1054. These four 'H NMR spectra were recorded from different isomers with molecular formulaCsH,CIO. They all contain a carbonyl group. Determine the structure of the different isomers.01050105109876543.109101066901091051876532-a8765110987652221032161032 1260

1st spectra: A single peak suggests that all 9 hydrogens are equivalent. With some trial and error,…

Answered: 10 5 4. These four 'H NMR spectra were…

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

Chapter17: Carboxylic Acids

Section: Chapter Questions

Problem 17.17P

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Use the 1H NMR and IR data to determine the structure of each compound.
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The H NMR spectra corresponds to a hydrocarbon with the molecular formula C6H₁4. Deduce the structure from the spectra. Draw the unknown compound. 'H NMR 12 H Select Draw Rings More 1.5 1.4 2 H 1.5 1.0 0.5 0.0 300-MHz H NMR spectrum ppm (8) 0.9 0.8 (CH3)4Si G G Q2E
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Compound A has molecular formula C7H7X. Its ¹H-NMR spectrum shows a singlet at 2.26 ppm and two doublets, one at 6.95 ppm and one at 7.28 ppm. The singlet has an integral of three and the doublets each have an integral of two. Its 13C- NMR shows five signals. The mass spectrum of A shows a peak at m/z 170 and another peak at m/z = 172; the relative height of the two peaks is 1:1 respectively. - Identify what atom X is, explaining your reasoning - Identify Compound A, explaining your reasoning Compound A is treated with a mixture of nitric and sulfuric acids to generate Compound B. The ¹H-NMR spectrum of B shows two singlets, one at 2.52 pm and one at 8.13 ppm. The 13C-NMR spectrum of B shows five signals. The mass spectrum of B shows a peak at m/z = 260 and another peak at m/z = 262; the relative height of the two peaks is 1:1 respectively. - Identify compound B, explaining your reasoning Compound B is treated with sodium ethoxide to generate compound C. The ¹H-NMR spectrum of C shows…
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16-17 Propylamine (bp 48°C), ethylmethylamine (bp 37°C), and trimethylamine (bp 3°C) are constitutional isomers with the molecular formula C3HgN. Account for the fact that trimethylamine has the lowest boiling point of the three and propylamine has the highest.
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Consider a molecule with the molecular formula of C6H13NO. The IR and ¹H NMR spectra are shown below. Draw the structure that best fits this data. www 20000 Wawa (¹ 400 TH A 30000 + 15000 100 500 1.1

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