Organic Chemistry
12th Edition
ISBN: 9781118875766
Author: T. W. Graham Solomons, Craig B. Fryhle, Scott A. Snyder
Publisher: WILEY
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Chapter 23, Problem 20P
Propose structures for
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Alpha-phellandrene, C₁0H₁6, is a pleasant-smelling chiral compound that may be obtained from the leaves
of eucalyptus trees. It has the molecular formula C₁0H16 and a UV absorption maximum at 263 nm. On
hydrogenation with a palladium catalyst, an achiral hydrocarbon with the formula C₁0H20 is obtained.
Ozonolysis of alpha-phellandrene, followed by treatment with zinc and acetic acid, produces the
following two compounds:
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CH3C-CH
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Propose a structure for alpha-phellandrene.
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HCCH₂CHCH
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CH(CH3)2
• You do not have to consider stereochemistry.
You do not have to explicitly draw H atoms.
• In cases where there is more than one answer, just draw one.
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There are several isomeric alcohols and ethers of molecular formula C5H12O. Two of these exhibit the following 1H-NMR spectra. Propose a structure for each of the isomers.
Isomer A: δ = 0.92 (t, 7.8 Hz, 3 H), 1.20 (s, 6H), 1.49 (q, 7.8 Hz, 2H), 1.85 (s, 1H) ppm
Isomer B: δ = 1.19 (s, 9 H), 3.21 (s, 3H) ppm
Compounds Y and Z are isomers with the molecular formula C10H12O. The IR spectrum of each compound shows a strong absorption band near 1710 cm−1 . The 1H NMR spectra of Y and Z are given below. Propose structures for Y and Z.
Chapter 23 Solutions
Organic Chemistry
Ch. 23 - Prob. 1PPCh. 23 - Prob. 2PPCh. 23 - Prob. 3PPCh. 23 - Prob. 4PPCh. 23 - Prob. 5PPCh. 23 - Prob. 6PPCh. 23 - Prob. 7PPCh. 23 - Prob. 8PPCh. 23 - Prob. 9PPCh. 23 - Prob. 10PP
Ch. 23 - Prob. 11PPCh. 23 - Prob. 12PPCh. 23 - Prob. 13PPCh. 23 - Prob. 14PCh. 23 - 23.15 How would you transform tetradecanal into...Ch. 23 - Prob. 16PCh. 23 - Prob. 17PCh. 23 - When limonene (Section 23.3) is heated strongly,...Ch. 23 - Gadoleic acid (C20H38O2), a fatty acid that can be...Ch. 23 - 23.20 -Phellandrene and -phellandrene are isomeric...Ch. 23 - Prob. 21PCh. 23 - Prob. 22PCh. 23 - Prob. 23PCh. 23 - The initial steps of a laboratory synthesis of...Ch. 23 - Prob. 25PCh. 23 - Prob. 26PCh. 23 - Prob. 27PCh. 23 - 2. The biosynthesis of fatty acids is accomplished...Ch. 23 - Prob. 3LGPCh. 23 - Prob. 4LGP
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- Treatment of a hydrocarbon A (molecular formula C9H18) with Br2 in the presence of light forms alkyl halides B and C, both having molecular formula C9H17Br. Reaction of either B or C with KOC(CH3)3 forms compound D (C9H16) as the major product. Ozonolysis of D forms cyclohexanone and acetone. Identify the structures of A–D.arrow_forwardCompounds 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_forwardAn unknown compound has a molecular formula of C4H6O2. Its IR spectrum shows absorptions at 3095, 1762, 1254, and 1118 cm -1. It exhibits the following signals in its 1H NMR spectrum (ppm): 2.12 (singlet,3H), 4.55 (doublets of doublets, 1H), 4.85 (doublet of doublets, 1H), 7.25 (doublets of doublets, 1H); and the following signals in its 13C NMR spectrum (ppm): 20.8, 100.4, 141.2, 168.0. Draw the structure of the unknown compoundarrow_forward
- Compound 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_forwardIn an aqueous solution containing sodium bicarbonate, aniline reacts quickly withbromine to give 2,4,6-tribromoaniline. Nitration of aniline requires very strong conditions,however, and the yields (mostly m-nitroaniline) are poor.(a) What conditions are used for nitration, and what form of aniline is present under theseconditions?arrow_forward
- 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_forwardReaction of (CH3)3CCHO with (C6H5)3P=C(CH3)OCH3, followed by treatment with aqueous acid, affords R (C7H14O). R has a strong absorption in its IR spectrum at 1717 cm−1 and three singlets in its 1H NMR spectrum at 1.02 (9 H), 2.13 (3 H), and 2.33 (2 H) ppm. What is the structure of R? We will learn about this reaction in Chapter 18.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.arrow_forward
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