Organic Chemistry
6th Edition
ISBN: 9781936221349
Author: Marc Loudon, Jim Parise
Publisher: W. H. Freeman
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Chapter 20, Problem 20.5P
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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.
An unknown compound C3H2NCl shows moderately strong IR absorptions around 1650 cm-1 and 2200 cm-1. Its NMR spectrum consists of two doublets (J = 14 Hz) at δ 5.9 and δ 7.1. Propose a structure consistent with this data?
The 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.
Chapter 20 Solutions
Organic Chemistry
Ch. 20 - Prob. 20.1PCh. 20 - Prob. 20.2PCh. 20 - Prob. 20.3PCh. 20 - Prob. 20.4PCh. 20 - Prob. 20.5PCh. 20 - Prob. 20.6PCh. 20 - Prob. 20.7PCh. 20 - Prob. 20.8PCh. 20 - Prob. 20.9PCh. 20 - Prob. 20.10P
Ch. 20 - Prob. 20.11PCh. 20 - Prob. 20.12PCh. 20 - Prob. 20.13PCh. 20 - Prob. 20.14PCh. 20 - Prob. 20.15PCh. 20 - Prob. 20.16PCh. 20 - Prob. 20.17PCh. 20 - Prob. 20.18PCh. 20 - Prob. 20.19PCh. 20 - Prob. 20.20PCh. 20 - Prob. 20.21PCh. 20 - Prob. 20.22PCh. 20 - Prob. 20.23PCh. 20 - Prob. 20.24PCh. 20 - Prob. 20.25PCh. 20 - Prob. 20.26PCh. 20 - Prob. 20.27APCh. 20 - Prob. 20.28APCh. 20 - Prob. 20.29APCh. 20 - Prob. 20.30APCh. 20 - Prob. 20.31APCh. 20 - Prob. 20.32APCh. 20 - Prob. 20.33APCh. 20 - Prob. 20.34APCh. 20 - Prob. 20.35APCh. 20 - Prob. 20.36APCh. 20 - Prob. 20.37APCh. 20 - Prob. 20.38APCh. 20 - Prob. 20.39APCh. 20 - Prob. 20.40APCh. 20 - Prob. 20.41APCh. 20 - Prob. 20.42APCh. 20 - Prob. 20.43APCh. 20 - Prob. 20.44APCh. 20 - Prob. 20.45APCh. 20 - Prob. 20.46APCh. 20 - Prob. 20.47APCh. 20 - Prob. 20.48APCh. 20 - Prob. 20.49APCh. 20 - Prob. 20.50APCh. 20 - Prob. 20.51APCh. 20 - Prob. 20.52APCh. 20 - Prob. 20.53APCh. 20 - Prob. 20.54APCh. 20 - Prob. 20.55APCh. 20 - Prob. 20.56APCh. 20 - Prob. 20.57APCh. 20 - Prob. 20.58APCh. 20 - Prob. 20.59APCh. 20 - Prob. 20.60AP
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- The 1H-NMR spectrum of compound R, C6H14O, consists of two signals: d 1.1 (doublet) and d 3.6 (septet) in the ratio 6:1. Propose a structural formula for compound R consistent with this informationarrow_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_forwardThe compound has a molecular formula of C5H10O and the following 13C spectrum. A proton NMR was measured, but the data was lost except for the fact that it showed three distinct signals: a singlet, a doublet, and a septet, all of which appear between 0 and 3 ppm. Identify the compound, and draw its 1H NMR.arrow_forward
- 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_forwardidentify the compound with molecular formula C2H6O that gives this 1H NMR spectrum.arrow_forwardThe 1H-NMR spectrum of Compound C shows five signals – δ 2.38 (1H, dt), 2.72 (1H, dt), 5.34 (1H, t), 5.49 (2H, ddd), 6.27 (2H, dd) ppm. Its 13C-NMR spectrum has four signals – δ 26, 58, 127, 129 ppm. In the compound’s mass spectrum, the M+1 peak appears at m/z = 115. An M+2 peak, whose intensity is roughly one-third that of the M+1 peak, also appears. Suggest a structure for this compound.arrow_forward
- 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.arrow_forwardA molecule of the molecular formula C5H11Br gives rise to the NMR spectrum below. When reacted with NaOH and water, it forms a product which by NMR has 2 protons 1H at 5.4 ppm and 1H at 5.5 ppm each having a 3J coupling of 17 Hz. (other protons also present) The product also has an IR stretch at 1550 cm 1. Provide the structures of the starting material and product. NaOH H,0 C;H„Br 6H triplet 4H quintet 1H quintet 10 8 Ppmarrow_forwardPredict the ¹H NMR spectrum of diethoxymethane.arrow_forward
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