A compound displays a prominent peak at 3400 cm-¹ in its IR spectrum. The mass spectrum has a molecular ion with m/z of 74. The base peak is at m/z = 45. Draw a structure that best fits this data. Select to Draw
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- 115 Problem 9 3458 D-H) IR Spectrum (CO souton 4000 3000 2000 znrl-3 1600 1200 V (cm') 800 100 80 50 Mass Spectrum Ho Gh Move blanc M 118 ( 1S) No significant UV absorption above 220 nm c Le 103 CeH1402 40 80 120 160 200 240 280 m/e 13C NMR Spectrum (100 O MHz CDCI, solution) -14-14 -zare DEPT CH CHt CH solvent proton decoupied 40 0 8 (ppm) 160 120 80 200 'H NMR Spectrum (200 MHz, CDCI, solution) Exchanges with D20 TMS 1 8 (ppm) 4. 3 7 6. 8. 10 98 Sof base peak2964 ーSOCT 1170 296 Please provide a structure consistent with the following IR, 13C NMR, and 'H NMR spectra. For full credit assign at least two bands in the IR and assign all protons in the 'H NMR/13C NMR spectrum. Problem 1. Formula: CSH100 00- 09 20- 1.4 1.2 with 0.6 PPM 3.0 2.8 2.6 2.4 2.0 1.8 1.6 0.8Problem 27 of 44 A compound displays a prominent peak at 3400 cm1 in its IR spectrum. The mass spectrum has a molecular ion with m/z of 74. The base peak is at m/z = 45. Draw a structure that best fits this data. Select to Draw Submit Version: 3.122.7 + 4631 production Tap here for additional resources
- 115 Problem 9 3458 R-H) IR Spectrum (CO soktiory 4000 zn+l=3 3000 2000 1600 1200 800 v (cm') 100 80 Mass Spectrum 60 Mate banch M 118 1S) 40 No significant UV absorption above 220 nm 20- 103 CeH1402 40 80 120 160 200 240 280 m/e 13C NMR Spectrum (100 O MHz CDCI, solution) -1-14 =zaro DEPT CH CH, CH proton decoupled solvent 200 160 120 80 40 8 (ppm) 'H NMR Spectrum (200 MHz. CDCI, solution) Exchanges with D20 TMS 9. 4. 3 8 (ppm) 98 ead aseps 10Problem 7 2249 IR Spectrum (KBr disc) 4000 3000 2000 1600 1200 800 v (cm') 100 Mass Spectrum 80 53 60 No significant UV 40 M+' = 80 absorption above 220 nm 40 20 C4HẠN2 160 m/e 40 80 120 200 240 280 T 13C NMR Spectrum (100.0 MHz, CDCI, solution) DEPT CHat CH, сн+ proton decoupled solvent 200 160 120 80 40 8 (ppm) H NMR Spectrum (400 MHz, CDCI, solution) TMS 10 1 8 (ppm) 9 7 6. 4 % of base peak 5Problem 8 114 IR Spectrum (quid fim) 4000 3000 2000 1600 1200 V (cm') 800 100 80 Mass Spectrum 57 More branch 60 E Co M 114 ( 1%) No significant UV 40 absorption above 220 nm 20 99 CBH18 2メ8k8-み8) 16 18-(6) 40 80 120 160 200 240 280 m/e 13C NMR Spectrum (50 O MHz. CDCI, solution) DEPT CH, CH,t CHt KAS ASym. 1500 solvent proton decoupled CAl3 Sym o 8(ppm) 340 200 160 120 80 40 1H NMR Spectrum (200 MHz. CDCIg solution) TMS 2 8. 7 6. 5 4 8 (ppm) 10 Sof base peak
- Problem 2: Provide 3 potential formula's and draw 1 possible structure based on the following information a) m/z = 82 0.8 0.6 0.4 0.2 TTTTTTT 3000 2000 1000 Wavenumber (cm-1) b) m/z =141 IR= 2200 strong, sharp, 1700 strong and sharpQUESTION 21 Elemental analysis determined an unknown compound to contain 40.0 % carbon, 6.7 % hydrogen. The unknown compound had an M+ = 60 m/z; three key IR absorbances: strong at 1710, strong below 3000, and broad from 2500 to 3500. The proton NMR showed an absorbance at 1.7 ppm (lowest whole number ratio of H =3), and an absorbance at 11.5 ppm (lowest whole number ratio of H = 1). The compound was able to turn blue litmus paper red, and neutralize a base. What is this compound? %3D O CH3CO2H O Formaldehyde О но C = CH2 %3D НО O CH2(OH)CHO O CH(OH)=CH(OH)8 7) Chemical Formula: Cal₁IN IR: weak peak at 3400cm-¹ 1H SH Unknowns for 1H NMR Final Report 6 16+2=18+1=19-11=8/2=4 NH2 HH 8) Chemical Formula: C8H1204 IR: strong peak 1720cm-1, peak at 1650cm-1 PPM 2H > 3H 4-14 H H H h+1=2 2 16-17 8x2=16+2=18-12=6/2= 0
- Problem 6 C-H IR Spectrum quid fm گرد 1716 4000 3000 2000 1600 1200 112 800 (cm") 100 80 43 Mass Spectrum UV Spectrum 60 Amas 289 nm (log, 1.4) solvent methan CHO2 40 80 120 160 200 240 280 m/e 1C NMR Spectrum (100 M. COC, solution) DEPT CH OHt ont proton decoupled Nvert 200 160 120 80 40 8 (ppm) H NMR Spectrum (200 M CDC, soluion) TMS 3. 10 8. 8 (ppm)Question 4 of 18 The typical range of observed peaks in a 19C NMR spectrum is A) 0-8 ppm B) 0-14 ppm C) 0-160 ppm D) 0-220 ppmStructure problem 4 Struktuurprobleem IR Spectrum (liquid film) 1686 4000 3000 2000 1600 1200 800 V (cm) 100 Mass Spectrum 105 80 77 60 M* 148 40 120 20 40 80 120 160 200 240 280 m/e 13C NMR Spectrum (50.0 MHz, CDCI, solution) DEPT CH CH,t CH expansion proton decoupled 135 130 ppm solvent 200 160 120 80 40 0 (ppm) 'H NMR Spectrum (200 MHz, CDCI, solution) expansion elle 3.0 2.0 1.0 ppm Integral TMS 10 9 8 7 6 4 2 1 8 (ppm) % of base peak