ProblemSet8_quiz5_

ProblemSet8_quiz5_ - Problem Set 5 (N MR Spectroscopy) S343...

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Unformatted text preview: Problem Set 5 (N MR Spectroscopy) S343 Summer 2008 Major topics covered for quiz 5: I Chemical equivalence o Diastereotopic atoms 0 Chemical shifi: tables and calculations 0 Splitting o Coupling constants 0 Tree diagrams 0 Interpretation ofNMR Here are some problems to help you practice: Smith text: Chapter 14 problems 5,6,13,20,21T 22, 23, 43, 50, 6], 63 Techniques in Organic Chemistry: Chapter 19: 1, 2, 3, 7 Additional problems: 1. Explain why 2-chloro—2-methylbutane has four '3 C peaks, but the peak at 20.6 ppm appears to be a doublet (20.9 and 20.4) upon more thorough investigation. PPM © . ’E‘hc J‘oltowing spectrum is n proton NMR of ally] ethyl ether. Assign the protons to the matching signais and explain the splitting based on these coupling constants: Jab: 01-12., Jug: [GI-135, Jim = 18 Hz, J“; = 5 1-17.. Ha / /O\ /CH3 C C H2 H2 Hb \ t1 \ C 3H 2H 2H 6.0 5.5 5.0 1i.5 4.0 PPM 3.5 i-5 1.4 1-2 1-5 PP” Prutun NHR I fit” { DJ The f’ollt'iwing spectrum is a proton NMR of3-mothylstyrcnc. Match the Signals fli‘ 5.2 ppm 5.8 ppm, and 6.7 ppm with protons A, 13. and C. Draw trot: diagrams to explain the coupling of ouch of those protons. {coupling constants: Jab: lOI-Iz, .1“: 161-13, Jim : 0 Hz. Ha Hb ? 5 4. 3 PPM 2 Proton NHH " Based on the IR, II—l NMR, and I3C NMR on the following page, predict the structure of the compound. Explain why no splitting is observed in signal A even though one might oxpoot it be couplod to another set ofoquivetiont protons. MF CS}; 100 13G- W86 8'5 3 :: TflF-‘fSflIT TFHC‘E 40 ‘3 I 1720. 51 B 4338 3559 3063 if! NWT”anme 19.0 a. 9.0 a. 5.0 7, pp 2.5 2.4 2.2 2.0 1.8 1.6 1.4 1.2 L0 0.3131314 Proton NHR 200 180 150 140 120 100 80 50 40 20 PPM 0 Earbun 13 NHH 5: The following spectrum represents one isomcrol‘ cl‘hylphcnol. Is it the 1.3-, 1,3", or 1,4 430mm"? Explain your answer based on ihc splitting, and integraliun observeclh1thc SpCCUlHn. OH Et 3H 2H EH ? 5 5 4 3 2 PPM i Prutnn NMFI The following three spectra represent the aromatic protons of 1.2—, 1,3~. 21nd LIL niEl‘Oamiline. Assign each speelrum to the appropriate compound and lhen match the appropriate protons will] each signal. (Hint: the nitre group is a strong eleclmn withdrawing group, which dcshields aclj:-1cent atoms 11)! induction.) NHg NH2 NHQ Ha N03 He Hh Hi Hk Hi, HG Hf N02 Hi H! HG Ha N02 (7) Based on ranges of coupling constants prnvidcd in class. estimate [he expected spiiuing (in Hz) observed between the lettered hydrogen atoms. (Give values 01' 12:11. 33:3: Jth 0133-) (a) 1‘11: (c) Ha> Hb Hb C1 C1 C143 Cl C! (-1.!) Ha Nb {6) Ha (F) OCH; I H“ (1 Cl “h (g) 11“ Ci-I3 (h) IL, Cl (i) H 1-1.. I‘Ib H1, @ Gruphicztfly analyze the following spccuum ol’ methyl vinyl thiocther. Assign the peaks and determine approximate coupling conslzmis For Jim Jul, and Jhc. Each Small square on the spectrum represents 5 Hz‘ (Him: Watch mum-them are 1WD signals that are partially overlapping.) Dclcrminc the structure of the following aromatic compound wiih formula Ch] IfiBl'O. The peak at about 5.6 ppm disappears when the sample is washed with 1330. .'_'_ ._,._ _ .'...'._. .;._._-_..1—-—:—- — -'-_-_-_._....__l:__... l .0 - u d :.‘ r - 5 . . . . ..l Ifi+8jmhmf It, LS It, ’7 LHD 0 73C __m_ _ _ if 75 MHz l 1. On the following page is an expansion of the aromatic section ofthe proton spectrum of the unknown you made in the Isomerization of Carvone experiment. Label peaks A-C as doublets. triplets, doublet ofdoublets, etc, and calculate the coupling constants. Match peaks A-C to the corresponding aromatic protons: CH3 —“"* H OH —+ H H +— HaC CH3 "H. 1W U —- -fi.. fl‘?fil .mfi.746 \——S.Tdfl Draw the SII'UCIUI‘C of [he product. Assign {he Signuis. Abbreviations in this section inciudc szsinglct. d:d0ublct. cfdzdouhlcl' of doublets, l: lriplet, q=quurleL pzpunlet, sxzscnci. cxch=cxchamgcuhlc A.GMHGO 'HNMR .i._i.I.. _11J.l. “ll... _ ...l."~.‘-.‘ . 1m: 1 l i 3 I 3 Split: p LIE] dd 5 [1 I “CNMR . .. . .5 i I. . .. , . : ._ .. I . I . ................. ................. .. 220 200 180 160 140 12F?PM 100 BID 60 £10 20 0 8. C5 HI}BI‘()3 'HNMR Li“ 1!” ,Ill._ _ _ _ 5 4 3 ......................... .. i .___| ................... __ ............ _. _ PPM I111; 1 l | 3 3 Spiil: dcl (Ed [3 s; d “CNMR 180 160 140 120 100 80 60 4O 20 gr CarHHO: 'HNMR Inlzl | 2 3 i l ’7‘ ’7 3 Splii‘: s s sx t (id dd 1 p cl “CNMR 220 i 200 180 1éo 1&0 1é0 100 I 30 I 60 40 20 I 0 ...
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This note was uploaded on 01/18/2012 for the course CHEM S343 taught by Professor Benburlingham during the Summer '11 term at Indiana.

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ProblemSet8_quiz5_ - Problem Set 5 (N MR Spectroscopy) S343...

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