EBK ORGANIC CHEMISTRY
6th Edition
ISBN: 8220103151757
Author: LOUDON
Publisher: MAC HIGHER
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Chapter 16, Problem 16.51AP
Interpretation Introduction
Interpretation:
The reason for the fact that the value
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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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?
Chapter 16 Solutions
EBK ORGANIC CHEMISTRY
Ch. 16 - Prob. 16.1PCh. 16 - Prob. 16.2PCh. 16 - Prob. 16.3PCh. 16 - Prob. 16.4PCh. 16 - Prob. 16.5PCh. 16 - Prob. 16.6PCh. 16 - Prob. 16.7PCh. 16 - Prob. 16.8PCh. 16 - Prob. 16.9PCh. 16 - Prob. 16.10P
Ch. 16 - Prob. 16.11PCh. 16 - Prob. 16.12PCh. 16 - Prob. 16.13PCh. 16 - Prob. 16.14PCh. 16 - Prob. 16.15PCh. 16 - Prob. 16.16PCh. 16 - Prob. 16.17PCh. 16 - Prob. 16.18PCh. 16 - Prob. 16.19PCh. 16 - Prob. 16.20PCh. 16 - Prob. 16.21PCh. 16 - Prob. 16.22PCh. 16 - Prob. 16.23PCh. 16 - Prob. 16.24PCh. 16 - Prob. 16.25PCh. 16 - Prob. 16.26PCh. 16 - Prob. 16.27PCh. 16 - Prob. 16.28PCh. 16 - Prob. 16.29PCh. 16 - Prob. 16.30PCh. 16 - Prob. 16.31PCh. 16 - Prob. 16.32PCh. 16 - Prob. 16.33PCh. 16 - Prob. 16.34PCh. 16 - Prob. 16.35APCh. 16 - Prob. 16.36APCh. 16 - Prob. 16.37APCh. 16 - Prob. 16.38APCh. 16 - Prob. 16.39APCh. 16 - Prob. 16.40APCh. 16 - Prob. 16.41APCh. 16 - Prob. 16.42APCh. 16 - Prob. 16.43APCh. 16 - Prob. 16.44APCh. 16 - Prob. 16.45APCh. 16 - Prob. 16.46APCh. 16 - Prob. 16.47APCh. 16 - Prob. 16.48APCh. 16 - Prob. 16.49APCh. 16 - Prob. 16.50APCh. 16 - Prob. 16.51APCh. 16 - Prob. 16.52APCh. 16 - Prob. 16.53APCh. 16 - Prob. 16.54APCh. 16 - Prob. 16.55APCh. 16 - Prob. 16.56APCh. 16 - Prob. 16.57APCh. 16 - Prob. 16.58APCh. 16 - Prob. 16.59APCh. 16 - Prob. 16.60APCh. 16 - Prob. 16.61APCh. 16 - Prob. 16.62APCh. 16 - Prob. 16.63APCh. 16 - Prob. 16.64APCh. 16 - Prob. 16.65APCh. 16 - Prob. 16.66APCh. 16 - Prob. 16.67APCh. 16 - Prob. 16.68AP
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- 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.arrow_forwardIdentify the C-H out-of-plane bending vibrations in the infrared spectrum of 4- methylcyclohexene. What structural information can be obtained from these bands?arrow_forwardA hydrocarbon, compound B, has molecular formula C6H6, and gave an NMR spectrum with two signals: delta 6.55 pm and delta 3.84 pm with peak ratio of 2:1. When warmed in pyridine for three hr, compound B quantitatively converts to benzene. Mild hydrogenation of B yielded another compound C with mass spectrum of m/z 82. Infrared spectrum showed no double bonds; NMR spectrum showed one broad peak at delta 2.34 ppm. With this information, address the following questions. a) How many rings are in compound C? b) How many rings are probably in B? How many double bonds are in B? c) Can you suggest a structure for compounds B and C? d) In the NMR spectrum of B, the up-field signal was a quintet, and the down field signal was a triplet. How must you account for these splitting patterns?arrow_forward
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