北京化工大学：《有机化学》课程教学资源（习题与答案）Chapter 10 Using Nuclear Magnetic Resonance Spectroscopy to Deduce Structure.pdf_大学文库

10-9 Carbon-13 Nuclear Magnetic Resonance The utility of another magnetic nucleus in NMR. Keys to the Chapter 10-2. Defining Spectroscopy Spectroscopy is mainly physics. In spite of that, the material in this section is really very basic and quite understandable: Spectroscopy detects the absorption of energy by molecules. The stake that the organic chemist has in this is also simple. Determining the structure of a molecule requires that the chemist be able to interpret spectroscopically observed energy absorptions in terms of structural features of an unknown molecule. This chapter describes the physical phenomena associated with nuclear magnetic resonance (NMR). It then describes their logical implications as applied to identifying structural features in a molecule from a nuclear magnetic resonance “spectrum.” Don’t be frightened. The end result—the ability to interpret spectroscopic data in terms of molecular structure—is actually one of the easiest skills to acquire in this course. It’s almost fun! 10-3. Hydrogen Nuclear Magnetic Resonance In the upcoming text sections the physical basis for the utility of nuclear magnetic resonance in organic chemistry is presented. This text section describes a couple of concepts that are likely to be unfamiliar to you. The first is the idea that a magnet can align with an external magnetic field in more than one way. Most of us are familiar with bar magnet compasses, which orient themselves along the Earth’s magnetic field in one direction only. Nuclear magnets are actually similar, except for the fact that the reorientation energies are so small that energy quantization becomes significant. Like the compass, the nuclear magnet does have one energetically preferred alignment (the spin for a proton). Less favored alignments (spin states) are very close in energy to the preferred one on a nuclear scale, however. So, unlike the macroscopic compass, the microscopic nuclei can commonly be observed in less favorable, higher energy orientations in a magnetic field. The second new idea is that of resonance. This is the term that describes the absorption of exactly the correct amount of quantized energy to cause a species in a lower energy state to move to a higher energy state. For nuclear magnets, this is commonly described as a spin flip ( spin state to spin state for a proton, the nucleus of the hydrogen atom). The amount of energy involved depends on the identity of the nucleus and the size of the external magnet. The text section describes these relationships in detail. It is the observation of resonance energy absorptions by magnetic nuclei at certain values of magnetic field strength and energy input (in the form of radio waves) that constitutes the physical basis for nuclear magnetic resonance spectroscopy. 10-4. The Hydrogen Chemical Shift The normal NMR spectrum can provide four important pieces of structural information about an unknown molecule. The first two pieces of information are derived from the fact that hydrogens in different chemical environments display separate resonance lines in a high resolution 1 H NMR spectrum. Thus, first, by counting the number of resonance signals in the spectrum, one knows the number of sets of hydrogen atoms in different chemical environments contained in the molecule. Second, the actual position of each resonance signal is characteristic of certain kinds of chemical environments, e.g., it can imply proximity to a specific type of functional group, or attachment to a certain type of atom. The phenomenon responsible for this is the shielding of a magnetic nucleus under observation by the nearby electrons in the molecule. How this comes about physically is described in detail. On a typical NMR spectrum, resonance signals to the left of the chart result from less strongly shielded (deshielded) hydrogens, whereas signals to the right are representative of more 176 • Chapter 10 USING NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY TO DEDUCE STRUCTURE 1559T_ch10_175-198 10/30/05 18:09 Page 176

the six hydrogens of ethane. There are some minor complications that might require a review of the material toward the end of Chapter 5, but on the whole these procedures are not difficult. 10-6. Integration Integration provides the third major piece of information available from NMR: the relative number of hydrogens responsible for each separate NMR signal. The integration is measured electronically by the NMR spectrometer and plotted directly on the spectrum. It tells you whether a given NMR signal is due to a single hydrogen or some number of chemical-shift-equivalent hydrogens in the molecule. The integration is critical for the correct interpretation of NMR spectra. 10-7. Spin–Spin Splitting Because many atomic nuclei are magnets, the NMR signal of a nucleus under observation can, in principle, be affected by the presence of other nearby magnetic nuclei. These neighboring nuclear magnets can align with or against the external magnetic field, so their fields can add to or subtract from the field generated by the NMR machine. The result is a slight change in the resonance line position for the nucleus under observation and is called spin–spin coupling or spin–spin splitting. Again, for spectral interpretation purposes, the theory, presented in the text (although it is not really complicated at all), is secondary to understanding the meaning of the phenomenon in terms of molecular structure. Thus it is often sufficient to rely on two simple rules: 1. Spin–spin splitting is not observed between chemical-shift-equivalent hydrogens. 2. The signal of a hydrogen with N neighboring hydrogens will be split into N 1 lines (“N 1 rule”). There are two important qualifications to rule 2: a. N 1 lines is a minimum. There may be more (Section 10-8). b. In determining N, you count only neighbors whose chemical shifts are different from that of the hydrogen whose signal is being considered (because of rule 1). A little careful observation of the text examples will help you get used to the consequences of spin–spin splitting in its most common forms. Table 10-5 and Figures 10-16, 21, and 22 nicely illustrate these situations. 10-8. Spin–Spin Splitting: Complications The splitting rules outlined in the previous section are idealized for two conditions that are only rarely met in any NMR spectrum. For these rules to hold exactly: First, all the signals must be separated from one another by a distance much greater than the coupling constant within each of their patterns (i.e., J ). Second, all the coupling constants (J values) associated with any hydrogen must be identical in size, even if that hydrogen is coupled to more than one group of neighboring hydrogens. If either of these conditions is not fulfilled, the spectrum won’t look exactly as you might expect. Fortunately, a lot of the time these conditions are approached, especially for hydrogens near electronegative atoms or functional groups. You will therefore have to be aware of the possible effects of “non-first-order” situations; but, for the most part, you won’t have to worry a whole lot about them. 10-9. Carbon-13 Nuclear Magnetic Resonance This is an extension of NMR spectroscopy that is now widely used. There are two reasons. First, modern NMR instruments make these spectra much easier to obtain than originally was the case. Second, the spectra contain useful information that is very easy to interpret, especially under conditions of broad-band hydrogen decoupling, which “wipes out” the spin–spin splitting by neighboring hydrogens. The results are spectra that contain only singlets, that is, a single line for each carbon or group of chemically equivalent carbons. Given such a spectrum, you can quickly determine whether or not it corresponds to a proposed structure simply by counting the lines in the spectrum. Of course, more extensive information is also available from the 13C 178 • Chapter 10 USING NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY TO DEDUCE STRUCTURE 1559T_ch10_175-198 10/30/05 18:09 Page 178

spectrum if desired, in the form of the carbon chemical shifts (Table 10-6), the proton splittings of the “undecoupled” spectrum (e.g., Figure 10-30), or the “DEPT” spectra (Figure 10-33). Solutions to Problems 21. To do this, you need to tell the difference between frequencies, , in units of s1 , and wavenumbers, ˜, in units of cm1 . Section 10-2 shows how they are related: c/ and ˜ 1/ , so c˜, or ˜ /c. For AM radio ( 106 s 1 ), ˜ 106 /(31010) ≈ 3 105 cm1 ; and for FM and TV ( 108 s 1 ), ˜ 108 /(3 1010) ≈ 3 103 cm1 . All these are well to the right end of the chart, very low in energy relative to most of the forms of electromagnetic radiation on the chart. 22. The conversion formulas are 1/˜ and c/ (Section 10-2). (a) 1/(1050 cm1 ) 9.5 104 cm 9.5 m (b) 510 nm 5.1 105 cm; (3 1010 cm s1 )/(5.1 105 cm) 5.9 1014 s 1 (c) 6.15 m 6.15 104 cm; ˜ 1/(6.15 104 cm) 1.63 103 cm1 (d) c˜ (3 1010 cm s1 )(2.25 103 cm1 ) 6.75 1013 s 1 23. Use E 28,600/ (Section 10-2), and use the equations 1/˜ and c/. Be sure to convert the units of to nm before calculating E, though! (a) 1/750 1.33 103 cm 1.33 104 nm (1 cm 102 m, and 1 nm 109 m, or 1 cm 107 nm), so E (2.86 104 )/(1.33 104 ) 2.15 kcal mol1 (b) 1/2900 3.45 104 cm 3.45 103 nm, so E (2.86 104 )/(3.45 103 ) 8.29 kcal mol1 (c) 350 nm (given), so E (2.86 104 )/350 82 kcal mol1 (d) 3 1010/(8.8 107 ) 3.4 102 cm 3.4 109 nm, so E (2.86 104 )/(3.4 109 ) 8.4 106 kcal mol1 (e) 7 102 nm, so E (2.86 104 )/(7 102 ) 4.1 105 kcal mol1 24. Only the value of is needed to calculate E. Use E 28,600/ , together with c/. (a) (3 1010 cm s1 )/(9 107 s 1 ) 333 cm 3.33 109 nm, so E (2.86 104 )/ (3.33 109 ) 8.59 106 kcal mol1 (b) E 4.76 105 kcal mol1 25. (a) (b) Like (a), but with an additional signal at 90 MHz (1 H). (c) This will show all the signals present in both (a) and (b). In addition, signals for 79Br and 81Br will be present (at 22.5 and 24.3 MHz, respectively). At 84,600 gauss the positions of all lines will be at frequencies 4 greater than at 21,150 gauss. For example, a 1 H signal will be at 360 MHz. CFCl3 84.6 19F Solutions to Problems • 179 1559T_ch10_175-198 10/30/05 18:09 Page 179

32. (a) The spectrum shows two signals, at 1.1 and 3.3. The 1.1 signal is for 9 equivalent hydrogens, and the 3.3 signal is for 2 equivalent hydrogens. A good way to get nine equivalent hydrogens is with a (CH3)3C group. The other two hydrogens must be on a separate, single carbon, because the (CH3)3C group contains all but one of the five carbons in the formula. The downfield location of these two hydrogens suggests their carbon is attached to the C1. So, (b) Somewhat similar: two signals, 1.9 and 3.8. The signal for six equivalent hydrogens is probably due to two methyl groups on the same carbon (CH3OCOCH3). The two-hydrogen signal can only be a OCH2O, because there are only four carbons in the molecule. So, 33. (a) The spectrum has two signals in a 31 intensity ratio. The molecule has eight hydrogens, so there must be one group of six equivalent H’s and another of two equivalent H’s (62 31). Two equivalent CH3’s and a CH2 account for all but the two oxygen atoms in the formula. The larger signal at 3.3 is just right for hydrogens on carbons attached to an oxygen. The downfield location for the small signal ( 4.4) is consistent with attachment of carbon to more than one oxygen. Putting it all together we get (b) Again two signals, but now in a 91 intensity ratio. Reasoning as in (a), there are three equivalent CH3’s, each attached to an oxygen, and a CH attached to more than one oxygen, as indicated by its downfield chemical shift ( 4.9). The only consistent structure is then (c) Two signals that are equal in intensity imply two different groups, each with six equivalent H’s. The signal at 3.1 could imply two equivalent CH3OO groups, and the signal at 1.2 suggests two equivalent CH3 groups not attached to oxygen. These all add up to C4H12O2, leaving one carbon unaccounted for in the formula of the molecule (C5H12O2). The fifth carbon can be used to connect the other four groups: (CH3O)2C(CH3)2, which is the answer. By comparison, 1,2-dimethoxyethane has two signals in a 32 ( 64) ratio, and they are both in a region consistent with H’s on a carbon attached to a single oxygen, as the structure CH3OCH2CH2OCH3 requires. 4.9 (CH3O)3CH 3.3 CH3 O CH2 O CH3 4.4 Equivalent, at 3.3 (2) Br Br Br CH3 C C CH3 CH2 CH3 CH3 CH2 1.9 3.8 3.3 as a plausible structure. (CH3)3C CH 2 Cl Cl (CH3)3C 1.1 CH2 Solutions to Problems • 183 1559T_ch10_175-198 10/30/05 18:09 Page 183