EXPERIMENT 11
IDENTIFYING COMPOUNDS BY IR AND NMR
rev 7/09
GOAL
This week we will study NMR (Nuclear Magnetic Resonance) and IR (Infrared
Spectrophotometry), two important pieces of modern laboratory instrumentation that chemists and
other scientists use routinely.
INTRODUCTION
Nuclear Magnetic Resonance, NMR
This week you will observe molecules interacting with radio-frequency energy in a technique
called nuclear magnetic resonance, NMR. The amount of structural information that can be
deduced from NMR spectra is unrivaled by other methods. When placed in a magnetic field, the
nuclei of many atoms exhibit multiple states, and will absorb or emit energy when changing from
one state to another. The energy difference between two of these nuclear states is detected in
NMR. We observe only one isotope at a time because different isotopes absorb in very different
portions of the spectrum. Different atoms of the same isotope will absorb slightly different
amounts of energy when they are in different environments.
H
O
H
H
C
H
C
H
H
C
H
H
Figure 1: 13C NMR spectrum and structure of isopropanol
13
C NMR
Figure 1 shows the 13C NMR spectrum of isopropanol. Each peak in the spectrum represents a
different type of carbon atom environment. The group of peaks at 77 ppm is due to the CDCl3
solvent, so we ignore this and have only two peaks and thus two types of carbon atoms in
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isopropanol. Although it has a total of three carbon atoms, the two end carbons in isopropanol
are identical and show up in the same NMR peak.
13
C NMR Summary
1. number of peaks = number of types of C atoms
2. peak position tells what the C atom is attached to (see Table 1)
3. ignore the CDCl3 solvent peak at 77 ppm
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H NMR:
Now let’s switch nuclei and look at 1H NMR. Figure 2a shows the 1H NMR spectrum of
isopropanol. Remember that we are considering only the hydrogen atoms this time. 1H NMR is
more complicated than 13C NMR, but rewards us with more information. Just as the number of
peaks in the 13C NMR told the number of types of carbon atoms, the number of peaks in the 1H
NMR tells us the number of types of hydrogen atoms. The spectrum also tells us how many
hydrogen atoms of each type are present, and how many hydrogen atoms are on the neighboring
carbon atoms.
Look closely at the spectrum in Figure 2a. It shows three peaks, representing the three types of
hydrogen atoms in isopropanol. These peaks aren’t as simple as the ones in our 13C spectrum.
The peak at 1.25 ppm is split into two closely spaced peaks; this unit is called a doublet. The
peak at 3.95 ppm is split into seven closely spaced peaks; this unit is called a septet. For the
moment, ignore this extra splitting. Look at the structure of isopropanol in Figure 2a. You
should be able to convince yourself that this structure has three types of hydrogen atoms: six
identical hydrogen atoms from the end -CH3 groups, one hydrogen on the central carbon atom,
and one hydrogen on the oxygen. So the three types of hydrogen match up with the spectrum’s
three peaks, but which peak corresponds to which type of hydrogen? As with the 13C NMR
spectrum, the positions of the peaks tell us what they hydrogen atoms are attached to. The
presence of double bonds and more electronegative elements such as oxygen or halogens moves
peaks further to the left. While peak position provides useful information, we will focus on area
and splitting to identify peaks.
One way to tell which peak goes with which atoms is to look at the area under the curve for each
peak. The integrated area of a peak is proportional to the number of hydrogen atoms it
represents. Our structure has six hydrogen atoms of one type, so the peak that corresponds to
these hydrogens should have an integrated area six times larger than the other peaks. Just from
looking at the spectrum, you can see that the peak at 1.25 is much larger than the others and
probably corresponds to the six -CH3 type hydrogen atoms. The small numbers that appear
directly below each peak are the integrated areas of those peaks as measured by the NMR
instrument. We make a ratio of these areas and then reduce the ratio. From our spectrum, the
areas are 31.7 : 37.7 : 190.1 which reduces to approximately 1:1:6. We now know that the peak
at 1.25 ppm is due to six identical hydrogen atoms, so this must be the six -CH3 type hydrogen
atoms.
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Figure 2a: Structure and 1H NMR Spectrum of Isopropanol
Figure 2b: Expanded Septet
H
O
H
H
C
H
C
H
H
C
H
H
How can we decide which of the two remaining hydrogen atoms gives rise to each of the two
other peaks? Now we look at the splitting. Splitting tells us how many hydrogen atoms are on
the nearest neighbor carbon atom. For every “n” hydrogen atoms on immediate neighbor carbon
atoms, a peak will be split into “n+1” parts. Look at the structure of isopropanol again and focus
on the hydrogen atom on the central carbon. That central carbon sees three hydrogen atoms on
the carbon immediately to its left and three additional hydrogen atoms on the carbon
immediately to its right, for a total of six hydrogen atoms on the immediate neighbor carbon
atoms. Six neighbors will split the peak for this central hydrogen atom into seven parts, forming
a septet. So, we now know that the peak centered at 3.95 ppm is due to hydrogen attached to the
central carbon atom because its splitting matches with the number of neighbor hydrogen atoms.
By process of elimination the broad peak at 2.45 ppm must be due to the hydrogen attached to
the oxygen.
Does the splitting work out consistently for the other atoms? Why is the peak at 1.25 split into a
doublet? Recall that we assigned this peak to the six -CH3 type hydrogen atoms. Look back at
the structure of isopropanol. Here the immediate neighbor carbon is the central carbon atom. It
has only one hydrogen on it, so the -CH3 peak will be split in two, forming a doublet. What
about the hydrogen on the oxygen? Its peak is a singlet, i.e. it remains unsplit. Hydrogen atoms
attached to oxygen atoms almost never show splitting, so this is to be expected.
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1.
2.
3.
4.
5.
6.
H NMR Summary
number of peaks = number of types of H atoms
peak position tells what the H atom is attached to (but we won’t use this today)
area under a peak is proportional to the number of H atoms of that type
splitting is caused by H atoms on immediately neighboring carbon atoms
“n” neighbors give a splitting into “n+1” peaks
hydrogen atoms bonded to oxygen atoms don’t show splitting
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Infrared Spectroscopy, IR
The vibrational state of a molecule may be studied by infrared spectroscopy. The bonds in a
molecule bend and stretch almost as if they were small springs stretching and contracting
constantly. Just as some springs are easier to stretch than others, different types of bonds vibrate at
different frequencies. The characteristic frequency of a given type of bond corresponds to the
energy that the bond will absorb. Table 2 lists some common types of bonds and the energies that
they absorb.
As you can see, the amount of energy required to cause such vibrations falls in the infrared (IR)
region of the electromagnetic spectrum. This means that by observing which energies of light a
molecule absorbs in the IR, we can determine which Bond
Absorption Range, cm-1
types of bonds the molecule contains. For example, type
if a molecule absorbs light at 1750 cm-1 we could
O–H
3600-3200 (usually broad)
conclude that it contains a C=O bond. Certain
C–H
3300-2800
absorption bands also have characteristic shapes.
B–H
2650-2300
The O-H absorption is usually very broad, so a
C
≡
N
2260-2220
small sharp band at 3400 cm-1 may not be an O-H
C=O
1780-1630 (usually strong)
stretch even though it is the correct region. A small
Table 2: IR frequencies
peak in this area may mean only that the sample or
holder used was a little wet. (H2O has an O-H stretch!)
HAZARDS
Although the unknowns are not particularly hazardous, you should still take reasonable care to
avoid prolonged contact of these compounds with the skin or prolonged inhalation of the vapors.
The unused portion of the unknowns should be kept in their original containers and returned to the
instructor.
The NMR instrument contains a very strong magnet. Persons with pacemakers and metal
prosthetic devices can also be harmed, and should not approach closer than five feet from the
magnet until it has been determined that a closer approach is not harmful. (The printers, screen,
and keyboard are not magnetic and are six feet from the magnet.)
PROCEDURE
You will be working with an assigned partner. You should work as a team on all aspects of this
experiment and turn in a single lab report for your team. Each person should do the normal pre-lab
entry in your own notebook, but once in lab all data will be recorded directly in the data tables at
the end of this experiment. Finish your report before leaving lab--staple your report, NMR spectra,
IR spectrum, and pre-lab pages together, and turn this in before leaving lab.
Part 1: Learning about NMR
Large organic molecules are often composed of small, familiar units connected in new ways. By
recognizing these small units, we can simplify our understanding of the larger molecules. On the
data sheet provided, determine what the 1H and 13C NMR spectra of the listed organic units will
look like. Your unknown for Part 2 will include some of these units, so you will use your Part 1
analysis to identify your Part 2 unknown. Use the description of the iso-propyl group from the
introduction as a guide as you predict the number and type of peaks expected for n-propyl, ethyl,
and methyl groups on the data sheet.
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Part 2 Obtaining your NMR Spectrum
Before going to the NMR room (Trexler 162) prepare your sample in the regular lab room. Attach
a new tip to the automatic pipetter that is set for 40 L (0.040 mL) and pipet this volume of your
NMR unknown into a clean NMR tube. Record the number of the unknown. Then place the NMR
tube in the measuring tube and add the NMR solvent, CDCl3, until the liquid level is between the
lines. This places 0.7 to 0.8 mL in the tube, the necessary volume. Be sure to close the stock
bottle of CDCl3. Cap, mix, and take this sample to the NMR in room 162 where the instructor will
help you record the 13C and 1H NMR spectra.
Part 3 Identifying your NMR Unknown
Follow the directions on the report sheet pages.
Part 4: Learning about Infrared Spectroscopy (IR)
Two IR spectra are given on your data sheets. Use Table 2 to identify the bond types that
correspond to bands above 1630 cm-1 in each spectrum.
Part 5: Obtaining your IR Spectrum and Identifying your Unknown
Take your “IR Unknown” to room 572. An instructor will demonstrate the proper way to prepare
your sample and operate the Perkin-Elmer infrared spectrophotometer. Once you have your
spectrum, see which of the types of bonds listed in Table 2 your molecule contains. This narrows
down the possible unknown identities. Now compare your spectrum with the spectra of the
knowns that contain the same types of bonds. Spectra of the knowns will be provided in the lab.
On your data sheet record the types of bonds that you conclude are in your unknown along with the
position of the absorptions that lead to your conclusions. Finally record the identity of your “IR
Unknown.”
LABORATORY REPORT
The Laboratory Report for this experiment consists of completing the following pages in this
manual. Attach your NMR and IR spectra and then turn in your report before leaving lab. Be sure
that you have not skipped any questions or parts of questions as you have skipped around doing the
various parts of this experiment.
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LABORATORY REPORT FOR
Identifying Compounds by IR and NMR
Team Members: ____________________________
Date __________
____________________________
Part 1: Learning about NMR
1. Complete the table predicting only the number of peaks (number of types of carbon) for 13C NMR. For
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H NMR predict the number of peaks (number of types of hydrogen). Then for each of these peaks, predict
the splitting multiplicities (from the number of neighbors), and the peak area (number of hydrogen atoms of
that type). Use the introduction as your guide, especially the Summary Boxes on Pages 2 and 3. The
open bond in each structure may be ignored; it has no effect on the spectra you will see. The entry for the
iso-propyl group and part of the entry for n-propyl have been completed for you. Use these as a check of
your understanding.
H
13
H
H
C
C
H
C
H
C number of types of C =
2
1
H number of types of H =
2
Splitting and area for each:
-doublet of area 6
-septet of area 1
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C number of types of C =
3
H
H
iso-propyl
H
H
C
H
H
H
C
C
1
H number of types of H =
3
H
H
Splitting and area for each:
-
n-propyl
H
H
C
H
H
13
C
1
H
C number of types of C =
H number of types of H =
Splitting and area for each:
ethyl
2. The 13C NMR spectrum of 2-methyl propane (see structure at right and model in lab)
has two peaks. Explain why. How many peaks do you expect for its 1H NMR spectrum?
Give the splitting multiplicity and area for each 1H peak. Explain your reasoning.
H
H
H
C
H
C
H
C
H
C
H
H
H
H
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3. Consider the structure at right. How many different types of carbon atoms are
present? This is equal to the total number of peaks expected in the 13C spectrum.
H H OH
H
C C C C H
H
H H
Number of types of carbon atoms and total peaks expected in 13C spectrum = _________
Part 2: Obtaining your NMR Spectrum
Follow the directions in the procedure section.
Record your unknown number ______
Part 3: Identifying your NMR Unknown
1.
Examine the 13C NMR spectrum of your unknown. Recall that the peaks near 77 ppm are just our
CDCl3 solvent and will be ignored. List the peaks below (you may not need all the spaces).
Peak
1
2
3
4
5
Position (in ppm)
How many total carbon environments (peaks) did you find? ________
2
Examine the 1H NMR spectrum for your unknown. List the peaks below. Recall that the solvent
peak may show at 7.2 ppm and should be ignored. Also ignore tiny impurities. Your unknown may
have fewer peaks than there are lines in the table. After you record the integrated peak areas as
given on the spectrum, reduce the ratio of areas to the smallest, whole number proportion.
Peak
Position
(in ppm)
Splitting (singlet,
doublet, etc.)
Integrated Peak Area (shown
below peak on spectrum)
Reduced Ratio
of Areas
1
2
3
4
5
7
3.
Now compare the table you just completed to the table you filled out in Part 1. Decide whether
your unknown contains an ethyl, n-propyl, or iso-propyl group. Focus your attention mostly on the
“splitting” and “reduced ratio of areas” columns. Remember that your unknown spectrum may
have some extra peaks because your unknown compound may contain something in addition to the
ethyl, n-propyl, or iso-propyl group.
Which of the three groups is present in your unknown? Explain your reasoning clearly.
4.
Go to the posted listing of possible NMR unknowns and find the page that corresponds to the
number of carbon environments you counted for your unknown (in Part 3, Step 1). Notice that each
possible unknown contains either an ethyl, n-propyl, or iso-propyl group along with some additional
groups. Since you just decided above which of these groups is present in your unknown, you
should be able to decide which of the compounds is your unknown. Copy its structure and name
below.
Part 4: Learning about Infrared Spectroscopy (IR)
Use Table 2 on Page 4 to assign bond types that correspond to bands above 1630 cm-1 in the two IR spectra
shown below.
Spectrum #1
Band position (cm-1)
Bond type
_______________
________
_______________
________
_______________
________
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Spectrum #2
Band position (cm-1)
Bond type
_______________
________
_______________
________
_______________
________
Part 5: Obtaining your IR Spectrum and Identifying your Unknown
Follow the directions in the Procedure Section
“IR Unknown” number _______________
Band position, cm -1
Bond type in
your compound
_______________
_______________
_______________
_______________
_______________
_______________
________________
_______________
Identity of Unknown
_______________
ATTACH your IR spectrum to your lab report
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Summary:
IR and NMR both help chemists learn about a molecule’s structure. Think about how the information we
get about structure from each is similar and different. Some of the techniques chemists use are simple
and fast but provide limited information. Other techniques are harder or longer but provide more
information. Think about how NMR and IR fit in these descriptions. Write a paragraph comparing IR
and NMR for the information provided about structures and also for ease of use versus amount of
information provided.
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