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- Identify the compound with molecular formula C4H9BrO that gives the following 1H NMR spectrum.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.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:
- 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.Identify each of the following compounds from its molecular formula and its 13C NMR spectrum:Following are 1H-NMR spectra for compounds G, H, and I, each with the molecular formula C5H12O. Each is a liquid at room temperature, is slightly soluble in water, and reacts with sodium metal with the evolution of a gas. (a) Propose structural formulas of compounds G, H, and I. (b) Explain why there are four lines between 0.86 and 0.90 for compound G. (c) Explain why the 2H multiplets at 1.5 and 3.5 for compound H are so complex.
