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Compound Y
Figure 1 The IR spectrum of compound Y (Problem 34)
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- 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_forwardThe '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.arrow_forwardA compound with molecular formula C8H8O3 exhibits the following IR, 1H NMR and 13C NMR spectra. Deduce the structure of this compound.arrow_forward
- A compound with molecular formula C3H8O produces a broad signal between 3200 and 3600 cm³¹ in its IR spectrum and produces two signals in its 13C NMR spectrum. Draw the structure of the compound. Draw Your Solutionarrow_forwardDetermine the structure of an alcohol with molecular formula C5H12O that exhibits the following signals in its 13C NMR spectra: a. Broadband decoupled: 73.8 δ, 29.1δ, and 9.5 δ b. DEPT-90: 73.8 δ c. DEPT-135: positive signals at 73.8 δ, and 9.5 δ, and negative signal at 29.1δ,arrow_forwardCompound 2 has molecular formula C6H12. It shows three signals in the 1H-NMR spectrum, one at 0.96 ppm, one at 2.03 ppm, and one at 5.33 ppm. The relative integrals of these three signals are 3, 2, and 1, respectively. Provide structure for compound 2, explain how you reached your conclusion.arrow_forward
- 3. Propose a structure for an organic compound with molecular formula C3H1404 given the following 'H NMR and IR spectra. Draw the structure only, no need to interpret the spectra. 1H NMR Spectrum IR Spectrum Triplet Singlet d (6H) 8 (4H) 8 (4H) 1740 cm1 Quartetarrow_forwardDeduce the structure of a compound with molecular formula C5H100 that exhibits the following ¹H and ¹³C NMR spectra. IH NMR CNMR 150 10 Structure A Structure B Structure C Structure D 24 1C 100 B 2H H 20 50 10 D 311arrow_forwardQ2: The proton NMR spectrum is shown for a compound with formula C5H9NO4. The infrared spectrum displays strong bands at 1750 and 1562 cm-1 and a medium-intensity band at 1320 cm-1. The normal carbon- 13 and the DEPT experimental results are tabulated. Draw the structure of this compound. Normal Carbon DEPT-135 DEPT-90 14 ppm Positive No peak No peak 16 Positive 63 Negative No peak 83 Positive Positive 165 No peak No peak Proton spectrum C;H,NO4 0.92 2.01 3.00 3.00 5.0 4.5 4.0 3.5 3.0 25 2.0 1.5 10 0.5 0.0arrow_forward
- Given the following 13C NMR signals, construct a structure for the unknown compounds. A. Molecular formula: C6H14O DEPT-135 (positive): 14.1 δ DEPT-135 (negative): 22.7 δ, 25.3 δ, 31.8 δ, 32.2 δ, 62.8 δ B. Molecular formula: C7H12O2 Broadband decoupled: 19.1 δ, 28.0 δ, 70.5 δ, 129.0 δ, 129.8 δ, 165.8 δ DEPT-90: 28.0 δ, 129.8 δ DEPT-135 (positive): 19.1 δ, 28.0 δ, 129.8 δ DEPT-135 (negative): 70.5 δ, 129.0 δarrow_forwardQ3: The proton NMR spectrum of a compound with formula C5H100 is shown. The DEPT experimental results are tabulated. The infrared spectrum shows medium-sized bands at 2968, 2937, 2880, 2811, and 2711 cm1 and strong bands at 1728 cm-1. Draw the structure of this compound. Normal Carbon DEPT-135 DEPT-90 11.35 ppm No peak No peak No peak Positive 12.88 Positive 23.55 Negative 47.78 Positive Positive 205.28 Positive Positive (C=0) T m 240 2.35 2.30 2.25 2.20 0.75 0.91 1.04 1.01 2.90 2.92 9.6 9'5 24 23 2.2 2.1 2.0 1.9 1.8 1.7 1.6 14 1.3 1.2 1.1 10 0.9 2890.2 \ 2888.0 690.2 0'889 F189¬ 669.6 667.8arrow_forwardWhat is the structure of the compound if its formula is C5H10O, it exhibits 13C-NMR signals Broadband decoupled: 207, 45, 28, and 12 ppm DEPT-90: no signals DEPT-135: positive signals at 28 and 12 ppm; negative signals at 45 and 16 ppm.arrow_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning