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Propose structures consistent with each set of data: (a) a hydrocarbon with a molecular ion at
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- The 'H NMR spectrum of compound A (C3H100) has four signals: a multiplet at 8 = 7.25-7.32 ppm (5 H), a singlet at d = 5.17 ppm (1 H), a quartet at d = 4.98 ppm (1 H), and a doublet at ô = 1.49 ppm (3 H). There are 6 signals in its 13C NMR spectrum. The IR spectrum has a broad absorption in the -3200 cm-1 region. Compound A reacts with KMNO4 in a basic solution followed by acidification to give compound B with the molecular formula C7H6O2. Draw structures for compounds A and B.arrow_forwardIndicate 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_forward
- Deduce the structure of compound C.arrow_forwardA 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_forward
- (b) Determine the structure of an organic compound with a molecular ion at m/z 73 and absorption in its IR at 3300 (doublet) and 1680 cmarrow_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_forwardHow could 1H NMR spectroscopy be used to distinguish among isomers A, B, and C?arrow_forward
- chromophores 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_forwardDetermine 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_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning