IR Spectrum(KBr disc)4000100806040&20www.30004410سحر4013C NMR Spectrum(100 0 MHz. D₂0 solution)DEPT CH₂ CH₁ CHproton decoupled200¹H NMR Spectrum(400 MHz. D₂0 solution)Note: there are 3 protonswhich exchange withthe D₂0 solvent3.9974 M 8980120expansions20003.7 ppm161620V (cm¹)m/e16016001601.4 ppm1L8 72001200120800Mass SpectrumC3H7NO₂240 280H₂O and HODin solvent16 5804Problem 20No significant UVabsorption above 220 nm340121260 8 (ppm)08 (ppm)1109

Answered: Image /qna-images/answer/4a1a5962-07cf-47ee-aa90-dcd33893138e.jpg

IR Spectrum (KBr disc) 4000 100 80 3000 44 40 13C NMR Spectrum (100 0 MHz. D₂0 solution) 74 M 89 80 DEPT CH₂ CH₁ CH proton decoupled 200 ¹H NMR Spectrum (400 MHz. D₂0 solution) Note: there are 3 protons which exchange with the D₂0 solvent 3.9 expansions 2000 3.7 ppm 1620 16 V (cm) 1600 120 160 m/e 160 1.4 ppm 200 1200 120 800 Mass Spectrum C3H7NO₂ 240 280 H₂O and HOD in solvent 80 Problem 20 No significant UV absorption above 220 nm 80 40

IR Spectrum (KBr disc) 4000 100 80 3000 44 40 13C NMR Spectrum (100 0 MHz. D₂0 solution) 74 M 89 80 DEPT CH₂ CH₁ CH proton decoupled 200 ¹H NMR Spectrum (400 MHz. D₂0 solution) Note: there are 3 protons which exchange with the D₂0 solvent 3.9 expansions 2000 3.7 ppm 1620 16 V (cm) 1600 120 160 m/e 160 1.4 ppm 200 1200 120 800 Mass Spectrum C3H7NO₂ 240 280 H₂O and HOD in solvent 80 Problem 20 No significant UV absorption above 220 nm 80 40

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

(f) C10H13NO2 4000 100 80 60 20 % of base peak 3284 43 3000 40 10 IR Spectrum (KBr disc) proton decoupled 13C NMR Spectrum (50.0 MHz, CDCI, solution) DEPT CH₂ CH₁ CH 200 9 80 ¹H NMR Spectrum (200 MHz, CDCI, solution) exchanges with D₂O on warming 108 le 137 M+* 179 120 160 2000 8 1658 v (cm¹) 1600 m/e 1 160 if L 7 120 expansion 1200 6 200 240 280 7.6 800 Mass Spectrum C10H13 NO₂ 5 solvent 80 l 7.0 4 ppm UV Spectrum max 250 nm (log₁0€ 3.1) A max 287 nm (log₁0 2.2) solvent: chloroform 3 40 2 0 1 8 (ppm) TMS 0 8 (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)
4000 IR Spectrum quid film) 100 80 60 40 YI 2248 41 10 3000 43 40 proton decoupled "C NMR Spectrum (150 MHz CDC, solution) 68 DEPT CH, CH, 1 CHT 9 200 ¹H NMR Spectrum (600 MHz, CDC, solution) M83 (<1%) 80 120 2000 v (cm) 8 m/e 160 160 2.3 L 7 1600 2.2 200 ryper 1200 6 ▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬ 120 Mass Spectrum 240 21 2.0 expansions (not to scale) 800 I CsH₂N 280 5 expansion 80 4 270 255 256 255 expansion No significant UV absorption above 220 nm 27.0 245 250 255 1.0 3 40 L 2 0 8 (ppm) 1 1 TMS L 1 0 8 (ppm) 10 ¹H-¹C me-HSQC Spectrum (600 MHz, CDC, solution) Black cross-peaks CH and CH Red cross-peaks CH₂ ppm 20 ¹H-"C HMBC Spectrum (600 MHz, CDC), solution) -22 -24 -26 -28 116 118 120 122 124 3.0 2.8 2.6 2.4 2.2 2.0 1.8 1.6 1.4 1.2 1.0 0.8 0.6 ppm ppm 20 22 -24 26 28 -116 118 -120 122 -124 3.0 2.8 2.6 2.4 2.2 2.0 1.8 1.6 1.4 1.2 1.0 0.8 0.6 ppm IR: List bands and possible/probable structural units responsible Band (cm³) Possible…
A compound has the molecular formula: C4H8O2 and gives the following H-1 NMR spectrum. Give the functional groups responsible for the signals, which integrate as 2H, 3H and 3H (downfield to upfield). nmrsim presentation 1 1 C: Bruker TopSpin3.5p17\examdata elle 3.0000 SODEŹ quartet is CH₂, singlet is CH3 and triplet is OCH3. quartet is CH3, singlet is OCH₂ and triplet is OCH3 quartet is CH₂, singlet is OCH3 and triplet is CH3. quartet is OCH₂, singlet is CH3 and triplet is CH3. quartet is OCH₂, singlet is CH₂ and triplet is CH3. POSE (ppm) 10 [rel]
For AIME3, sketch the room temperature 1H NMR spectrum and that observed at -50°C. Explain the observations.
4000 100 80 60 40 20 3410 tele 3000 40 10 43 ¹3C NMR Spectrum (50.0 MHz, CDCI, solution) proton decoupled 6.0 DEPT CH, CH₂ 1 CH 1 9 71 80 200 ¹H NMR Spectrum (200 MHz, CDCI, solution) expansion ORGANIC STRUCTURES FROM SPECTRA 5.5 8 IR Spectrum (liquid film) 2000 M 86 (1%) 120 V (cm) m/e 160 1600 7 160 200 5.0 ppm 6 120 1200 J Mass Spectrum 240 800 5 CsH10O 280 solvent 80 4 exchanges with D.O 1 L 3 40 No significant UV absorption above 220 nm 2 E Q 115 08 (ppm) 1 TMS L 0 ō (ppm) IR: List bands and possible/probable structural units responsible Band (cm³) Possible stretches/functional groups MS: Molecular Formula, 13C NMR: 1 2 3 4 5 6 Chem. Shift (8) Type of C (sp), sp², carbonyl, etc.) ¹H NMR: 1 Chem. Shift (8) Multiplicity (singlet, doublet, etc.) IHD(DU)= 2 3 4 5 6 If you need more space, add any additional signal info here: (Show calculation) # of attached H's # of identical C's in signal Probable Type of H (methyl, aromatic, etc.) # of identical H's in signal
(c) 'H NMR spectrum; Iron and Chlorine are NMR inactive, 31P, I = ½ 100% abundant. PCI3 Cll. PCI3
CHEM 241L. Exp#12 NMR Spectre 3. Compound 3 is a low boiling liquid (bp-58°C) that reacts with Br. in CCl. to form a 1,1,2,2 tetra-bromide. Elemental Analysis: C-48.36%, II - 4.06%; contains halogen "C Spectral Data: 80.4ppm (doublet), 68.3ppm (singlet), 34.7ppm (triplet) Mass Spectrom Absorbence Intonally 48 'H NMR Infrared Spectrom 4000 3000 2500 2000 74 70 80 90 100 110 120 130 140 150 160 170 100 mie Wave Number, cm .1 1300 1200 1100 1000 mon 1600 Wavelength, microne Compound 3 Structure Ppin
9 ¹H-NMR Spectra 2 M.F. = C₁₁H₁2O 15 15 -00 8 15 15 7 1 S 6 O) 15 15 5 PPM 4 21 -♡ 3 2 3) 0
4000 100 80 60 20 % of base peak 3376 40 10 3000 59 13C NMR Spectrum (50.0 MHz, CDCI, solution) 173 proton decoupled 200 I 9 80 ¹H NMR Spectrum (200 MHz, CDCI, solution) DEPT CH₂ CH₂ CH4 IR Spectrum (liquid film) 2000 M+88 (<1%) 120 1600 v (cm¹) m/e 160 160 I I 8 7 mmm 1 120 200 240 6 1200 . Mass Spectrum L 800 1 5 C5H₁20 280 solvent L 80 4 problem 78 3 No significant UV absorption above 220 nm 40 Il 08 (ppm) 1 2 1 TMS 0 8 (ppm)
3000 15MHz carbon-13 NMR spectrum CDCi, solution off-resonance decoupled proton decoupled 200 In spectrum (nujol mull) 135 28 2000 100MHz proton NMR spectrum CDCi, solution 1600 V (cm-1) wayang 147 40 60 80 100 120 140 160 180 200 220 240 260 260 300 320 340 m/e 10% expansion at 130ppm 1670 mt 150. 2H 1200 800 mass spectrum 209 3H 124 180 160 140 120 100 80 60 40 UV spectrum a 14.999mg/10ml hexane 0.150mg/10ml hexane 1cm b 0.5cm cells 250 300 200 problem 3H 2 (nm) 20 0 350 6 (ppm) 8 (ppm)
13C NMR Spectrum (100.0 MHz, D20 solution) CgHgNO3 DEPT proton decoupled 200 160 120 80 40 O 8 (ppm) 'H NMR Spectrum (400 MHz, D20 solution) expansions Note: there are 4 protons which exchange with the D20 solvent 4.3 4.1 ppm 3.6 3.4 ppm 1.3 1.1 ppm HаО and HOD in solvent 8 7 6 4 3 2 1 8 (ppm) 10 9 Determine the structure of the compound.