北京化工大学：《有机化学》课程教学资源（双语习题与答案）Chapter 11 Alkenes; Infrared Spectroscopy and Mass Spectrometry.pdf_大学文库

200 • Chapter 11 ALKENES; INFRARED SPECTROSCOPY AND MASS SPECTROMETRY Keys to the Chapter 11-1 through 11-4. Nomenclature and Physical Properties Little needs to be added to the text descriptions for these two topics. The nomenclature rules are straightforward. Again, a small number of common names are still in use and must be learned. However, the systematic nomenclature is logical and easy to master. Note that alkenes, like cyclic alkanes, have two distinct “sides,” and therefore substituents may be either cis or trans to each other. For alkenes, however, the cis and trans designations should be restricted to molecules with exactly two substituents, one on each of the two doubly bonded carbons. If more substituents are present, the E,Z nomenclature system should always be applied. The two sides of alkenes are due to the nature of the four-electron double bond. In the simplest picture, we assign two of its electrons to a basic, garden-variety bond between the atoms. The other two electrons are then placed in two parallel p orbitals, overlapping “sideways” to form the bond. This -type overlap prevents the carbons at each end of the double bond from rotating with respect to one another. Ethene, therefore, is a perfectly flat molecule, and, in general, the carbons of the alkene functional group and all the atoms attached to them will lie in a plane, with the electrons above and below. One other significant consequence of the enforced planarity of double bonds and the cis-trans relationships of attached groups is seen in the NMR spectra of alkenes. A molecule’s alkene hydrogens do not all have to be chemical-shift equivalent. When they aren’t, coupling will be observable, sometimes leading to very complicated patterns as a result of J values that vary widely as a function of the structural relationships between the hydrogens involved (see Table 11-2). Figure 11-11 illustrates this feature. Even in complex spectra, however, you will still be able to derive the information you need for structure determination as long as you remember to look separately for the four basic pieces of information the spectrum contains: number of signals, chemical shift of each one, integration, and splitting patterns. If the splitting is too complicated to interpret, you can still use the other three pieces of data to come up with an answer. 11-5. Infrared Spectroscopy Once the most important spectroscopic technique, infrared spectroscopy is now used to complement NMR data. The IR technique helps confirm the presence or absence of common functional groups in a molecule. It is most diagnostic for the following: HO, CqN, CqC, CPO, and CPC. Different types of COH bonds can be readily identified, helping to confirm information obtained by NMR. Although occasionally the detailed data in Table 11-4 and in the text may be necessary to solve a problem, for the most part you will only need to look for bands in certain general regions of the IR spectrum, much the same way you have learned to divide up the NMR spectrum into rather general segments (e.g., alkane COH and alkene COH). The following illustration, derived from the data in Table 11-4, shows these regions. For example, a compound exhibiting a strong band somewhere between 1680 and 1800 cm1 contains a CPO group. However, IR data tells us about both the presence and the absence of functional groups in a molecule. Don’t neglect the usefulness of the latter! For instance, a molecule lacking absorption between 3200 and 3700 cm1 cannot be an alcohol. Combining information from a molecular formula with NMR and IR data often permits complete determination of the structure of an unknown molecule. The text problems give you opportunities to practice. 1559T_ch11_199-219 11/02/05 21:42 Page 200

Keys to the Chapter • 201 11-6 and 11-7. Mass Spectrometry The kind of information available from mass spectrometry falls into two categories. First, the m/z value for the molecular ion provides information useful in calculating the molecular formula of the molecule. Second, the lower molecular weight fragments that appear in the mass spectrum contain clues concerning structural features of the molecule in question. Be sure that you understand how to extract these kinds of information from mass spectral data. 11-8. Degrees of Unsaturation When you are faced with the problem of coming up with a reasonable structure to match some spectroscopic or chemical data, it is possible to waste a lot of time writing answers that are incompatible with the molecular formula of the molecule. Determining the number of rings bonds ahead of time (the degree of unsaturation) can make solving these problems go much more smoothly: You automatically know whether or not you need to consider these structural elements as possible parts of an unknown molecule. In practice, you should try to reconcile your IR and NMR data with the degree of unsaturation before you start writing down possible structures. For example, if the IR and NMR indicate the absence of bonds (no IR bands around 1650 cm1 ; no NMR signals downfield of about 5) but the formula indicates one degree of unsaturation, then your answer must contain one ring. On the other hand, if the IR and NMR do show such signals in a molecule with exactly one degree of unsaturation, then the unsaturation must be a bond and a ring cannot be present. Thus by the process of elimination, you get closer to the right answer more quickly. Try it yourself, beginning with Problem 28. In some cases you will be given an additional piece of data: the result of hydrogenation. Use it to distinguish bonds from rings, because, in general, after hydrogenation the bonds will be gone, but the rings will still be there. The degree of unsaturation will be given in the answers in this study guide. 11-9. Hydrogenation: Relative Stability of Double Bonds The stability order of different kinds of alkenes is a well-established feature of this compound class: More substituted alkenes are more stable than less substituted ones, and trans are more stable than cis. This topic does not exist in isolation, however. In fact, it has important consequences for both reactions that form alkenes as well as reactions that alkenes undergo. Learn this stability order. You will need to use it later. 11-10. Preparation of Alkenes: Elimination Revisited This is a review of the material from Sections 7-6 and 7-7. There are two new considerations. First, many haloalkanes can give rise to several alkenes upon elimination, each with the double bond in a different position in the carbon chain. These products arise when there are several different -hydrogens that can be lost in the elimination process together with the leaving group. The rule to remember is as follows: All E1 and, with one main exception, all E2 processes tend to produce the most highly substituted, most stable alkene (Saytzev elimination). The major exception is that very bulky bases will favor production of the least substituted, least stable alkene in E2 processes (Hofmann elimination). The second new consideration relates to stereochemistry. As was briefly mentioned in Chapter 7, the E2 elimination mechanism strongly prefers an anti conformation between the leaving group and the -hydrogen being removed. The result is that E2 eliminations will tend to give alkenes arising from the best available anti conformation. E1 eliminations are not as restricted and will simply tend to give the most stable alkene (i.e., trans in preference to cis) as the major product. Stereochemistry is important when considering the use of elimination reactions for alkene synthesis. Certain kinds of haloalkanes possess only one reactive conformation for E2 elimination (see Problem 48) and will therefore give only a single stereoisomer upon reaction. This can be very useful. El eliminations, however, are more prone to yield mixtures of stereoisomeric products. 11-11. Alkenes by Dehydration of Alcohols Again, this is mainly a review of earlier material (Chapters 7 and 9). Note that, unlike the situation with basepromoted E2 eliminations, under the reaction conditions for alcohol dehydration, the usual result is formation 1559T_ch11_199-219 11/02/05 21:42 Page 201

