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Propose structures consistent with each set of data: (a) a hydrocarbon with a molecular ion at
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- Indicate two basic differences that exist between the spectra of 1H y 13C in NMR.arrow_forwardTreatment of 2-methylpropanenitrile [(CH3)2CHCN] withCH3CH2CH2MgBr, followed by aqueous acid, affords compound V, whichhas molecular formula C7H14O. V has a strong absorption in its IRspectrum at 1713 cm−1, and gives the following 1H NMR data: 0.91(triplet, 3 H), 1.09 (doublet, 6 H), 1.6 (multiplet, 2 H), 2.43 (triplet, 2 H), and2.60 (septet, 1 H) ppm. What is the structure of V?arrow_forwardDeduce the structure of compound C.arrow_forward
- A compound (L) with the molecular formula C9H10 reacts with bromine and gives an IR absorption spectrum that includes the following absorption peaks: 3035 cm ¹(m), 3020 cm ¹(m), 2925 cm ¹(m), 2853 cm ¹(w), 1640 cm ¹1(m), 990 cm ¹(s), 915 cm ¹(s), 740 cm ¹(s), 695 cm ¹(s). The ¹H NMR spectrum of L consists of: Doublet 3.1 ppm (2H) Multiplet 5.1 ppm Multiplet 7.1 (5H) ppm Multiplet 4.8 ppm Multiplet 5.8 ppm The UV spectrum shows a maximum at 255 nm. Propose a structure for compound L. Edit Drawing harrow_forwardIdentify the structures of isomers H and I (molecular formula C8H11N).a.Compound H: IR absorptions at 3365, 3284, 3026, 2932, 1603, and 1497 cm−1b.Compound I: IR absorptions at 3367, 3286, 3027, 2962, 1604, and 1492 cm−1arrow_forwardTreatment 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?arrow_forward
- How could 1H NMR spectroscopy be used to distinguish among isomers A, B, and C?arrow_forwardchromophores as (b) Identify the indicated NMR entiotopic/diastereotopic/homotopic magnetically equivalent or non-equivalent. as well as CH₂ H CH3 HOCHI HH₂C (c) A tri-substituted benzene possessing one bromine and two methoxy substituents exhibit three aromatic resonances at d = 6.40, 6.46, and 7.41 ppm. Identify the substitution pattern.arrow_forwardPropose a structure given the 1H and 13C NMR spectra of the unknown compound. Assign chemical shifts to corresponding hydrogen and carbon atoms Molecular Formula: C5H10O3arrow_forward
- Determine the most likely structure of a compound with the formula C9H12 if it gave an H NMR spectrum consisting of: A doublet at d 1.25, a septet at d 2.90 and a multiplex at d 7.25arrow_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 C has molecular formula C5H8O. The IR, mass, 1H-NMR, and 13C-NMR spectra are shown below. Suggest a structure for C and explain your reasoning.arrow_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning