Ch11 problems - 488 4 Alkenyl hydrogens and carbons appear...

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Unformatted text preview: 488 4. Alkenyl hydrogens and carbons appear at low field in 1H NMR (5 = 4.65.7 ppm) and 1’i‘c NMR (3 2 100— 140 ppm) experiments, respectively. Jam.Is is larger than Jcis, germinal is very small, and Jauwc is variable but small. 5. Infrared spectroscopy measures vibrational excita- tion. The energy of the incident radiation ranges from about 1 to 10 kcal mol"1 (A a 25—161 ,um; 17 = 600“ 4000 cm”). Characteristic peaks are observed for cer— tain functional groups, a consequence of stretching, bending, and other modes of vibration, and their com- bination. Moreover, each molecule exhibits a character— istic infrared spectral pattern in the fingerprint region below 1500 cm_l. 6. Alkanes show IR bands characteristic of CeH bonds in the range from 2840 to 3000 cm”. The C=C stretch- ing absorption for alkenes is in the range from 1620 to 1680 cm‘ 1, that for the alkenyl CWH bond is about 3100 cm‘ 1. Bending modes sometimes give useful peaks be- low 1500 cm’l. Alcohols are usually characterized by a broad peak for the 0—H stretch between 3200 and 3550 cm“. 7. Mass spectrometry is a technique for ionizing mole cules and separating the resulting ions magnetically by molecular mass. Because the ionizing beam has high en- ergy, the ionized molecules also fragment into smaller particles, all of which are separated and recorded as the mass spectrum of a compound. High-resolution maSS Spectral data allow determination of molecular formu- Emblems @Draw the structures of the molecules with the following names. Chaplin t1 ALKENES; INFRARED SPECTROSCOPY AND MASS SPECTROMETRY 10. 11. (a) 4,4—Dichioro-trans-2-octene (c) 5-Methyl-cis—3 ~hexen-1mol (e) (E )-3 -Meth0xy—2—methy1—2-buten— 1 —01 27; - ame each of the following molecules in accord with the IUPAC system of nomenclature. :1 “”1 M (e) . The relative stability of isomeric alkenes can be estub}: las from exact mass values. The presence of certain 91, ements (such as Cl, Br) can be detected by their is”, topic patterns. The presence of fragment-ion signals ;-., I mass spectra can be used to deduce the structure of: molecule. . Degree of unsaturation (number of rings + numberm 1r bonds) is calculated from the molecular formula by u. ing the equation ' Hsat W Hacrua] I Degree of unsaturation = 2 where HS,“ = 2m: + 2 - nx + nN (disregard oxygen and sulfur). u | lished by comparing heats of hydrogenation. It decreases - with decreasing substitution; trans isomers are more su. . ble than cis. — Elimination of haioalkanes (and other alkyl derivatrgiiill may follow the Saytzev rule (nonbulky base, intefiui i alkene formation) or the Hofmann rule (bulky base..l§ti"rl| minal alkene formation). Trans alkenes as products dominate over cis alkenes. Elimination is stereospecl ' I as dictated by the anti transition state. Dehydration of alcohols in the presence of strongmf‘l usually leads to a mixture of products, with the masts-fl. ble alkene being the major constituent. (b) (Z)—4~Bromo-2—iodo-2—pentene (d) (R)-1,3-Dichlorocycloheptene C1 CF3 (f) A/kCl \ innit-111m: 25.2 (CH2), 41.9 (CH). 48.5 (CH2), 135.2 (CH). (Hint: This one is difficult. flit- mgtecule has one double bond. How many rings must it have?) I l .15th from both ordinary and DEPT 13C NMR spectra for several compounds with the F“ Win'lk'lfl CSHJD are given here. Deduce a structure for each compound. (a) 25.3 (CH2); tilt} [3.3 ((3-13), 17.1 (CI-I3), 25.5 (CH3), 118.7 (CH), 131.7 (Cqummm); (c) 12.0 (CH3), If: .1: (CH3), 20.3 (CH2), 122.8 (CH), 132.4 (CH). _ ,_ 1.5111 the Hooke‘s law equation, would you expect the C—X bonds of common haloal- EQ- 5W5.“ = Cl, Br, I) to have IR bands at higher or lower wavenumbers than are typical "Iii? mg between carbon and lighter elements (e.g., oxygen)? .flnvflt each of the following IR frequencies into micrometers. ii 11211 cm-1 (c=0) (b) 1650 cm—1 (C=C) (c) 3300 cmml (0—H) "tilt-$90- Cmrl (alkene bend) (e) 1100 our1 (C—O) (f) 2260 cm"1 (GEN) L 'Lnlth each of the following structures with the IR data that correspond best. ’15:? _ ,fihhm'iations; w, weak; m, medium; 5, strong; br, broad. (a) 905 (s), 995 (m), " "5mm (m), 1640 (111), 2850—2980 (s), 3090 (In), 3400 (s, br) em“; (b) 2840 (s), :11an (5) cm"; (c) 1665 (m), 2890—2990 (s), 3030 (m) cm_1; (d) 1040 (m), ”glam—2930 (s), 3300 (5, br) cm“. ‘0 :- . WOH Won A B c D 35'“.'- ~ to a are five}; . 1%; 111.] have just entered the chemistry stockroom to look for several isomeric bromopen- nee. There are three bottles on the shelf marked CsHllBr, but their labels have fallen I iiifi'. The NMR machine is broken, so you devise the following experiment in an attempt '-!tr:dtttrmine which isomer is in which bottle: You first treat a sample of the contents in march bottle with NaOH in aqueous ethanol, and then you determine the IR spectrum of each. product or product mixture. Here are the results: " polarity {Ll re possible: I N O 111 C5H11Br isomer in bottle A 3—H) IR bands at 1660, 2850—3020, and 3350 cm-1 N OH , CHI {ii} CSHHBr isomer in bottle B a—> IR bands at 1670 and 2850—3020 cm—1 ‘2 / ' - .. N OH , C=C\ ill"; C5H11Br isomer in bottle C 8—) IR bands at 2850—2960 and 3350 cm 1 H fifll'i'r'liat do the data tell you about each product or product mixture? ill} Suggest possible structures for the contents of each bottle. 'Z, 1" organic compound exhibits IR spectrum F, on p. 494. From the group of structures tiltomu, choose the one that matches the spectrum best. {v Yr Jr _< : H M66113 ldS are givfi'lF H each case. H 1; (b) 04% ; 25.8 (CH2)._n..; | 2), 125.7 (CHE (Cquatemfiryh' Problems 493 494 013le ll AlKENES: INFRARED SPECTROSCOPY AND MASS SPECTRflMETRY Relative abundance lb 100 amuse -tl 13%]- each stn 1mm and I ' I /\ !I ll WE I J (Ethan-am th ,jmlscuiar ft {in cxsflzso Transmittance (9%) | laconic, data: 0 .. 4000 3500 3000 2500 2000 1500 1000 600 cm-‘ll' I Wavenumber | lg Emolocu F I these data. .hflmhe three compounds hexane, 2-methylpentane and 3- -methylpentane correspond to thta ree mass spectra shown below. Match each compound with the spectrum that best fit its structure on the basis of the fragmentation pattern. H 100 50 0 2IOIIII [III'5. 40 60 80 “'1'.- C m/z miz m/z _ 080.:signBOO as many peaks as you can in the mass spectrum of 1—bromopropane (Figure 11' ."' 40. The following table lists selected mass~spectral data for three isomeric alcohols with {1135!r " I muia C5H120. 0n the basis of the peak positions and intensities, suggest structures for' of the three isomers. A dash means that the peak is very weak or absent entirely. Relative Peak Intensities m/z Isomer A Isomer B 15°13‘5- ss M+ _ _ a 87 (M — 1)+ 2 2 d 5 =:. 73 (M _ 15V ‘ 7 53 .,;}suiausn 70 (M a 159* 38 3 Hayse- has 55 (M — 15 — MN 60 17 33 . : __msun v 45 (M — 43)+ 5 100 10 . . mien; In 42 (M— 18L28)+ 100 4 5-._ :' 0 7T Problems 495 Wnnine the molecular formulas corresponding to each of the following structures. '. reach Sanctum, calculate the number of degrees of unsaturation from the molecular . _-,. 'li'nnnllm and evaluate whether your calculations agree with the structures shown. l or \ g, (b) /\/ (c) (a) HO NM 0/\ -- -- ‘ r) (g) \ (h) I I I. :3' iculfllc the degree 0f unsaturation that corresponds to each of the following ‘ifphEEfllllflr fermulas, (a) C7H12; (b) C3H7N02; (C) CesCls; (d) C10H22011; (e) CsHmS; jfliicisflesoz- :. 13;" n-iwmflfl]fl]1 with an exact molecular mass of 96.0940 exhibits the following spectro- | "'"mic data: 1H NMR 5 = 1.3 (m, 2 H), 1.7 (in, 4 H), 2.2 (m, 4 H), and 4.8 (quin, '-.1‘..=- 3 Hz, 2 H) ppm; 13C NMR 5 = 26.8, 28.7, 35.7, 106.9, and 149.7 ppm. The [R | Mum is shown below (spectrum G). Hydrogenation furnishes a product with an jgfiségqmolecular mass of 98.1096. Suggest a structure for the compound consistent with mass- data. 'Iflfl l l I | l'. .j;l'l' a so .'-'- E- ___ g 3072 (Figure -'11"!‘1- "I 11— ols with the 'g‘:'. 1649 ictures form-:51: rely. ‘ ' 888 || 0 150111.. _. :_ n 4000 3500 3000 2500 2000 1500 1000 600 curl _. | Wavenurnber _G 55 II __. -'1 3 | #11111: isolation of a new form of molecular carbon, C50, was reported in 1990. The 00 fight-“mm has the shape of a soccer ball of carbon atoms and possesses the nickname 1 | _t bllfl‘kyhuil" (you don’t want to know the IUPAC name). Hydrogenation produces a 33 ' :hll'fimfiatbuu with the molecular formula C60H36. How many degrees of unsaturation 10 - .1117; Present in C60? In CGDHM? Does the hydrogenation result place limits on the 6 _ Mull-5:55 of 11- bonds and rings in “buckyball”? (More on can is found in Chemical highligm 15-1.) 496 Ehapter ll AIKENES; INFRARED SPECTROSCOPY AND MASS SPECTROMETRY @lace the alkenes in each group in order of increasing stability of the double bond and . increasing heat of hydrogenation. " "- th H3C\ /CH3 H3C\ ““5“ (a) cszcm /C=C\ /c= CH2 _ :2: was rig tc H3C CH3 H3C rib: ' 'tls sho (b) H\C_C/H H\C——C/CH(CH3)2 H\ C/H / _ \ / _ \ /C"— \ .. H3C eaten?)2 ch H (CH3)2CH omega“ | (£0 0:] 0:) I h.) CH3 CH; CH "I . . . H C H C H C ,_', 'fihe‘ of tiles (d) 3 \é 3 fi 3 \6 (9) III 0 ‘1‘. l_" -.I-_ r lllld i . | . filhin in t 46. Write the structures of as many simple alkenes as you can that, upon catalytic hydra; ’ III-[Will genation with H; over Pt, will give as the product (a) 2-methylbutane; (b) 23- , dimethylbutane; (c) 3,3-dimethylpentane; (d) 1,1,4-trirnethylcyclohexane. In each cases-h which you have identified more than one alkene as an answer, rank the alkenes in err-titty, of stability. I .‘H @The reaction between 2-bromobutane and sodium ethoxide in ethanol gives rise to that] E2 products. What are they? Predict their relative amounts. I 48. at key structural feature distinguishes haloalkanes that give more than one stercd‘__ mer on E2 elimination (e.g., 2-bromobutane, Problem 47) from those that give only Hi. I 7 single isomer exclusively (e.g., 2-bromo-3-methylpentane, Section 11-10)? @Write the most likely major product(s) of each of the following haloalkanes with sodiit - I C ethoxide in ethanol or potassium tert—bntoxide in 2-methyl-2-propanol (tert—butyl ' , _ j alcohol). (a) Chloromethane; (b) l-bromopentane; (c) 2-bromopentane; (d) l-chloro— i ' . l-methylcyclohexane; (e) (1«bromoethyD~cyclopentane; (f) (2R,3R)—2—ch]oro- . 3-ethylhexane; (g) (212,3S)-2-chloro—3-ethylhexane; (h) (ZS,3R)—2—chloro-3-ethy1hexan‘~‘3 50. Referring to the data in Chapter Integration Problem 11-24, predict how the rate oil IIIII E2 reaction between 1-brornopropane and sodium ethoxide in ethanol would comP= with those of the three substrates discussed in that problem under the same reactinj} I conditions. conformation required for E2 elimination. (See the structures labeled “StereospecifieiLfi in the E2 Reaction of 2-Bromo-3-methylpentane” on p. 481.) Are the reactive conform"- tions also the most stable conformations? Explain. .. . I @Draw Newman projections of the four stereoisomers of 2-bromo—3—methylpentane in M. 52. Refen-ing to the answer to Problem 34 of Chapter 7, predict (qualitatively) the relatiWT' | amounts of isomeric alkenes that are formed in the elimination reactions shown. 53. Referring to the answers to Problem 28 of Chapter 9, predict (qualitatively) the relatfr'fi. yields of all the alkenes formed in each reaction. Compare and contrast the major products of dehydrohalogenation of 2—chloro- . .Illiifintiesg w 4-methylpentane with (a) sodium ethoxide in ethanol and (b) potassium tart—butoxideih'Ir ' ' 2-methyl-2—propanol (tert—butyl alcohol). Write the mechanism of each process. N63?l consider the reaction of 4-methyl-2—pentanol with concentrated H2804 at 130°C. and a the ir ”Jim-if“ these _ Problems 497 jouble bum] |. . 1““ its producds) and the mechanism of its (their) formation with those from the m. .. Wmhflmggnnfinns in (a) and (b). (Hint: The dehydration gives as its major product a 1-4-1 5w“. that is not observed in the dehydrohalogenations.) l'rlfldfl to Problem 53 of Chapter 7, write the structure of the alkene that you would 't'iD be formed as the major product from E2 elimination of each of the chlorinated is shown. CH2 CH2 ”1';"'Eq'hy'l'll'clnhflcm is Il'lore stable than methylenecyclohexane (A, in the margin), but A 2| .4.‘ ”cm-cycl‘opropanc (B) is more stable than 1-methyl-cyclopropene. Explain. ' ' products of bimolecular elimination from each of the following isomeric halo— A B compounds. Br H (b) Br H ’ “ C H5 “ c H I& 6 CfiHS/gg 6 5 H’ CH3 ch .- . min in detail the differences between the mechanisms giving rise to the following .s.._-.-_ - Experimntal results. ' In each $113 9H3 alkenes ii In “L I Cl I '. " .- Nat ‘OCH2CH1, CHSCHZOH ~—--—__t.._> res rise tut | N _ 'l I cntcng)2 CH(CH3)2 ,( M- 100% it give on? I'. )? ' ., : CH3 ¢H3 . p;_ I i nes With s will -. ~ - Na+ -ocnzcnj, CH3CH20H 2rt—butyl‘_ _ _ + d) l-Chl'filllI-‘f: I C] Q)“; m , ._ anon?)2 CH(CH3)2 CH(CH3)2 -6 y. ._ 25% 75% w the rate»? W0111d 00 ___A You have just been named president of the famous perfume company, - ,ame rea -'_'. is “R” Us. Searching for a hot new item to market, you run across a bottle labeled Lilli“ Biol-1200, which contains a liquid with a wonderfully sweet rose aroma. You want lpentane iii," .-,- ill “with so you set out to elucidate its structure. Do so from the following data. (i) 1H :reospetl - - fill-1m: clear signals at 8 = 0.94 (d, J= 7 Hz, 3 H), 1.63 (S, 3 H), 1.71 (s, 3 H), 3.68 (t, :tive cont. ' " TI“? 7 Hz, 2 H), 5.10 (t, J = 6 Hz, 1 H) ppm; the other 8 H have overlapping absorptions in gully-range a = 1.3—2.2 ppm. (ii) 13C NMR (1H decoupled): s = 60.7, 125.0, 130.9 ppm; _ 4W9“ other signals are upfield of 3 = 40 ppm. (iii) IR: 13 = 1640 and 3350 cm“. (iv) Ox- ') the re ‘ 'l'l'illlflfln with buffered PCC (Section 8-6) gives a compound with the molecular formula l'jfiql’iiso. Its spectra show the following changes compared with the starting material: 1H y) the rail“ }, 5‘.- . 1 signal at 8 = 3.68 ppm is gone, but a new signal is seen at 5 = 9.64 ppm; 13C "'“ pflpfls: signal at 8 = 60.7 ppm is gone, replaced by one at 6 = 202.1 ppm; IR: loss of W ElilEfiD-l at 13 = 3350 cm‘l; new peak at fi = 1728 cm‘l. (v) Hydrogenation gives C10H220. OH [010- .1 _ Eitllltntiual with that formed on hydrogenation of the natural product geraniol (see margin). Geraniol ’rt-butogt . . I . scess. Next r-fiumtl ll‘ld information in Table 11-4, match up each set of the following IR signals with 130°C, art! I (t. “'33 “I" these naturally occurring compounds: camphor; menthol; chrysanthemic ester; ...
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