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Propose a structure consistent with each set of data.
a. Compound A:
b. Compound B:
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Chapter 15 Solutions
ORGANIC CHEMISTRY
- Compound A, MW 86, shows an IR absorption at 1730 cm-1 and a very simple 1H NMR spectrum with peaks at 9.7 (1 H, singlet) and 1.2 (9 H, singlet). Propose a structure for A.arrow_forwardCompound 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_forwardPropose a structure of compound C (molecular formula C10H12O) consistent with the following data. C is partly responsible for the odor and flavor of raspberries. Compound C: IR absorption at 1717 cm-1arrow_forward
- An unknown compound has a molecular formula of C,H,O. Its IR spectrum shows prominent absorptions at 2980, 2960, and 1718 cm . It exhibits the following signals in its H NMR spectrum (ppm): 1.06 (triplet, 3H), 2.12 (singlet, 3H), 2.45 (quartet, 2H); and the following signals in its ¹3C NMR spectrum ( ppm): 7.6, 29.5, 36.8, 208.8. Draw the structure of the unknown compound. Click and drag to start drawing a structure. 0 X 0:0arrow_forwardTreatment 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) ppmarrow_forwardAn unknown compound has a molecular formula of C4H6O2. Its IR spectrum shows absorptions at 3095, 1762, 1254, and 1118 cm -1. It exhibits the following signals in its 1H NMR spectrum (ppm): 2.12 (singlet,3H), 4.55 (doublets of doublets, 1H), 4.85 (doublet of doublets, 1H), 7.25 (doublets of doublets, 1H); and the following signals in its 13C NMR spectrum (ppm): 20.8, 100.4, 141.2, 168.0. Draw the structure of the unknown compoundarrow_forward
- A compound of molecular formula C4H8O2 shows no IR peaks at 3600–3200 or 1700 cm-1. It exhibits one singlet in its 1H NMR spectrum at 3.69 ppm, and one line in its 13C NMR spectrum at 67 ppm. What is the structure of this unknown?arrow_forwardWhen compound A (C5H12O) is treated with HBr, it forms compound B (C5H11Br). The 1H NMR spectrum of compound A has a 1H singlet, a 3Hdoublet, a 6H doublet, and two 1H multiplets. The 1H NMR spectrum of compound B has a 6H singlet, a 3H triplet, and a 2H quartet. Identifycompounds A and B.arrow_forwardIdentify products A and B from the given 1H NMR data. Treatment of CH2=CHCOCH3 with one equivalent of HCl forms compound A. A exhibits the following absorptions in its 1H NMR spectrum: 2.2 (singlet, 3H), 3.05 (triplet, 2 H), and 3.6 (triplet, 2 H) ppm. What is the structure of A?arrow_forward
- 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?arrow_forwardIdentify 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_forwardThymol (molecular formula C10H14O) is the major component of the oil ofthyme. Thymol shows IR absorptions at 3500–3200, 3150–2850, 1621, and1585 cm−1. The 1H NMR spectrum of thymol is given below. Propose apossible structure for thymol.arrow_forward
Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning