Concept explainers
Interpretation:
The structure for the compound C is to be proposed.
Concept introduction:
DEPT stands for distortion less enhancement by polarization transfer. DEPT
The carbon signals in a DEPT spectrum are classified as
Presence of the electronegative compounds like halogens or oxygen or sulfur elements will downfield the signal i.e., the chemical shift will occur at higher δ than the unsubstituted carbons.
The tollens test is used to distinguish between
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Organic Chemistry
- Benzonitrile (C6H5CN) is reduced to two different products depending on the reducing agent used. Treatment with lithium aluminum hydride followed by water forms K, which has a molecular ion in its mass spectrum at 107 and the following IR absorptions: 3373, 3290, 3062, 2920, and 1600 cm-1. Treatment with a milder reducing agent forms L, which has amolecular ion in its mass spectrum at 106 and the following IR absorptions: 3086, 2820, 2736, 1703, and 1600 cm-1. L shows fragments in its mass spectrum at m/z = 105 and 77. Propose structures for K and L and explain how you arrived at your conclusions.arrow_forwardTreatment of compound E (molecular formula C4H8O2) with excessCH3CH2MgBr yields compound F (molecular formula C6H14O) afterprotonation with H2O. E shows a strong absorption in its IR spectrum at1743 cm−1. F shows a strong IR absorption at 3600−3200 cm−1. The 1HNMR spectral data of E and F are given. What are the structures of E andF?Compound E signals at 1.2 (triplet, 3 H), 2.0 (singlet, 3 H), and 4.1 (quartet, 2 H) ppm Compound F signals at 0.9 (triplet, 6 H), 1.1 (singlet, 3 H), 1.5 (quartet, 4H), and 1.55 (singlet, 1 H) ppmarrow_forwardReaction of (CH3)2CO with LiCCH followed by H2O gives compound X, which has a molecular ion in the mass spectrum at 84. It also has prominent absorptions in the IR spectrum at 3600-3200, 3303, 2938, and 2120 cm-1. The proton NMR shows a singlet at 1.53 (6H), a singlet at 2.37 (1H) and a singlet at 2.43 (1H). What is the structure for compound X?arrow_forward
- (a) The IR spectrum below shows the important absorption bands for compound A (C7H1402). Its 'H NMR spectrum shows only singlets at &H 1.3, 2.3, 3.9 and 8.5 with the integration ratio of 9:2:2:1. Using all the spectral data, deduce the structure of compound A. 100 50- 2898 2986 1718 3446 1150 0- 4000 3000 2000 1500 1000 500 wavenumber 5. Transmittancearrow_forward(c) Compound C has a molecular formula of C6H14O. The IR spectrum shows strong absorption bands at 2980, 2850 and 1100 cm-1. The 'H NMR spectrum shows three singlets at &H 1.25 (9H), 3.23 (3H) and 3.79 (2H). Based on the spectral data given, deduce the structure of compound C.arrow_forwardAs reaction of (CH3)2CO with LIC≡CH followed by H2O affords compound D, which has a molecular ion in its mass spectrum at 84 and prominent absorptions in its IR spectrum at 3600−3200, 3303, 2938, and 2120 cm−1. D shows the following 1H NMR spectral data: 1.53 (singlet, 6 H), 2.37 (singlet, 1 H), and 2.43 (singlet, 1 H) ppm. What is the structure of D?arrow_forward
- Reaction of (CH3)3CCHO with (C6H5)3P=C(CH3)OCH3, followed by treatment with aqueous acid, affords R (C7H14O). R has a strong absorption in its IR spectrum at 1717 cm−1 and three singlets in its 1H NMR spectrum at 1.02 (9 H), 2.13 (3 H), and 2.33 (2 H) ppm. What is the structure of R? We will learn about this reaction in Chapter 18.arrow_forwardTreatment of compound E (molecular formula C4H8O2) with excess CH3CH2MgBr yields compound F (molecular formula C6H14O) after protonation with H2O. E shows a strong absorption in its IR spectrum at 1743 cm-1. F shows a strong IR absorption at 3600–3200 cm-1. The 1H NMR spectral data of E and F are given. What are the structures of E and F?Compound E signals at 1.2 (triplet, 3 H), 2.0 (singlet, 3 H), and 4.1 (quartet, 2 H) ppmCompound F signals at 0.9 (triplet, 6 H), 1.1 (singlet, 3 H), 1.5 (quartet, 4 H), and 1.55 (singlet, 1 H) ppmarrow_forwardA compound of formula C6H10O2 shows only two absorptions in the proton NMR: a singlet at 2.67 ppm and a singlet at2.15 ppm. These absorptions have areas in the ratio 2:3. The IR spectrum shows a strong absorption at 1708 cm-1. Proposea structure for this compound.arrow_forward
- Compound C has the molecular formula C5H8O. The IR, 1H, 13C, and DEPT NMR spectra of this compound are shown below. (a) Calculate the double bond equivalent of compound C and briefly explain what the values obtains represents. Interpret the IR spectrum. (b) Based on the information provided, determine the structure of compound D.arrow_forwardComplete the following answer using the data provided here: • Molecular formula: C5H10O2 • Important IR data (cm-1): 3150-3010 (many); 1740; 1248; 1050 • All 'H NMR data (ppm, splitting, integration): 3.66 ppm (s), 3; 2.20 ppm (t), 2; 1.80 ppm (m), 2; 0.98 ppm (t), 3 Which of the following best fit this spectroscopic data? OH H. A C D Earrow_forward(b) Compound C has the molecular formula of C-H12. The IR spectrum shows the important absorption bands at 2950, 2830 and 2250 cm-1 while the 'H NMR spectrum shows two singlets at dH 1.24 (9H) and ôH 1.80 (3H). Based on these spectral data, deduce the structure of compound C.arrow_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning