15597.eh10.175-19810/30/0518:09Page175 EQA 10 Using Nuclear Magnetic Resonance Spectroscopy to Deduce Structure Ho s anyon spectroscopy,n Outline of the Chapter 10-1 Chemic e as te us)methods of structure elucidation 10-2 Defining Spectroscopy Or,now for something just a little bit different. 1o3地2ode人enios of spectroscopy. 10-4 The Hydrogen Chemical Shift The first two pieces of information available from an NMR spectrum. en you 10-7 Spin-Spin Splitting ehpsf ad of f 10-8 Spin-Spin Splitting:Complications The complications get a bit more complicated. 175
10 Using Nuclear Magnetic Resonance Spectroscopy to Deduce Structure In this chapter we address the question that faces anyone trying to identify the molecular structure of a substance, namely, “What is it?” To put it another way, this chapter begins to answer the obvious question that you may already have been thinking after the first nine chapters of this book, namely, “How does anyone really know that all those molecules are what we say they are?” In the “olden days,” tedious indirect identification methods had to be used, and some are described in the text. Nowadays these questions are answered through the use of spectroscopy, a technique that serves as the “eyes” of an organic chemist with respect to the structures of molecules. The most important and widely used type of spectroscopy, nuclear magnetic resonance (NMR), is described in this chapter. Outline of the Chapter 10-1 Physical and Chemical Tests The “classical” (translate as “tedious”) methods of structure elucidation. 10-2 Defining Spectroscopy Or, now for something just a little bit different. 10-3 Hydrogen Nuclear Magnetic Resonance An example of fairly simple physics put to very good use. A general overview of NMR spectroscopy. 10-4 The Hydrogen Chemical Shift The first two pieces of information available from an NMR spectrum. 10-5 Chemical Shift Equivalence How to tell when you have it. 10-6 Integration The third piece of information available from an NMR spectrum. 10-7 Spin–Spin Splitting A complication that gives rise to the fourth and, perhaps, most useful kind of information present in an NMR spectrum. 10-8 Spin–Spin Splitting: Complications The complications get a bit more complicated. 175 1559T_ch10_175-198 10/30/05 18:09 Page 175
1559T_ch10_175-19810/30/0518:09Pa9e176 ⊕ EQA 176.chapter 10 USING NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY TO DEDUCE STRUCTURE 10-9 Corbon-13 Nuclear Reso tility of another magnet Keys to the Chapter 10-2.Defining Spectroscopy to interpret spectroscopically observed energy absorptions in terms of structural features of an unkno nuclear magnetic resonance"spectrum.Don't be frightened.The end result-the ability to interpret spectro This text section describes a couple of concepts that st is the that a magnet can algn an e nt th than one one direction only.Nuclear magnets are actually similar,except for the fact that the are so small th ergy qua ntization become the compass the nuclear magnet does have close in enerav microscopic nuclei can commonly be observed in less favorable,higher cnergy orientations in a magnetic The second new idea is that of resonance.This is the term that describes the absorption of exactly the correct amount of quantized nergy to cause a species in a lowe r energy state to move to a higher energy tate.For nuc (spin state to B spin state or a pro r magnetic resonance spectroscopy 10-4 The emical Shift ce lines in a high resolutionHN hus.first,by c ironm physically is described in detail chart resul from less strongly shielded (des
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
