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?

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?

Organic Chemistry

8th Edition

ISBN:9781305580350

Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote

Publisher:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote

Chapter13: Nuclear Magnetic Resonance Spectroscopy

Section: Chapter Questions

Problem 13.17P: Following are 1H-NMR spectra for compounds G, H, and I, each with the molecular formula C5H12O. Each...

See similar textbooks

Similar questions

Following are 1H-NMR spectra for compounds G, H, and I, each with the molecular formula C5H12O. Each is a liquid at room temperature, is slightly soluble in water, and reacts with sodium metal with the evolution of a gas. (a) Propose structural formulas of compounds G, H, and I. (b) Explain why there are four lines between 0.86 and 0.90 for compound G. (c) Explain why the 2H multiplets at 1.5 and 3.5 for compound H are so complex.
Ketones undergo a reduction when treated with sodium borohydride, NaBH4. What is the structure of the compound produced by reaction of 2-butanone with NaBH4 if it has an IR absorption at 3400 cm-1 and M+=74 in the mass spectrum?
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.
What is the structure of an unknown compound with molecular formula C6H15N that gives the following 1H NMR absorptions: 0.9 (singlet, 1 H), 1.10 (triplet, 3 H), 1.15 (singlet, 9 H), and 2.6 (quartet, 2 H) ppm?
A 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?
What is the structure of a compound of molecular formula C1H1402 that shows a strong IR absorption at 3150-2850 cm1 and gives the following 'H NMR absorptions: 1.4 (triplet, 6 H), 4.0 (quartet, 4 H), and 6.8 (singlet, 4 H) ppm?
Compound B has molecular formula C9H12. It shows five signals in the 1H-NMR spectrum - a doublet of integral 6 at 1.22 ppm, a septet of integral 1 at 2.86 ppm, a singlet of integral 1 at 5.34 ppm, a doublet of integral 2 at 6.70 ppm, and a doublet of integral 2 at 7.03 ppm. The 13C-NMR spectrum of B shows six unique signals (23.9, 34.0, 115.7, 128.7, 148.9, and 157.4). Identify B and explain your reasoning.
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.
Compound A has molecular formula C7H7X. Its 1H-NMR spectrum shows a singlet at 2.26 ppm and two doublets, one at 6.95 ppm and one at 7.28 ppm. The singlet has an integral of three and the doublets each have an integral of two. Its 13C-NMR shows five signals. The mass spectrum of A shows a peak at m/z = 170 and another peak at m/z = 172; the relative height of the two peaks is 1:1 respectively. - Identify what atom X is, explaining your reasoning - Identify Compound A, explaining your reasoning 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 reasoning Compound B is treated with sodium ethoxide to generate compound C. The 1H-NMR spectrum of C shows…
Compound A has molecular formula C7H7X. Its 1H-NMR spectrum shows a singlet at 2.25 ppm and two doublets, one at 7.28 ppm and one at 7.39 ppm. The singlet has an integral of three and the doublets each have an integral of two. The mass spectrum of A shows a peak at m/z = 126 and another peak at m/z = 128; the relative height of the two peaks is 3:1 respectively. - Identify what atom X is, explaining your reasoning - Identify Compound A, explaining your reasoning
The H1H1 NMR spectra corresponds to an alcohol with the molecular formula C5H12O. Deduce the structure from the spectra
Compound X (molecular formula C10H120) was treated with NH2NH2, ¯OH to yield compound Y (molecular formula C10H14). Match the 1H NMR spectra of X and Y to the corresponding structures of X and Y. Compound NH2NH2 Compound 'H NMR of X 6 H OH Y 1 H 5H 8. 6. 4 ppm or H NMR of Y 6 H 2H 5H 1 H multiplet multiplet 8. 6. 4. 3. 1 nnm 2. 2. 3, O: