Organic Chemistry
6th Edition
ISBN: 9781936221349
Author: Marc Loudon, Jim Parise
Publisher: W. H. Freeman
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Chapter 19, Problem 19.6P
Interpretation Introduction
Interpretation:
The structure of
Concept introduction:
Spectroscopy method is used to identify the structure of the molecule. It is based on the interactions between matter and
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The 13C NMR spectrum of 1-bromo-3-chloropropane contains peaks at δ 30, δ 35, and δ 43. Assign these signals to the appropriate carbons.
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Draw the H1 NMR spectra of ethylcyclopropane. Draw the chemical structure and predict the proton splitting and chemical shifts.
Chapter 19 Solutions
Organic Chemistry
Ch. 19 - Prob. 19.1PCh. 19 - Prob. 19.2PCh. 19 - Prob. 19.3PCh. 19 - Prob. 19.4PCh. 19 - Prob. 19.5PCh. 19 - Prob. 19.6PCh. 19 - Prob. 19.7PCh. 19 - Prob. 19.8PCh. 19 - Prob. 19.9PCh. 19 - Prob. 19.10P
Ch. 19 - Prob. 19.11PCh. 19 - Prob. 19.12PCh. 19 - Prob. 19.13PCh. 19 - Prob. 19.14PCh. 19 - Prob. 19.15PCh. 19 - Prob. 19.16PCh. 19 - Prob. 19.17PCh. 19 - Prob. 19.18PCh. 19 - Prob. 19.19PCh. 19 - Prob. 19.20PCh. 19 - Prob. 19.21PCh. 19 - Prob. 19.22PCh. 19 - Prob. 19.23PCh. 19 - Prob. 19.24PCh. 19 - Prob. 19.25PCh. 19 - Prob. 19.26PCh. 19 - Prob. 19.27PCh. 19 - Prob. 19.28PCh. 19 - Prob. 19.29PCh. 19 - Prob. 19.30PCh. 19 - Prob. 19.31PCh. 19 - Prob. 19.32PCh. 19 - Prob. 19.33PCh. 19 - Prob. 19.34PCh. 19 - Prob. 19.35PCh. 19 - Prob. 19.36PCh. 19 - Prob. 19.37PCh. 19 - Prob. 19.38PCh. 19 - Prob. 19.39PCh. 19 - Prob. 19.40APCh. 19 - Prob. 19.41APCh. 19 - Prob. 19.42APCh. 19 - Prob. 19.43APCh. 19 - Prob. 19.44APCh. 19 - Prob. 19.45APCh. 19 - Prob. 19.46APCh. 19 - Prob. 19.47APCh. 19 - Prob. 19.48APCh. 19 - Prob. 19.49APCh. 19 - Prob. 19.50APCh. 19 - Prob. 19.51APCh. 19 - Prob. 19.53APCh. 19 - Prob. 19.54APCh. 19 - Prob. 19.55APCh. 19 - Prob. 19.56APCh. 19 - Prob. 19.57APCh. 19 - Prob. 19.58APCh. 19 - Prob. 19.59APCh. 19 - Prob. 19.60APCh. 19 - Prob. 19.61APCh. 19 - Prob. 19.62APCh. 19 - Prob. 19.63APCh. 19 - Prob. 19.64APCh. 19 - Prob. 19.65APCh. 19 - Prob. 19.66APCh. 19 - Prob. 19.67APCh. 19 - Prob. 19.68APCh. 19 - Prob. 19.69APCh. 19 - Prob. 19.70APCh. 19 - Prob. 19.71APCh. 19 - Prob. 19.72AP
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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.arrow_forwardAssign as many resonances as you can to specific carbon atoms in the 13C NMR spectrum of ethyl benzoate.arrow_forwardThe 1H-NMR spectrum of compound B,C7H14O , consists of the following signals: δ 0.9 (t, 6H), 1.6 (sextet, 4H), and 2.4 (t, 4H). Draw the structural formula of compound B.arrow_forward
- Shown below are carbon NMR spectra for various isomers of C5H11Br. Assign structures for the compound that would produce each spectrum.arrow_forwardCompound 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.arrow_forwardCompound 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.arrow_forward
- The following NMR spectra were obtained from a compound with the molecular formula C4H6O. Use this information to predict its structure.arrow_forwardThe H1H1 NMR spectrum shown corresponds to an unknown compound with the molecular formula C6H10C6H10. There are no strong IR bands between 2100 and 2300 or 3250 and 3350 cm−1. Deduce and draw the structure of the molecule that corresponds to the spectrum.arrow_forward1Compound 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.arrow_forward
- 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.arrow_forwardCompound 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.arrow_forwardThe '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.arrow_forward
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