CHEM 2243 Lab: IR NMR Unknown
Purpose: In this virtual lab experiment, you will explore the use of mass spectroscopy, IR spectroscopy, and 1H and 13C NMR spectroscopy to identify an unknown organic compound.
Introduction: In the first part of this virtual lab experiment, you will use Beyond Labz to record the mass spectrum, IR spectrum, 1H NMR spectrum, and 13C NMR spectrum for a known compound, ethyl 2-phenylacetate. You will then completely analyze the spectra.
O
O
ethyl
2-
phenylacetate
In the second part, you will identify an unknown organic compound given its molecular formula, IR spectrum, 1H NMR spectrum, and 13C NMR spectrum.
Piece of Information What it tells you
IR spectrum | Functional group(s) in molecule. Particularly useful in spotting C=O, O-H, alkenes or aromatic rings. |
Molecular Formula | Use this to determine the degrees of unsaturation (# of pi bonds and/or rings) |
13C NMR spectrum | Number of different carbons in the molecule and what type of carbons (sp3, C=C, C=O, etc.) you have. |
1H NMR spectrum | Number of different types of hydrogens in the molecule Integration tells you the relative # of H’s in the signal. Multiplicity gives you information about # of H neighbors. Chemical shift helps determine electronic environment of H’s. |
A molecule’s degrees of unsaturation is the SUM of the pi bonds and rings that are in the molecule. A molecule with two degrees of unsaturation has either two pi bonds, two rings or one pi bond and one ring. This can help determine the structure of an unknown.
IR Spectroscopy:
We can identify functional groups using infrared (IR) spectroscopy. IR radiation is in the same range of frequencies (4000-600 cm-1) as the stretching and bending vibrations of bonds in organic molecules. When the frequency of the radiation exactly matches the frequency of the bond, the molecule absorbs energy, resulting in an absorption peak in the IR spectrum. Most functional groups show absorption bands in the region of 4000-1400 cm-1.
See the handouts: Characteristic Infrared Absorption Peaks and IR Tutorial UCLA for help identifying the peaks.
Mass Spectrometry:
We can get some information about the structure of a molecule from examining its mass spectrum. In a mass spectrometer, a sample is vaporized, and then ionized. The most common method of ionization is electron ionization (EI). In EI, the vaporized compound is bombarded with a beam of high energy electrons (typically around 70 eV). When the electron beam first collides with the vaporized compound, the compound loses one electron. This initially formed ion, the molecular ion or M+, has the same mass as the starting compound and so can be used to determine the molecular weight. The signals that are displayed in the mass spectrum are reported as their mass to charge ratio (m/z). In EI mass spectroscopy, the charge is +1 so we can read this number as the mass of the ion. The ions formed inside the mass spectrometer are unstable and often fragment before reaching the detector. Only positively charged species are recorded in the mass spectrum. The positively charged fragments that reach the detector in the greatest abundance tend to be the fragments that are the most stable and/or are formed in the greatest quantity. Functional groups tend to fragment in somewhat predictable ways to form stabilized cations.
Refer to the lab handout for experiment 4, and sections 13.1-13.9 in Organic Chemistry by Bruice (8th edition), for additional help in analyzing the mass spectrum.
Nuclear Magnetic Resonance (NMR) Spectroscopy:
Any nuclei that contain an odd number of protons or an odd number of neutrons can be studied by nuclear magnetic resonance (NMR) spectroscopy, but organic chemists tend to focus on carbon and hydrogen. Organic chemists use NMR to identify the carbon-hydrogen and carboncarbon framework of an organic compound. For 1H NMR, there are 4 key pieces of information that we can get from the spectrum:
- The number of signals- number of chemically equivalent hydrogens
- The chemical shift of the signals- type of proton(s)
- The integration of the signals- number of protons that produce the signal
- The splitting of the signals- # of protons on adjacent carbon atoms
- The number of signals. Chemically equivalent protons are protons that are in the same environment, and each set of chemically equivalent protons produces its own signal in the 1H NMR spectrum. For the relatively simple organic molecules we will be looking at, we can assume that hydrogens that are attached to the same carbon atom are chemically equivalent, unless there is restricted rotation- from a pi-bond or a ring. In the figure shown below, the chemically equivalent hydrogens are designated by the same letter.
