1Compound 1 has molecular formula C7H16. It shows three signals in the 1H-NMR spectrum, one at 0.85 ppm, one at 1.02 ppm, and one at 1.62 ppm. The relative integrals of these three signals are 6, 1, and 1, respectively. Compound 2 has molecular formula C7H14. It shows three signals in the 1H-NMR spectrum, one at 0.98 ppm, one at 1.36 ppm, and one at 1.55 ppm. The relative integrals of these three signals are 3, 2, and 2, respectively. Propose structures for compounds 1 and 2, explaining how you reach your conclusion.

1Compound 1 has molecular formula C7H16. It shows three signals in the 1H-NMR spectrum, one at 0.85 ppm, one at 1.02 ppm, and one at 1.62 ppm. The relative integrals of these three signals are 6, 1, and 1, respectively. Compound 2 has molecular formula C7H14. It shows three signals in the 1H-NMR spectrum, one at 0.98 ppm, one at 1.36 ppm, and one at 1.55 ppm. The relative integrals of these three signals are 3, 2, and 2, respectively. Propose structures for compounds 1 and 2, explaining how you reach your conclusion.

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

Chapter13: Nuclear Magnetic Resonance Spectroscopy

Section13.7: Chemical Shift

Problem 13.5P: Following are two constitutional isomers with the molecular formula C4H8O2. (a) Predict the number...

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The 13C-NMR spectrum of 3-methyl-2-butanol shows signals at 17.88 (CH3), 18.16 (CH3), 20.01 (CH3), 35.04 (carbon-3), and 72.75 (carbon-2). Account for the fact that each methyl group in this molecule gives a different signal.
Following are two constitutional isomers with the molecular formula C4H8O2. (a) Predict the number of signals in the 1H-NMR spectrum of each isomer. (b) Predict the ratio of areas of the signals in each spectrum. (c) Show how you can distinguish between these isomers on the basis of chemical shift.
Propose structures for compounds that fit the following descriptions: (a) A hydrocarbon with seven lines in its 13C NMR spectrum (b) A six-carbon compound with only five lines in its 13C NMR spectrum (c) A four-carbon compound with three lines in its 13C NMR spectrum
Compound 1 has molecular formula C7H16. It shows three signals in the 1H-NMR spectrum, one at 0.85 ppm, one at 1.02 ppm, and one at 1.62 ppm. The relative integrals of these three signals are 6, 1, and 1, respectively. Compound 2 has molecular formula C7H14. It shows three signals in the 1H-NMR spectrum, one at 0.98 ppm, one at 1.36 ppm, and one at 1.55 ppm. The relative integrals of these three signals are 3, 2, and 2, respectively. Propose structures for compounds 1 and 2, explaining how you reach your conclusion.
Compound 1 has molecular formula C6H12. It shows three signals in the 1H-NMR spectrum, one at 0.96 ppm, one at 2.03 ppm, and one at 5.33 ppm. The relative integrals of these three signals are 3, 2, and 1, respectively. Compound 2 has molecular formula C7H15Br. It shows two signals in the 1H-NMR spectrum, one at 1.08 ppm and one at 1.59 ppm. The relative integrals of these two signals are 3 and 2, respectively. Propose structures for compounds 1 and 2, explaining how you reach your conclusion.
Compound 2 has molecular formula C6H12. It shows three signals in the 1H-NMR spectrum, one at 0.96 ppm, one at 2.03 ppm, and one at 5.33 ppm. The relative integrals of these three signals are 3, 2, and 1, respectively. Provide structure for compound 2, explain how you reached your conclusion.
Compound B has molecular formula C9H12. It shows five signals in the 1H-NMR spectrum - a doublet of integral 6 at 1.22 ppm, a septet of integral 1 at 2.86 ppm, a singlet of integral 1 at 5.34 ppm, a doublet of integral 2 at 6.70 ppm, and a doublet of integral 2 at 7.03 ppm. The 13C-NMR spectrum of B shows six unique signals (23.9, 34.0, 115.7, 128.7, 148.9, and 157.4). Identify B and explain your reasoning.
Compound 1 has molecular formula C7H15Cl. It shows two signals in the 1H-NMR spectrum, one at 1.08 ppm and one at 1.59 ppm. The relative integrals of these two signals are 3 and 2, respectively. Propose structures for compound 1, explaining how you reach your conclusion.
Compound A, with molecular formula C4H9Cl, shows two signals in its 13C NMR spectrum. Compound B, an isomer of compound A, shows foursignals, and in the proton-coupled mode, the signal farthest downfield is a doublet. Identify compounds A and B.
Compound A, with molecular formula C4H9Cl, shows two signals in its 13C NMR spectrum. Compound B, an isomer of compound A, shows four signals, and in the proton-coupled mode, the signal farthest downfield is a doublet. Identify compounds A and B
The 'H NMR spectrum of compound A (C3H100) has four signals: a multiplet at 8 = 7.25-7.32 ppm (5 H), a singlet at d = 5.17 ppm (1 H), a quartet at d = 4.98 ppm (1 H), and a doublet at ô = 1.49 ppm (3 H). There are 6 signals in its 13C NMR spectrum. The IR spectrum has a broad absorption in the -3200 cm-1 region. Compound A reacts with KMNO4 in a basic solution followed by acidification to give compound B with the molecular formula C7H6O2. Draw structures for compounds A and B.
Predict the number of signals in the 1H-NMR spectrum of each isomer.