IDENTIFYING COMPOUNDS BY UV-VIS AND GC-MS rev 5/11

Transcription

1 EXPERIMENT 9 IDENTIFYING OMPOUNDS BY UV-VIS AND G-MS rev 5/11 GOAL hemists use many types of modern instrumentation to identify compounds. This week you will use two types of ultraviolet-visible spectrophotometers, and the G-MS, gas chromatograph-mass spectrometer. INTRODUTION Laboratory instruments allow us to extend our powers of observation and quantify minute differences between molecules. For example, while our eyes might tell us that two compounds are slightly different shades of red, a laboratory spectrophotometer would be able to identify the specific wavelengths at which those compounds absorb visible light. Such specific information supplied by various laboratory instruments can be used to determine what is in a sample, how much of that substance is present, separate the components of a mixture, or even cause a chemical reaction to occur. Many common instruments are spectrophotometers, that is, instruments that measure the amount of light that a sample absorbs or emits at a particular wavelength. Before we can understand these instruments, however, we must first understand the relationship between light and energy. The visible light that we can see is only a small part of the electromagnetic spectrum. Different portions of the spectrum vary by wavelength and energy content. The energy of one photon of any type of light may be calculated from the relationships E = h ν and E = h c / λ where E is energy per photon, h is Planck's constant, (nu) is the frequency, c is the speed of light, and (lambda) is the wavelength. Notice that energy is inversely proportional to wavelength. In other words, as wavelength increases, energy decreases. Red light has the longest wavelength of all visible light, but has the lowest energy content. A chemical substance will absorb only certain characteristic amounts of energy and thus certain wavelengths of light. The substance absorbs these particular wavelengths/energies of light because they correspond to the energy differences between various states of the molecule. For example, in Figure 1 we see a molecule that has two states, one with higher energy than the other. The molecule will absorb light with an energy that exactly matches the energy gap between the two states shown. The molecule is not able to absorb light that is slightly higher or slightly lower in energy, only the light with the correct Figure 1: Energy transition between two states energy to match the gap between the states. Because the state of a molecule can be changed in many ways, and because we have many ways of observing these changes, a great number of different types of spectroscopy have been developed. Ultraviolet-Visible Spectroscopy, UV-Vis The energy needed to move an electron from one orbital to another is equal to energies of light in the visible and ultraviolet portions of the electromagnetic spectrum. When an electron moves from a higher energy orbital to a lower energy orbital, the difference in energy between the two orbitals must be released. One of 1

2 Absorbance the ways that the molecule may release this energy is by emitting light with an energy equal to the energy difference between the two orbitals. (You may recall that Bohr used this concept to develop his model of the atom.) In order for an electron to move from a lower to a higher energy orbital it must gain energy, and the molecule may get this energy by absorbing light with that specific energy content. Since most molecules have energy gaps between orbitals that fall in the visible or ultraviolet (UV) regions of the spectrum, studying UV-visible spectra of a molecule provides information on the characteristic energy gaps between molecular orbitals. This information in turn can be used to help identify the molecule. The human eye is able to distinguish among wavelengths in the visible region. When all the visible colors - red, orange, yellow, green, blue, indigo, and violet - are present, the eye sees white. If any color is missing, the eye sees that color's complement, the mixture of colors that are left. Figure 2 shows a small complementary color wheel. olors directly opposite each other are complements. For example, red and green are complements. If red is missing from a mixture of visible light, the eye will see the light as green. This means that green-colored compounds appear that way because red is missing from the light they reflect. In other words, they must absorb light in the red region. A compound that absorbs light around 470 nm is absorbing blue light and will appear orange to orange-yellow to the human eye. Figure 2: omplementary olor Wheel In addition to the qualitative information, quantitative values can also be derived from a UV-visible spectrum. The amount of light that any solution will absorb is proportional to the concentration of that solution. This is expressed mathematically as Beer's Law: A = abc where A is the absorbance of the solution, a is the absorptivity constant, b is the thickness of the sample (width of the cuvette), and c is the concentration of the solution. The concentration of any solution can be calculated from its absorbance if we know the values of a and b. The value of b is normally 1.00 cm. When we don t know the value of a, we can find the concentration of an unknown solution by comparing its absorbance to that of solutions with known concentrations. Typically we will prepare three or more solutions of known concentrations, measure the absorbance of each, and then plot a calibration curve of absorbance vs. concentration, such as the one shown here. After measuring the absorbance of the unknown solution, its concentration can be read from the graph. For example, if the solution of unknown concentration had an absorbance of 0.60, we would consult the calibration curve, and find a concentration of 0.18 M. alibration urve for Spec oncentration, M Figure 3: alibration urve Graph 2

3 Abundance Mass Spectrometry, MS Although mass spectrometry can be a standalone technique, our instrument couples gas chromatography with mass spectrometry in an instrument called a G-MS. As you saw in an earlier experiment, gas chromatography separates a mixture, allowing each component to come off the end of the chromatography column at a different time. This results in a chromatogram such as the one in Figure 4 in which a mixture has been separated into five major components. One way to identify these components is to send the material coming off the chromatography column directly into a mass spectrometer m/e 1 Time Figure 4: Gas hromatogram In the mass spectrometer, molecules are turned into +1 ions by removing one electron. Then the mass spectrometer measures the exact mass of each ion that reaches its detector. The mass spectrum of 3 l is shown in Figure 5. The x-axis shows the masses of the particles and the y-axis shows how many particles of a given mass were detected. ow can one compound give multiple mass peaks? Let s figure out where these peaks are coming from. Imagine that we have l molecules. The molar mass of 3 l that you add up off the periodic table is 50.5, but that is an average. Approximately 75 of those 100 molecules contain 35 l and have a mass of 50. The remaining 25 molecules contain 37 l and have a mass of 52. Some of these 3 l molecules make it to the detector. intact giving peaks with masses at 52 and 50. Figure 5: Mass Spectrum of hloromethane Figure 6: Fragmentation of hloromethane (M) 100 l + Many of our l molecules, however, don t make it to the detector intact. onditions inside the mass spectrometer cause some bonds to break, creating fragments that then travel to the detector. Some of the 3 l molecules will break a -l bond, others will break a - bond. As seen in the fragmentation scheme in Figure 6, this creates fragments with masses of 15, 35, 37, 49, and 51. The peak at 50 is considered the molecular ion and often labeled (M) because it represents the intact molecule with its most abundant mass. When we try to interpret the mass spectrum of an unknown, the molecular ion peak is extremely helpful since it is approximately equal to the molar mass of the unknown compound. A significant M+2 peak is almost always due to the presence of l or Br in the molecule. One l atom gives an M+2 peak about 1/3 as tall as M. One Br atom gives an M+2 peak nearly equal to M in height. A peak at M+1 is primarily due to 13 (In nature, 1.1% of all carbon atoms are 13 ). The number of carbon atoms in the molecule can be calculated from the height of M+1 relative to the M peak. l mass = 50 or 52 l mass = 35 or 37 mass = 49 or 51 mass

4 Number of carbon atoms = (M+1 height as a percentage of M) / 1.1 Thus if the M+1 peak is 10% as tall as M, then the number of carbon atoms in the molecule equals 10/1.1 = 9. The masses of the fragments from a spectrum allow us (or a computer) to figure out what the original, intact molecule must have look like. The way that a molecule fragments in a mass spectrometer is predictable to some extent. Molecules don t normally fall totally apart into individual atoms, but rather break at a couple of natural weak points. By looking at a molecule s structure or using a model, we can often see where the molecule is most likely to break apart. onsider the structure of acetophenone shown in Figure 7 and imagine how it might fragment. The ring of six carbon atoms is very common in organic chemistry and quite stable. This is not likely to fragment, but the larger groups hanging off of the ring will be more easily broken off. In particular, it is very common for a molecule to fragment directly next to a =O group. We can imagine this molecule fragmenting in at least two ways: the bond between the ring and the =O group might break, or the bond between the =O and the 3 might break. Such a fragmentation pattern will give us peaks at 120, 105, 77, 43, and 15. Of course, not every possible fragmentation pattern will be observed, and some bonds will be more likely to break than others will, so these peaks are not all expected to be of equal intensity. Figure 7: Fragmentation of Acetophenone O + O 3 mass = mass = 77 mass = 120 O + 3 mass = 15 mass = 105 Mass Spectrum Summary 1. Molecular ion (M) gives mass of the intact molecule 2. M+2 peak indicates presence of l if M+2 height is 1/3 M height Br if M+2 height is approximately equal to M height 3. M+1 peak indicates number of carbon atoms # = (M+1 height as a percentage of M) / Molecules fragment at key point, breaking off groups e.g. next to rings, =O groups, etc. 4

5 AZARDS 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 kept in their original containers and returned to the instructor PROEDURE 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. You each 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, spectra, graph, and G-MS printouts together, and turn this in before leaving lab. Your instructor will tell you in what order to do the various parts of this experiment. Part 1: Scanning Visible Spectroscopy Take your unknown labeled Scanning Visible Unknown to room 472 as directed by the instructor, who will demonstrate the correct use of the scanning UV-visible spectrophotometer. Run the visible spectrum of your compound and measure the absorbance at the highest point of each peak (the absorbance at each max ). Print your spectrum. Identify your unknown by comparing its max (the wavelength where its absorbance is greatest) to those given in the list of possible unknowns posted in lab. Part 2: UV-Visible at a Fixed Wavelength the Spec-20 Find the concentration of your unknown labeled Spec-20 Unknown using a Spec-20 in the regular lab room. onsult the Appendix for detailed instructions on the use of a Spec-20 (download the file from the usual web site). Take three clean, dry beakers to the reagent bench and get 5 ml of each of the Spec-20 standards. Be sure that you record the concentration of each solution that is given on the dispenser and also the wavelength at which measurements are to be made. Following the instructions from the Appendix, record the absorbance of each of the standards, and also that of your Spec-20 Unknown. Be sure that you do not accidentally dilute any of your solutions by using wet glassware. Do not attempt to dry out the Spec- 20 cuvette, but rather rinse the cuvette with a small volume of your solution and then discard this rinse solution down the drain. (These solutions are not hazardous.) On graph-ruled paper (a page from your lab notebook), draw a calibration curve similar to the one in Figure 3 using the absorbance and concentration data for the solutions with known concentrations. Also include the point (0,0) because you set the absorbance to zero with distilled water. onsult the Appendix for correct graphing techniques (download the file on graphing!). Your instructor will grade the graphs strictly by the criteria listed there. Read the concentration of your Spec-20 Unknown from the calibration curve and record this value on the data sheet. Part 3: The olors of Visible Light by Spec-20 Use the Spec-20 in the regular lab room to find the colors that correspond to wavelengths in the visible portion of the spectrum. Obtain a special Spec-20 cuvette that contains an angled piece of chalk. Place this in the Spec-20 and leave the sample cover open. Set the wavelength control to 700 nm. Record the color of the light that you see on the chalk. Now change the wavelength by 10 nm. Record the color you see. ontinue recording the colors of light that you see for wavelengths between 750 and 400 nm. Also record the wavelength of both the longest and shortest wavelengths that were visible to your eye. 5

6 Part 4: Questions on UV-Visible Spectroscopy Answer all the questions on the data sheets. Part 5: Learning to Interpret Mass Spectra arefully read pages 3 and 4. Note the summary box on page 4. Use this information to complete the questions on the data sheet. Models are available in the lab to help you visualize the molecules. Part 6: Using the G-MS Go to the G-MS in Room 472 as directed by your instructor, who will guide your work for this part. Because the G is time-consuming to run and you have already run this same instrument in the rime Solving hromatography experiment, you will spend your time doing a computerized analysis of a sample that has already been run and saved as a data file. Be sure to staple the G-MS printouts to your report. With the help of the instructor, do the following. 1. view the parts of the instrument 2. at the computer, position the cursor on each of the major peaks 3. retrieve the mass spectrum for each major peak 4. do a computerized library search to identify each major peak 5. record this information on your report sheet (Questions 1 and 3) 6. get a packet of printouts including the chromatogram and the identifications of the major peaks 7. answer the remaining questions Note: The instructor may choose to have you do the data analysis on the four computers in the entrance annex to our regular lab. LABORATORY REPORT Turn in one joint lab report for your team before leaving lab. Staple together the data sheets that follow with your UV-Visible spectrum, calibration curve, G-MS printout pages, and preliminary notebook entry pages. 6

7 LABORATORY REPORT FOR IDENTIFYING OMPOUNDS WIT UV-VISIBLE AND G-MS Team Members: Date Part 1: Scanning UV-Visible Spectroscopy Scanning visible unknown number olor of Scanning visible unknown solution max, nm (Your unknown may have only one peak, or several peaks.) Absorbance at max (If your spectrum has more than one max, record the absorbance at each) Identity of unknown ATTA your visible spectrum to your lab report Part 2: UV-Visible at a Fixed Wavelength the Spec-20 Spec-20 Unknown number Wavelength for measurements Spec-20 Data: Solution Identity oncentration, M Absorbance Blank 0 0 Standard #1 Standard #2 Standard #3 Your Spec-20 Unknown XXX oncentration of your Spec-20 Unknown (read from the graph you made, and include units) ATTA your calibration curve to your lab report 7

8 Part 3: The olors of Visible Light by Spec-20 Wavelength olor Wavelength olor Wavelength olor 750 nm 630 nm 510 nm 740 nm 620 nm 500 nm 730 nm 610 nm 490 nm 720 nm 600 nm 480 nm 710 nm 590 nm 470 nm 700 nm 580 nm 460 nm 690 nm 570 nm 450 nm 680 nm 560 nm 440 nm 670 nm 550 nm 430 nm 660 nm 540 nm 420 nm 650 nm 530 nm 410 nm 640 nm 520 nm 400 nm Longest wavelength visible to your eye : nm (first student) and nm (second student) Shortest wavelength visible to your eye : nm (first student) and nm (second student) (list values for both partners since you may get differing results) 8

9 Part 4: Questions on UV-Visible Spectroscopy Answer the following using complete sentences. Note that to answer some of these questions, you must have already completed the parts above. 1. onsult the schematic diagram of a Spec-20 that is given in the Appendix as you answer these questions. a. Based upon the schematic diagram, describe the path that light travels through the spectrophotometer. b. A relatively small portion of the light produced by the source eventually reaches the detector. onsult the schematic diagram and then describe two reasons for this. c. Describe two ways in which the scanning spectrophotometer you used in Rm 472 is different from the Spec What wavelength(s) was/were max for your Scanning Visible Unknown? onsult your results from part 3. What color(s) corresponded to this/these wavelength(s)? Are these the same color as your unknown? Explain! 3. What is the wavelength of the highest energy light visible to your eyes in part 3? Be careful! 9

10 Abundance 4. ompare your results for part 3 with those of two other teams. Did you get exactly the same answers? Did you find the same longest and shortest visible wavelengths? What conclusions do you draw from this? 5. In part 3, you could not see any light striking the chalk at the highest and lowest wavelengths. Does this mean that no light was reaching the chalk? Explain. Part 5: Learning to Interpret Mass Spectra (onsult pages 3-4, and especially the summary box on Page 4, for help with these questions.) 1. The mass spectrum at right is for n-butane, the structure shown. Imagine how this molecule might break apart. Draw structures of the fragments responsible for the peaks at masses 15, 29, 43, and m/e 10

11 Abundance Abundance 2. The mass spectrum at right is for a compound containing only carbon, hydrogen, and fluorine. The M+1 peak is about 1% of M. (This gives specific information! heck the introduction.) Use the molecular ion (M) mass to deduce the formula for the compound. Explain your reasoning. Draw structures of the fragments responsible for the peaks at masses 15, 19, and m/e M(34) 3. The organic compound whose mass spectrum is shown contains only three elements. Determine the molecular formula of this compound. Start by considering what the height of the M+2 peak tells you. The M+1 peak is just under 7% of the height of the M peak. You do not need to determine the structure of this compound, just its molecular formula. M(112) M m/e 11

12 4. The first thing that happens inside the mass spectrometer is that an electron is removed from each molecule, making a +1 ion. (It is easier to measure the mass of an ion than a neutral molecule, and the detector is sensitive to ions, not neutral molecules). Does the lost electron significantly affect the measured mass? Why or why not? Part 6: Using the G-MS 1. According to the chromatogram, how many major components are in the sample? Explain. 2. Suppose that the sample was known to contain two more major components than you just counted. Give one possible explanation for this discrepancy. 3. With the instructor s help, use the computer s software to identify each of the major components by matching their mass spectra to the library of known mass spectra. omplete the table below, recording the best identification given by the computer, but also comment on the quality of each identification. The computer supplies a numerical rating for the quality of the match between a peak s mass spectrum and the known mass spectrum. A quality of 100 is a perfect match. The computer also gives several alternative compounds to choose from. Peak retention time Best computer ID of the ompound omments on Quality of ID 12

13 Your instructor will give you a packet of printouts that includes the chromatogram and a best match identification for each compound in our mixture. Each of these best match pages shows two mass spectra. The one at the top is the mass spectrum of a component of our sample, labeled with its retention time. The mass spectrum at the bottom of the page is the spectrum of a pure sample of the compound identified as the best match identification. Use these printouts to answer the following questions. 1. a. Find the printout with the best match for the peak at 3.9 minutes. Give the name and structure for this compound. b. Think about how this structure might fragment, then draw the fragments that correspond to the mass spectrum peaks at 114, 85, and 71. c. Note the four evenly spaced peaks at 85, 71, 57, and 43. What constant mass difference separates adjacent pairs of these peaks? What small bit of the compound s structure does that represent? 2. a. Find the printout with the best match for the peak at 5.9 minutes. Give the name and structure for this compound. b. The molecular ion peak (M) is at 156. What does the peak at 158 tell you? 13

14 c. Think about how the structure of this compound might fragment, then draw the structure of the fragment at a. Find the printout with the best match for the peak at 7.81 minutes. Give the name and structure for this compound. b. Think about how the structure of this compound might fragment, then draw structures for the fragments that correspond to mass spectrum peaks at 136, 105, and 77. c. ompare the mass spectra and structures for the peaks at 7.56, 7.81, and 10.2 minutes. What mass spectrum peaks are common to all three? What common structural feature explains this? 14