Organic Chemistry (6th Edition)
6th Edition
ISBN: 9781260119107
Author: Janice Gorzynski Smith
Publisher: McGraw Hill Education
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Chapter 15, Problem 54P
Thymol (molecular formula
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Thymol (molecular formula C10H14O) is the major component of the oil of thyme. Thymol shows IR absorptions at 3500–3200, 3150–2850, 1621, and 1585 cm−1. The 1H NMR spectrum of thymol is given below. Propose a possible structure for thymol.
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.
You are provided an unknown sample with the molecular formula C9H10O. After running the NMR, you obtain the following 1H and 13C spectra. Use this information to infer the identity of your unknown. Draw the structure and name the compound
Chapter 15 Solutions
Organic Chemistry (6th Edition)
Ch. 15.2 - Prob. 1PCh. 15.2 - Problem 17.2 What orbitals are used to form the...Ch. 15.3 - Prob. 4PCh. 15.3 - Problem-17.5 What is the structure of propofol,...Ch. 15.6 - Prob. 7PCh. 15.8 - Prob. 8PCh. 15.8 - Prob. 11PCh. 15.8 - Prob. 12PCh. 15.8 - Problem 17.16 Rank the following compounds in...Ch. 15.8 - Problem 17.17 Draw the seven resonance structures...
Ch. 15 - 17.23 Name each compound and state how many lines...Ch. 15 - Prob. 21PCh. 15 - Prob. 22PCh. 15 - 17.27 Give the IUPAC name for each compounds.
a....Ch. 15 - 17.28 Draw a structure corresponding to each...Ch. 15 - 17.29 a. Draw the 14 constitutional isomers of...Ch. 15 - Prob. 26PCh. 15 - Prob. 27PCh. 15 - 17.38
How many electrons does C contain?
How...Ch. 15 - Prob. 36PCh. 15 - 17.40 Explain the observed rate of reactivity of...Ch. 15 - 17.41 Draw a stepwise mechanism for the following...Ch. 15 - Prob. 39PCh. 15 - 17.43 Draw additional resonance structures for...Ch. 15 - Prob. 41PCh. 15 - Prob. 42PCh. 15 - 17.46 Which compound in each pair is the stronger...Ch. 15 - 17.47 Treatment of indene with forms its...Ch. 15 - Prob. 45PCh. 15 - 17.49 Draw the conjugate bases of pyrrole and...Ch. 15 - 17.50 a. Explain why protonation of pyrrole occurs...Ch. 15 - Prob. 48PCh. 15 - Prob. 49PCh. 15 - 17.53 How many signals does each compound...Ch. 15 - 17.54 Which of the diethylbenzene isomers (ortho,...Ch. 15 - 17.55 Propose a structure consistent with each...Ch. 15 - 17.56 Propose a structure consistent with each...Ch. 15 - 17.57 Thymol (molecular formula ) is the major...Ch. 15 - 17.58 You have a sample of a compound of molecular...Ch. 15 - 17.59 Explain why tetrahydrofuran has a higher...Ch. 15 - 17.60 Rizatriptan (trade name Maxalt) is a...Ch. 15 - 17.61 Zolpidem (trade name Ambien) promotes the...Ch. 15 - 17.62 Answer the following questions about...Ch. 15 - 17.63 Stanozolol is an anabolic steroid that...Ch. 15 - Prob. 61P
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- 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_forwardThe H1H1 NMR spectra corresponds to an alcohol with the molecular formula C5H12O. Deduce the structure from the spectraarrow_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
- Compound 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_forwardThe '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_forward
- Draw 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_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_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_forward
- The 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_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_forwardCompound 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_forward
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