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Absorbance Compound #14 is a low-melting solid (melting point 48-49 °C). Utilizing the Mass, IR, and 'H NMR spectra, along with "C NMR data, determine the structure of this compound. Elemental Analysis: C, 85.69; H, 5.53; O, 8.78. Intensity 105 182 (0.8569)= 182(0.0553) 182 (0.0878) = C HoO 182 155.9 12.01 = 13 = 10.06 = 10 1.007 15.9 15.9 1.005 $0 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 m/e Wave Number, cm*1 4000 3000 2500 2000 1500 1300 1200 1100 1000 900 800 700 DOU = 2(13) +2-10== "C NMR Spectral Data: singlet, 187.0 ppm singlet, 137.8 ppm doublet, 132.2 ppm doublet, 130.1 ppm doublet, 128.2 ppm 10 11 1$ 14 15 Wavelength, microns ppm $

Absorbance Compound #14 is a low-melting solid (melting point 48-49 °C). Utilizing the Mass, IR, and 'H NMR spectra, along with "C NMR data, determine the structure of this compound. Elemental Analysis: C, 85.69; H, 5.53; O, 8.78. Intensity 105 182 (0.8569)= 182(0.0553) 182 (0.0878) = C HoO 182 155.9 12.01 = 13 = 10.06 = 10 1.007 15.9 15.9 1.005 $0 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 m/e Wave Number, cm*1 4000 3000 2500 2000 1500 1300 1200 1100 1000 900 800 700 DOU = 2(13) +2-10== "C NMR Spectral Data: singlet, 187.0 ppm singlet, 137.8 ppm doublet, 132.2 ppm doublet, 130.1 ppm doublet, 128.2 ppm 10 11 1$ 14 15 Wavelength, microns ppm $

Chemistry
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...
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Struggling how to identify a compound based on carbon NMR spectra
Absorbance Compound #14 is a low-melting solid (melting point 48-49 °C). Utilizing the Mass, IR, and 'H NMR spectra, along with "C NMR data, determine the structure of this compound. Elemental Analysis: C, 85.69; H, 5.53; O, 8.78. Intensity 105 182 (0.8569)= 182(0.0553) 182 (0.0878) = C HoO 182 155.9 12.01 = 13 = 10.06 = 10 1.007 15.9 15.9 1.005 $0 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 m/e Wave Number, cm*1 4000 3000 2500 2000 1500 1300 1200 1100 1000 900 800 700 DOU = 2(13) +2-10== "C NMR Spectral Data: singlet, 187.0 ppm singlet, 137.8 ppm doublet, 132.2 ppm doublet, 130.1 ppm doublet, 128.2 ppm 10 11 1$ 14 15 Wavelength, microns ppm $
Transcribed Image Text:Absorbance Compound #14 is a low-melting solid (melting point 48-49 °C). Utilizing the Mass, IR, and 'H NMR spectra, along with "C NMR data, determine the structure of this compound. Elemental Analysis: C, 85.69; H, 5.53; O, 8.78. Intensity 105 182 (0.8569)= 182(0.0553) 182 (0.0878) = C HoO 182 155.9 12.01 = 13 = 10.06 = 10 1.007 15.9 15.9 1.005 $0 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 m/e Wave Number, cm*1 4000 3000 2500 2000 1500 1300 1200 1100 1000 900 800 700 DOU = 2(13) +2-10== "C NMR Spectral Data: singlet, 187.0 ppm singlet, 137.8 ppm doublet, 132.2 ppm doublet, 130.1 ppm doublet, 128.2 ppm 10 11 1$ 14 15 Wavelength, microns ppm $
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