For GC, GC-MS, NMR, IR, and UV/VIS analysis, we currently have only one or two instruments of each type to serve the 150 students in each section of this course. Therefore, it is important that these expensive instruments be utilized as fully as possible during the class period when you will be synthesizing and analyzing your products in the second half of the semester. In order to increase the use of these instruments, Chem 35/36/36B/36H TA's will hold office hours in the Instrument Room (206 Whitmore). This provides additional help and equipment availability over 40 hours a week. A complete schedule will be posted on the doors of 206 Whitmore. 

Instrumentation is available by signing up on the appropriate sign-up sheets located conveniently near each instrument. Please be considerate of your colleagues by arriving on time and working efficiently. In its generalized form, the system works as follows: 

1. When you have a product to be analyzed, prepare it for analysis according to the specific instructions in each of the instrument sections below. IR analysis sample preparation cannot be done ahead of time, but all others can. 

2. Sign-up for the appropriate instrument during a convenient time slot. Include your name, class, section and desk number. (i.e. Joan Smith, Chem 36.2, desk #88) If you are going to run spectra outside your normal class period, you must prepare your sample ahead of time and label it properly. You MUST leave the sample in the Instrument Room. The laboratory doors are locked outside of the class period and you will be unable to obtain your sample. The instrument TA will not have the key to let you in. 

3. When your turn comes, use the instrument immediately! Instrument time is precious. If you are really tied up and can't use the instrument immediately, cross out your name and inform the Instrument Room TA. 

4. If you are running a GC, obtain the appropriate GC Analysis Conditions Sheet, located on the table near the GC's in room 206 and fill it out completely. This must be included in your report. 

Gas chromatographic (GC) analysis is a very sensitive method for separating small amounts of liquid or gaseous mixtures and quantitating the amount of each of the separated components. It should not be used for samples that have not been distilled or vacuum transferred because they may contain materials which cannot be vaporized sufficiently to be carried through the column. These remain on the GC column for a long time, ruin the performance of the GC, and mess up subsequent GC analyses. Brief (and good) descriptions of gas chromatography and of the instruments themselves are given in Williamson's Microscale and Macroscale Organic Experiments, Ch. 20, p 268-270. (Copies can be found in the library, the instrument room or in the tutorial room, 211 Whitmore.) You need to read this section before running your analysis so that you understand how the analysis is to be conducted. There are three instruments in Rm 206 which are set up for different analyses: 

1. A Hewlett-Packard 5980 gas chromatograph (GC #1) with an flame ionization detector (FID) and temperature programmable oven. 

2. A Hewlett-Packard 5980 gas chromatograph (GC #2) with an flame ionization detector (FID) and temperature programmable oven. 

3. Varian gas chromatograph (GC #4) which has a column packed with a 3% OV-1 stationary phase that is used for preparative gas chromatography in which the peaks eluting from the detector are collected and isolated. This is not temperature programmable. This instrument is not used often, so please inform the instrument room TA that you will need to use the instrument at least 30 minutes before you need to run the sample. 

The generalized analysis procedure involves these steps: 

1. Prepare the sample - There are three ways of preparing a GC sample depending on whether your product is a liquid (usually obtained by distillation) or a gaseous mixture: 

a. Solutions of organic liquid mixtures in dichloromethane are the most common form for injecting samples. Products, which may be a single pure component or a mixture of two or more compounds, should be diluted by placing two or three drops of the neat material into a "Shorty" vial and adding enough dichloromethane to fill the vial half to two-thirds full. Since GC columns are damaged by moisture, add enough anhydrous sodium sulfate to cover the bottom of the vial to a depth of 2 to 3 mm and this will ensure that all traces of water are removed. The sodium sulfate can be left in the sample since it won't go into the fine needle of the microliter syringe used to inject the sample. 

b. Neat or pure liquids are often injected directly without dilution when they contain volatile components that are difficult to separate from a solvent, for example the mixture of methylbutenes in Exp 20.1. Also, if the GC is used to separate and collect the purified component of a mixture, neat liquids are usually injected. 

c. Gases are normally collected over water (see exp't 20 for details). Do not remove the septum-capped collection tube from the beaker of water but bring the two together to the GC room. A gas syringe is inserted through the rubber septum, 50 microliters of the gas sample is withdrawn and then injected into the GC. 

2. Sign-up for an instrument - Once you have a solution, liquid, or gas sample ready for GC analysis, fill out a GC Analysis Conditions Sheet obtainable in room 206 between GC #1 and GC#2. Check with the Compendium of GC Conditions, in order to predict how many and at what retention time you will observe your peaks. Use GC #1 or #2, unless specifically instructed to use the Varian, GC #3 and use CH₂Cl₂ as a solvent unless directed otherwise as well. Unfortunately, due to cuts in staff in the mass spectrometry facility, it is likely that we will not be able to run mass spectra of solids for all samples submitted by Chem 431W students. 

4. Inject the Sample and Record the Chromatogram - This is normally done with the assistance of an instructor. 

5. Document the GC Analysis Conditions - Forms like the one in Figure 11.1 will be found near the GC. These must be filled out completely and attached to your gas chromatogram for inclusion in your final report. 

6. Clean-Up - Rinse out the syringe by filling it with dichloromethane and squirting it onto a tissue. Clean and return any sample collection tubes, if used. Write the date and time on your GC analysis. 

Questions Frequently asked about Interpretation of Chromatograms 

1. How do I identify each peak in the chromatogram? 

Each experiment that deals with chromatography has a corresponding entry in the Compendium of GC Conditions that is located by each chromatograph in the Instrument Room. All peaks that could be identified have been identified and labeled on the sample chromatogram. Often, identification was made by injecting a sample whose identity was known and finding the peak in the mixture that had a matching retention time. 

Since samples vary slightly and the reproducibility between various runs is not completely 100.0%, there can be some variation in the appearance and retention times of peaks. If the oven temperature programming rate is higher or lower than reported in the compendium, then the retention times for all of the peaks will be consistently shorter or consistently longer. Chemists primarily look for similarities in the pattern of peaks, not for perfect matches in retention times. 

The first peak that comes through is most likely dichloromethane, the solvent you used to dilute your sample before injection. There will often be some additional peaks that are rather low in intensity, that are difficult or impossible to assign. As long as these impurities are minor, say less than 10% of the total area, (not including the solvent) they can be ignored. 

2. How do I calculate the percent of product in the mixture? 

The printout from the electronic recorder-integrator gives the retention time, the peak area, and the % area for each peak. The peak area is proportional to the amount of each component in the mixture. The area % value is calculated by summing up all of the peaks, and since the solvent, dichloromethane is the overwhelming majority of the sample, it will have the largest % area 

For example, a sample composed of a mixture of A and B was diluted in dichloromethane and analyzed by GC to give areas of 3263300, 27073, and 2958 for dichloromethane, component A, and component B. The total area would be 3,293,331, and the % of the three would be 99.0%, 0.82%, and 0.09%, respectively. (Keep in mind that this is roughly the mole percent; the detector response is not exactly the same for all compounds. For higher accuracy, this varying response 

Other less convenient methods of quantifying the relative amounts of resolved components in the chromatogram is by cutting out the peaks and weighing the individual triangle-shaped traces or by multiplying the height of the peak times the width of the peak at half-height. The first method assumes a fairly uniform thickness of paper used for plotting the chromatogram; if the peaks are quite narrow, the errors in following the lines with the scissors are large. The second method assumes a nice symmetrical peak, but the difficulties in accurately measuring the width contribute to large errors. The bottom line: for capillary GC, where the peak widths are narrow, the height of the peak and the electronic-integrator areas are used; for packed column GC with pen plotters, the height x width-at-half-height or cut and weigh methods must be used. 

Gas Chromatographic-Mass Spectrometric (GC-MS) analysis is one of the most powerful analysis methods available today. It combines the "mixture-separating" power of gas chromatography with the "compound-identifying" power of mass spectrometry. As each component of a liquid mixture comes out of the chromatography column, it enters the mass spectrometer, which is scanning continuously and recording a gas chromatographic trace. The hundreds of mass spectra acquired in a GC-MS run are stored in a computer and these can be subsequently accessed and the mass spectra of all components can be examined, computer matched with a library of thousands of spectra, and printed. 

Consult the Instrument Room Instructor to obtain help in analyzing your sample by GC-MS. All samples run on the GC-MS system must be first analyzed on either the HP 5980 or Finnigan 9500 gas chromatograph to ascertain that the sample contains the desired components, is approximately the right concentration, and is not contaminated with an overwhelming amount of impurities. By screening out unsuitable samples, we want to minimize excessive wear on the GC-MS, which can cost hundreds of dollars to repair. After analyzing the sample by gas chromatography, show the chromatogram produced to an instructor before carrying out the analysis on the GC-MS. The same sample used for GC analysis should be the one used for GC-MS analysis. Only inject 0.2 to 0.5 mL of the CH₂Cl₂ solution. See the ends of the sections on gas chromatography and mass spectrometry for explanations of how to interpret the data from GC-MS. 

The generalized procedure involves these steps: 

1. Sign-up for IR analysis - Once you have a pure sample for IR analysis sign-up for the appropriate instrument. If you have a liquid sample sign-up for IR #1 or if you are running a solid sample, sign-up for IR #2. 

2. Prepare the sample. - There are two ways of preparing an IR sample depending on whether it is a solid or a liquid: 

Alternate Method of Obtaining IR Spectra. If you are not getting a good IR spectrum, consult with an instrument room assistant. They may suggest you analyze your solid using a KBr disc. This technique requires a little extra time, but produces very good results. When you sign-up, weigh out between 70 and 120 mg KBr (found on the Common Shelf in a dessicator jar) and place it in a small vial. Add one mg (a small crystal) of your solid sample. When your turn comes, have someone assist you with the preparation of the KBR disc. 

b. Liquids are run as a thin film between two salt plates (sodium chloride discs). The plates are very moisture sensitive, can be easily broken, and are quite expensive. Therefore, they should be handled with great care, avoiding all contact with water or "wet" samples! Under no circumstances should samples containing even traces of water be placed on the salt disks as they will dissolve! Liquids don't require any preparation prior to actually running the IR analysis.When it is your turn, take your sample to the IR room and carry out the analysis starting at step 1 of the instruction sheet found on the liquid sample prep table. 

3. Run and Plot the Sample Spectrum. Specific instructions for sample introduction and instrument operation are given in a special handouts which can be found next to the IRs in the Instrument Room. Follow the step-by-step instructions on the sheets on the solid or liquid preparation tables and by each IR's computer. 

4. While the spectrum is plotting, notify the next person on the sign-up sheet that the IR is available. The next IR analysis can be started while your spectrum is still plotting. This will help fully utilize the available in class time for sample analysis. 

5. Clean up. The DRIFT sample cups or liquid sample salt plates must be carefully cleaned and placed in the proper dessicator according to the instruction sheets on the sample prep tables. 

Questions Frequently Asked Regarding the Interpretation of IR Spectra 

1. Where do I find a correlation chart for the various functional groups? 

Refer to any large-sized organic chemistry textbook. 

2. I have several peaks in the spectrum that I cannot assign. Why? 

Be sure to evaluate the strongest IR absorbances for any starting materials or solvents that you used, since these are the most likely candidates for any extra peaks in the spectrum. (This of course, is why you are asked to write a spectral features comparison in your PreLab exercises.) Sometimes, during the course of the procedure, you've collected solvent from the distillation instead of the liquid product you should have collected. Or, a lower-boiling starting material or solvent distilled first. 

Be sure that your spectral interpretation includes an evaluation of the purity of the sample and, if present, the identification of significant contaminants. This is, after all, the purpose of analyzing the product! 

The FT-IR instruments we use in Organic are sensitive enough that they often detect the carbon dioxide in the air surrounding the sample. This shows up as a distinct peak around 2360 cm^-1, and often distinguished from "real" peaks because it is split into two peaks. 

If the carbon dioxide peak is really strong and product peaks barely rise above the baseline, you need to use more sample for obtaining the spectrum. The software automatically makes the strongest peak present fit the spectrum; a really strong carbon dioxide peak means there was more carbon dioxide between the source and detector than there was sample in the pellet! 

Use the relative intensities of the peaks for a qualitative evaluation of the purity of the sample and to evaluate the nature of the contaminant, if present. If the strongest absorption expected for the starting material is but a tiny blip compared to the strong absorbances for product, use that to justify the purity of your sample. A large absorbance due to starting material compared to product and your melting point is far below expected? Sounds like you need to redo the experiment. 

For any molecule having a total of N atoms, there are (3 x N) ^-6 modes of vibration possible, (3 x N) - 5 for linear molecules. A molecule such as benzene, C₆H₆, has 12 atoms, so (3 x 12) - 6 = 30 modes of vibration. However, some modes are not IR active, some modes have energies so close to each other that they overlap and appear as one broadened absorption, so usually there are fewer strong absorptions than this equation predicts. Compare the variety of modes of bending available for CH₂ groups, yet from the table, only one absorption is expected for bending. 

On the other hand, there are numerous weak absorptions that result from various additive, subtractive, and multiples of the energies of the absorptions that are present in the molecule. These are not easily interpreted, and not useful at all except in the case of the overtone absorptions for various isomers of aromatic compounds. 

Your primary focus should be to assign any peaks that logically represent some segment of the molecule you are analyzing. Draw a structure for the molecule, and find as much evidence that justifies what you've drawn as possible. 

For example, if you were analyzing 2-butanol, you'd hope to see evidence for 

a) aliphatic C-H stretching 

b) CH₃ bending 

c) CH₂ bending 

d) C-O stretching 

e) O-H stretching 

A correlation chart would be necessary to predict the position that each of these groups would show absorbance. 

If you were analyzing t-butanol , all of the above should be present of course except for CH₂ bending, which, not present in your pretty drawing, you'd expect its absence in the IR. 

The bottom line is that often, the absence of peaks is more important than the presence of peaks. 

3. Does an IR spectrum prove the identity of an organic substance? 

Yes and no. 

Yes in that every organic compound has a unique IR absorption pattern, and if you had access to absolutely every IR spectrum of every known substance, then one could make a perfect match (although computer libraries of spectra do this much faster). 

No, in that the expected patterns can be very similar for similar structural isomers. One might be able to distinguish 2-butanol from 1-butanol based on the IR if spectra for each could be comparedone could hand-wave and decide that the spectrum with a stronger CH₃ bending would be 2-butanol, since its structural formula shows two methyl groups (although they are not equivalent) compared to 1-butanol which has only one. For simple molecules, this kind of qualitative interpretation is easy. 

4. Why is the carbonyl absorption frequency the first peak to consider in deducing the molecular structure? Why does its position vary? 

First, it is often the strongest absorption in the spectrum, if present, and is often in the middle of the spectrum. Second, the identification of several functional groups depends on correctly identification of the carbonyl absorption. Variety here is good! The carbonyl group is very sensitive to the type of functional group it is part of, and the local connectivity within the molecule. 

Conjugation or hydrogen bonding with solvent tend to lower the energy of the carbonyl stretch, whereas substitution by more electronegative substituents or a decrease in the <CCC bond angle about the carbonyl (as in cyclobutanone or cyclopentanone) tends to raise the energy. 

So, if two possible structures are being considered, one where the carbonyl is conjugated with an aromatic ring, and one where it is not, and the C=O stretch is much lower than expected for that functional group, then the former arrangement is the correct interpretation. 

5. Where do I start interpreting an IR spectrum? 

Start at the high energy endthis is where the first evidence of functional groups is most easily recognized. As a rule of thumb, peaks above 1350 cm^-1 are most likely to be reliable. 

Peaks below 1350 cm^-1 tend to be group frequencies and less easily assignable. In fact, the absence of a peak in this 1350 and lower region is more important than the presence of an absorption. 

Think about all of the data you have collected up to this point for the sample: 

· History of the compound (i.e., compound W was treated with liquid reagents X and Y to give Z, isolated as a solid; this means that W is a strong contender as a contaminant, perhaps X or Y) 

· Formula or molecular weight (if given/known) 

· Melting or boiling point 

· Solubility 

· Refractive index 

· Other analysis data from instrumentation 

Generic Approach to Inspecting an IR Spectrum 

Begin by looking for the presence or absence of a carbonyl group which is often a prominent peak around 1700 cm^-1 or a little over. If it is present, evaluate other areas for other functional groups, such as the C-H stretch that is unique in the structure of aldehydes, or if this is not there, must conclude the compound is a ketone Other possibilities are amides, acids, or esters. 

If there is no carbonyl, look for a strong O-H stretch, indicative of alcohols or phenols. No O-H, it must be an ether. (Note again, that an alcohol has an O-H and a C-O stretch; ethers have only C-O). 

Check now for C=C double bonds or aromatic rings at about 1650 for alkenes and 1650 to 1450 cm^-1 for aromatic rings. Since these typically are rather weak, you should also check the C-H stretching region for the presence of aromatic/olefinic C-H stretching (above 3000 cm^-1) or aliphatic C-H stretching (below 3000 cm^-1). Ideally, for alkenes/aromatics, you should see both C=C stretching and C-H stretching, too. 

Check for triple bonds, C=N at 2250, sharp or C=C weak at 2150cm^-1. 

Check for nitro group absorptions, of which there should be two, one at 1600-1500 and another at 1390-1300 cm^-1. 

If none of the above are present, it must be a hydrocarbon or halogenated hydrocarbon. 

Draw structures for the molecule, using all of the positive evidence available. 

If there exists possibilities of stereoisomers or geometric isomers, check those appropriate regions using Tables 1, 2, and 3 on the previous page. These tables show the patterns expected for various alkenes or aromatics. (s = strong, m= medium, and w= weak) To determine positional isomers or stereoisomers, the patterns observed (or absent) are more important than the wavenumber positions, which can vary. 

An extremely valuable analytical tool in modern structure determination is mass spectrometry. A mass spectrum often provide you with the molecular weight of the compound, and the interpretation of the fragmentation pattern will, with experience, give you further information about the structure of the compound. For the analysis of organic liquid products, see Section B above on GC-MS. Unfortunately, we do not yet have in the Instrument Room an instrument for MS analysis of solids, but solid products will be run for you in the graduate research MS facility. The departmental instruments are a KRATOS MS-9/50 and KRATOS MS-50 double-focusing magnetic mass spectrometers. The latter is used for FAB-MS analyses, which is used when a sample is ionic or too polar to be analyzed by the normal electron ionization MS method. 

To obtain a mass spectrum of your solid sample, place a very small amount (1 to 2 mg) of your pure solid material in a plastic "snap-top" sample vial. Complete one of the MS Analysis Request Forms available on the MS table in the instrument room. Be sure to fill out the form completely, including the mp or bp entry or the sample will not be analyzed. Tape the vial to the form with "Scotch" tape and place it in the "Mass Spec Analysis - To Be Run" box on the same table where you found the form. The spectrum should be analyzed within one week from submittal. Check regularly for your mass spectrum in the "Completed Mass Spec Analyses " box on this same table. Be sure to turn in your product with your final report for mass spec analysis. 

Questions Frequently asked about Interpretation of Mass Spectra 

1. How is the mass spectrum obtained? 

In electron ionization (EI) mass spectrometry, a very small amount of sample is vaporized and the gaseous sample bombarded with energetic electrons. When the electron impacts the molecule, it knocks an electron out of the molecule, yielding a positive ion. The electron may be ejected from a pi bond or from a non-bonding pair of electrons if present, or from a sigma bond if there are no pi or non-bonding electrons in the molecule. Many molecular ions have so much energy that they fragment immediately at certain bonds. After ionization, the molecular and fragment ions are accelerated in a beam which passes into a magnetic field (magnetic mass spectrometer) or varying electric field (quadrupole mass spectrometer) where ions are separated by their mass to charge ratio, m/z. Since z is usually 1, we often just talk about separation by mass (m/1=m). The mass range is scanned and the ion signal intensity at each ion mass is recorded to produce a mass spectrum, a plot of the ion masses on the x-axis versus the relative amounts of each on the y-axis. The largest peak (most abundant ion) is arbitrarily set to 100% and is called the base peak. How the molecule fragments depends on the structure of the molecule. Different molecules will have different fragmentation behavior yielding different and unique mass spectra. 

For almost all the spectra you will be interpreting, the mass range scanned starts at about mass 35 and extends to the highest m/z detected. 

2. What important information may be derived from a mass spectrum? 

Careful interpretation of the mass spectra provides two important pieces of structural information: 

b) Structural features. Fragment ion masses provide clues about the presence of certain atomic groups in the molecule such as methyl (15), ethyl (29), propyl (43), butyl (57), phenyl (77), methoxyl (31), ethoxyl (45), hydroxyl (17 or often 18 because water is lost), chlorine (35 & 37, two isotopes with characteristic intensity patterns), bromine (79 & 81, two isotopes with characteristic intensity patterns), etc. Detailed information is provided below. 

Bear in mind that it is very difficult to interpret mass spectra of mixtures, because it's hard to tell whether a certain mass ion peak is a molecular ion or a fragment from a heavier molecular ion. 

3. What are some general rules about interpreting a mass spectrum? 

Unfortunately, the mass spectral interpretation section in McMurry is rather meager, and the topic is not covered in Williamson. Therefore, study the steps given below and you should be able to successfully interpret your mass spectrum. 

First, locate the molecular ion in the spectra and mark it with an M⁺ or better, M. You normally know what M⁺ should be because you just synthesized the material or you have some other clue as to its identity. One immutable rule about molecular weights of organic compounds is that they are always even except in the case of nitrogen-containing compounds. In this case, if the molecule contains an odd number of nitrogens, it will have an odd molecular weight; if even, an even molecular weight. All mass spectra shown below have even M⁺s because they don't contain N, except for Mass Spectrum # 8, which is an amine and contains 1 (an odd number) nitrogen. and therefore has an odd M⁺. 

Aromatic compounds and ring structures, as exemplified by 1-tetralone in Mass Spectrum #1 shown above, usually have an intense M⁺ because it is more difficult to break aromatic bonds and there is resonance stabilization of the positive charge. 

While the molecular ion is normally the highest mass peak in the mass spectrum, you will often see smaller peaks one and two mass units higher than the M⁺, for example the peaks at m/z 147 and 148 in Mass Spectrum #1. These are due to the fact that each carbon in a molecule has a 1.1% chance of being a ¹³C instead of a ¹²C, since all carbon in the world is 98.9% ¹²C and 1.1% ¹³C. If a molecule has 10 carbons, the ¹³C peak one unit higher in mass will be 11% (10 x 1.1%) of the ¹²C peak. This is observed for tetralone (C₁₀H₁₀O) in Spectrum #1. 

Alcohols, however, fragment easily and thus have very weak, or as in the case of Mass Spectrum #2 below, unobserved molecular ions. In cases where you have no idea what the molecular weight should be, you can work backwards from the fragment ions by adding the weights of common molecular pieces, like methyl, to fragment ions. In the example below, if we add 15 (for CH₃) to 101 we get 116; if we add 18 (for H₂O) we get 116; and if we add 1 (for H) to 115 we get 116. This is strong evidence that the molecular weight of the compound is 116, even though no molecular ion can be detected. 

Be aware that if your sample is not pure or if there is contamination in the mass spectrometer from another compound, you may get peaks at masses higher than your molecular weight and may choose an incorrect M⁺. This will make correct interpretation of the mass spectrum almost impossible. If you are unsure about which peak is your molecular ion, talk with an instructor. 

Next look for any chlorine or bromine isotope patterns. These are two mass units apart and have a definite ratio depending on the number of halogens as shown below: 

These unique patterns allow you to quickly tell the number of chlorines or bromines in the compound as demonstrated by the two spectra shown below: 

Some common neutral losses are: 

M-15 = loss of methyl (M-CH₃) 

M-17 = loss of hydroxyl (M-OH) 

M-18 = loss of water (M-H₂O) 

Mass Spectrum # 5 shows the losses of methyl and water. Mass Spectrum # 6 shows loss of hydroxyl. 

In the spectrum of the aliphatic aldehyde, hexanal, shown below, the peak at m/z 71 could arise from either loss of the aldehyde group, CHO, or an ethyl group, C₂H₅. It is not possible to tell which is correct without some further analysis. 

M-45 = loss of ethoxyl (M-OC₂H₅) 

The loss of ethoxyl is shown below in the mass spectrum of the ethyl ester of the fatty acid, hexadecanoic acid (stearic acid) yielding the ion at m/z 239. 

Aliphatic carboxylic acids tend to show weak molecular ion peaks and exhibit the various peaks similar to esters. Characteristic of any acid with g-hydrogens is a peak at m/z 60 (radical cation CH₂=C(OH)₂), formed by rearrangement. 

M-79/81 = loss of bromine (M-Br) 

See Mass Spectrum # 3 and Mass Spectrum # 4 above for examples of the loss of these halogens. Iodine and fluorine are also lost readily. For many molecules, a peak is observed that involves the loss of H-Cl (M-36 and 38), H-Br (M-80 and 82) or H-I (M-128) 

M-42 = loss of ketene (M-CH₂=C=O) 

In Mass Spectrum # 11 below, a loss of 42 derives from the fragmentation of steroid ring A of the fertility hormone progesterone to eliminate a molecule of ketene (CH₂=C=O) and yield the ion at m/z 272. 

As you would predict, compounds with aliphatic chains containing four or more carbons will show this loss. 

Below is the mass spectrum of hexane. We see a fairly intense (~30%) molecular ion, M⁺, at m/z 86. Major fragments are observed at mz/ 71, 57, 43 which differ from each other by 14 mass units. This difference of 14 is equivalent to a -CH₂- group, i.e., these arise by fragmentation of the hexane molecule at various points along the carbon backbone as shown. 

Carbonyl-containing compounds usually give strong molecular ion peaks. Cleavage at the carbonyl produces the acylium ions stabilized by resonance, and these are usually quite intense. 

Certainly not; just about every compound will show a lot of peaks in the lowest molecular mass regions. Be sure to identify the molecular ion, the base peak (that is, the highest intensity peak in the entire plot), and any peaks that are easily correlated to any structural features of the compound. Mass spectra of mixtures are truly complicated, nearly impossible to interpret unless you know in great detail the history of the sample in question (solvents used in its synthesis, starting materials, recrystallization solvents, etc.). For this reason, it essential to check the purity of your sample by melting point or some other means prior to submitting it for analysis by mass spectrometry. 

5. What are some other types of ionization methods that can help to confirm molecular weights or to analyze very polar, involatile molecules? 

Some samples fragment so badly by electron impact ionization that it is difficult to determine their molecular weights. In this case, a "gentler" ionization method called chemical ionization (CI) is often used. Here, the sample is ionized in the gas phase by protonation, so the molecular ion is observed at one mass unit higher (sometimes called an MH⁺ peak). The result is a more intense molecular ion peak than obtained by EI with very little fragmentation. 

Polar, involatile molecules such as proteins, or charged molecules, such as amino acids, will not evaporate when heated in a vacuum to provide enough molecules in the gas phase for electron or chemical ionization. In this case, they can be ionized by either Fast Atom Bombardment (FAB) ionization or Electro-Spray Ionization (ESI). In FAB, the sample is placed in a drop of a high boiling liquid such as glycerol, and ions are "sputtered" from this droplet by a beam of energetic xenon atoms. In ESI, the sample is dissolved in a volatile solvent and sprayed into a vacuum at high voltage. The solvent evaporates from the charged spray droplets, leaving charged molecules that then enter the mass analyzer of the mass spectrometer. Both of these are "gentle" so that minimal fragmentation occurs. 

"One good spectrum is worth 10 (1000, 10⁶, an infinite number of?) bad spectra!!" 

The two NR-80 NMR instruments were purchased with funds provided by the Howard Hughes Foundation. Another extremely useful characterization technique in modern chemistry is nuclear magnetic resonance spectroscopy. Proton (¹H) NMR spectra often enable you to determine the exact number of protons in your compound, the number and arrangement of neighboring protons relative to a given proton, and the chemical environment in which each proton is located. Williamson's Macroscale and Microscale Organic Experiments has much useful information in Ch. 14 on the theory and practice of NMR. Copies can be found in Davey library, the instrument room (206 Whitmore) and the tutorial room (211 Whitmore.) 

There are two NMR instruments in the CRC Instrument Room. You will acquire an NMR spectrum yourself in a "self-training" mode using CALIOPE (Computer-Assisted Lab Instrument Operation and Principles Explanation) tutorial program on the NMR computer. An instrument room assistant will be available to help. 

1. Sample Prepartion - Careful sample preparation is critical. Most NMR analysis problems are due to sloppy sample preparation. 

a. NMR Tubes - This is the long, thin tube with plastic cap that is part of your desk equipment. To clean used tubes thoroughly, rinse them out well with acetone and place them upside down in a beaker or flask to dry for at least 24 hr. Otherwise, acetone impurity peaks are likely to show up in your spectrum. (You can blow acetone out of the tube by connecting your 1/16" teflon tubing from your Red kit to the nitrogen line and pushing this to the bottom of your NMR tube. Do not dry in an oven as this "warps" the tube. 

1.Deuterium (²H) eliminates interfering solvent peaks from the spectrum that would be huge if protium (¹H) were present. 

2.The FT-NMR "locks on" to the deuterium nuclear magnetic resonance to maintain a very stable magnetic field. 

Place 20 to 30 mg of a solid product or 2 drops of a liquid product into a clean vial, reaction tube, test tube, or 10-mL Erlenmeyer flask. Add approximately 0.5 mL of deuterochloroform, CDCl₃, located on the Hooded shelf under "Chloroform, Deuterated". Agitate the mixture to dissolve a solid. Everything must go into solution. Even small amounts of solid particles will produce very poor quality spectra. Look carefully at your sample solution; if it looks cloudy or has particles visible to the eye, you need to remove those particles. To remove all insoluble particles, filter the solution through a filter pipet directly into the NMR tube. (See Chapter 4 in this Lab Guide for specific instructions on using filter pipets.) Rinse out the vial, tube, or flask and filter pipet by adding an additional 0.5 mL of CDCl₃ to it, swirling, and transferring this to the NMR tube. Finally, add additional CDCl₃ to the tube to bring the liquid level up to about 4 cm (1.5 inch), and cap the tube. 

If your sample is insoluble in CDCl₃, try adding one or two drops of deutero-DiMethylSulfOxide (deutero-DMSO, also called Dimethylsulfoxide, Deuterated, available from the stockroom.) Mix vigorously. Virtually all compounds dissolve in this mixture. If some material remains undissolved, filter it off using a filter pipet as described above. Avoid using pure d-DMSO; spectra are ugly with this solvent. 

c. Sign-up for NMR Analysis - NMR tube tags can be found on a table in Room 206. Push your NMR sample tube through a pre-punched hole in the tube tag as directed. Do not use tape or sticky labels on NMR tubes as this gunks up the spinner. Take the labeled tube and completed form to the Instrument Room (206), place the sample in the "To Be Run" rack. When your turn comes, use the instrument immediately! Instrument time is precious with large classes and only two NMR spectrometers. If you are really tied up and can't use the instrument immediately, make sure the next person is notified so that s/he can utilize the instrument. If the NMR is backed-up, you may need to sign-up for additional time outside of the regular period. This will unable you to take your time, get one-on-one help from a TA and may provide a greater understanding of NMR. 

2. Acquiring the NMR Spectrum on the NR-80 FT-NMR 

Run the sample according to the instructions in theCALIOPE (Computer-Assisted Lab Instrument Operation and Principles Explanation) tutorial program on the NMR computer which covers theses steps: 

a. Putting the Sample in the Instrument 

b. Locking on the Deuterated Solvent 

c. Shimming to Optimize the sample signal 

d. Sample signal (FID) Acquisition 

e. Fourier Transformation 

f. Phasing 

g. Integrating 

h. Plotting 

1. Why are there extra peaks in my spectrum? 

This first question is one rationale for asking you to discuss comparisons of spectral features of starting material and product in your PreLab exercises. Correct interpretation of the spectrum obtained is a good first step in deducing what may have gone wrong in a synthetic experiment. 

2. Why are there peaks in the spectrum that cannot be attributed to starting material or to product? 

Any deuterated solvent has some trace of non-deuterated solvent, as well. The most common of these is deuterochloroform, CDCl₃, which has a trace of CHCl₃, which shows up at about 7.26 or 7.27 ppm. Deuterated acetone, deuterated water, deuterodimethylsulfoxide, or any deuterated solvent all contain a trace of the protio-compound and will show absorptions (usually rather small compared to the peaks due to solute) in the chemical shift range expected for those types of hydrogens. 

A peak at the farthest right-hand side of the spectrum of course is TMS, tetramethylsilane, which the operator uses to calibrate the spectrum, and assigns the chemical shift as 0.0 ppm; all other shifts are measured as shifts downfield of TMS. 

When TMS is not present, the spectrum is calibrated relative to the signal of the residual non-deuterated solvent; spectra obtained in deuterochloroform are often assigned 7.24 or 7.25 ppm. 

There is often an extra peak between 2 and 3 ppm, usually a short one; this is due to water. If the peak is weak, it may be due to traces that are in the NMR solvent. Polar organics that come into contact with aqueous solutions or are exposed to atmospheric moisture for any length of time will contain varying amounts of moisture. (This is proof of the need to distill polar organic solvents from drying agents when used for extremely water-sensitive reagents!) If the peak is large, and your procedure involved precipitating a product from an aqueous solvent, then some water adhered to the solid. Similar problems arise from distilling liquids from aqueous reaction mixtures. 

3. How to tell is a peak is too large? 

Compare it to peaks you know are due to product or compare integrations. 

4. Why does the chemical shift for water vary between different sample in the same solvent or from one solvent to another? 

Protons (hydrogens) attached to oxygen or nitrogen atoms often have a variable range for chemical shifts. The variability is due to varying degrees of hydrogen bonding with the NMR solvent or as the concentration of the solute varies. This can readily be observed by taking a series of spectra of the compound and adding slightly more sample to the NMR solvent each time. 

In addition, under normal experimental conditions, quite often there is no coupling seen for these hydrogens. The reason for this is that these hydrogens are often undergoing rapid exchange with each other, with the NMR solvent, or with traces of water that may be present. Quite literally, before the hydrogen is able to couple to other hydrogens in the molecule, it chemically is transferred to some other species in the solvent. When exchange occurs between a molecule containing hydroxyl group(s) such as acids or alcohols, the hydrogen of the hydroxyl quickly becomes a deuterium, which does not show resonance in the region of the spectrum that data is being collected. This exchange comes to equilibration within seconds (essentially upon dissolving the solute in the NMR solvent). 

Non-aqueous solvents that were used in the synthesis or recrystallization may also show up in the spectrum, especially if the solvent has a relatively high boiling point. Toluene (bp 110°C) is a commonly used high-boiling solvent that frequently shows up in the spectrum. This is proof of the need for ample drying time for solids obtained from crystallization. 

Sometimes, integration of peaks makes it easy to eliminate a peak from thoughtful consideration. For example, if the molecule of interest should show integral ratios of 5 to 3 to 2 and the peak you're looking at integrates as 1 hydrogen, or some non-whole number ratio, then it must be something else. 

5. Why does a peak fall outside of the predicted range? 

Remember to check for starting material or to consider that the peak is due to something else altogether. Remember also that effects that cause shifts are additive. 

The following discussion shows how to rationalize an inexplicable peak at 4.6 ppm for benzyl chloride. 

CH₃ groups are expected to absorb at 1.0 ppm if bonded to another sp³ carbon; if bonded to an aromatic ring, they feel some of the same deshielding that the ring hydrogens feel, so they are shifted to about 2.2 to 2.5 ppm. 

CH₃ groups absorb at 1.0 ppm if bonded to another sp³ carbon; if bonded to chlorine, the methyl hydrogens absorb at 3.05 ppm. 

In benzyl chloride, the CH₂ feels deshielding due to the ring and deshielding due to chlorine; therefore, you'd expect the chemical shift to be greater than 3.05 ppm. Indeed, the CH₂ signal appears at 4.6 ppm for benzyl chloride. 

Similarly, the CH₂ group of aliphatic ethers absorb at 3.5 to 3.8 ppm; the CH₂ group of benzylic ethers absorb at 3.9 to 4.3 ppm, the proximity to the aromatic ring further deshielding the hydrogens. 

6. Why aren't there integrations in the spectrum? 

The operator didn't integrate the peaks. Quite often, a pure product has only one kind of hydrogen and there is no ratio of areas to be compared; or all of the hydrogens present are in the aromatic shrubbery region of the spectrum, so integration was determined to be of little value. 

Sometimes the peaks are so close together or overlap so much that the integration covers two different signals. This is common for hydrocarbons that contain several CH₂ groups that are in very similar chemical environments. 

7. Why doesn't my spectrum look exactly like the reference spectrum? 

Could be different spectrometer strengths (although the chemical shifts should be the samesee earlier discussion on chemical shifts) which help to resolve multiplets that have very small coupling constants. So, your spectrum may show a triplet which is split into tiny doublets, whereas the reference spectrum shows only a triplet. This may work the other way around, too. Purities may differ also, especially with respect to water content, which will greatly affect the chemical shift of compounds with hydroxyl, carboxyl, or amine groups. 

Apparently some of the hydrogens have very similar chemical shifts and there is some overlap. A simple example is n-pentane (CH₃CH₂CH₂CH₂ CH_3.) Those CH₂ groups are very similar in their shifts and may overlap; instead of seeing three kinds of hydrogen in 6 to 4 to 2 ratios, the methyls are a well-resolved triplet (area of 6) and the CH₂ are a broadened multiplet (area of 6). Similar problems arise with longer alkane chains as in hexane, heptane, fatty acid chains, or hydrogens on aromatic rings. 

9. How are FT-NMR spectrometers different from continuous wave spectrometers? 

The spectrometer we've described throughout this handout are continuous wave spectrometers, where the radiofrequency energy is held constant, and the magnetic field strength is varied. Fourier Transform Spectrometers (FT) work by exciting all magnetically active nuclei at once and measuring the decay curve as energy is released as the nuclei relax back to the ground state. Each different type of hydrogen contributes to the interference pattern, similar to the way different notes contribute to a chord. Computer programs can work backwards and deduce the absorption frequency of each hydrogen that contributed to the total signal. All aspects of chemical shift, splitting, number of unique hydrogens, etc., are all the same, for FT-NMR and for continuous wave. 

10. What advantages do FT-NMR spectrometers offer over continuous wave instruments? 

Spectra of nuclei that are in low abundance or samples that are very dilute can be scanned rapidly and repeatedly until a decent spectrum is obtained. Summation in this way with continuous wave spectrometers would take forever; with FT-NMR, each individual scan can be done in a few seconds, so that hundreds or thousands of scans can be obtained in just an hour. 

The spectra obtained by FT methods have a better signal-to-noise ratio, thus signals due to trace chemicals that barely peep above the baseline do so much clearly. This is also handy if you have extremely small amounts of the compound you are trying to analyze. 

With FT-NMR, a person can easily take a lot of spectra and store the data in the computer to be processed later. This is important when say a reaction is monitored by NMR spectroscopy and the operator cannot wait a long time between data aquisitions. 

11. When can the height of the peak be used as an approximate value for the integration? 

When the peak is narrow and well resolved from the baseline. With organic compounds, this often is the case. However, if the peak is broadened and flat, to the extent that a significant percentage of the peak area lies just above the baseline, a separately determined integration is more dependable. 

12. How do I start interpreting a spectrum of a known compound whose source is known? 

In the fortunate event that you are asked to interpret a spectrum of a known compound or an isolated product(?) from a reaction where the reagents that were involved were known, then interpreting a spectrum is very easy. 

Start by drawing the structures for the organic starting materials and products. Try to evaluate the number of different hydrogens that are present in each structure. Now, using any correlation table for NMR spectra, predict the ranges you'd expect for each hydrogen present. 

Look at your spectrum now, and begin to make some matches in the chemical shifts expected for each type of hydrogen. Label the absorptions in your spectrum, either by drawing a line from the hydrogen(s) that generated the signal to the signal plotted on the spectrum, or by labeling each type of hydrogen in the drawn structure with a letter and writing the same letter over the peak(s). 

Be sure to identify the source of splitting of each resonance, and reconcile these with the structures that you've drawn. For example, if there is a triplet in the spectrum, you should indicate the identity of the signal and also that the splitting is due to two equivalent hydrogens on an adjacent carbon. Finally, if the integration is plotted on the spectrum, reconcile the ratio(s) of the integral heights with expected integral ratio(s). 

13. How do I start interpreting a spectrum for a compound when the only information given is the molecular formula? 

Shown below is a spectrum of a compound whose formula is C₉H₁₀O₂. 

First, calculate the units of unsaturation. A saturated molecule with nine carbons would contain (9x2)+2=20 hydrogens, whereas this molecule contains 10. 20-10 = 10 hydrogens, divide by 2 gives 5 units of unsaturation; that is, the whole molecule contains a total of 5 rings and/or double bonds. 

Note: Remember that oxygens do not change the number of hydrogens in a saturated organic compound, nor does sulfur. Since halogens form one bond to carbon as does hydrogen, then the total number of hydrogens and halogens in the saturated compound is the same as the parent hydrocarbon. For every nitrogen in a formula, add one hydrogen for the number of hydrogens in the saturated compound. 

There are three kinds of hydrogens in the molecule, and since they are all singlets, then there is no coupling taking place in the molecule, and the correct structure will show no non-equivalent hydrogens on adjacent carbon atoms, except in the aromatic ring. Since an aromatic ring accounts for four units of unsaturation, an aromatic compound is a possibility. This is apparent from the chemical shift of the singlet at 7.3 ppm. Since there are 5, the ring is monosubstituteddisubstitution would give 4 aromatic hydrogens. 

The singlet at 2.1 ppm is most likely a methyl group, since all three hydrogens are equivalent; it must also be bonded to a carbon that has no hydrogens since there is no splitting. 2.1 ppm is rather far downfield for a methyl bonded to an sp³ carbon, is not quite far enough for CH₃ bonded to and oxygen, so at this point, perhaps bonded to a carbonyl carbon. 

The singlet at 5.1 ppm looks like a methylene group (integration = 2), this time is farther downfield than CH₂ bonded to oxygen are expected, so it looks like it could be a CH₂ bonded to an aromatic ring and to an oxygen. Let's propose a structure. 

What if you have no formula and the spectrum is of a true unknown compound? Peaks at intermediate chemical shifts (2.5-4.0) are often a clue to the presence of oxygen or nitrogen. Previous work with the unknown may lead you to conclusions about the acid/base properties of the substance. A basic organic substance may indicate an amine group; weak acid properties, a carboxylic acid, etc. An IR spectrum is very helpful, in that it helps to establish the identity of any functional groups that may be present in the molecule. 

Notice that, given the formula of the compound, we did not try to interpret the data for the problem above in terms of carbons bonded to nitrogen or halogens. Likewise, if there were only one oxygen, then it would not make sense to try to interpret the structure as an acid or ester, as these require two oxygens (as our example had). Always check to make sure your structure has the correct number of units of unsaturation. If the molecular formula were C₉H₂₀O₂ (0 units of unsaturation), then we could not draw structures with rings, carbonyls, or carbon-carbon double bonds. Always check the units of unsaturation and the functional group(s) that are involved! 

Also, be sure that the splitting pattern (if present) makes sense with the structure you've drawn. In the problem worked out above, we did not draw a molecule where the CH₂ and CH₃ were next to each other as in an ethyl group. In this arrangement, the hydrogens would split each other and the CH₂ would appear as a quartet and the CH₃ as a triplet, which were not observed in the spectrum shown. 

Remember that splitting gives clues to the connectivity, and that whereas the presence of splitting assures the presence of non-equivalent hydrogens on adjacent carbons, non-equivalent hydrogens on adjacent carbons do not always show observable splitting patterns. 

Remember also that the ratio of integrals is the nearest whole number ratio and may not be the actual number of hydrogens in the molecule. An NMR spectrum of butane shows two kinds of hydrogens in the molecule in a ratio of 2 to 3, but in actuality, there are 10 hydrogens in the molecule, so there's really 4 H of one kind and 6 H of another. 

How could you know there's something funny going on? Draw any hydrocarbon that contains only C and H or C, H, and O, and count the hydrogensit's always even!! (So is the molecular weight.) 

Which brings us to the last bit of chemical triviathe odd nitrogen rule. Draw any compound with 1, 3, or 5 nitrogens and count the number of hydrogensit's odd, and so is the molecular weight. This is called the "Odd Nitrogen Rule". Odd molecular weight, odd number of nitrogens. 

14. Where can I find an NMR correlation chart? 

In Williamson, Chapter 14, or any large-sized organic textbook, like McMurray. Williamson's textbook can be found in the instrument room, tutorial room or Davey library. 

15. How can I identify common NMR solvent impurities? 

Common solvent impurities in CDCl₃ are found in the figure on the next page. It is important to identify impurity peaks so that you can ignore them in interpreting your NMR spectrum. 

There is no sign-up sheet for the polarimeter because only a few experiments involve its use and it is therefore usually available. If you need to use the instrument for an extended period of time, for example Exp't 205, please leave a note on the instrument for the courtesy of your classmates. 

You can determine the optical rotation of an optically active product using the Perkin-Elmer Model 241 digital polarimeter located in the Instrument Room. This instrument is accurate to 0.002°. It should be turned on 15 minutes before running your sample. Make sure the pushbutton marked Na/CONT on the righthand unit of the instrument is depressed. 

1. Prepare your Sample Solution - Carefully weigh and dissolve as much of your product as possible in exactly 2 mL of water or methanol, whichever solvent the sample is most soluble in. The higher the concentration, the better the reading. The sample must be perfectly clear, not cloudy. Ask yourself, is this solution as clear as pure drinking water, like that from Rock Springs Water Co.? If it's not, there's a good chance that the polarimeter will not give a good reading. If it contains suspended material, use a desktop centrifuge to remove the particles. Sometimes, even after centrifuging, the particles are still present. In this case, you need to place a small wad of cotton or glass wool in a Pasteur pipette, a little layer of celite (Filter-Aid) and filter the solution. 

2. Fill the Sample Cell - Handle the sample cell with great care! It is worth $700! Remove the sample cell from its storage box and set on 3 or 4 paper towels on the benchtop with the small holes in the frosted glass ends facing upward. Do not get these confused with the water jacket hose connections which are not used. Draw all of your sample into a Pasteur pipette and then fill the cell from either end. If you are having trouble filling the cell without getting air bubbles, there are two approaches. First, squeeze the bulb of the pipette and draw up as much sample as possible; hold the cell at an angle and start filling from the lower of the two fill holes. As you squeeze the bulb to push more and more sample into the cell, you need to gradually bring the cell to the horizontal position. Another approach is to use one of your spotting capillaries as a guide: let the sample solution drain from the pipette along the capillary while the end of the capillary is in the fill hole. This allows liquid to flow into the cell by capillary action and lets air or air bubbles to escape. You must avoid getting any air bubbles into the cell as this will make a reading impossible. Your 2-mL solution should just about fill the cell to overflowing. 

3. Determine the Observed Rotation - Set the INTEGRATION SEC switch to 1 or 5. Press and release the ZERO push-button on the left side of the polarimeter and note that the digital read-out goes to zero (or very near to zero). Carefully pick up the cell and look through the small, clear window on the end to check that the cell's light path is bubble free. Open the sample compartment and set the cell on the support rods parallel to them. Close the cell lid. The electronics will now rotate the internal polarizers until maximum energy is received by the detector, and then display the sign and amount of this rotation in the digital readout. It may take the instrument several minutes to reach a "fairly" steady-state value in this reading, and it will still probably vary quite a lot in the thousandths of a degree digit. Average 10 to 20 readings and record this in your book. If you see that the instrument continues to scan to the negative values, back to positive, to negatives again, and so on, that is a clue that your sample is too cloudy. In this event, there is so little light reaching the detector that it is having difficulty finding the angle at which the maximum light reaches the detector. In this case, see above for clarifying your sample. NOTE: Do not expect to see a number for the rotation. You are measuring only the rotation for your dilute solution. This value will be used to calculate the specific rotation, taking into account your sample concentration as shown below. 

5. Calculating the Specific Rotation - The specific rotation, , is calculated using the equation:  

By manipulating this equation you can determine the amount of solute needed based on the specific rotation value, measured rotation, and volume of the cell you are using. For example, when you dissolve 142 mg of lactose in 1.5 mL of water, that was isolated from milk and measure the rotation , the average value calculated was 8.77°. To calculate the specific rotation of your lactose sample, 

You can determine the refractive index of a liquid using the refractometer located in the Instrument Room. The refractive index of a substance is an intrinsic property just as is the boiling point, but a somewhat more sophisticated instrument is required to measure it. It is defined as the ratio of the speed of light through a vacuum to the speed of light through the substance. 

You will need 5 or 6 drops (about 0.25 mL) of liquid to determine a refractive index. 

1. If necessary, remove the plastic cover from the refractometer. Turn the front control dial to the n_D setting. 

2. At the back of the refractometer you will see a light on the end of an arm which can be moved up and down. Move the light so that it is not over the glass prism on the back of the refractometer. 

3. On the right side of the top glass prism, you will see a small metal handle protruding. Gently pull up on this. The top prism will open on a hinge on its left side by which it is attached to the lower prism. If there is a tissue under it, remove and discard it. 

4. Before you analyze your product, make sure the refractometer is operational and practice using it by measuring the refractive index of a pure standard such as methanol (n = 1.3290) or toluene (n= 1.4960). 

5. Using a Pasteur pipette, place four or five drops of liquid on the lower prism. Be sure not to scratch the prism with the pipet tip. 

6. Carefully close the top prism back onto the bottom prism and press the handle down firmly so the top prism snaps into place and is firmly contacting the lower prism. 

7. Move the light arm up so it is within 1" of the prism assembly. 

8. Look through the eyepiece and gently turn the large wheel on the right side clockwise or counterclockwise until you see a white/black demarcation in your field of view -black on the bottom- like the one shown below. If it is black on the top, turn the big wheel fully one way or the other to the stop and you should see a white/black demarcation with black at the bottom. 

Dark field must be on the bottom. 

10. Open the prism assembly by raising the top prism as before, and clean the prism off with a tissue wetted with methanol. Dry it with a dry tissue. 

11. Now measure the refractive index of your sample by repeating steps 5 through 10. 

12. Leave a tissue between the closed prisms and cover the instrument. 

0.0004 x 3.8 = .0015 n= 1.3177 + .0015 = 1.3292 

This is close to the literature, n= 1.3290, for methanol. 

The composition of a mixture may be determined from the index of refraction, assuming a linear relationship between the refractive index and concentration (percent molar composition). The refractive index for sec-butyltoluene is n= 1.4860 and for toluene, n= 1.4960; therefore, a 50/50 mole percent mixture of sec-butyltoluene and toluene would have an n= 1.4910; a 25/75 mixture an n= 1.4935, etc. 

Ultraviolet-visible spectroscopy is generally less useful than other spectroscopies in providing structural information, but can be useful in accurate measurements of concentration. The UV spectrum is dependent on the electronic structure of the molecule, and is usually limited to conjugated systems such as a,b-unsaturated carbonyl compounds and aromatic rings. (See the brief but informative section on UV-VIS spectroscopy and spectral interpretation in Chapter 15 of Williamson.) If no instructions on sample preparation are given in your assigned experiment, dissolve about one mg of sample in 10 mL of 95% ethanol. This may be too dilute or too concentrated to get a measurement in the desired absorbance range, and you will need to adjust the concentration accordingly. You will be running your spectra on a modern Hewlett Packard 5842 Diode Array UV-Visible spectrophotometer (and associated data system) located in Rm 206. Follow the instructions to be found near the instrument titled "The Care and Feeding of the UV/Vis". 

Be sure that the solvent for the sample solution and the solvent used in the background subtraction are from exactly the same bottle. Traces of contaminant may be present in one and not the other, leading to weird UV-VIS spectra. 

Cleaning Up - Rinse out the quartz cell with 95% ethanol or methanol and put it back in the appropriate box. 

Interpretation of UV/Vis spectra is NOT at all like interpreting IR, NMR, or MS, where spectral features can be associated with certain functional groups or carbon skeleton substructural features. In UV/Vis spectra, the absorption maxima (l_max) are normally assigned to conjugated systems which can include either the whole molecule or the portion containing the conjugated system. Thus, the longest wavelength absorbances are usually not associated with a C=C or a C=O but with the combination assuming they are adjacent. 

1. INTRODUCTION 

When ultraviolet radiation passes through a sample of a compound, some of the ultraviolet and/or visible light is absorbed. How much of the light and the wavelengths that are absorbed are a function of the structure of the molecule(s) involved, the pathlength of the sample solution, and the concentration of the sample in the solution. The absorptions are caused by electronic excitation from the electronic ground state (g.s.) to the lowest electronically excited state (e.s.). The energy of the electronic transition(s) are quantized, meaning that there is a discrete energy difference between the two states, and only light of energy (wavelength) corresponding to that energy jump will be absorbed. 

A plot of absorption versus wavelength is referred to as an UV-VIS spectrum. 

For beta-carotene, the difference in energy between the ground state and the excited state is 4.4 x 10^-19 Joule. Use E=hn and c=ln to calculate the wavelength of the absorption and predict the color that will be observed. 

Answer: n = E/h = 4.4 x 10^-19 J/6.626 x 10^-34 J sec^-1 = 6.6 x 10¹⁴ Hz 

l= c/n = 3 x 10⁸ m sec^-1/6.6 x 10¹⁴ sec^-1 = 4.5 x 10^-7 m x 1 x 10⁹ nm/1 m = 452 nm (orange) 

There are two basic designs for UV-VIS spectrophotometers. (Figures are adapted from Organic Chemistry, 2nd Edition, by S. Ege, p. 718. (D. C. Heath and Company). The UV-VIS spectrometer used in the Organic Instructional Labs is a single-beam instrument with a diode array detector. 

Since UV-VIS spectroscopy involves transitions of bonding and non-bonding electrons from bonding orbitals to higher-energy anti-bonding orbitals, a short review of bonding in organic compounds is helpful. From molecular orbital theory (MO), linear combinations of atomic orbitals (A.O.) can be used to represent bonding (addition) or antibonding (subtraction). The number of A.O.'s that are used equal the number of M.O.'s that can be made. When two orbitals that are symmetrical about the bond axis interact constructively, they form a M.O. that is also symmetrical about the bond axis, with a higher probability density that lies between the two 

Additive (bonding) combinations and Subtractive (antibonding) combinations of atomic orbitals to give molecular orbitals. Note that the number of a.o. used equals the number of m.o. that result. 

p orbitals can also interact with each other constructively or destructively, by sideways overlap. Like the p orbitals from which linear combinations can be made, these result in molecular orbitals which are not symmetrical with respect to rotation around the bonding axis. Constructive combination results in a p bonding orbital (pi bonding), which is shaped like a pair of kidney beans above and below the axis lying between the two atoms of the bond. Destructive addition of the two p orbitals results in a p* (pi star antibonding) antibonding orbital. Since this is not a direct overlap bonding mode, the overlap of electron density is not as effective as in sigma bonds; therefore, it represents an energy level not quite as low as a sigma bond, and likewise the antibonding combination is not as high in energy as a sigma antibonding orbital. Population of the p* orbitals explains why F₂ has only a single bond and Ne₂, with filled antibonding orbitals, contains no bond. 

The relative energies of these orbitals are reflected by the chemistry that molecules exhibit. Molecules with only sigma bonds (alkanes) are rather unreactivethe bonding orbitals are very low in energy, and will not interact with other reagents very readily. 

Molecules with p bonds (alkenes, alkynes) are more reactive in comparisonthey have p electrons, which are at higher energies, and are more available for donation to Lewis acid reagents. Finally, molecules with nonbonding electrons are more reactive, and the lone pair of the molecule often is the source of initial interaction with other chemical reagents that lead to reaction, such as protonation by a catalyst or when the molecule with the non-bonding electrons behaves as a reactive nucleophile. 

3. TYPES OF ELECTRONIC TRANSITIONS 

Molecules that have a variety of filled bonding (and unfilled antibonding) orbitals leads to a variety of possible electronic excitations and therefore absorptions. For example, electrons in an n level can be excited to either p* or to s* levels. Likewise, p electrons can be excited to s* orbital. (p to s* or p to n are not often seen, since these are forbidden due to changes in symmetry.) Finally, s electrons can be excited to either s* or to p* orbitals. 

As we would predict from the figure shown, the energy for the different types of transitions vary. The larger the energy gap, the higher the energy of the photon absorbed must be to effect that transition. 

Alkanes contain only single bonds, so only s to s* transitions are possible. These are very high energy, therefore have very high energy, and usually cannot be observed using the commercially available spectrometers. (The 100 to 200 nm range is referred to as "vacuum ultraviolet; to observe transitions in this range of wavelength, the sample must be in a vacuum, as atmospheric gases absorb in this region, also. 

Molecules which contain only single bonds but also have N, O, S, F, Cl, Br, or I contain sigma bonding orbitals and non-bonding orbitals, so s to s* and n to s* transitions are possible. 

Molecules that contain carbon-carbon p bonds (alkenes or alkynes) can show p to p* transitions. 

Finally, molecules with carbonyl groups (aldehydes, ketones, esters, acids, or amides) or carbon-carbon pi bonds (alkenes or alkynes) can also show n to p* transitions. Actually, these transitions are the most often studied, partly because they are easily observed in the ranges available on commercial UV-VIS spectrometers, and because they are very sensitive to substitution by other fragments in the molecule. 

Notice that not all transitions are possible; some are forbidden and never observed, such as s > p, s> n, p> n, or p > s*. Some, such as n > p* are forbidden by selection rules, but are often observed, although usually weak. For example, 4-methyl-3-penten-2-one, a conjugated enone, absorbs at 235 nm (p>p*) and at 320 nm (n>p*). Similarly, 3-buten-2-one absorbs at 219 nm (e = 3600, p>p*) and at 324 nm (e = 24, n>p*). 

Often, a plot of the UV-VIS spectrum of an organic compound shows a broadened band of absorption, rather than a discrete sharp absorption. If transitions are quantized, why do they absorb over a broad range of wavelenths? Molecules in the ground state and in the excited state can exist at various vibrational levels; although the excitation of a molecule in the ground state V=1 level to the excited state V=1 level is quantized, so is the jump to the V=2, V=3, etc. vibrational levels of the excited state, and each jump represents a slightly different energy. The variety of energies associated with the simple electronic transition is further smoothed out by the various rotational energy levels underlying each vibrational level. The tiptop of the absorption band represents the energy of the electronic transition from the most likely vibrational level and rotational level of the electronic ground state to the most likely levels of the electronically excited state. The bottom line: unless you artifically remove some of the available energy transitions, the spectra appear as broad bands, not peaks. 

Solvent choice can similarly affect the appearance of the spectrum. Non-polar solvents will not interact with solute molecules effectively, therefore having a similar effect on the spectral appearance as cooling the sample solution. Polar solvents interact more strongly, and fine structure will disappear. 

5. BEER-LAMBERT LAW 

The absorbance, A, is also equal to the log of the ratio of the incident radiation (I_o) to the light passing through the sample cell (I). 

The amount of light absorbed is a function of the concentration of the solute, c (in units of Molarity, moles per liter), the pathlength of the cell, b (in centimeters, the most common width of curvettes being 1.0 cm), and the molar absorptivity, e (in units of liter per mole-cm) commonly called the extinction coefficient. 

Absorbance, A, is measured by the instrument; the concentration and pathlength is determined by the operator of the instrument; and the molar absorptivity e is a property of the chemical, influenced largely by the size of the portion of the molecule that absorbs and whether the transition is forbidden or allowed. Forbidden transitions have a low probability of occurring, so have smaller absorptivities say 100 to 1000. 

The actual spectrum is seldom reproduced for the purpose of reporting data; rather, the wavelength(s) of the absorption maxima (l_max) are reported, along with e, the extinction coefficient, a numerical value of the intensity of the absorptions. In Chem 36 we usually only report l_max's and not e's. 

Instruments can be designed with either single-beam or with double-beam arrangements. In the Instrument Lab, the instrument is a single beam, which means that for each sample, a background spectrum is obtained using a cuvette filled with pure solvent. The sample (dissolved in the same solvent used for background) is then placed in the cuvette. When the spectrum for the sample + solution is obtained, the computer subtracts the background from the obtained spectrum to give a spectrum of only the sample. (Other instruments pass light simultaneously through a cuvette of solvent and a cuvette of sample + solvent, and subtract the spectrum of the reference cell.) 

As a method of determining the structure of an unknown compound, UV-VIS is probably the least useful of all the common spectrophotometric methods. It is, however, a powerful method for monitoring the concentration of known compounds or as a detector for other separation methods, such as liquid chromatography. In determining the structure of an unknown compound, its real value is apparent when it is used in combination with some other analytical technique, especially NMR or IR spectroscopy. Knowing the functional group(s) present and certainly the history of the compound in question (know the reagents used to produce the sample and what else has been done to the sample up to this point) can help you interpret the data that is obtained from the UV spectrum. 

Sample Problem: 

A sample of a solute, concentration of 1x10^-4 M,shows two absorptions, one at 300 nm (A = 0.50) and another at 250 nm (A = 1.25). Calculate the molar absorptivity for each absorption, assuming the cell pathlength were 1 cm. If you were monitoring water for the presence of this compound, which wavelength would be more sensitive for measuring small quantities of the chemical? 

similarly, e = 12500 for the 250 nm absorption. Since the compound of interest absorbs more strongly at 250 nm, then small concentrations may be more accurately measured than at 300 nm. 

6. CHOICE OF SOLVENT AND CUVETTE 

The solvent should be free of absorptions in the region where the solute of interest is to be measured. Common solvents and the maximum wavelengths where they can be used without interfering with solute absorptions are 

water 190 nm n-hexane 200 nm 

ethanol (95%) 205 nm isooctane 195 nm 

cyclohexane 195 nm methanol 205 nm 

Anhydrous ethanol cannot be used for obtaining UV-VIS spectra because the residual water is removed by azeotropic distillation with benzene; measureable traces of benzene still remain in the ethanol, and the cutoff for benzene is at much longer wavelengths. 

Since glass absorbs in the lower wavelengths of interest, most cuvettes are made of quartz, which makes them expensive. Quartz is transparent from 200 nm through the visible portion of the spectrum. Even with a quartz cuvette, absorptions below 200 nm are impossible to detect due to absorption by atmospheric oxygen and carbon dioxide. Vacuum techniques are necessary for measuring absorptions below 200 nm, hence the region 100 nm to 200 nm is often referred to as "vacuum ultraviolet". If only visible spectra are of interest, glass is adequate, and is often used for the test tubes in which sample solutions are analyzed (colorimetric analysis) using the "Spec 20". 

Different solvents can strongly influence the appearance of the spectrum of a compound. A good example is that of hexane vs ethanol: ethanol is a hydrogen-bonding, polar solvent, and thus will interact strongly with the solute molecules; this has the effect of making the energy differences between the different vibrational states even smaller. Hexane, on the other hand, does not hydrogen bond, and the spectrum of the compound will show a lot of fine structure, similar to the appearance of the spectrum if it were obtained in the gaseous state. 

7. TRENDS IN ABSORPTION SPECTRA 

Although the absorption of ultraviolet and visible light effects the excitation of electrons from ground states to excited states, the energy difference between the two states is determined by the nuclei that the bonding electrons are holding together in the first place. Ultimately, the nature of the atoms held together determine the appearance of the UV-VIS absorption spectrum, and the appearance of the spectrum can be used to deduce the connectivity in an unknown compound. 

A group of atoms that give rise to an absorption is called a chromophore; in other words, these are functional groups or recognizeable fragments that give rise to absorption. Auxochromes are additional substituents that increase the intensity and possibly the wavelength of the absorption. Different compounds with the same chromophore will differ in their absorption spectrum due to the different auxochromes that are present. 

As we've said before, the absorptivity depends on the size of the molecular system that gives rise to the absorption. One of the simplest comparisons is between the simplest alkene, ethene and the simplest conjugated diene, 1,3-butadiene. 

In ethene, each of the two carbon atoms are sp² hybridized, the pure unhybridized p orbitals of each combining to give p and p* orbitals. Ethene absorbs at 171 nm (vacuum ultraviolet), a very high energy transition, in the process an electron is excited from the p to the p* orbital. 

In 1,3-butadiene, there are four carbon atoms, again, all are sp² hybrids. Since there are 4 

The importance of conjugation is shown by comparing the wavelength of absorption for dienes that are not conjugated. 1,4-pentadiene is an example of an isolated diene, and absorbs (178 nm) more like ethene than like 1,3-butadiene. 

If we continue this pattern for longer conjugated systems, the energy difference keeps getting smaller and smaller, and the wavelength of absorption longer and longer. For 1,3,5-hexatriene, the l _max appears at 274 nm. In 1,3,5,7-octatetraene, there are four conjugated C=C units, and absorption occurs at 290 nm. For really long conjugated systems like beta-carotene (11 double bond units, conjugated) the absorption wavelengths are so long that they are in the visible light region. Remember, when visible light is absorbed, the light reflected is the color of the chemical or object. So, in the earlier sample problem, carotene absorbs at 452 nm, well within the blue-violet region of the visible spectrum. 

The bottom line: the longer the conjugated system, the lower the energy required for exciting the electrons from the highest occupied orbital to the lowest unoccupied orbital. 

Following the pattern shown for the simplest of the polyenes, draw the energy levels of the molecular orbitals for 1,3,5,7-octatetraene. How many p molecular orbitals and how many p* orbitals should there be? Why is the wavelength of the p to p* transition longer than in 1,3,5-hexatriene? 

Answer: You should show 4 p bonding levels below 4 p* antibonding levels, with the bottom four levels filled. The energy difference between the highest occupied orbital and the lowest unoccupied orbital is smaller than it is in 1,3,5-hexatriene, so the energy it takes to excite a bonding electron is smaller, thus the wavelength of the light corresponding to that transition is longer. 

If the wavelength of absorption is large enough, then the compound may show color. Remember that color is due to light reflected, which is the incident light minus the absorbed light. When a compound absorbs at 400 nm (violet), then the light reflected (appearance) is yellow. 

Wavelength of Color of Light Observed 

Light Absorbed Absorbed Color 

400 nm violet yellow 

450 nm blue orange 

500 nm blue-green red 

530 nm yellow-green red-violet 

550 nm yellow violet 

600 nm orange-red blue-green 

700 nm red green 

The pattern also follows for enones, or carbonyl-containing compounds where the C=O is conjugated with a C=C group. For comparison, acetone shows an n > p* transition at l_max 280 nm (e = 15, n>p*) and another absorption at 189 nm e = 900, p>p*). 3-buten-2-one, a simple enone, shows the same n > p*transition at 324 nm (e = 24), as well as another absorption at 219 nm (e= 3600) for the p to p* transition. Conjugation therefore results in longer wavelengths (lower energy) of analogous transitions as well as increasing the extinction coefficient (molar absorptivity) of that transition. 

Problems: 

A molecule known to have a carbonyl group and a C=C double bond shows p to p* transition at 240 nm and a n to p* transition at 310 nm. Is the C=O and C=C conjugated or not? 

Answer: Most likely, this represents a conjugated molecule, since the wavelengths of absorption are comparable to that of 3-buten-2-one. 

A sample is known to be either 1,3-cyclohexadiene or 1,4-cyclohexadiene. How could UV-VIS spectroscopy determine which isomer is present? What kind of transition(s) will be observed for either of the two? Compared to simpler alkenes, about what wavelength would each one absorb? 

Answers: The 1,3-isomer is conjugated, therefore an absorption around 217 nm is expected. The 1,4 isomer contains isolated C=C, thus absorbs at wavelengths seen for simple olefins, such as 171 or 178 nm (vacuum UV). In either case, the absorption is due to a p> p* transition; in the conjugated molecule, this energy difference is smaller, resulting in a longer wavelength of absorption. 

a. 1,5-hexadiene 

b. benzene 

c. toluene 

d. 2-butanone 

e. 1-butene 

Answer: b, c, d: b and c are conjugated "trienes"; ketone d shows a weak n>p* at about 280 nm; a and e only have isolated C=C 

Another broad class of auxochromes include those that extend the conjugation via resonance structures with the chromophore. These include atoms that have non-bonding lone pairs, such as N, O, or S. Examples are shown below. By extending the conjugation system, the wavelength of absorption increases and the molar absorptivity increases, as well. 

8. Predicting the Absorptions for Aromatic Compounds 

Although there is extensive information for predicting the wavelength of absorption of various conjugated alkenes or enones, these really do not apply to the kinds of products that you make in Chem 36. Aromatic compounds typically are difficult to predict the absorption wavelengths, as the additive effect of multiple substituents is hard to predict. However, there are some reliable empirical rules regarding aromatic compounds that contain a carbonyl group conjugated with the ring, and can predict the appearance of various isomers of the same compound. These values allow you to predict the wavelength of the primary absorption to within about + 5 nm. We take a generic structure of a benzoyl derivative as the parent compound. 

If R is H, then the parent compound absorbs at 250 nm. 

If R is OH or OR, then the parent absorbs at 230 nm. 

Increments for each additional substitutent: 

Substituent Location Add: 

alkyl or ring o, m 3 

p 10 

OH or OR where o,m 7 

R is any alkyl group p 25 

Cl o,m 0 

p 10 

Br o,m 2 

p 15 

NH₂ o,m 13 

p 58 

NH-(C=O) CH₃ o, m 20 

p 45 

NHCH₃ p 73 

N(CH₃)₃ o,m 20 

p 85 

Example: Predict the wavelength at which 2,3-dichlorobenzoic acid would absorb UV-VIS light. 

Answer: 

The parent, benzoic acid (that is, R is OH) absorbs at 230 nm. The chlorine on the 2 position (ortho) contributes nothing to the absorbance, the chlorine on the 3 position (meta to the carboxylic acid group) also contributes nothing. Prediction then is that 2,3-dichlorobenzoic acid absorbs at 230 nm. 

Answer: The parent, acetophenone, absorbs at 246 nm. An ortho or meta amino groups shifts the absorbance to 13 + 246 = 259 nm; a para amino group shifts the absorbance to 58 + 246 = 304 nm. Apparently, the unknown is the para isomer. 

Example: How to assign a peak to the type of transition that causes the absorption? 

Just looking at a spectrum, how can you assign an absorption? First, s > s* or n > s* of saturated molecules are usually at such high energies (low wavelengths, below 200) that they are often unobserved; the same is true for the p > p* transitions of simple alkenes that are not conjugated. 

Unsaturated molecules with oxygen or nitrogen undergo n > p* are often observed, such as in carbonyl compounds. However, these are forbidden, thus low intensity. 

Conjugated dienes show absorption in the 214 to 280 nm range as shown in the examples on the following page. 

Here are some examples of data interpretation you are likely to encounter in Chem 36 

1. The concentration of vanillin in an ethanol solution was 2.73 x 10^-6 M. The UV-VIS spectrum of vanillin (Figure 1) shows four significant absorptions were observed, at the following wavelengths and absorptions: 232 nm (0.204), 280 nm (0.172), 312 nm (0.172) and 352 nm (0.125). 

Calculate the molar absorptivity of each absorption, using A = ebc (Note that the annotation labels some "peaks" that are just tiny noise signals in the spectrumnot really peaks!) 

Answer: A = ebc; rearranging, e = A/bc for the absorption at 232 nm: 

e = 0.204 = 7.47 x 10⁴ 

1cm x 2.73 x 10^-6 M 

similarly, the peak at 280 nm has e of 6.30 x 10⁴, the peaks at 312 has an e of 6.30 x 10⁴, and the peak at 352 has an e of 4.58 x 10⁴. 

2. Predict theprimary band of absorption for methyl anthranilate and compare your prediction with the actual value observed in the UV-VIS spectrum Figure 2. 

The UV-VIS spectrum shows three significant peaks, at 230, 248, and 338 nm. 

Answer: The parent molecule is the methyl ester of benzoic acid, which absorbs at 230 nm. There is also an NH₂ group on the ortho position, so an increment of 13 nm is added to give 243 nm. There is an absorption at 248, so the prediction is pretty close. Notice that this is not the lmax value; aromatics show at least two bands of absorption; the calculated predicted absorption is the shorter wavelength absorption of the two. 

Answer: p to p*. The n to p* has a lower e value, very weak, often not observed.