Concept explainers
Deduce the structures of compounds
Compound
Compound
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ORGANIC CHEMISTRY
- Compound I (C11H14O2) is insoluble in water, aqueous acid, and aqueous NaHCO3, but dissolves readily in 10% Na2CO3 and 10% NaOH. When these alkaline solutions are acidified with 10% HCl, compound I is recovered unchanged. Given this information and its 1H-NMR spectrum, deduce the structure of compound I.arrow_forwardAn 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 compoundarrow_forwardIdentify the structure of compound C (molecular formula C11H15NO2), which has an IR absorption at 1699 cm−1 and the 1H NMR spectrum shown below.arrow_forward
- 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.arrow_forwardThymol (molecular formula C10H14O) is the major component of the oil of thyme. Thymol shows IR absorptions at 3500–3200, 3150–2850, 1621, and 1585 cm−1. The 1H NMR spectrum of thymol is given below. Propose a possible structure for thymol.arrow_forwardIdentify the structures of D and E, isomers of molecular formulaC6H12O2, from their IR and 1H NMR data. Signals at 1.35 and 1.60 ppm inthe 1H NMR spectrum of D and 1.90 ppm in the 1H NMR spectrum of Eare multipletsarrow_forward
- 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 attemptarrow_forwardA compound has the molecular formula C6H12O2. Its IR spectrum shows a strong absorption band near 1740 cm-1. Its 1H NMR spectrum consists of two singlets at δ 1.2 and δ 3.6. Which is the most likely structure of the compound?arrow_forwardDraw 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.arrow_forward
- Identify the structures of D and E, isomers of molecular formulaC6H12O2, from their IR and 1H NMR data. Signals at 1.35 and 1.60 ppm inthe 1H NMR spectrum of D and 1.90 ppm in the 1H NMR spectrum of Eare multiplets.arrow_forwardA 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?arrow_forwardPhenacetin is an analgesic compound having molecular formula C10H13NO2. Once a common component in over-the-counter pain relievers such as APC (aspirin, phenacetin, caffeine), phenacetin is no longer used because of its liver toxicity. Deduce the structure of phenacetin from its 1H NMR and IR spectra.arrow_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning