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Problemset8answers - Problem Set 5 Answers Smith Text See...

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Unformatted text preview: Problem Set 5 Answers Smith Text: See solutions manual Techniques in Organic Chemistry: I. The calculated chemical shift values are in the parentheses. a. CsH] 1C1 is 1-chloro-2,2-dimethylpropane. e. C5H1202 is 4—hydroxy-4-methyl—2—pentanone. 3.33 / 1.26 (1 2) OH O 2 '18 \\‘M/ (21) \ 2.63 (2.7) d. C5H100 is 3-methyl—butanal. 2-21‘1 H O (1.9) / 9.77 (95-105. from Table 19.2) \\\““ H X 0.98 2.31 (1.0) (2.3) e. C5HgO is 3-buten-2-ol. The assignments of the 5.19 ppm and 5.06 ppm protons cannot be made on the basis of calculated chemical shifts; however, the relative sizes of the cis— and trans-proton—proton coupling constants allow the assignments (see Table 19.6). / 2.50 H‘ 5.19 El43eg\ OH (5.27) \\“ H / HK 25502555} 1 27/ H‘\ I (1.4) 5.90 (5.92) 2. dq, 7.0ppm H O \ s, 3.7 ch OCHa dd, 1.9ppm H dq, 5.85ppm 3. 6H, d, 1.2ppm H30><CH3 HO H 1H, s, 2.0me 1H, 513.40 ppm The lH NMR signal at 2.0 ppm, which exchanges with D20 to form HOD at 4.6 ppm, is the 0—H of 2-propanol. This proton at oxygen splits with the methine proton at C—2 if chemical exchange of the O~H becomes slow. An increasing concentration of 2-propanol leads to greater intermolecular hydrogen bonding and a downfield shift of the proton on oxygen. The two methyl groups comprise the doublet at 1.2 ppm, While the septet at 4.04 ppm is the methine proton at C-2. 7. The molecular formula indicates two Double Bond Equivalents; because one double- bond equivalent is a carbonyl group, the other one may be a C=C. That assignment would be consistent with the Vinyl signal (ii-I) at 5.06 ppm. C3H140 also has three singlet methyl groups; the chemical shift of one methyl group (2.10 ppm) suggests that it is attached to the carbonyl carbon. The chemical shifts of the other two methyl groups are consistent with their being allylic; there is a small allylic splitting in the 1.65 ppm methyl peak, which suggests that it is trans to the vinyl proton (Table 19.6). Two methylene groups appear at 2.25 ppm and 2.45 ppm. The C3H|40 compound is 6-mothyl—S-heptene- 2—0ne. Additional Problems: 1. The two methyl groups are actually not equivalent because they are diastereotopic. They have very similar chemical shifts, so they nearly overlap. CI 20.9 22.6 20.4 K97 Proton NMR Handout #2 ® The renewing spectrum is a prolou NMR of ally! ethyl ether. Assign the protons to lhc matching signals and explain Lhe spiitting based on [has 0. coupiing constants: .iuhz [III-12,1“: 101412.le = [8 Hz, Jul: 5 [‘12. ”WW 1.5 1.4 1.2 1.0 PM The Following spectrum is a proton NMR of 3~mclhylstyrene Match the signals at 5.2 ppm, 5.8 ppm, and 6.7 ppm with protons A, B, and C. Draw to explain the coupling ol'each of these .1“: 16142, Jim = 0 Hz. H H a tree: diagt‘mns protons. [coupling constants: Jan: till-1:5, Hb : ~15 \ 7 5 ti, 3 PPM 2 Proton NHH ‘17. Based on the IR, ]1-! NMR, and ”C NMR on the following page, predict the structure of the compound. Explain why no splitting is observed in signal A even though one might expect it be coupled to another set of equivalent protons. 0 H E CHECHZCHQ CH3 LT_J .94 "J— Poir- alclflhyclé‘ 2: 0‘2 H2: 30 flo+0b5€Fv E". The following speck-um represents one isomer ol‘cthylphcno]. Is it the 1.2—. J,3-, or 1,4-isomcr'? Explain your answer based on the splitting and integration observed in the spectrum. OH \ _ “iv-fix / Et 3H 2H 2H 7 B 5 4 3 EPPH 1 Prutun NHH In+€jm+w1 “ W‘Hq {34’0“} 7%??? H j)” H are 22:39st 0F [O’OTE’WL‘ 14‘ A m 7L$jia+r3 +0 2 béof/CPU‘E-jc H14 SF/‘iz’zay 1%? ‘T’NJ Luca? Vé’r’LSc; 513‘“ The following three spcclra represent Ihc aromatic protons of l.2-, 1.3-. and 1.4, nizronnilinc. Assign each spectrum to {he appropriale compound and than match the appropriate protons with each signal. (Hint: the nitro gmup is :1 strong electron withdrawing gmUp. which dcshiclds adjacent atoms by induction.) H.1 Noa He Hh H1 Hg Ht: ' Hd H, N02 H] H, 'p. Based on ranges of coupling constants provided in class, cslimalc {he expected splitting (in Hz) observed between the lettered i'lydmgen atoms. (Give values of J:llh Jun: Jbuv [=th (a) H:1 C1 Hg, C1 ([1) IIn III}; Cl C] (s) H“ cu.1 C113 ”1: "a TAB: .L JAB: (9-:on 3‘ — [1-1% H: Ac; (b) 1‘1: Cl (e) 0%} Ha Cl [-11, CH; C! I“! t. (C) 171.1 C] (U OCH; (1) Ha Cl HI: Cl .1-1a II LI 1"!) Ha EEEEEEEE EEE EEEE “E H E E'“fEE EE :3 ‘2. Graphically analyze the l’olluwing Spectrum of methyl vinyl thioether. Assign the peaks and determine approximate coupling constants f or Jmt Jed. and Jim. Each small square on the. spectrum represents 5 Hz. {I—Iinl: Watch 0111—-—therc are two signals that are partially overlapping.) fi-gw ' H1 ." EEE THEME E E ' . EEE EEEEEEEEE 3%; 5:5! EEEE'E EE. E'EEEEEE' E'fEEEEE "' 'E‘: ' EEEEEEEE. EEEEEE' '"E E 'EE EE EEEEEEE -EEE IEE:EEEE EEE '.E.EEEE..EEEEE 5E ‘EEEEEEEEE EEEEEEEEEEEEEEEE. 13.2. E - E “E "EEEEE E EEEEEE E 5E. EEEEE ’- EEE LE EEEE E'E EEEEEE EEEEE EEEEEEEEEEEEEEE EEEEEE EEEEE €555 EEEEEE E E EEEEEEEEEEEE EEEEEEEEEEEEE Egg. EEEEEEEEEEEEEEEE" E...::E !W‘LJ‘EE' .:_._______-:EEEEEE :mm. _____;_::: m: azngw .Emaanimanaa ': C 5.0T.“— 4.0 H-Ll—h .0 IQ ‘3. Determine the siruclure of the foilowing urornulic compound with formqu CgHsBrO. The peak at about 5,6 ppm disappears when the sample is washed with 030. OH Br '9. Determine the structure of the i‘oilewing compound wilh formula Cal-17310. :L 0 CW G": L'E} ‘ 9 9 LUCICI “‘1 2’sz 0% :9 as o m ‘3‘” {l [I ” UN 3? C) ‘H "J '3 9 D“ (a E l 3: 9+ N N N . \5 ' Q a \ J. W '5 gt NE E3 Kw_ _w. _ “M—__..__fltw_—__ _._— ...__ ........ _ __. —T 0‘? 1 _ f, —~—~_ ...... ”___ W (:9 H __ f-uxrss" 6.? .‘—- W _Wrm—‘— ENG] -"’)‘J _‘Wflmmwfl .746 Ev) ( ' 5 .142 “__‘"___ __.-.4:..6_‘UB1 .. fw-‘W'r- __.—w#——.__—-»W“—‘—W' """"" _ L— E . U75 G‘j ‘ MW/ tgl¥5.QHHQO ‘HNMR Int 1 Spun p ”CNMR 2‘3 zan255 CI 3.69 1.5'.‘T 0.90 29.3 0?. 50.5 58.7 30.5 11.6 115 a. C5HQBI'02 ‘HNMR .llu_.”l]_.l“.l_.— _ .. 5 I 4'1 I :I’: I 2I WM Int: 1 '1 I 3 Split: dd dd p s 13 CNMR 1é0 I 1éo 14o 1&0 I 160 I ab WM Answers: r r 4.20 2521 49.5 c) O 1-79 4.5?:4.42 Y 2” 6m \‘T‘Z/ O O 25.? 12 C G} C?H1403 lHNMR 5 ' 4'; ' '3 ' 5 4 mm Int.1 1 1 '2 1 1 2 2 3 Split: 5 5 5x I; dd dd t p d ”CNMR WM Answers-1 4 .81 H O OH O 352 1.35 OH 439 62.9 m. 25.4 OH 1-15 2.73243 2.15 3.50 235 52.6 33.9" 62.3 ...
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