The "C NMR of ethyl phenylacetate is shown below. Sketch the DEPT-90 and DEPT-135 4. spectra directly below it. 200 180 160 140 120 100 80 60 40 20 ppm c-43-24 DEPT-90 DEPT-135
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- Draw the structure of the compound identified by the simulated 'H NMR and ¹3C NMR spectra. The molecular formula of the compound is C₁0H₁2O. (Blue numbers next to the lines in the 'H NMR spectra indicate the integration values.) ¹H NMR 1H 2H 2H 2H 2H 3H IT 111 10 8 6 8 (ppm) 13C NMR III. 220 200 180 160 140 60 40 20 100 8 (ppm) Deduce the structure from the spectra. Select Draw More CHO 2 Ć Rings 120 80 Erase Q 2 QWhich set of H's in the phenacetin analog shown below correspond to HA/HA' in the 1H NMR spectrum shown below? Phenacetin1HNMR Chem235.12.fid 7.9 7.8 7.7 7.6 HA/HA' HB/HB' N A/A' 7.5 11 Group 1 one H3C 10 B 7.2 7.1 7.0 F1 (ppm) C/C' ZI 6.8 6.7 6.6 6.5 6 f1 (ppm) 5 [Choose ] [Choose ] CH3 Group 2 900 -800 700 -600 -500 400 -300 -200 100 -0 > -3800 3600 3400 3200 -3000 2800 2600 2400 -2200 -2000 1800 -1600 1400 - 1200 -1000 -800 600 -400 -200 -2002. (Chapter 13 - Q58b) The compound whose H NMR spectrum is shown has the molecular formula C7H7B1. Follow the following questions to predict the unknown structure. Chem. shift Rel. area 2.31 1.50 701 1.00 7.35 1.00 TMS 10 7 6 O ppm 4 3 2 Chemical shift (8) e20s Cenge leaming 2(a) Degree of the unsaturation of this compound is= 2(b) The two distinct peaks in the aromatic region of the H NMR indicate that compound is .disubstituted = 2(c) The splitting pattern of the peak at 2.31 ő is = 2(d) The group that corresponds to the splitting pattern in 1(c) is = 2(e) This compound has the plane of symmetry Yes or No = 2(f) The name of the unknown compound = Intensity
- 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)Give structures for the compounds below that show one 1H NMR signal each. C2H6O C4H9Cl c) C4H8O2Propose a structure for an amine with molecular formula C6H7N by interpreting the given "H and 13c NMR data below. 1H NMR Ō 2.35 ppm Ō 7.10 ppm Ō 8.46 ppm singlet (3) doublet (2) doublet (2) 13C NMR 2, 4 1,5 6.00 5.00- 4.00 3.00- 2.00- 1.00- 0.00- 200 180 160 140 120 100 80 60 40 20 8/ ppm
- Propose structure for the ester, C11H14O2 compound that has the following C NMR spectra. CDC33 200 150 100 5 0 ppmPropose the structure that corresponds to the following H NMR spectra 1H-NMR, CDCI; Solvent, Molecular Formula: C3H1002 3.5 3.0 2.5 2.0 1.5 1.0 PpmIndicate the shift of carbon bonded to an alcohol (-CH₂-OH) (in ppm) in a ¹³C NMR spectrum. 20 90 60 135 O210
- Deduce the structure of a compound with molecular formula C5H100 that exhibits the following ¹H and ¹³C NMR spectra. IH NMR CNMR 150 10 Structure A Structure B Structure C Structure D 24 1C 100 B 2H H 20 50 10 D 31113C NMR Spectrum (50.0 MHz, CDCI, solution) DEPT C3H,NO2 solvent proton decoupled 200 160 120 80 40 8 (ppm) 1H NMR Spectrum (200 MHz, CDCI, solution) expansion 4.0 3.0 2.0 1.0 ppm TMS 10 7 6. 4 3 2 1 8 (ppm) Determine the structure of the compoundA compound has the molecular formula: C4H8O2 and gives the following C-13 NMR spectrum. Provide the most likely functional groups for each signal. nmrsim presentation 1 1 C:Bruken Topspin3.5pl7 examdata -170.7658 150 100 d 170.8 (O CH3) ; 60.4 (OCH2); 20.8, 14.1 (2 x C=0) 8170.8 (0 CH₂); 60.4 (OCH₂); 20.8, 14.1 (2 x CH3). 8 170.8 (CH3); 60.4 (C-0); 20.8, 14.1 (2 x OCH₂) 8170.8 (C=0); 60.4 (OCH₂); 20.8, 14.1 (2 x CH3). 8170.8 (C) ; 60.4 (O CH2); 20.8, 14.1 (2 x O CH3) -60.4293 50 -20.7902 -14.1385 [ppm] [+]