(a) The 1H-NMR spectrum of compound A, C7H15 Cl, consists of the following signals: 8 1.1 (s, 9H) and 1.6 (s, 6H). Draw the structural formula of compound A. • You do not have to consider stereochemistry. · Explicitly draw all H atoms. 百 Sn ? ChemDoodle
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- 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.(b) The 1H-NMR spectrum of compound B, C5 H10 O2, consists of the following signals: 8 1.2 (d, 6H), 2.0 (s, 3H), and 5 (septet, 1H); contains an ester. Draw the structural formula of compound B. • You do not have to consider stereochemistry. • Explicitly draw all H atoms. ୫) C n ? ChemDoodle ® Show HintThe '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.
- ) The H-NMR spectrum of compound B, C7H₁4 O, consists of the following signals: 8 0.9 (t, 6H), 1.6 (sextet, 4H), and 2.4 (t, 4H). Draw the structural formula of compound B. . You do not have to consider stereochemistry. Explicitly draw all H atoms. .The 1H-NMR spectrum of compound R, C6H14O, consists of two signals: d 1.1 (doublet) and d 3.6 (septet) in the ratio 6:1. Propose a structural formula for compound R consistent with this informationCompound A has molecular formula C5H10O. It shows three signals in the 1H-NMR spectrum - a doublet of integral 6 at 1.1 ppm, a singlet of integral 3 at 2.14 ppm, and a quintet of integral 1 at 2.58 ppm. Suggest a structure for A and explain your reasoning.
- Compound X of the molecular formula C7H10 has the 13C NMR spectrum (5 signals) shown below. On treatment with excess H2/Pt (catalytic hydrogenation), X is converted to methylcyclohexane. Propose a structure for X and justify your reasoning by clearly labeling each carbon signal and write out the reaction. 200 180 160 140 120 100 80 60 40 20Following are 1H-NMR spectra for compounds B (C6H12O2) and C (C6H10O). Upon warming in dilute acid, compound B is converted to compound C. Deduce the structural formulas for compounds 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?
- Draw a structure for the compound, C3H5Br, that fits the following 'H NMR data: d 2.32 (3H, singlet) ô 5.35 (1H, broad singlet) ô 5.54 (1H, broad singlet) You do not have to consider stereochemistry. You do not have to explicitly draw H atoms. In cases where there is more than one answer, just draw one. C opy aste C ChemDoodleA hydrocarbon, compound B, has molecular formula C6H6, and gave an NMR spectrum with two signals: delta 6.55 pm and delta 3.84 pm with peak ratio of 2:1. When warmed in pyridine for three hr, compound B quantitatively converts to benzene. Mild hydrogenation of B yielded another compound C with mass spectrum of m/z 82. Infrared spectrum showed no double bonds; NMR spectrum showed one broad peak at delta 2.34 ppm. With this information, address the following questions. a) How many rings are in compound C? b) How many rings are probably in B? How many double bonds are in B? c) Can you suggest a structure for compounds B and C? d) In the NMR spectrum of B, the up-field signal was a quintet, and the down field signal was a triplet. How must you account for these splitting patterns?Two compounds with the molecular formula C5H10O have the following 1H and 13C NMR data. Both compounds have a strong IR absorption band in the 1710–1740-cm_1 region. Elucidate the structure of these two compounds and interpret the spectra. (10) (a) 1H NMR: d 2.55 (septet, 1H), 2.10 (singlet, 3H), 1.05 (doublet, 6H); 13C NMR: d 212.6, 41.5, 27.2, 17.8 (b) 1H NMR: d 2.38 (triplet, 2H), 2.10 (singlet, 3H), 1.57 (sextet, 2H), 0.88 (triplet, 3H); 13C NMR: d 209.0, 45.5, 29.5, 17.0, 13.2