Concept explainers
Interpretation: The structures and appropriate spectra for each structure are to be proposed. The cleavage which is responsible for the base peak at m/z 44 in the mass spectrum of A and the prominent peak at m/z 58 in the mass spectrum of B is to be described.
Concept introduction: NMR spectroscopy is a technique used to determine a unique structure of the compounds. It identifies the carbon-hydrogen bonding of an organic compound. A hydrogen atom is called as a proton in the NMR spectroscopy. IR spectroscopy is used to identify the functional groups present in the given compounds.
To determine: The structures and appropriate spectra for each structure and the cleavage which is responsible for the base peak at m/z 44 in the mass spectrum of A and the prominent peak at m/z 58 in the mass spectrum of B.
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ORGANIC CHEMISTRY MASTERINGCHEM ACCESS
- 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_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 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 reasoningarrow_forward
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- Do not give handwriting solution.arrow_forwardCompound A has a molecular formula C5H11NO and a HNMR spectrum as such: 8.05 (s, 1H), 3.04 (s, 3H), 2.67 (t, 2H), 1.55 (m, 2H), 0.91 (t, 3H). The CNMR shows one peak above 160ppm. When hydrolyzed with H3O+, two compounds are formed B and C after aqueous workup. Draw the structures of A, B, and C.arrow_forwardPropose a structure for each of the following two isomers with formula C6H14 given their 1H-NMR spectra. Isomer A: δ = 0.84 (d, 12 H), 1.39 (septet, 2H) ppm Isomer B: δ = 0.84 (t, 3 H), 0.86 (t, 9H), 1.22 (q, 2H) ppmarrow_forward
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- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning