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A solution of acetone
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ORGANIC CHEMISTRY
- When the 1HNMR spectrum of an alcohol is run in dimethylsulfoxide (DMSO) solvent rather than in chloroform, exchange of the Ο-H proton is slow and spin-spin splitting is seen between the Ο-H proton and C-H protons on the adjacent carbon. What spin multiplicities would you expect for the hydroxyl protons in the following alcohols? (a) 2-Methyl-2-propanol (b) Cyclohexanol (c) Ethanol (d) 2-Propanol (e) Cholesterol (f) 1-Methylcyclohexanolarrow_forward10- If 3-Bromo-1-phenyl-1-propene shows a complex NMR spectrum in which thevinylic proton at C2 is couples with both the C1 vinylic proton (J = 16 Hz) and the C3methylene protons (J = 8 Hz) Draw a tree diagram for the C2 proton signal.arrow_forwardThe 13C NMR spectrum of 1-bromo-3-chloropropane contains peaks at δ 30, δ 35, and δ 43. Assign these signals to the appropriate carbons.arrow_forward
- Deduce the structure of compound C.arrow_forwardMass spectrometry of an unknown compound revealed a molecular ion at m/z 100.09. The IR spectrum, the ¹³C NMR spectrum, and the ¹H NMR spectrum are shown below. Draw the structure of this compound. 4000 220 3000 200 180 160 2000 M 1500 Wavenumbers (cm-¹) 140 120 100 1H 80 6H 1000 40 3H 20 ppm 500arrow_forwardCgH,NO2 IR |1* Spectrum IR = NMR Spectrum- 3 THSarrow_forward
- The 13C-NMR spectrum of 3-methyl-2-butanol shows signals at d 17.88 (CH3), 18.16 (CH3), 20.01 (CH3), 35.04 (carbon-3), and 72.75 (carbon-2). Account for the fact that each methyl group in this molecule gives a different signal.arrow_forward4. You find a bottle on the shelf only labelled C3HSO. The sample was subjected to IR analysis yielding the following spectrum. www 1.0 0.8 0.6 0.2 0.0 3500 3000 2500 2000 1500 1000 500 Wavenumbers (cm-1) i. Deduce the structure of the compound stored in the bottle and explain your answer. Explain what is the difference between the IR spectrum of an alcohol and a carboxylic acid. ii. Transmitancearrow_forwardDraw the structure of the compound identified by the simulated 'H NMR and C NMR spectra. The molecular formula of the compound is C,0H,O. 12 (Blue numbers next to the lines in the 'H NMR spectra indicate the integration values.) H NMR 1H 2H 2H 2H 2H|| 3H 10 8. 4 (dd) g 13C NMR 220 200 180 160 140 120 100 80 60 40 20 8 (ppm) Deduce the structure from the spectra. Select Draw Rings More Erasearrow_forward
- Indicate two basic differences that exist between the spectra of 1H y 13C in NMR.arrow_forwardIR Spectrum (CCl4 solution) U 4000 3000 100 80 60 40 20 JL 40 80 13C NMR Spectrum (100.0 MHz, CDCI, solution) % of base peak DEPT CH₂ CH₂ CH proton decoupled 200 ¹H NMR Spectrum (200 MHz, CDCI, solution) 10 9 8 2000 1693 v (cm¹) 155/157 120 160 m/e 160 7 1600 183/185 7.8 6 your 1200 120 800 Mass Spectrum C8H7OBr 280 solvent 80 M+= 198/200 200 240 expansion سلا 5 7.2 ppm 4 3 Problem 100 UV Spectrum max 258 nm (log₁0€ 4.2) solvent: ethanol 40 2 0 1 206 8 (ppm) TMS L 0 8 (ppm) 189arrow_forwardThe chemical formula is C8H12O. What is it’s structure and degrees of unsaturation? What important functional groups are in the spectra, and what peaks correspond with which hydrogens and carbons in the structure predicted.arrow_forward
- Macroscale and Microscale Organic ExperimentsChemistryISBN:9781305577190Author:Kenneth L. Williamson, Katherine M. MastersPublisher:Brooks Cole