Organic Chemistry
12th Edition
ISBN: 9781118875766
Author: T. W. Graham Solomons, Craig B. Fryhle, Scott A. Snyder
Publisher: WILEY
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Chapter 14, Problem 14PP
Practice Problem 14.14
Explain how
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Combined Spectra
14.46
Identify the C3 H5 Br isomers on the basis of the following information:
(a) Isomer A has the 'H NMR spectrum shown in 4 Figure 14.49.
(b) Isomer B has three peaks in its 13C NMR spectrum: 8 32.6 (CH2 ); 118.8 (CH2 ); and 134.2 (CH).
(c) Isomer C has two peaks in its 1°C NMR spectrum: 8 12.0 (CH2) and 16.8 (CH). The peak at lower
field is only half as intense as the one at higher field.
10
9.
8
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6.
5
4
3
2
1
Chemical shift (8, ppm)
(a) The 'H-NMR spectrum of cyclobutanone shows two signals - signal A at 3.00 ppm and signal B at 1.95 ppm. Give the multiplicity of each signal. cyclobutanone
(b) When cyclobutanone is treated with D20 and NaOD, the only signal observable in the 1H-NMR is a singlet at 2.00 ppm. Explain why this is the case. [Note: Deuterium atoms do not display signals in the TH-NMR spectrum]
13. (a) Compound C undergoes a reaction with SOC12 (or PCls) to
yield compound D (Molecular Formula C₁0H₁0OCIBr)
which has the following spectral data:
Compound D: IR: 1685 cm¹; ¹H NMR: 8 7.84 (d, J = 8 Hz,
2H), 7.60 (d, J = 8 Hz, 2H), 3.65 (t, J = 7 Hz, 2H), 3.18 (t, J
= 7 Hz, 2H), 2.25 (pentet, J = 7 Hz, 2H) ppm; ¹³C NMR: d
28, 36, 45, 128, 130, 133, 137, 197 ppm; EI MS m/z: 200,
198(1:1), 185, 183 (1:1). Identify compound C from the
spectral data of compound B and justify your observation.
Chapter 14 Solutions
Organic Chemistry
Ch. 14 - PRACTICE PROBLEM 14.1 Provide a name for each of...Ch. 14 - Prob. 2PPCh. 14 - Prob. 3PPCh. 14 - Practice Problem 14.4 Apply the polygon-and-circle...Ch. 14 - Practice Problem 14.5 Apply the polygon-and-circle...Ch. 14 - Practice Problem 14.6 1,3,5-Cycloheptatriene is...Ch. 14 - Prob. 7PPCh. 14 - Prob. 8PPCh. 14 - Practice Problem 14.9 In 1967 R. Breslow (of...Ch. 14 - Prob. 10PP
Ch. 14 - Practice Problem 14.11 In addition to a signal...Ch. 14 - PRACTICE PROBLEM 14.12
Azulene has an appreciable...Ch. 14 - Practice Problem 14.13 (a) The -Sh group is...Ch. 14 - Practice Problem 14.14
Explain how NMR...Ch. 14 - PRACTICE PROBLEM 14.15 Four benzenoid compounds,...Ch. 14 - Prob. 16PCh. 14 - Write structural formulas and give acceptable...Ch. 14 - Prob. 18PCh. 14 - Prob. 19PCh. 14 - Prob. 20PCh. 14 - Which of the hydrogen atoms shown below is more...Ch. 14 - 14.22 The rings below are joined by a double bond...Ch. 14 - Prob. 23PCh. 14 - 14.24 (a) In 1960 T. Katz (Columbia University)...Ch. 14 - Prob. 25PCh. 14 - Prob. 26PCh. 14 - 14.27 5-Chloro-1,3-cyclopentadiene (below)...Ch. 14 - Prob. 28PCh. 14 - Furan possesses less aromatic character than...Ch. 14 - 14.30 For each of the pairs below, predict...Ch. 14 - Assign structures to each of the compounds A, B,...Ch. 14 - Prob. 32PCh. 14 - Give a structure for compound F that is consistent...Ch. 14 - Prob. 34PCh. 14 - Prob. 35PCh. 14 - The IR and 1H NMR spectra for compound X(C8H10)...Ch. 14 - Prob. 37PCh. 14 - Prob. 38PCh. 14 - 14.39 Given the following information, predict the...Ch. 14 - Consider these reactions: The intermediate A is a...Ch. 14 - Prob. 41PCh. 14 - Compound E has the spectral features given below....Ch. 14 - Draw all of the molecular orbitals for...Ch. 14 - Prob. 1LGPCh. 14 - Prob. 2LGPCh. 14 - 3. The NMR signals for the aromatic hydrogens of...Ch. 14 - Prob. 4LGPCh. 14 - Prob. 5LGP
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- 13. (a) Compound C undergoes a reaction with SOC1₂ (or PC15) to yield compound D (Molecular Formula C₁0H₁0OCIBr) which has the following spectral data: Compound D: IR: 1685 cm¹; ¹H NMR: 8 7.84 (d, J = 8 Hz, 2H), 7.60 (d, J=8 Hz, 2H), 3.65 (t, J = 7 Hz, 2H), 3.18 (t, J = 7 Hz, 2H), 2.25 (pentet, J = 7 Hz, 2H) ppm; ¹3C NMR: d 28, 36, 45, 128, 130, 133, 137, 197 ppm; EI MS m/z: 200, 198(1:1), 185, 183 (1:1). Identify compound C from the spectral data of compound B and justify your observation. (b) Differentiate between the following: i) COSY and NOESY ii) COSY 90 and COSY 45arrow_forward13. (a) Compound C undergoes a reaction with SOC12 (or PC15) to yield compound D (Molecular Formula C₁0H10OCIBr) which has the following spectral data: Compound D: IR: 1685 cm ¹; ¹H NMR: 87.84 (d, J = 8 Hz, 2H), 7.60 (d, J = 8 Hz, 2H), 3.65 (t, J = 7 Hz, 2H), 3.18 (t, J = 7 Hz, 2H), 2.25 (pentet, J = 7 Hz, 2H) ppm; ¹³C NMR: d 28, 36, 45, 128, 130, 133, 137, 197 ppm; EI MS m/z: 200, 198(1:1), 185, 183 (1:1). Identify compound C from the spectral data of compound B and justify your observation.arrow_forward2. Explain in detail how you would distinguish between the following set of constitutional isomers using 13C-NMR. U W Tarrow_forward
- Propose a structure consistent with each set of spectral data: a. C4H8Br2: IR peak at 3000–2850 cm−1; NMR (ppm): 1.87 (singlet, 6 H) 3.86 (singlet, 2 H) b.C3H6Br2: IR peak at 3000–2850 cm−1; NMR (ppm): 2.4 (quintet) 3.5 (triplet) c. C5H10O2: IR peak at 1740 cm−1; NMR (ppm): 1.15 (triplet, 3 H) 2.30 (quartet, 2 H) 1.25 (triplet, 3 H) 4.72 (quartet, 2 H) d.C3H6O: IR peak at 1730 cm−1; NMR (ppm): 1.11 (triplet) 2.46 (multiplet) 9.79 (triplet)arrow_forwardAlthough benzene itself absorbs at 128 ppm in its 13C NMR spectrum, the carbons of substituted benzenes absorb either upeld or downeld from this value depending on the substituent. Explain the observed values for the carbon ortho to the given substituent in the monosubstituted benzene derivatives X and Y.arrow_forwardAn unknown compound has a molecular formula of C3H6O2. Its IR spectrum shows a very strong and broad band at 2980 and a strong sharp peak at 1716 cm-1. It exhibits the following signals in its 1H NMR spectrum (ppm): 1.21 (triplet, 3H), 2.48 (quartet, 2H), 11.7 (singlet, 1H); and the following signals in its 13C NMR spectrum (ppm): 8.9, 27.6, 181.5. Draw the structure of the unknown compound.arrow_forward
- Although benzene itself absorbs at 128 ppm in its 13C NMR spectrum, the carbons of substituted benzenes absorb either upfield or downfield from this value depending on the substituent. Explain the observed values for the carbon ortho to the given substituent in the monosubstituted benzene derivatives X and Y.arrow_forwardCompound A has prominent infrared absorptions at 1050, 1786, and 1852 cm and shows a single absorption in the proton NMR spectrum at 8 3.00. When heated gently with methanol, compound B, C,H,O,, is obtained. Compound B has IR absorptions at 2500–3000 (broad), 1730, and 1701 cm-, and its proton NMR spectrum in D,0 consists of resonances at & 2.7 (complex splitting) and & 3.7 (a singlet) in the intensity ratio 4:3. Give the structures A and B, omitting stereochemistry.arrow_forwardA and B, isomers of molecular formula C3H5Cl3, are formed by the radical chlorination of a dihalide C of molecular formula C3H6Cl2. a.Identify the structures of A and B from the following 1H NMR data: Compound A: singlet at 2.23 and singlet at 4.04 ppm Compound B: doublet at 1.69, multiplet at 4.34, and doublet at 5.85 ppm b.What is the structure of C?arrow_forward
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