Question 1. 2D-NMR: C 6 H 10 O 2. (20 points) Integrations show signals 3H 1 & 5, 2H for signal 4, and 1H each for signals 2 and 3. - Draw the structure. - Assign the hydrogens to signals 1 5 (that is, draw the number next to the H in the structure). - In addition to the lines drawn for signals A and B, draw horizontal and vertical lines to indicate all the significant off-diagonal peaks and their associations. Structure (E)-H 3 C-CH=CH-CO-O-CH 2 -CH 3 3 J, A: H 3 C(1)-C(2)H= 3 J, B: O-C(4)H 2 -C(5)H 3 3 J, C: -C(2)H=C(3)H- 4 J, D: H 3 C(1)-CH=C(3)H- Need to see FOUR squares for full credit. -1-
Question 2. 2D-NMR reported by Higham et al. JOC 2007, asap. (20 points) The 2D H-Cosy NMR spectra of isomers A and B are shown on the following pages. Assign which isomer goes with which spectrum and explain briefly. Include the appropriate lines in the spectra. Spectrum 1 because The Hs of interest are at low chemical shift because they all are part of an anionic system. The phenyls are at high chemical shift; deshielded by P-cation. Isomer A Two Hs couple strongly with each other (red). The is one H without strong coupling; that H couples only weakly and only with the low- H (green diagonal peak and green highlighted off-diagonal crosspeak asymmetries). Spectrum 2 because The Hs of interest are at low chemical shift because they all are part of an anionic system. The phenyls are at high chemical shift; deshielded by P-cation. Isomer B Three Hs and the central H couples strongly with the two others (red, green). The weak cross-peaks (blue arrows) are due to 4 J coupling. -2-
Spectrum 1-3-
Spectrum 2-4-
Question 3. Structure Analysis I. Which Pyrimidine Base is it? (Very easy 20 points) Mark significant peaks in IR & O NH 2 MS spectra and (next page) assign all NMR peaks as best as possible. Thymine N H NH O Cytosine N H N O NH 2 signals M(Cyt) = 111-5-
Cytosine: Only the alkene Hs are seen as doublets. All H at N exchange in D 2 O. Cannot be Thymine: There is no methyl signal. Four signals. There are four different Cs in Cytosine. Any reason for two Cs in T to be isochronous? Not really. And there is no methyl C!! Assignments: = 168.10 ppm, int. = 149, C(NH 2 ); 159.91, 138, C(O); 144.05, 925,C(NH); 95.79, 1000, C(C). -6-
Question 4. Structure Analysis II. Which Isomer of C 6 H 4 BrF? (20 points) 60 MHz 60 MHz 300 MHz Hz ppm Int. 672.56 7.510 629 672.00 7.504 200 670.94 7.492 87 670.25 7.484 200 667.63 7.455 651 667.06 7.449 211 665.75 7.434 269 665.38 7.430 243 663.44 7.408 935 660.88 7.380 239 660.25 7.373 145 658.50 7.353 917 655.25 7.317 105 632.56 7.063 99 629.38 7.028 928 627.00 7.001 230 624.31 6.971 103 622.56 6.952 199 621.06 6.935 1000 620.25 6.926 753 618.69 6.909 246 617.00 6.890 66 614.19 6.858 184 612.50 6.839 193 611.94 6.833 648 608.63 6.796 55 This is about bromofluorobenzene. Ortho, Meta or Para-Structure? Explain. 300 MHz spectrum shows two signals and strongly suggests para. Signal A at 7.42 ppm: H close to Br. Coupled to H (d, 3 J(H A,H B )) and to F (d) with smaller 4 J(H,F). Signal B at 6.94 ppm. H close to F. Coupled to other H (d, 3 J(H A,H B )) and to F (d) with larger 3 J(H,F). Use tables of increment systems to make the case for para! Still: The spectra clearly do show more hyperfine structure than is expected for two dd systems. To be magnetically equivalent, two nuclei must be chemically equivalent and they must couple to the same nuclei (as opposed to nuclei that are the same just chemically). Think about that! -7-
Question 5. Structure Analysis III. C 4 -Compound with M = 216. (20 points) H-NMR, IR and MS spectra are provided below. Draw the correct structure on the right. Indicate in the spectra what you conclude from each spectrum that helped to make the case for this structure or against an alternative structure. Structure 1,2-dibromobutane Exclude isomers 1,1-, 2,2-, 2,3-, and 1,4-isomers: simpler H-NMR. 1,3-isomer shows no methyl as triplet. Key information: m/z 135 and 137 show that there is one Br in the fragment. One Br was lost; 135 + 79 = 214, 135 + 81 = 216, 137 + 79 = 216, 137 + 81 = 218. Total mass 214 (with 12 C, 79 Br), mass of C 4 Br 2 = 206 ==> C 4 H 8 Br 2 ==> Saturated and Acyclic. Note: The molecular mass actually is 216 because on average every molecule contains one 79 Br and one 79 Br. H-NMR: Messy. Suggests an ethyl group. Strategy needed: Exclude all other isomers because they all have simpler H-NMR spectra. -8-
4.2: CH; 3.6-3.9 ppm: different Hs in CH 2 Br; 1.8-2.2 ppm: different Hs in CH 2 ; 1 pm: CH 3. -9-