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Compound W
Figure 20.6 The
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- Treatment of compound C (molecular formula C9H12O) with PCC affords D (molecular formula C9H10O). Use the 1H NMR and IR spectra of D to determine the structures of both C and D.arrow_forwardThe NMR spectra for compound 1 were acquired in a 7.5 mg / 0.6 mL solution ofCDCl3 . The 1H and 13C peaks are also listedbelow. Provide a full analysis of the NMR spectra for compound 1. correct assignment of NMR spectra of both 13C spectra. correct rationalisation of 13C spectrum1H NMR (400 MHz, CDCl3) δ 7.73 (d, J = 9.5 Hz, 1H), 7.56 (ddd, J = 8.5, 7.5, 1.6 Hz, 1H),7.51 (dd, J = 7.5, 1.6 Hz, 1H), 7.36 (d, J = 8.5 Hz, 1H), 7.30 (dd, J = 8.5, 7.5 Hz, 1H), 6.45(d, J = 9.5 Hz, 1H).13C NMR (101 MHz, CDCl3) δ 160.79, 154.09, 143.43, 131.85, 127.87, 124.44, 118.86,116.94, 116.74.Note: There are two carbon peaks in the 13C spectrum that are so close together that they are not differentiable at the resolution in this experiment. you should be able to assign these peaks to one of two carbon atoms in 1.arrow_forward08) The NMR spectra of the two isomeric compounds with formula C3H5ClO2 are shown in letters a and b. Low-field protons appearing in the NMR spectrum around 12.1 and 11.5 ppm, respectively, are shown highlighted. Draw the structures of the isomers.arrow_forward
- Name the following compounds A and B. How could you distinguish these two molecules by using 1H NMR and IR techniques? Propose an analytical technique to determine the iron content of these compounds. Calculate the mass percentages of C and H of compound B (C: 12.01 g/mol; H: 1.008 g/mol; Fe: 55.845 g/mol).arrow_forwardLl.128.arrow_forwardIdentify the structures of isomers H and I (molecular formula C8H11N).a.Compound H: IR absorptions at 3365, 3284, 3026, 2932, 1603, and 1497 cm−1b.Compound I: IR absorptions at 3367, 3286, 3027, 2962, 1604, and 1492 cm−1arrow_forward
- The HNMR , CNMR , 2D NMR SPECTRA OF COMPOUND M1 ARE ATTACHED. WHAT IS STRUCTURE OF IT ? IS IT SYMMETRICAL REGARDS TO ITS CNMR SPECTRUM ? WHAT IS THE STRUCTURE SUGGESTION AND WHY ?arrow_forwardCompound X (molecular formula C10H120) was treated with NH2NH2, ¯OH to yield compound Y (molecular formula C10H14). Match the 1H NMR spectra of X and Y to the corresponding structures of X and Y. Compound NH2NH2 Compound 'H NMR of X 6 H OH Y 1 H 5H 8. 6. 4 ppm or H NMR of Y 6 H 2H 5H 1 H multiplet multiplet 8. 6. 4. 3. 1 nnm 2. 2. 3, O:arrow_forwardA molecule of the molecular formula C5H11Br gives rise to the NMR spectrum below. When reacted with NaOH and water, it forms a product which by NMR has 2 protons 1H at 5.4 ppm and 1H at 5.5 ppm each having a 3J coupling of 17 Hz. (other protons also present) The product also has an IR stretch at 1550 cm 1. Provide the structures of the starting material and product. NaOH H,0 C;H„Br 6H triplet 4H quintet 1H quintet 10 8 Ppmarrow_forward
- Analysis of a sweet-smelling, neutral compound of carbon, hydrogen and oxygen produced the following results: %C = 54.5; %H = 9.1. From its mass spectrum, the molecular ion had a mass/charge ratio of 88. Its infra-red spectrum showed a prominent peak at 1735 cm–1. Figure below shows the NMR spectrum of the compound. Which of the following is this compound? A. 3-hydroxybutanalB. Methyl propionateC. Ethyl acetateD. 2-methylpropanoic acidE. Methoxyacetonearrow_forwardAn organic compound B with formula C6H14O has the following: IR Spectroscopy 2974 cm-1, 1080 cm-1 Mass Spectrometry 102 (M+), 87, 73 1H NMR Spectroscopy Eight signals at δ 1.10 (d, 3H), 1.13 (dd, 3H), 1.14 (dd, 3H), 1.59 (ddq, 1H), 1.60 (ddq, 1H), 3.19 (ddq, 1H), 3.51 (dq, 1H), 3.50 (dq, 1H). Compound B is obtained by the reaction of compound A with NaH followed by CH3CH2Br. The stereochemistry of A is "S" Using this information, deduce a plausible structure for Compound A with correct stereochemistry.arrow_forwardA molecule of the molecular formula C5H11Br gives rise to the NMR spectrum below. When reacted with NaOH and water, it forms a product which by NMR has 2 protons 1H at 5.4 ppm and 1H at 5.5 ppm each having a J coupling of 17 Hz. (other protons also present) The product also has an IR stretch at 1550 cm1. Provide the structures of the starting material (1 pt) and product (2 pts) NaOH H20 C5H1,Br 6H triplet 4H quintet 1H quintet 10 8 7 6. 4 3 1 HSP-06-347 ppmarrow_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage LearningMacroscale and Microscale Organic ExperimentsChemistryISBN:9781305577190Author:Kenneth L. Williamson, Katherine M. MastersPublisher:Brooks Cole