Chemistry 605 (Reich) FIRST HOUR EXAM Thur, March 3, 2011 Question/Points R-10A_II R-10B_/15 R-10C_/23 R-10D_125 R-10E_/30 Total_/100 Hi Average Median AB BC CD 87 74 79 82 55 40 Distribution from grade list (average: 74.5; count: 43) i.........i.........i.........i.........i.........i.........i.........i.........i.........i.........i 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. Problem R-10A (C6H80). Below are given the 13C NMR spectra of two stereoisomers of 3-methyl-pent-2-ene-4-yn-1 -ol. Assign structures, and assign the signals by writing the 8 values next to the appropriate carbons on each structure (Source: Aldrich Spectra Viewer). y-effect, CH3 should be upfield 22.9 137.5 119.4 51 82.2 y-effect, C=C should be upfield E-isomer Z-isomer R-02G- 1 R-02G- 2 _L 180 160 140 120 100 ppm 80 60 40 20 Briefly explain the basis for your assignment. Be specific. y-lnteraction across double bonds causes upfield shifts (vs H at one of the positions). Thus in the E-isomer the CH3 would be upfield ca 5 ppm compared to the Z-isomer. Similarly, 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, but this is not a predictable y-effect. 15 Problem R-10B (C9H20O). Identify the compound whose dC NMR spectrum is given below Problem R-10B (C9H20O). 75 MHz 13C NMR Spectrum in CDCI3. Source: ASV/Reich 10/41 HINT: the longest chain is 6 carbons (it is a substituted hexane) A G D ^ B v SOH F E There are three "terminating" groups (A-Me, C-tBu, G-CH20H), the branch point is a CH carbon (B). So, basically, you have to fit two CH2 groups (E and F) into the framework below: t CH2OH Only 4 ways to do this and still have a hexane _L E D C B A _I_i_L 200 180 160 140 120 100 ppm 80 60 40 20 2 (a) DBE_°_. What functional groups might be present? C"0H or C"°"C (alcohol> ether) (b) What does the signal at 8 61.1 tell you about the structure? From the H-count (only 19 H on carbon), we know we have an alcohol (also absence of two downfield carbons) 62.6-2.5 = 60.1 The shift really requires: -CH2-CH2-OH 62.6 70.1 67 *OH *0' (c) Plausible structures? Circle your choice. Assign the signals by placing the appropriate letters (A-G) on the structure. -2.1 + 2a + 4ß + 2Yc + 2°(4°) + 2°(2°) -2.1 + 2a + 3ß + 3Yc + r0H 2°(3°) + 2°(2°) = -2.1 + 2(9.1) + 4(9.4) - 2(2.5) - 7.5 - 0 = 41.2 = -2.1 + 2(9.1) + 3(9.4) - 3(2.5) -5-0 = 31.8 Obs: 51.3 ^ \ Obs: 51.3 \ / I 70.1 + y = 67.6 , . , ^ , . . . . . Obs:61.1 -CH2-CH2-OH -C(CH3)3 -CH3 -CH A \C / I y JL G ^OH F -OH 23.9 ^ 28.9 2.1 + 2a + 3ß + 3yc + y0H + 2°(3°) + 2°(2°) 2.1 + 2(9.1) + 3(9.4) -3(2.5) - 5 - 2.5 - 0 = 29.3 Calc F: -2.1 + 2a + 5p + y + y0H + 2°(3°) + 2°(4°) + 80H = -2.1 + 2(9.1) + 5(9.4) -2.5 - 5 - 2.5 - 7.5 = 45.6 (d) Do a chemical shift calculation for the carbon to which you have assigned the signal F (8 51.3) using the Grant-Cheney parameters. Calc F: -2.1 + 2a + 5ß + y + 2°(3°) + 2°(4°) + 80H = -2.1 + 2(9.1) + 5(9.4) -2.5 - 2.5 - 7.5 = 50.6 Obs: 51.3 Calc B: -2.1 + 3a + 2ß + 3y + y0H + 3°(2°) + 3°(2°) = -2.1 + 3(9.1) + 2(9.4) - 3(2.5) - 5 - 2(3.7) = 24.1 Obs: 25.9 23 Problem R-10C (Cgsh^OgSSi) A 500 MHz 1H spectra is provided. (a) The structure of R-10C is given below. All of the important signals in the 1H NMR spectrum are labeled (A, B, C etc). Assign the proton signals by placing appropriate labels on the structure. For parts (b), (c) and (d), identify the couplings (e.g. for (b): JGX = 22 Hz, JGY = 32 Hz) Common errors: switch A/B, H/J, E/F. Decide between A and B with broadening of B by 4 J to CH3. Select H for a better match with coupling to G, and leaning (weak). Select E for matched coupling with G, and size of coupling. 8 m-Me -0.12 o-SPh -0.06 -0.16 o-Me -0.20 m-SPh -0.09 To make assignments use: 1. Chemical shifts (type of proton 2. Integrations 3. Multiplicity 4. J-values Z = 23 Hz End-to-end = 23 Hz -3 for no J ID JGH = 9 Hz Jgh =1 Hz JGE = 6 Hz JGF = 1 Hz 8 (b) The multiplet at 8 2.9 (G) is shown above. How many other protons are coupled to this one?_4Draw a coupling tree for G and report the coupling constants i ■ ■ ■ ■ i ■ ■ ■■ i ■ ■■ ■ i ■■ ■■ i ■ ■ ■ ■ i ■ ■ ■ ■ i ■■ ■ ■ i ■ ■■ ■ i ■■ ■■ i ■ ■■ ■ i ■ ■ ■ ■ i ■■ ■■ i ■ ■■ ■ i ■■ ■■ i ■ ■■ ■ i ■■ ■ ■ i Hz 80 70 60 50 40 30 20 10 0 16 lines - therefore coupled to 4 protons Jmh = 9 Hz UUl_/U 17 Hz Jmh = 7 Hz = 13 Hz This is AB of ABXY, AB coupled equally to X and Y 7 (c) What kind of pattern is the multiplet at 8 2.5 (H)? ^x (AMX) Draw a coupling tree on the multiplet, label it, and report J values. -3 for no J ID 5 (d) 5 Pt. BONUS question (don't do unless you have spare time): What kind of pattern is the multiplet at 8 2.2 (J)? _AB of ABXY>aw a coupling tree on the multiplet, label it, and report Jvalues with assignments. This is basically an AB pattern (diastereotopic CH2), each peak of which is split into a dd from coupling to C and D Problem R-10C ^Hg^SSi) 500 MHz 1H NMR Spectrum in CDCI3 (Source: Margaret K. Jones/Burke 10/19) _l_I_I_I_I_I_I_I_L_ 5.51 2.95 2.90 2.10 0.98 1.08 f I UL JU H UUUl_. _l_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_l_ 2.55 2.50 2.45 2.40 5.30 j_i_i_i_i_i_i_ 2.20 2.15 30 20 10 0 Hz US. 1r AB CDEF GHIJKL M J_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_I_L. 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0 ppm 25 Problem R-10D (C8H7N02). Determine the structure of R-10D from the 1H NMR spectrum provided. 2 (a) DBE 6 (b) Analyze the 1H NMR signals. For each of the signals listed below report multiplicity and coupling constants to the extent the signals are amenable to first order analysis, and the part structure each corresponds to. 10 D 5.43, 1H, d, J = 10 H .H B 5.88, 1H,d, J = 17 6.75, 1H, dd, J H = 17,10 SN^íJ\H 7.48, 1H, t, J = 8 7.69, 1H,dt, J = 8, ca 1 8.07 1H,ddd, J = 8, 2, 1 Q 8.22 1H, t, J = 1.5 (probably dd) These together require a vinyl group (CH=CH2) with no other coupling The coupling between the gem-vinyl protons is too small to resolve These together require a meta-substituted benzene (c) Give the structure of R-10D. If more than one structure is possible, circle your best choice. 13 ^N02 HE^ HD Hp ppm 30 Problem R-10E (C8H1603). Determine the structure of R-10E from the 1H NMR, 13C NMR and IR spectra provided. 1 (a) DBE_ (b) What information can you obtain from the IR spectrum (give frequency and peak assignment). 3 3500-3600 cm1 OH stretch 1730 cm1 Carbonyl stretch - probably an ester (c) Interpret the 13C NMR spectrum, showing any part structures that can be identified, and the corresponding 5 values. Very likely 4 CH3 4* carbon (not bearing an O) disappears in DEPT must be -CH2-OR Must be a '^OR (d) The signal at 5 2.7 in the H NMR spectrum disappears when the sample is shaken with D20, and the signal 8 3.9 becomes a 1:3:3:1 quartet. What does this tell you about the structure? ,H 2.7 D?0 O' ,H 2.7 H 'CHo H H 'CHo Also possible: 13 HC Pentet 3.9 " Quartet n 3.9 (e) Analyze the 1H NMR spectrum. For each of the groups of signals marked on the spectrum, report the multiplet structure in the standard format (e.g., 0.0 8, dtd, J = 0.0, 0.0, 0.0 Hz, 2H) and any part structure you could obtain from the signal(s). I_l 5 1.15, 1H,d, J = 7Hz CH3-6 A 5 1.17, 1.18 - possible two methyl singlets, or isopropyl (J = 5 Hz - a little small) 5 1.27, 3H, t, J = 7Hz CH^-CHp- 5 2.70, 1H,t, J = 7Hz -O-H H 5 3.86, 1H, pentet, J = 7 Hz CH3-C-OH 5 4.17, 2H,q, J = 7 Hz -O-CJi-CHg (f) Give your answer below. If more than one structure fits the data, draw them, but indicate your best choice by circling the structure 10 0 II OH I <^ / 10 OH O O' DEPT 135 Problem R-10E (C8H1603). 13C NMR Spectrum in CDCI3. Source: Kevin Jantzi/Reich 10/31 8 14.13 17.68 19.76 22.40 46.98 60.65 72.47 177.80 Normal '.........i.........'.........i.........'.........i.........i.........i.........'.........i.........'.........i.........'.........i.........i.........i.........i.........i.........'.........i.........'.........i" 200 180 160 140 120 100 80 ppm 60 40 20 0 Problem R-10E (C8H1603). IR Spectrum Neat Source: Nicolet FT-IR f_i_i_I_I_I_i_I_I_I_i_i_I_I_i_I_I_I_i_I_I_I_i_I_I_i_ 1600 1400 1200 1000 800 600 3500-3600 cm"1 OH stretch 1730 cm1 Ester C=0 stretch This one from EX-2-10/11 Problem R-10G (C8H8Br2). This problem requires you to analyze the signals at 8 4.1 and 8 5.2. You are given the structure. (a) Do a "first order" analysis of the two multiplets shown below. Draw a coupling tree, and estimate couplings. What type of pattern is this? Br H* ■■- ct> o ^ o6 c\i od c\i lo lo lo lo od oo od cO