5. Propose a structure for a compound, with molecular formula C8H19N, that fits the following spectroscopic data: IR: 3400 cm-¹ (single peak) ¹H NMR: 0.98 (triplet, 6H), 1.1 8 (singlet, 1H, exchanges with D₂O), 1.38 (sextet, 4H), 1.4 8 (quintet, 4H), 2.68 (triplet, 4H) 13C NMR: 13.7, 20.5, 30.4, 49.2 Explain your reasoning by assigning the peaks.

5. Propose a structure for a compound, with molecular formula C8H19N, that fits the following spectroscopic data: IR: 3400 cm-¹ (single peak) ¹H NMR: 0.98 (triplet, 6H), 1.1 8 (singlet, 1H, exchanges with D₂O), 1.38 (sextet, 4H), 1.4 8 (quintet, 4H), 2.68 (triplet, 4H) 13C NMR: 13.7, 20.5, 30.4, 49.2 Explain your reasoning by assigning the peaks.

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

Chapter21: Benzene And The Concept Of Aromaticity

Section: Chapter Questions

Problem 21.26P: Compound I (C11H14O2) is insoluble in water, aqueous acid, and aqueous NaHCO3, but dissolves readily...

See similar textbooks

Similar questions

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.
Propose a structure for the compound that fits the following description. C10H14 7.18 δ, (4H, broad singlet) 2.70 δ (4H, quartet J = 7 Hz) 1.20 δ (6H. triplet J = 7 Hz) IR: 745 cm-1 7(a) IR absorption indicate the compound is .........-disubstituted = 7(b) The name of the compound is =
Following are IR and 1H-NMR spectra of compound D. The mass spectrum of compound D shows a molecular ion peak at m/z 136, a base peak at m/z 107, and other prominent peaks at m/z 118 and 59. Q.) Propose structural formulas for ions in the mass spectrum at m/z 118, 107, and 59.
A solution of acetone [(CH3)2C=O] in ethanol (CH3CH2OH) in the presence of a trace of acid was allowed to stand for several days, and a new compound of molecular formula C7H16O2 was formed. The IR spectrum showed only one major peak in the functional group region around 3000 cm−1, and the 1H NMR spectrum is given here. What is the structure of the product?
Treatment of anisole (CH3OC6H5) with Cl2 and FeCl3 forms P, which has peaks in its mass spectrum at m/z = 142 (M), 144 (M + 2), 129, and 127. P has absorptions in its IR spectrum at 3096–2837 (several peaks), 1582, and 1494 cm-1. Propose possible structures for P.
Treatment of benzaldehyde (C6H5CHO) with Zn(Hg) in aqueous HCl forms a compound Z that has a molecular ion at 92 in its mass spectrum. Z shows absorptions at 3150–2950, 1605, and 1496 cm-1 in its IR spectrum. Give a possible structure for Z.
Propose a structure for the compound that fits the following description. C10H14 7.0 δ, (4H, broad singlet) 2.85 δ (1H, septet J = 8 Hz) 1.20 δ (6H, doublet J = 8 Hz) 2.28 δ (3H, singlet) IR: 825 cm-1 5(a) Degree of the unsaturation of this compound is= , 5(b) Two peaks at 2.85 δ and 1.20 δ indicate that the compound has .......... group = 4(c) A peak at 2.28 δ indicates that the compound has ............. group= 4(d) IR absorption indicates the compound is .........-disubstituted = 4(e) The name of the compound is =
Compounds B and C are isomers with molecular formula C5H9BrO2. The 1H NMR spectrum of compounds B and C are shown below. The IR spectrum corresponding to compound B showed strong absorption bands at 1739, 1225, and 1158 cm-1, while the spectrum corresponding to compound C have strong bands at 1735, 1237, and 1182 cm-1. 1.Based on the information provided, determine the structure of compounds B and C. 2.Assign all peaks in 1H NMR spectrum of compounds B and C.
compound with the molecular formula C7H9N exhibits IR bands at 3450 cm-1 (medium, doublet), and 855 cm-1 (strong) and shows the following major mass spectral signals (m/z): 106 (base); 107 (M+, about 70% of base), 91 (40% of base), and 77 (about 20% of base). Deduce a reasonable structure from this data.
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?
C9H10O2: IR absorption at 1718 cm–1. Propose a structure with data given.
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.