Interpreting Mass Spectra What kinds of information can we get from a mass spectrum? The most obvious information is the molecular weight of the sample, which in itself can be invaluable. If we were given samples of hexane (MW = 86), 1-hexene (MW = 84), and 1-hexyne (MW = 82), for example, mass spectrometry would easily distinguish them. Some instruments, called double-focusing mass spectrometers , have two magnetic sectors in their mass analyzers, giving these spectrometers have such high resolution that they provide mass measurements accurate to 5 ppm, or about 0.0005 amu, making it possible to distinguish between two formulas with the same nominal mass. For example, both C 5 H 12 and C 4 H 8 O have MW = 72, but they differ slightly beyond the decimal point: C 5 H 12 has an exact mass of 72.0939 amu, whereas C 4 H 8 O has an exact mass of 72.0575 amu. A high-resolution instrument can easily distinguish between them. Note, however, that exact mass measurements refer to molecules with specific isotopic compositions. Thus, the sum of the exact atomic masses of the specific isotopes in a molecule is measured—1.007 83 amu for 1 H, 12.000 00 amu for 12 C, 14.003 07 amu for 14 N, 15.994 91 amu for 16 O, and so on—rather than the sum of the average atomic masses of elements, as found on a periodic table. Unfortunately, not every compound shows a molecular ion in its electron-impact mass spectrum. Although M + is usually easy to identify if it’s abundant, some compounds, such as 2,2-dimethylpropane, fragment so easily that no molecular ion is observed ( ). In such cases, alternative “soft” ionization methods that don’t use electron bombardment can prevent or minimize fragmentation (see Section 12.4). Mass spectrum of 2,2-dimethylpropane (C 5 H 12 ; MW = 72). No molecular ion is observed when electron-impact ionization is used. What do you think is the formula and structure of the M + peak at m / z = 57? Knowing the molecular weight makes it possible to narrow considerably the choices of molecular formula. For example, if the mass spectrum of an unknown compound shows a molecular ion at m / z = 110, the molecular formula is likely to be C 8 H 14 , C 7 H 10 O, C 6 H 6 O 2 , or C 6 H 10 N 2 . There are always a number of molecular formulas possible for all but the lowest molecular weights, and a computer can easily generate a list of the choices. A further point about mass spectrometry, noticeable in the spectra of both propane ( ) and 2,2-dimethylpropane ( ), is that the peak for the molecular ion is not at the highest m/z value. There is also a small peak at M + 1 due to the presence of different isotopes in the molecules. Although 12 C is the most abundant carbon isotope, a small amount (1.10% natural abundance) of 13 C is also present. Thus, a certain percentage of the molecules analyzed in the mass spectrometer are likely to contain a 13 C atom, giving rise to the observed M + 1 peak. In addition, a small amount of 2 H (deuterium; 0.015% natural abundance) is present, making a further contribution to the M + 1 peak. Mass spectrometry would be useful even if molecular weight and formula were the only information that could be obtained, but in fact it provides much more. For one thing, the mass spectrum of a compound serves as a kind of “molecular fingerprint.” Every organic compound fragments in a unique way depending on its structure, and the likelihood of two compounds having identical mass spectra is small. Thus, it’s sometimes possible to identify an unknown by computer-based matching of its mass spectrum to one of the more than 785,061 searchable spectra recorded in a database called the Registry of Mass Spectral Data . It’s also possible to derive structural information about a molecule by interpreting its fragmentation pattern. Fragmentation occurs when the high-energy cation radical flies apart by spontaneous cleavage of a chemical bond. One of the two fragments retains the positive charge and is a carbocation, while the other fragment is a neutral radical. Not surprisingly, the positive charge often remains with the fragment that is best able to stabilize it. In other words, a relatively stable carbocation is often formed during fragmentation. For example, 2,2-dimethylpropane tends to fragment in such a way that the positive charge remains with the tert -butyl group. 2,2-Dimethylpropane therefore has a base peak at m/z = 57, corresponding to C 4 H 9 + ( ). Because mass-spectral fragmentation patterns are usually complex, it’s often difficult to assign structures to fragment ions. Most hydrocarbons fragment in many ways, as demonstrated by the mass spectrum of hexane in . The hexane spectrum shows a moderately abundant molecular ion at m / z = 86 and fragment ions at m/z = 71, 57, 43, and 29. Since all the carbon–carbon bonds of hexane are electronically similar, all break to a similar extent, giving rise to the observed mixture of ions. Mass spectrum of hexane (C 6 H 14 ; MW = 86). The base peak is at m / z = 57, and numerous other ions are present. shows how the hexane fragments might arise. The loss of a methyl radical (CH 3 , M = 15) from the hexane cation radical (M + = 86) gives rise to a fragment of mass 86 – 15 = 71; the loss of an ethyl radical (C 2 H 5 , M = 29) accounts for a fragment of mass 86 – 29 = 57; the loss of a propyl radical (C 3 H 7 , M = 43) accounts for a fragment of mass 86 – 43 = 43; and the loss of a butyl radical accounts for a fragment of mass 29. With practice, it’s sometimes possible to analyze the fragmentation pattern of an unknown compound and work backward to a structure that is compatible with the data. Fragmentation of hexane in a mass spectrometer. We’ll see in the next section and in later chapters that specific functional groups, such as alcohols, ketones, aldehydes, and amines, show specific kinds of mass spectral fragmentations that can be interpreted to provide structural information. Using Mass Spectra to Identify Compounds Assume that you have two unlabeled samples, one of methylcyclohexane and the other of ethylcyclopentane. How could you use mass spectrometry to tell them apart? The mass spectra of both are shown in . Mass spectra of unlabeled samples A and B for . Strategy Look at the possible structures and decide on how they differ. Then think about how any of these differences in structure might give rise to differences in mass spectra. Methyl cyclohexane, for instance, has a –CH 3 group, and ethylcyclopentane has a –CH 2 CH 3 group, which should affect the fragmentation patterns. Solution Both mass spectra show molecular ions at M + = 98, corresponding to C 7 H 14 , but they differ in their fragmentation patterns. Sample A has its base peak at m / z = 69, corresponding to the loss of a CH 2 CH 3 group (29 mass units), but B has a rather small peak at m / z = 69. Sample B shows a base peak at m / z = 83, corresponding to the loss of a CH 3 group (15 mass units), but sample A has only a small peak at m / z = 83. We can therefore be reasonably certain that A is ethylcyclopentane and B is methylcyclohexane. Problems

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Mass Spectrometry of Some Common Functional Groups As each functional group is discussed in future chapters, mass-spectral fragmentations characteristic of that group will be described. As a preview, though, we’ll point out some distinguishing features of several common functional groups. Alcohols Alcohols undergo fragmentation in a mass spectrometer by two pathways: alpha cleavage and dehydration . In the α -cleavage pathway, a C–C bond nearest the hydroxyl group is broken, yielding a neutral radical plus a resonance-stabilized, oxygen-containing cation. This type of fragmentation is seen in the spectrum of 2-pentanol in . Mass spectrum of 2-pentanol. In the dehydration pathway, water is eliminated, yielding an alkene radical cation with a mass 18 amu less than M + . For simplicity, we have drawn the dehydration below as an E2-type process. Often the hydrogen that is lost is not beta to the hydroxyl. Only a small peak from dehydration is observed in the spectrum of 2-pentanol ( ). Amines The nitrogen rule of mass spectrometry says that a compound with an odd number of nitrogen atoms has an odd-numbered molecular weight. The logic behind the rule comes from the fact that nitrogen is trivalent, thus requiring an odd number of hydrogen atoms. The presence of nitrogen in a molecule is often detected simply by observing its mass spectrum. An odd-numbered molecular ion usually means that the unknown compound has one or three nitrogen atoms, and an even-numbered molecular ion usually means that a compound has either zero or two nitrogen atoms. Aliphatic amines undergo a characteristic α cleavage in a mass spectrometer, similar to that observed for alcohols. A C–C bond nearest the nitrogen atom is broken, yielding an alkyl radical and a resonance-stabilized, nitrogen-containing cation. The mass spectrum of triethylamine has a base peak at m / z = 86, which arises from an alpha cleavage resulting in the loss of a methyl group ( ). Mass spectrum of triethylamine. Halides The fact that some elements have two common isotopes gives their mass spectra a distinctive appearance. Chlorine, for example, exists as two isotopes, 35 Cl and 37 Cl, in roughly a 3 : 1 ratio. In a sample of chloroethane, three out of four molecules contain a 35 Cl atom and one out of four has a 37 Cl atom. In the mass spectrum of chloroethane ( we see the molecular ion (M) at m / z = 64 for ions that contain a 35 Cl and another peak at m / z = 66, called the M + 2 peak, for ions containing a 37 Cl. The ratio of the relative abundance of M : M + 2 is about 3 : 1, a reflection of the isotopic abundances of chlorine. Mass spectrum of chloroethane. In the case of bromine, the isotopic distribution is 50.7% 79 Br and 49.3% 81 Br. In the mass spectrum of 1-bromohexane ( ) the molecular ion appears at m / z = 164 for 79 Br-containing ions and the M + 2 peak is at m / z = 166 for 81 Br-containing ions. The ions at m / z = 135 and 137 are informative as well. The two nearly equally large peaks tell us that the ions at those m / z values still contain the bromine atom. The peak at m / z = 85, on the other hand, does not contain bromine because there is not a large peak at m / z = 87. Mass spectrum of 1-bromohexane. Carbonyl Compounds Ketones and aldehydes that have a hydrogen on a carbon three atoms away from the carbonyl group undergo a characteristic mass-spectral cleavage called the McLafferty rearrangement . The hydrogen atom is transferred to the carbonyl oxygen, a C–C bond between the alpha and beta carbons is broken, and a neutral alkene fragment is produced. The charge remains with the oxygen-containing fragment. In addition, ketones and aldehydes frequently undergo α cleavage of the bond between the carbonyl carbon and the neighboring carbon to yield a neutral radical and a resonance-stabilized acyl cation. Because the carbon neighboring the carbonyl carbon is called the alpha carbon, the reaction is called an alpha cleavage. (To be more general about neighboring positions in carbonyl compounds, Greek letters are used in alphabetical order: alpha, beta, gamma, delta, and so on.) The mass spectrum of butyrophenone illustrates both alpha cleavage and the McLafferty rearrangement ( ). Alpha cleavage of the propyl substituent results in the loss of C 3 H 7 = 43 mass units from the precursor ion at m / z = 148 to give the fragment ion at m / z = 105. A McLafferty rearrangement of butyrophenone results in the loss of ethylene, C 2 H 4 = 28 mass units, from the precursor leaving the ion at m / z = 120. Mass spectrum of butyrophenone. Identifying Fragmentation Patterns in a Mass Spectrum The mass spectrum of 2-methyl-3-pentanol is shown in . What fragments can you identify? Mass spectrum of 2-methyl-3-pentanol, for . Strategy Calculate the mass of the molecular ion, and identify the functional groups in the molecule. Then write the fragmentation processes you might expect, and compare the masses of the resultant fragments with the peaks present in the spectrum. Solution 2-Methyl-3-pentanol, an open-chain alcohol, has M + = 102 and might be expected to fragment by α cleavage and by dehydration. These processes would lead to fragment ions of m / z = 84, 73, and 59. Of the three expected fragments, dehydration is not observed (no m / z = 84 peak), but both α cleavages take place ( m / z = 73, 59). Problems

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Mass Spectrometry in Biological Chemistry: Time-of-Flight (TOF) Instruments MS analyses of sensitive biological samples rarely use magnetic sector ionization. Instead, they typically use either electrospray ionization ( ESI ) or matrix-assisted laser desorption ionization ( MALDI ), typically linked to a time-of-flight ( TOF ) mass analyzer. Both ESI and MALDI are soft ionization methods that produce charged molecules with little fragmentation, even with sensitive biological samples of very high molecular weight. In an ESI source, as a sample solution exits the tube, it is subjected to a high voltage that causes the droplets to become charged. The sample molecules gain one or more protons from charged solvent molecules in the droplet. The volatile solvent quickly evaporates, giving variably protonated sample molecules (M + H n n + ). In a MALDI source, the sample is adsorbed onto a suitable matrix compound, such as 2,5-dihydroxybenzoic acid, which is ionized by a short burst of laser light. The matrix compound then transfers the energy to the sample and protonates it, forming M + H n n + ions. Following ion formation, the variably protonated sample molecules are electrically focused into a small packet with a narrow spatial distribution, and the packet is given a sudden kick of energy by an accelerator electrode. As each molecule in the packet is given the same energy, E = mv 2 /2, it begins moving with a velocity that depends on the square root of its mass, v = 2 E / m . Lighter molecules move faster, and heavier molecules move slower. The analyzer itself—the drift tube —is simply an electrically grounded metal tube inside which the different charged molecules become separated as they move at different velocities and take different amounts of time to complete their flight. The Time of Flight technique is considerably more sensitive than the magnetic sector alternative, and protein samples of up to 100 kilodaltons (100,000 amu) can be separated with a mass accuracy of 3 ppm. shows a MALDI–TOF spectrum of chicken egg-white lysozyme, MW = 14,306.7578 daltons. Biochemists generally use the unit dalton , abbreviated Da, instead of amu, although the two are equivalent (1 dalton = 1 amu). MALDI–TOF mass spectrum of chicken egg-white lysozyme. The peak at 14,306.7578 daltons (amu) is due to the monoprotonated protein, M + H + , and the peak at 28,614.2188 daltons is due to an impurity formed by dimerization of the protein. Other peaks at lower m / z values are various protonated species, M + H n n + .

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Spectroscopy and the Electromagnetic Spectrum Infrared, ultraviolet, and nuclear magnetic resonance spectroscopies differ from mass spectrometry in that they are nondestructive and involve the interaction of molecules with electromagnetic energy rather than with an ionizing source. Before beginning a study of these techniques, however, let’s briefly review the nature of radiant energy and the electromagnetic spectrum. Visible light, X rays, microwaves, radio waves, and so forth are all different kinds of electromagnetic radiation. Collectively, they make up the electromagnetic spectrum , shown in . The electromagnetic spectrum is arbitrarily divided into regions, with the familiar visible region accounting for only a small portion, from 3.8 × 10 –7 m to 7.8 × 10 –7 m in wavelength. The visible region is flanked by the infrared and ultraviolet regions. The electromagnetic spectrum covers a continuous range of wavelengths and frequencies, from radio waves at the low-frequency end to gamma ( γ ) rays at the high-frequency end. The familiar visible region accounts for only a small portion near the middle of the spectrum. Electromagnetic radiation is often said to have dual behavior. In some respects, it has the properties of a particle, called a photon , yet in other respects it behaves as an energy wave. Like all waves, electromagnetic radiation is characterized by a wavelength , a frequency , and an amplitude ( ). The wavelength, λ (Greek lambda), is the distance from one wave maximum to the next. The frequency, ν (Greek nu), is the number of waves that pass by a fixed point per unit time, usually given in reciprocal seconds (s –1 ), or hertz, Hz (1 Hz = 1 s –1 ). The amplitude is the height of a wave, measured from midpoint to peak. The intensity of radiant energy, whether a feeble glow or a blinding glare, is proportional to the square of the wave’s amplitude. Electromagnetic waves are characterized by a wavelength, a frequency, and an amplitude. (a) Wavelength ( λ ) is the distance between two successive wave maxima. Amplitude is the height of the wave measured from the center. (b)–(c) What we perceive as different kinds of electromagnetic radiation are simply waves with different wavelengths and frequencies. Multiplying the wavelength of a wave in meters (m) by its frequency in reciprocal seconds (s –1 ) gives the speed of the wave in meters per second (m/s). The rate of travel of all electromagnetic radiation in a vacuum is a constant value, commonly called the “speed of light” and abbreviated c. Its numerical value is defined as exactly 2.997 924 58 × 10 8 m/s, usually rounded off to 3.00 × 10 8 m/s. Wavelength × Frequency = Speed λ ( m ) × ν ( s − 1 ) = c ( m/s ) λ = c v or v = c λ Just as matter comes only in discrete units called atoms, electromagnetic energy is transmitted only in discrete amounts called quanta . The amount of energy ε corresponding to 1 quantum of energy (1 photon ) of a given frequency ν is expressed by the Planck equation ε = h v = h c λ where h = Planck’s constant (6.62 × 10 –34 J ‧ s = 1.58 × 10 –34 cal ‧ s). The Planck equation says that the energy of a given photon varies directly with its frequency ν but inversely with its wavelength λ . High frequencies and short wavelengths correspond to high-energy radiation such as gamma rays; low frequencies and long wavelengths correspond to low-energy radiation such as radio waves. Multiplying ε by Avogadro’s number N A gives the same equation in more familiar units, where E represents the energy of Avogadro’s number (one “mole”) of photons of wavelength λ : E = N A h c λ = 1.20 × 10 − 4 kJ/mol λ ( m ) or 2.86 × 10 − 5 kcal/mol λ ( m ) When an organic compound is exposed to a beam of electromagnetic radiation, it absorbs energy of some wavelengths but passes, or transmits, energy of other wavelengths. If we irradiate the sample with energy of many different wavelengths and determine which are absorbed and which are transmitted, we can measure the absorption spectrum of the compound. An example of an absorption spectrum—that of ethanol exposed to infrared radiation—is shown in . The horizontal axis records the wavelength, and the vertical axis records the intensity of the various energy absorptions in percent transmittance. The baseline corresponding to 0% absorption (or 100% transmittance) runs along the top of the chart, so a downward spike means that energy absorption has occurred at that wavelength. An infrared absorption spectrum for ethanol, CH 3 CH 2 OH. A transmittance of 100% means that all the energy is passing through the sample, whereas a lower transmittance means that some energy is being absorbed. Thus, each downward spike corresponds to an energy absorption. The energy a molecule gains when it absorbs radiation must be distributed over the molecule in some way. With infrared radiation, the absorbed energy causes bonds to stretch and bend more vigorously. With ultraviolet radiation, the energy causes an electron to jump from a lower-energy orbital to a higher-energy one. Different radiation frequencies affect molecules in different ways, but each provides structural information when the results are interpreted. There are many kinds of spectroscopies, which differ according to the region of the electromagnetic spectrum used. We’ll look at three: infrared spectroscopy, ultraviolet spectroscopy, and nuclear magnetic resonance spectroscopy. Let’s begin by seeing what happens when an organic sample absorbs infrared energy. Correlating Energy and Frequency of Radiation Which is higher in energy, FM radio waves with a frequency of 1.015 × 10 8 Hz (101.5 MHz) or visible green light with a frequency of 5 × 10 14 Hz? Strategy Remember the equations ε = hν and ε = hc / λ, which say that energy increases as frequency increases and as wavelength decreases. Solution Since visible light has a higher frequency than radio waves, it is higher in energy. Problems

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Infrared Spectroscopy In infrared (IR) spectroscopy , the IR region of the electromagnetic spectrum covers the range from just above the visible (7.8 × 10 –7 m) to approximately 10 –4 m, but only the midportion from 2.5 × 10 –6 m to 2.5 × 10 –5 m is used by organic chemists ( . Wavelengths within the IR region are usually given in micrometers (1 μ m = 10 –6 m), and frequencies are given in wavenumbers rather than in hertz. The wavenumber v ˜ is the reciprocal of wavelength in centimeters and is therefore expressed in units of cm –1 . Wavenumber: v ˜ ( cm − 1 ) = 1 λ ( cm ) Thus, the useful IR region is from 4000 to 400 cm –1 , corresponding to energies of 48.0 kJ/mol to 4.80 kJ/mol (11.5–1.15 kcal/mol). The infrared and adjacent regions of the electromagnetic spectrum. Why does an organic molecule absorb some wavelengths of IR radiation but not others? All molecules have a certain amount of energy and are in constant motion. Their bonds stretch and contract, atoms wag back and forth, and other molecular vibrations occur. Some of the kinds of allowed vibrations are shown below: The amount of energy a molecule contains is not continuously variable but is quantized. That is, a molecule can stretch or bend only at specific frequencies corresponding to specific energy levels. Take bond stretching, for example. Although we usually speak of bond lengths as if they were fixed, the numbers given are really averages. In fact, a typical C–H bond with an average bond length of 110 pm is actually vibrating at a specific frequency, alternately stretching and contracting as if there were a spring connecting the two atoms. When a molecule is irradiated with electromagnetic radiation, energy is absorbed if the frequency of the radiation matches the frequency of the vibration. The result of this energy absorption is an increased amplitude for the vibration; in other words, the “spring” connecting the two atoms stretches and compresses a bit further. Since each frequency absorbed by a molecule corresponds to a specific molecular motion, we can find what kinds of motions a molecule has by measuring its IR spectrum. By interpreting these motions, we can find out what kinds of bonds (functional groups) are present in the molecule. IR spectrum → What molecular motions? → What functional groups?

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Interpreting Infrared Spectra The complete interpretation of an IR spectrum is difficult because most organic molecules have dozens of different bond stretching and bending motions, and thus have dozens of absorptions. On the one hand, this complexity is a problem because it generally limits the laboratory use of IR spectroscopy to pure samples of fairly small molecules—little can be learned from IR spectroscopy about large, complex biomolecules. On the other hand, this complexity is useful because an IR spectrum acts as a unique fingerprint of a compound. In fact, the complex region of the IR spectrum, from 1500 cm –1 to around 400 cm –1 , is called the fingerprint region . If two samples have identical IR spectra, they are almost certainly identical compounds. Fortunately, we don’t need to interpret an IR spectrum fully to get useful structural information. Most functional groups have characteristic IR absorption bands that don’t change much from one compound to another. The C=O absorption of a ketone is almost always in the range 1680 to 1750 cm –1 ; the O–H absorption of an alcohol is almost always in the range 3400 to 3650 cm –1 ; the C=C absorption of an alkene is almost always in the range 1640 to 1680 cm –1 ; and so forth. By learning where characteristic functional-group absorptions occur, it’s possible to get structural information from IR spectra. lists the characteristic IR bands of some common functional groups. Characteristic IR Absorptions of Some Functional Groups Functional Group Absorption (cm –1 ) Intensity Alkane C–H 2850–2960 Medium Alkene =C–H 3020–3100 Medium C=C 1640–1680 Medium Alkyne ≡C–H 3300 Strong C≡C 2100–2260 Medium Alkyl halide C–Cl 600–800 Strong C–Br 500–600 Strong Alcohol O–H 3400–3650 Strong, broad C–O 1050–1150 Strong Arene C–H 3030 Weak Aromatic ring 1660–2000 Weak 1450–1600 Medium Amine N–H 3300–3500 Medium C–N 1030–1230 Medium Carbonyl compound C═O 1670–1780 Strong Aldehyde 1730 Strong Ketone 1715 Strong Ester 1735 Strong Amide 1690 Strong Carboxylic acid 1710 Strong Carboxylic acid O–H 2500–3100 Strong, broad Nitrile C≡N 2210–2260 Medium Nitro NO 2 1540 Strong Look at the IR spectra of hexane, 1-hexene, and 1-hexyne in to see an example of how IR spectroscopy can be used. Although all three IR spectra contain many peaks, there are characteristic absorptions of C═C and C≡C functional groups that allow the three compounds to be distinguished. Thus, 1-hexene shows a characteristic C═C absorption at 1660 cm –1 and a vinylic =C–H absorption at 3100 cm –1 , whereas 1-hexyne has a C≡C absorption at 2100 cm –1 and a terminal alkyne ≡C – H absorption at 3300 cm –1 . IR spectra of (a) hexane, (b) 1-hexene, and (c) 1-hexyne. Spectra like these are easily obtained from sub-milligram amounts of material in a few minutes using commercially available instruments. It helps in remembering the position of specific IR absorptions to divide the IR region from 4000 cm –1 to 400 cm –1 into four parts, as shown in . The four regions of the infrared spectrum: single bonds to hydrogen, triple bonds, double bonds, and fingerprint. The region from 4000 to 2500 cm –1 corresponds to absorptions caused by N–H, C–H, and O–H single-bond stretching motions. N–H and O–H bonds absorb in the 3300 to 3600 cm –1 range; C–H bond stretching occurs near 3000 cm –1 . The region from 2500 to 2000 cm –1 is where triple-bond stretching occurs. Both C≡N and C≡C bonds absorb here. The region from 2000 to 1500 cm –1 is where double bonds ( C═O , C═N , and C═C ) absorb. Carbonyl groups generally absorb in the range 1680 to 1750 cm –1 , and alkene stretching normally occurs in the narrow range of 1640 to 1680 cm –1 . The region below 1500 cm –1 is the fingerprint portion of the IR spectrum. A large number of absorptions due to a variety of C–C, C–O, C–N, and C–X single-bond vibrations occur here. Why do different functional groups absorb where they do? As noted previously, a good analogy is that of two weights (atoms) connected by a spring (a bond). Short, strong bonds vibrate at a higher energy and higher frequency than do long, weak bonds, just as a short, strong spring vibrates faster than a long, weak spring. Thus, triple bonds absorb at a higher frequency than double bonds, which in turn absorb at a higher frequency than single bonds. In addition, C–H, O–H, and N–H bonds vibrate at a higher frequency than bonds between heavier C, O, and N atoms. Distinguishing Isomeric Compounds by IR Spectroscopy Acetone (CH 3 COCH 3 ) and 2-propen-1-ol ( H 2 C═ CHCH 2 OH ) are isomers. How could you distinguish them by IR spectroscopy? Strategy Identify the functional groups in each molecule, and refer to . Solution Acetone has a strong C=O absorption at 1715 cm –1 , while 2-propen-1-ol has an –OH absorption at 3500 cm –1 and a C=C absorption at 1660 cm –1 . Problems

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Infrared Spectra of Some Common Functional Groups As each functional group is discussed in future chapters, the spectroscopic properties of that group will be described. For the present, we’ll point out some distinguishing features of the hydrocarbon functional groups already studied and briefly preview some other common functional groups. We should also point out, however, that in addition to interpreting absorptions that are present in an IR spectrum, it’s also possible to get structural information by noticing which absorptions are not present. If the spectrum of a compound has no absorptions at 3300 and 2150 cm –1 , the compound is not a terminal alkyne; if the spectrum has no absorption near 3400 cm –1 , the compound is not an alcohol; and so on. Alkanes The IR spectrum of an alkane is fairly uninformative because no functional groups are present and all absorptions are due to C–H and C–C bonds. Alkane C–H bonds show a strong absorption from 2850 to 2960 cm –1 , and saturated C–C bonds show a number of bands in the 800 to 1300 cm –1 range. Since most organic compounds contain saturated alkane-like portions, most organic compounds have these characteristic IR absorptions. The C–H and C–C bands are clearly visible in the three spectra shown previously in . Alkenes Alkenes show several characteristic stretching absorptions. Vinylic =C–H bonds absorb from 3020 to 3100 cm –1 , and alkene C═C bonds usually absorb near 1650 cm –1 , although in some cases their peaks can be rather small and difficult to see clearly when the alkene is symmetric, or nearly so. Both absorptions are visible in the 1-hexene spectrum in b. Alkenes have characteristic =C–H out-of-plane bending absorptions in the 700 to 1000 cm –1 range, thereby allowing the substitution pattern on a double bond to be determined ( ). For example, monosubstituted alkenes such as 1-hexene show strong characteristic bands at 910 and 990 cm –1 , and 1,1-disubstituted alkenes ( R 2 C═ CH 2 ) have an intense band at 890 cm –1 . C–H out-of-plane bending vibrations for substituted alkenes. Alkynes Alkynes show a C≡C stretching absorption at 2100 to 2260 cm –1 , an absorption that is much more intense for terminal alkynes than for internal alkynes. Terminal alkynes such as 1-hexyne also have a characteristic ≡C–H stretching absorption at 3300 cm –1 ( c). This band is diagnostic for terminal alkynes because it is fairly intense and quite sharp. Aromatic Compounds Aromatic compounds, such as benzene, have a weak C–H stretching absorption at 3030 cm –1 , just to the left of a typical saturated C–H band. In addition, they have a series of weak absorptions in the 1660 to 2000 cm –1 range and a series of medium-intensity absorptions in the 1450 to 1600 cm –1 region. These latter absorptions are due to complex molecular motions of the entire ring. The C–H out-of-plane bending region for benzene derivatives, between 650 to 1000 cm –1 , gives valuable information about the ring’s substitution pattern, as it does for the substitution pattern of alkenes ( ). C–H out-of-plane bending vibrations for substituted benzenes. The IR spectrum of phenylacetylene, shown in Figure 12.29 at the end of this section, gives an example, clearly showing the following absorbances: ≡C–H stretch at 3300 cm –1 , C–H stretches from the benzene ring at 3000 to 3100 cm –1 , C═C stretches of the benzene ring between 1450 and 1600 cm –1 , and out-of-plane bending of the ring’s C–H groups, indicating monosubstitution at 750 cm –1 . Alcohols The O–H functional group of alcohols is easy to spot. Alcohols have a characteristic band in the range 3400 to 3650 cm –1 that is usually broad and intense. Hydrogen bonding between O–H groups is responsible for making the absorbance so broad. If an O–H stretch is present, it’s hard to miss this band or to confuse it with anything else. Cyclohexanol ( ) is a good example. IR spectrum of cyclohexanol. Amines The N–H functional group of amines is also easy to spot in the IR, with a characteristic absorption in the 3300 to 3500 cm –1 range. Although alcohols absorb in the same range, an N–H absorption band is much sharper and less intense than an O–H band. Primary amines (R–NH 2 ) have two absorbances—one for the symmetric stretching mode and one for the asymmetric mode ( ). Secondary amines (R 2 N–H) only have one N–H stretching absorbance in this region. IR spectrum of cyclohexylamine. Carbonyl Compounds Carbonyl functional groups are the easiest to identify of all IR absorptions because of their sharp, intense peak in the range 1670 to 1780 cm –1 . Most important, the exact position of absorption within this range can often be used to identify the exact kind of carbonyl functional group—aldehyde, ketone, ester, and so forth. ALDEHYDES Saturated aldehydes absorb at 1730 cm –1 ; aldehydes next to either a double bond or an aromatic ring absorb at 1705 cm –1 . The C–H group attached to the carbonyl is responsible for the characteristic IR absorbance for aldehydes at 2750 and 2850 cm –1 ( ). Although these are not very intense, the absorbance at 2750 cm –1 is helpful when trying to distinguish between an aldehyde and a ketone. The IR spectrum of benzaldehyde. KETONES Saturated open-chain ketones and six-membered cyclic ketones absorb at 1715 cm –1 . Ring strain stiffens the C═O bond, making five-membered cyclic ketones absorb at 1750 cm –1 and four-membered cyclic ketones absorb at 1780 cm –1 , about 20 to 30 cm –1 lower than the corresponding saturated ketone. ESTERS Saturated esters have a C═O absorbance at 1735 cm –1 and two strong absorbances in the 1300 to 1000 cm –1 range from the C–O portion of the functional group. Like other carbonyl functional groups, esters next to either an aromatic ring or a double bond absorb at 1715 cm –1 , about 20 to 30 cm –1 lower than a saturated ester. Predicting IR Absorptions of Compounds Where might the following compounds have IR absorptions? Strategy Identify the functional groups in each molecule, and then check to see where those groups absorb. Solution (a) Absorptions: 3400 to 3650 cm –1 (O–H), 3020 to 3100 cm –1 (=C–H), 1640 to 1680 cm –1 ( C═C ). This molecule has an alcohol O–H group and an alkene double bond. (b) Absorptions: 3300 cm –1 ( ≡C–H ), 2100 to 2260 cm –1 ( C≡C ), 1735 cm –1 ( C═O ). This molecule has a terminal alkyne triple bond and a saturated ester carbonyl group. Identifying Functional Groups from an IR Spectrum The IR spectrum of an unknown compound is shown in . What functional groups does the compound contain? IR spectrum for . Strategy All IR spectra have many absorptions, but those useful for identifying specific functional groups are usually found in the region from 1500 cm –1 to 3300 cm –1 . Pay particular attention to the carbonyl region (1670 to 1780 cm –1 ), the aromatic region (1660 to 2000 cm –1 ), the triple-bond region (2000 to 2500 cm –1 ), and the C–H region (2500 to 3500 cm –1 ). Solution The spectrum shows an intense absorption at 1725 cm –1 due to a carbonyl group (perhaps an aldehyde, –CHO), a series of weak absorptions from 1800 to 2000 cm –1 characteristic of aromatic compounds, and a C–H absorption near 3030 cm –1 , also characteristic of aromatic compounds. In fact, the compound is phenylacetaldehyde. Problems X-Ray Crystallography The various spectroscopic techniques described in this and the next two chapters are enormously important in chemistry and have been fine-tuned to such a degree that the structure of almost any molecule can be found. Nevertheless, wouldn’t it be nice if you could simply look at a molecule and “see” its structure with your eyes? Determining the three-dimensional shape of an object around you is easy—you just look at it, let your eyes focus the light rays reflected from the object, and let your brain assemble the data into a recognizable image. If the object is small, you use a microscope and let the microscope lens focus the visible light. Unfortunately, there is a limit to what you can see, even with the best optical microscope. Called the diffraction limit, you can’t see anything smaller than the wavelength of light you are using for the observation. Visible light has wavelengths of several hundred nanometers, but atoms in molecules have dimensions on the order of 0.1 nm. Thus, to see a molecule—whether a small one in the laboratory or a large, complex enzyme with a molecular weight in the tens of thousands—you need wavelengths in the 0.1 nm range, which corresponds to X rays. Let’s say that we want to determine the structure and shape of an enzyme or other biological molecule. The technique used is called X-ray crystallography . First, the molecule is crystallized (which often turns out to be the most difficult and time-consuming part of the entire process) and a small crystal of 0.4 to 0.5 mm on its longest axis is glued to the end of a glass fiber. The fiber and attached crystal are then mounted in an instrument called an X-ray diffractometer, which consists of a radiation source, a sample positioning and orienting device that can rotate the crystal in any direction, a detector, and a controlling computer. Once mounted in the diffractometer, the crystal is irradiated with X rays, usually so-called Cu K α radiation with a wavelength of 0.154 nm. When the X rays strike the enzyme crystal, they interact with electrons in the molecule and are scattered into a diffraction pattern which, when detected and visualized, appears as a series of intense spots against a null background. The structure of human muscle fructose-1,6-bisphosphate aldolase, as determined by X-ray crystallography. (credit: modification of work Protein Data Bank, 1ALD. PDB ID: 1ALD, Gamblin, S.J. Davies, G.J. Grimes, J.M. Jackson, R.M. Littlechild, J.A. Watson, H.C. (1991) J. Mol. Biol. 219: 573-576, CC BY 1.0.) Manipulation of the diffraction pattern to extract three-dimensional molecular data is a complex process, but the final result is an electron-density map of the molecule. Because electrons are largely localized around atoms, any two centers of electron density located within bonding distance of each other are assumed to represent bonded atoms, leading to a recognizable chemical structure. So important is this structural information for biochemistry that an online database of approximately 145,000 biological substances has been created. Operated by Rutgers University and funded by the U.S. National Science Foundation, the Protein Data Bank (PDB) is a worldwide repository for processing and distributing three-dimensional structural data for biological macromolecules. We’ll see how to access the PDB in the Chapter 26 Chemistry Matters . absorption spectrum amplitude base peak cation radical electromagnetic spectrum frequency, ν hertz, Hz, infrared (IR) spectroscopy MALDI mass spectrometry (MS) McLafferty Rearrangement photon precursor peak quadrupole mass analyzer Time of Flight (TOF) wavelength, λ wavenumber, v ˜ Summary Finding the structure of a new molecule, whether a small one synthesized in the laboratory or a large protein found in living organisms, is central to the progression of chemistry and biochemistry. The structure of an organic molecule is usually determined using spectroscopic methods, including mass spectrometry and infrared spectroscopy. Mass spectrometry (MS) tells the molecular weight and formula of a molecule; infrared (IR) spectroscopy identifies the functional groups present in the molecule. In small-molecule mass spectrometry, molecules are first ionized by collision with a high-energy electron beam. The ions then fragment into smaller pieces, which are magnetically sorted according to their mass-to-charge ratio ( m / z ). The ionized sample molecule is called the molecular ion, M + , and measurement of its mass gives the molecular weight of the sample. Structural clues about unknown samples can be obtained by interpreting the fragmentation pattern of the molecular ion. Mass-spectral fragmentations are usually complex, however, and interpretation is often difficult. In biological mass spectrometry, molecules are protonated using either electrospray ionization (ESI) or matrix-assisted laser desorption ionization (MALDI), and the protonated molecules are separated by time-of-flight (TOF) mass analysis. Infrared spectroscopy involves the interaction of a molecule with electromagnetic radiation . When an organic molecule is irradiated with infrared energy, certain frequencies are absorbed by the molecule. The frequencies absorbed correspond to the amounts of energy needed to increase the amplitude of specific molecular vibrations such as bond stretching and bending. Since every functional group has a characteristic combination of bonds, every functional group has a characteristic set of infrared absorptions. For example, the terminal alkyne ≡C–H bond absorbs IR radiation of 3300 cm –1 , and the alkene C═C bond absorbs in the range 1640 to 1680 cm –1 . By observing which frequencies of infrared radiation are absorbed by a molecule and which are not, it’s possible to determine the functional groups a molecule contains. Additional Problems Visualizing Chemistry Mass Spectrometry Infrared Spectroscopy General Problems

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Why This Chapter? NMR spectroscopy is an invaluable aid in carrying out the design and synthesis of new drugs. (credit: modification of work by Unknown/Pxhere, CC0 1.0) Nuclear magnetic resonance (NMR) spectroscopy has far-reaching applications in many scientific fields, particularly in chemical structure determination. Although we’ll just give an overview of the subject in this chapter, focusing on NMR applications with small molecules, more advanced NMR techniques are also used in biological chemistry to study protein structure and folding. Nuclear magnetic resonance (NMR) spectroscopy is the most valuable spectroscopic technique available to organic chemists. It’s the method of structure determination that organic chemists usually turn to first. We saw in the chapter on Structure Determination: Mass Spectrometry and Infrared Spectroscopy that mass spectrometry gives a molecule’s formula and infrared spectroscopy identifies a molecule’s functional groups. Nuclear magnetic resonance spectroscopy complements these other techniques by mapping a molecule’s carbon–hydrogen framework. Taken together, MS, IR, and NMR make it possible to determine the structures of even very complex molecules. Technique Examines Mass spectrometry Molecular size and formula Infrared spectroscopy Functional groups present NMR spectroscopy Map of carbon–hydrogen framework

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Nuclear Magnetic Resonance Spectroscopy Many kinds of nuclei behave as if they were spinning about an axis, somewhat as the earth spins daily. Because they’re positively charged, these spinning nuclei act like tiny magnets and can interact with an external magnetic field, denoted B 0 . Not all nuclei act this way, but fortunately for organic chemists, both the proton ( 1 H) and the 13 C nucleus do have spins. The more common 12 C isotope, however, does not have nuclear spin. (In speaking about NMR, the words proton and hydrogen are often used interchangeably, since a hydrogen nucleus is just a proton.) Let’s see what the consequences of nuclear spin are and how we can use the results. In the absence of an external magnetic field, the spins of magnetic nuclei are oriented randomly. When a sample containing these nuclei is placed between the poles of a strong magnet, however, the nuclei adopt specific orientations, much as a compass needle orients in the earth’s magnetic field. A spinning 1 H or 13 C nucleus can orient so that its own tiny magnetic field is aligned either with (parallel to) or against (antiparallel to) the external field. The two orientations don’t have the same energy, however, and aren’t equally likely. The parallel orientation is slightly lower in energy by an amount that depends on the strength of the external field, making this spin state very slightly favored over the antiparallel orientation ( ). (a) Nuclear spins are oriented randomly in the absence of an external magnetic field but (b) have a specific orientation in the presence of an external field, B 0 . Some of the spins (red) are aligned parallel to the external field while others (blue) are antiparallel. The parallel spin state is slightly lower in energy and therefore favored. If the oriented nuclei are irradiated with electromagnetic radiation of the proper frequency, energy absorption occurs and the lower-energy spin state “flips” to the higher-energy state. When this spin-flip occurs, the magnetic nuclei are said to be in resonance with the applied radiation—hence the name nuclear magnetic resonance. The exact frequency necessary for resonance depends both on the strength of the external magnetic field, the identity of the nucleus, and the electronic environment of the nucleus. If a very strong magnetic field is applied, the energy difference between the two spin states is larger and higher-frequency (higher-energy) radiation is required for a spin-flip. If a weaker magnetic field is applied, less energy is required to effect the transition between nuclear spin states ( ). The energy difference Δ E between nuclear spin states depends on the strength of the applied magnetic field. Absorption of energy with frequency ν converts a nucleus from a lower spin state to a higher spin state. (a) Spin states have equal energies in the absence of an applied magnetic field but (b) have unequal energies in the presence of a magnetic field. At ν = 200 MHz, ∆ E = 8.0 × 10 –5 kJ/mol (1.9 × 10 –5 kcal/mol). (c) The energy difference between spin states is greater at larger applied fields. At ν = 500 MHz, ∆ E = 2.0 × 10 –4 kJ/mol. In practice, superconducting magnets that produce enormously powerful fields up to 23.5 tesla (T) are sometimes used, but field strengths in the range of 4.7 to 7.0 T are more common. At a magnetic field strength of 4.7 T, so-called radiofrequency (rf) energy in the 200 MHz range (1 MHz = 10 6 Hz) brings a 1 H nucleus into resonance, and rf energy of 50 MHz brings a 13 C nucleus into resonance. At the highest field strength currently available in commercial instruments (23.5 T), 1000 MHz energy is required for 1 H spectroscopy. These energies needed for NMR are much smaller than those required for IR spectroscopy; 200 MHz rf energy corresponds to only 8.0 × 10 –5 kJ/mol versus the 4.8 to 48 kJ/mol needed for IR spectroscopy. 1 H and 13 C nuclei are not unique in their ability to exhibit the NMR phenomenon. All nuclei with an odd number of protons ( 1 H, 2 H, 14 N, 19 F, 31 P, for example) and all nuclei with an odd number of neutrons ( 13 C, for example) show magnetic properties. Only nuclei with even numbers of both protons and neutrons ( 12 C, 16 O, 32 S) do not give rise to magnetic phenomena ( ). The NMR Behavior of Some Common Nuclei Magnetic nuclei Nonmagnetic nuclei 1 H 12 C 13 C 16 O 2 H 32 S 14 N 19 F 31 P Problems

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The Nature of NMR Absorptions From the description thus far, you might expect all 1 H nuclei in a molecule to absorb energy at the same frequency and all 13 C nuclei to absorb at the same frequency. If so, we would observe only a single NMR absorption signal in the 1 H or 13 C spectrum of a molecule, a situation that would be of little use. In fact, the absorption frequency is not the same for all 1 H or all 13 C nuclei. All nuclei in molecules are surrounded by electrons. When an external magnetic field is applied to a molecule, the electrons moving around nuclei set up tiny local magnetic fields of their own. These local magnetic fields act in opposition to the applied field so that the effective field actually felt by the nucleus is a bit weaker than the applied field. B effective = B applied - B local In describing this effect of local fields, we say that nuclei experience shielding from the full effect of the applied field by the surrounding electrons. Because each chemically distinct nucleus in a molecule is in a slightly different electronic environment, each nucleus is shielded to a slightly different extent and the effective magnetic field felt by each is slightly different. These tiny differences in the effective magnetic fields experienced by different nuclei can be detected, and we thus see a distinct NMR signal for each chemically distinct 13 C or 1 H nucleus in a molecule. As a result, an NMR spectrum effectively maps the carbon–hydrogen framework of an organic molecule. With practice, it’s possible to read this map and derive structural information. shows both the 1 H and the 13 C NMR spectra of methyl acetate, CH 3 CO 2 CH 3 . The horizontal axis shows the effective field strength felt by the nuclei, and the vertical axis indicates the intensity of absorption of rf energy. Each peak in the NMR spectrum corresponds to a chemically distinct 1 H or 13 C nucleus in the molecule. Note that NMR spectra are formatted with the zero absorption line at the bottom, whereas IR spectra are formatted with the zero absorption line at the top; . Note also that 1 H and 13 C spectra can’t be observed simultaneously on the same spectrometer because different amounts of energy are required to spin-flip the different kinds of nuclei. The two spectra must be recorded separately. (a) The 1 H NMR spectrum and (b) the proton-decoupled 13 C NMR spectrum of methyl acetate, CH 3 CO 2 CH 3 . The small peak labeled “TMS” at the far right of each spectrum is a calibration peak, as explained in the next section. The 13 C NMR spectrum of methyl acetate in b shows three peaks, one for each of the three chemically distinct carbon atoms in the molecule. The 1 H NMR spectrum in a shows only two peaks, however, even though methyl acetate has six hydrogens. One peak is due to the CH 3 C═O hydrogens, and the other to the −OCH 3 hydrogens. Because the three hydrogens in each methyl group have the same electronic environment, they are shielded to the same extent and are said to be equivalent. Chemically equivalent nuclei always show the same absorption. The two methyl groups themselves, however, are not equivalent, so the two sets of hydrogens absorb at different positions. The operation of a basic NMR spectrometer is illustrated in . An organic sample is dissolved in a suitable solvent (usually deuteriochloroform, CDCl 3 , which has no hydrogens) and placed in a thin glass tube between the poles of a magnet. The strong magnetic field causes the 1 H and 13 C nuclei in the molecule to align in one of the two possible orientations, and the sample is irradiated with rf energy. If the frequency of the rf irradiation is held constant and the strength of the applied magnetic field is varied, each nucleus comes into resonance at a slightly different field strength. A sensitive detector monitors the absorption of rf energy, and its electronic signal is then amplified and displayed as a peak. Schematic operation of a basic NMR spectrometer. A thin glass tube containing the sample solution is placed between the poles of a strong magnet and irradiated with rf energy. NMR spectroscopy differs from IR spectroscopy ( – ) in that the timescales of the two techniques are quite different. The absorption of infrared energy by a molecule giving rise to a change in vibrational amplitude is an essentially instantaneous process (about 10 –13 s), but the NMR process is much slower (about 10 –3 s). This difference in timescales between IR and NMR spectroscopy is analogous to the difference between cameras operating at very fast and very slow shutter speeds. The fast camera (IR) takes an instantaneous picture and freezes the action. If two rapidly interconverting species are present, IR spectroscopy records the spectrum of both. The slow camera (NMR), however, takes a blurred, time-averaged picture. If two species interconverting faster than 10 3 times per second are present in a sample, NMR records only a single, averaged spectrum, rather than separate spectra of the two discrete species. Because of this blurring effect, NMR spectroscopy can be used to measure the rates and activation energies of very fast chemical processes. In cyclohexane, for example, a ring-flip ( ) occurs so rapidly at room temperature that axial and equatorial hydrogens can’t be distinguished by NMR; only a single, averaged 1 H NMR absorption is seen for cyclohexane at 25 °C. At –90 °C, however, the ring-flip is slowed down enough that two absorption peaks are visible, one for the six axial hydrogens and one for the six equatorial hydrogens. Knowing the temperature and the rate at which signal blurring begins to occur, it’s possible to calculate that the activation energy for the cyclohexane ring-flip is 45 kJ/mol (10.8 kcal/mol). 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Chemical Shifts NMR spectra are displayed on charts that show the applied field strength increasing from left to right ( ). Thus, the left part of the chart is the low-field, or downfield , side, and the right part is the high-field, or upfield , side. Nuclei that absorb on the downfield side of the chart require a lower field strength for resonance, implying that they have less shielding. Nuclei that absorb on the upfield side require a higher field strength for resonance, implying that they have more shielding. The NMR chart. The downfield, deshielded side is on the left, and the upfield, shielded side is on the right. The tetramethylsilane (TMS) absorption is used as reference point. To define the position of an absorption, the NMR chart is calibrated and a reference point is used. In practice, a small amount of tetramethylsilane [TMS; (CH 3 ) 4 Si] is added to the sample so that a reference absorption peak is produced when the spectrum is run. TMS is used as reference for both 1 H and 13 C measurements because in both cases it produces a single peak that occurs upfield of other absorptions normally found in organic compounds. The 1 H and 13 C spectra of methyl acetate in have the TMS reference peak indicated. The position on the chart at which a nucleus absorbs is called its chemical shift . The chemical shift of TMS is set as the zero point, and other absorptions normally occur downfield, to the left on the chart. NMR charts are calibrated using an arbitrary scale called the delta ( δ ) scale , where 1 δ equals 1 part-per-million (1 ppm) of the spectrometer operating frequency. For example, if we were measuring the 1 H NMR spectrum of a sample using an instrument operating at 200 MHz, 1 δ would be 1 part per million of 200,000,000 Hz, or 200 Hz. If we were measuring the spectrum using a 500 MHz instrument, 1 δ = 500 Hz. The following equation can be used for any absorption: δ = Observed chemical shift (number of Hz away from TMS) Spectrometer frequency in MHz Although this method of calibrating NMR charts may seem complex, there’s a good reason for it. As we saw earlier, the rf frequency required to bring a given nucleus into resonance depends on the spectrometer’s magnetic field strength. But because there are many different kinds of spectrometers with many different magnetic field strengths available, chemical shifts given in frequency units (Hz) vary from one instrument to another. Thus, a resonance that occurs at 120 Hz downfield from TMS on one spectrometer might occur at 600 Hz downfield from TMS on another spectrometer with a more powerful magnet. By using a system of measurement in which NMR absorptions are expressed in relative terms (parts per million relative to spectrometer frequency) rather than absolute terms (Hz), it’s possible to compare spectra obtained on different instruments. The chemical shift of an NMR absorption in δ units is constant, regardless of the operating frequency of the spectrometer. A 1 H nucleus that absorbs at 2.0 δ on a 200 MHz instrument also absorbs at 2.0 δ on a 500 MHz instrument. The range in which most NMR absorptions occur is quite narrow. Almost all 1 H NMR absorptions occur from 0 to 10 δ downfield from the proton absorption of TMS, and almost all 13 C absorptions occur from 1 to 220 δ downfield from the carbon absorption of TMS. Thus, there is a likelihood that accidental overlap of nonequivalent signals will occur. The advantage of using an instrument with higher field strength (say, 500 MHz) rather than lower field strength (200 MHz) is that different NMR absorptions are more widely separated at the higher field strength. The chances that two signals will accidentally overlap are therefore lessened, and interpretation of spectra becomes easier. For example, two signals that are only 20 Hz apart at 200 MHz (0.1 ppm) are 50 Hz apart at 500 MHz (still 0.1 ppm). Problems

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Chemical Shifts in 1 H NMR Spectroscopy As mentioned previously, differences in chemical shifts are caused by the small local magnetic field of electrons surrounding different nuclei. Nuclei that are more strongly shielded by electrons require a higher applied field to bring them into resonance so they absorb on the right side of the NMR chart. Nuclei that are less strongly shielded need a lower applied field for resonance so they absorb on the left of the NMR chart. Most 1 H chemical shifts fall within the range 0 to 10 δ, which can be divided into the five regions shown in . By remembering the positions of these regions, it’s often possible to tell at a glance what kinds of protons a molecule contains. Regions of the 1 H NMR Spectrum shows the correlation of 1 H chemical shift with electronic environment in more detail. In general, protons bonded to saturated, sp 3 -hybridized carbons absorb at higher fields, whereas protons bonded to sp 2 -hybridized carbons absorb at lower fields. Protons on carbons that are bonded to electronegative atoms, such as N, O, or halogen, also absorb at lower fields. Correlation of 1 H Chemical Shifts with Environment Type of hydrogen Diagram Chemical shift ( δ ) Reference Si(C H 3 ) 4 0 Alkyl (primary) —C H 3 0.7–1.3 Alkyl (secondary) —C H 2 — 1.2–1.6 Alkyl (tertiary) 1.4–1.8 Allylic 1.6–2.2 Methyl ketone 2.0–2.4 Aromatic methyl Ar—C H 3 2.4–2.7 Alkynyl — C ≡ C — H 2.5–3.0 Alkyl halide 2.5–4.0 Alcohol 2.5–5.0 Alcohol, ether 3.3–4.5 Vinylic 4.5–6.5 Aryl Ar— H 6.5–8.0 Aldehyde 9.7–10.0 Carboxylic acid 11.0–12.0 Predicting Chemical Shifts in 1 H NMR Spectra Methyl 2,2-dimethylpropanoate (CH 3 ) 3 CCO 2 CH 3 has two peaks in its 1 H NMR spectrum. What are their approximate chemical shifts? Strategy Identify the types of hydrogens in the molecule, and note whether each is alkyl, vinylic, or next to an electronegative atom. Then predict where each absorbs, using if necessary. Solution The –OCH 3 protons absorb around 3.5 to 4.0 δ because they are on carbon bonded to oxygen. The (CH 3 ) 3 C– protons absorb near 1.0 δ because they are typical alkane-like protons. Problems

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Integration of 1 H NMR Absorptions: Proton Counting Look at the 1 H NMR spectrum of methyl 2,2-dimethylpropanoate in . There are two peaks, corresponding to the two kinds of protons, but the peaks aren’t the same size. The peak at 1.2 δ, due to the (CH 3 ) 3 C– protons, is larger than the peak at 3.7 δ, due to the –OCH 3 protons. The 1 H NMR spectrum of methyl 2,2-dimethylpropanoate. Integrating the peaks in a stair-step manner shows that they have a 1 : 3 ratio, corresponding to the ratio of the numbers of protons (3 : 9) responsible for each peak. Modern instruments give a direct digital readout of relative peak areas. The area under each peak is proportional to the number of protons causing that peak. By electronically measuring, or integrating , the area under each peak, it’s possible to measure the relative numbers of the different kinds of protons in a molecule. Modern NMR instruments provide a digital readout of relative peak areas, but an older, more visual method displays the integrated peak areas as a stair-step line, with the height of each step proportional to the area under the peak, and therefore proportional to the relative number of protons causing the peak. For example, the two steps for the peaks in methyl 2,2-dimethylpropanoate have a 1 : 3 (or 3 : 9) height ratio when integrated—exactly what we expect because the three −OCH 3 protons are equivalent and the nine (CH 3 ) 3 C– protons are equivalent. Problems

Integration of1H NMR Absorptions: Proton Counting

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Spin–Spin Splitting in 1 H NMR Spectra In the 1 H NMR spectra we’ve seen thus far, each different kind of proton in a molecule has given rise to a single peak. It often happens, though, that the absorption of a proton splits into multiple peaks, called a multiplet . For example, in the 1 H NMR spectrum of bromoethane shown in , the –CH 2 Br protons appear as four peaks (a quartet ) centered at 3.42 δ and the –CH 3 protons appear as three peaks (a triplet ) centered at 1.68 δ . The 1 H NMR spectrum of bromoethane, CH 3 CH 2 Br. The –CH 2 Br protons appear as a quartet at 3.42 δ , and the –CH 3 protons appear as a triplet at 1.68 δ . Called spin–spin splitting , multiple absorptions of a nucleus are caused by the interaction, or coupling , of the spins of nearby nuclei. In other words, the tiny magnetic field produced by one nucleus affects the magnetic field felt by a neighboring nucleus. Look at the –CH 3 protons in bromoethane, for example. The three equivalent –CH 3 protons are neighbored by two other magnetic nuclei—the two protons on the adjacent –CH 2 Br group. Each of the neighboring –CH 2 Br protons has its own nuclear spin, which can align either with or against the applied field, producing a tiny effect that is felt by the –CH 3 protons. There are three ways in which the spins of the two –CH 2 Br protons can align, as shown in . If both proton spins align with the applied field, the total effective field felt by the neighboring –CH 3 protons is slightly larger than it would be otherwise. Consequently, the applied field necessary to cause resonance is slightly reduced. Alternatively, if one of the –CH 2 Br proton spins aligns with the field and one aligns against the field, there is no effect on the neighboring –CH 3 protons. (This arrangement can occur in two ways, depending on which of the two proton spins aligns which way.) Finally, if both –CH 2 Br proton spins align against the applied field, the effective field felt by the –CH 3 protons is slightly smaller than it would be otherwise, and the applied field needed for resonance is slightly increased. The origin of spin–spin splitting in bromoethane. The nuclear spins of neighboring protons, indicated by horizontal arrows, align either with or against the applied field, causing the splitting of absorptions into multiplets. Any given molecule has only one of the three possible alignments of –CH 2 Br spins, but in a large collection of molecules, all three spin states are represented in a 1 : 2 : 1 statistical ratio. We therefore find that the neighboring –CH 3 protons come into resonance at three slightly different values of the applied field, and we see a 1 : 2 : 1 triplet in the NMR spectrum. One resonance is a little above where it would be without coupling, one is at the same place it would be without coupling, and the third resonance is a little below where it would be without coupling. In the same way that the –CH 3 absorption of bromoethane is split into a triplet, the –CH 2 Br absorption is split into a quartet. The three spins of the neighboring –CH 3 protons can align in four possible combinations: all three with the applied field, two with and one against (three ways), one with and two against (three ways), or all three against. Thus, four peaks are produced for the –CH 2 Br protons in a 1 : 3 : 3 : 1 ratio. As a general rule, called the n + 1 rule , protons that have n equivalent neighboring protons show n + 1 peaks in their NMR spectrum. For example, the spectrum of 2-bromopropane in shows a doublet at 1.71 δ and a seven-line multiplet, or septet, at 4.28 δ . The septet is caused by splitting of the –CHBr– proton signal by six equivalent neighboring protons on the two methyl groups ( n = 6 leads to 6 + 1 = 7 peaks). The doublet is due to signal splitting of the six equivalent methyl protons by the single –CHBr– proton ( n = 1 leads to 2 peaks). Integration confirms the expected 6 : 1 ratio. The 1 H NMR spectrum of 2-bromopropane. The –CH 3 proton signal at 1.71 δ is split into a doublet, and the –CHBr– proton signal at 4.28 δ is split into a septet. Note that the distance between peaks is the same in both multiplets. Note also that the two outer peaks of the septet are small enough to be nearly missed. The distance between peaks in a multiplet is called the coupling constant and is denoted J . Coupling constants are measured in hertz and generally fall in the range 0 to 18 Hz. The exact value of the coupling constant between two neighboring protons depends on the geometry of the molecule, but a typical value for an open-chain alkane is J = 6 to 8 Hz. The same coupling constant is shared by both groups of hydrogens whose spins are coupled and is independent of spectrometer field strength. In bromoethane, for instance, the –CH 2 Br protons are coupled to the –CH 3 protons and appear as a quartet with J = 7 Hz. The –CH 3 protons appear as a triplet with the same J = 7 Hz coupling constant. Because coupling is a reciprocal interaction between two adjacent groups of protons, it’s sometimes possible to tell which multiplets in a complex NMR spectrum are related to each other. If two multiplets have the same coupling constant, they are probably related, and the protons causing those multiplets are therefore adjacent in the molecule. The most commonly observed coupling patterns and the relative intensities of lines in their multiplets are listed in . Note that it’s not possible for a given proton to have five equivalent neighboring protons. (Why not?) A six-line multiplet, or sextet, is therefore found only when a proton has five nonequivalent neighboring protons that coincidentally happen to be coupled with an identical coupling constant J . Some Common Spin Multiplicities Number of equivalent adjacent protons Multiplet Ratio of intensities 0 Singlet 1 1 Doublet 1 : 1 2 Triplet 1 : 2 : 1 3 Quartet 1 : 3 : 3 : 1 4 Quintet 1 : 4 : 6 : 4 : 1 6 Septet 1 : 6 : 15 : 20 : 15 : 6 : 1 Spin–spin splitting in 1 H NMR can be summarized by three rules. RULE 1 Chemically equivalent protons don’t show spin–spin splitting. Equivalent protons may be on the same carbon or on different carbons, but their signals don’t split. RULE 2 The signal of a proton with n equivalent neighboring protons is split into a multiplet of n + 1 peaks with coupling constant J . Protons that are farther than two carbon atoms apart don’t usually couple, although they sometimes show weak coupling when they are separated by a π bond. RULE 3 Two groups of protons coupled to each other have the same coupling constant, J . The spectrum of para -methoxypropiophenone in further illustrates these three rules. The downfield absorptions at 6.91 and 7.93 δ are due to the four aromatic-ring protons. There are two kinds of aromatic protons, each of which gives a signal that is split into a doublet by its neighbor. The –OCH 3 signal is unsplit and appears as a sharp singlet at 3.84 δ. The –CH 2 – protons next to the carbonyl group appear at 2.93 δ in the region expected for protons on carbon next to an unsaturated center, and their signal is split into a quartet by coupling with the protons of the neighboring methyl group. The methyl protons appear as a triplet at 1.20 δ in the usual upfield region. The 1 H NMR spectrum of para -methoxypropiophenone. Assigning a Chemical Structure from a 1 H NMR Spectrum Propose a structure for a compound, C 5 H 12 O, that fits the following 1 H NMR data: 0.92 δ (3 H, triplet, J = 7 Hz), 1.20 δ (6 H, singlet), 1.50 δ (2 H, quartet, J = 7 Hz), 1.64 δ (1 H, broad singlet). Strategy It’s best to begin solving structural problems by calculating a molecule’s degree of unsaturation (we’ll see this again in Worked Example 13.4). In the present instance, a formula of C 5 H 12 O corresponds to a saturated, open-chain molecule, either an alcohol or an ether. To interpret the NMR information, let’s look at each absorption individually. The three-proton absorption at 0.92 δ is due to a methyl group in an alkane-like environment, and the triplet-splitting pattern implies that the CH 3 is next to a CH 2 . Thus, our molecule contains an ethyl group, CH 3 CH 2 –. The six-proton singlet at 1.20 δ is due to two equivalent alkane-like methyl groups attached to a carbon with no hydrogens, (CH 3 ) 2 C, and the two-proton quartet at 1.50 δ is due to the CH 2 of the ethyl group. All 5 carbons and 11 of the 12 hydrogens in the molecule are now accounted for. The remaining hydrogen, which appears as a broad one-proton singlet at 1.64 δ, is probably due to an OH group, since there is no other way to account for it. Putting the pieces together gives the structure: 2-methyl-2-butanol. Solution Problems

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1 H NMR Spectroscopy and Proton Equivalence Because each electronically distinct hydrogen in a molecule has its own unique absorption, one use of 1 H NMR is to find out how many kinds of electronically nonequivalent hydrogens are present. In the 1 H NMR spectrum of methyl acetate shown previously in a, for instance, there are two signals, corresponding to the two kinds of nonequivalent protons present, CH 3 C═O protons and –OCH 3 protons. For relatively small molecules, a quick look at the structure is often enough to decide how many kinds of protons are present and thus how many NMR absorptions might appear. If in doubt, though, the equivalence or nonequivalence of two protons can be determined by comparing the structures that would be formed if each hydrogen were replaced by an X group. There are four possibilities. One possibility is that the protons are chemically unrelated and thus nonequivalent. If so, the products formed on substitution of H by X would be different constitutional isomers. In butane, for instance, the –CH 3 protons are different from the –CH 2 – protons. They therefore give different products on substitution by X than the –CH 2 protons and would likely show different NMR absorptions. A second possibility is that the protons are chemically identical and thus electronically equivalent. If so, the same product would be formed regardless of which H is substituted by X. In butane, for instance, the six –CH 3 hydrogens on C1 and C4 are identical, would give the identical structure on substitution by X, and would show an identical NMR absorption. Such protons are said to be homotopic . The third possibility is a bit more subtle. Although they might at first seem homotopic, the two –CH 2 – hydrogens on C2 in butane (and the two –CH 2 – hydrogens on C3) are in fact not identical. Substitution by X of a hydrogen at C2 (or C3) would form a new chirality center, so different enantiomers ( ) would result, depending on whether the pro- R or pro- S hydrogen had been substituted ( ). Such hydrogens, whose substitution by X would lead to different enantiomers, are said to be enantiotopic . Enantiotopic hydrogens, even though not identical, are nevertheless electronically equivalent and thus have the same NMR absorption. The fourth possibility arises in chiral molecules, such as ( R )-2-butanol. The two –CH 2 – hydrogens at C3 are neither homotopic nor enantiotopic. Because substitution of a hydrogen at C3 would form a second chirality center, different diastereomers ( ) would result, depending on whether the pro- R or pro- S hydrogen had been substituted. Such hydrogens, whose substitution by X leads to different diastereomers, are said to be diastereotopic . Diastereotopic hydrogens are neither chemically nor electronically equivalent. They are completely different and would likely show different NMR absorptions. Problems

1H NMR Spectroscopy and Proton Equivalence

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More Complex Spin–Spin Splitting Patterns In the 1 H NMR spectra we’ve seen so far, the chemical shifts of different protons have been distinct and the spin–spin splitting patterns have been straightforward. It often happens, however, that different kinds of hydrogens in a molecule have accidentally overlapping signals. The spectrum of toluene (methylbenzene) in , for example, shows that the five aromatic ring protons give a complex, overlapping pattern, even though they aren’t all equivalent. The 1 H NMR spectrum of toluene, showing the accidental overlap of the five nonequivalent aromatic ring protons. Yet another complication in 1 H NMR spectroscopy arises when a signal is split by two or more nonequivalent kinds of protons, as is the case with trans -cinnamaldehyde, isolated from oil of cinnamon ( ). Although the n + 1 rule predicts splitting caused by equivalent protons, splittings caused by nonequivalent protons are more complex. The 1 H NMR spectrum of trans -cinnamaldehyde. The signal of the proton at C2 (blue) is split into four peaks—a doublet of doublets—by the two nonequivalent neighboring protons. To understand the 1 H NMR spectrum of trans -cinnamaldehyde, we have to isolate the different parts and look at the signal of each proton individually. The five aromatic proton signals (black in ) overlap into a complex pattern with a large peak at 7.42 δ and a broad absorption at 7.57 δ . The aldehyde proton signal at C1 ( red ) appears in the normal downfield position at 9.69 δ and is split into a doublet with J = 6 Hz by the adjacent proton at C2. The vinylic proton at C3 ( green ) is next to the aromatic ring and is therefore shifted downfield from the normal vinylic region. This C3 proton signal appears as a doublet centered at 7.49 δ . Because it has one neighbor proton at C2, its signal is split into a doublet, with J = 12 Hz. The C2 vinylic proton signal ( blue ) appears at 6.73 δ and shows an interesting, four-line absorption pattern. It is coupled to the two nonequivalent protons at C1 and C3 with two different coupling constants: J 1-2 = 6 Hz and J 2-3 = 12 Hz. A good way to understand the effect of multiple coupling, such as that occurring for the C2 proton of trans -cinnamaldehyde, is to draw a tree diagram , like that in . The diagram shows the individual effect of each coupling constant on the overall pattern. Coupling with the C3 proton splits the signal of the C2 proton in trans -cinnamaldehyde into a doublet with J =12 Hz. Further coupling with the aldehyde proton then splits each peak of the doublet into new doublets with J = 6 Hz, and we therefore observe a four-line spectrum for the C2 proton. A tree diagram for the C2 proton of trans -cinnamaldehyde shows how it is coupled to the C1 and C3 protons with different coupling constants. One further trait evident in the cinnamaldehyde spectrum is that the four peaks of the C2 proton signal are not all the same size. The two left-hand peaks are somewhat larger than the two right-hand peaks. Such a size difference occurs whenever coupled nuclei have similar chemical shifts—in this case, 7.49 δ for the C3 proton and 6.73 δ for the C2 proton. The peaks nearer the signal of the coupled partner are always larger, and the peaks farther from the signal of the coupled partner are always smaller. Thus, the left-hand peaks of the C2 proton multiplet at 6.73 δ are closer to the C3 proton absorption at 7.49 δ and are larger than the right-hand peaks. At the same time, the right-hand peak of the C3 proton doublet at 7.49 δ is larger than the left-hand peak because it is closer to the C2 proton multiplet at 6.73 δ . This skewing effect on multiplets can often be useful because it tells where to look in the spectrum to find the coupled partner: look in the direction of the larger peaks. Problems

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Uses of 1 H NMR Spectroscopy NMR is used to help identify the product of nearly every reaction run in the laboratory. For example, we said in that hydroboration–oxidation of alkenes occurs with non-Markovnikov regiochemistry to yield the less highly substituted alcohol. With the help of NMR, we can now prove this statement. Does hydroboration–oxidation of methylenecyclohexane yield cyclohexylmethanol or 1-methylcyclohexanol? The 1 H NMR spectrum of the reaction product is shown in a. The spectrum shows a two-proton peak at 3.40 δ , indicating that the product has a –CH 2 – group bonded to an electronegative oxygen atom (–CH 2 OH). Furthermore, the spectrum shows no large three-proton singlet absorption near 1 δ , where we would expect the signal of a quaternary –CH 3 group to appear. ( b) gives the spectrum of 1-methylcyclohexanol, the alternative product.) Thus, it’s clear that cyclohexylmethanol is the reaction product. (a) The 1 H NMR spectrum of cyclohexylmethanol, the product from hydroboration–oxidation of methylenecyclohexane, and (b) the 1 H NMR spectrum of 1-methylcyclohexanol, the possible alternative reaction product. Problems

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13 C NMR Spectroscopy: Signal Averaging and FT–NMR In some ways, it’s surprising that carbon NMR is even possible. After all, 12 C, the most abundant carbon isotope, has no nuclear spin and can’t be seen by NMR. Carbon-13 is the only naturally occurring carbon isotope with a nuclear spin, but its natural abundance is only 1.1%. Thus, only about 1 of every 100 carbon atoms in an organic sample can be observed by NMR. The problem of low abundance has been overcome, however, by the use of signal averaging and Fourier-transform NMR ( FT–NMR ). Signal averaging increases instrument sensitivity, and FT–NMR increases instrument speed. The low natural abundance of 13 C means that any individual NMR spectrum is extremely “noisy.” That is, the signals are so weak that they are cluttered with random background electronic noise, as shown in a. If, however, hundreds or thousands of individual runs are added together by a computer and then averaged, a greatly improved spectrum results ( b). Background noise, because of its random nature, increases very slowly as the runs are added, while the nonzero signals stand out clearly. Unfortunately, the value of signal averaging is limited when using the method of NMR spectrometer operation described in , because it takes about 5 to 10 minutes to obtain a single spectrum. Thus, a faster way to obtain spectra is needed if signal averaging is to be used. Carbon-13 NMR spectra of 1-pentanol, CH 3 CH 2 CH 2 CH 2 CH 2 OH. Spectrum (a) is a single run, showing the large amount of background noise. Spectrum (b) is an average of 200 runs. In the method of NMR spectrometer operation described in , the rf frequency is held constant while the strength of the magnetic field is varied so that all signals in the spectrum are recorded sequentially. In the FT–NMR technique used by modern spectrometers, however, all the signals are recorded simultaneously. A sample is placed in a magnetic field of constant strength and is irradiated with a short pulse of rf energy that covers the entire range of useful frequencies. All 1 H or 13 C nuclei in the sample resonate at once, giving a complex, composite signal that is mathematically manipulated using so-called Fourier transforms and then displayed in the usual way. Because all resonance signals are collected at once, it takes only a few seconds rather than a few minutes to record an entire spectrum. Combining the speed of FT–NMR with the sensitivity enhancement of signal averaging is what gives modern NMR spectrometers their power. Literally thousands of spectra can be taken and averaged in a few hours, resulting in sensitivity so high that a 13 C NMR spectrum can be obtained from less than 0.1 mg of sample and a 1 H spectrum can be recorded from only a few micro grams. One further question needs to be answered before moving forward with our discussion of 13 C NMR. Why is spin–spin splitting seen only for 1 H NMR? Why is there no splitting of carbon signals into multiplets in 13 C NMR? After all, you might expect that the spin of a given 13 C nucleus would couple with the spin of an adjacent magnetic nucleus, either 13 C or 1 H. No coupling of a 13 C nucleus with nearby carbons is seen because their low natural abundance makes it unlikely that two 13 C nuclei will be adjacent. No coupling of a 13 C nucleus with nearby hydrogens is seen because 13 C spectra are normally recorded using broadband decoupling . At the same time that the sample is irradiated with a pulse of rf energy to cover the carbon resonance frequencies, it is also irradiated by a second band of rf energy covering all the hydrogen resonance frequencies. This second irradiation makes the hydrogens spin-flip so rapidly that their local magnetic fields average to zero and no coupling with carbon spins occurs.

13C NMR Spectroscopy: Signal Averaging and FT–NMR

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Characteristics of 13 C NMR Spectroscopy At its simplest, 13 C NMR makes it possible to count the number of different carbon atoms in a molecule. Look at the 13 C NMR spectra of methyl acetate and 1-pentanol shown previously in b and b. In each case, a single sharp resonance line is observed for each different carbon atom. Most 13 C resonances are between 0 and 220 ppm downfield from the TMS reference line, with the exact chemical shift of each 13 C resonance dependent on that carbon’s electronic environment within the molecule. shows the correlation of chemical shift with environment. Chemical shift correlations for 13 C NMR. The factors that determine chemical shifts are complex, but it’s possible to make some generalizations from the data in . One trend is that a carbon atom’s chemical shift is affected by the electronegativity of nearby atoms. Carbons bonded to oxygen, nitrogen, or halogen absorb downfield (to the left) of typical alkane carbons. Because electronegative atoms attract electrons, they pull electrons away from neighboring carbon atoms, causing those carbons to be deshielded and to come into resonance at a lower field. Another trend is that sp 3 -hybridized carbons generally absorb from 0 to 90 δ , while sp 2 carbons absorb from 110 to 220 δ . Carbonyl carbons (C=O) are particularly distinct in 13 C NMR and are always found at the low-field end of the spectrum, from 160 to 220 δ . shows the 13 C NMR spectra of 2-butanone and para -bromoacetophenone and indicates the peak assignments. Note that the C=O carbons are at the left edge of the spectrum in each case. Carbon-13 NMR spectra of (a) 2-butanone and (b) para -bromoacetophenone. The 13 C NMR spectrum of para -bromoacetophenone is interesting in several ways. Note particularly that only six carbon absorptions are observed, even though the molecule contains eight carbons. para -bromoacetophenone has a symmetry plane that makes ring carbons 4 and 4′, and ring carbons 5 and 5′ equivalent. (Remember from that aromatic rings have two resonance forms.) Thus, the six ring carbons show only four absorptions in the range 128 to 137 δ . A second interesting point about both spectra in is that the peaks aren’t uniform in size. Some peaks are larger than others even though they are one-carbon resonances (except for the two 2-carbon peaks of para -bromoacetophenone). This difference in peak size is a general feature of broadband-decoupled 13 C NMR spectra, and explains why we can’t integrate 13 C NMR spectra in the same way we integrate the resonances in a 1 H NMR spectrum. The local environment of each carbon atom determines not only its chemical shift but also the time it takes for the nuclei to return to their equilibrium state after receiving a pulse of rf radiation and flipping their spins. Quaternary carbons, regardless of their hybridization state or substituents, typically give smaller resonances than primary, secondary, or tertiary carbons. Predicting Chemical Shifts in 13 C NMR Spectra At what approximate positions would you expect ethyl acrylate, H 2 C═CHCO 2 CH 2 CH 3 , to show 13 C NMR absorptions? Strategy Identify the distinct carbons in the molecule, and note whether each is alkyl, vinylic, aromatic, or in a carbonyl group. Then predict where each absorbs, using as necessary. Solution Ethyl acrylate has five chemically distinct carbons: two different C=C, one C=O, one O–C, and one alkyl C. From , the likely absorptions are The actual absorptions are at 14.1, 60.5, 128.5, 130.3, and 166.0 δ . Problems