Figure 6.1 Determining chemically equivalent sets of hydrogens. (From Organic Chemistry by Bruice, 8th Ed.)
- The chemical shift of the signals. The chemical shift of a signal indicates the environment the hydrogen is in. Hydrogens that are in electron-rich environments are shielded from the applied magnetic field and appear at a lower ppm value. Hydrogens that are in electron-poor environments are deshielded from the applied magnetic field and appear at a higher ppm value. In general, proximity to an electronegative atom deshields hydrogen atoms. Hydrogens that are attached to sp2-hybridized carbons are also deshielded.
The 1H NMR spectrum can be roughly divided into the 7 regions shown in figure 6.2 (one is empty). This is a good, quick, guide to the types of hydrogens that are present in a molecule.
Figure 6.2 Seven regions of the 1H NMR spectrum. (From Organic Chemistry by Bruice, 8th Ed.) A more detailed chart of approximate 1H NMR chemical shifts is shown below. A proton that is affected by more than one functional group will show cumulative effects.
Figure 6.3 Approximate 1H NMR chemical shifts. (From Organic Chemistry by Bruice, 8th Ed.)
- The integration of the signals. The area under each signal is proportional to the number of chemically equivalent hydrogen atoms that produce the signal. The spectrometer will often show the relative integration in arbitrary units, and you will then need to calculate the number of hydrogens for each signal. From the units given, divide all by the smallest number, and then if necessary, multiply all values by the same number to get a ratio of whole numbers. Add up the values to ensure it matches the number of hydrogen atoms expected in the molecule.
Figure 4.4 Solving for the integral ratio in a 1H NMR spectrum.
In Beyond Labz, the numbers above the peaks in the 1H NMR spectrum are the peak number (they are counting the number of peaks from left to right). Below the spectrum is a table that has those same peak numbers and the height of the integral trace. Calculate the integration as shown above using these height values. The data in Beyond Labz is real data, and thus will show some slight experimental error.
- The splitting of the signals. Splitting of signals is caused by protons bound to adjacent carbon atoms. The N + 1 rule describes the splitting of the signals, where N is equal to the number of hydrogens bonded to adjacent carbons.
Figure 6.5 How to describe signals in 1H NMR spectrum.
For 13C NMR, only the number of signals and the chemical shift need to be considered. A table of approximate chemical shift values for 13C NMR is shown in figure 6.6.
Figure 6.6 Approximate 13C NMR chemical shifts. (From Organic Chemistry by Bruice, 8th Ed.)
Figure 6.7 The qualitative analysis virtual lab.
Virtual Lab Instructions (video instructions also available). Help is available by clicking on the bell on the stockroom counter.
Part 1: Spectral analysis of ethyl 2-phenyl acetate
- Open the organic chemistry lab on the Beyond Labz platform. Select “Qualitative Analysis” at the top left of the screen to open the Qualitative Analysis virtual lab.
- Select “Esters” from the chalkboard, then click and drag (or double-click) the reagent bottle for the ethyl 2-phenylacetate to add it to the flask.
- Drag the flask to the cork ring support on the lab bench.
- Record the IR spectrum by clicking on the IR spectrometer and dragging the salt plate icon to the flask. This will display the IR spectrum on the screen. You can type the name of the compound on the spectrum and click save to save it to your lab notebook. Click ok to close the spectrum.
- Record the mass spectrum by clicking on the mass spectrometer and dragging the sample vial icon to the flask. This will display the mass spectrum on the screen. You can type the name of the compound on the spectrum and click save to save it to your lab notebook. To view the m/z values, mouse over each peak- the first number is the m/z value. Click ok to close the spectrum.
- Record the 1H NMR spectrum by clicking on the NMR spectrometer and dragging the sample tube to the flask. This will display the 1H NMR spectrum on the screen. You can type the name of the compound on the spectrum and click save to save it to your lab notebook. Click ok to close the spectrum.
- By default, the NMR is set to record 1H- change to 13C by clicking the window on the NMR spectrometer. Record the 13C NMR spectrum by clicking on the NMR spectrometer and dragging the sample tube to the flask. This will display the 13C NMR spectrum on the screen. You can type the name of the compound on the spectrum and click save to save it to your lab notebook. To view the chemical shift values (in ppm) mouse over each peak- the first number is the chemical shift. Click ok to close the spectrum.
- Your virtual experiment is now complete!
Part 2: Unknown identification (Beyond Labz not needed for this part).
Identify an unknown organic compound given its molecular formula, IR spectrum, 1H NMR spectrum, and 13C NMR spectrum. Your unknown will be assigned by your instructor and the spectra are on D2L.
Assignment:
(Labeled IR spectrum for ethyl 2-phenylacetate and solved example appear at the end of this handout.
Points (total = 20)
- 8 pts ethyl 2-phenylacetate results
- 12 pts unknown results
Ethyl 2-phenylacetate Results (8 pts total) Complete the following tables and answer any questions.
Infrared spectrum (1.5 pt):
Absorbance peak (cm-1) | Bond(s)/functional group |
3090-3034
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2983-2874
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1739
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1605
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1H NMR spectrum (2 pts):
Signal | ~ Chemical shift in ppm | Height | Integration | Splitting |
1 |
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5
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Show the calculations for the integrations (1 pt).
Draw 2-phenylacetate and show which signals correspond to each hydrogen by labeling the H’s using the #s 1-5 from the above table. (1.5 pts)
13C NMR spectrum (1 pts):
Signal | ~ Chemical shift in ppm |
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(0.5 pt) Which signal corresponds to the carbonyl carbon of ethyl 2-phenylacetate? (Give the chemical shift in ppm.)
(0.5 pt) Which signal corresponds to the carbon with a single bond to oxygen (not the carbonyl carbon)? (Give the chemical shift in ppm.)
Unknown Sample (12 pts)
What is your Unknown Letter? (0.5 pt)
Calculate the degrees of unsaturation in your molecule. Show your work. (1 pt)
Infrared spectrum (2 pts): (List all relevant peaks. You may not use all of the columns)
Absorbance peak (cm-1) | Bond(s)/Functional Group |
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What are the functional groups in the molecule?
1H NMR spectrum (2 pts): (You may not use all of the columns)
Signal | ~ Chemical shift in ppm | Integration | Splitting |
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13C NMR spectrum (1.5 pts):
Signal | ~ Chemical shift in ppm |
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Unknown Identification and Correlation with NMR Spectra (5 pts)
Draw the structure of the unknown and indicate which sets of hydrogens correspond to which signal in the 1H NMR spectrum (3 pts).
Redraw the structure of the unknown and indicate which carbon atoms correspond to which signal in the 13C NMR spectrum (2 pts).
EXAMPLE Ethyl 2-phenylacetate IR spectrum
Solved example: methyl propionate (spectra appear after tables)
O
CH3CH2COCH3
methyl propionate Infrared spectrum:
Absorbance peak (cm–[1]) | Bond(s) | |
2985-2850 | sp3 C-H | |
1741 | C=O, saturated ester | |
1207 | C-O, ester |
Mass spectrum:
m/z value | Charged Fragment | Fragment Lost | ||
88 |
| O CH3CH2COCH3 |
| N/A |
59 |
O COCH3 | ·CH2CH3 | ||
57 |
O CH3CH2C | ·OCH3 |
Signal | ~ Chemical shift in ppm | Height | Integration | Splitting |
1 | 3.69
| 170.11 | 3 H | singlet |
2 | 2.28
| 118.59 | 2 H | quartet |
3 | 1.15 | 183.07 | 3 H | triplet |
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Show the calculations for the integrations.
170.11
118.59
≈
1.5
×
2
=
3
118.59
118.59
=
1
×
2
=
2
183.07
118.59
≈
1.5
×
2
=
3
Draw the structure of methyl propionate and indicate which sets of hydrogens correspond to which signal in the 1H NMR spectrum.
O
CH3CH2COCH3
- 2 1
13C NMR spectrum:
Signal | ~ Chemical shift in ppm |
1 | 174.8
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2 | 51.9
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3 | 23.7
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4 | 9.4
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Draw the structure of methyl propionate and indicate which carbon atoms correspond to which signal in the 13C NMR spectrum.
O
CH3CH2COCH3
- 3 1 2
H NMR spectrum: ↑