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
1559T_ch11_199-21911/02/0521:43Pa9e204 ⊕ EQA 204.Chapter 11 ALKENES:INFRARED SPECTROSCOPY AND MASS SPECTROMETRY olecule everything is accounted for,and the two CH H2C=CH-CH2-CH2OH (e)Ha=6+2 unsaturation=(6-4)/2=1bond or ring 1.7 (doublet.3 H):CHa,next to CH =5.7 (quartet,1 H):alkene CH,next to CH3 29.The multi t8=231s rtet for the 0 CH se of the three neighbors,the CH s additi 心8hg in th TPi6mtpiae In other words.lines 5 and6overlap to give one tall line in the center of the pattem.and lines7and overlap to give the oth 30.(a)Yes.1-Butene>trans-2-butene(which should be zero) (b)No (e)Yes.cistrans(which,again,is zero)
As in (c), with only four carbons in the molecule everything is accounted for, and the two CH2 groups must be attached to each other; thus, H2CPCHOCH2OCH2OH (e) Hsat 6 2 2 6; degrees unsaturation (6 4)/2 1 bond or ring. 1.7 (doublet, 3 H): CH3, next to CH 5.7 (quartet, 1 H): alkene CH, next to CH3 All that’s left is one C and two Cl’s, so the pieces, and two Cl’s, combine to give the answer, CH3OCHPCCl2. 29. The multiplet at 2.3 is a quartet for the C2 CH2 group, because of the three neighbors, the CH2 of C1 and the CH of C3. The signal shows additional very fine splitting, indicating that there is a small coupling to the alkene hydrogens on C4 (J 1 Hz). The multiplet for the hydrogen signal at 5.7 can be interpreted in detail. It corresponds to the CH at C3, which is split by the alkane CH2 on one side and the alkene CH2 on the other. The result is a pattern that can be described as a doublet (for trans alkene coupling) of doublets (for cis alkene coupling) of triplets (for coupling to the alkane CH2), 12 lines total, of which 10 are seen. If we construct a plausible splitting diagram we find that the two “missing” lines are actually “hidden” under the tallest lines in the center of the pattern: In other words, lines 5 and 6 overlap to give one tall line in the center of the pattern, and lines 7 and 8 overlap to give the other. 30. (a) Yes. 1-Butene trans-2-butene (which should be zero) (b) No (c) Yes. cis trans (which, again, is zero) 1 2 3 4 5 6 7 8 9 10 11 12 Triplet splitting J 7 Hz Cis doublet splitting J 10 Hz Trans doublet splitting J 16 Hz CH3 CH C 204 • Chapter 11 ALKENES; INFRARED SPECTROSCOPY AND MASS SPECTROMETRY 1559T_ch11_199-219 11/02/05 21:43 Page 204
15597ch11199-21911/02/0521:43Pag0205 EQA Solutions o Problems205 31.Use the carbon types together with the chemical shifts to choose between alterative possibilities. hee must be two of each.ToCH plusCH-H-.which can only combin CH=CH H2C-CH Cyclobutene (1 bond and 1 ring) and therefore.mu CHa-CH=CH-C-H(2 bonds).You do not have enough information aboutC NMR to determine the stereochemistry. (e)H.=8+2=10:degrees of unsaturation =(10-8)/2=1. 1362581390112 CH3-CH2-CH=CH2 two alkene carbons.of which only one(=125.7)has an H on it. The pieces:(2xCH--CH2---CH-C.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 CH: CH、 CH2-OH C=C H CH2-OH H HO-CH2 C-C45 H You do not have the information to tell which of the three is the actual compound. ()ly(d an alkene CH2,whereas=149.2 is an alkene C lacking hydrogens. What do you have so far?The molecule has the picce CH2=C.leaving three C's and six H's to make up the formula.which must still contain one more element of unsaturation (a ring?)
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
1559T_ch11_199-21911/3/0516:02Pa9e206 EQA 206.Chapter 11 ALKENES:INFRARED SPECTROSCOPY AND MASS SPECTROMETRY Because the highfield signals are triplets,these can only be CH groups:three of them Combining CHa=C with three CHa's can only give CH2= (f)H.=14+2 16:degrees of unsaturation =(16-10)/2 =3,or 1 T bond and 2 rings Again,be careful.No ow ther are four signals,bu 1352).Be an this signal must represent two equivalent alkene CH groups: -CH-CH-.So you have at least wo CH 's.an alk e CH.and CH-CH-.for a total to groups aready identificd n order to keep the NMR spectrum as simples it is.In her words ⊕ -CH2- -CH2- -CH- c-- -CH=CH- Each isa reasonable possibility (the secondone..is actually correct) 32.Hat =10+2=12:degrees of unsaturation=(12-10)/2 =1. (a)The ony way for five carbons to be equivalent is to make a ring:is the answer. (b)Three CH,'s,and a-CH=C: CH;-CH=( (c)Two CH3's.one CH2.and-CH=CH-:CH3-CH2-CH=CH-CH3 is the answer (stereochemistry is ambiguous)
Because the highfield signals are triplets, these can only be CH2 groups: three of them. Combining with three CH2’s can only give (f) Hsat 14 2 16; degrees of unsaturation (16 10)/2 3, or 1 bond and 2 rings. Again, be careful. Now there are four signals, but seven carbons in the molecule. Upfield, there are two different kinds of CH2’s ( 25.2 and 48.5) and one kind of CH ( 41.9). There is one kind of alkene carbon ( 135.2). Because a double bond must connect two alkene carbons, this signal must represent two equivalent alkene CH groups: OCHPCHO. So you have at least two CH2’s, an alkane CH, and OCHPCHO, for a total of C5H7. So two carbons and three H’s are still required: One more CH2 and one more CH would do, and these each must be equivalent to groups already identified in order to keep the NMR spectrum as simple as it is. In other words, A here are the pieces you have for the molecule: 2 equivalent OCH2O’s, 2 equivalent OCH’s, a A single unique OCH2O, and the OCHPCHO group, for a total of C7H10. How do you put this all together? Remembering that symmetry can make groups equivalent, you can write these groups in symmetrical arrangements and connect them in a trial-and-error manner. Each is a reasonable possibility (the second one, norbornene, is actually correct). 32. Hsat 10 2 12; degrees of unsaturation (12 10)/2 1. (a) The only way for five carbons to be equivalent is to make a ring: is the answer. (b) Three CH3’s, and a (c) Two CH3’s, one CH2, and OCHPCHO: CH3OCH2OCHPCHOCH3 is the answer (stereochemistry is ambiguous). CH3 CH3 CH3 CH C CH C : CH2 CH2 CH2 CH CH HC CH or or CH2 CH2 The 31.1 signal accounts for the C two equivalent CH2 groups (circled). CH2 CH2 CH2 C 206 • Chapter 11 ALKENES; INFRARED SPECTROSCOPY AND MASS SPECTROMETRY 1559T_ch11_199-219 11/3/05 16:02 Page 206
15597.ch11.199-21911/02/0521:43Page207 ⊕ Solutions o Problems27 33.Lower,because the vibrational frequency varies inversely with the square of the"reduced"mass 34.10,.000=um.(a5.81μm:(b)6.06m()3.03um(d11.24μm(e)9.09umf)4.42μm 35.A-(b)(saturated alkane) B-(d)(alcohol band at 3300 cm-) C-(a)(alkene band at 1640 cm-in addition to alcohol band) D-(e)(alkene band at 1665 cm 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 (b)A possibilities:CHaCHBrCH.CH-CH3.CH;CHBrCH(CH3)2.or CH;CH2CHBrCH2CHs (CH)CHCH.CHBr or CH,CH.CH(but probably not (CH)CCH,Br (too hindered to give S2 reaction) 37.Process of elimination:we begin by noticing the of absorptions in certain regions of the mand con d3300 rups cannot (bands near 2100 and 3300 cm)and no C-C(1680 cm).All that are bonds.The