H NMR (300 mMHz, CDC13, 21C) 0.92 ppm (t, 3H, J = 7Hz), 1.20 (s, 6H), 1.50 (q, 2H, J = 7 Hz), 1.64 (broad singlet, 1H). Select the correct structure for this molecule. A CD B OH OH of HO D E
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- Compound CsH12 gives the following H-NMR. Draw the structure of the compound. Draw a box around the structure you want graded. 5.00 1.00 6.04 70 6.5 6.0 4.0 3.5 3.0 25 20 1.5 A student was adding bromine across the double bond of 2-butene to make 2,3-dibromobutane. After taking the NMR, the student discovered they didn't get the product expected. Based on the NMR, what product did they obtain? Draw a box around your answer. 1.00 2.01 |2.00 3.00Draw the structure of the compound identified by the simulated 'H NMR and C NMR spectra. The molecular formula of the compound is C,0H,O. 12 (Blue numbers next to the lines in the 'H NMR spectra indicate the integration values.) H NMR 1H 2H 2H 2H 2H|| 3H 10 8. 4 (dd) g 13C NMR 220 200 180 160 140 120 100 80 60 40 20 8 (ppm) Deduce the structure from the spectra. Select Draw Rings More EraseFollowing is a 1H-NMR spectrum of 2-butanol. Explain why the CH2 protons appear as a complex multiplet rather than as a simple quintet.
- 4. A structure of molecular formula C5H₁2O has the following NMR: ¹H NMR (300 MHz, CDC13, 21C) 80.92 ppm (t, 3H, J = 7 Hz), 1.20 (s, 6H), 1.50 (q, 2H, J = 7Hz), 1.64 (broad singlet, 1H). What is the chemical shift of each of the three lines in the triplet signal at 0.92 ppm? Propose a reasonable structure given the ¹H NMR data.500 20 8.0 Interpret the 'H-NMR spectrum and assign the correct structure. Please be sure to explain your work. C₂H₁, CL 3.0 7.0 400 4.0 6.0 5.0 -B00 5.0 6.0 4.0 200 7.0 E 3.0 8.0 MA 20 100 9.0 1.0 10.0 (ppm) 가 O Hz 08 (ppm)13C NMR Spectrum (50.0 MHz, CDCI, solution) DEPT proton decoupled 200 ¹H NMR Spectrum (200 MHz, CDCI, solution) C₂H₂ NO ₂ expansion 3.0 160 2.0 120 1.0 ppm 10 9 8 7 6 Determine the structure of the compound 5 solvent 80 4 40 3 2 08 (ppm) 1 TMS 0 8 (ppm)
- Draw the structure of the compound identified by the simulated 'H NMR and 13C NMR spectra. The molecular formula of the compound is C1,H120. (Blue numbers next to the lines in the 'H NMR spectra indicate the integration values.) Η NMR 1H 2H 2H 2H 2H 3H| 10 8 4 2 8 (ppm) 13C NMR 220 200 180 160 140 120 100 80 60 40 20 d (ppm)-1 -1 An acidic liquid has a broad band at 3300 cm and a sharp band at 1710 cm in the infrared spectrum. One mole of this unknown reacts with one mole of NaOH. The ¹H NMR spectrum shows three peaks in the ratio 2:2: 1. Microanalysis gave C, 33.2%; H, 4.6%; C1, 32.7 %. Deduce the structure of the unknown. Spell out the full name of the compound. 1-chloroproponoic acid6. The following compound is a carboxylic acid that contains a bromine atom: C,H,02Br. The peak at 10.97 ppm was moved onto the chart (which runs only from 0-8 ppm) for clarity. What is the structure of the compound? 400 300 200 100 CPS H NMR 60 MHz Integral = 3 CaH;O>Br Integral = 1 Integral = 1 Integral = 2 8.0 7.0 6.0 5.0 3.0 2.0 1.0 O PPM
- A low-resolution mass spectrum of 2-methyl-2-butanol (MW 88) shows 16 peaks. The molecular ion is absent. Account for the formation of peaks at m/z 73, 70, 59, and 55 and propose a structural formula for each cation.The 1H NMR spectra of two carboxylic acids with molecular formula C3H5O2Cl are shown below. Identify the carboxylic acids. (The “offset” notation means that the farthest-left signal has been moved to the right by the indicated amount to fit on the spectrum; thus, the signal at 9.8 ppm offset by 2.4 ppm has an actual chemical shift of 9.8 + 2.4 = 12.2 ppm.)The 1H NMR spectra of two carboxylic acids with molecular formula C3H5O2Cl are shown below. Identify the carboxylic acids. (The “offset” notation means that the farthest-left signal has been moved to the rightby the indicated amount to fit on the spectrum; thus, the signal at 9.8 ppm offset by 2.4 ppm has an actual chemical shift of 9.8 + 2.4 = 12.2 ppm.)