Chemistry 605 (Reich) FIRST HOUR EXAM Thur, March 3, 2011 Question/Points R-10A R-10B R-10C R-10D R-10E /7 /15 /23 /25 /30 AB BC CD 82 55 40 Hi Average Median 87 74 79 Total /100 Distribution from grade list (average: 74.5; count: 43) 4 3 Number 2 1 0 0 10 20 30 40 50 Grade 60 70 80 90 100 Grading Copy Name If you place answers anywhere else except in the spaces provided, (e.g. on the spectra or on extra pages) clearly indicate this on the answer sheets. CH3 should be upfield 17. C≡C should be upfield Z-isomer R-02G.9 119. Thus in the E-isomer the CH 3 would be upfield ca 5 ppm compared to the Z-isomer.7 Problem R-10A (C6H8O).2 81.9 OH 58. but this is not a predictable γ-effect. Assign structures.2 137. .2 85.1 3 E-isomer 1 R-02G137. Below are given the 13C NMR spectra of two stereoisomers of 3-methyl-pent-2-ene-4-yn-1-ol.4 137.4 119.9 20 0 Briefly explain the basis for your assignment.6 17.5 R-10A-2 180 160 140 120 119.5 OH γ-effect.4 20 m m st ve 180 160 140 120 100 ppm 80 60 40 20 0 82. the first C≡C carbon would be upfield in the Z-isomer compared to the E For some reason the terminal acetylene carbon also moves a lot between isomers.9 75.4 81.9 61. 4 γ-Interaction across double bonds causes upfield shifts (vs H at one of the positions).4 100 ppm 80 60 40 22. and assign the signals by writing the δ values next to the appropriate carbons on each structure (Source: Aldrich Spectra Viewer).1 137.8 61. Be specific.8 82.1 2 c (w w e R-10A-1 119.1 22.6 85. SImilarly.4 58. γ-effect.9 75. 3(2.9 C G D F B OH = -2.1 30.2 Obs: 51.5) .1 + 2α + 5β + γ + γOH + 2°(3°) + 2°(4°) + δOH = -2.1 51.5 .5 = 50.1 + 2α + 5β + γ + 2°(3°) + 2°(4°) + δOH = -2.4) -2.1 + 2(9. Assign the signals by placing the appropriate letters (A-G) on the structure. 75 MHz 13C NMR Spectrum in CDCl3. -2.5 .5 = 45.6 Obs: 61. What functional groups might be present? C-OH or C-O-C (alcohol.5 .0 = 29.8 Obs: 51.5 .6 .5 = 60. the branch point is a CH carbon (B).6 (d) Do a chemical shift calculation for the carbon to which you have assigned the signal F (δ 51.9 OH OH OH -2. So.9 3 .3 42.1 28.0 = 31.4) .7 HINT: the longest chain is 6 carbons (it is a substituted hexane) C D F A G B E q OH There are three "terminating" groups (A-Me.1 tell you about the structure? From the H-count (only 19 H on carbon).1) + 2(9.15 Problem R-10B (C9H20O).0 = 41.5) .5 .1) + 5(9. Source: ASV/Reich 10/41 13 C NMR spectrum is given below.3 Calc B: -2.3 Calc F: -2.4) . we know we have an alcohol (also absence of two downfield carbons) The shift really requires: 4 -CH2-CH2-OH O 62. basically.4) -3(2.1) + 3(9.1 + 2α + 3β + 3γC + ΓOH 2°(3°) + 2°(2°) = -2.3 A Pieces: 6 -CH2-CH2-OH -C(CH3)3 -CH3 -CH 23.1 + 3α + 2β + 3γ + γOH + 3°(2°) + 3°(2°) = -2.3(2.2(3.5) .7) = 24. you have to fit two CH2 groups (E and F) into the framework below: Me t-Bu C H CH2OH Only 4 ways to do this and still have a hexane t t t s d q G F E D CBA 200 180 160 140 120 100 ppm 80 60 40 20 0 2 (a) DBE 0 .1 + 2α + 3β + 3γC + γOH + 2°(3°) + 2°(2°) = -2. Calc F: -2. 61.5 .6 Obs: 51. Identify the compound whose Problem R-10B (C9H20O).1 + 2(9.1) + 3(9.1 + 2(9.2.5 .2(2.1) + 5(9.2. G-CH2OH).4) -2.1) + 4(9.3 31.1 67 OH (c) Plausible structures? Circle your choice. C-tBu.1 Obs: 25.1 + γ = 67.5 .1 + 2α + 4β + 2γC + 2°(4°) + 2°(2°) -2.3) using the Grant-Cheney parameters.5) .1 + 2(9. ether) (b) What does the signal at δ 61.9 22.2.4) .7.7.1 OH OH 62.6 70.1 + 3(9.5 .3 70.0 25.1 + 2(9.5 .2.7. C etc).therefore coupled to 4 protons 0 H JHH = 17 Hz J J 80 70 60 50 H 40 30 20 10 JHH = 17 Hz JHG = 7 Hz JHG = 9 Hz JJJ = 13 Hz JJC = 7 Hz JJD = 1 Hz This is AB of ABXY.g.29 K D Me3Si H C J H H F L H G CH3 I A H S O E O H Common errors: switch A/B. To make assignments use: 1. This is basically an AB pattern (diastereotopic CH 2). and size of coupling. (a) The structure of R-10C is given below.20 m-SPh -0. (c) and (d). How many other protons are coupled to this one? coupling tree for G and report the coupling constants Hz 16 lines . E/F. J-values Σ = 23 Hz End-to-end = 23 Hz -3 for no J ID 4Draw a G JGH = 9 Hz JGH = 7 Hz JGE = 6 Hz JGF = 1 Hz 8 8 (b) The multiplet at δ 2. Select H for a better match with coupling to G. Decide between A and B with broadening of B by 4J to CH3.5 (H)? and report J values. Integrations 3. label it. B. identify the couplings (e. JGY = 32 Hz) m-Me -0. All of the important signals in the 1H NMR spectrum are labeled (A.09 -0. H/J. For parts (b). each peak of which is split into a dd from coupling to C and D .23 Problem R-10C (C23H34O2SSi) A 500 MHz 1H spectra is provided. -3 for no J ID ABX (AMX) Draw a coupling tree on the multiplet. Multiplicity 4.2 (J)? AB of ABXY Draw a coupling tree on the multiplet.9 (G) is shown above. Assign the proton signals by placing appropriate labels on the structure. for (b): JGX = 22 Hz. Chemical shifts (type of proton 2. AB coupled equally to X and Y 7 (c) What kind of pattern is the multiplet at δ 2. 5 _ (d) 5 Pt. and report J values with assignments. and leaning (weak). BONUS question (don't do unless you have spare time): What kind of pattern is the multiplet at δ 2. Select E for matched coupling with G.06 -0.16 o-Me -0.12 o-SPh -0. label it. 5 4.15 1.90 2.0 .50 2.20 0.09 5.0 4.5 C 6.0 6.5 B 7.5 M 0. Jones/Burke 10/19) 30 20 10 0 Hz 9.98 2.89 A 7.5 L 1.20 J B A HE O S D Me3Si M H J C H L 2.10 1.40 6.67 E G H 5.08 1.98 1.30 2.45 2.0 D E F 5.51 2.0 ppm 3.26 2.15 1.0 H I J 2.0 0.95 CH3 I 2.5 5.Problem R-10C (C23H34O2SSi) 500 MHz 1H NMR Spectrum in CDCl3 (Source: Margaret K.55 2.5 2.5 G 3.0 K 1.10 O H K HG H F 2. Determine the structure of R-10D from the 1H NMR spectrum provided. 1 G 8. For each of the signals listed below report multiplicity and coupling constants to the extent the signals are amenable to first order analysis.69. d.25 Problem R-10D (C8H7NO2). H A B 5.dt. J = 10 5. d. HG NO2 HE HD HF 13 . and the part structure each corresponds to. t. 1H. If more than one structure is possible. 2 (a) DBE 6 (b) Analyze the 1H NMR signals. J = 17. t.22 1H. 1H. 2. J = 8 E 7. ddd.43.07 1H. 1H. J = 8. ca 1 These together require a meta-substituted benzene F 8.5 (probably dd) (c) Give the structure of R-10D. J = 8. J = 1. circle your best choice. 1H.88. 1H. 10 H H H These together require a vinyl group (CH=CH2) with no other coupling H The coupling between the gem-vinyl protons is too small to resolve 10 D 7. dd.48. J = 17 H H C 6.75. 4 6.5 A 5. 10 Hz 8.15 1.02 1.04 G F 10 9 8 E D 7 C 6 B A 5 ppm 4 3 2 1 0 . J = 17.7 E 7.18 5.9 2.7 H H H d.8 B 5.9 7.Problem R-03D (C8H7NO2) Assign proton signals 300 MHz 1H NMR Spectrum in CDCl3 Source: Aldrich Spectral Viewer/Reich H Y X Y H H H H H H 30 20 10 0 H Hz Y X H X H H H dd.6 7.0 7.00 2.2 G 8.5 D 7.8 7.6 5. J = 17 Hz 5.7 5.4 1.1 F 8. J = 10 Hz d.8 C 6. 27. H δ 1. and the signal δ 3.0 δ.a little small) δ 1. t.0. J = 7 Hz -O-CH2-CH3 (f) Give your answer below. J = 7Hz CH3-CH2δ 2.80 Very likely 4 CH3 4* carbon (not bearing an O) disappears in DEPT must be -CH2-OR OH Must be a O H OR (d) The signal at δ 2. pentet. 2H. 0. 1H..17.98 60.15.68 19. 1.0.13 17. 13C NMR and IR spectra provided. but indicate your best choice by circling the structure O 10 10 OH OH O O O 6 . d.17.18 . 1H. What does this tell you about the structure? D H 2.probably an ester 30 (c) Interpret the 13C NMR spectrum. 1 (a) DBE (b) What information can you obtain from the IR spectrum (give frequency and peak 2 assignment). J = 7 Hz -O-H δ 3.70. 1H. δ 14.9 Quartet Pentet 3. q. showing any part structures that can be identified. draw them. If more than one structure fits the data.47 177. 0. or isopropyl (J = 5 Hz . J = 7 Hz H CH3-C-OH δ 4.0 Hz.possible two methyl singlets. Determine the structure of R-10E from the 1H NMR. t.9 (e) Analyze the 1H NMR spectrum. J = 7Hz A 6 B C D E CH3-C δ 1.7 O D2O O O Also possible: 4 CH3 CH3 CH2 HC H H H 3.86.7 in the 1H NMR spectrum disappears when the sample is shaken with D 2O.7 H 2.g. and the corresponding δ values. 3H. 2H) and any part structure you could obtain from the signal(s). 0. dtd.9 becomes a 1:3:3:1 quartet. For each of the groups of signals marked on the spectrum.40 46.76 5 22.Problem R-10E (C8H16O3). report the multiplet structure in the standard format (e.65 72. 3 3500-3600 cm-1 OH stretch 1730 cm-1 Carbonyl stretch . J = 0. 47 177. Source: Kevin Jantzi/Reich 10/31 δ 14.40 46.68 19. IR Spectrum Neat Source: Nicolet FT-IR X 3800 3600 3400 3200 3000 2800 2600 2400 2200 2000 1800 1600 1400 1200 1000 800 600 .65 72.98 60.DEPT 135 Problem R-10E (C8H16O3).13 17.76 22.80 Normal x x 200 180 160 140 120 100 ppm 80 60 40 20 0 Problem R-10E (C8H16O3). 13 C NMR Spectrum in CDCl3. 5 2.5 3.0 6.5 5.0 ppm 3.30 1.10 2.5 6.00 1.5 0.15 1. 1 H NMR Spectrum in CDCl3.0 0.Problem R-10E ((C8H16O3).20 4.10 3.0 2.80 x B x A C 2.5 7.5 4.0 .0 1.85 ppm 3.5 1.20 ppm 1.05 x x E D C B A 7.44 O O OH x x E 4.0 5. Source: Joe Langenhan/Gellman 10/31 30 20 10 0 Hz 13.90 x D 3.0 4.70 ppm 1.25 1.15 ppm 4.33 x 1. 9 1218.1 c.55 Δνab.1 and δ 5.39 (-) 1219.8 1202.6 Br HX Br HB HA Hz X JBX JAX "AMX" solution (approximate Solution 1) JAB B A JAB 6 3 5 4 1 JBX 8 2 5 JAX 7 5.10 4.0 1532.063 4. (a) Do a "first order" analysis of the two multiplets shown below.9 1192.20 1 2 3 4 5 6 7 8 15 1202.6.5.5 c+ ± δ+/2 = 1216. show them both.015 4. 1199.10 4.= (5+3)/2 = 1215.99 (+) +10.6 1209. Use appropriate criteria to distinguish the two.6 +4.9 c.8 1202. What type of pattern is this? 1228.6 Solution 1 is correct: 1.47 (-) +22.9 1548.09 1204. If more than one solution is possible. which never happens 2.00 3.9 1538.= δ.15 5.7 1223. Intensity calculation fits better for Solution 1 .36 1210.034 Solution 1 Solution 2 c+ = (6+4)/2 = 1207.4 1218.7 1223. Use the frequencies shown above. You are given the structure.04 1213.± δ-/2 = 1221.5 1216.8 1228.322 4.3 1212.90 to 10. This problem requires you to analyze the signals at δ 4.85 Δνab+ = δ+ = sqrt((8-2)(6-4)) = 17.93 (-) -6.6 1212.1 1542. Show your work.4 1192.31 14. Solution 2 has one negative 3J.05 4.95 ppm ppm (b) Do an accurate (quantitative) analysis.77 0. and estimate couplings. and draw the proper coupling tree on the spectra below.044 4.69 2.= sqrt((7-1)(5-3)) = 11. and tabulate your data in an easily readable form.20 This one from EX-2-10/11 Problem R-10G (C8H8Br2).995 0. 1209.1 JBX JAX νA νB νA νB JAB JAX JBX νA νB νAB i10 = i11 δA δB 9.3 1212.2.6 JBX 1199. 5.4 Solution 1 Solution 2 JAX 1221. Draw a coupling tree. 0 4.95 8 7 6 5 4 ppm 3 2 1 0 .3 1212.Problem R-10G (C8H8Br2) 300 MHz 1H NMR Spectrum in CDCl3 (Source: Wayne Goldenberg/Reich 11/31) 30 20 Hz 10 0 1228.8 1202.9 1548.7 1223.10 4.15 ppm 1532.4 1218.4 5.00 0.10 4.20 5.1 1542.6 3.00 ppm 1192.05 4.99 5.9 1538.95 2.