Compound 67 is a high-boiling liquid (boiling point 180° C) that reacts with I, in aqueous base to give a yellow precipitate of CHL, (the jodoform test). The compound is readily prepared by the condensation of two moles of ethyl acetate in the presence of ethanol/ ethoxide anion. The Mass, IR, and 'H NMR spectra, along with 13C NMR data, are given below. Elemental Analysis: C, 55.37; H, 7.74; O, 36.88. Mass Spectrum 143 88 130 115 60 70 100 110 120 130 140 150 160 170 180 190 200 30 40 50 80 90 m/e Infrared Spectrum Wave Number, cm -1 1500 1300 1200 1100 1000 800 700 4000 3000 2500 2000 006 Absorbance Intensity

Compound 67 is a high-boiling liquid (boiling point 180° C) that reacts with I, in aqueous base to give a yellow precipitate of CHL, (the jodoform test). The compound is readily prepared by the condensation of two moles of ethyl acetate in the presence of ethanol/ ethoxide anion. The Mass, IR, and 'H NMR spectra, along with 13C NMR data, are given below. Elemental Analysis: C, 55.37; H, 7.74; O, 36.88. Mass Spectrum 143 88 130 115 60 70 100 110 120 130 140 150 160 170 180 190 200 30 40 50 80 90 m/e Infrared Spectrum Wave Number, cm -1 1500 1300 1200 1100 1000 800 700 4000 3000 2500 2000 006 Absorbance Intensity

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

Is it possible to distinguish between the following compounds using the spectroscopic techniques covered in CHE331? If so, discuss how. If not, discuss why.Benzonitrile vs Benzaldehyde(a) UV spectroscopy(b) IR spectroscopy(c) Mass spectrometry(d) 13C NMR spectroscopy(e) 1H NMR spectroscopy
In the presence of a small amount of acid, a solution of acetaldehyde (CH3CHO) in methanol (CH3OH) was allowed to stand and a new compound L was formed. L has a molecular ion in its mass spectrum at 90 and IR absorptions at 2992 and 2941 cm−1. L shows three signals in its 13C NMR at 19, 52, and 101 ppm. The 1H NMR spectrum of L is given below. What is the structure of L?
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.
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?
Benzonitrile (C6H5CN) is reduced to two different products depending on the reducing agent used. Treatment with lithium aluminum hydride followed by water forms K, which has a molecular ion in its mass spectrum at 107 and the following IR absorptions: 3373, 3290, 3062, 2920, and 1600 cm-1. Treatment with a milder reducing agent forms L, which has amolecular ion in its mass spectrum at 106 and the following IR absorptions: 3086, 2820, 2736, 1703, and 1600 cm-1. L shows fragments in its mass spectrum at m/z = 105 and 77. Propose structures for K and L and explain how you arrived at your conclusions.
Compound 1, was found to consist of C,H,N, and Cl with an elemental analysis of 61.1% C, 2.9% H, and 10.2% N. IR Spectra displayed absorbances at 2215, 1605. Deduce the structure of this compound. (there is only one Cl)
Ethers are not easily differentiated by their infrared spectra, but they tend to form predictable fragments in the mass spectrum. The following compounds give similar butdistinctive mass spectra.O Obutyl propyl ether butyl isopropyl etherBoth compounds give prominent peaks at m>z 116, 73, 57, and 43. But one compoundgives a distinctive strong peak at 87, and the other compound gives a strong peak at 101.Determine which compound gives the peak at 87 and which one gives the peak at 101.Propose fragmentations to account for the ions at m>z 116, 101, 87, and 73.
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.
Reaction of butanenitrile (CH3CH2CH2CN) with methylmagnesium bromide (CH3MgBr), followed by treatment with aqueous acid, forms compound G. G has a molecular ion in its mass spectrum at m/z = 86 and a base peak at m/z = 43. G exhibits a strong absorption in its IR spectrum at 1721 cm−1 and has the 1H NMR spectrum given below. What is the structure of G?
Reaction of pentanoyl chloride (CH3CH2CH2CH2COCl) with lithium dimethyl cuprate [LiCu(CH3)2] forms a compound J that has a molecular ion in its mass spectrum at 100, as well as fragments at m/z = 85, 57, and 43 (base). The IR spectrum of J has strong peaks at 2962 and 1718 cm-1. Propose a structure for J.
Provide a detailed analysis of IR, Mass, 1H NMR and 13C NMR data for the following compound.
Shown are the H NMR spectra for 2 isomeric compounds of the formula C5H10O. The IR spectrum of both have an absorption in the region of 1700 to 1730 cm-1. Provide the structure for each compound, and which hydrogen atoms give rise to the peaks in each spectrum. The peak at 7.27 ppm can be ignored, and the red numbers are integration values.