strongly absorptions do not appear in th e sample spectra 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 mas s spectrum.Then try to predict mostkeytofngmcs3sagmitnpinciplaprfercnC orming moThe three】 mA shows a base peak(CH)and other significant ions withm56.41.29 e迎sC时怎C方 spectrum A,wh e molecular ion is more intens Spectrum n C shows a base peak at m/=43 (C:H )The molecular ion is weak,but m/=71 (CH)is Now dhereestructuresand the bonds in the moecular n most likelytoframen in ech Hexame:[CH.CHCHCH CHCH.]" 29 43) CH CH-CH-CH2+(m/=57) Spectrum B seems the best match because of the prominent molecular ion
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
1559T_ch11_199-21911/02/0521:43Pa9e208 ⊕ EQA 208.Chapter 11 ALKENES:INFRARED SPECTROSCOPY AND MASS SPECTROMETRY 2-Methylpentane:[CH,-CH-CH,CH,CH3]+.- Cleavage will occur mainly about the CH to give secondary cations.The best match is spectrum C. 3-Methylpemtane:[CH,(w/:=57) Fragmentation occurs at the indicated bond to form mainly sec-butyl cation and,to a lesser extent,ethyl ha lae to in tha processes do not usually dominate the spectrum. 39.Major peaks: m/k43 from M-B m/41 from M-HBr-H Minor peaks: from M-CH from M-HBr m/29 (CH,CH-CH2)* (CH-CH) from M-Br-CHz (CH 40.Compound is saturated (see Section 11-6).Try to use the general guidelines that intense fragment es or are This is most likely if the remaining fragment is a very stable cation.for example )CH CH,CH2- -C-OH CHCH-COH CHs CH3 CH 73 Looking at the rest of the spectrum.the base peak is at m/59.or(M-29).loss of CHCHa CH31+. CHCH2+C-OH 一(CH32COH+CH,CH2 /59 This is,all together,good evidence for isomer C being 2-methyl-2-butanol,as shown
208 • Chapter 11 ALKENES; INFRARED SPECTROSCOPY AND MASS SPECTROMETRY CH3 A 2-Methylpentane: [CH3OCHOCH2CH2CH3]G? 88n [CH3CHCH3]G (m/z 43) h h [CH3CH2CH2CHCH3]G (m/z 71) Cleavage will occur mainly about the CH to give secondary cations. The best match is spectrum C. CH3 A 3-Methylpentane: [CH3CH2OCHOCH2CH3]G? 88n [CH3CH2CHCH3]G (m/z 57) h h Fragmentation occurs at the indicated bond to form mainly sec-butyl cation and, to a lesser extent, ethyl cation (m/z 29). Spectrum A fits best. As is typical in mass spectrometry, the large amount of energy imparted to molecules in the process induces rearrangements and other modes of fragmentation as well. Fortunately, ions arising from such processes do not usually dominate the spectrum. 39. Major peaks: m/z 43 (CH3CH2CH2) from MOBr m/z 41 (CH2CHPCH2) from MOHBrOH Minor peaks: m/z 109 (CH2CH2 81Br) from MOCH3 m/z 107 (CH2CH2 79Br)} m/z 42 (CH3CHPCH2) from MOHBr m/z 29 (CH3CH2) from MOBrOCH2 m/z 28 (CH2PCH2) from MOBrOCH3 m/z 27 (CH2PCH) from MOBrOCH3OH 40. Compound is saturated (see Section 11-6). Try to use the general guidelines that intense fragment peaks either result from the loss of relatively stable neutral species or are due to relatively stable cations. So, looking at the high intensity m/z 73 peak for isomer C, it corresponds to (M 15), or loss of CH3. This is most likely if the remaining fragment is a very stable cation, for example, Looking at the rest of the spectrum, the base peak is at m/z 59, or (M 29), loss of CH3CH2. This is, all together, good evidence for isomer C being 2-methyl-2-butanol, as shown. Isomer B also has a peak at m/z 73 for loss of CH3. Its base peak (m/z 45) corresponds to loss of 43, or CH3CH2CH2. These signals are what you might expect for 2-pentanol. CH3CH2 C OH CH3 CH3 (CH3)2COH m/z 59 CH3CH2 CH3CH2 C OH CH3 CH3 CH3 CH3CH2COH m/z 73 cation, stabilized by oxygen lone pair. CH3 1559T_ch11_199-219 11/02/05 21:43 Page 208
