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- A and B are isomeric dicarbonyl compounds of the molecular formula C5H&O2. The 'H NMR spectrum of A contains a singlet at 2.05 ppm and another singlet at 5.40 ppm. The 'H NMR spectrum of B contains three signals: a singlet at 2.3 ppm, a triplet at 1.10 ppm and a quartet at 2.70 ppm. Suggest structures for A and B and draw them in their respective boxes below. 1st attemptA 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?The 1H-NMR spectrum of Compound D of molecular formula C10H12O shows three singlets – δ 2.20 (6H, s), 4.86 (4H), 7.10 (2H) ppm. Its 13C-NMR spectrum has five signals – 20, 74, 127, 135, 146 ppm. Suggest a structure for this compound.
- Thymol (molecular formula C10H14O) is the major component of the oil ofthyme. Thymol shows IR absorptions at 3500–3200, 3150–2850, 1621, and1585 cm−1. The 1H NMR spectrum of thymol is given below. Propose apossible structure for thymol.Following are IR and 1H-NMR spectra of compound D. The mass spectrum of compound D shows a molecular ion peak at m/z 136, a base peak at m/z 107, and other prominent peaks at m/z 118 and 59. Q.) Propose structural formulas for ions in the mass spectrum at m/z 118, 107, and 59.Propose a structural formula for compound A, C4H10O, consistent with the following 1H-NMR and IR spectra. Please assign all the appropriate peaks in the IR and NMR spectra and provide a short narrative describing what structural information each piece of data provided.
- 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 informationThe '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 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.
- 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?An unknown compound has a molecular formula of C4H6O2. Its IR spectrum shows absorptions at 3095, 1762, 1254, and 1118 cm -1. It exhibits the following signals in its 1H NMR spectrum (ppm): 2.12 (singlet,3H), 4.55 (doublets of doublets, 1H), 4.85 (doublet of doublets, 1H), 7.25 (doublets of doublets, 1H); and the following signals in its 13C NMR spectrum (ppm): 20.8, 100.4, 141.2, 168.0. Draw the structure of the unknown compoundDraw the structure of a compound, C4H8O3 that exhibits IR absorptions at 1710 and 2500-3000 cm-1 and thefollowing 1H NMR signals: 11.1 (1H, singlet), 4.14 (2H, singlet), 3.63 (2H, quartet), 1.26 (3H, triplet) ppm.