3458 Problem 9 115 IR Spectrum (CO solution 4000 100 282 2 20 80 % of base peak W 3000 2000 V (cm) 59 103 1600 40 1200 600 1400 Move branch M-118 (<1%) Mass Spectrum Ho Gh C6H1402 40 80 120 160 200 240 280 m/e 13C NMR Spectrum (100 0 MHz CDCI, solution) <- 4 =1400 2n+1=3 No significant UV absorption above 220 nm 2x61214-2)+(x2) =(12+2-12+2/14 = 12+2= 12 +4 114-14 =2016 DEPT CH CH CH proton decoupled solvent L T 200 160 120 80 40 'H NMR Spectrum (200 MHz, CDCI, solution) Exchanges with D₂O ୮ L 1451427 =148 0 8 (ppm) TMS 2 L 10 9 8 7 6 5 4 3 2 1 0 8 (pp

3458 Problem 9 115 IR Spectrum (CO solution 4000 100 282 2 20 80 % of base peak W 3000 2000 V (cm) 59 103 1600 40 1200 600 1400 Move branch M-118 (<1%) Mass Spectrum Ho Gh C6H1402 40 80 120 160 200 240 280 m/e 13C NMR Spectrum (100 0 MHz CDCI, solution) <- 4 =1400 2n+1=3 No significant UV absorption above 220 nm 2x61214-2)+(x2) =(12+2-12+2/14 = 12+2= 12 +4 114-14 =2016 DEPT CH CH CH proton decoupled solvent L T 200 160 120 80 40 'H NMR Spectrum (200 MHz, CDCI, solution) Exchanges with D₂O ୮ L 1451427 =148 0 8 (ppm) TMS 2 L 10 9 8 7 6 5 4 3 2 1 0 8 (pp

Organic Chemistry

9th Edition

ISBN:9781305080485

Author:John E. McMurry

Publisher:John E. McMurry

Chapter13: Structure Determination: Nuclear Magnetic Resonance Spectroscopy

Section13.SE: Something Extra

Problem 29AP

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Resolves into 13C NMR Spectrum (50.0 MHz, CDCI, solution) two signals at higher field DEPT CgH11Br Resolves into two signals at higher field expansions proton decoupled solvent 35.0 34.0 ppm 200 160 120 80 40 O 8 (ppm) 1H NMR Spectrum (200 MHz, CDCI, solution) expansion 3.0 2.5 ppm TMS 10 9 8 7 6 5 4 3 2 1 8 (ppm) Determine the structure of the compound based on the above information
Unrounded Rounded ε∗,L/μmol 0.0259825 0.0260 Heres the data if needed: TZ # Concentration Absorbance at 430 nmnm 1 33.6480 0.931 2 25.2360 0.757 3 16.8420 0.210 4 8.41200 0.137 5 4.20600 0.122
Myriam 1718 1. 4000 3000 1600 1200 800 V (cm¹) 100 Mass Spectrum 43 119 80 60 JIK M+* 40 176 20 161 C12 H16 O 40 80 200 280 120 160 m/e 13C NMR Spectrum (50.0 MHz, CDCI, solution) expansion 188 IR Spectrum (liquid film) % of base peak DEPT CH₂ CH₂ CH proton decoupled L 200 ¹H NMR Spectrum (200 MHz, CDCI, solution) J 1 9 10 I 8 2000 160 the 1 7 6 240 120 130 130 125 ppm 125 ppm L 5 expansion 1 80 1 4 solvent Problem 99 UV Spectrum max 262 nm (log₁0 € 2.5) solvent methanol 40 t I 3 1 2 4 08 (ppm) TMS 1 0 8 (ppm) 1 1 205
100 08 60 40 CH3(CH,),COOH 20 hexanoic acid 1711 1600 wavenumber (cm) 4000 3500 3000 2500 2000 1800 1400 1200 1000 800 600 tTAANSE-TTANCE
Molecular formula: C10H1802 ¹H spectrum 1H, t J = 6.2 Hz 13C spectrum 180 160 2H, d J = 6.2 Hz 140 120 PPM 100 PPM 80 3H, s 60 3H, s 9H, s 40 20
7.) From the following sets of data, please propose a reasonable structure for the unknown organic compound. This molecule is entirely composed of C, H, and N only. Please show all of your work and feel free to write on the spectra any clues as to the structure. 4000 3000 2000 1000 S DO Infrared spectrum of unknown. KBr salt plate sample. 100 MS-NU-6396 80 60 40 20 - 10 20 30 40 50 60 70 80 90 100 110 m/z Mass spectrum of unknown. Molecular ion = 83.1 Relative Intensity
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)
IR Spectrum (CCl4 solution) U 4000 3000 100 80 60 40 20 JL 40 80 13C NMR Spectrum (100.0 MHz, CDCI, solution) % of base peak DEPT CH₂ CH₂ CH proton decoupled 200 ¹H NMR Spectrum (200 MHz, CDCI, solution) 10 9 8 2000 1693 v (cm¹) 155/157 120 160 m/e 160 7 1600 183/185 7.8 6 your 1200 120 800 Mass Spectrum C8H7OBr 280 solvent 80 M+= 198/200 200 240 expansion سلا 5 7.2 ppm 4 3 Problem 100 UV Spectrum max 258 nm (log₁0€ 4.2) solvent: ethanol 40 2 0 1 206 8 (ppm) TMS L 0 8 (ppm) 189
The spectra (IR, ¹H NMR, ¹³C NMR and MS) of an unknown compound are shown in the following spetta. Based on these spectra, determine the structure of the unknown compound. Explain your answers clearly for each spectrum. Absorbance/% IR 2000 1500 1000 500 50- 100+ 4000 11 1H NMR 3000 Integration 9 10 2500 7.95 7.25 U 2 2 8 7 Wavenumbers/cm-1 6 5 ppm 4.30 2 3 2.35 3 2 1.35 3 T 1 O
опанит 1715 1600 IR Spectrum 100 4000 (liquid film) 43 80 75 60 M 40 20 % of base peak 10 3000 40 C 13C NMR Spectrum (20.0 MHz, CDCI solution) proton decoupled 9 80 200 ¹H NMR Spectrum (100 MHz, CDCI, solution) 101 8 2000 v (cm¹) M** 132 120 160 m/e 160 7 6 1200 120 Mass Spectrum 200 240 280 800 C-H 5 C6H1203 solvent L 80 4 -CH3 No significant UV absorption above 220 nm 3 H-C-H u -CH3 40 2 59 TMS 0 1 8 (ppm) TMS 0 8 (ppm)
Molecular Formula: C₁1H15O₂N ¹H spectrum 160 2H, d J = 7.5 Hz 13C spectrum 140 2H, d J = 7.5 Hz 120 100 5 PPM 80 PPM 60 1H, quintet J = 7 Hz 3 1 6H. triplet J = 7 Hz 4H, quintet J = 7 Hz 40 20
43 184 IR Spectrum (Iquid film) 4000 3000 100 80 60 40 20 40 13C NMR Spectrum (100.0 MHz, CDCI, solution) 9 80 DEPT CH₂ CH₂4 CHA proton decoupled 200 ¹H NMR Spectrum (400 MHz CDCI, solution) expansion 7.5 7.3 ppm 8 P 10 SAMSUNG Galaxy A53 5G 2000 1600 V (cm) 105 M+- 134 120 160 m/e 160 09b0f2ca-eac1-4... 1200 T L 120 2.8 2.7 ppm 7 6 ww 800 Mass Spectrum C10H14 200 240 280 T 1 5 solvent " 80 expansions 1.8 1.7 ppm 1 4 3:54 م 0 *1 470 صفحة Problem 96 UV Spectrum max 260 nm (log10=2.5) PE 40 1.45 1.35 ppm 3 2 Problem 97 08 (ppm) expansion 1.05 0.95 TMS I 0 8 (ppm) 1 202 تمشح بـ 100 جيجابايت تخزين على OneDrive cloud لمدة 6 أشهر 185.