problemset3b-2-answers

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Problem Set 3 1 H NMR Spectroscopy CHEM 393 Dr. H.M. Muchall 1. The following 1 H NMR spectrum was recorded on a 60 MHz spectrometer. It shows three signals. The molar mass of the hydrocarbon is 120 g/mol. a) (1 point) What causes the unidentified, unintegrated signal? TMS b) (1 point) What are the chemical shifts of signals a and b in ppm? c) (4 points) Label signals a and b with all necessary information. Which compound is it? d) (3 points) Calculate and evaluate the chemical shift for all non-equivalent protons. a: δ 1 H = 7.27 – 2·0.14 – 0.17 = 6.82 ppm (6.7 ppm) ok b: δ 1 H = 0.23 + 1.85 + 0 = 2.08 ppm (2.2 ppm) ok e) (2 points) Give a closely related isomeric compound and reason why it is not a proper solution. there would be 2 methyl signals in 2:1 ratio, the aromatic protons would probably show their different δ and coupling, the chemical shifts would be different s 1H C 6 H 3 – no coupling probably s 3H CH 3 –EW pay close attention to the TMS position 120/13 = 9 + 3/13, C 9 H 12 ; U = 4 -0.3 ppm => a 6.7 ppm, b 2.2 ppm 4.0 cm 1.3 cm if you do not provide all labels, you will lose out on 3 points! evaluation in this question is versus the experimental values

2. Predict the 1 H NMR spectra for the following compounds. Include chemical shift (with evaluation), integration and multiplicity. Give proper drawings that consider the intensity of the lines within a multiplet. a) (4 points) b) (5 points) a: CH 2 (a): t, 4 H, δ 1 H = 0.23 + 2.53 + 0.47 = 3.23 ppm ok Cl R CH 2 (b): quint, 2 H, δ 1 H = 0.23 + 0.47 + 0.47 ? = 1.17 ppm (will be too low) R R b: CH 3 (a): d, 3 H, δ 1 H = 0.23 + 0.47 ? = 0.70 ppm (will be too low) R CH: q, 1 H, δ 1 H = 0.23 + 0.47 + 2.53 + 1.70 = 4.93 ppm (cannot evaluate) R Cl COR CH 3 (b): s, 3 H, δ 1 H = 0.23 + 1.70 = 1.93 ppm ok COR 3. For each set of 1 H NMR data, suggest a structure that is consistent with the data. a) (2 points) C 3 H 5 Cl 3 : 2.20 ppm, 3H; 4.02 ppm, 2H CH 3 -EW -CH 2 -EW U = 0 b) (2 points) C 7 H 8 O: 2.43 ppm, 1H; 4.58 ppm, 2H; 7.28 ppm, 5H -OH -CH 2 -O C 6 H 5 - U = 4 Cl Cl O Cl Cl Cl Cl OH a b a b evaluation in this question is versus the available increments you do not have rules for methine (CH) protons, so you have to make do with simply adding three increments; this is not particularly predictive! integration here is provided as area under the peak; you can also draw a steptrace

4. (11 points) Two isomeric ketones show the following 1 H NMR spectra. Identify the compounds. s 3H CH 3 –EW s 2H C–CH 2 –EW m 5H C 6 H 5 – t 3H CH 3 –CH 2 q 2H CH 3 –CH 2 –EW m 5H C 6 H 5 – O O “C”: it is important to indicate “no coupling” seeing that the aromatic region is a mess, and without more explicit instructions, you can treat both sets of signals as one, just like in the spectrum above if you do not provide all labels, you will lose out on 9 points! two sets of protons here in both labels: I need to be able to tell which belongs to the signal; underline! 1 point 4.5 points 1 point 4.5 points

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CH 3 H H H O H CH 3 H H O or H cis H trans 3 J 11 - 18 Hz 5. (10 points) The following multiplets are due to protons A, M, and X. Determine the signal multiplicity, the coupling constants J AM , J AX and J MX as well as the number of protons in each group (take the sum of the height of the lines as an integral). Classify the systems as AMX or AM 2 X. 10 Hz a) b) 6. (10 points) Identify the compound that shows the following 1 H NMR spectrum. Provide full labels for all signals. 2.5 cm A M X dd d d A M X dt dd dt int.: 2:2:2 or 1:1:1 => AMX b 0.5 cm J AM = 2 Hz a 0.25 cm J AX = 1 Hz J MX = 0 Hz int.: 2:4:2 or 1:2:1 => AM 2 X a 0.75 cm J AM = 3 Hz c 1.0 cm J AX = 4 Hz b 0.25 cm J MX = 1 Hz a a b b a a b b c c U = 2 C=C, C=O O C C H H 10 ppm d 1H dq 1H dd 1H dd 3H C C H CH 3 H C=C C H C H CH 3 C C H C H H 2.6 cm = 100 Hz 0.4 cm = 15 Hz remember that an integration records the total area under a signal (for example, under the three peaks of a triplet); therefore, in line spectra you need to sum up the heights of all lines that belong to a signal the red lines indicate which distances you have taken for your analysis and need to be provided! 1/2 point for every piece of information we will ignore the unresolved quartet broadening the distances in cm given here might be a bit off due to the pdf conversion 2 points