1559Tch10175-19810/30/0518:09Page177 EQA Keys to he Chapler·177 Deshielded hydrogens Shielded hydroge 甜 Field strength Chemical shift thefield-independent m ppm.Table 10-2 shows typica General Regions of the NMR Spectrum 风 B N 10.09.3 150 Chemical shift Hydrogen typ 0-1.5 Ppm alkane-type hydrogens 1.5-30pg oasettocarton-conthaining 3.045ppm hydrogens on carbons attached to electronegative atoms 4.5-6.0ppm alkene-type hydrogens 6.0-9.5pp benzene-type hydrogens 9.5-10.0ppm hydrogens of aldehyde group With thisas a basis.ady to start NMR spectra.For the intial problems of tis typ come up that displays the correct number of approxmately the observed place hmrin th text section.If you can.you have probably picked a seblestuctre for the This section ical shifts due to chemical eauivalence.The simplest examples of this are the four hydrogens of methane or
strongly shielded ones. Deshielded hydrogens require lower external magnetic fields for resonance, whereas shielded ones require higher fields. So we have NMR spectra that have the following qualitative relationships: As described in the text, a resonance signal’s position is measured as a field-independent chemical shift, which has units of parts per million (ppm) of the total applied field, often called units. These read from right to left, with the usual hydrogen spectrum covering a range from 0 to 10 ppm. Table 10-2 shows typical chemical shifts for common types of hydrogens. There is a lot here, but for most purposes all you really need to know are the types of hydrogens that resonate in several general regions of the NMR spectrum. Region Chemical shift Hydrogen type A 0–1.5 ppm alkane-type hydrogens B 1.5–3.0 ppm hydrogens on carbons next to carbon-containing functional groups C 3.0–4.5 ppm hydrogens on carbons attached to electronegative atoms D 4.5–6.0 ppm alkene-type hydrogens E 6.0–9.5 ppm benzene-type hydrogens F 9.5–10.0 ppm hydrogens of aldehyde group With this as a basis, you are ready to start interpreting NMR spectra. For the initial problems of this type, simply count the number of signals in the spectrum and note the position of each one. Then see if you can come up with a structure that displays the correct number of signals in approximately the observed places, using the material in the text section. If you can, you have probably picked a sensible structure for the unknown molecule. 10-5. Chemical Shift Equivalence This section presents detailed procedures for determining which hydrogens in a molecule have identical chemical shifts due to chemical equivalence. The simplest examples of this are the four hydrogens of methane or Deshielded hydrogens Shielded hydrogens Keys to the Chapter • 177 1559T_ch10_175-198 10/30/05 18:09 Page 177
1559T_ch10_175-19810/30/0518:09Pa9e178 ⊕ EQA 178.Chapter 10 USING NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY TO DEDUCE STRUCTURE 10-6.Integration Integration pr hle gnal is due to a singl hydrogen or som enumber of chemical-shift-equivalent hydrogens in the molecule.The integrtionisritical for the correct interpretation of NMR spectra 10-7.Spin-Spin gnets.the NMR signal of a nucleus under observation can.in principle be affected by the presence of other nearby magneti uclei.These neighboring nuclear in add to o d generate observation and is called spinin,for spectral interpretation purposes the o und anding simple rulcs: ith There are two important qualifications to rule .may be mre (10-) whose chem 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 10-16.2 Compiationieaiadfcpondiomstalatomyarynr in any NMR spectrum.For these rules to hold exactly:First,all the signals must be separated fro n one an anc much great each of the c..Av > d with that hydrogen is coupled to more than one group of neighboring hydrogens.If either of these conditions isno to worry a whole lot about them. 10-9. Carbe n-13 Nuclear Magnetic Re onance s.First. NMR instruments make thes easier to obtain the case second the contain useful information that is very easy to interpret.especially under conditions of broad-band hydroger ngby neighb e spectra tha ngemRCeptdococehcrencnihoaaaknaesyby such a spe ponds to a p
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
15597ch10175-19810/30/0518:09Page179 EQA Solutions to Problems179 spectrum (e.g..Pigure spectra Solutions to Problems 21.To do this.you need to tell the difference between frequencies,v.in units of sand wavenumbers, on 10-2 sh 6 EM =1013×100= relative to most of the forms of electromagnetic radiation on the chart. 22.The conversion formulas are=1/p andv=c(Section 10-2). (a)λ=1/1050cm-)=9.5×10-4cm=9.5um d)510nm=5.1×10-5cmp=(3×100cms-5.1×10-5cm=5.9×104s (g6.15μm=6.15×10-4cm:=1/6.15×10-4cm)=1.63×103cm (dv=c=(6×10cms-2.25×103cm-=6.75×103s1 23.Use AE=28.600/A(Section 10-2).and use the equations=1/and =c/v.Be sure to convert the units of入to nm before calculating△E.though (a)=1750=1.33×10-3cm=1.33×10nm1 ☒ (c)λ=350nm(given).s0△E=(2.86×10/350=82 keal mol- @A10280=34×1cm=34×10m0△E=6×104×10- (e)=7×10-2m,s0△E=(2.86×107×10)=4.1×10 kcal mol 24.Only the value of vis needed to calculate AE.Use AE=28,600/A.together with=c/v. (b)AE=4.76 X 10-5 kcal mol-1 25.(a1℉ H0=21,150G 84.6 22.68.827.340←MHz (b)Like (a),but with an additional signal at 90 MHz (H). s will be at frequencies 4X greater than at 21.150 gauss.For example.a'H signal will be at 360 MHz
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 /(3 1010) ≈ 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
1559T_ch10_175-19810/30/0518:09Pa9e180 ⊕ EQA 180 Chapter 10 USING NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY TO DEDUCE STRUCTURE 27.(a)Divide by300:307/300=1.02:617/300=2.06:683/300=2.28ppm CH3-C-CH2-C(CH3)3 8=206 228 102 28.(a)(CH)20 O more electronegative than N.so hydrogens in the ether are less shielded (b)CHaCOCH Hydrogens on carbons next to electronegative atoms are downfield relative to those next to double-bonded functional groups (Table 10-2) Closer to the electronegative atom (d)(CHa)2S-O Attach (a)Three-the hydrogens cis to the Br (H)are different from the hydrogens trans to the Br(H) Br Ha (b)One-all four H's are equivalent (c)Three (d)Two (e)Three BrBr 30.Chemical shifts have been estimated from the values in Table 10-2.with adjustments for nearby functional groups,and are approximate. (a)2 signals:CH3-CH2-CH2-CHs 0913 (b)2signals: CH,-CHBr-CH3 1.5
26. In (c) the high resolution spectrum around 22.6 MHz will show two 13C resonance signals, because this molecule contains two nonidentical carbon atoms. 27. (a) Divide by 300: 307/300 1.02; 617/300 2.06; 683/300 2.28 ppm. (b) At 90 MHz: 307 (90/300) 92 Hz; 617 (90/300) 185 Hz; 683 (90/300) 205 Hz. At 500 MHz: 307 (500/300) 512 Hz; 617 (500/300) 1028 Hz; 683 (500/300) 1138 Hz. (c) 28. (a) (CH3)2O O more electronegative than N, so hydrogens in the ether are less shielded (b) Hydrogens on carbons next to electronegative atoms are downfield relative to those next to double-bonded functional groups (Table 10-2) (c) Closer to the electronegative atom (d) (CH3)2SPO Attachment to oxygen increases the electron-withdrawing nature of the sulfur, deshielding the hydrogens in the sulfoxide more than in the sulfide 29. The number of signals equals the number of nonequivalent sets of hydrogens in the molecule. In the answers below, each nonequivalent hydrogen or set of hydrogens is labeled a, b, c, etc. (a) Three—the hydrogens cis to the Br (Hb) are different from the hydrogens trans to the Br (Hc) (b) One—all four H’s are equivalent (c) Three (d) Two (e) Three 30. Chemical shifts have been estimated from the values in Table 10-2, with adjustments for nearby functional groups, and are approximate. (a) (b) 2 signals: CH3 CHBr CH3 1.5 3.8 2 signals: CH3 CH2 CH2 CH3 0.9 1.3 Hb Br Hc Ha Br Br Br Ha Ha Br Hb Hb Br Br Ha Ha Hb Hc Hb Hb Hc Hc Br Ha CH3CH2CH2OH O CH3COCH3 CH3 C CH2 C(CH3)3 O 2.06 2.28 1.02 180 • Chapter 10 USING NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY TO DEDUCE STRUCTURE 1559T_ch10_175-198 10/30/05 18:09 Page 180
15597.ch10.175-19810/30/0518:09Page181 ⊕ Solutions to Problems181 (c)3 signals:H-O-CH2-CCl(CH3)2 Variable 4.0 1.4 (d)4 signals:(CH3)2CH-CH2-CH3 09151309 (f)3signals:(CH,CH2)CH (g)4 signals:CH3-O-CH2-CH2-CH3 ↑↑↑ 381710 （)2 signals:: CH2 c=0 1.5↑ 2 (i)3 signals:CH3-CH2-( =0 H-9.5 1220 34一CH,0CH ①4 signals:CH,-CH-C-cH,←o 0可 Chemical Highlight 10-3for further details). CH-回 (a)(CH)CBr-CH-CH,BrCH-CH-CH-CHs (CH:)CH-CHz-CHzBr ↑
(c) (d) (e) (f ) (g) (h) (i) (j) 31. As in the previous problem, the chemical shifts are approximations. Chemical shifts in boxes are most useful for distinguishing structures. Integrations are given in parentheses. Signals marked with asterisks (*) will be complicated as a result of the presence of a chiral carbon in the molecule (see Chemical Highlight 10-3 for further details). (a) 1.5 1.8 1.1 (6) (2) (3) (CH3)2CBr CH2 CH3 1.1 2.0 (1) 1.5 (2)* 0.9 (3) (3) CH3 CH3 BrCH2 CH CH2 3.5 (2)* CH3 CH3 CH3 CH3 CH CH3O 4 signals: 4.0 0.9 3.4 1.4 C 3 signals: CH3 CH2 2.0 9.5 1.2 C H O CH2 CH2 O 2 signals: 1.5 2.4 CH2 C 4 signals: CH3 O CH2 3.4 3.8 1.7 1.0 CH2 CH3 3 signals: (CH3CH2)3CH 0.9 1.3 1.5 2 signals: (CH3)3C NH2 1.3 Variable 4 signals: (CH3)2CH CH2 CH3 0.9 1.5 1.3 0.9 3 signals: H O CH2 CCl(CH3)2 Variable 4.0 1.4 Solutions to Problems • 181 0.9 1.6 1.8 (6) (1) (2) 3.5 (2) (CH3)2CH CH2 CH2Br 1559T_ch10_175-198 10/30/05 18:09 Page 181
1559T_ch10_175-19810/30/0518:09Pa9e182 ⊕ EQA 182.Chapter 10 USING NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY TO DEDUCE STRUCTURE Com The firs e has n cmpndqent mtopwhereh (b)Cl-CH2-CH2-CH2-CH2-OH 哥日受 CCH2←B☑a CH,-CH-CH2-OH (CH3)2CCl-CH2-OH methyl signal in one but no in the other. CH3←156) The last two compounds are readily identified.The second of thes (e)ClCH,CBrCH-(CH3)2 signal of signals.only the third one has downfield ofwith aih to with the same integration ratios.Only minor differences in chemical CH←126) shifts will be present. CICH.-CHCBr(CH,)z CICH-C(CH )CHBrCH CICH2-CHBrC(CHs)s ↑个 恩“回
Compounds are readily distinguished by their NMR spectra: The first one has no signals downfield of 2, whereas the others do. The latter will also show different numbers of signals. The second compound has two nonequivalent methyl groups, whereas the third has two identical methyls. (b) Compounds are again distinguishable. The number and integration of signals downfield of 3 distinguishes the last compound. The other two are distinguished by the presence of an upfield methyl signal in one but not in the other. (c) 4.2 (1) ClCH2 CHBrC(CH3)3 4.0 (2)* 1.1 (9) 1.5 (3) 1.3 (6)* ClCH2C(CH3)2CHBrCH3 3.7 (2)* 4.0 (1) 3.7 (2)* 1.5 (6)* 2.0 (1) CH3 ClCH2 CHCBr(CH3)2 1.2 (3) 4.0 (2)* 1.1 (6)* 2.0 (1) CH3 ClCH2CBrCH (CH3)2 1.5 (3) Var (1) 4.0 (2) (CH3)2CCl CH2 OH 1.5 (6) 2.0 (1) Var (1) 3.6 (2)* (2)* OH ClCH2 CH3 CH CH2 1.1 (3) 3.7 3.7 1.7 1.6 (2) (2) (2) Var (1) 3.6 (2) Cl CH2 CH2 CH2 CH2 OH 182 • Chapter 10 USING NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY TO DEDUCE STRUCTURE The last two compounds are readily identified. The second of these contains three equivalent methyl groups, giving a single signal of intensity 9. Although the other three compounds all have four signals, only the third one has two signals downfield of 3 (with a total integration of 3). The first two compounds will be difficult to distinguish by NMR—they each have the same number of signals with the same integration ratios. Only minor differences in chemical shifts will be present. 1559T_ch10_175-198 10/30/05 18:09 Page 182
1559r.ch10.175-19810/30/0518:09Page183 EQA Solutions to Problems183 32.(a)The spectrum shows two signals,at=1.1 and 3.3.The=1.1 signal is for 9 equivalent single carbon.because the (H The downfield location of these two hydrogens suggests their carbon is attached to the CI.So. (CH3);C-+-CH2-+-Cl(CH3);C-CH2-Cl 33 as a plausible structure. (b)Somewhat similar:two signals.=19 and 3.8.The signal for six signal can only be a-CH2-because there are only four carbons in the molecule.So. ，←3.8 CH-C-CH +-CH-+(2x)-Br CH- equivalent CH for all but the two oxygen atoms in the formula.The large signal at3.3 is just right for hydrogens on carbons attached to an oxygen.The downfield xygen. CH3-O-CH2-O-CH3 4.4 (b)Again two signals,but now in a 9:I intensity ratio.Reasoning as in (a).there are three equivalent (CHOCH 334.9 二 @that are intimply differento.cach wit油出 imply two used to connect the thr four groups:(CH,O)C(CH nal whic answe in a region consist CH.OCH2CH2OCHsrequires
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
1559T_ch10_175-19810/30/0518:09Pa9e184 EQA 184.Chapter 10 USING NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY TO DEDUCE STRUCTURE 3H'sin another (functional group,based on the2.1 chemical shift).It is simples CH3(3×)CH,Co which adds up to CHO.One additional C is needed.but nothing else.If we draw all these pieces with the possible bonds they can form we have -CH3-CH3-CH3 -c- 1.2 Attaching the first CH,to the CO will explain its chemical shift None of the other three CH's can be directly on the CO because(1)the chemical shifts are Now the only possible final step is to attach the three CH,'s: 0 CH-c-C(CH)s This is the answer (b)Both show twe ing to nine H's and thre that three of the CHa's are again on the extra C atom:that is.we have as our pieces 一C(CH -0 For isomer 1.-CH,is at 8=2.0 and-C(CH)is at 8=1.5.The-C(CHa)is downfield relative to the ketone in (a). The CH in isomer2 must be attached to the extra with the rest of the molecule similar to the ketone in(a).This compound is an ester. CH-0-C-C(CH2)
34. (a) There are two signals in a 3:1 intensity ratio. There are 12 H’s in the formula, so there are 9 H’s in one location (not adjacent to a functional group, according to the 1.2, high-field shift) and 3 H’s in another (close to a functional group, based on the 2.1 chemical shift). It is simplest to start with CH3 groups as pieces of a possible structure. There is also a CO group, because the molecule is a ketone. So we have so far CH3 (3 )CH3 CO which adds up to C5H12O. One additional C is needed, but nothing else. If we draw all these pieces with the possible bonds they can form we have Attaching the first CH3 to the CO will explain its chemical shift: None of the other three CH3’s can be directly on the CO because (1) the chemical shifts are wrong and (2) there would be no place left to attach the rest of the pieces. So, attach the unbonded C atom instead: Now the only possible final step is to attach the three CH3’s: (b) Both show two signals in a 3:1 ratio, corresponding to nine H’s and three H’s again. Let’s assume again that the pieces are the same as in (a), plus the extra O atom. Let’s also assume that three of the CH3’s are again on the extra C atom; that is, we have as our pieces For isomer 1, OCH3 is at 2.0 and OC(CH3)3 is at 1.5. The OC(CH3)3 is downfield relative to the ketone in (a). For isomer 2, OCH3 is at 3.6, far downfield relative to the ketone in (a), but the OC(CH3)3 is at 1.2, almost identical to the ketone in (a). The CH3 in isomer 2 must be attached to the extra O, with the rest of the molecule similar to the ketone in (a). This compound is an ester. CH3 O O C C(CH3)3 CH3 C(CH3)3 O O C CH3 O C C(CH3)3 This is the answer. CH3 O C C CH3 O C CH3 2.1 1.2 CH3 CH3 CH3 C O C 184 • Chapter 10 USING NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY TO DEDUCE STRUCTURE 1559T_ch10_175-198 10/30/05 18:09 Page 184
