1559ch11199-21911/02/0521:42Page199 EQA 11 Alkenes;Infrared Spectroscopy and Mass Spectrometry Instead.its from special characteristics of electrons in so-caledbonds.The properties of a gencra Most of the reactions ar ones you have aready s because the maior methods of alkene synthescs are the am limnion actions of alcohols and haloalkanes that were presented in Chapersand Only some seful tool for ualitative identification of functional Outline of the Chapter 11-1 Nomenclature 11-2 Structure and Bonding in Ethene 11-3 Physical Properties of Alkenes 11-4 Nuclear Magnetic Resonance of Alkenes molecular characterization. 11-8 Degrees of Unsaturation tion for solving structure problems 11-9 Hydrogenation:Relative Stability of Double Bonds Comparing alkenes and alkanes. 11-10 Preparation of Alkenes:Elimination Revisited 11-11 Alkenes by Dehydration of Alcohols Mostly review material in these two sections
11 Alkenes; Infrared Spectroscopy and Mass Spectrometry In Chapters 11 and 12 we return to the presentation of a new functional group: the carbon–carbon double bond. This functional group differs from those seen so far in that it lacks strongly polarized covalent bonds. Instead, its reactivity arises from special characteristics of electrons in so-called bonds. The properties of these electrons and their consequences are discussed in the next chapter. Chapter 11 is restricted to a general description of alkenes as a compound class and a presentation of methods of preparation of double bonds. Most of the reactions are ones you have already seen because the major methods of alkene syntheses are the same elimination reactions of alcohols and haloalkanes that were presented in Chapters 7 and 9. Only some finer details have been added. This chapter also introduces infrared spectroscopy, a useful tool for qualitative identification of functional groups, and mass spectrometry, the best method for determining molecular composition. Outline of the Chapter 11-1 Nomenclature 11-2 Structure and Bonding in Ethene 11-3 Physical Properties of Alkenes 11-4 Nuclear Magnetic Resonance of Alkenes 11-5 Infrared Spectroscopy Another useful spectroscopic technique. 11-6 and 11-7 Mass Spectrometry A different sort of technique for molecular characterization. 11-8 Degrees of Unsaturation More information for solving structure problems. 11-9 Hydrogenation: Relative Stability of Double Bonds Comparing alkenes and alkanes. 11-10 Preparation of Alkenes: Elimination Revisited 11-11 Alkenes by Dehydration of Alcohols Mostly review material in these two sections. 199 1559T_ch11_199-219 11/02/05 21:42 Page 199
1559T_ch11_199-21911/02/0521:42Pa9e200 EQA 200.Chapter 11 ALKENES:INFRARED SPECTROSCOPY AND MASS SPECTROMETRY Keys to the Chapter ward.Again.a small number of common names are still in use and must be learned.However.the systematic kancs ave two nd t carbo subs nts are present.the system shou .be we sign two of its electrons to a basic,garden-variety bond be veen the atoms.The other two atoms attached to them will lie in a plane,with theelectrons above and below. ty or do the cis-transre of. be chemical-shift equivalent.When they aren't couping will be observable,sometimes leading to verycom however you will still he able to derive the informatio f signals oney o theopluing四6oop 11-5 Infrared Spectroscopy sed to data The ir technique helns confirm the or abse pa It is most diagnosti for the following:HO. CN.CEC.C O.and C Different types of C -H bon 14 d i to look for bands in certain general regions of the IR spectrum.much the sa e up the nts (e.g.,alkane somewhere between 1680 and 1800 cm contains a C O group How er,IR data tells us about both the pre sence and the absence of functional groups in a IR data often permits complete determination of the structure of an unknown molecule.The text problems give you opportunities to practice Regions of the Infrared Spectrum c Fingerprint region 40003500300025002000180016801500 1000 Wavenumber (cm-)
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
15597ch11199-21911/02/0521:42Pag0201 EQA Keys o the Chapter·201 11-6 and 11-7. Mass Spectror netry falls int 11-8.De ree ration up with a reasonable structure to match formula of the mole ning the numbe of rings+bon In practice,you should try to reconcile your IR and NMR data with the degree of unsaturation before you of unsaturation,then your answer must contain one ring.On the other hand,if the IR and NMR do shov signa n a molecule with exac y one degree of n,then th be a bon Trone begining h pbIn me celbehen pie sult of u give in the answers in this study guide. stablished feature of this compound class:Mon 心心 form alkene 11-10.Pr eparation of Alkenes: Revisited in the carbon chain.The in th one main e e the most highly substituted.most stable alkene (Say eliminatio n).The major exception is that very bulky bases will favor production of the least substituted was bricfly mentioned in Chanter 7.the E2 elimination mechanism strongly prefers ani conformation between the leaving group and the B-hydroger eing removed Et is to give alken m the be Ge.trans in preference to cis)as the maior product.Stereochemistr is important when considering the use For E2 reactions for alke Cert inds of haloa nes po sess only one reactive conto products. 11-11.Alkenes by Dehydration of Alcohols Again,thi
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
1559T_ch11_199-21911/02/0521:43Page202 ⊕ EQA 202.Chapter 11 ALKENES:INFRARED SPECTROSCOPY AND MASS SPECTROMETRY of the most stable alkene (the ther 0m100% CHsCH-CH-CHaBrK-OCCH(CH.COH CH:CH2CH=CH2 CHCH-CHBICHCH,CH-CH-CH2+CHCH-CHCH Major Minor,cis and trans CH.CH-CHBCH CH,CH.CH-CHCHCH-CHCH Maior cis and trans Either 1-or 2-butanol CHCH.CH-CH+CH,CH-CHCH Additional information pertaining to this process is presented in Chapter 12. Solutions to Problems HO、 OCH (HO (d) (e) 27.(a)cis-or Z-2-Pentene (b)3-Ethyl-1-pentene (c)trans-or E-6-Chloro-5-hexen-2-ol (d)Z-1-Bromo-2-chloro-2-fluoro-1-iodoethene (Priorities are I Br on CI and Cl F on C2.) (e)Z-2-Ethyl-5.5.5-trifluoro-4-methyl-2-penten-1-ol (f)1.1-Dichloro-1-butene (g)-1.2-Dimethoxypropene (h)-2.3-Dimethyl-3-heptene 28.(a)H.at =8+2-1=9:degrees of unsaturation =(9-7)/2 =1 bond or ring present.The functional group 8=4.0(s,2 H):CH2,most hed to Cl 1 H):Two alkene hydrogens
of the most stable alkene (the thermodynamic product). Alcohol dehydrations are susceptible to rearrangement processes. A classic example is encountered in attempted syntheses of terminal alkenes such as 1-butene. The only 100% reliable method is base-promoted E2 elimination of a suitable 1-butyl compound (e.g., 1-bromobutane, 1-butyl tosylate). Any other method will give mixtures: Additional information pertaining to this process is presented in Chapter 12. Solutions to Problems 26. (a) (b) (c) (d) (e) 27. (a) cis- or Z-2-Pentene (b) 3-Ethyl-1-pentene (c) trans- or E-6-Chloro-5-hexen-2-ol (d) Z-1-Bromo-2-chloro-2-fluoro-1-iodoethene (Priorities are I Br on C1 and Cl F on C2.) (e) Z-2-Ethyl-5,5,5-trifluoro-4-methyl-2-penten-1-ol (f ) 1,1-Dichloro-1-butene (g) Z-1,2-Dimethoxypropene (h) Z-2,3-Dimethyl-3-heptene (i) 1-Ethyl-6-methylcyclohexene. This name is better than 2-ethyl-3-methylcyclohexene, because the first number is smaller. 28. (a) Hsat 8 2 1 9; degrees of unsaturation (9 7)/2 1 bond or ring present. The integrated intensities reveal the pieces. 1.8 (s, 3 H): CH3, attached to an unsaturated functional group 4.0 (s, 2 H): CH2, most likely attached to Cl 4.9 and 5.1 (singlets, each 1 H): Two alkene hydrogens HO OCH3 Cl HO Cl Cl Cl I Br Conc. H2SO4, Either 1- or 2-butanol Minor Major, cis and trans CH3CH2CH CH2 CH3CH CHCH3 NaOCH2CH3, CH3CH2OH CH3CH2CHBrCH3 Minor Major, cis and trans CH3CH2CH CH2 CH3CH CHCH3 K OC(CH3)3, (CH3)3COH CH3CH2CHBrCH3 Major Minor, cis and trans CH3CH2CH CH2 CH3CH CHCH3 K OC(CH3)3, (CH3)3COH The only elimination product CH3CH2CH2CH2Br CH3CH2CH CH2 202 • Chapter 11 ALKENES; INFRARED SPECTROSCOPY AND MASS SPECTROMETRY 1559T_ch11_199-219 11/02/05 21:43 Page 202
1559r.ah11.199-21911/02/0521:43Page203 Solutions o Problems203 Thus,you have CH--CHa-Cl,and C=C attached to two H's There are three ways to attach the four groups around the double bond. CH C=C C=C H CH-CI CICH c-C-H 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 =C couplings H are typically very small whereas cis and trans H-C-C-H couplings are large):it is the correct answer (b)H =2.1 (s,3H):CH,next to an unsaturated functional group al H downfield of the ypical of a terminal ethenvl the 65.9 signal suggests an adjacent CHa [compare Figure 11-11(b) --so far:CHO.leaving a Cand an O to add CH3-C-O-CH2-CH=CH2 =16(ho kely CHs CH bond or ring 8=4.3 (quintet w/fi e splitting.I H):Probably CH-OH,four immediate neighbors to the CH edoublct and oe ider doblct.Heach):Alkeme ydrognse cis and one trans to a third alkene hydrogen 7hc9 4.3)fragme as the only CH一CHOH-CH=CH (d)Same formula as (c).so again I bond or ring. 37(uriplet.2 H):Almost ce ertainly Ch H =5.2 (multiplet.2H):and 5.7 (multiplet,1 H):Alkene hydrogens.HC-CH-again
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