Solutions to Problems • 203 Thus, you have CH3O, OCH2OCl, and attached to two H’s. There are three ways to attach the four groups around the double bond. In the first two compounds, all the NMR signals should show substantial couplings. Only the third compound will show a spectrum as simple as A (remember that couplings are typically very small whereas cis and trans HOCPCOH couplings are large); it is the correct answer. (b) Hsat 10 2 12; degrees of unsaturation (12 8)/2 2 bonds and/or rings. The NMR shows the following: 2.1 (s, 3 H): CH3, next to an unsaturated functional group 4.5 (d, 2 H): CH2, attached to oxygen, split by one H 5.3 and 5.9 (m, 2H and 1H): OCHPCH2, the internal H downfield of the other two, typical of a terminal ethenyl group; the extensive splitting of the 5.9 signal suggests an adjacent CH2 [compare Figure 11-11(b) in the text] The pieces are CH3O and CH2PCHOCH2OOO so far: C4H8O, leaving a C and an O to add in, and one more bond (a ring would be impossible). So let them be giving the final solution: (c) Hsat 8 2 10; degrees of unsaturation (10 8)/2 1 bond or ring. 1.3 (doublet, 3 H): Most likely CH3OCH 1.6 (broad singlet, 1 H): Perhaps OH? 4.3 (quintet w/fine splitting, 1 H): Probably CHOOH, four immediate neighbors to the CH (not including the OH) 5.1 and 5.3 (one narrow doublet and one wider doublet, 1 H each): Alkene hydrogens, one cis and one trans to a third alkene hydrogen 5.9 (multiplet, 1 H): The third alkene hydrogen, in a H2CPCHO fragment The methyl group (at 1.3) must be attached to the CH (at 4.3), giving as the only possible solution CH3OCHOHOCHPCH2 (d) Same formula as (c), so again 1 bond or ring. 1.4 (broad singlet, 1 H): Perhaps OH again? 2.3 (quartet w/fine splitting, 2 H): A CH2 with three immediate neighbors 3.7 (triplet, 2 H): Almost certainly CH2OCH2OOH (no splitting to OH) 5.2 (multiplet, 2H): and 5.7 (multiplet, 1 H): Alkene hydrogens, H2CPCHO again C O CH3 O CH2 CH CH2 C O, C H H C C CH CH2Cl 3 H H C C CH2Cl CH3 H H C C ClCH2 CH3 H H C C 1559T_ch11_199-219 11/02/05 21:43 Page 203

31. Use the carbon types together with the chemical shifts to choose between alternative possibilities. (a) Hsat 8 2 10; degrees of unsaturation (10 6)/2 2 bonds or rings present. 30.2 is a CH2 group; 136.0 is an alkene CH. Because those alone only add up to C2H3, there must be two of each. Two OCH2O’s, plus OCHPCHO, which can only combine to make (b) Hsat 8 2 10; degrees of unsaturation (10 6)/2 2 again. 18.2 is a CH3, not attached to the oxygen; 134.9 and 153.7 are alkene CH’s; 193.4 is in the CPO region, O B and therefore, must be a OCH group because it is a doublet. The answer is therefore O B CH3OCHPCHOCOH (2 bonds). You do not have enough information about 13C NMR to determine the stereochemistry. (c) Hsat 8 2 10; degrees of unsaturation (10 8)/2 1. (d) Hsat 10 2 12; degrees of unsaturation (12 10)/2 1. This one has two CH3 groups ( 17.6 and 25.4), a CH2 downfield enough ( 58.8) to be attached to the O, and two alkene carbons, of which only one ( 125.7) has an H on it. The pieces: (2)CH3O, OCH2OOO, There is one H unlocated. Because it is not attached to one of the carbons, it must be on the oxygen. So, the possible answers are You do not have the information to tell which of the three is the actual compound. (e) Notice that there are only four signals, but there are five carbons. Be careful. Hsat 10 2 12; degrees of unsaturation (12 8)/2 2 now. 15.8 and 31.1 are CH2 groups; 103.9 is an alkene CH2, whereas 149.2 is an alkene C lacking hydrogens. What do you have so far? The molecule has the piece leaving three C’s and six H’s to make up the formula, which must still contain one more element of unsaturation (a ring?). CH2 C , CH3 CH3 CH2 C C H OH CH3 CH2 CH3 C C OH H CH2 CH3 CH3 C C HO H CH C . CH3 CH2 CH CH2 Answer directly available from the carbon types. 13.6 25.8 139.0 112.1 H2C CH2 CH CH Cyclobutene (1 bond and 1 ring) Solutions to Problems • 205 1559T_ch11_199-219 11/02/05 21:43 Page 205

Solutions to Problems • 207 33. Lower, because the vibrational frequency varies inversely with the square of the “reduced” mass involving the atoms about the bond. So, bonds involving heavier atoms have lower energies associated with vibrational excitation. Typically, ˜ COCl 700 cm1 , ˜ COBr 600 cm1 , and ˜ COI 500 cm1 . 34. 10,000/˜ m. (a) 5.81 m; (b) 6.06 m; (c) 3.03 m; (d) 11.24 m; (e) 9.09 m; (f ) 4.42 m. 35. A—(b) (saturated alkane) B—(d) (alcohol band at 3300 cm1 ) C—(a) (alkene band at 1640 cm1 in addition to alcohol band) D—(c) (alkene band at 1665 cm1 but no alcohol band) 36. (i) Both alkene (1660) and alcohol (3350) products have formed. (ii) Only alkene (1670) forms. (iii) Only alcohol (3350) forms. (a) Conclusions: Isomer C is probably a primary bromoalkane, which gives a primary alcohol product (SN2). Isomer B is probably a tertiary bromoalkane, which gives only alkene as product (E2). Isomer A is probably a secondary bromoalkane, which gives a mixture of SN2 and E2 products. (b) A possibilities: CH3CHBrCH2CH2CH3, CH3CHBrCH(CH3)2, or CH3CH2CHBrCH2CH3 B possibilities: (CH3)2CBrCH2CH3 (only tertiary isomer) C possibilities: CH3CH2CH2CH2CH2Br, (CH3)2CHCH2CH2Br, or CH3CH2CH(CH3)CH2Br, but probably not (CH3)3CCH2Br (too hindered to give SN2 reaction) 37. Process of elimination: we begin by noticing the absence of absorptions in certain regions of the spectrum, and conclude that the corresponding functional groups cannot be in the actual molecule. So, no OOH (no strong broad band around 3300 cm1 ), no CPO (nothing around 1700 cm1 or so), no CqCOH (bands near 2100 and 3300 cm1 ), and no CPC (1680 cm1 ). All that are left as possibilities are the alkane and the ether. The spectrum shows strong bands between 1000 and 1200 cm1 , strongly suggesting the presence of COO bonds. The fact that such absorptions do not appear in the sample spectra for alkanes in the chapter (see Figures 11-14 and 15) confirms that the correct answer is the ether, . 38. Begin by noting the masses of the most prominent ions in each mass spectrum. Then try to predict how each of the alkanes might be most likely to fragment, using as a guiding principle a preference for forming more rather than less stable carbocations upon bond cleavage. The three compounds are constitutional isomers, all with the molecular formula C6H14 and a molecular weight of 86. Spectrum A shows a base peak at m/z 57 (C4H9) and other significant ions with m/z 56, 41, 29 (C2H5) and 27. The molecular ion at m/z 86 is weak. Spectrum B also shows a base peak at m/z 57 (C4H9). The peak at m/z 43 (C3H7) is bigger than in spectrum A, while that at m/z 29 (C2H5) is smaller. The molecular ion is more intense. Spectrum C shows a base peak at m/z 43 (C3H7). The molecular ion is weak, but m/z 71 (C5H11) is prominent in this spectrum. Now consider the three structures and the bonds in the molecular ion most likely to fragment in each: CH3CH2G (m/z 29) Hexane: [CH3CH2OCH2OCH2OCH2CH3]G? 88n CH3CH2CH2G (m/z 43) hhh CH3CH2CH2CH2G (m/z 57) Cleavage is most favorable between CH2 groups (avoiding formation of methyl cation), but since the best cations that you can get are only primary—and not very stable—fragmentation as a whole will be less likely. Spectrum B seems the best match because of the prominent molecular ion. O      1559T_ch11_199-219 11/02/05 21:43 Page 207