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- Compound X (molecular formula C10H120) was treated with NH2NH2, ¯OH to yield compound Y (molecular formula C10H14). Match the 1H NMR spectra of X and Y to the corresponding structures of X and Y. Compound NH2NH2 Compound 'H NMR of X 6 H OH Y 1 H 5H 8. 6. 4 ppm or H NMR of Y 6 H 2H 5H 1 H multiplet multiplet 8. 6. 4. 3. 1 nnm 2. 2. 3, O:Identify the structures of isomers A and B (molecular formula C9H10O). Compound A: I R peak at 1742 cm−1; 1H NMR data (ppm) at 2.15 (singlet, 3 A: H), 3.70 (singlet, 2 H), and 7.20 (broad singlet, 5 H). Compound B: I R peak at 1688 cm−1; 1H NMR data (ppm) at 1.22 (triplet, 3 B: H), 2.98 (quartet, 2 H), and 7.28–7.95 (multiplet, 5 H).An unknown compound X has the molecular formula C6H140. Compound at 3000 cm. The 'H NMR spectral data of compound X is given below. W Here PPM values are given in delta symbol. absorption 8 ratio triplet 1.0 3 singlet 1.1 quartet singlet 1.5 3.5 3 623 6 O C) III O A) I O D) IV O B) II I 11 OH t III IV
- Identify A and B, isomers of molecular formula C3H4Cl2, from the given 1H NMR data: Compound A exhibits peaks at 1.75 (doublet, 3 H, J = 6.9 Hz) and 5.89 (quartet, 1 H, J = 6.9 Hz) ppm. Compound B exhibits peaks at 4.16 (singlet, 2 H), 5.42 (doublet, 1 H, J = 1.9 Hz), and 5.59 (doublet, 1 H, J = 1.9 Hz) ppm.Identify the structures of isomers A (molecular formula C9H10O).Compound A: I R peak at 1742 cm−1; 1H NMR data (ppm) at 2.15 (singlet, 3H), 3.70 (singlet, 2 H), and 7.20 (broad singlet, 5 H).How could 1H NMR spectroscopy be used to distinguish among isomers A, B, and C?
- Treatment 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?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?The 1H NMR spectra of two compounds, each with molecular formula C11H16, are shown here. Identify the compounds.
- When compound A (C5H12O) is treated with HBr, it forms compound B (C5H11Br). The 1H NMR spectrum of compound A has a 1H singlet, a 3Hdoublet, a 6H doublet, and two 1H multiplets. The 1H NMR spectrum of compound B has a 6H singlet, a 3H triplet, and a 2H quartet. Identifycompounds A and B.What is the 1H NMR data (chemical shift, integration, multiplicty) of hte compound shown below?Reaction of C6H5CH2CH2OH with CH3COCl affords compound W, whichhas molecular formula C10H12O2. W shows prominent IR absorptions at3088–2897, 1740, and 1606 cm−1. W exhibits the following signals in its1H NMR spectrum: 2.02 (singlet), 2.91 (triplet), 4.25 (triplet), and 7.20–7.35(multiplet) ppm. What is the structure of W?