Organic Chemistry
6th Edition
ISBN: 9781936221349
Author: Marc Loudon, Jim Parise
Publisher: W. H. Freeman
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Chapter 17, Problem 17.53AP
Interpretation Introduction
Interpretation:
The compounds,
Concept introduction:
Diels Alder reaction is the
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Compound A, C8H10O, has the IR and 1H NMR spectra shown. Propose a structure consistent with the observed spectra, and label each peak in the NMR spectrum. Note that the absorption at 5.5 î disappears when D2O is added.
Treatment of compound E (molecular formula C4H8O2) with excess CH3CH2MgBr yields compound F (molecular formula C6H14O) after protonation with H2O. E shows a strong absorption in its IR spectrum at 1743 cm-1. F shows a strong IR absorption at 3600–3200 cm-1. The 1H NMR spectral data of E and F are given. What are the structures of E and F?Compound E signals at 1.2 (triplet, 3 H), 2.0 (singlet, 3 H), and 4.1 (quartet, 2 H) ppmCompound F signals at 0.9 (triplet, 6 H), 1.1 (singlet, 3 H), 1.5 (quartet, 4 H), and 1.55 (singlet, 1 H) ppm
Identify products A and B from the given 1H NMR data.
Treatment of acetone [(CH3)2C=O] with dilute aqueous base forms B. Compound B exhibits four singlets in its 1H NMR spectrum at 1.3 (6 H), 2.2 (3 H), 2.5 (2 H), and 3.8 (1H) ppm. What is the structure of B?
Chapter 17 Solutions
Organic Chemistry
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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- Compounds A and B are isomers having the molecular formula C4H8O3. Identify A and B on the basis of their 1H NMR spectra.Compound A: δ 1.3 (3H, triplet); 3.6 (2H, quartet); 4.1 (2H, singlet); 11.1 (1H, broad singlet)Compound B: δ 2.6 (2H, triplet); 3.4 (3H, singlet); 3.7 (2H triplet); 11.3 (1H, broad singlet)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_forwardTreatment of compound C (molecular formula C4H8O) with C6H5MgBr, followed by H2O, affords compound D (molecular formula C10H14O). Compound D has a strong peak in its IR spectrum at 3600–3200 cm−1. The 1H NMR spectral data of C and D are given. What are the structures of C and D? Compound C signals at 1.3 (singlet, 6 H) and 2.4 (singlet, 2 H) ppm Compound D signals at 1.2 (singlet, 6 H), 1.6 (singlet, 1 H), 2.7 (singlet, 2 H), and 7.2 (multiplet, 5 H) ppmarrow_forward
- Compound A is treated with a mixture of nitric and sulfuric acids to generate Compound B. The 1H-NMR spectrum of B shows two singlets, one at 2.52 pm and one at 8.13 ppm. The 13C-NMR spectrum of B shows five signals. The mass spectrum of B shows a peak at m/z = 260 and another peak at m/z = 262; the relative height of the two peaks is 1:1 respectively. - Identify compound B, explaining your reasoningarrow_forwardPropose a structure for compound X (molecular formula C6H12O2), which gives a strong peak in its IR spectrum at 1740 cm−1. The 1H NMR spectrum of X shows only two singlets, including one at 3.5 ppm. The 13C NMR spectrum is given below. Propose a structure for X.arrow_forwardCompound A has molecular formula C5H10O. It shows three signals in the 1H-NMR spectrum - a doublet of integral 6 at 1.1 ppm, a singlet of integral 3 at 2.14 ppm, and a quintet of integral 1 at 2.58 ppm. Suggest a structure for A and explain your reasoning.arrow_forward
- There are several isomeric alcohols and ethers of molecular formula C5H12O. Two of these exhibit the following 1H-NMR spectra. Propose a structure for each of the isomers. Isomer A: δ = 0.92 (t, 7.8 Hz, 3 H), 1.20 (s, 6H), 1.49 (q, 7.8 Hz, 2H), 1.85 (s, 1H) ppm Isomer B: δ = 1.19 (s, 9 H), 3.21 (s, 3H) ppmarrow_forwardDeduce the structures of compounds A and B, two of the major components of jasmine oil, from the given data. Compound A: C9H10O2; IR absorptions at 3091–2895 and 1743 cm-1; 1H NMR signals at 2.06 (singlet, 3 H), 5.08 (singlet, 2 H), and 7.33 (broad singlet, 5 H) ppm. Compound B: C14H12O2; IR absorptions at 3091–2953 and 1718 cm-1; 1H NMR signals at 5.35 (singlet, 2 H) and 7.26–8.15 (multiplets, 10 H) ppm.arrow_forwardTreatment of benzoic acid (C6H5CO2H) with NaOH followed by 1-iodo-3methylbutane forms H. H has a molecular ion at 192 and IR absorptions at 3064, 3035, 2960−2872, and 1721 cm−1. Propose a structure for H.arrow_forward
- Identify the structure of compound C (molecular formula C11H15NO2), which has an IR absorption at 1699 cm−1 and the 1H NMR spectrum shown below.arrow_forwardThe IR spectrum of compound A with a molecular formula of C5H12O is shown below. Compound A is oxidized to give compound B, a ketone with a molecular formula of C5H10O. When compound A is heated with H2SO4, compounds C and D are obtained. Considerably more D is obtained than C.Reaction of compound C with O3, followed by treatment with dimethyl sulfide, gives two products: formaldehyde and compound E, with a molecular formula of C4H8O. Reaction of compound D with O3, followed by treatment with dimethyl sulfide, gives two products: compound F, with a molecular formula of C3H6O, and compound G, with a molecular formula of C2H4O. What are the structures of compounds A through G?arrow_forwardCompound A has molecular formula C5H8Br4 but shows only one singlet in the 1H-NMR spectrum. Suggest a structure for A and explain your reasoning.arrow_forward
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