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DEPT 13 C NMR Spectroscopy Numerous techniques developed in recent years have made it possible to obtain enormous amounts of information from 13 C NMR spectra. Among these techniques is one called DEPT–NMR , for distortionless enhancement by polarization transfer, which makes it possible to distinguish between signals due to CH 3 , CH 2 , CH, and quaternary carbons. That is, the number of hydrogens attached to each carbon in a molecule can be determined. A DEPT experiment is usually done in three stages, as shown in for 6-methyl-5-hepten-2-ol. The first stage is to run an ordinary spectrum (called a broadband-decoupled spectrum) to locate the chemical shifts of all carbons. Next, a second spectrum called a DEPT-90 is run, using special conditions under which only signals due to CH carbons appear. Signals due to CH 3 , CH 2 , and quaternary carbons are absent. Finally, a third spectrum called a DEPT-135 is run, using conditions under which CH 3 and CH resonances appear as positive signals, CH 2 resonances appear as negative signals—that is, as peaks below the baseline—and quaternary carbons are again absent. DEPT–NMR spectra for 6-methyl-5-hepten-2-ol. Part (a) is an ordinary broadband-decoupled spectrum, which shows signals for all eight carbons. Part (b) is a DEPT-90 spectrum, which shows only signals for the two CH carbons. Part (c) is a DEPT-135 spectrum, which shows positive signals for the two CH and three CH 3 carbons and negative signals for the two CH 2 carbons. Putting together the information from all three spectra makes it possible to tell the number of hydrogens attached to each carbon. The CH carbons are identified in the DEPT-90 spectrum, the CH 2 carbons are identified as negative peaks in the DEPT-135 spectrum, the CH 3 carbons are identified by subtracting the CH peaks from the positive peaks in the DEPT-135 spectrum, and quaternary carbons are identified by subtracting all peaks in the DEPT-135 spectrum from the peaks in the broadband-decoupled spectrum. Assigning a Chemical Structure from a 13 C NMR Spectrum Propose a structure for an alcohol, C 4 H 10 O, that has the following 13 C NMR spectral data: Broadband decoupled 13 C NMR: 19.0, 31.7, 69.5 δ ; DEPT-90: 31.7 δ ; DEPT-135: positive peak at 19.0 δ , negative peak at 69.5 δ . Strategy As noted in , it usually helps with compounds of known formula but unknown structure to calculate the compound’s degree of unsaturation. In the present instance, a formula of C 4 H 10 O corresponds to a saturated, open-chain molecule. To gain information from the 13 C data, let’s begin by noting that the unknown alcohol has four carbon atoms, yet has only three NMR absorptions, which implies that two of the carbons must be equivalent. Looking at chemical shifts, two of the absorptions are in the typical alkane region (19.0 and 31.7 δ ), while one is in the region of a carbon bonded to an electronegative atom (69.5 δ )—oxygen in this instance. The DEPT-90 spectrum tells us that the alkyl carbon at 31.7 δ is tertiary (CH); the DEPT-135 spectrum tells us that the alkyl carbon at 19.0 δ is a methyl (CH 3 ) and that the carbon bonded to oxygen (69.5 δ ) is secondary (CH 2 ). The two equivalent carbons are probably both methyls bonded to the same tertiary carbon, (CH 3 ) 2 CH–. We can now put the pieces together to propose a structure: 2-methyl-1-propanol. Solution Problems

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Uses of 13 C NMR Spectroscopy The information derived from 13 C NMR spectroscopy is extraordinarily useful for structure determination. Not only can we count the number of nonequivalent carbon atoms in a molecule, we can also get information about the electronic environment of each carbon and find how many protons are attached to each. As a result, we can address many structural questions that go unanswered by IR spectroscopy or mass spectrometry. Here’s an example: how do we know that the E2 reaction of an alkyl halide follows Zaitsev’s rule ( )? Does treatment of 1-chloro-1-methylcyclohexane with a strong base give predominantly the trisubstituted alkene 1-methylcyclohexene or the disubstituted alkene methylenecyclohexane? 1-Methylcyclohexene will have five sp 3 -carbon resonances in the 20 to 50 δ range and two sp 2 -carbon resonances in the 100 to 150 δ range. Methylenecyclohexane, however, because of its symmetry, will have only three sp 3 -carbon resonance peaks and two sp 2 -carbon peaks. The spectrum of the actual reaction product, shown in , clearly identifies 1-methylcyclohexene as the product of this E2 reaction. The 13 C NMR spectrum of 1-methylcyclohexene, the E2 reaction product from treatment of 1-chloro-1-methylcyclohexane with base. Problems Magnetic Resonance Imaging (MRI) As practiced by organic chemists, NMR spectroscopy is a powerful method of structure determination. A small amount of sample, typically a few milligrams or less, is dissolved in a small amount of solvent, the solution is placed in a thin glass tube, and the tube is placed into the narrow (1–2 cm) gap between the poles of a strong magnet. Imagine, though, that a much larger NMR instrument were available. Instead of a few milligrams, the sample size could be tens of kilograms; instead of a narrow gap between magnet poles, the gap could be large enough for a whole person to climb into so that an NMR spectrum of body parts could be obtained. That large instrument is exactly what’s used for magnetic resonance imaging (MRI) , a diagnostic technique of enormous value to the medical community. Like NMR spectroscopy, MRI takes advantage of the magnetic properties of certain nuclei, typically hydrogen, and of the signals emitted when those nuclei are stimulated by radiofrequency energy. Unlike what happens in NMR spectroscopy, though, MRI instruments use data manipulation techniques to look at the three-dimensional location of magnetic nuclei in the body rather than at the chemical nature of the nuclei. As noted, most MRI instruments currently look at hydrogen, present in abundance wherever there is water or fat in the body. The signals detected by MRI vary with the density of hydrogen atoms and with the nature of their surroundings, allowing identification of different types of tissue and even allowing the visualization of motion. For example, the volume of blood leaving the heart in a single stroke can be measured, and heart motion can be observed. Soft tissues that don’t show up well on X-ray images can be seen clearly, allowing diagnosis of brain tumors, strokes, and other conditions. This technique is also valuable in diagnosing damage to knees or other joints and is a noninvasive alternative to surgical explorations. Several types of atoms in addition to hydrogen can be detected by MRI, and the applications of images based on 31 P atoms are being explored. This approach holds great promise for studies of metabolism. If you’re a runner, you really don’t want this to happen to you. The MRI of this left knee shows bone formation in the tissue in a patient with osteochondroma. (credit: “Osteochondroma MRI” by Michael R Carmont, Sian Davies, Daniel Gey van Pittius and Robin Rees/Wikimedia Commons, CC BY 2.0) chemical shift coupling coupling constant ( J ) delta ( δ ) scale DEPT-NMR diastereotopic downfield enantiotopic FT–NMR homotopic integration multiplet n + 1 rule nuclear magnetic resonance (NMR) spectroscopy shielding spin–spin splitting upfield Summary Nuclear magnetic resonance spectroscopy , or NMR , is the most valuable of the numerous spectroscopic techniques used for structure determination. Although we focused in this chapter on NMR applications with small molecules, more advanced NMR techniques are also used in biological chemistry to study protein structure and folding. When magnetic nuclei, such as 1 H and 13 C, are placed in a strong magnetic field, their spins orient either with or against the field. On irradiation with radiofrequency (rf) waves, energy is absorbed and the nuclei “spin-flip” from the lower energy state to the higher energy state. This absorption of rf energy is detected, amplified, and displayed as an NMR spectrum. Each electronically distinct 1 H or 13 C nucleus in a molecule comes into resonance at a slightly different value of the applied field, thereby producing a unique absorption signal. The exact position of each peak is called the chemical shift . Chemical shifts are caused by electrons setting up tiny local magnetic fields that shield a nearby nucleus from the applied field. The NMR chart is calibrated in delta units ( δ ) , where 1 δ = 1 ppm of spectrometer frequency. Tetramethylsilane (TMS) is used as a reference point because it shows both 1 H and 13 C absorptions at unusually high values of applied magnetic field. The TMS absorption occurs on the right-hand ( upfield ) side of the chart and is arbitrarily assigned a value of 0 δ . 13 C spectra are run on Fourier-transform NMR ( FT–NMR ) spectrometers using broadband decoupling of proton spins so that each chemically distinct carbon shows a single unsplit resonance line. As with 1 H NMR, the chemical shift of each 13 C signal provides information about a carbon’s chemical environment in the sample. In addition, the number of protons attached to each carbon can be determined using the DEPT–NMR technique. In 1 H NMR spectra, the area under each absorption peak can be electronically integrated to determine the relative number of hydrogens responsible for each peak. In addition, neighboring nuclear spins can couple , causing the spin–spin splitting of NMR peaks into multiplets . The NMR signal of a hydrogen neighbored by n equivalent adjacent hydrogens splits into n + 1 peaks (the n + 1 rule ) with coupling constant J . Additional Problems Visualizing Chemistry Chemical Shifts and NMR Spectroscopy 1 H NMR Spectroscopy 13 C NMR Spectroscopy General Problems

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Why This Chapter? Saffron, derived from stigmas of the saffron crocus, is the world’s most expensive spice. Its color is caused by the presence of alternating single and double bonds. (credit: modification of work “Welcome to My World! ~ Huron River Watershed” by j van cise photos/Flickr, CC BY 2.0) Conjugated compounds of many different sorts are common in nature. Many of the pigments responsible for the brilliant colors of fruits, flowers, and even animals have numerous alternating single and double bonds. β -Carotene, for instance, the orange pigment responsible for the color of carrots and an important source of vitamin A, is a conjugated polyene with 11 double bonds. Conjugated enones (alk ene + ket one ) are common structural features of many biologically important molecules such as progesterone, the hormone that prepares the uterus for implantation of a fertilized ovum. Cyclic conjugated molecules such as benzene are a major field of study in themselves. In this chapter, we’ll look at some of the distinctive properties of conjugated molecules and at the reasons for those properties. Most of the unsaturated compounds we looked at in the chapters on Alkenes: Structure and Reactivity and Alkenes: Reactions and Synthesis had only one double bond, but many compounds have numerous sites of unsaturation. If the different unsaturations are well separated in a molecule, they react independently, but if they’re close together, they may interact. In particular, compounds that have alternating single and double bonds—so-called conjugated compounds—have some distinctive characteristics. The conjugated diene 1,3-butadiene, for instance, has some properties quite different from those of the nonconjugated 1,4-pentadiene.

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