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Thymol (molecular formula
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Organic Chemistry
- Compound I (C11H14O2) is insoluble in water, aqueous acid, and aqueous NaHCO3, but dissolves readily in 10% Na2CO3 and 10% NaOH. When these alkaline solutions are acidified with 10% HCl, compound I is recovered unchanged. Given this information and its 1H-NMR spectrum, deduce the structure of compound I.arrow_forwardCompound B of molecular formula C9H19N shows a noteworthy infrared absorption at 3300 cm-1. Its 1H-NMR spectrum shows three singlets – δ 1.0 (6H), 1.1 (12H), 1.4 (1H) ppm. Its 13C-NMR spectrum has four signals – δ 25, 28, 41, 64 ppm. Suggest a structure for this compound. Please show work.arrow_forwardA ¹H NMR spectrum is shown for a molecule with the molecular formula of C10H12O4. Draw the structure that best fits this data. 1H 11 10 2H 2H U 7 1H 1H ΤΗ 1Η ppm Qarrow_forward
- 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_forwardPredict the ¹H NMR spectrum of diethoxymethane.arrow_forwardDraw the structure of molecular formula C8H10O that produced the 1H NMR spectra shown below. The IR spectrum does not show a broad absorbance at 3300 cm–1 or a strong absorbance at 1710 cm–1.arrow_forward
- The 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_forwardCompound B of molecular formula C9H19N shows a noteworthy infrared absorption at 3300 cm-1. Its 1H-NMR spectrum shows three singlets – δ 1.0 (6H), 1.1 (12H), 1.4 (1H) ppm. Its 13C-NMR spectrum has four signals – δ 25, 28, 41, 64 ppm. Suggest a structure for this compound.arrow_forwardThe 1H-NMR spectrum of Compound D of molecular formula C10H12O shows three singlets – δ 2.20 (6H, s), 4.86 (4H), 7.10 (2H) ppm. Its 13C-NMR spectrum has five signals – 20, 74, 127, 135, 146 ppm. Suggest a structure for this compound.arrow_forward
- Compound 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_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_forwardA ¹H NMR spectrum is shown for a molecule with the molecular formula of C9H1002. Draw the structure that best fits this data. 11 10 1H 9 2H 8 2H 7 6 2H 5 4 3 2 3H ppm Qarrow_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning