17.49 Identify the lettered compounds in the following reaction scheme. Compounds F, G, and K are isomers of molecular formula C13H18O. How could 1H NMR spectroscopy distinguish these three compounds from each other? OH OH OH F G K 17.49 Compounds F, G, and K are all alcohols with aromatic rings so there will be many similarities in their proton NMR spectra. These compounds, however, will show differences in absorptions due to the CH protons on the carbon bearing the OH group. F has a CH₂OH group, which will give a singlet in the 3–4 ppm region of the spectrum. G is a 3º alcohol that has no protons on the C bonded to the OH group so it will have no peak in the 3-4 ppm region of the spectrum. K is a 2° alcohol that will give a doublet in the 3–4 ppm region of the spectrum for the CH proton on the carbon with the OH group.
17.49 Identify the lettered compounds in the following reaction scheme. Compounds F, G, and K are isomers of molecular formula C13H18O. How could 1H NMR spectroscopy distinguish these three compounds from each other? OH OH OH F G K 17.49 Compounds F, G, and K are all alcohols with aromatic rings so there will be many similarities in their proton NMR spectra. These compounds, however, will show differences in absorptions due to the CH protons on the carbon bearing the OH group. F has a CH₂OH group, which will give a singlet in the 3–4 ppm region of the spectrum. G is a 3º alcohol that has no protons on the C bonded to the OH group so it will have no peak in the 3-4 ppm region of the spectrum. K is a 2° alcohol that will give a doublet in the 3–4 ppm region of the spectrum for the CH proton on the carbon with the OH group.
Organic Chemistry
8th Edition
ISBN:9781305580350
Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote
Publisher:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote
Chapter21: Benzene And The Concept Of Aromaticity
Section: Chapter Questions
Problem 21.25P: Compound H (C8H6O3) gives a precipitate when treated with hydroxylamine in aqueous ethanol and a...
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Why do only the immediately adjacent H's show up in the number of peaks? Are there normally peaks for the H's that are 2-3 carbons away?

Transcribed Image Text:17.49 Identify the lettered compounds in the following reaction scheme. Compounds F, G, and K are isomers of molecular
formula C13H18O. How could 1H NMR spectroscopy distinguish these three compounds from each other?
OH
OH
OH
F
G
K
17.49 Compounds F, G, and K are all alcohols with aromatic rings so there will be many similarities in
their proton NMR spectra. These compounds, however, will show differences in absorptions due
to the CH protons on the carbon bearing the OH group. F has a CH₂OH group, which will give a
singlet in the 3–4 ppm region of the spectrum. G is a 3º alcohol that has no protons on the C
bonded to the OH group so it will have no peak in the 3-4 ppm region of the spectrum. K is a 2°
alcohol that will give a doublet in the 3–4 ppm region of the spectrum for the CH proton on the
carbon with the OH group.
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