H,CO A в H. D Proton NMR: 7.1 (doublet, 2H), 6.9 (doublet, 2H), 4.0 (quartet, 2H), 3.5 (singlet, 2H), 2.2 (singlet, 3H), 1.3 (triplet, 3H) Carbon-13 NMR: 205, 157, 130, 126, 114, 65, 49, 30, 15 IR (below): Note that none of the choices has an O-H group, so the peak at 3500 must be for water impurity? LOD TRANSMETTANCEI

H,CO A в H. D Proton NMR: 7.1 (doublet, 2H), 6.9 (doublet, 2H), 4.0 (quartet, 2H), 3.5 (singlet, 2H), 2.2 (singlet, 3H), 1.3 (triplet, 3H) Carbon-13 NMR: 205, 157, 130, 126, 114, 65, 49, 30, 15 IR (below): Note that none of the choices has an O-H group, so the peak at 3500 must be for water impurity? LOD TRANSMETTANCEI

Organic Chemistry

9th Edition

ISBN:9781305080485

Author:John E. McMurry

Publisher:John E. McMurry

Chapter13: Structure Determination: Nuclear Magnetic Resonance Spectroscopy

Section13.SE: Something Extra

Problem 66GP: A 13C NMR spectrum of commercially available 2,4-pentanediol, shows five peaks at 23.3, 23.9, 46.5,...

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Compound K: molecular ion at 88; IR peak at 3600-3200 cm-1; 1H NMR data (ppm) at 0.9 (triplet, 3H), 1.2 (singlet, 6H), 1.5 (quartet, 2H), and 1.6 (singlet, 1H). Propose a structure.
Reaction of C6H5CH2CH2OH with CH3COCl affords compound W, whichhas molecular formula C10H12O2. W shows prominent IR absorptions at3088–2897, 1740, and 1606 cm−1. W exhibits the following signals in its1H NMR spectrum: 2.02 (singlet), 2.91 (triplet), 4.25 (triplet), and 7.20–7.35(multiplet) ppm. What is the structure of W?
Identify the structures of D and E, isomers of molecular formulaC6H12O2, from their IR and 1H NMR data. Signals at 1.35 and 1.60 ppm inthe 1H NMR spectrum of D and 1.90 ppm in the 1H NMR spectrum of Eare multiplets.
Treatment of isobutene [(CH3)2C = CH2] with (CH3)3CLi forms a carbanion that reacts with CH2=O to form H after water is added to the reaction mixture. H has a molecular ion in its mass spectrum at m/z = 86, and shows fragments at 71 and 68. H exhibits absorptions in its IR spectrum at 3600–3200 and 1651 cm−1, and has the 1H NMR spectrum given below. Whatis the structure of H?
(a) The 'H-NMR spectrum of cyclobutanone shows two signals - signal A at 3.00 ppm and signal B at 1.95 ppm. Give the multiplicity of each signal. cyclobutanone (b) When cyclobutanone is treated with D20 and NaOD, the only signal observable in the 1H-NMR is a singlet at 2.00 ppm. Explain why this is the case. [Note: Deuterium atoms do not display signals in the TH-NMR spectrum]
Treatment of isobutene [(CH3)2C=CH2] with (CH3)3CLi forms a carbanionthat reacts with CH2=O to form H after water is added to the reactionmixture. H has a molecular ion in its mass spectrum at m/z = 86, andshows fragments at 71 and 68. H exhibits absorptions in its IR spectrumat 3600−3200 and 1651 cm−1, and has the 1H NMR spectrum given below.What is the structure of H?
ClCH₂CO₂H , elucidate the following spectra bellow
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When 2-bromo-3,3-dimethylbutane is treated with K+ −OC(CH3)3, a singleproduct T having molecular formula C6H12 is formed. When 3,3-dimethylbutan-2-ol is treated with H2SO4, the major product U has thesame molecular formula. Given the following 1H NMR data, what are thestructures of T and U? Explain in detail the splitting patterns observedfor the three split signals in T.1H NMR of T: 1.01 (singlet, 9 H), 4.82 (doublet of doublets, 1 H, J = 10, 1.7 Hz), 4.93 (doublet of doublets, 1 H, J = 18, 1.7 Hz), and 5.83 (doublet ofdoublets, 1 H, J = 18, 10 Hz) ppm1H NMR of U: 1.60 (singlet) ppm Additional problems on the spectroscopy of alkenes are given in Chapters A–C:Mass spectrometry: A.16b, A.20, A.23Infrared spectroscopy: B.5, B.7(A), B.12c, B17a, B.18cNuclear magnetic resonance spectroscopy: C.12a; C.15d, e; C.29d; C.32d;C.37; C.38d, f; C.43i, j; C.44; C.45; C.49d, f; C.50b; C.51c; C.55
Because 31P has an odd mass number, 31P nuclei absorb in the NMR and, in many ways, these nuclei behave similarly to protons in NMR spectroscopy. With this in mind, explain why the 1H NMR spectrum of methyl dimethylphosphonate, CH3PO(OCH3)2, consists of two doublets at 1.5 and 3.7 ppm.
Phenacetin, a substance formerly used in over-the-counter headache remedies, has the formula C10H13NO2. Phenacetin is neutral and does not dissolve in either acid or base. When warmed with aqueous NaOH, phenacetin yields an amine, C8H11NO, whose 1H NMR spectrum is shown. When heated with HI, the amine is cleaved to an aminophenol, C6H7NO. What is the structure of phenacetin, and what are the structures of the amine and the aminophenol?
3-Chlorocyclopropene, on treatment with AgBF4, gives a precipitate of AgCl and a stable solution of a product that shows a single 1H NMR absorption at 11.04 Î´. What is a likely structure for the products, and what is its relation to HĂ¼ckelâ€™s rule?