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- A and B, isomers of molecular formula C3H5Cl3, are formed by the radical chlorination of a dihalide C of molecular formula C3H6Cl2.a. Identify the structures of A and B from the following 1H NMR data:Compound A: singlet at 2.23 and singlet at 4.04 ppmCompound B: doublet at 1.69, multiplet at 4.34, and doublet at 5.85 ppmb. What is the structure of C?You have a sample of a compound of molecular formula C11H15NO2, which has a benzene ring substituted by two groups, (CH3)2N− and −CO2CH2CH3, and exhibits the given 13C NMR. What disubstituted benzene isomer corresponds to these 13C data?You have a sample of a compound of molecular formula C11H15NO2, which has a benzene ring substituted by two groups, (CH3)2N – and – CO2CH2CH3, and exhibits the given 13C NMR. What disubstituted benzene isomer corresponds to these 13C data?
- Treatment of butan-2-one (CH3COCH2CH3) with strong base followed byCH3I forms a compound Q, which gives a molecular ion in its massspectrum at 86. The IR (> 1500 cm−1 only) and 1H NMR spectra of Q aregiven below. What is the structure of Q?A and B, isomers of molecular formula C3H5Cl3, are formed by the radical chlorination of a dihalide C of molecular formula C3H6Cl2. a.Identify the structures of A and B from the following 1H NMR data: Compound A: singlet at 2.23 and singlet at 4.04 ppm Compound B: doublet at 1.69, multiplet at 4.34, and doublet at 5.85 ppm b.What is the structure of C?These are the 1H RMN and IR spectrums of an A compound with formula C10H13NO2. When A reacts with NaOH in water and heat, B compound is formed, with formula C10H11NO. What are the structures of A and B?
- Treatment of butan-2-one (CH3COCH2CH3) with strong base followed by CH3I forms a compound Q, which gives a molecular ion in its mass spectrum at 86. The IR (> 1500 cm−1 only) and 1H NMR spectrum of Q are given below. What is the structure of Q?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 1H NMR data, what are the structures of T and U? Explain in detail the splitting patterns observed for the three split signals in T.1H 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) ppm1H NMR of U: 1.60 (singlet) ppmCompound X (molecular formula C10H12O) was treated with NH2NH2, −OH to yield compound Y (molecular formula C10H14). Based on the 1HNMR spectra of X and Y given below, what are the structures of X and Y?
- Treatment of butan-2-one (CH3COCH2CH3) with strong base followed by CH3I forms a compound Q, which gives a molecular ion in its mass spectrum at 86. The IR (> 1500 cm−1 only) and 1H NMR spectra of Q are given below. What is the structure of Q?Choose the structure corresponding to the given 1H and 13C NMR spectraWhen 2-bromo-3,3-dimethylbutane is treated with K+ −OC(CH3)3, a singleproduct T having molecular formula C6H12 is formed. When 3,3-dimethylbutan-2-ol is treated with H2SO4, the major product U has thesame molecular formula. Given the following 1H NMR data, what are thestructures of T and U? Explain in detail the splitting patterns observedfor the three split signals in T.1H 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 ofdoublets, 1 H, J = 18, 10 Hz) ppm1H NMR of U: 1.60 (singlet) ppm Additional problems on the spectroscopy of alkenes are given in Chapters A–C:Mass spectrometry: A.16b, A.20, A.23Infrared spectroscopy: B.5, B.7(A), B.12c, B17a, B.18cNuclear magnetic resonance spectroscopy: C.12a; C.15d, e; C.29d; C.32d;C.37; C.38d, f; C.43i, j; C.44; C.45; C.49d, f; C.50b; C.51c; C.55