Chapter_12_solutions - Structure Determination: NMR...

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Unformatted text preview: Structure Determination: NMR Spectroscopy 263 Solutions to Problems 1 2.1 1.20 x 10-4 kJ/mol 7t(inm) s K_C=M;V=137MHZ v v 3.0x103m/s 7t = _—e——e— : 1.60m 1.87x103Hz .4 E = 1.20 x11060kllmol : 75 x 10-5 kymol 1.87 x 108 Hz Compare this value with E = 8.0 x 10'5 kJ/mol for 1H. It takes slightly less energy to spin-flip a l9F nucleus than to spinmflip a 1H nucleus. 1 2 . 2 b H lCHs a 0:0 / \ cH CI 2—Chloropropene has three kinds of protons. Protons b and c differ because one is cis to the chlorine and the other is trans. 1 2 . 3 H Observed chemical shift (in Hz) 200 MHZ 1454 HZ . 610 HZ _ F = .2 = M = .0 f C Cl 200MHz 7 751m CHC13 (b) 8 200 3 55 or H3 693 Hz 1060 Hz M = d 2 ——~—-———* I 3.46 5 for CH3OH ( ) 5 200 MHZ 5.30 5 for CH2C12 r-s :3 \_/ 00 | (C) 8 : 200MHz 1 2 .4 (a) 5 fl Observed chemical shift (# Hz away from TMS) Spectrometer frequency in MHZ Units of 3 are parts per million. In this problem, 5 = 2.1 ppm Observed chemical shift 200 (MHz) 420 Hz 2 Observed chemical shift 2.1 ppm = (b) If the 1H NNIR spectrum of acetone were recorded at 500 MHz, the position of absorption would still be 2.1 5 because measurements given in ppm or 5 units are independent of the operating frequency of the Nh/[R spectrometer. (c) 2.15 : pbser‘gegocmflsm; Observed chemical shift : lOSOHz 1 Structure Determination: NMR Spectroscopy 265 12. 7 lglethyl propanoate has 4 unique carbons, and each one absorbs in a specific region of the C sppptrum. The absorption (4) has the lowest value of 6 and occurs in the —CH3 region of the C spectrum. Absorption (3) occurs in the —CH2— region. The methyl group (1) is next to an electronegative atom and absorbs downfield from the other two absorptions. The carbonyl carbon (2) absorbs far downfield. 5 (ppm) Assignment I? 9.3 4 CH30H2COCH3 27.6 3 4 3 2 1 51.4 1 174.6 2 1 2 . 8 Strategy: The top spectrum shows all eight 13C NMR peaks. The middle spectrum (DEPT~90) shows only peaks due to CH carbons. From the DEPT-90 spectrum, the absorption at 124 6 can be assigned to the vinyl carbon (5), and the absorption at 68 6 can be assigned to the —OH carbon (2). The DEPT—135 spectrum shows all but the quaternary carbon (6), which appears in the top spectrum at 132 8. The top half of the DEPT—135 spectrum shows absorptions due to CH3 carbons and CH carbons (which we have already identified). The 3 remaining peaks on the top of the DEPT—135 spectrum are due to methyl groups. Although we haven’t learned enough to identify these peaks, the peak at 23 8 is due to carbon (1). The other two peaks arise from carbons (7) (26 5) and (8) (18 5). The bottom half of the DEPT~135 shows the two CH2 carbons. Carbon (3) absorbs at 39 5 (negative), and carbon (4) absorbs at 24 5 (negative). Solution: w 6 \ 4 2 7 5 3 1 6—Methy1hept—5—en—2—ol Carbon Chemical Shift (5) 1 23 68 39 (negative) 24 (negative) 124 132 26 18 OOxJQM-p-LAI‘J 12.9 Strategy: Identify the carbons as CH3, CH2, CH or quaternary, and use Figure 12.7 to find approximate values for chemical shifts. (When an actual spectrum is given, it is easier to assign the carbons to the chemical shifts.) Remember: DEPT—90 spectra identify CH carbons, and DEPT — 135 spectra identify CH3 carbons (positive peaks), CH carbons (positive peaks already identified), and CH3 carbons (negative peaks). Quaternary carbons are identified in the broadband—decoupled spectrum, in which all peaks appear. 266 Chapter 12 Solution: Carbon Chemical Shifi (6) DEPT-90? DEPT—J35? 1 10—30 no yes (positive) H3g\ O O 2 30—50 no yes (negative) A/b 3 160—230 no no 6 5\4 3 2 1 4 l 10—} 50 yes yes (positive) 5 1 10—150 no no 6 1040 no yes (positive) 7 50k90 no yes (positive) 1 2.10 Strategy: Always start this type of problem by calculating the degree of unsaturation of 12.11 the unknown compound. C1 11116 has 4 degrees of unsaturation. Since the unknown hydrocarbon is aromatic, a benzene ring accounts for all four degrees of unsaturation. Next, look for elements of symmetry. Although the molecular formula indicates ll carbons, only 7 peaks appear in the 13C NMR spectrum, indicating a plane of symmetry. Four of the 7 peaks are due to aromatic carbons, indicating a benzene ring that is probably monosubstituted. (Prove to yourself that a monosubstituted benzene ling has 4 different kinds of carbons). Solution:The DEPT~90 spectrum shows that 3 of the kinds of carbons in the aromatic ring are CH carbons. The positive peaks in the DEPT—135 spectrum include these three peaks, along with the peak at 29.5 5, which is due to a CH3 carbon. The negative peak in the DEPT-135 spectrum is due to a CH2 carbon. Two peaks remain unidentified and are thus quaternary carbons; one of them is aromatic. At this point, the unknown structure is a monosubstituted benzene ring with a substituent that contains CH2, C, and CH3 carbons. A structure for the unknown compound that satisfies all data: CH3 I OCH2—?—CH3 CH3 I? CH30H2CH20H2C : CH2 2-Bromohex—l—ene CH3CH20H20H2CECH + HBr ————a-— 01‘ CH30H20H20H20H = CH Br ? l—Bromohex— 1 —ene The two possible products are easy to distinguish by using 13C NMR. 2-Bromohex-l—ene, the actual product formed, shows no peaks in its DEPT-90 13C NMR spectrum because it has no CH carbons. The other possible product, l—bromohex—l—ene, shows 2 peaks in its DEPT—90 spectrum. Structure Determination: NMR Spectroscopy 267 1 2.. 1 2 Strategy: First, check for protons that are unrelated (none appear in this problem). Next, look for molecules that already have chirality centers. Replacement of a —CH2- proton by X in (d) and replacement of a -CH3 proton in (e) produces a second chirality Center, and the two possible replacement products are diastereorners. Thus, the indicated protons in (d) and (e) are diastereotopic. Finally, for the other molecules, mentally replace each of the two hydrogens in the indicated set with X, a different group. In (a), the resulting products are enantiomers, and the protons are enantiotopic. Replacement of the protons in (b) produces two chirality centers (the carbon bearing the hydroxyl group is now chiral) and the indicated protons are diastereotopic. Replacement of one of the methyl protons in each of the groups in (c) produces a pair of double—bond isomers that are diastereomers; these protons are diastereotopic. The protons in (f) are homotopic. Solution: (a) enantiotopic (b) diastereotopic (c) diastereotopic \ / —>H H —> H3C\ [H H H —> H OH lc=c\ —" H30 CH3 0 (d) diastereotopic (e) diastereotopic (f) homotopic \ / o —" H 0 H H O H —)- H Br +- 12. 13 Kinds 0 non- Kinds ofnori- Compound equivalent protons Compound equivalent protons (21) 1 2 (b) 4 1 2 I3 4 CH3OCH20HCH3 (C) 1 2 3 (d) 3 a CH30H2CH2N02 3 H 4 4 1 H CH3 268 Chapter 12 Kinds ofnon- Kinds of non- Compound equivalent protons Compound equivalent protons (e) 3 4 5 (1°) 1 2 . 2 1 H H t ' 30\ I (:Hsc\H2 : /CH20H3 3 = i 1 /2 \5 arc : C\3 CH3CH2 H H : H The two vinylic protons plane of are nonequivalent. symmetry 1 2 .14 4 3 HO ‘H I * 002‘ *020 (S)-Malate 2 \1 H H / \ diastereotopic Because (S)—malate already has a chirality center (starred), the two protons next to it are diastereotopic and absorb at different values. The 1H NMR spectrum of(S)—malate has four absorptions. 1 2. I 5 Compound 5 Kind of proton (a) Cyclohexane 1.43 secondary alkyI (b) CH3COCH3 2.17 methyl ketone (C) Benzene 7.37 aromatic (d) CH20I2 5.30 protons adjacent to two halogens (6) OHCCHO 9.70 aldehyde (f) (CH3)3N 2.12 methyl protons adjacent to nitrogen 1 2 . I 6 Proton 5 Kind of proton 5 4 1 1.0 primaly alkyl 5 H T 2 1 2 1.8 allylic H Geo/CHcha 3 6.1 vinylic 7 5 3:1 4 6.3 vinylic (different from proton 3) CH30 8 H 5 7 . 2 aromatic H 6 6.8 aromatic 7 3.8 ether This compound has seven different kinds of protons. Notice that the two protons labeled 5 are equivalent to each other, as are the two protons labeled 6, because of rotation around the bond joining the aromatic ring and the side chain. Structure Determination: NMR Spectroscopy 269 HSCQ’CHS p-Xylene There are two absorptions in the 1H NMR spectrum of p—xylene. The four ring protons absorb at 7.0 5, and the six methyl-group protons absorb at 2.3 5. The peak ratio of methyl protonszring protons is 3:2. 12.17 5 12.18 Number of Compound Proton Adjacent Protons Splitting (a) 1 2 CHBF2CH3 1 3 quartet 2 1 doublet (b) 1 2 a . CH3OCH20H28r l 0 Singlet 2 2 triplet 3 2 triplet (C) 1 2 t 4 quintet (d) 1 H30 0 l l doublet 1 la” 3 4 2 6 septet CHSCHCOCHECHs 3 3 quartet 4 2 triplet A - .---«-r—: ‘ .“qeti'r‘mz smug:- 12.21 12.22 Structure Determination: NMR Spectroscopy 271 1 H I 3 Ce: /CH23r C 2 I H (B—3—Bromo-Lpheny1prop— l —ene Coupling of the C2 proton to the Cl vinylic proton occurs with J = 16 Hz and causes the signal of the C2 proton to be split into a doublet. The C2 proton is also coupled to the two C3 protons with J = 8 Hz. This splitting causes each leg of the C2 proton doublet to be split into a triplet, producing six lines in all. Because of the size of the coupling constants, two of the lines coincide, and a quintet is observed. J1_2 = 16 H2 J2_3 = 8 HZ CH3 CH3 CI H HCI or c: H CH3 _..—.’. Focus on the 1H NMR methyl group absorption. In the first product. the methyl group signal is unsplit; in the other product, it appears as a doublet. In addition, the second product shows a downfield absorption in the 2.5 5 — 4.0 5 region due to the proton bonded to a carbon that is also bonded to an electronegative atom. If you were to take the 1H NMR spectrum of the reaction product, you would find an unsplit methyl group, and you could conclude that the product was l-chloro—l—methylcyclohexane. 272 Chapter 12 Visualizing Chemistry 12.23 (a) (b) 3 4 O “‘3'” H H H +— 1 II 3/ \ l H» H c c H c , o 3 “‘0’ \CHZCI ‘c’ / \ I 5 \2 —> H30 H 2 l. doublet H ‘— 2. septet l. singlet 3. singlet 2. doublet 3. doublet 4. doublet 5. triplet 1 2.24 The compound has 5 different types of carbons and 4 different types of hydrogens. 13C 1H 3 3 0 H H 0 H H 1 ll \ / 10 g \C/ HO 2 C C 5 H 4 3 \C/3\O/4\CH3 3 \C/ \0/ \CH3 /\ 2/\ H CI H Cl 13c 1 5 4 2 3 200 150 100 50 0 Chemical Shift (8) Structure Determination: NMR Spectrosc0py 273 6 4 2 0 Chemical Shift (5) ; 1 2 .25 If you assign R,S configurations to the two carbons bonded to the methyl group, it is apparent that cis—l ,2—dimethylcyclohexane is a meso compound.When the cyclohexane E ring undergoes a rin —flip, the ring passes through an intermediate that has a plane of ' symmetry. Both the 3C NMR spectrum and the 1H NMR spectrum show 4 peaks. 13C CH 4 2 3 4 2 CH 3 3 12. 26 (a) Because cysteine has a chirality center, the indicated protons are diastereotopic. (b) Imagine replacing first one, then the other, of the indicated protons with a substituent X. The two resulting compounds would be enantiomers. (=1) diastereotopic (b) \ / H H H2NH \ / enantiotopic 274 Chapter 12 Additional Problems 12.27 12.28 1 2.29 (a) Since the symbol "8" indicates ppm downfield from TMS, chloroform absorbs at 7.3 5 _ Observed chemical shift (in ‘ 200 MHz (b) 4.78 5 (a) 2.18 s (c) 7.52 5 6 x 300 MHZ 7- Observed chemical shift (in Hz) (21) 6301112 1035 HZ 1890112 (d) 23”) HZ ppm. “3) 5 c @Eszrsashsslissw — Spectrometer frequency in MHZ chemical shift . . . = -—--—~--r~—- . = '1 1 ft 7 3 ppm 360 MHZ , 7 3 ppm x 360 MHZ chemica SH 2600 Hz 2 chemical shift (c) The value of 5 is still 7.3 because the chemical shift measured in 5 is independent of the operating frequency of the spectrometer. (at) (b) H3C\ /CH3 4 3 1 3 2 1 C=C CH3 “20—055, CH3 / \ o o=c c\ “30 CH3 CH3 \ ’ CH3 1 H2C—CH2 1 4 3 3 2 13C: 2 absorptions 13C: 5 absorptions 1H: 3 absorptions 1H; [absorption (at room temperature) (at room temperature) II It 1 H 2 CH3CCH3 (CH3)3CCCH3 (CH3)3CCCH3 1 2 3 4 13C: 2 absorptions 1H: 1 absorption 13C: 4 absorptions 1H: 2 absorptions (6) (1”) H30 CH3 H30 CH3 5 13C: 3 absorptions 1H: 2 absorptions 13C: 3 absorptions 1H: 2 absomtions Structure Determination: NMR Spectroscopy 275 12.31—12.32 Number of 13 C Absorprions Carbons Showing Peaks in DEPT-135 l3C NMR Spectrum ---~———____________._____.__ No Peeks Compound Positive Peaks Negative Peaks (a) 1 CH H 1 r 0 . H\ 3 I CH3 4 carbons 1,- carbons 3,4 3 3 4 4 (b) 1 2 3 CH30H206H3 3 carbons 1,3 carbon 2 (C) 1 (13H3 6 carbons 1,3 carbons 4,5,6 carbon 2 1 1 CH3 4 4 5 5 6 (d) 3 CH3 6 carbons l,3,4,6 carbon 5 carbon 2 654|21 CHSCHZCHCEECH (e) i 1 H30 CH3 4 carbons 1,2 carbons 3,4 2 2 H-- —~H 3 3 4 4 (1") 2 1 O 4 carbons 2,3,4 carbon 1 3 40: 3 12.33 A nucleus that absorbs at 6.50 5 is less shielded than a nucleus that absorbs at 3.20 5 and thus requires a weaker applied field to come into resonance. A shielded nucleus feels a smaller effective field, and a stronger applied field is needed to bring it into resonance. 276 Chapter 12 12.34 (a) enantiotopic (b) diastereotoPic (c) diastereotopic % w H G —>- H H <— —> H H <2— H' H H. H 7' \ Refer to Problem 12.12 for help. The protons in (c) are diastereotopic because the molecules that result from replacement of the indicated hydrogens are diaslereomers (prove it to yourself with models). 1 2 . 3 5 Kinds of non— Kinds ofnou- Cor-npound equivalent protons Compound equivalent protons (a) 1 1 0)) H30 CH3 1 2 3 4 2 2 4 4 3 3 4 (C) (d) 3 H 1 1 2 4 I 1 6 C H | 2 2 H B 4 2 1 1 5 (e) 3 H 2 i 5 H C 4 5 “cl? \COOCHZCHa 1 H 1 2 . 3 6 (a) homotopic (b) enantiotopic (c) diastereotopic H +— H 4— . ‘ > I l > H C 3 CH2 : : H3C +- H <— H *— 1 2. 3 7 Lowest Chemical Shift ———-1- Highest Chemical Shift CH4 < Cyclohexane < CHsCOCH3 < CHECI2,HQC=CH2 < Benzene 0.23 1.43 2.17 5.30 5.33 7.37 Structure Determination: 1 2 . 3 8 Number Peak Compound of peaks Assignment (fl) 1 2 (CH3)30H 2 1 2 (b) O 3 l 1 2 || 3 2 CHBCHchCHB 3 (C) 2 1 H CH3 2 1 \ x 2 0:0 1 / \2 H30 H 1 2 .3 9 Peak Splitting 0 ' Assignment Patten: 1 2 || 3 4 CH30H2000H(CH3)2 I triplet 2 quartet 3 septet 4 doublet NMR Spectroscopy 277 Splitting Partem doublet (9H) multiplet (dectet) (1H) tn'plet (3H) quartet (2H) singlet (3H) doublet (6H) quartet (2H) (3H) (2H) (1H) (6H) 12 .40 First, check each isomer for structural differences that are obviously recognizable in the 1H NMR spectrum. If it is not possible to pick out distinguishing features immediately, it may be necessary to sketch an approximate spectrum of each isomer for comparison. (a) CH3CH=CHCH2CH3 has two vinylic protons with chemical shifts at 5.4 — 5.5 5. Because ethylcyclopropane shows no signal in this region, it should be easy to distinguish one isomer from the other. (b) CH3CI’IZOCH’2CH3 has two kinds of protons, and its 'H NMR spectrum consists of two peaks — a triplet and a quartet. CH3OCH2CI-12CH3 has four different types of protons, and its spectrum is more complex. In particular, the methyl group bonded to oxygen shows an unsplit singlet absorption. .c _ ._.....A. “Wyow M.WJW.—,_w u 278 Chapter 12 (c) Each compound shows three peaks in its 1H NMR spectrum. The ester, however, shows a downfield absorption due to the —CI'12— hydrogens next to oxygen. No comparable peak shows in the spectrum of the ketone. O CH JDI 3 0 CH2C H:3 0 ll CH3CHQCCH3 8 6 4 2 0 8 6 4 2 0 Chemical Shiit(8) Chemical Shift (8) (1) Each isomer contains four different kinds of protons — two kinds of methyl protons and two kinds of vinylic protons. For the first isomer, the methyl peaks are both singlets, whereas for the second isomer, one peak is a singlet and one is a doublet. 12.41 (a) (b) (C) O Q E] O 12.42 H CH3 CH3 H CH3 H H H H H c CH 3 . . 3 ring—flip T nng'fllp H30 cis—l,3-Dimethylcyclohexane H30 H H CH3 H CH3 CH3 H trans—1,3—Dimethylcyclohexane cis-1,3—Dimethylc clohexane is a meso compound. Because of symmetry, it shows 5 absorptions in its C NMR spectrum. trans—1,3—Dimethylcyclohexane exists as a pair of enantiomers. At room temperature, both enantiomers undergo ting-flips at a rate that is faster than the time frame of an NMR spectrum and that averages absorptions due to nonequivalent carbons. Likegthe cis isomer, the racemic mixture of trans enantiomers shows 5 absorptions in its ‘C spectrum. Structure Determination: NMR Spectroscopy 279 12 .43 (a),(b) C3H60 contains one double bond or ring. Possible structures for C3H60 include: “2? —Cl7H2 O I \ Cyclic ether Cyclic ether Ether, double bond 0 0 H20 \ , OH II 1| H2O: CHCH20H H (IV on H CH3CCH3 CHaCHQCH 2 Alcohol, double bond Cyclic alcohol Ketone (acetone) Aldehyde (c) Saturated ketones absorb at 1715 cmCl in the infrared. Only the last two compounds above show an infrared absorption in this region. (d) Since the unknown compound of this problem is a ketone and shows only one 1H NMR absorption (in the methyl ketone region), it must be acetone. 1 2.44 Either 1H NMR or 13C NMR can be used to distinguish among these isomers. In either case, it is first necessary to find the number of different kinds of protons or carbon atoms. Kinds of Kinds ofCarbon Number of 1H Number of 13C Compound Protons atoms NMR peaks NM R peaks H2?—(I3H2 1 1 l 1 H20 — CH2 H20=CHCHZCH3 5 4 5 4 CH3CH=CHCH3 2 2 2 2 (CH3)2C 2 CH2 2 I 3 2 3 [3C NMR is the simplest method for identifying these compounds because each isomer differs in the number of absorptions in its 13C NMR spectrum. 1H NMR can also be used to distinguish among the isomers because the two isomers that show two 1H NMR peaks differ in their splitting patterns. 280 Chapter 12 l 2 . 4 5 I Number of Distinguishing 0 Peaks Absorptions 13C 7 Two vinylic peaks lH 5 Unsplit vinylic peak, relative area .1 CH3 0 H 13C 5 One Vinylic peak Split Vinylic peak, relative area 2 The two isomers have different numbers of peaks in both 1H NMR and 13C NMR. In addition, the distinguishing absorptions in the vinylic region of both the 1H and 13C spectra make it possible to identify each isomer by its NMR spectrum. The kctone IR absorption of 3-methylcyclohex-2-enone occurs near 1690 cm" because the double bond is next to the ketone group. The ketone IR absorption of 3-cyclopentenyl methyl ketone occurs near 1715 cm"1, the usual position for ketone absorption. 3-Methylcyclohex—2—enone shows a UV absorption because its double bonds are conjugated. 1 2.46 The unknown compound has no degrees of unsaturation and has two different kinds of hydrogens. The unknown compound is BrCHZCH2CH2Br. 1 2 .4 7 (a) (b) 1 0 H30 H 1 3 2 \ / (CH3)20HCCH3 /C :C\ 2, 3 1 z 0.95 5 (isopropyl group) Br H 2 = 2.10 5 (methyl ketone) 1 = 2.32 6 (methyl group attached 3 = 2.43 5(isopropy1 group) to double bond) 2,3 = 5.35 6, 5.54 5 (vinylic H) 1 2 .48 Possible structures for C4H7C102 are CH3CI-12C02CH2C1 and ClCH2C02CH2CH3. Chemical shift data can distinguish between them. 0 0 II II CH3CH2COCH2CI ClCHgCOCHgCHs A e In A, the protons attached to the carbon bonded to both oxygen and chlorine (—OCH’ZICI) absorb far downfield (5.0 — 6.0 6). Because no signal is present in this region of the H NMR spectrum given, the unknown must be B. In addition, the quartet absorbing at 4.3 8 is typical of a CH; group next to an electronegative atom and coupled with a methyl group. -“ .5 Structure Determination: NMR Spectroscopy 281 12.49 a) b (AC g () C(CA) (c) (d) 2 1 3 3 3 3 O CH CH CH Br \Cfic/ 3 2 II 1 2 2 2 / -— \ BFCHECH2CCH3 Cl 012-120 2 }4 1 = 2.18 5 (allylic) 1: 1.30 5 (saturated) 1: 2.11 5 (next to C=O) l 2 2.15 5 2 = 4.16 5 (H,C1 2 = 7.30 5 (aromatic) 2 = 3.52 5 (next to 2 = 2.75 5 (benzylic) bonded to same C) C=O, Br) 3 : 3.38 5 (H.Br 3 =2 5.71 5 (vinylic) 3 = 4.40 5 (H,Br bonded to same C) The Eisomer is bonded to same C) 4 = 7.22 5 (aromatlc) also a satisfactory answer In (b) and (cl), the aromatic ring hydrogcns accidentally have the same chemical shift. 12.50 (a) (b) (CH3)2CHCH2Br CH3cl3HCH20H2CI C! 12.51 3% 1 ‘1 u 282 Chapter 12 1 2 . 5 2 Carbon 5 (Lyn-n) H O 1 14 5 H I 3 C \ 2 1 2 61 6 4 OCH2CH3 i 166 7 5 S ‘ 127L133 (4 oaks 5 Ethyl benzoate g p ) 1 2 .5 3 Compound A (4 multiple bonds and/or rings) must be symmetrical because it exhibits only six peaks in its 13C NMR spectrum. Saturated carbons account for two of these peaks (8 = 15, 28 ppm), and unsaturated carbons account for the other four (5 = 119, 129, 13], 143 DPT“)- 1H NMR shows a triplet (3 H at 1.1 6), and a quartet (2 H at 2.5 6), indicating the presence of an ethyl group. The other signals (4 H at 6.9 — 7.3 5 are due to aromatic protons. Br @CH2CH3 A 12.5 4 (a) (b) (C) 3 CH3 20H28r CHSCH2CHQCCH3 O 0/ Br 1 2.5 5 The peak in the mass spectrum at iii/z = 84 is probably the molecular ion of the unknown compound and corresponds to a formula of C6H12 — one double bond or ring. The base peak at m/z = 55, corresponds to the loss of an ethyl group. 13C NMR shows three different kinds of carbons and indicates a symmetrical hydrocarbon. The absorption at 132 5 is due to a vinylic carbon atom. A reasonable structure for the unknOWn is hex—3—ene. The data do not distinguish between Cis and trans isomers. CHSCHQCH = CHCHECHS Hex—3—ene Structure Determination: NMR Spectroscopy 283 1 2 .5 6 Compound A, a hydrocarbon having M+ = 96, has "the formula C7H12, indicating two degrees of unsaturation. Because it reacts with B113, Compound A contains a double bond. From the broadband decoupled 13C NMR spectrum, we can see that C711” is symmetrical, since it shows only five peaks. The DEPT~135 spectrum of Compound A indicates three different CH2 carbons, one :CHg carbon and one "C: carbon; the last two carbons are shown to be spZ—hybridized by their chemical shifts. In the DEPT—135 spectrum of Compound B, the absorptions due to the double bond carbons have been replaced by a CH carbon and a CH2 carbon bonded to ‘ an electronegative group. 1 1 1 3 2 CH2 3 2 4 1. 8H3, THF 4 H ———--—-—-—~——-)- 5 3 2. H202, _OH 5 3 4 4 Compound A Compound B 1 106.9 5 1 68.2 8 2 149.7 5 2 40.5 5 3 35.7 6 3,4,5 29.9 6, 26.9 3, 26.1 6 4, 5 26.8 8, 28.7 5 12.57 The IR absorption indicates that C is an alcohol. From NY“, we can arrive at a molecular formula of C511 100, which indicates one degree of unsaturation. The broadbandu decoupled spectrum shows five peaks; two are due to a double bond, and one is due to a carbon bonded to an electronegative atom (O). The DEPT spectra show that C contains 4 CH2 carbons and one CH carbon, and that C has a monosubstituted double bond. HOCHQCH2CH2CH=CH2 is a likely structure for C. 12.58 Compound D is very similar to Compound C. The DEPT spectra make it possible to distinguish between the isomers. D has 2 CH carbons, one CH3 carbon, and 2 CH2 carbons, and, like C, has a monosubstituted double bond. The peak at 74.4 5 is due to a secondary alcohol. C|)H CHSCHchCH =CH2 Compound I) 1 2.5 9 Compound E, C7Hl203, has two degrees of unsaturation and has two equivalent carbons ' because its broadband-decoupled spectrum shows only 6 peaks. Two carbons absorb in the vinylic region of the spectrum; because one is a CH carbon and the other is a CHg carbon, E contains a monosubstitutcd double bond. The peak at 165.8 5, not seen in the DEPT spectra, is due to a carbonyl group. Carbon 6 (ppm) 1 1 19.1 (I). ?H3 2 28.0 e 5 4 3 2 1 4 165.8 5 129.8 Compound E 6 129.0 WWWWm__~m_mmu .u. . m.-. 284 Chapter 12 12.60 .3 2 3 2 1 HBr 4 H 4 ——-)- 1 4 Br 3 2 3 2 Compound F Compound G Carbon 6 (ppm) Carbon 5 (ppm) 1 132.4 1 56.0 2 32.2 2 39.9 29.3 27.7 3’4 27.6 3’4 { 25.1 1 2.61 Make a model of one enantiomer of 3~methylbutan—2—ol and orient it as a staggered Newman projection along the C2—C3 bond. The S enantiomer is pictured. H H30 CH3 I H3C OH H Because of the chirality center at C2, the two methyl groups at the front of the projection are diastereotopic. Since the methyl groups aren't equivalent, their carbons show slightly different signals in the 13C NMR. 1 2.62 Commercial pentane-2,4—diol is a mixture of three stereoisomers: (RR), (5,5), and (RS). The meso isomer shows three signals in its 13C NMR spectrum. Its diastereomers, the RR and 5,5 enantiomeric pair, also show three signals, but two of these signals occur at different 8 values from the meso isomer. This is expected, because diastereOmers differ in physical and chemical properties. One resonance from the meso compound accidentally overlaps with one signal from the enantiomeric pair. 1 2.63 The product (W: 88) has the formula C4H302. the IR absorption indicates that the product is an ester. The 1H NMR shows an ethyl group and an —OCH3 group. O O CH CH Cor-I CH3OH CH CH COCH ————-—---—-—> 3 2 H+ catalyst 3 2 3 12 .64 The product is a methyl ketone. O _ 1. CH3MgBr ll CH3CHC : N ‘—"""““_;_"""—"’ | 2. H3O | ...
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Chapter_12_solutions - Structure Determination: NMR...

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