有机化学（英文原版）：Chapt13

487 CHAPTER 13 SPECTROSCOPY U ntil the second half of the twentieth century, the structure of a substance—a newly discovered natural product, for example—was determined using information obtained from chemical reactions. This information included the identification of functional groups by chemical tests, along with the results of experiments in which the substance was broken down into smaller, more readily identifiable fragments. Typical of this approach is the demonstration of the presence of a double bond in an alkene by cat- alytic hydrogenation and subsequent determination of its location by ozonolysis. After considering all the available chemical evidence, the chemist proposed a candidate struc- ture (or structures) consistent with the observations. Proof of structure was provided either by converting the substance to some already known compound or by an indepen- dent synthesis. Qualitative tests and chemical degradation have been supplemented and to a large degree replaced by instrumental methods of structure determination. The most prominent methods and the structural clues they provide are: ? Nuclear magnetic resonance (NMR) spectroscopy tells us about the carbon skeleton and the environments of the hydrogens attached to it. ? Infrared (IR) spectroscopy reveals the presence or absence of key functional groups. ? Ultraviolet-visible (UV-VIS) spectroscopy probes the electron distribution, espe- cially in molecules that have conjugated H9266 electron systems. ? Mass spectrometry (MS) gives the molecular weight and formula, both of the molecule itself and various structural units within it. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website As diverse as these techniques are, all of them are based on the absorption of energy by a molecule, and all measure how a molecule responds to that absorption. In describing these techniques our emphasis will be on their application to structure determination. We’ll start with a brief discussion of electromagnetic radiation, which is the source of the energy that a molecule absorbs in NMR, IR, and UV-VIS spectroscopy. 13.1 PRINCIPLES OF MOLECULAR SPECTROSCOPY: ELECTROMAGNETIC RADIATION Electromagnetic radiation, of which visible light is but one example, has the properties of both particles and waves. The particles are called photons, and each possesses an amount of energy referred to as a quantum. In 1900, the German physicist Max Planck proposed that the energy of a photon (E) is directly proportional to its frequency (H9263). E H11005 hv The SI units of frequency are reciprocal seconds (s H110021 ), given the name hertz and the symbol Hz in honor of the nineteenth-century physicist Heinrich R. Hertz. The constant of proportionality h is called Planck’s constant and has the value h H11005 6.63 H11003 10 H1100234 J H11080 s Electromagnetic radiation travels at the speed of light (c H11005 3.0 H11003 10 8 m/s), which is equal to the product of its frequency H9263 and its wavelength H9261: c H11005 vH9261 The range of photon energies is called the electromagnetic spectrum and is shown in Figure 13.1. Visible light occupies a very small region of the electromagnetic spec- trum. It is characterized by wavelengths of 4 H11003 10 H110027 m (violet) to 8 H11003 10 H110027 m (red). 488 CHAPTER THIRTEEN Spectroscopy 10 0 Infrared Ultra- violet 10 –2 10 2 10 4 10 6 10 8 10 10 10 12 10 20 10 18 10 8 10 6 10 4 Frequency (s –1 ) Wavelength (nm) 400 500 600 750 nm X-ray Microwave Radio frequency Gamma ray Ultra- violet 10 16 V isible Infrared Visible region 10 12 10 14 10 10 Highest energy Lowest energy 700 “Modern” physics dates from Planck’s proposal that en- ergy is quantized, which set the stage for the develop- ment of quantum mechanics. Planck received the 1918 No- bel Prize in physics. FIGURE 13.1 The electromagnetic spectrum. (From M. Silberberg, Chemistry, 2d edition, WCB/McGraw-Hill, 2000, p. 260.) Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website When examining Figure 13.1 be sure to keep the following two relationships in mind: 1. Frequency is inversely proportional to wavelength; the greater the frequency, the shorter the wavelength. 2. Energy is directly proportional to frequency; electromagnetic radiation of higher frequency possesses more energy than radiation of lower frequency. Depending on its source, a photon can have a vast amount of energy; gamma rays and X-rays are streams of very high energy photons. Radio waves are of relatively low energy. Ultraviolet radiation is of higher energy than the violet end of visible light. Infrared radiation is of lower energy than the red end of visible light. When a molecule is exposed to electromagnetic radiation, it may absorb a photon, increasing its energy by an amount equal to the energy of the photon. Molecules are highly selective with respect to the frequencies that they absorb. Only photons of certain specific frequencies are absorbed by a molecule. The particular photon energies absorbed by a molecule depend on molecular structure and can be measured with instruments called spectrometers. The data obtained are very sensitive indicators of molecular structure and have revolution- ized the practice of chemical analysis. 13.2 PRINCIPLES OF MOLECULAR SPECTROSCOPY: QUANTIZED ENERGY STATES What determines whether or not a photon is absorbed by a molecule? The most impor- tant requirement is that the energy of the photon must equal the energy difference between two states, such as two nuclear spin states, two vibrational states, or two elec- tronic states. In physics, the term for this is resonance—the transfer of energy between two objects that occurs when their frequencies are matched. In molecular spectroscopy, we are concerned with the transfer of energy from a photon to a molecule, but the idea is the same. Consider, for example, two energy states of a molecule designated E 1 and E 2 in Figure 13.2. The energy difference between them is E 2 H11002 E 1 , or H9004E. In nuclear magnetic resonance (NMR) spectroscopy these are two different spin states of an atomic nucleus; in infrared (IR) spectroscopy, they are two different vibrational energy states; in ultraviolet-visible (UV-VIS) spectroscopy, they are two different electronic energy states. Unlike kinetic energy, which is continuous, meaning that all values of kinetic energy are available to a molecule, only certain energies are possible for electronic, vibra- tional, and nuclear spin states. These energy states are said to be quantized. More of the molecules exist in the lower energy state E 1 than in the higher energy state E 2 . Exci- tation of a molecule from a lower state to a higher one requires the addition of an incre- ment of energy equal to H9004E. Thus, when electromagnetic radiation is incident upon a molecule, only the frequency whose corresponding energy equals H9004E is absorbed. All other frequencies are transmitted. Spectrometers are designed to measure the absorption of electromagnetic radiation by a sample. Basically, a spectrometer consists of a source of radiation, a compartment containing the sample through which the radiation passes, and a detector. The frequency of radiation is continuously varied, and its intensity at the detector is compared with that at the source. When the frequency is reached at which the sample absorbs radiation, the detector senses a decrease in intensity. The relation between frequency and absorption is plotted on a strip chart and is called a spectrum. A spectrum consists of a series of peaks at particular frequencies; its interpretation can provide structural information. Each type of spectroscopy developed independently of the others, and so the format followed in presenting the data is different for each one. An NMR spectrum looks different from an IR spectrum, and both look different from a UV-VIS spectrum. 13.2 Principles of Molecular Spectroscopy: Quantized Energy States 489 E 2 E 1 H9004E H11005 E 2 H11002 E 1 H11005 h FIGURE 13.2 Two energy states of a molecule. Absorp- tion of energy equal to E 2 H11002 E 1 excites a molecule from its lower energy state to the next higher state. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website With this as background, we will now discuss spectroscopic techniques individu- ally. NMR, IR, and UV-VIS spectroscopy provide complementary information, and all are useful. Among them, NMR provides the information that is most directly related to molecular structure and is the one we shall examine first. 13.3 INTRODUCTION TO 1 H NMR SPECTROSCOPY Nuclear magnetic resonance spectroscopy depends on the absorption of energy when the nucleus of an atom is excited from its lowest energy spin state to the next higher one. We should first point out that many elements are difficult to study by NMR, and some can’t be studied at all. Fortunately though, the two elements that are the most common in organic molecules (carbon and hydrogen) have isotopes ( 1 H and 13 C) capable of giv- ing NMR spectra that are rich in structural information. A proton nuclear magnetic res- onance ( 1 H NMR) spectrum tells us about the environments of the various hydrogens in a molecule; a carbon-13 nuclear magnetic resonance ( 13 C NMR) spectrum does the same for the carbon atoms. Separately and together 1 H and 13 C NMR take us a long way toward determining a substance’s molecular structure. We’ll develop most of the general principles of NMR by discussing 1 H NMR, then extend them to 13 C NMR. The 13 C NMR discussion is shorter, not because it is less important than 1 H NMR, but because many of the same principles apply to both techniques. Like an electron, a proton has two spin states with quantum numbers of H11001 and H11002 . There is no difference in energy between these two nuclear spin states; a proton is just as likely to have a spin of H11001 as H11002 . Absorption of electromagnetic radiation can only occur when the two spin states have different energies. A way to make them different is to place the sample in a magnetic field. A proton behaves like a tiny bar mag- net and has a magnetic moment associated with it (Figure 13.3). In the presence of an external magnetic field H5108 0 , the state in which the magnetic moment of the nucleus is aligned with H5108 0 is lower in energy than the one in which it opposes H5108 0 . 1 2 1 2 1 2 1 2 490 CHAPTER THIRTEEN Spectroscopy H11545 H11545 H11545 H11545 H11545 H11545 H11545 H11545 H11545 H11545 (a) No external magnetic field (b) Apply external magnetic field H5108 0 H5108 0 FIGURE 13.3 (a) In the absence of an external magnetic field, the nuclear spins of the protons are randomly oriented. (b) In the presence of an external magnetic field H5108 0 , the nuclear spins are oriented so that the resulting nuclear magnetic moments are aligned either parallel or antiparallel to H5108 0 . The lower energy orientation is the one parallel to H5108 0 and there are more nuclei that have this orientation. Nuclear magnetic resonance of protons was first detected in 1946 by Edward Purcell (Harvard) and by Felix Bloch (Stanford). Purcell and Bloch shared the 1952 Nobel Prize in physics. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website As shown in Figure 13.4, the energy difference between the two states is directly proportional to the strength of the applied field. Net absorption of electromagnetic radi- ation requires that the lower state be more highly populated than the higher one, and quite strong magnetic fields are required to achieve the separation necessary to give a detectable signal. A magnetic field of 4.7 T, which is about 100,000 times stronger than earth’s magnetic field, for example, separates the two spin states of 1 H by only 8 H11003 10 H110025 kJ/mol (1.9 H11003 10 H110025 kcal/mol). From Planck’s equation H9004E H11005 hH9263, this energy gap cor- responds to radiation having a frequency of 2 H11003 10 8 Hz (200 MHz) which lies in the radio frequency (rf) region of the electromagnetic spectrum (see Figure 13.1). PROBLEM 13.1 Most of the NMR spectra in this text were recorded on a spec- trometer having a field strength of 4.7 T (200 MHz for 1 H). The first generation of widely used NMR spectrometers were 60-MHz instruments. What was the mag- netic field strength of these earlier spectrometers? The response of an atom to the strength of the external magnetic field is different for different elements, and for different isotopes of the same element. The resonance fre- quencies of most nuclei are sufficiently different that an NMR experiment is sensitive only to a particular isotope of a single element. The frequency for 1 H is 200 MHz at 4.7 T, but that of 13 C is 50.4 MHz. Thus, when recording the NMR spectrum of an organic compound, we see signals only for 1 H or 13 C, but not both; 1 H and 13 C NMR spectra are recorded in separate experiments with different instrument settings. PROBLEM 13.2 What will be the 13 C frequency setting of an NMR spectrome- ter that operates at 100 MHz for protons? The essential features of an NMR spectrometer, shown in Figure 13.5, are not hard to understand. They consist of a magnet to align the nuclear spins, a radiofrequency (rf) transmitter as a source of energy to excite a nucleus from its lowest energy state to the next higher one, a receiver to detect the absorption of rf radiation, and a recorder to print out the spectrum. Frequency of electromagnetic radiation (s H110021 or Hz) Magnetic field (T) Energy difference between nuclear spin states (kJ/mol or kcal/mol) is proportional to is proportional to 13.3 Introduction to 1 H NMR Spectroscopy 491 H5108 0 H5108 0 ' E 1 E 1 ' E 2 E 2 ' No energy difference in nuclear spin states in absence of external magnetic field Nuclear magnetic moment antiparallel to H5108 0 Nuclear magnetic moment parallel to H5108 0 Increasing strength of external magnetic field ?E'?E FIGURE 13.4 An external magnetic field causes the two nuclear spin states to have different energies. The difference in energy H9004E is proportional to the strength of the applied field. The Sl unit for magnetic field strength is the tesla (T), named after Nikola Tesla, a contemporary of Thomas Edison and who, like Edison, was an inventor of electrical devices. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website It turns out though that there are several possible variations on this general theme. We could, for example, keep the magnetic field constant and continuously vary the radiofrequency until it matched the energy difference between the nuclear spin states. Or, we could keep the rf constant and adjust the energy levels by varying the magnetic field strength. Both methods work, and the instruments based on them are called continuous wave (CW) spectrometers. Many of the terms we use in NMR spectroscopy have their origin in the way CW instruments operate, but CW instruments are rarely used anymore. CW-NMR spectrometers have been replaced by a new generation of instruments called pulsed Fourier-transform nuclear magnetic resonance (FT-NMR) spectrometers. FT-NMR spectrometers are far more versatile than CW instruments and are more com- plicated. Most of the visible differences between them lie in computerized data acquisi- tion and analysis components that are fundamental to FT-NMR spectroscopy. But there is an important difference in how a pulsed FT-NMR experiment is carried out as well. Rather than sweeping through a range of frequencies (or magnetic field strengths), the sample is irradiated with a short, intense burst of radiofrequency radiation (the pulse) that excites all of the protons in the molecule. The magnetic field associated with the new orientation of nuclear spins induces an electrical signal in the receiver that decreases with time as the nuclei return to their original orientation. The resulting free-induction decay (FID) is a composite of the decay patterns of all of the protons in the molecule. The free-induction decay pattern is stored in a computer and converted into a spectrum by a mathematical process known as a Fourier transform. The pulse-relaxation sequence takes only about a second, but usually gives signals too weak to distinguish from back- ground noise. The signal-to-noise ratio is enhanced by repeating the sequence many times, then averaging the data. Noise is random and averaging causes it to vanish; sig- nals always appear at the same place and accumulate. All of the operations—the inter- val between pulses, collecting, storing, and averaging the data and converting it to a spectrum by a Fourier transform—are under computer control, which makes the actual taking of an FT-NMR spectrum a fairly routine operation. 492 CHAPTER THIRTEEN Spectroscopy Magnet rf input oscillator rf output signal amplifier NMR Spectrum rf output receiver rf input coil Sample tube H5108 0 FIGURE 13.5 Diagram of a nuclear magnetic resonance spectrometer. (From S. H. Pine, J. B. Hendrickson, D. J. Cram, and G. S. Hammond, Organic Chemistry, 4th edition, McGraw-Hill, New York, 1980, p. 136.) Richard R. Ernst of the Swiss Federal Institute of Technol- ogy won the 1991 Nobel Prize in chemistry for devis- ing pulse-relaxation NMR techniques. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Not only is pulsed FT-NMR the best method for obtaining proton spectra, it is the only practical method for many other nuclei, including 13 C. It also makes possible a large number of sophisticated techniques that have revolutionized NMR spectroscopy. 13.4 NUCLEAR SHIELDING AND 1 H CHEMICAL SHIFTS Our discussion so far has concerned 1 H nuclei in general without regard for the envi- ronments of individual protons in a molecule. Protons in a molecule are connected to other atoms—carbon, oxygen, nitrogen, and so on—by covalent bonds. The electrons in these bonds, indeed all the electrons in a molecule, affect the magnetic environment of the protons. Alone, a proton would feel the full strength of the external field, but a pro- ton in an organic molecule responds to both the external field plus any local fields within the molecule. An external magnetic field affects the motion of the electrons in a mole- cule, inducing local fields characterized by lines of force that circulate in the opposite direction from the applied field (Figure 13.6). Thus, the net field felt by a proton in a molecule will always be less than the applied field, and the proton is said to be shielded. All of the protons of a molecule are shielded from the applied field by the electrons, but some are less shielded than others. Sometimes the term “deshielded,” is used to describe this decreased shielding of one proton relative to another. The more shielded a proton is, the greater must be the strength of the applied field in order to achieve resonance and produce a signal. A more shielded proton absorbs rf radiation at higher field strength (upfield) compared with one at lower field strength (downfield). Different protons give signals at different field strengths. The dependence of the resonance position of a nucleus that results from its molecular environment is called its chemical shift. This is where the real power of NMR lies. The chemical shifts of various protons in a molecule can be different and are characteristic of particular struc- tural features. Figure 13.7 shows the 1 H NMR spectrum of chloroform (CHCl 3 ) to illustrate how the terminology just developed applies to a real spectrum. Instead of measuring chemical shifts in absolute terms, we measure them with respect to a standard—tetramethylsilane (CH 3 ) 4 Si, abbreviated TMS. The protons of TMS are more shielded than those of most organic compounds, so all of the signals in a sam- ple ordinarily appear at lower field than those of the TMS reference. When measured using a 100-MHz instrument, the signal for the proton in chloroform (CHCl 3 ), for exam- ple, appears 728 Hz downfield from the TMS signal. But since frequency is proportional to magnetic field strength, the same signal would appear 1456 Hz downfield from TMS on a 200-MHz instrument. We simplify the reporting of chemical shifts by converting them to parts per million (ppm) from TMS, which is assigned a value of 0. The TMS need not actually be present in the sample, nor even appear in the spectrum in order to serve as a reference. Chemical shift (H9254) H11005H1100310 6 Thus, the chemical shift for the proton in chloroform is: H9254H11005 H1100310 6 H11005 7.28 ppm When chemical shifts are reported this way, they are identified by the symbol H9254 and are independent of the field strength. 1456 Hz H11002 0 Hz 200 H11003 10 6 Hz position of signal H11002 position of TMS peak spectrometer frequency 13.4 Nuclear Shielding and 1 H Chemical Shifts 493 H5108 0 C H FIGURE 13.6 The induced magnetic field of the elec- trons in the carbon–hydro- gen bond opposes the exter- nal magnetic field. The resulting magnetic field experienced by the proton and the carbon is slightly less than H5108 0 . The graphic that begins this chapter is an electrostatic potential map of tetramethylsi- lane. Learning By Modeling con- tains models of (CH 3 ) 4 Si and (CH 3 ) 4 C in which the greater electron density at the carbons and hydrogens of TMS is appar- ent both in the electrostatic po- tential and in the calculated atomic charges. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website PROBLEM 13.3 The 1 H NMR signal for bromoform (CHBr 3 ) appears at 2065 Hz when recorded on a 300-MHz NMR spectrometer. (a) What is the chemical shift of this proton? (b) Is the proton in CHBr 3 more shielded or less shielded than the proton in CHCl 3 ? NMR spectra are usually run in solution and, although chloroform is a good sol- vent for most organic compounds, it’s rarely used because its own signal at H9254 7.28 ppm would be so intense that it would obscure signals in the sample. Because the magnetic properties of deuterium (D H11005 2 H) are different from those of 1 H, CDCl 3 gives no sig- nals at all in an 1 H NMR spectrum and is used instead. Indeed, CDCl 3 is the most com- monly used solvent in 1 H NMR spectroscopy. Likewise, D 2 O is used instead of H 2 O for water-soluble substances such as carbohydrates. 13.5 EFFECTS OF MOLECULAR STRUCTURE ON 1 H CHEMICAL SHIFTS Nuclear magnetic resonance spectroscopy is such a powerful tool for structure determi- nation because protons in different environments experience different degrees of shield- ing and have different chemical shifts. In compounds of the type CH 3 X, for example, the shielding of the methyl protons increases as X becomes less electronegative. Inas- 494 CHAPTER THIRTEEN Spectroscopy Problem 13.3 in the preced- ing section was based on the chemical shift difference be- tween the proton in CHCl 3 and the proton in CHBr 3 and its relation to shielding. 0.01.02.03.04.05.06.07.08.09.010.0 Chemical shift (δ, ppm) Tetramethylsilane (TMS) δ 0 ppm H±CCl 3 δ 7.28 ppm Upfield Increased shielding Downfield Decreased shielding FIGURE 13.7 The 200-MHz 1 H NMR spectrum of chloroform (HCCl 3 ). Chemical shifts are mea- sured along the x-axis in parts per million (ppm) from tetramethylsilane as the reference, which is assigned a value of zero. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website much as the shielding is due to the electrons, it isn’t surprising to find that the chemi- cal shift depends on the degree to which X draws electrons away from the methyl group. A similar trend is seen in the methyl halides, in which the protons in CH 3 F are the least shielded (H9254 4.3 ppm) and those of CH 3 I (H9254 2.2 ppm) are the most. The deshielding effects of electronegative substituents are cumulative, as the chem- ical shifts for various chlorinated derivatives of methane indicate: PROBLEM 13.4 There is a difference of 4.6 ppm in the 1 H chemical shifts of CHCl 3 and CH 3 CCl 3 . What is the chemical shift for the protons in CH 3 CCl 3 ? Explain your reasoning. Vinyl protons in alkenes and aryl protons in arenes are substantially less shielded than protons in alkanes: One reason for the decreased shielding of vinyl and aryl protons is related to the directional properties of the induced magnetic field of the H9266 electrons. As Figure 13.8 shows, the induced magnetic field due to the H9266 electrons is just like that due to elec- trons in H9268 bonds; it opposes the applied magnetic field. However, all magnetic fields close upon themselves, and protons attached to a carbon–carbon double bond or an aro- matic ring lie in a region where the induced field reinforces the applied field, which decreases the shielding of vinyl and aryl protons. A similar, although much smaller, effect of H9266 electron systems is seen in the chem- ical shifts of benzylic and allylic hydrogens. The methyl hydrogens in hexamethylben- zene and in 2,3-dimethyl-2-butene are less shielded than those in ethane. Chemical shift (H9254), ppm: H H HH H H Benzene 7.3 C H H H H C Ethylene 5.3 CH 3 CH 3 Ethane 0.9 Chemical shift (H9254), ppm: CHCl 3 Chloroform (trichloromethane) 7.3 CH 2 Cl 2 Methylene chloride (dichloromethane) 5.3 CH 3 Cl Methyl chloride (chloromethane) 3.1 Increased shielding of methyl protons Decreasing electronegativity of attached atom Chemical shift of methyl protons (H9254), ppm: CH 3 F Methyl fluoride 4.3 CH 3 OCH 3 Dimethyl ether 3.2 (CH 3 ) 3 N Trimethylamine 2.2 CH 3 CH 3 Ethane 0.9 13.5 Effects of Molecular Structure on 1 H Chemical Shifts 495 CC H H H H H H H H HH H5108 0 (a) (b) H5108 0 FIGURE 13.8 The induced magnetic field of the H9266 elec- trons of (a) an alkene and (b) an arene reinforces the applied fields in the regions where vinyl and aryl protons are located. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Table 13.1 collects chemical-shift information for protons of various types. Within each type, methyl (CH 3 ) protons are more shielded than methylene (CH 2 ) protons, and methylene protons are more shielded than methine (CH) protons. These differences are small—only about 0.7 ppm separates a methyl proton from a methine proton of the same type. Overall, proton chemical shifts among common organic compounds encompass a range of about 12 ppm. The protons in alkanes are the most shielded, and O±H pro- tons of carboxylic acids are the least shielded. Chemical shift (H9254), ppm: CH 3 CH 3 CH 3 H 3 C H 3 C H 3 C Hexamethylbenzene 2.2 C CH 3 CH 3 H 3 C H 3 C C 2,3-Dimethyl-2-butene 1.7 496 CHAPTER THIRTEEN Spectroscopy TABLE 13.1 Chemical Shifts of Representative Types of Protons *Approximate values relative to tetramethylsilane; other groups within the molecule can cause a proton signal to appear outside of the range cited. ? The chemical shifts of protons bonded to nitrogen and oxygen are temperature- and concentration- dependent. Type of proton H±C±R W W H±C±C?C W W H±C±CPN W W H±CPC± H±C±C± W W O X H±C± O X H±C±Ar W W H±Ar H±C?C ± ± W Type of proton H±C±NR W W H±C±Cl W W H±C±Br W W H±C±O W W H±OC± O X H±NR H±OAr H±OR 0.9–1.8 Chemical shift (H9254), ppm* 1.6–2.6 2.1–2.5 2.1–3 2.5 2.3–2.8 4.5–6.5 6.5–8.5 9–10 2.2–2.9 Chemical shift (H9254), ppm* 3.1–4.1 2.7–4.1 3.3–3.7 1–3 ? 0.5–5 ? 6–8 ? 10–13 ? Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website The ability of an NMR spectrometer to separate signals that have similar chemi- cal shifts is termed its resolving power and is directly related to the magnetic field strength of the instrument. Two closely spaced signals at 60 MHz become well separated if a 300-MHz instrument is used. (Remember, though, that the chemical shift H9254, cited in parts per million, is independent of the field strength.) 13.6 INTERPRETING PROTON NMR SPECTRA Analyzing an NMR spectrum in terms of a unique molecular structure begins with the information contained in Table 13.1. By knowing the chemical shifts characteristic of various proton environments, the presence of a particular structural unit in an unknown compound may be inferred. An NMR spectrum also provides other useful information, including: 1. The number of signals, which tells us how many different kinds of protons there are. 2. The intensity of the signals as measured by the area under each peak, which tells us the relative ratios of the different kinds of protons. 3. The multiplicity, or splitting, of each signal, which tells us how many protons are vicinal to the one giving the signal. Protons that have different chemical shifts are said to be chemical-shift-non- equivalent (or chemically nonequivalent). A separate NMR signal is given for each chemical-shift-nonequivalent proton in a substance. Figure 13.9 shows the 200-MHz 1 H NMR spectrum of methoxyacetonitrile (CH 3 OCH 2 CN), a molecule with protons in two different environments. The three protons in the CH 3 O group constitute one set, the two 13.6 Interpreting Proton NMR Spectra 497 0.01.02.03.04.0 Chemical shift (δ, ppm) 6.07.08.09.010.0 5.0 3H 2H CH 3 CH 2 NPCCH 2 OCH 3 FIGURE 13.9 The 200-MHz 1 H NMR spectrum of methoxyacetonitrile (CH 3 OCH 2 CN). Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website protons in the OCH 2 CN group the other. These two sets of protons give rise to the two peaks that we see in the NMR spectrum and can be assigned on the basis of their chem- ical shifts. The protons in the OCH 2 CN group are connected to a carbon that bears two electronegative substituents (O and CPN) and are less shielded than those of the CH 3 O group, which are attached to a carbon that bears only one electronegative atom (O). The signal for the protons in the OCH 2 CN group appears at H9254 4.1 ppm; the signal corre- sponding to the CH 3 O protons is at H9254 3.3 ppm. Another way to assign the peaks is by comparing their intensities. The three equiv- alent protons of the CH 3 O group give rise to a more intense peak than the two equiva- lent protons of the OCH 2 CN group. This is clear by simply comparing the heights of the peaks in the spectrum. It is better, though, to compare peak areas by a process called integration. This is done electronically at the time the NMR spectrum is recorded, and the integrated areas are displayed on the computer screen or printed out. Peak areas are proportional to the number of equivalent protons responsible for that signal. It is important to remember that integration of peak areas gives relative, not absolute, proton counts. Thus, a 3:2 ratio of areas can, as in the case of CH 3 OCH 2 CN, correspond to a 3:2 ratio of protons. But in some other compound a 3:2 ratio of areas might correspond to a 6:4 or 9:6 ratio of protons. PROBLEM 13.5 The 200-MHz 1 H NMR spectrum of 1,4-dimethylbenzene looks exactly like that of CH 3 OCH 2 CN except the chemical shifts of the two peaks are H9254 2.2 ppm and H9254 7.0 ppm. Assign the peaks to the appropriate protons of 1,4- dimethylbenzene. Protons are equivalent to one another and have the same chemical shift when they are in equivalent environments. Often it is an easy matter to decide, simply by inspec- tion, when protons are equivalent or not. In more difficult cases, mentally replacing a proton in a molecule by a “test group” can help. We’ll illustrate the procedure for a sim- ple case—the protons of propane. To see if they have the same chemical shift, replace one of the methyl protons at C-1 by chlorine, then do the same thing for a proton at C-3. Both replacements give the same molecule, 1-chloropropane. Therefore the methyl protons at C-1 are equivalent to those at C-3. If the two structures produced by mental replacement of two different hydrogens in a molecule by a test group are the same, the hydrogens are chemically equivalent. Thus, the six methyl protons of propane are all chemically equivalent to one another and have the same chemical shift. Replacement of either one of the methylene protons of propane generates 2-chloro- propane. Both methylene protons are equivalent. Neither of them is equivalent to any of the methyl protons. The 1 H NMR spectrum of propane contains two signals: one for the six equiva- lent methyl protons, the other for the pair of equivalent methylene protons. PROBLEM 13.6 How many signals would you expect to find in the 1 H NMR spec- trum of each of the following compounds? (a) 1-Bromobutane (c) Butane (b) 1-Butanol (d) 1,4-Dibromobutane CH 3 CH 2 CH 3 Propane ClCH 2 CH 2 CH 3 1-Chloropropane CH 3 CH 2 CH 2 Cl 1-Chloropropane 498 CHAPTER THIRTEEN Spectroscopy Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (e) 2,2-Dibromobutane (g) 1,1,4-Tribromobutane (f) 2,2,3,3-Tetrabromobutane (h) 1,1,1-Tribromobutane SAMPLE SOLUTION (a) To test for chemical-shift equivalence, replace the pro- tons at C-1, C-2, C-3, and C-4 of 1-bromobutane by some test group such as chlo- rine. Four constitutional isomers result: Thus, separate signals will be seen for the protons at C-1, C-2, C-3, and C-4. Bar- ring any accidental overlap, we expect to find four signals in the NMR spectrum of 1-bromobutane. Chemical-shift nonequivalence can occur when two environments are stereochem- ically different. The two vinyl protons of 2-bromopropene have different chemical shifts. One of the vinyl protons is cis to bromine; the other trans. Replacing one of the vinyl protons by some test group, say, chlorine, gives the Z isomer of 2-bromo-1-chloro- propene; replacing the other gives the E stereoisomer. The E and Z forms of 2-bromo- 1-chloropropene are stereoisomers that are not enantiomers; they are diastereomers. Pro- tons that yield diastereomers on being replaced by some test group are described as diastereotopic. The vinyl protons of 2-bromopropene are diastereotopic. Diastereotopic protons can have different chemical shifts. Because their environments are similar, how- ever, this difference in chemical shift is usually small, and it sometimes happens that two diastereotopic protons accidentally have the same chemical shift. Recording the spec- trum on a higher field NMR spectrometer is often helpful in resolving signals with sim- ilar chemical shifts. PROBLEM 13.7 How many signals would you expect to find in the 1 H NMR spec- trum of each of the following compounds? (a) Vinyl bromide (d) trans-1,2-Dibromoethene (b) 1,1-Dibromoethene (e) Allyl bromide (c) cis-1,2-Dibromoethene (f) 2-Methyl-2-butene SAMPLE SOLUTION (a) Each proton of vinyl bromide is unique and has a chem- ical shift different from the other two. The least shielded proton is attached to the carbon that bears the bromine. The pair of protons at C-2 are diastereotopic with respect to each other; one is cis to bromine while the other is trans to bromine. There are three proton signals in the NMR spectrum of vinyl bromide. Their observed chemical shifts are as indicated. H9254 5.7 ppm H9254 5.8 ppmH9254 6.4 ppm C Br H HH C H9254 5.3 ppm H9254 5.5 ppm C Br H H 3 CH C 2-Bromopropene CH 3 CH 2 CH 2 CHBr W Cl 1-Bromo-1- chlorobutane CH 3 CH 2 CHCH 2 Br W Cl 1-Bromo-2- chlorobutane CH 3 CHCH 2 CH 2 Br W Cl 1-Bromo-3- chlorobutane ClCH 2 CH 2 CH 2 CH 2 Br 1-Bromo-4- chlorobutane 13.6 Interpreting Proton NMR Spectra 499 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website When enantiomers are generated by replacing first one proton and then another by a test group, the pair of protons are enantiotopic with respect to one another. The meth- ylene protons at C-2 of 1-propanol, for example, are enantiotopic. Replacing one of these protons by chlorine as a test group gives (R)-2-chloro-1-propanol; replacing the other gives (S)-2-chloro-1-propanol. Enantiotopic protons have the same chemical shift, regardless of the field strength of the NMR spectrometer. At the beginning of this section we noted that an NMR spectrum provides struc- tural information based on chemical shift, the number of peaks, their relative areas, and the multiplicity, or splitting, of the peaks. We have discussed the first three of these fea- tures of 1 H NMR spectroscopy. Let’s now turn our attention to peak splitting to see what kind of information it offers. 13.7 SPIN–SPIN SPLITTING IN NMR SPECTROSCOPY The 1 H NMR spectrum of CH 3 OCH 2 CN (see Figure 13.9) discussed in the preceding section is relatively simple because both signals are singlets; that is, each one consists of a single peak. It is quite common though to see a signal for a particular proton appear not as a singlet, but as a collection of peaks. The signal may be split into two peaks (a doublet), three peaks (a triplet), four peaks (a quartet), or even more. Figure 13.10 shows the 1 H NMR spectrum of 1,1-dichloroethane (CH 3 CHCl 2 ), which is characterized by a doublet centered at H9254 2.1 ppm for the methyl protons and a quartet at H9254 5.9 ppm for the methine proton. The number of peaks into which the signal for a particular proton is split is called its multiplicity. For simple cases the rule that allows us to predict splitting in 1 H NMR spectroscopy is Multiplicity of signal for H a H11005 n H11001 1 where n is equal to the number of equivalent protons that are vicinal to H a . Two pro- tons are vicinal to each other when they are bonded to adjacent atoms. Protons vicinal to H a are separated from H a by three bonds. The three methyl protons of 1,1- dichloroethane are vicinal to the methine proton and split its signal into a quartet. The single methine proton, in turn, splits the methyl protons’ signal into a doublet. The physical basis for peak splitting in 1,1-dichloroethane can be explained with the aid of Figure 13.11, which examines how the chemical shift of the methyl protons is affected by the spin of the methine proton. There are two magnetic environments for the methyl protons: one in which the magnetic moment of the methine proton is paral- lel to the applied field, and the other in which it is antiparallel to it. When the magnetic This proton splits the signal for the methyl protons into a doublet. These three protons split the signal for the methine proton into a quartet. Cl H CH 3 C Cl Enantiotopic protons C CH 3 CH 2 OH H H H C CH 2 OH C CH 2 OH 1-Propanol CH 3 Cl H (R)-2-Chloro-1-propanol CH 3 Cl (S)-2-Chloro-1-propanol 500 CHAPTER THIRTEEN Spectroscopy Enantiotopic protons can have different chemical shifts in a chiral solvent. Be- cause the customary solvent (CDCl 3 ) used in NMR mea- surements is achiral, this phenomenon is not observed in routine work. More complicated splitting patterns conform to an ex- tension of the “n H11001 1” rule and will be discussed in Sec- tion 13.11. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website moment of the methine proton is parallel to the applied field, it reinforces it. This decreases the shielding of the methyl protons and causes their signal to appear at slightly lower field strength. Conversely, when the magnetic moment of the methine proton is antiparallel to the applied field, it opposes it and increases the shielding of the methyl protons. Instead of a single peak for the methyl protons, there are two of approximately equal intensity: one at slightly higher field than the “true” chemical shift, the other at slightly lower field. Turning now to the methine proton, its signal is split by the methyl protons into a quartet. The same kind of analysis applies here and is outlined in Figure 13.12. The methine proton “sees” eight different combinations of nuclear spins for the methyl 13.7 Spin–Spin Splitting in NMR Spectroscopy 501 0.01.02.03.04.0 Chemical shift (δ, ppm) 6.07.08.0 6.4 6.0 5.6 2.4 2.0 9.0 5.0 CH 3 Cl 2 CH Cl 2 CHCH 3 FIGURE 13.10 The 200-MHz 1 H NMR spectrum of 1,1-dichloroethane, showing the methine proton as a quartet and the methyl protons as a doublet. The peak multiplicities are seen more clearly in the scale-expanded insets. Cl W W CH 3 H5108 0 H±C±Cl Spin of methine proton reinforces H5108 0 ; a weaker H5108 0 is needed for resonance. Methyl signal appears at lower field. Spin of methine proton shields methyl protons from H5108 0 . Methyl signal appears at higher field. Cl W W CH 3 H±C±Cl FIGURE 13.11 The magnetic moments (blue arrows) of the two possible spin states of the methine proton affect the chemical shift of the methyl protons in 1,1- dichloroethane. When the magnetic moment is parallel to the external field H5108 0 (green arrow), it adds to the external field and a smaller H5108 0 is needed for resonance. When it is antiparallel to the external field, it subtracts from it and shields the methyl protons. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website protons. In one combination, the magnetic moments of all three methyl protons rein- force the applied field. At the other extreme, the magnetic moments of all three methyl protons oppose the applied field. There are three combinations in which the magnetic moments of two methyl protons reinforce the applied field, whereas one opposes it. Finally, there are three combinations in which the magnetic moments of two methyl protons oppose the applied field and one reinforces it. These eight possible combina- tions give rise to four distinct peaks for the methine proton, with a ratio of intensities of 1:3:3:1. We describe the observed splitting of NMR signals as spin–spin splitting and the physical basis for it as spin–spin coupling. It has its origin in the communication of nuclear spin information between nuclei. This information is transmitted by way of the electrons in the bonds that intervene between the nuclei. Its effect is greatest when the number of bonds is small. Vicinal protons are separated by three bonds, and coupling between vicinal protons, as in 1,1-dichloroethane, is called three-bond coupling or vic- inal coupling. Four-bond couplings are weaker and not normally observable. A very important characteristic of spin–spin splitting is that protons that have the same chemical shift do not split each other’s signal. Ethane, for example, shows only a single sharp peak in its NMR spectrum. Even though there is a vicinal relationship between the protons of one methyl group and those of the other, they do not split each other’s signal because they are equivalent. PROBLEM 13.8 Describe the appearance of the 1 H NMR spectrum of each of the following compounds. How many signals would you expect to find, and into how many peaks will each signal be split? (a) 1,2-Dichloroethane (d) 1,2,2-Trichloropropane (b) 1,1,1-Trichloroethane (e) 1,1,1,2-Tetrachloropropane (c) 1,1,2-Trichloroethane SAMPLE SOLUTION (a) All the protons of 1,2-dichloroethane (ClCH 2 CH 2 Cl) are chemically equivalent and have the same chemical shift. Protons that have the same chemical shift do not split each other’s signal, and so the NMR spectrum of 1,2-dichloroethane consists of a single sharp peak. 502 CHAPTER THIRTEEN Spectroscopy There are eight possible combinations of the nuclear spins of the three methyl protons in CH 3 CHCl 2 . These eight combinations cause the signal of the CHCl 2 proton to be split into a quartet, in which the intensities of the peaks are in the ratio 1:3:3:1. 3 J ab 3 J ab 3 J ab FIGURE 13.12 The methyl protons of 1,1-dichloroethane split the signal of the methine pro- ton into a quartet. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Coupling of nuclear spins requires that the nuclei split each other’s signal equally. The separation between the two halves of the methyl doublet in 1,1-dichloroethane is equal to the separation between any two adjacent peaks of the methine quartet. The extent to which two nuclei are coupled is known as the coupling constant J and in simple cases is equal to the separation between adjacent lines of the signal of a particular pro- ton. The three-bond coupling constant 3 J ab in 1,1-dichloroethane has a value of 7 Hz. The size of the coupling constant is independent of the field strength; the separation between adjacent peaks in 1,1-dichloroethane is 7 Hz, irrespective of whether the spec- trum is recorded at 200 MHz or 500 MHz. 13.8 SPLITTING PATTERNS: THE ETHYL GROUP At first glance, splitting may seem to complicate the interpretation of NMR spectra. In fact, it makes structure determination easier because it provides additional information. It tells us how many protons are vicinal to a proton responsible for a particular signal. With practice, we learn to pick out characteristic patterns of peaks, associating them with particular structural types. One of the most common of these patterns is that of the ethyl group, represented in the NMR spectrum of ethyl bromide in Figure 13.13. 13.8 Splitting Patterns: The Ethyl Group 503 0.01.02.03.04.0 Chemical shift (δ, ppm) 6.07.08.09.0 3.6 3.5 3.4 3.3 1.80 1.70 1.60 5.0 CHCl 3 TMS BrCH 2 CH 3 CH 3 CH 2 FIGURE 13.13 The 200-MHz 1 H NMR spectrum of ethyl bromide, showing the characteristic triplet–quartet pattern of an ethyl group. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website In compounds of the type CH 3 CH 2 X, especially where X is an electronegative atom or group, such as bromine in ethyl bromide, the ethyl group appears as a triplet–quartet pattern. The methylene proton signal is split into a quartet by coupling with the methyl protons. The signal for the methyl protons is a triplet because of vicinal coupling to the two protons of the adjacent methylene group. We have discussed in the preceding section why methyl groups split the signals due to vicinal protons into a quartet. Splitting by a methylene group gives a triplet cor- responding to the spin combinations shown in Figure 13.14 for ethyl bromide. The rel- ative intensities of the peaks of this triplet are 1:2:1. PROBLEM 13.9 Describe the appearance of the 1 H NMR spectrum of each of the following compounds. How many signals would you expect to find, and into how many peaks will each signal be split? (a) ClCH 2 OCH 2 CH 3 (b) CH 3 CH 2 OCH 3 (c) CH 3 CH 2 OCH 2 CH 3 (d) p-Diethylbenzene (e) ClCH 2 CH 2 OCH 2 CH 3 SAMPLE SOLUTION (a) Along with the triplet–quartet pattern of the ethyl group, the NMR spectrum of this compound will contain a singlet for the two protons of the chloromethyl group. Table 13.2 summarizes the splitting patterns and peak intensities expected for cou- pling to various numbers of protons. Split into triplet by two protons of adjacent methylene group Split into quartet by three protons of methyl group Singlet; no protons vicinal to these; therefore, no splitting ClCH 2 CH 3 O CH 2 Br CH 3 CH 2 These three protons split the methylene signal into a quartet. These two protons split the methyl signal into a triplet. 504 CHAPTER THIRTEEN Spectroscopy There are four possible combinations of the nuclear spins of the two methylene protons in CH 3 CH 2 Br. 3 J ab 3 J ab These four combinations cause the signal of the CH 3 protons to be split into a triplet, in which the intensities of the peaks are in the ratio 1:2:1. FIGURE 13.14 The methyl- ene protons of ethyl bro- mide split the signal of the methyl protons into a triplet. TABLE 13.2 Splitting Patterns of Common Multiplets Number of equivalent protons to which nucleus is coupled 1 2 3 4 5 6 1:1 1:2:1 1:3:3:1 1:4 :6:4 :1 1:5:10:10:5:1 1:6:15:20:15:6:1 Intensities of lines in multiplet Doublet Triplet Quartet Pentet Sextet Septet Appearance of multiplet The intensities correspond to the coefficients of a binomial expansion (Pascal’s triangle). Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 13.10 Splitting Patterns: Pairs of Doublets 505 0.01.02.03.04.0 Chemical shift (δ, ppm) 6.07.08.09.010.0 5.0 CH CH 3 H W W Cl 4.04.14.24.34.4 H 3 C±C±CH 3 1.41.61.8 FIGURE 13.15 The 200-MHz 1 H NMR spectrum of iso- propyl chloride, showing the doublet–septet pattern of an isopropyl group. 13.9 SPLITTING PATTERNS: THE ISOPROPYL GROUP The NMR spectrum of isopropyl chloride (Figure 13.15) illustrates the appearance of an isopropyl group. The signal for the six equivalent methyl protons at H9254 1.5 ppm is split into a doublet by the proton of the H±C±Cl unit. In turn, the H±C±Cl proton sig- nal at H9254 4.2 ppm is split into a septet by the six methyl protons. A doublet–septet pat- tern is characteristic of an isopropyl group. 13.10 SPLITTING PATTERNS: PAIRS OF DOUBLETS We often see splitting patterns in which the intensities of the individual peaks do not match those given in Table 13.2, but are distorted in that the signals for coupled protons “lean” toward each other. This leaning is a general phenomenon, but is most easily illus- trated for the case of two nonequivalent vicinal protons as shown in Figure 13.16. H 1 ±C±C±H 2 The appearance of the splitting pattern of protons 1 and 2 depends on their coupling con- stant J and the chemical shift difference H9004H9263 between them. When the ratio H9004H9263/J is large, two symmetrical 1:1 doublets are observed. We refer to this as the “AX” case, using two This proton splits the signal for the methyl protons into a doublet. These six protons split the methine signal into a septet. H CH 3 CH 3 C Cl Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website letters that are remote in the alphabet to stand for signals well removed from each other on the spectrum. Keeping the coupling constant the same while reducing H9004H9263 leads to a steady decrease in the intensity of the outer two peaks with a simultaneous increase in the inner two as we progress from AX through AM to AB. At the extreme (A 2 ), the two protons have the same chemical shift, the outermost lines have disappeared, and no split- ting is observed. Because of its appearance, it is easy to misinterpret an AB pattern as a quartet, rather than the pair of skewed doublets it really is. The skewed AB pattern is clearly visible in the 1 H NMR spectrum of 2,3,4- trichloroanisole (Figure 13.17). In addition to the singlet at H9254 3.9 ppm for the protons of the ±OCH 3 group, we see doublets at H9254 6.8 and H9254 7.3 ppm for the two protons of the aromatic ring. Doublet H9254 7.3 ppm Doublet H9254 6.8 ppm Singlet H9254 3.9 ppm OCH 3 Cl Cl Cl HH 2,3,4-Trichloroanisole 506 CHAPTER THIRTEEN Spectroscopy Chemical shift difference much larger than coupling constant AX A 2 AM Same chemical shift; no splitting AB J J JJ J J FIGURE 13.16 The appear- ance of the splitting pattern of two coupled protons depends on their coupling constant J and the chemical shift difference H9004H9263 between them. As the ratio H9004H9263/J de- creases, the doublets be- come increasingly distorted. When the two protons have the same chemical shift, no splitting is observed. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website A similar pattern can occur with geminal protons (protons bonded to the same carbon). Geminal protons are separated by two bonds, and geminal coupling is referred to as two- bond coupling ( 2 J) in the same way that vicinal coupling is referred to as three-bond coupling ( 3 J). An example of geminal coupling is provided by the compound 1-chloro- 1-cyanoethene, in which the two hydrogens appear as a pair of doublets. The splitting in each doublet is 2 Hz. Splitting due to geminal coupling is seen only in CH 2 groups and only when the two protons have different chemical shifts. All three protons of a methyl (CH 3 ) group are equivalent and cannot split one another’s signal, and, of course, there are no protons geminal to a single methine (CH) proton. 13.11 COMPLEX SPLITTING PATTERNS All the cases we’ve discussed so far have involved splitting of a proton signal by cou- pling to other protons that were equivalent to one another. Indeed, we have stated the splitting rule in terms of the multiplicity of a signal as being equal to n H11001 1, where n is equal to the number of equivalent protons to which the proton that gives the signal is coupled. What if all the vicinal protons are not equivalent? Figure 13.18a shows the signal for the proton marked ArCH a ?CH 2 in m-nitrostyrene, which appears as a set of four peaks in the range H9254 6.7–6.9 ppm. These four peaks are in fact a “doublet of doublets.” The proton in question is unequally Doublet Doublet 2 J H11005 2 Hz C H Cl CNH C 1-Chloro-1-cyanoethene 13.11 Complex Splitting Patterns 507 0.01.02.03.04.0 Chemical shift (δ, ppm) 6.07.08.09.010.0 5.0 TMS 7.0 6.67.2 6.87.4 Cl OCH 3 W W W HH ± W ± Cl Cl FIGURE 13.17 The 200-MHz 1 H NMR spectrum of 2,3,4- trichloroanisole, illustrating the splitting of the ring pro- tons into a pair of doublets that “lean” toward each other. The protons in 1-chloro-1- cyanoethene are diaste- reotopic (Section 13.6). They are nonequivalent and have different chemical shifts. Re- member, splitting can only occur between protons that have different chemical shifts. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website coupled to the two protons at the end of the vinyl side chain. The size of the vicinal coupling constant between protons trans to each other on a double bond is normally larger than that between cis protons. In this case the trans coupling constant is 16 Hz and the cis coupling constant is 12 Hz. Thus, as shown in Figure 13.18b, the signal is split into a doublet with a spacing of 16 Hz by one vicinal proton, and each line of this doublet is then split into another doublet with a spacing of 12 Hz. PROBLEM 13.10 In addition to the proton marked H a in m-nitrostyrene in Fig- ure 13.18, there are two other vinylic protons. Assuming that the coupling con- stant between the two geminal protons in ArCH?CH 2 is 2 Hz and the vicinal cou- pling constants are 12 Hz (cis) and 16 Hz (trans), describe the splitting pattern for each of these other two vinylic hydrogens. The “n H11001 1 rule” should be amended to read: When a proton H a is coupled to H b , H c , H d , etc., and J ab HS11005 J ac , HS11005 J ad , etc., the original signal for H a is split into n H11001 1 peaks by n H b protons, each of these lines is further split into n H11001 1 peaks by n H c pro- tons, and each of these into n H11001 1 lines by n H d protons, etc. Bear in mind that because of overlapping peaks, the number of lines actually observed can be less than that expected on the basis of the splitting rule. PROBLEM 13.11 Describe the splitting pattern expected for the proton at (a) C-2 in (Z )-1,3-dichloropropene (b) SAMPLE SOLUTION (a) The signal of the proton at C-2 is split into a doublet by coupling to the proton cis to it on the double bond, and each line of this dou- blet is split into a triplet by the two protons of the CH 2 Cl group. C-2 in CH 3 CHCH O X W Br 508 CHAPTER THIRTEEN Spectroscopy 12 Hz 12 Hz 16 Hz H a (b)(a) O 2 N H a C C H H 6.76.86.9 You will find it revealing to construct a splitting diagram similar to that of Figure 13.18 for the case in which the cis and trans H±C?C±H coupling con- stants are equal. Under those circumstances the four-line pattern simplifies to a triplet, as it should for a proton equally coupled to two vici- nal protons. FIGURE 13.18 Splitting of a signal into a doublet of doublets by unequal cou- pling to two vicinal protons. (a) Appearance of the signal for the proton marked H a in m-nitrostyrene as a set of four peaks. (b) Origin of these four peaks through successive splitting of the signal for H a . Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 13.12 1 H NMR SPECTRA OF ALCOHOLS The hydroxyl proton of a primary alcohol RCH 2 OH is vicinal to two protons, and its sig- nal would be expected to be split into a triplet. Under certain conditions signal splitting of alcohol protons is observed, but usually it is not. Figure 13.19 presents the NMR spec- trum of benzyl alcohol, showing the methylene and hydroxyl protons as singlets at H9254 4.7 and 2.5 ppm, respectively. (The aromatic protons also appear as a singlet, but that is because they all accidentally have the same chemical shift and so cannot split each other.) The reason that splitting of the hydroxyl proton of an alcohol is not observed is that it is involved in rapid exchange reactions with other alcohol molecules. Transfer of a pro- ton from an oxygen of one alcohol molecule to the oxygen of another is quite fast and effectively decouples it from other protons in the molecule. Factors that slow down this exchange of OH protons, such as diluting the solution, lowering the temperature, or increasing the crowding around the OH group, can cause splitting of hydroxyl resonances. The chemical shift of the hydroxyl proton is variable, with a range of H9254 0.5–5 ppm, depending on the solvent, the temperature at which the spectrum is recorded, and the concentration of the solution. The alcohol proton shifts to lower field strength in more concentrated solutions. This proton splits signal for proton at C-2 into a doublet. C H Cl CH 2 Cl H 12 C These protons split signal for proton at C-2 into a triplet. Proton at C-2 appears as a doublet of triplets. 13.12 1 H NMR Spectra of Alcohols 509 0.01.02.03.04.0 Chemical shift (δ, ppm) 6.07.08.09.010.0 5.0 TMS CH 2 OH W FIGURE 13.19 The 200-MHz 1 H NMR spectrum of benzyl alcohol. The hydroxyl proton and the methylene protons are vicinal but do not split each other because of the rapid intermolecular exchange of hydroxyl protons. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website An easy way to verify that a particular signal belongs to a hydroxyl proton is to add D 2 O. The hydroxyl proton is replaced by deuterium according to the equation: Deuterium does not give a signal under the conditions of 1 H NMR spectroscopy. Thus, replacement of a hydroxyl proton by deuterium leads to the disappearance of the OH peak. Protons bonded to nitrogen and sulfur also undergo exchange with D 2 O. Those bound to carbon normally do not, and so this technique is useful for assigning the pro- ton resonances of OH, NH, and SH groups. 13.13 NMR AND CONFORMATIONS We know from Chapter 3 that the protons in cyclohexane exist in two different envi- ronments: axial and equatorial. The NMR spectrum of cyclohexane, however, shows only a single sharp peak at H9254 1.4 ppm. All the protons of cyclohexane appear to be equiva- lent in the NMR spectrum. Why? The answer is related to the very rapid rate of ring flipping in cyclohexane. One property of NMR spectroscopy is that it is too slow a technique to “see” the indi- vidual conformations of cyclohexane. What NMR sees is the average environment of the protons. Since chair–chair interconversion in cyclohexane converts each axial proton to an equatorial one and vice versa, the average environments of all the protons are the same. A single peak is observed that has a chemical shift midway between the true chem- ical shifts of the axial and the equatorial protons. The rate of ring flipping can be slowed down by lowering the temperature. At tem- peratures on the order of H11002100°C, separate signals are seen for the axial and equatorial protons of cyclohexane. 13.14 13 C NMR SPECTROSCOPY We pointed out in Section 13.3 that both 1 H and 13 C are nuclei that can provide useful structural information when studied by NMR. Although a 1 H NMR spectrum helps us infer much about the carbon skeleton of a molecule, a 13 C NMR spectrum has the obvi- ous advantage of probing the carbon skeleton directly. 13 C NMR spectroscopy is analo- gous to 1 H NMR in that the number of signals informs us about the number of differ- ent kinds of carbons, and their chemical shifts are related to particular chemical environments. However, unlike 1 H, which is the most abundant of the hydrogen isotopes (99.985%), only 1.1% of the carbon atoms in a sample are 13 C. Moreover, the intensity of the signal produced by 13 C nuclei is far weaker than the signal produced by the same number of 1 H nuclei. In order for 13 C NMR to be a useful technique in structure deter- mination, a vast increase in the signal-to-noise ratio is required. Pulsed FT-NMR pro- vides for this, and its development was the critical breakthrough that led to 13 C NMR becoming the routine tool that it is today. H x H y H x H y RCH 2 OH RCH 2 ODD 2 O DOHH11001H11001 510 CHAPTER THIRTEEN Spectroscopy Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website To orient ourselves in the information that 13 C NMR provides, let’s compare the 1 H and 13 C NMR spectra of 1-chloropentane (Figures 13.20a and 13.20b, respectively). The 1 H NMR spectrum shows reasonably well defined triplets for the protons of the CH 3 13.14 13 C NMR Spectroscopy 511 0.01.02.03.04.0 Chemical shift (δ, ppm) 6.07.08.09.010.0 5.0 TMS ClCH 2 CH 2 CH 2 CH 2 CH 3 (a) 020406080 Chemical shift (δ, ppm) 120140160180200 100 ClCH 2 CH 2 CH 2 CH 2 CH 3 CDCl 3 (b) FIGURE 13.20 (a) The 200- MHz 1 H NMR spectrum and (b) the 13 C NMR spectrum of 1-chloropentane. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website and CH 2 Cl groups (H9254 0.9 and 3.55 ppm, respectively). The signals for the six CH 2 pro- tons at C-2, C-3, and C-4 of CH 3 CH 2 CH 2 CH 2 CH 2 Cl, however, appear as two unresolved multiplets at H9254 1.4 and 1.8 ppm. The 13 C NMR spectrum, on the other hand, is very simple: a separate, distinct peak is observed for each carbon. Notice, too, how well-separated these 13 C signals are: they cover a range of over 30 ppm, compared with less than 3 ppm for the proton signals of the same compound. In general, the window for proton signals in organic molecules is about 12 ppm; 13 C chemical shifts span a range of over 200 ppm. The greater spread of 13 C chemical shifts makes it easier to interpret the spectra. PROBLEM 13.12 How many signals would you expect to see in the 13 C NMR spectrum of each of the following compounds? (a) Propylbenzene (d) 1,2,4-Trimethylbenzene (b) Isopropylbenzene (e) 1,3,5-Trimethylbenzene (c) 1,2,3-Trimethylbenzene SAMPLE SOLUTION (a) The two ring carbons that are ortho to the propyl sub- stituent are equivalent and so must have the same chemical shift. Similarly, the two ring carbons that are meta to the propyl group are equivalent to each other. The carbon atom para to the substituent is unique, as is the carbon that bears the substituent. Thus, there will be four signals for the ring carbons, designated w, x, y, and z in the structural formula. These four signals for the ring carbons added to those for the three nonequivalent carbons of the propyl group yield a total of seven signals. 13.15 13 C CHEMICAL SHIFTS Just as chemical shifts in 1 H NMR are measured relative to the protons of tetramethyl- silane, chemical shifts in 13 C NMR are measured relative to the carbons of tetra- methylsilane as the zero point of the chemical-shift scale. Table 13.3 lists typical chemical-shift ranges for some representative types of carbon atoms. In general, the factors that most affect 13 C chemical shifts are: 1. The hybridization of carbon 2. The electronegativity of the groups attached to carbon Both can be illustrated by comparing the chemical shifts of the designated carbon in the compounds shown. (The numbers are the chemical shift of the indicated carbon in parts per million.) sp 3 -Hybridized carbons are more shielded than sp 2 as the chemical shifts for C-2 in pen- tane versus 1-pentene and C-1 in 1-butanol versus butanal demonstrate. The effect of substituent electronegativity is evident when comparing pentane with 1-butanol and 23 Pentane 138 1-Pentene 61 OH 1-Butanol 202 O Butanal PropylbenzeneCH 2 CH 2 CH 3 w xy z x y 512 CHAPTER THIRTEEN Spectroscopy Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 1-pentene with butanal. Replacing the methyl group in pentane by the more electroneg- ative oxygen deshields the carbon in 1-butanol. Likewise, replacing C-1 in 1-pentene by oxygen deshields the carbonyl carbon in butanal. PROBLEM 13.13 Consider carbons x, y, and z in p-methylanisole. One has a chemical shift of H9254 20 ppm, another has H9254 55 ppm, and the third H9254 157 ppm. Match the chemical shifts with the appropriate carbons. sp-Hybridized carbons are a special case; they are less shielded than sp 3 but more shielded than sp 2 -hybridized carbons. 13.16 13 C NMR AND PEAK INTENSITIES Two features that are fundamental to 1 H NMR spectroscopy—integrated areas and split- ting patterns—are not very important in 13 C NMR. Although it is a simple matter to integrate 13 C signals, it is rarely done because the observed ratios can be more misleading than helpful. The pulsed FT technique that is standard for 13 C NMR has the side effect of distorting the signal intensities, especially for carbons that lack attached hydrogens. Examine Figure 13.21 which shows the 13 C spectrum of 3-methylphenol (m-cresol). Notice that, contrary to what we might expect for a compound with seven peaks for seven different carbons, the intensities of these peaks are not nearly the same. The two least intense signals, those at H9254 140 and H9254 157 ppm, correspond to carbons that lack attached hydrogens. PROBLEM 13.14 To which of the compounds of Problem 13.12 does the 13 C NMR spectrum of Figure 13.22 belong? OCH 3 H 3 C xzy 13.16 13 C NMR and Peak Intensities 513 TABLE 13.3 Chemical Shifts of Representative Carbons *Approximate values relative to tetramethylsilane. 0–35 15–40 25–50 30–40 65–90 Chemical shift (H9254) ppm* 100–150 110–175 20–40 25–50 35–50 50–65 110–125 Chemical shift (H9254) ppm* 160–185 190–220 Type of carbon R 2 C?CR 2 RCH 3 R 2 CH 2 R 3 CH R 4 C RCPCR Hydrocarbons Type of carbon RCOH O X RCOR O X and RCH O X RCR O X and RCH 2 Br RCH 2 Cl RCH 2 NH 2 RCH 2 OH RCPN RCH 2 ORand Functionally substituted carbons Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 514 CHAPTER THIRTEEN Spectroscopy 020406080100120140160180200 Chemical shift (δ, ppm) CDCl 3 W CH 3 OH W FIGURE 13.21 The 13 C NMR spectrum of m-cresol. Each of the seven carbons of m-cresol gives a separate peak. Integrating the spec- trum would not provide use- ful information because the intensities of the peaks are so different, even though each one corresponds to a single carbon. 020406080 Chemical shift (δ, ppm) 120140160180200 100 CDCl 3 FIGURE 13.22 The 13 C NMR spectrum of the unknown compound of Problem 13.14. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 13.17 13 C ± 1 H COUPLING You may have noticed another characteristic of 13 C NMR spectra—all of the peaks are singlets. With a spin of H11006 , a 13 C nucleus is subject to the same splitting rules that apply to 1 H, and we might expect to see splittings due to 13 C± 13 C and 13 C± 1 H couplings. We don’t. Why? The lack of splitting due to 13 C± 13 C coupling is easy to understand. 13 C NMR spectra are measured on samples that contain 13 C at the “natural abundance” level. Only 1% of all the carbons in the sample are 13 C, and the probability that any molecule con- tains more than one 13 C atom is quite small. Splitting due to 13 C± 1 H coupling is absent for a different reason, one that has to do with the way the spectrum is run. Because a 13 C signal can be split not only by the protons to which it is directly attached, but also by protons separated from it by two, three, or even more bonds, the number of splittings might be so large as to make the spectrum too complicated to interpret. Thus, the spectrum is measured under conditions, called broadband decoupling, that suppress such splitting. In addition to pulsing the sample by a radiofrequency tuned for 13 C, the sample is continuously irradiated by a second rf transmitter that covers the entire frequency range for all the 1 H nuclei. The effect of this second rf is to decouple the 1 H spins from the 13 C spins, which causes all the 13 C signals to collapse to singlets. What we gain from broadband decoupling in terms of a simple-looking spectrum comes at the expense of some useful information. For example, being able to see split- ting corresponding to one-bond 13 C± 1 H coupling would immediately tell us the num- ber of hydrogens directly attached to each carbon. The signal for a carbon with no attached hydrogens (a quaternary carbon) would be a singlet, the hydrogen of a CH group would split the carbon signal into a doublet, and the signals for the carbons of a CH 2 and a CH 3 group would appear as a triplet and a quartet, respectively. Although it is possible, with a technique called off-resonance decoupling, to observe such one-bond couplings, identifying a signal as belonging to a quaternary carbon or to the carbon of a CH, CH 2 , or CH 3 group is normally done by a method called DEPT, which is described in the next section. 13.18 USING DEPT TO COUNT THE HYDROGENS ATTACHED TO 13 C In general, a simple pulse FT-NMR experiment involves the following stages: 1. Equilibration of the nuclei between the lower and higher spin states under the influ- ence of a magnetic field 2. Application of a radiofrequency pulse to give an excess of nuclei in the higher spin state 3. Acquisition of free-induction decay data during the time interval in which the equi- librium distribution of nuclear spins is restored 4. Mathematical manipulation (Fourier transform) of the data to plot a spectrum The pulse sequence (stages 2–3) can be repeated hundreds of times to enhance the signal- to-noise ratio. The duration of time for stage 2 is on the order of milliseconds, and that for stage 3 is about 1 second. Major advances in NMR have been made by using a second rf transmitter to irra- diate the sample at some point during the sequence. There are several such techniques, of which we’ll describe just one, called “distortionless enhancement of polarization transfer,” abbreviated as DEPT. 1 2 13.18 Using DEPT to Count the Hydrogens Attached to 13 C 515 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website In the DEPT routine, a second transmitter excites 1 H, and this affects the appear- ance of the 13 C spectrum. A typical DEPT experiment is illustrated for the case of 1-phenyl-1-pentanone in Figure 13.23. In addition to the normal spectrum shown in Fig- 516 CHAPTER THIRTEEN Spectroscopy 020406080 Chemical shift (δ, ppm) (a) 120140160180200 100 CH C CH CH CH 2 CH 2 CH 3 HH HH CCH 2 CH 2 CH 2 CH 3 H ± ± ± ± ±± CH 2 O X O X C 20406080 Chemical shift (δ, ppm) (b) 120140160180200 100 0 CH CH CH CH 3 HH HH CCH 2 CH 2 CH 2 CH 3 H ± ± ± ± ±± CH 2 CH 2 CH 2 O X FIGURE 13.23 13 C NMR spectra of 1-phenyl-1-pen- tanone. (a) Normal spec- trum. (b) DEPT spectrum recorded using a pulse se- quence in which CH 3 and CH carbons appear as positive peaks, CH 2 carbons as nega- tive peaks, and carbons without any attached hydro- gens are nulled. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website ure 13.23a, four more spectra are run using prescribed pulse sequences. In one (Figure 13.23b), the signals for carbons of CH 3 and CH groups appear normally, whereas those for CH 2 groups are inverted and those for C without any attached hydrogens are nulled. In the others (not shown) different pulse sequences produce combinations of normal, nulled, and inverted peaks that allow assignments to be made to the various types of car- bons with confidence. 13.18 Using DEPT to Count the Hydrogens Attached to 13 C 517 MAGNETIC RESONANCE IMAGING L ike all photographs, a chest X-ray is a two- dimensional projection of a three-dimensional object. It is literally a collection of shadows pro- duced by all the organs that lie between the source of the X-rays and the photographic plate. The clear- est images in a chest X-ray are not the lungs (the cus- tomary reason for taking the X-ray in the first place) but rather the ribs and backbone. It would be desir- able if we could limit X-ray absorption to two dimen- sions at a time rather than three. This is, in fact, what is accomplished by a technique known as computer- ized axial tomography, which yields its information in a form called a CT (or CAT) scan. With the aid of a computer, a CT scanner controls the movement of an X-ray source and detector with respect to the patient and to each other, stores the X-ray absorption pat- tern, and converts it to an image that is equivalent to an X-ray photograph of a thin section of tissue. It is a noninvasive diagnostic method, meaning that surgery is not involved nor are probes inserted into the patient’s body. As useful as the CT scan is, it has some draw- backs. Prolonged exposure to X-rays is harmful, and CT scans often require contrast agents to make cer- tain organs more opaque to X-rays. Some patients are allergic to these contrast agents. An alternative technique was introduced in the 1980s that is not only safer but more versatile than X-ray tomography. This technique is magnetic resonance imaging, or MRI. MRI is an application of nuclear magnetic reso- nance spectroscopy that makes it possible to examine the inside of the human body using radiofrequency radiation, which is lower in energy (see Figure 13.1) and less damaging than X-rays and requires no imag- ing or contrast agents. By all rights MRI should be called NMRI, but the word “nuclear” was dropped from the name so as to avoid confusion with nuclear medicine, which involves radioactive isotopes. Although the technology of an MRI scanner is rather sophisticated, it does what we have seen other NMR spectrometers do; it detects protons. Thus, MRI is especially sensitive to biological materials such as water and lipids that are rich in hydrogen. Figure 13.24 shows an example of the use of MRI to detect a brain tumor. Regions of the image are lighter or darker according to the relative concentration of pro- tons and to their environments. Using MRI as a substitute for X-ray tomography is only the first of what are many medical applica- tions. More lie on the horizon. If, for example, the rate of data acquisition could be increased, then it would become possible to make the leap from the equivalent of still photographs to motion pictures. One could watch the inside of the body as it works— see the heart beat, see the lungs expand and con- tract—rather than merely examine the structure of an organ. FIGURE 13.24 A magnetic resonance image of a section of a brain that has a tumor in the left hemisphere. The image has been computer-enhanced to show the tumor and the surrounding liquid in different shades of red, fatty tissues in green, the normal part of the brain in blue, and the eyeballs in yellow. (Photograph courtesy of Simon Fraser Science Photo Library, Newcastle upon Tyne.) Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 13.19 INFRARED SPECTROSCOPY Before the advent of NMR spectroscopy, infrared (IR) spectroscopy was the instrumen- tal method most often applied to determine the structure of organic compounds. Although NMR spectroscopy, in general, tells us more about the structure of an unknown com- pound, IR still retains an important place in the chemist’s inventory of spectroscopic methods because of its usefulness in identifying the presence of certain functional groups within a molecule. Infrared radiation is the portion of the electromagnetic spectrum (see Figure 13.1) between microwaves and visible light. The fraction of the infrared region of most use for structure determination lies between 2.5 H11003 10 H110026 m and 16 H11003 10 H110026 m in wavelength. Two units commonly employed in infrared spectroscopy are the micrometer and the wave number. One micrometer (H9262m) is 10 H110026 m, and infrared spectra record the region from 2.5 H9262m to 16 H9262m. Wave numbers are reciprocal centimeters (cm H110021 ), so that the region 2.5–16 H9262m corresponds to 4000–625 cm H110021 . An advantage to using wave numbers is that they are directly proportional to energy. Thus, 4000 cm H110021 is the high-energy end of the scale, and 625 cm H110021 is the low-energy end. Electromagnetic radiation in the 4000–625 cm H110021 region corresponds to the sepa- ration between adjacent vibrational energy states in organic molecules. Absorption of a photon of infrared radiation excites a molecule from its lowest, or ground, vibrational state to a higher one. These vibrations include stretching and bending modes of the type illustrated for a methylene group in Figure 13.25. A single molecule can have a large number of distinct vibrations available to it, and infrared spectra of different molecules, like fingerprints, are different. Superposability of their infrared spectra is commonly offered as proof that two compounds are the same. 518 CHAPTER THIRTEEN Spectroscopy Stretching: Symmetric Antisymmetric Bending: In plane In plane Out of plane Out of plane FIGURE 13.25 Stretch- ing and bending vibrations of a methylene unit. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website A typical infrared spectrum, such as that of hexane in Figure 13.26, appears as a series of absorption peaks of varying shape and intensity. Almost all organic compounds exhibit a peak or group of peaks near 3000 cm H110021 due to carbon–hydrogen stretching. The peaks at 1460, 1380, and 725 cm H110021 are due to various bending vibrations. Infrared spectra can be recorded on a sample regardless of its physical state—solid, liquid, gas, or dissolved in some solvent. The spectrum in Figure 13.26 was taken on the neat sample, meaning the pure liquid. A drop or two of hexane was placed between two sodium chloride disks, through which the infrared beam is passed. Solids may be dissolved in a suitable solvent such as carbon tetrachloride or chloroform. More com- monly, though, a solid sample is mixed with potassium bromide and the mixture pressed into a thin wafer, which is placed in the path of the infrared beam. In using infrared spectroscopy for structure determination, peaks in the range 1600–4000 cm H110021 are usually emphasized because this is the region in which the vibra- tions characteristic of particular functional groups are found. The region 1300–625 cm H110021 is known as the fingerprint region; it is here that the pattern of peaks varies most from compound to compound. Table 13.4 lists the frequencies (in wave numbers) associated with a variety of groups commonly found in organic compounds. 13.19 Infrared Spectroscopy 519 Like NMR spectrometers, some IR spectrometers oper- ate in a continuous-sweep mode, whereas others em- ploy pulse Fourier-transform (FT-IR) technology. All the IR spectra in this text were ob- tained on an FT-IR instru- ment. TABLE 13.4 Infrared Absorption Frequencies of Some Common Structural Units Single bonds Double bonds Triple bonds 3200–3600 2500–3600 Frequency, cm H115461 3350–3500 1620–1680 3310–3320 3000–3100 2850–2950 1200 1025–1200 910, 990 890 665–730 960–980 790–840 1710–1750 1700–1725 1800–1850 and 1740–1790 1770–1815 1730–1750 1680–1700 2100–2200 2240–2280 730–770 and 690–710 735–770 750–810 and 680–730 790–840 Structural unit Frequency, cm H115461 Structural unit ±O±H (alcohols) N±H ± ± ±O±H (carboxylic acids) C?C ± ± ± ± C?O ± ± sp C±H sp 2 C±H sp 3 C±H Aldehydes and ketones sp 2 C±O sp 3 C±O Stretching vibrations Carboxylic acids Acid anhydrides Acyl halides Esters Amides ±CPC± ±CPN RCH?CH 2 R 2 C?CH 2 cis-RCH?CHRH11032 trans-RCH?CHRH11032 R 2 C?CHRH11032 Bending vibrations of diagnostic value Alkenes: Monosubstituted Ortho-disubstituted Meta-disubstituted Para-disubstituted Substituted derivatives of benzene: Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website To illustrate how structural features affect infrared spectra, compare the spectrum of hexane (Figure 13.26) with that of 1-hexene (Figure 13.27). The two are quite dif- ferent. In the C±H stretching region of 1-hexene, there is a peak at 3095 cm H110021 , whereas all the C±H stretching vibrations of hexane appear below 3000 cm H110021 . A peak or peaks above 3000 cm H110021 is characteristic of a hydrogen bonded to sp 2 -hybridized carbon. The IR spectrum of 1-hexene also displays a peak at 1640 cm H110021 corresponding to its C?C stretching vibration. The peaks near 1000 and 900 cm H110021 in the spectrum of 1-hexene, absent in the spectrum of hexane, are bending vibrations involving the hydrogens of the doubly bonded carbons. Carbon–hydrogen stretching vibrations with frequencies above 3000 cm H110021 are also found in arenes such as tert-butylbenzene, as shown in Figure 13.28. This spectrum also contains two intense bands at 760 and 700 cm H110021 , which are characteristic of monosub- stituted benzene rings. Other substitution patterns, some of which are listed in Table 13.4, give different combinations of peaks. In addition to sp 2 C±H stretching modes, there are other stretching vibrations that appear at frequencies above 3000 cm H110021 . The most important of these is the O±H stretch of alcohols. Figure 13.29 shows the IR spectrum of 2-hexanol. It contains a broad peak at 3300 cm H110021 ascribable to O±H stretching of hydrogen-bonded alcohol groups. In dilute solution, where hydrogen bonding is less and individual alcohol molecules are pres- ent as well as hydrogen-bonded aggregates, an additional peak appears at approximately 3600 cm H110021 . Carbonyl groups rank among the structural units most readily revealed by IR spec- troscopy. The carbon–oxygen double bond stretching mode gives rise to a very strong peak 520 CHAPTER THIRTEEN Spectroscopy Wave number, cm H110021 Transmittance (%) CH 3 CH 2 CH 2 CH 2 CH 2 CH 3 H±C FIGURE 13.26 The infrared spectrum of hexane. All of the calculated vibra- tional frequencies given on Learning By Modeling are too high. For example, the C?C stretching frequency of 1-hexene observed at 1640 cm H110021 is calculated to be at 1857 cm H110021 . Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Wave number, cm H110021 Transmittance (%) Ar±H Monosubstituted benzene C(CH 3 ) 3 H±C ± FIGURE 13.27 The infrared spectrum of 1-hexene. FIGURE 13.28 The infrared spectrum of tert-butylbenzene. Wave number, cm H110021 Transmittance (%) CH 2 ?CHCH 2 CH 2 CH 2 CH 3 C?C±H C?C CH 2 ?C± H±C Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 522 CHAPTER THIRTEEN Spectroscopy Wave number, cm H110021 Transmittance (%) CH 3 CH 2 CH 2 CH 2 CHCH 3 O±H H±C OH W in the 1650–1800 cm H110021 region. This peak is clearly evident in the spectrum of 2-hexanone, shown in Figure 13.30. The position of the carbonyl peak varies with the nature of the substituents on the carbonyl group. Thus, characteristic frequencies are associated with aldehydes and ketones, amides, esters, and so forth, as summarized in Table 13.4. PROBLEM 13.15 Which one of the following compounds is most consistent with the infrared spectrum given in Figure 13.31? Explain your reasoning. In later chapters, when families of compounds are discussed in detail, the infrared frequencies associated with each type of functional group will be described. 13.20 ULTRAVIOLET-VISIBLE (UV-VIS) SPECTROSCOPY The main application of UV-VIS spectroscopy, which depends on transitions between electronic energy levels, is in identifying conjugated H9266 electron systems. Much greater energies separate vibrational states than nuclear spin states, and the energy differences between electronic states are greater yet. The energy required to OH Phenol CCH 3 O X Acetophenone COH O X Benzoic acid CH 2 OH Benzyl alcohol FIGURE 13.29 The infrared spectrum of 2-hexanol. The C?O stretching fre- quency in 2-hexanone ap- pears at 1720 cm H110021 . To view this vibration on Learning By Modeling, select the calcu- lated value of 1940 cm H110021 . Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Wave number, cm H110021 Transmittance (%) CH 3 CCH 2 CH 2 CH 2 CH 3 C?O H±C O X Wave number, cm H110021 Transmittance (%) FIGURE 13.30 The infrared spectrum of 2-hexanone. FIGURE 13.31 The infrared spectrum of the unknown compound in Problem 13.15. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website promote an electron from one electronic state to the next lies in the visible and ultravi- olet range of the electromagnetic spectrum (see Figure 13.1). We usually identify radia- tion in the UV-VIS range by its wavelength in nanometers (1 nm H11005 10 H110029 m). Thus, the visible region corresponds to 400–800 nm. Red light is the low-energy (long wavelength) end of the visible spectrum, violet light the high-energy (short wavelength) end. Ultra- violet light lies beyond the visible spectrum with wavelengths in the 200–400-nm range. Figure 13.32 shows the UV spectrum of the conjugated diene cis,trans-1,3-cyclooc- tadiene, measured in ethanol as the solvent. As is typical of most UV spectra, the absorp- tion is rather broad and is often spoken of as a “band” rather than a “peak.” The wave- length at an absorption maximum is referred to as the H9261 max of the band. There is only one band in the UV spectrum of 1,3-cyclooctadiene; its H9261 max is 230 nm. In addition to H9261 max ,UV-VIS bands are characterized by their absorbance (A), which is a measure of how much of the radiation that passes through the sample is absorbed. To correct for concentration and path length effects, absorbance is converted to molar absorptivity (H9280) by dividing it by the concentration c in moles per liter and the path length l in cen- timeters. H9280H11005 Molar absorptivity, when measured at H9261 max , is cited as H9280 max . It is normally expressed without units. Both H9261 max and H9280 max are affected by the solvent, which is therefore included when reporting UV-VIS spectroscopic data. Thus, you might find a literature reference expressed in the form Figure 13.33 illustrates the transition between electronic energy states responsible for the 230-nm UV band of cis-trans-1,3-cyclooctadiene. Absorption of ultraviolet radi- ation excites an electron from the highest occupied molecular orbital (HOMO) to the lowest unoccupied molecular orbital (LUMO). In alkenes and polyenes, both the HOMO cis, trans-1,3-Cyclooctadiene H9261 ethanol max 230 nm 2630H9280 ethanol max A c H11554 l 524 CHAPTER THIRTEEN Spectroscopy An important enzyme in bio- logical electron transport called cytochrome P450 gets its name from its UV absorp- tion. The “P” stands for “pig- ment” because it is colored, and the “450” corresponds to the 450-nm absorption of one of its derivatives. Molar absorptivity used to be called the molar extinc- tion coefficient. Molar absorptivity ( ε ) 2000 1000 0 200 240220 260 280 Wavelength, nm FIGURE 13.32 The ultravio- let spectrum of cis,trans-1,3- cyclooctadiene. LUMO HOMO Most stable electron configuration Electron configuration of excited state 2 ψ 1 ψ 2 ψ 1 ψ ?E = hv 4 ψ * 3 ψ * 4 ψ * 3 ψ * FIGURE 13.33 The H9266 → H9266* transition in cis,trans-1,3- cyclooctadiene involves exci- tation of an electron from the highest occupied molec- ular orbital (HOMO) to the lowest unoccupied molecu- lar orbital (LUMO). Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website and LUMO are H9266-type orbitals (rather than H9268); the HOMO is the highest energy H9266 orbital and the LUMO is the lowest energy H9266* orbital. Exciting one of the H9266 electrons from a bonding H9266 orbital to an antibonding H9266* orbital is referred to as a H9266→H9266* transition. PROBLEM 13.16 H9261 max for the H9266→H9266* transition in ethylene is 170 nm. Is the HOMO–LUMO energy difference in ethylene greater than or less than that of cis,trans-1,3-cyclooctadiene? The HOMO–LUMO energy gap and, consequently, H9261 max for the H9266→H9266* transition varies with the substituents on the double bonds. The data in Table 13.5 illustrate two substituent effects: adding methyl substituents to the double bond, and extending conju- gation. Both cause H9261 max to shift to longer wavelengths, but the effect of conjugation is the larger of the two. Based on data collected for many dienes it has been found that each methyl substituent on the double bonds causes a shift to longer wavelengths of about 5 nm, whereas extending the conjugation causes a shift of about 36 nm for each additional double bond. PROBLEM 13.17 Which one of the C 5 H 8 isomers shown has its H9261 max at the longest wavelength? A striking example of the effect of conjugation on light absorption occurs in lycopene, which is one of the pigments in ripe tomatoes. Lycopene has a conjugated sys- tem of 11 double bonds and absorbs visible light. It has several UV-VIS bands, each characterized by a separate H9261 max . Its longest wavelength absorption is at 505 nm. Many organic compounds such as lycopene are colored because their HOMO- LUMO energy gap is small enough that H9261 max appears in the visible range of the spectrum. 13.20 Ultraviolet-Visible (UV-VIS) Spectroscopy 525 TABLE 13.5 Absorption Maxima of Some Representative Alkenes and Polyenes* Compound Ethylene 2-Methylpropene 1,3-Butadiene 4-Methyl-1,3-pentadiene 2,5-Dimethyl-2,4-hexadiene (2E,4E,6E)-2,4,6-Octatriene (2E,4E,6E,8E)-2,4,6,8-Decatetraene (2E,4E,6E,8E,10E)-2,4,6,8,10-Dodecapentaene 170 188 217 234 241 263 299 326 H9261 max (nm) H 2 C?CH 2 H 2 C?C(CH 3 ) 2 H 2 C?CHCH?CH 2 H 2 C?CHCH?C(CH 3 ) 2 (CH 3 ) 2 C?CHCH?C(CH 3 ) 2 CH 3 CH?CHCH?CHCH?CHCH 3 CH 3 CH?CH(CH?CH) 2 CH?CHCH 3 CH 3 CH?CH(CH?CH) 3 CH?CHCH 3 Structure *The value of H9261 max refers to the longest wavelength H9266￡H9266* transition. Lycopene Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website All that is required for a compound to be colored, however, is that it possess some absorption in the visible range. It often happens that a compound will have its H9261 max in the UV region but that the peak is broad and extends into the visible. Absorption of the blue-to-violet components of visible light occurs, and the compound appears yellow. A second type of absorption that is important in UV-VIS examination of organic compounds is the n→H9266* transition of the carbonyl (C?O) group. One of the electrons in a lone-pair orbital of oxygen is excited to an antibonding orbital of the carbonyl group. The n in n→H9266* identifies the electron as one of the nonbonded electrons of oxygen. This transition gives rise to relatively weak absorption peaks (H9280 max H11021 100) in the region 270–300 nm. The structural unit associated with the electronic transition in UV-VIS spectroscopy is called a chromophore. Chemists often refer to model compounds to help interpret UV- VIS spectra. An appropriate model is a simple compound of known structure that incor- porates the chromophore suspected of being present in the sample. Because remote sub- stituents do not affect H9261 max of the chromophore, a strong similarity between the spectrum of the model compound and that of the unknown can serve to identify the kind of H9266 electron system present in the sample. There is a substantial body of data concerning the UV-VIS spectra of a great many chromophores, as well as empirical correlations of sub- stituent effects on H9261 max . Such data are helpful when using UV-VIS spectroscopy as a tool for structure determination. 13.21 MASS SPECTROMETRY Mass spectrometry differs from the other instrumental methods discussed in this chap- ter in a fundamental way. It does not depend on the absorption of electromagnetic radi- ation but rather examines what happens when a molecule is bombarded with high-energy electrons. If an electron having an energy of about 10 electronvolts (10 eV H11005 230.5 kcal/mol) collides with an organic molecule, the energy transferred as a result of that collision is sufficient to dislodge one of the molecule’s electrons. We say the molecule AB has been ionized by electron impact. The species that results, called the molecular ion, is positively charged and has an odd number of electrons—it is a cation radical. The molecular ion has the same mass (less the negligible mass of a single electron) as the molecule from which it is formed. Although energies of about 10 eV are required, energies of about 70 eV are used. Electrons this energetic not only cause ionization of a molecule but impart a large amount of energy to the molecular ion, enough energy to break chemical bonds. The molecular ion dissipates this excess energy by dissociating into smaller fragments. Dissociation of a cation radical produces a neutral fragment and a positively charged fragment. Ionization and fragmentation produce a mixture of particles, some neutral and some positively charged. To understand what follows, we need to examine the design of an electron-impact mass spectrometer, shown in a schematic diagram in Figure 13.34. The sample is bombarded with 70-eV electrons, and the resulting positively charged ions (the A H11001 B Cation radical A H11001 Cation H11001 Radical B e H11002 Electron 2e H11002 Two electrons H11001H11001 A H11001 B Cation radicalMolecule AB 526 CHAPTER THIRTEEN Spectroscopy Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website molecular ion as well as fragment ions) are directed into an analyzer tube surrounded by a magnet. This magnet deflects the ions from their original trajectory, causing them to adopt a circular path, the radius of which depends on their mass-to-charge ratio (m/z). Ions of small m/z are deflected more than those of larger m/z. By varying either the mag- netic field strength or the degree to which the ions are accelerated on entering the ana- lyzer, ions of a particular m/z can be selectively focused through a narrow slit onto a detector, where they are counted. Scanning all m/z values gives the distribution of pos- itive ions, called a mass spectrum, characteristic of a particular compound. Modern mass spectrometers are interfaced with computerized data-handling sys- tems capable of displaying the mass spectrum according to a number of different for- mats. Bar graphs on which relative intensity is plotted versus m/z are the most common. Figure 13.35 shows the mass spectrum of benzene in bar graph form. The mass spectrum of benzene is relatively simple and illustrates some of the infor- mation that mass spectrometry provides. The most intense peak in the mass spectrum is called the base peak and is assigned a relative intensity of 100. Ion abundances are 13.21 Mass Spectrometry 527 If necessary, heater vaporizes sample Electron beam knocks electrons from atoms Electric field accelerates particles toward magnetic region Magnetic field separates particles according to their mass-to-charge ratio Sample enters chamber Electron source Charged particle beam Lightest particles in sample Detector Heaviest particles in sample 2 3 4 5 1 Z H11001 X H11001 Y H11001 Magnet FIGURE 13.34 Diagram of a mass spectrometer. Only positive ions are detected. The cation X H11001 has the lowest mass-to-charge ratio, and its path is deflected most by the magnet. The cation Z H11001 has the highest mass-to-charge ratio, and its path is deflected least. (Adapted, with per- mission, from M. Silberberg, Chemistry, 2d edition, WCB/McGraw-Hill, New York, 2000, p. 56.) FIGURE 13.35 The mass spectrum of benzene. The peak at m/z H11005 78 corresponds to the C 6 H 6 molecular ion. 78 10 20 30 40 50 60 70 80 90 m/z 100 110 120 130 140 150 0 20 40 60 80 100 Relative intensity Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website proportional to peak intensities and are reported as intensities relative to the base peak. The base peak in the mass spectrum of benzene corresponds to the molecular ion (M H11001 ) at m/z H11005 78. Benzene does not undergo extensive fragmentation; none of the fragment ions in its mass spectrum are as abundant as the molecular ion. There is a small peak one mass unit higher than M H11001 in the mass spectrum of ben- zene. What is the origin of this peak? What we see in Figure 13.35 as a single mass spec- trum is actually a superposition of the spectra of three isotopically distinct benzenes. Most of the benzene molecules contain only 12 C and 1 H and have a molecular mass of 78. Smaller proportions of benzene molecules contain 13 C in place of one of the 12 C atoms or 2 H in place of one of the protons. Both these species have a molecular mass of 79. Not only the molecular ion peak but all the peaks in the mass spectrum of benzene are accompanied by a smaller peak one mass unit higher. Indeed, since all organic com- pounds contain carbon and most contain hydrogen, similar isotopic clusters will appear in the mass spectra of all organic compounds. Isotopic clusters are especially apparent when atoms such as bromine and chlorine are present in an organic compound. The natural ratios of isotopes in these elements are Figure 13.36 presents the mass spectrum of chlorobenzene. There are two prominent mo- lecular ion peaks, one at m/z 112 for C 6 H 5 35 Cl and the other at m/z 114 for C 6 H 5 37 Cl. The peak at m/z 112 is three times as intense as the one at m/z 114. PROBLEM 13.18 Knowing what to look for with respect to isotopic clusters can aid in interpreting mass spectra. How many peaks would you expect to see for the molecular ion in each of the following compounds? At what m/z values would these peaks appear? (Disregard the small peaks due to 13 C and 2 H.) (a) p-Dichlorobenzene (c) p-Dibromobenzene (b) o-Dichlorobenzene (d) p-Bromochlorobenzene 79 Br 81 Br H11005 100 97.5 35 Cl 37 Cl H11005 100 32.7 1 H 1 H 1 H 1 H 1 H 1 H 93.4% (all carbons are 12 C) Gives M H11001 78 2 H 1 H 1 H 1 H 1 H 1 H 0.1% (all carbons are 12 C) Gives M H11001 79 6.5% (* H11005 13 C) Gives M H11001 79 1 H 1 H 1 H 1 H 1 H 1 H * H H H H H H Benzene H11001 e H11002 Electron Molecular ion of benzene H11001 H H H H H H H11001 Two electrons 2e H11002 528 CHAPTER THIRTEEN Spectroscopy Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website SAMPLE SOLUTION (a) The two isotopes of chlorine are 35 Cl and 37 Cl. There will be three isotopically different forms of p-dichlorobenzene present. They have the structures shown as follows. Each one will give an M H11001 peak at a different value of m/z. Unlike the case of benzene, in which ionization involves loss of a H9266 electron from the ring, electron-impact-induced ionization of chlorobenzene involves loss of an elec- tron from an unshared pair of chlorine. The molecular ion then fragments by carbon–chlorine bond cleavage. The peak at m/z 77 in the mass spectrum of chlorobenzene in Figure 13.36 is attributed to this fragmentation. Because there is no peak of significant intensity two atomic mass units higher, we know that the cation responsible for the peak at m/z 77 cannot contain chlorine. Some classes of compounds are so prone to fragmentation that the molecular ion peak is very weak. The base peak in most unbranched alkanes, for example, is m/z 43, which is followed by peaks of decreasing intensity at m/z values of 57, 71, 85, and so on. These peaks correspond to cleavage of each possible carbon–carbon bond in the mol- ecule. This pattern is evident in the mass spectrum of decane, depicted in Figure 13.37. The points of cleavage are indicated in the following diagram: Many fragmentations in mass spectrometry proceed so as to form a stable carbo- cation, and the principles that we have developed regarding carbocation stability apply. CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 3 M H11001 142 43 57 71 85 99 113 127 H11001Cl Chlorobenzene e H11002 H11001 Cl Molecular ion of chlorobenzene Cl Chlorine atom H11001 Phenyl cation m/z 77 35 Cl 35 Cl m/z 146 35 Cl 37 Cl m/z 148 37 Cl 37 Cl m/z 150 13.21 Mass Spectrometry 529 112 114 10 20 30 40 50 60 70 80 90 m/z 100 110 120 130 140 150 0 20 40 60 80 100 Relative intensity FIGURE 13.36 The mass spectrum of chlorobenzene. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 530 CHAPTER THIRTEEN Spectroscopy 142 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 0 20 40 60 80 100 Relative intensity m/z FIGURE 13.37 The mass spectrum of decane. The peak for the molecular ion is extremely small. The most prominent peaks arise by fragmentation. GAS CHROMATOGRAPHY, GC/MS, AND MS/MS A ll of the spectra in this chapter ( 1 H NMR, 13 C NMR, IR, UV-VIS, and MS) were obtained using pure substances. It is much more common, however, to encounter an organic substance, either formed as the product of a chemical reaction or iso- lated from natural sources, as but one component of a mixture. Just as the last half of the twentieth cen- tury saw a revolution in the methods available for the identification of organic compounds, so too has it seen remarkable advances in methods for their sepa- ration and purification. Classical methods for separation and purifica- tion include fractional distillation of liquids and re- crystallization of solids, and these two methods are routinely included in the early portions of laboratory courses in organic chemistry. Because they are capa- ble of being adapted to work on a large scale, frac- tional distillation and recrystallization are the pre- ferred methods for purifying organic substances in the pharmaceutical and chemical industries. Some other methods are more appropriate when separating small amounts of material in laboratory-scale work and are most often encoun- tered there. Indeed, it is their capacity to deal with exceedingly small quantities that is the strength of a number of methods that together encompass the various forms of chromatography. The first step in all types of chromatography involves absorbing the sam- ple onto some material called the stationary phase. Next, a second phase (the mobile phase) is allowed to move across the stationary phase. Depending on the properties of the two phases and the components of the mixture, the mixture is separated into its compo- nents according to the rate at which each is removed from the stationary phase by the mobile phase. In gas chromatography (GC), the stationary phase consists of beads of an inert solid support coated with a high-boiling liquid, and the mobile phase is a gas, usually helium. Figure 13.38 shows a typical gas chromatograph. The sample is injected by Syringe to inject sample Heated injection block Septum Carrier gas Flow rate control valve Chromatography column (coiled to conserve space in oven) Oven heater and fan Heated detector block Output to recorder FIGURE 13.38 Diagram of a gas chromatograph. When connected to a mass spectrometer as in GC/MS, the effluent is split into two streams as it leaves the column. One stream goes to the detector, the other to the mass spectrometer. (Adapted, with permission, from H. D. Durst and G. W. Gokel, Experimental Organic Chemistry, 2nd ed., McGraw-Hill, New York, 1987.) Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 13.21 Mass Spectrometry 531 Alkylbenzenes of the type C 6 H 5 CH 2 R undergo cleavage of the bond to the benzylic car- bon to give m/z 91 as the base peak. The mass spectrum in Figure 13.39 and the fol- lowing fragmentation diagram illustrate this for propylbenzene. Although this cleavage is probably driven by the stability of benzyl cation, evidence has been obtained suggesting that tropylium cation, formed by rearrangement of benzyl cation, is actually the species responsible for the peak. CH 2 91 CH 2 CH 3 M H11001 120 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 0 20 40 60 80 100 Relative intensity 120 91 m/z FIGURE 13.39 The mass spectrum of propylbenzene. The most intense peak is C 7 H 7 H11001 . The structure of tropylium cation is given in Section 11.20. syringe onto a heated block where a stream of he- lium carries it onto a coiled column packed with the stationary phase. The components of the mixture move through the column at different rates. They are said to have different retention times. Gas chro- matography is also referred to as gas–liquid partition chromatography, because the technique depends on how different substances partition themselves be- tween the gas phase (dispersed in the helium carrier gas) and the liquid phase (dissolved in the coating on the beads of solid support). Typically the effluent from a gas chromato- graph is passed through a detector, which feeds a sig- nal to a recorder whenever a substance different from pure carrier gas leaves the column. Thus, one determines the number of components in a mixture by counting the number of peaks on a strip chart. It is good practice to carry out the analysis under differ- ent conditions by varying the liquid phase, the tem- perature, and the flow rate of the carrier gas so as to ensure that two substances have not eluted together and given a single peak under the original condi- tions. Gas chromatography can also be used to iden- tify the components of a mixture by comparing their retention times with those of authentic samples. In gas chromatography/mass spectrometry (GC/MS), the effluent from a gas chromatograph is passed into a mass spectrometer and a mass spectrum is taken every few milliseconds. Thus gas chromatog- raphy is used to separate a mixture, and mass spec- trometry used to analyze it. GC/MS is a very powerful analytical technique. One of its more visible applica- tions involves the testing of athletes for steroids, stimulants, and other performance-enhancing drugs. These drugs are converted in the body to derivatives called metabolites, which are then excreted in the urine. When the urine is subjected to GC/MS analysis, the mass spectra of its organic components are iden- tified by comparison with the mass spectra of known metabolites stored in the instrument’s computer. Us- ing a similar procedure, the urine of newborn infants is monitored by GC/MS for metabolite markers of ge- netic disorders that can be treated if detected early in life. GC/MS is also used to detect and measure the concentration of halogenated hydrocarbons in drink- ing water. Although GC/MS is the most widely used ana- lytical method that combines a chromatographic sep- aration with the identification power of mass spec- trometry, it is not the only one. Chemists have coupled mass spectrometers to most of the instru- ments that are used to separate mixtures. Perhaps the ultimate is mass spectrometry/mass spectrome- try (MS/MS), in which one mass spectrometer gener- ates and separates the molecular ions of the compo- nents of a mixture and a second mass spectrometer examines their fragmentation patterns! Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website PROBLEM 13.19 The base peak appears at m/z 105 for one of the following compounds and at m/z 119 for the other two. Match the compounds with the appropriate m/z values for their base peaks. Understanding how molecules fragment upon electron impact permits a mass spec- trum to be analyzed in sufficient detail to deduce the structure of an unknown compound. Thousands of compounds of known structure have been examined by mass spectrome- try, and the fragmentation patterns that characterize different classes are well docu- mented. As various groups are covered in subsequent chapters, aspects of their frag- mentation behavior under conditions of electron impact will be described. 13.22 MOLECULAR FORMULA AS A CLUE TO STRUCTURE As we have just seen, interpreting the fragmentation patterns in a mass spectrum in terms of a molecule’s structural units makes mass spectrometry much more than just a tool for determining molecular weights. Nevertheless, even the molecular weight can provide more information than you might think. Compare, for example, heptane and cyclopropyl acetate. Heptane and cyclopropyl acetate have different molecular formulas but have the same molecular weight—at least to a first approximation. Because we normally round off mo- lecular weights to whole numbers, both have a molecular weight of 100 and both have a peak for their molecular ion at m/z 100 in a typical mass spectrum. Recall, however, that mass spectra contain isotopic clusters that differ according to the isotopes present in each ion. Using the exact values for the major isotopes of C, H, and O, we calculate exact masses of m/z of 100.1253 and 100.0524 for the molecular ions of heptane (C 7 H 16 ) and cyclopropyl acetate (C 5 H 8 O 2 ), respectively. As similar as these values are, it is pos- sible to distinguish between them using a high-resolution mass spectrometer. What this means is that the exact mass of a molecular ion can usually be translated into a unique molecular formula. Once we have the molecular formula, it can provide information that limits the amount of trial-and-error structure writing we have to do. Consider, for example, hep- tane and its molecular formula of C 7 H 16 . We know immediately that the molecular for- mula belongs to an alkane because it corresponds to C n H 2nH110012 . What about a substance with the molecular formula C 7 H 14 ? This compound can- not be an alkane but may be either a cycloalkane or an alkene, because both these classes of hydrocarbons correspond to the general molecular formula C n H 2n . Any time a ring or a double bond is present in an organic molecule, its molecular formula has two fewer hydrogen atoms than that of an alkane with the same number of carbons. The relationship between molecular formulas, multiple bonds, and rings is referred to as the index of hydrogen deficiency and can be expressed by the equation: CH 3 (CH 2 ) 5 CH 3 Heptane (C 7 H 16 ) Cyclopropyl acetate (C 5 H 8 O 2 ) CH 3 CO O CH 2 CH 3 CH 3 CH 3 CH 2 CH 2 CH 3 CH 3 CH 3 CHCH 3 CH 3 532 CHAPTER THIRTEEN Spectroscopy You can’t duplicate these molecular weights for C 7 H 16 and C 5 H 8 O 2 by using the atomic weights given in the periodic table. Those values are for the natural- abundance mixture of iso- topes. The exact values are 12.00000 for 12 C, 1.00783 for 1 H, and 15.9949 for 16 O. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Index of hydrogen deficiency H11005 (C n H 2nH110012 H11002 C n H x ) where C n H x is the molecular formula of the compound. A molecule that has a molecular formula of C 7 H 14 has an index of hydrogen defi- ciency of 1: Index of hydrogen deficiency H11005 (C 7 H 16 H11002 C 7 H 14 ) Index of hydrogen deficiency H11005 (2) H11005 1 Thus, the compound has one ring or one double bond. It can’t have a triple bond. A molecule of molecular formula C 7 H 12 has four fewer hydrogens than the corre- sponding alkane. It has an index of hydrogen deficiency of 2 and can have two rings, two double bonds, one ring and one double bond, or one triple bond. What about substances other than hydrocarbons, 1-heptanol [CH 3 (CH 2 ) 5 CH 2 OH], for example? Its molecular formula (C 7 H 16 O) contains the same carbon-to-hydrogen ratio as heptane and, like heptane, it has no double bonds or rings. Cyclopropyl acetate (C 5 H 8 O 2 ), the structure of which was given at the beginning of this section, has one ring and one double bond and an index of hydrogen deficiency of 2. Oxygen atoms have no effect on the index of hydrogen deficiency. A halogen substituent, like hydrogen is monovalent, and when present in a mo- lecular formula is treated as if it were hydrogen for counting purposes. How does one distinguish between rings and double bonds? This additional piece of information comes from catalytic hydrogenation experiments in which the amount of hydrogen consumed is measured exactly. Each of a molecule’s double bonds consumes one molar equivalent of hydrogen, but rings are unaffected. For example, a substance with a hydrogen deficiency of 5 that takes up 3 moles of hydrogen must have two rings. PROBLEM 13.20 How many rings are present in each of the following com- pounds? Each consumes 2 moles of hydrogen on catalytic hydrogenation. (a) C 10 H 18 (d) C 8 H 8 O (b) C 8 H 8 (e) C 8 H 10 O 2 (c) C 8 H 8 CI 2 (f) C 8 H 9 ClO SAMPLE SOLUTION (a) The molecular formula C 10 H 18 contains four fewer hydrogens than the alkane having the same number of carbon atoms (C 10 H 22 ). Therefore, the index of hydrogen deficiency of this compound is 2. Since it con- sumes two molar equivalents of hydrogen on catalytic hydrogenation, it must have two double bonds and no rings. 13.23 SUMMARY Section 13.1 Structure determination in modern-day organic chemistry relies heavily on instrumental methods. Several of the most widely used ones depend on the absorption of electromagnetic radiation. Section 13.2 Absorption of electromagnetic radiation causes a molecule to be excited from its most stable state (the ground state) to a higher energy state (an excited state). 1 2 1 2 1 2 13.23 Summary 533 Other terms that mean the same thing as the index of hydrogen deficiency include elements of unsaturation, sites of unsaturation, and the sum of double bonds and rings. A more detailed discussion can be found in the May 1995 issue of the Journal of Chemical Education, pp. 245–248. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Spectroscopic method Transitions between Nuclear magnetic resonance Spin states of an atom’s nucleus Infrared Vibrational states Ultraviolet-visible Electronic states Mass spectrometry is not based on absorption of electromagnetic radia- tion, but monitors what happens when a substance is ionized by collision with a high-energy electron. 1 H Nuclear Magnetic Resonance Spectroscopy Section 13.3 In the presence of an external magnetic field, the H11001 and H11002 nuclear spin states of a proton have slightly different energies. Section 13.4 The energy required to “flip” the spin of a proton from the lower energy spin state to the higher state depends on the extent to which a nucleus is shielded from the external magnetic field by the molecule’s electrons. Section 13.5 Protons in different environments within a molecule have different chem- ical shifts; that is, they experience different degrees of shielding. Chem- ical shifts (H9254) are reported in parts per million (ppm) from tetramethylsi- lane (TMS). Table 13.1 lists characteristic chemical shifts for various types of protons. Section 13.6 In addition to chemical shift, a 1 H NMR spectrum provides structural information based on: Number of signals, which tells how many different kinds of protons there are Integrated areas, which tells the ratios of the various kinds of protons Splitting pattern, which gives information about the number of protons that are within two or three bonds of the one giving the signal Section 13.7 Spin-spin splitting of NMR signals results from coupling of the nuclear spins that are separated by two bonds (geminal coupling) or three bonds (vicinal coupling). In the simplest cases, the number of peaks into which a signal is split is equal to n H11001 1, where n is the number of protons to which the proton in question is coupled. Protons that have the same chemical shift do not split each other’s signal. Section 13.8 The methyl protons of an ethyl group appear as a triplet and the meth- ylene protons as a quartet in compounds of the type CH 3 CH 2 X. Section 13.9 The methyl protons of an isopropyl group appear as a doublet and the methine proton as a septet in compounds of the type (CH 3 ) 2 CHX. H H C Geminal hydrogens are separated by two bonds C HH C Vicinal hydrogens are separated by three bonds 1 2 1 2 534 CHAPTER THIRTEEN Spectroscopy Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Section 13.10 A doublet of doublets characterizes the signals for the protons of the type shown (where W, X, Y, and Z are not H or atoms that split H themselves). Section 13.11 Complicated splitting patterns can result when a proton is unequally cou- pled to two or more protons that are different from one another. Section 13.12 Splitting resulting from coupling to the O±H proton of alcohols is not normally observed, because the hydroxyl proton undergoes rapid inter- molecular exchange with other alcohol molecules, which “decouples” it from other protons in the molecule. Section 13.13 Many processes such as conformational changes take place faster than they can be detected by NMR. Consequently, NMR provides information about the average environment of a proton. For example, cyclohexane gives a single peak for its 12 protons even though, at any instant, 6 are axial and 6 are equatorial. 13 C Nuclear Magnetic Resonance Spectroscopy Section 13.14 13 C has a nuclear spin of H11006 but only about 1% of all the carbons in a sample are 13 C. Nevertheless, high-quality 13 C NMR spectra can be obtained by pulse FT techniques and are a useful complement to 1 H NMR spectra. Section 13.15 13 C signals are more widely separated from one another than proton sig- nals, and 13 C NMR spectra are relatively easy to interpret. Table 13.3 gives chemical shift values for carbon in various environments. Section 13.16 13 C NMR spectra are rarely integrated because the pulse FT technique distorts the signal intensities. Section 13.17 Carbon signals normally appear as singlets, but several techniques are available that allow one to distinguish among the various kinds of car- bons shown. Section 13.18 One of the special techniques for distinguishing carbons according to the number of their attached hydrogens is called DEPT. A series of NMR measurements using different pulse sequences gives normal, nulled, and inverted peaks that allow assignment of primary, secondary, tertiary, and quaternary carbons. 3 attached hydrogens (Primary carbon) C H H C H 2 attached hydrogens (Secondary carbon) C H C C H 1 attached hydrogen (Tertiary carbon) C C C C H no attached hydrogen (Quaternary carbon) C C C C C 1 2 C HH W CZ XY 13.23 Summary 535 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Infrared Spectroscopy Section 13.19 Infrared spectroscopy probes molecular structure by examining transitions between vibrational energy levels using electromagnetic radiation in the 625–4000-cm H110021 range. The presence or absence of a peak at a character- istic frequency tells us whether a certain functional group is pres- ent. Table 13.4 lists IR absorption frequencies for common structural units. Ultraviolet-Visible Spectroscopy Section 13.20 Transitions between electronic energy levels involving electromagnetic radiation in the 200–800-nm range form the basis of UV-VIS spectroscopy. The absorption peaks tend to be broad but are often useful in indicating the presence of particular H9266 electron systems within a molecule. Mass Spectrometry Section 13.21 Mass spectrometry exploits the information obtained when a molecule is ionized by electron impact and then dissociates to smaller fragments. Pos- itive ions are separated and detected according to their mass-to-charge (m/z) ratio. By examining the fragments and by knowing how classes of molecules dissociate on electron impact, one can deduce the structure of a compound. Mass spectrometry is quite sensitive; as little as 10 H110029 g of compound is sufficient for analysis. Section 13.22 A compound’s molecular formula gives information about the number of double bonds and rings it contains and is a useful complement to spec- troscopic methods of structure determination. PROBLEMS 13.21 Each of the following compounds is characterized by a 1 H NMR spectrum that consists of only a single peak having the chemical shift indicated. Identify each compound. (a) C 8 H 18 ; H9254 0.9 ppm (f) C 2 H 3 Cl 3 ; H9254 2.7 ppm (b) C 5 H 10 ; H9254 1.5 ppm (g) C 5 H 8 Cl 4 ; H9254 3.7 ppm (c) C 8 H 8 ; H9254 5.8 ppm (h) C 12 H 18 ; H9254 2.2 ppm (d) C 4 H 9 Br; H9254 1.8 ppm (i) C 3 H 6 Br 2 ; H9254 2.6 ppm (e) C 2 H 4 Cl 2 ; H9254 3.7 ppm 13.22 Each of the following compounds is characterized by a 1 H NMR spectrum that consists of two peaks, both singlets, having the chemical shifts indicated. Identify each compound. (a) C 6 H 8 ; H9254 2.7 ppm (4H) and 5.6 ppm (4H) (b) C 5 H 11 Br; H9254 1.1 ppm (9H) and 3.3 ppm (2H) (c) C 6 H 12 O; H9254 1.1 ppm (9H) and 2.1 ppm (3H) (d) C 6 H 10 O 2 ; H9254 2.2 ppm (6H) and 2.7 ppm (4H) 13.23 Deduce the structure of each of the following compounds on the basis of their 1 H NMR spectra and molecular formulas: (a) C 8 H 10 ; H9254 1.2 ppm (triplet, 3H) H9254 2.6 ppm (quartet, 2H) H9254 7.1 ppm (broad singlet, 5H) (b) C 10 H 14 ; H9254 1.3 ppm (singlet, 9H) H9254 7.0 to 7.5 ppm (multiplet, 5H) 536 CHAPTER THIRTEEN Spectroscopy Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (c) C 6 H 14 ; H9254 0.8 ppm (doublet, 12H) (f) C 4 H 6 Cl 2 ; H9254 2.2 ppm (singlet, 3H) H9254 1.4 ppm (heptet, 2H) H9254 4.1 ppm (doublet, 2H) (d) C 6 H 12 ; H9254 0.9 ppm (triplet, 3H) H9254 5.7 ppm (triplet, 1H) H9254 1.6 ppm (singlet, 3H) (g) C 3 H 7 ClO; H9254 2.0 ppm (pentet, 2H) H9254 1.7 ppm (singlet, 3H) H9254 2.8 ppm (singlet, 1H) H9254 2.0 ppm (pentet, 2H) H9254 3.7 ppm (triplet, 2H) H9254 5.1 ppm (triplet, 1H) H9254 3.8 ppm (triplet, 2H) (e) C 4 H 6 Cl 4 ; H9254 3.9 ppm (doublet, 4H) (h) C 14 H 14 ; H9254 2.9 ppm (singlet, 4H) H9254 4.6 ppm (triplet, 2H) H9254 7.1 ppm (broad singlet, 10H) 13.24 From among the isomeric compounds of molecular formula C 4 H 9 Cl, choose the one hav- ing a 1 H NMR spectrum that (a) Contains only a single peak (b) Has several peaks including a doublet at H9254 3.4 ppm (c) Has several peaks including a triplet at H9254 3.5 ppm (d) Has several peaks including two distinct three-proton signals, one of them a triplet at H9254 1.0 ppm and the other a doublet at H9254 1.5 ppm 13.25 Identify the C 3 H 5 Br isomers on the basis of the following information: (a) Isomer A has the 1 H NMR spectrum shown in Figure 13.40. Problems 537 0.01.02.03.04.0 Chemical shift (δ, ppm) 6.07.08.09.010.0 5.0 1 1 3 FIGURE 13.40 The 200-MHz 1 H NMR spectrum of isomer A of C 3 H 5 Br (Problem 13.25a). Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (b) Isomer B has three peaks in its 13 C NMR spectrum: H9254 32.6 ppm (CH 2 ); 118.8 ppm (CH 2 ); and 134.2 ppm (CH). (c) Isomer C has two peaks in its 13 C NMR spectrum: H9254 12.0 ppm (CH 2 ) and 16.8 ppm (CH). The peak at lower field is only half as intense as the one at higher field. 13.26 Identify each of the C 4 H 10 O isomers on the basis of their 13 C NMR spectra: (a) H9254 18.9 ppm (CH 3 ) (two carbons) (c) H9254 31.2 ppm (CH 3 ) (three carbons) H9254 30.8 ppm (CH) (one carbon) H9254 68.9 ppm (C) (one carbon) H9254 69.4 ppm (CH 2 ) (one carbon) (b) H9254 10.0 ppm (CH 3 ) H9254 22.7 ppm (CH 3 ) H9254 32.0 ppm (CH 2 ) H9254 69.2 ppm (CH) 13.27 Identify the C 6 H 14 isomers on the basis of their 13 C NMR spectra: (a) H9254 19.1 ppm (CH 3 ) (d) H9254 8.5 ppm (CH 3 ) H9254 33.9 ppm (CH) H9254 28.7 ppm (CH 3 ) (b) H9254 13.7 ppm (CH 3 ) H9254 30.2 ppm (C) H9254 22.8 ppm (CH 2 ) H9254 36.5 ppm (CH 2 ) H9254 31.9 ppm (CH 2 ) (e) H9254 14.0 ppm (CH 3 ) (c) H9254 11.1 ppm (CH 3 ) H9254 20.5 ppm (CH 2 ) H9254 18.4 ppm (CH 3 ) H9254 22.4 ppm (CH 3 ) H9254 29.1 ppm (CH 2 ) H9254 27.6 ppm (CH) H9254 36.4 ppm (CH) H9254 41.6 ppm (CH 2 ) 13.28 A compound (C 4 H 6 ) has two signals of approximately equal intensity in its 13 C NMR spec- trum; one is a CH 2 carbon at H9254 30.2 ppm, the other a CH at H9254 136 ppm. Identify the compound. 13.29 A compound (C 3 H 7 ClO 2 ) exhibited three peaks in its 13 C NMR spectrum at H9254 46.8 (CH 2 ), 63.5 (CH 2 ), and 72.0 ppm (CH). Excluding compounds that have Cl and OH on the same carbon, which are unstable, what is the most reasonable structure for this compound? 13.30 From among the compounds chlorobenzene, o-dichlorobenzene, and p-dichlorobenzene, choose the one that (a) Gives the simplest 1 H NMR spectrum (b) Gives the simplest 13 C NMR spectrum (c) Has three peaks in its 13 C NMR spectrum (d) Has four peaks in its 13 C NMR spectrum 13.31 Compounds A and B are isomers of molecular formula C 10 H 14 . Identify each one on the basis of the 13 C NMR spectra presented in Figure 13.41. 13.32 A compound (C 8 H 10 O) has the infrared and 1 H NMR spectra presented in Figure 13.42. What is its structure? 13.33 Deduce the structure of a compound having the mass spectrum and 1 H NMR spectrum pre- sented in Figure 13.43. 13.34 Figure 13.44 presents several types of spectroscopic data (IR, 1 H NMR, 13 C NMR, and mass spectra) for a particular compound. What is it? 538 CHAPTER THIRTEEN Spectroscopy Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Problems 539 20406080 Chemical shift (δ, ppm) (a) 120140160180200 100 0 CH Compound A CDCl 3 C CH 3 20406080 Chemical shift (δ, ppm) (b) 120140160180200 100 0 CH Compound B C C CH 3 CH 3 FIGURE 13.41 The 13 C NMR spectrum of (a) compound A and (b) compound B, isomers of C 10 H 14 (Problem 13.31). Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 540 CHAPTER THIRTEEN Spectroscopy (a) Wave number, cm H110021 Transmittance (%) (b) Chemical shift (δ, ppm) C 8 H 10 O 0.01.02.03.04.05.06.07.08.09.010.0 4.85.0 (ppm) 1.60 (ppm) 5 3 1 1 FIGURE 13.42 (a) Infrared and (b) 200-MHz 1 H NMR spectra of a compound C 8 H 10 O (Problem 13.32). Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Problems 541 (a) 10 20 30 40 50 60 70 80 90 100 110 120 130 134 140 150 0 20 40 60 80 100 Relative intensity m/z (b) Chemical shift (δ, ppm) 0.01.02.03.04.05.06.07.08.09.010.0 (ppm) 2.80 2.70 2.60 (ppm) 1.5 4 4 6 1.4 1.3 1.2 FIGURE 13.43 (a) Mass spectrum and (b) 200-MHz 1 H NMR spectrum of an unknown compound (Problem 13.33). Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 542 CHAPTER THIRTEEN Spectroscopy Wave number, cm H110021 Transmittance (%) (b) 100 20 40 60 80 0 86 Relative intensity (a) 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 m/z FIGURE 13.44 (a) Mass, (b) infrared, (c) 200-MHz 1 H NMR, and (d ) 13 C NMR spectra for the compound of Problem 13.34. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Problems 543 (c) Chemical shift (δ, ppm) 0.01.02.03.04.05.06.07.08.09.010.0 4 6 (ppm) 1.20 1.10 1.00 (ppm) 2.50 2.40 (d) Chemical shift (δ, ppm) 020406080100120140160180200 CH 3 CH 2 C Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 13.35 [18]-Annulene exhibits a 1 H NMR spectrum that is unusual in that in addition to a peak at H9254 8.8 ppm, it contains a second peak having a chemical shift H9254 of H110021.9 ppm. A negative value for the chemical shift H9254 indicates that the protons are more shielded than those of tetramethylsilane. This peak is 1.9 ppm upfield from the TMS peak. The high-field peak has half the area of the low- field peak. Suggest an explanation for these observations. 13.36 19 F is the only isotope of fluorine that occurs naturally, and it has a nuclear spin of H11006 . (a) Into how many peaks will the proton signal in the 1 H NMR spectrum of methyl fluoride be split? (b) Into how many peaks will the fluorine signal in the 19 F NMR spectrum of methyl fluoride be split? (c) The chemical shift of the protons in methyl fluoride is H9254 4.3 ppm. Given that the geminal 1 H± 19 F coupling constant is 45 Hz, specify the H9254 values at which peaks are observed in the proton spectrum of this compound at 200 MHz. 13.37 In general, the vicinal coupling constant between two protons varies with the angle between the C±H bonds of the H±C±C±H unit. The coupling constant is greatest when the protons are periplanar (dihedral angle H11005 0° or 180°) and smallest when the angle is approximately 90°. Describe, with the aid of molecular models, how you could distinguish between cis-1-bromo-2- chlorocyclopropane and its trans stereoisomer on the basis of their 1 H NMR spectra. 13.38 The H9266→H9266* transition in the UV spectrum of trans-stilbene (trans-C 6 H 5 CH?CHC 6 H 5 ) appears at 295 nm compared with 283 nm for the cis stereoisomer. The extinction coefficient H9280 max is approximately twice as great for trans-stilbene as for cis-stilbene. Both facts are normally inter- preted in terms of more effective conjugation of the H9266 electron system in trans-stilbene. Construct a molecular model of each stereoisomer, and identify the reason for the decreased effectiveness of conjugation in cis-stilbene. 13.39 31 P is the only phosphorus isotope present at natural abundance and has a nuclear spin of H11006 . The 1 H NMR spectrum of trimethyl phosphite, (CH 3 O) 3 P, exhibits a doublet for the methyl protons with a splitting of 12 Hz. (a) Into how many peaks is the 31 P signal split? (b) What is the difference in chemical shift (in hertz) between the lowest and highest field peaks of the 31 P multiplet? 13.40 We noted in section 13.13 that an NMR spectrum is an average spectrum of the conforma- tions populated by a molecule. From the following data, estimate the percentages of axial and equatorial bromine present in bromocyclohexane. (CH 3 ) 3 C Br H H9254 4.62 ppm (CH 3 ) 3 C H9254 3.81 ppm H Br H9254 3.95 ppm H Br 1 2 1 2 [18]-Annulene 544 CHAPTER THIRTEEN Spectroscopy The dependence of 3 J on di- hedral angle is referred to as the Karplus relationship af- ter Martin Karplus (Harvard University) who offered the presently accepted theoreti- cal treatment of it. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 13.41 Infrared spectroscopy is an inherently “faster” method than NMR, and an IR spectrum is a superposition of the spectra of the various conformations, rather than an average of them. When 1,2-dichloroethane is cooled below its freezing point, the crystalline material gives an IR spectrum consistent with a single species that has a center of symmetry. At room temperature, the IR spec- trum of liquid 1,2-dichloroethane retains the peaks present in the solid, but includes new peaks as well. Explain these observations. 13.42 Microwave spectroscopy is used to probe transitions between rotational energy levels in mol- ecules. (a) A typical wavelength for microwaves is 10 H110022 m, compared with 10 H110025 m for infrared radiation. Is the energy separation between rotational energy levels in a molecule greater or less than the separation between vibrational energy levels? (b) Microwave ovens cook food by heating the water in the food. Absorption of microwave radiation by the water excites it to a higher rotational energy state, and it gives off this excess energy as heat when it relaxes to its ground state. Why are vibrational and electronic energy states not involved in this process? 13.43 The peak in the UV-VIS spectrum of acetone [(CH 3 ) 2 C?O] corresponding to the n → H9266* transition appears at 279 nm when hexane is the solvent, but shifts to 262 nm in water. Which is more polar, the ground electronic state or the excited state? 13.44 A particular vibration will give an absorption peak in the infrared spectrum only if the dipole moment of the molecule changes during the vibration. Which vibration of carbon dioxide, the sym- metrical stretch or the antisymmetrical stretch, is “infrared-active”? 13.45 The protons in the methyl group shown in italics in the following structure are highly shielded and give a signal 0.38 ppm upfield from TMS. The other methyl group on the same car- bon has a more normal chemical shift of 0.86 ppm downfield from TMS. Why is the indicated methyl group so highly shielded? (Building a molecular model can help.) CH 3 CH 3 CH 3 CH 3 O?C?O ￠￡ Symmetrical stretch O?C?O ￡￡ Antisymmetrical stretch Problems 545 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website