Organic Chemistry
12th Edition
ISBN: 9781118875766
Author: T. W. Graham Solomons, Craig B. Fryhle, Scott A. Snyder
Publisher: WILEY
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Chapter 16, Problem 52P
Compound B
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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.
Compound B of molecular formula C9H19N shows a noteworthy infrared absorption at 3300 cm-1. Its 1H-NMR spectrum shows three singlets – δ 1.0 (6H), 1.1 (12H), 1.4 (1H) ppm. Its 13C-NMR spectrum has four signals – δ 25, 28, 41, 64 ppm. Suggest a structure for this compound.
The 1H-NMR spectrum of Compound D of molecular formula C10H12O shows three singlets – δ 2.20 (6H, s), 4.86 (4H), 7.10 (2H) ppm. Its 13C-NMR spectrum has five signals – 20, 74, 127, 135, 146 ppm. Suggest a structure for this compound.
Chapter 16 Solutions
Organic Chemistry
Ch. 16 - PRACTICE PROBLEM 16.1 (a) Give IUPAC substitutive...Ch. 16 - Prob. 2PPCh. 16 - Prob. 3PPCh. 16 - Practice Problem 16.4
Provide the reagents and...Ch. 16 - Prob. 5PPCh. 16 - Prob. 6PPCh. 16 - Prob. 7PPCh. 16 - Prob. 8PPCh. 16 - Prob. 9PPCh. 16 - Practice Problem 16.10
Shown below is the...
Ch. 16 - Prob. 11PPCh. 16 - Practice Problem 16.12
What product would be...Ch. 16 - Prob. 13PPCh. 16 - Practice Problem 16.14
Dihydropyran reacts readily...Ch. 16 - Practice Problem 16.15 Show how you might use...Ch. 16 - Practice Problem 16.16 (a) Show how you might...Ch. 16 - Practice Problem 16.17
In addition to...Ch. 16 - Practice Problem 16.18
Triphenylphosphine can be...Ch. 16 - Prob. 19PPCh. 16 - PRACTICE PROBLEM 16.20
Give the structure of the...Ch. 16 - PRACTICE PROBLEM 16.21 What would be the major...Ch. 16 - Prob. 22PCh. 16 - 16.23 Write structural formulas for the products...Ch. 16 - Give structural formulas for the products formed...Ch. 16 - 16.25 What products would be obtained when...Ch. 16 - Predict the major organic product from each of the...Ch. 16 - 16.27 Predict the major product from each of the...Ch. 16 - 16.28 Predict the major product from each of the...Ch. 16 - Prob. 29PCh. 16 - 16.30 Write detailed mechanisms for each of the...Ch. 16 - Prob. 31PCh. 16 - Prob. 32PCh. 16 - Show how you would convert benzaldehyde into each...Ch. 16 - 16.34 Show how ethyl phenyl ketone could be...Ch. 16 - Show how benzaldehyde could be synthesized from...Ch. 16 - Give structures for compounds AE. Cyclohexanol...Ch. 16 - Prob. 37PCh. 16 - Prob. 38PCh. 16 - Prob. 39PCh. 16 - Prob. 40PCh. 16 - Prob. 41PCh. 16 - Prob. 42PCh. 16 - 16.43 The structure of the sex pheromone...Ch. 16 - Provide reagents that would accomplish each of the...Ch. 16 - Write a detailed mechanism for the following...Ch. 16 - Prob. 46PCh. 16 - Dutch elm disease is caused by a fungus...Ch. 16 - Prob. 48PCh. 16 - Compounds W and X are isomers; they have the...Ch. 16 - Compounds Y and Z are isomers with the molecular...Ch. 16 - Compound A (C9H18O) forms a phenylhydrazone, but...Ch. 16 - Compound B (C8H12O2) shows a strong carbonyl...Ch. 16 - Prob. 53PCh. 16 - Prob. 54PCh. 16 - Prob. 55PCh. 16 - (a) What would be the frequencies of the two...Ch. 16 - Prob. 57PCh. 16 - Prob. LGP
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- A hydrocarbon, compound B, has molecular formula C6H6, and gave an NMR spectrum with two signals: delta 6.55 pm and delta 3.84 pm with peak ratio of 2:1. When warmed in pyridine for three hr, compound B quantitatively converts to benzene. Mild hydrogenation of B yielded another compound C with mass spectrum of m/z 82. Infrared spectrum showed no double bonds; NMR spectrum showed one broad peak at delta 2.34 ppm. With this information, address the following questions. a) How many rings are in compound C? b) How many rings are probably in B? How many double bonds are in B? c) Can you suggest a structure for compounds B and C? d) In the NMR spectrum of B, the up-field signal was a quintet, and the down field signal was a triplet. How must you account for these splitting patterns?arrow_forwardAn unknown compound C3H2NCl shows moderately strong IR absorptions around 1650 cm-1 and 2200 cm-1. Its NMR spectrum consists of two doublets (J = 14 Hz) at δ 5.9 and δ 7.1. Propose a structure consistent with this data?arrow_forwardCompound B of molecular formula C9H19N shows a noteworthy infrared absorption at 3300 cm-1. Its 1H-NMR spectrum shows three singlets – δ 1.0 (6H), 1.1 (12H), 1.4 (1H) ppm. Its 13C-NMR spectrum has four signals – δ 25, 28, 41, 64 ppm. Suggest a structure for this compound. Please show work.arrow_forward
- Compounds A and B are isomers having the molecular formula C4H8O3. Identify A and B on the basis of their 1H NMR spectra.Compound A: δ 1.3 (3H, triplet); 3.6 (2H, quartet); 4.1 (2H, singlet); 11.1 (1H, broad singlet)Compound B: δ 2.6 (2H, triplet); 3.4 (3H, singlet); 3.7 (2H triplet); 11.3 (1H, broad singlet)arrow_forwardCompound P has molecular formula C5H9ClO2. Deduce the structure of P from its 1H and 13C NMR spectra.arrow_forwardA and B are isomeric dicarbonyl compounds of the molecular formula C5H&O2. The 'H NMR spectrum of A contains a singlet at 2.05 ppm and another singlet at 5.40 ppm. The 'H NMR spectrum of B contains three signals: a singlet at 2.3 ppm, a triplet at 1.10 ppm and a quartet at 2.70 ppm. Suggest structures for A and B and draw them in their respective boxes below. 1st attemptarrow_forward
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