chlorine, or bromine and if so how many of each by comparing the pattern of peaks in the
mass spectrum to the patterns given below. (Be sure to check the printed list of peaks not just the spectra.) Each peak in the
patterns are separated by two mass units. Very small peaks that differ from the larger peaks by single mass units are not
shown below but will also be very important in determining the molecular formula of your unknown. If you determine that
the molecule contains chlorine or bromine draw them in your investigation notebook.
The most substantial contributors to the M + 1 peak are molecules of the molecular ion containing carbon-13. The odds
that a molecule contains a carbon-13 atom are dependent the number of carbons in the molecule and the percent of naturally
occurring carbon that is carbon-13. Since we know that 1.08% of carbon is carbon-13 and we can determine the odds that a
molecule contains a carbon-13 by comparing intensity of the M + 1 peak to the intensity of the molecular ion peak (M+ ) we
can calculate how many carbons are in our molecule. You can do the algebra if you like but the resulting formula relating the
number carbons in the molecule to the relative sizes of the M+ and M + 1 peaks is given below. Since the M + 1 peak is
usually a weak peak the error in its intensity can be quite substantial. As a result this calculation is really only an
approximation. Perform the calculation and record the calculated number of carbons in your investigation notebook.
# of carbons =
(intensity of the M + 1 peak)
+
[(intensity of the M peak) + (intensity of the M + 1 peak)]
x
1
0.0108
Add up the atomic weights of all the atoms you have determined are in the molecule so far including any functional
groups identified by Infrared Spectroscopy. If more than one isotope is common for any of these atoms use the lowest weight
isotope (e.g. use 35 for chlorine, and 79 for bromine). Subtract this number from the molecular weight of the compound as
determined from the molecular ion peak in the mass spectra. If your molecule contains halogens use the lowest m/e value
from the molecular cluster. The resulting number corresponds to the number of hydrogens in the molecule. If you get a
negative number reduce the number of carbons by one and recalculate. If the number of hydrogens plus halogens, minus
two, is more than twice the number of carbons then increase the number of carbons by one and recalculate.
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Degrees of Hydrogen Deficiency
Determine the number of rings plus π-bonds in your molecule by calculating the degrees of hydrogen deficiency.
Record this value in your investigation notebook. If the value comes out to four or higher there is a good chance your
compound contains an aromatic ring. Verify this by looking in the proton NMR for peaks around 7.0 ppm and the 13C-NMR
for peaks around 130 ppm. If you have an aromatic ring record this in your investigation notebook as well.
D.H.D. = (# of C x 2) + (# of N) + 2 - (# of H) - (# of halogens)
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Interpretation of 13C-NMR
Count the number of peaks in the 13C-NMR. (Note, the three equal sized peaks centered at 77 ppm are the carbon o f
CDCl3, the solvent; do not count these three peaks.) If this number is less than the number of carbons you determined the
molecule has then the molecule must possess symmetry making some of the carbons equivalent to each other. This
symmetry determination is particularly useful if your compound contains a benzene ring. Since you know benzene rings
have six carbons there should be six peaks around 130 ppm unless there is symmetry within the ring. This symmetry
determination can help you determine the substitution pattern on the ring. A few examples are given below. The rings
possessing a plane of symmetry will only give four peaks in the 13C-NMR instead of six.
X
X
X
X
X
Y
Y
X
Y
plane of symmetry
no symmetry
no symmetry
X
X
X
X
Y
no symmetry
Y
no symmetry
plane of symmetry
Y
X
X
X
Y
plane of symmetry
Y
X
X
no symmetry
no symmetry
Y
plane of symmetry
X
If the Infrared spectra indicated that there are particular functional groups in the molecule verify that the appropriate 13CNMR peaks exist. That is if the Infrared spectra indicated a carbonyl group there should be a 13C-NMR peak around 195
ppm. If the Infrared spectra indicated an alcohol there should be a 13C-NMR peak around 60 ppm. If this verification fails
recheck your Infrared interpretation. If the interpretation is correct you may need to rerun the Infrared spectrum.
Interpretation of 1H-NMR
Excluding obvious impurities and the TMS peak, add up all the integration numbers from the bottom of the spectra.
Divide this number by the number of hydrogens in the molecule. This number corresponds to the integration of one
hydrogen in arbitrary units. Using this value calculate the number of hydrogens producing each peak in the NMR. As a first
approximation, assume that signals integrating to three hydrogens are produced by CH3’s. For a signal that integrates to two
hydrogens there are three possibilities: a CH2, two identical CH (would have to see symmetry in 13C-NMR), or an NH2.
Signals that integrate to one are produced by CH, OH, or one of the hydrogens of a CH2 when the CH2 is not free to rotate.
Any signals that integrate to more than three imply that the molecule must have symmetry, which should also be observed in
the 13C-NMR. The integration of the peaks corresponding to the aromatic hydrogens (peaks at approximately 7.0 ppm)
provides information about the substitution of the benzene (five aromatic hydrogens implies mono substituted, four aromatic
hydrogens implies disubstituted, three aromatic hydrogens implies trisubstituted etc.).
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Determine the splitting pattern of each signal in the NMR. Using the N + 1 rule determine how many hydrogens are on
atoms adjacent to the hydrogens producing the signal (N is the number of hydrogens on adjacent atoms and N + 1 is the
number of peaks the signal is split into). For example a doublet implies one adjacent hydrogen, a triplet implies two adjacent
hydrogens, a quartet implies three adjacent hydrogens and a quintet implies four adjacent hydrogens. In your investigation
notebook draw out any fragments you can deduce from this analysis. Remember if you deduce that there is a CH2 next to a
CH somewhere in the NMR spectra you must account for a CH that can be next to a CH2 and that these two are the same
fragment of the molecule (do not count them twice).
The Final Structure
Using all the information you have in your investigation notebook (molecular formula, functional groups, existence of an
aromatic ring, symmetry, and the molecular fragments from 1H-NMR) propose possible structures that are supported by all
the evidence. Keep in mind that you cannot violate the octet rule when proposing your structures. There should be very few
possible structures that fit all the information you have collected so far. Check each proposed structure to insure they are
completely consistent with all the collected spectra, particularly the 1H-NMR and 13C-NMR including observed chemical
shifts.
Laboratory Report Hints
The data section should list the location and type of peaks seen in the spectra. You can also tape copies of your spectra into
your lab notebook. In the discussion, guide the reader through your reasoning in determining the structure of your unknown
and what you determined the structure to be. If there are any ambiguities you should mention them and what they may mean
to your final structural assignment.
The report should be written as if the reader DOES NOT has access to the original spectra. You must tell where peaks
are and what they look like. Avoid statements like “The peak labeled A in the 1H-NMR indicates the presence of aromatic
hydrogens.” The reader does not have the spectra so they cannot see the peak labeled A. A better way would be, “The two
doublets each integrating to two hydrogens at 7.3 ppm and 6.9 ppm in the 1H-NMR indicate the presence of aromatic
hydrogens.”
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Spectroscopy: Procedure
Hazards: Deuterochloroform is a possible carcinogen and an irritant. The
unknowns could have one or more of the following properties: flammable; can
cause skin/eye/respiratory track irritation; lachrymator. Acetone is flammable.
Wear goggles and gloves, work in the fume hood, and keep chemical containers
closed when not in use.
Day 1
1. Obtain a sample of an unknown compound from your TA. Record its number in
your lab notebook.
2. Prepare an NMR sample of your compound, using CDCl3 as the solvent. See
the handout on Canvas under PagesàTechniques for instructions.
3. Obtain an IR spectrum of the unknown. See the handout on Canvas under
PagesàTechniques for instructions.
Day 2
Your TA will provide you with NMR and MS data for your unknown. Copies of all
spectra can be taped into your notebook. Analyze your spectral data and
determine as best you can the structure of your unknown compound.
Refer to the Spectroscopy handout on Canvas under PagesàSpectroscopy for
additional information on spectroscopic techniques and interpretation of results.