Chapter 14: Q73P (page 527)
Chapter 14: Q73P (page 527)
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Get started for freeQuestion: Reaction of C6H5CH2CH2OH with CH3COCl affords compound W, which has molecular formula C10H12O2. W shows prominent IR absorptions at 3088–2897, 1740, and 1606cm-1 . W exhibits the following signals in its 1 H NMR spectrum: 2.02 (singlet), 2.91 (triplet), 4.25 (triplet), and 7.20–7.35 (multiplet) ppm. What is the structure of W? We will learn about this reaction in Chapter 22.
Question. 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 2941cm-1 . L shows three signals in its 13C-NMR at 19, 52, and 101 ppm. 1H-NMR spectrum of L is given below. What is the structure of L?
Question: Label the signals due to Ha ,Hb and Hc in the 1 H NMR spectrum of acrylonitrile (CH2=CHCN ). Draw a splitting diagram for the absorption due to the proton Hc.
Question. When 2-bromo-3,3-dimethylbutane is treated with K+- OC(CH3)3, a single product T having molecular formula C6H12 is formed. When 3,3-dimethylbutan-2-ol is treated with H2SO4 , the major product U has the same molecular formula. Given the following -NMR data, what are the structures of T and U? Explain in detail the splitting patterns observed for the three split signals in T. 1 H NMR of T: 1.01 (singlet, 9 H), 4.82 (doublet of doublets, 1 H, J = 10, 1.7 Hz), 4.93 (doublet of doublets, 1 H, J = 18, 1.7 Hz), and 5.83 (doublet of doublets, 1 H, J = 18, 10 Hz) ppm 1 H NMR of U: 1.60 (singlet) ppm.
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