Interpretation:
Diastereomers, chemically nonequivalent sets of protons, and
Concept introduction:
Nuclear magnetic resonance (NMR) is one of the most capable analytical techniques used for determining the functional groups and how the atoms are structured and arranged in a molecule.
When a compound containing protons or carbon-13 placed under very strong magnetic field and treated with
Nuclear magnetic resonance spectroscopy is a graph showing characteristic energy absorption frequencies and intensities of a compound under magnetic field.
When each of two hydrogens are replaced by a same group, then the compounds formed are enantiomers and the hydrogens are called enantiotopic hydrogens. Significance of enantiotopic hydrogens is that they give only one signal in
When each of two hydrogens are replaced by a same group, then the compounds formed are diastereomers and the hydrogens are called diastereotopic hydrogens. Diastereotopic hydrogens give us two different signals in
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ORGANIC CHEM. VOL.1+2-W/WILEYPLUS
- Predict the theoretical number of different NMR signals produced by each compound, and give approximate chemical shifts. Point out any diastereotopic relationships. (a) Ph¬CHBr¬CH2Br (b) vinyl chloridearrow_forward(c) Assign the correct structure for compound M with molecular formula C5H100, which has 'H-NMR signal at 8 = 5-64 (A), 5:53 (B), 4-23 (C), 1-92 (D), (D)is an exchangable singlet (with D20). . multiplets, (Ẹ) and (F) are doublets and a E 1:68 (E) and 1-24 (F)" in рpm. JAR = 16:3 Hz. (A), (B)_ and (C) are (A), TB). and (C) arearrow_forward) The H-NMR spectrum of compound B, C7H₁4 O, consists of the following signals: 8 0.9 (t, 6H), 1.6 (sextet, 4H), and 2.4 (t, 4H). Draw the structural formula of compound B. . You do not have to consider stereochemistry. Explicitly draw all H atoms. .arrow_forward
- In the 1H NMR spectra of 2-bromopropane (CH3)2CHBr and 1-bromopropane CH3CH2CH2Br, how many signals do you expect to see?arrow_forwardCompound 1 has molecular formula C7H15Cl. It shows two signals in the 1H-NMR spectrum, one at 1.08 ppm and one at 1.59 ppm. The relative integrals of these two signals are 3 and 2, respectively. Propose structures for compound 1, explaining how you reach your conclusion.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_forward
- 1) They were given a mysterious chemical by ASP Rinuprasad, coded as Compound X, which was believed to be a forbidden drug smuggled through KLIA airport. Little they know about the chemical from the convict. As competent officers, together they decided to run several spectroscopic and analytical tests to determine the molecular structure of Compound X. They managed to obtain the spectra and data of IR, 'H NMR (400 MHz in CDCI 3), 13C and DEPT (100.6 MHz in CDC13), COSY, HMQC, HMBC, mass spectrometry and elemental analyses. Determine the molecular structure and name of Compound X. IR SPECTRUM 2.5 100 3.0 4.0 5,0 6.0 7.0 8.0 9.0 10 12 15 2000 avenumber cm 1500 000 1000 1Η ΝMR 7.50 7.45 7.40 7.35 7.30 7.25 7.20 7.15 7.10 7.05 7.00 55 50 1.5 100 9.5 9.0 8.5 8.0 7.5 7.0 65 60 4.5 35 3.0 2.5 20 10 0.5arrow_forward14. Compound B has molecular formula C9H12. It shows five signals in the 1H-NMR spectrum - a doublet of integral 6 at 1.22 ppm, a septet of integral 1 at 2.86 ppm, a singlet of integral 1 at 5.34 ppm, a doublet of integral 2 at 6.70 ppm, and a doublet of integral 2 at 7.03 ppm. The 13C-NMR spectrum of B shows six unique signals (23.9, 34.0, 115.7, 128.7, 148.9, and 157.4). Identify B and explain your reasoning.arrow_forwardPlease show work 1. Compound A has molecular formula C5H10O. It shows three signals in the 1H-NMR spectrum - a doublet of integral 6 at 1.1 ppm, a singlet of integral 3 at 2.14 ppm, and a quintet of integral 1 at 2.58 ppm. Suggest a structure for A and explain your reasoning.arrow_forward
- Following is the 1H-NMR spectrum of compound O, molecular formula C7H12. Compound O reacts with bromine in carbon tetrachloride to give a compound with the molecular formula C7H12Br2. The 13C-NMR spectrum of compound O shows signals at d 150.12, 106.43, 35.44, 28.36, and 26.36. Deduce the structural formula of compound O.arrow_forwardUse the NMR tables & to predict the range (ppm) where each compound should absorb.arrow_forward13. Each of the following compounds will have one or more 'H-NMR signals which might be split by one or more neighboring non-equivalent protons. Identify the separate sets of equivalent protons and predict the splitting patterns and integrations. (It will be helpful to draw in all hydrogen atoms.) (a) CI 3H triplet (0.91 ppm), 2H sextet (1.46 ppm), broad 2H singlet (1.77 ppm), 2H triplet (2.65 ppm) (b) 14. For each of the molecules in the previous question, also predict the approximate chemical shift (ppm) of each signal. 3.5 15. Below is the 'H-NMR spectrum of propylamine, CH3CH₂CH₂NH₂. Draw the structure of this compound and assign each of the signals in the spectrum to the set of protons that produced it. 3.0 2.00 2.5 20 (c) 2.29 2.13 15 1.0 Br a 3.06 0.5arrow_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning