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Organic Chemistry Study Guide and Solutions
6th Edition
ISBN: 9781936221868
Author: Marc Loudon, Jim Parise
Publisher: W. H. Freeman
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Chapter 17, Problem 17.36AP
Interpretation Introduction
Interpretation:
The structural basis for the given observations is to be stated.
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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Indicate two basic differences that exist between the spectra of 1H y 13C in NMR.
The 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.
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Chapter 17 Solutions
Organic Chemistry Study Guide and Solutions
Ch. 17 - Prob. 17.1PCh. 17 - Prob. 17.2PCh. 17 - Prob. 17.3PCh. 17 - Prob. 17.4PCh. 17 - Prob. 17.5PCh. 17 - Prob. 17.6PCh. 17 - Prob. 17.7PCh. 17 - Prob. 17.8PCh. 17 - Prob. 17.9PCh. 17 - Prob. 17.10P
Ch. 17 - Prob. 17.11PCh. 17 - Prob. 17.12PCh. 17 - Prob. 17.13PCh. 17 - Prob. 17.14PCh. 17 - Prob. 17.15PCh. 17 - Prob. 17.16PCh. 17 - Prob. 17.17PCh. 17 - Prob. 17.18PCh. 17 - Prob. 17.19PCh. 17 - Prob. 17.20PCh. 17 - Prob. 17.21PCh. 17 - Prob. 17.22APCh. 17 - Prob. 17.23APCh. 17 - Prob. 17.24APCh. 17 - Prob. 17.25APCh. 17 - Prob. 17.26APCh. 17 - Prob. 17.27APCh. 17 - Prob. 17.28APCh. 17 - Prob. 17.29APCh. 17 - Prob. 17.30APCh. 17 - Prob. 17.31APCh. 17 - Prob. 17.32APCh. 17 - Prob. 17.33APCh. 17 - Prob. 17.35APCh. 17 - Prob. 17.36APCh. 17 - Prob. 17.37APCh. 17 - Prob. 17.38APCh. 17 - Prob. 17.39APCh. 17 - Prob. 17.40APCh. 17 - Prob. 17.41APCh. 17 - Prob. 17.42APCh. 17 - Prob. 17.43APCh. 17 - Prob. 17.44APCh. 17 - Prob. 17.45APCh. 17 - Prob. 17.46APCh. 17 - Prob. 17.47APCh. 17 - Prob. 17.48APCh. 17 - Prob. 17.49APCh. 17 - Prob. 17.50APCh. 17 - Prob. 17.51APCh. 17 - Prob. 17.52APCh. 17 - Prob. 17.53APCh. 17 - Prob. 17.54APCh. 17 - Prob. 17.55APCh. 17 - Prob. 17.56APCh. 17 - Prob. 17.57APCh. 17 - Prob. 17.58APCh. 17 - Prob. 17.59AP
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- Treatment of alcohol A (molecular formula C5H12O) with CrO3, H2SO4, and H2O affords B with molecular formula C5H10O, which gives an IR absorption at 1718 cm−1. The 1H NMR spectrum of B contains the following signals: 1.10 (doublet, 6 H), 2.14 (singlet, 3 H), and 2.58 (septet, 1 H) ppm. What are the structures of A and B?arrow_forwardPredict the structure of a compound based on this 13C NMR spectra. The chemical formula of this compound is C4H6O2. Briefly explain your answer.arrow_forwardYou have an unknown with an absorption at 1680 cm-1; it might be an amide, an isolated double bond, a conjugated ketone, aconjugated aldehyde, or a conjugated carboxylic acid. Describe what spectral characteristics you would look for to help youdetermine which of these possible functional groups might be causing the 1680 peak.arrow_forward
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- Sketch the 1H NMR spectra of the following compounds.arrow_forwardThe 1H NMR spectra of two carboxylic acids with molecular formula C3H5O2Cl are shown below. Identify the carboxylic acids. (The “offset” notation means that the farthest-left signal has been moved to the rightby the indicated amount to fit on the spectrum; thus, the signal at 9.8 ppm offset by 2.4 ppm has an actual chemical shift of 9.8 + 2.4 = 12.2 ppm.)arrow_forwardCompound X of the molecular formula C7H10 has the 13C NMR spectrum (5 signals) shown below. On treatment with excess H2/Pt (catalytic hydrogenation), X is converted to methylcyclohexane. Propose a structure for X and justify your reasoning by clearly labeling each carbon signal and write out the reaction. 200 180 160 140 120 100 80 60 40 20arrow_forward
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