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*10. Explain the patterns and intensities of the isopropyl group in isopropyl iodide. CH3 CH–I CH;

*10. Explain the patterns and intensities of the isopropyl group in isopropyl iodide. CH3 CH–I CH;

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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 pavia Spectroscopy fifth edition page 278 ques 10 and page 281 ques 19

Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Problems 281 *19. Following are the NMR spectra of three isomeric esters with the formula C,H14O2, all derived from propanoic acid. Provide a structure for each. 'H NMR C,H1402 300 MHz 2.00 1.95 1.90 1.85 1.97 1.92 1.04 2.89 5.85 3.5 3.0 2.5 1.5 1.0 (a) C,H14O2 1Η ΝMR 300 MHz 1.98 8.60 2.97 4.0 3.0 2.5 2.0 1.5 1.0 (b) Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 282 Nuclear Magnetic Resonance Spectroscopy • Part One: Basic Concepts ΙΗ ΝMR C,H14O2 300 MHz 1,33 1.30 1.31 1.35 IL 1.87 1.94 T TTT 4.0 3.5 2.5 2.0 05 (c) *20. The two isomeric compounds with the formula C10H14 have NMR spectra shown below. Make no attempt to interpret the aromatic proton area between 7.1 and 7.3 ppm except to determine the number of protons attached to the aromatic ring. Draw the structures of the isomers. C10H14
Transcribed Image Text:Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Problems 281 *19. Following are the NMR spectra of three isomeric esters with the formula C,H14O2, all derived from propanoic acid. Provide a structure for each. 'H NMR C,H1402 300 MHz 2.00 1.95 1.90 1.85 1.97 1.92 1.04 2.89 5.85 3.5 3.0 2.5 1.5 1.0 (a) C,H14O2 1Η ΝMR 300 MHz 1.98 8.60 2.97 4.0 3.0 2.5 2.0 1.5 1.0 (b) Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 282 Nuclear Magnetic Resonance Spectroscopy • Part One: Basic Concepts ΙΗ ΝMR C,H14O2 300 MHz 1,33 1.30 1.31 1.35 IL 1.87 1.94 T TTT 4.0 3.5 2.5 2.0 05 (c) *20. The two isomeric compounds with the formula C10H14 have NMR spectra shown below. Make no attempt to interpret the aromatic proton area between 7.1 and 7.3 ppm except to determine the number of protons attached to the aromatic ring. Draw the structures of the isomers. C10H14
have such small chemical shifts. *7. In benzaldehyde, two of the ring protons have resonance at 7.87 ppm, and the other three have resonance in the range from 7.5 to 7.6 ppm. Explain. *8. Make a three-dimensional drawing illustrating the magnetic anisotropy in 15,16-dihydro-15, 16-dimethylpyrene, and explain why the methyl groups are observed at -4.2 ppm in the 'H NMR spectrum. CH3- -CH 15,16-Dihydro-15,16-dimethylpyrene Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 278 Nuclear Magnetic Resonance Spectroscopy • Part One: Basic Concepts *9. Work out the spin arrangements and splitting patterns for the following spin system: НА Нв Cl-C- -C -Br ČI HB *10. Explain the patterns and intensities of the isopropyl group in isopropyl iodide. CH3 CH-I CH3 *11. What spectrum would you expect for the following molecule? CI H Cl H -C -С—Н CI Cl *12. What arrangement of protons would give two triplets of equal area? *13. Predict the appearance of the NMR spectrum of propyl bromide. *14. The following compound, with the formula C,H3O2, is an ester. Give its structure and assign the chemical shift values. 1000 500 250 100 400 300 200 100 O CPS 1H NMR | 60 MHz Integral = 3 50 Integral = 3 Integral = 2 8.0 7.0 6.0 5.0 4.0 3.0 2.0 1.0 O PPM Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Problems 279 *15. The following compound is a monosubstituted aromatic hydrocarbon with the formula C,H12- Give its structure and assign the chemical shift values. H NMR 60 MHz 400 300 200 100 O CPS Integral = 5 Integral = 6 Integral = 1
Transcribed Image Text:have such small chemical shifts. *7. In benzaldehyde, two of the ring protons have resonance at 7.87 ppm, and the other three have resonance in the range from 7.5 to 7.6 ppm. Explain. *8. Make a three-dimensional drawing illustrating the magnetic anisotropy in 15,16-dihydro-15, 16-dimethylpyrene, and explain why the methyl groups are observed at -4.2 ppm in the 'H NMR spectrum. CH3- -CH 15,16-Dihydro-15,16-dimethylpyrene Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 278 Nuclear Magnetic Resonance Spectroscopy • Part One: Basic Concepts *9. Work out the spin arrangements and splitting patterns for the following spin system: НА Нв Cl-C- -C -Br ČI HB *10. Explain the patterns and intensities of the isopropyl group in isopropyl iodide. CH3 CH-I CH3 *11. What spectrum would you expect for the following molecule? CI H Cl H -C -С—Н CI Cl *12. What arrangement of protons would give two triplets of equal area? *13. Predict the appearance of the NMR spectrum of propyl bromide. *14. The following compound, with the formula C,H3O2, is an ester. Give its structure and assign the chemical shift values. 1000 500 250 100 400 300 200 100 O CPS 1H NMR | 60 MHz Integral = 3 50 Integral = 3 Integral = 2 8.0 7.0 6.0 5.0 4.0 3.0 2.0 1.0 O PPM Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Problems 279 *15. The following compound is a monosubstituted aromatic hydrocarbon with the formula C,H12- Give its structure and assign the chemical shift values. H NMR 60 MHz 400 300 200 100 O CPS Integral = 5 Integral = 6 Integral = 1
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