Organic Chemistry
9th Edition
ISBN: 9781305080485
Author: John E. McMurry
Publisher: Cengage Learning
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Textbook Question
Chapter 15.2, Problem 4P
Pyridine is a flat, hexagonal molecule with bond angles of 120°. It undergoes substitution rather than addition and generally behaves like benzene. Draw a picture of the π orbitals of pyridine to explain its properties. Check your answer by looking ahead to section 15-5
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4E) Formic acid (HCO2H) and piperidine (C5H11N)
Chapter 15 Solutions
Organic Chemistry
Ch. 15.1 - Prob. 1PCh. 15.1 - Give IUPAC names for the following compounds:Ch. 15.1 - Prob. 3PCh. 15.2 - Pyridine is a flat, hexagonal molecule with bond...Ch. 15.3 - Prob. 5PCh. 15.4 - Draw the five resonance structures of the...Ch. 15.4 - Prob. 7PCh. 15.4 - Prob. 8PCh. 15.5 - Prob. 9PCh. 15.5 - Prob. 10P
Ch. 15.6 - Prob. 11PCh. 15.6 - How many electrons does each of the four nitrogen...Ch. 15.SE - Give IUPAC names for the following substances (red...Ch. 15.SE - All-cis cyclodecapentaene is a stable molecule...Ch. 15.SE - 1, 6-Methanonaphthalene has an interesting 1H NMR...Ch. 15.SE - Prob. 16VCCh. 15.SE - Azulene, an isomer of naphthalene, has a...Ch. 15.SE - Give IUPAC names for the following compounds:Ch. 15.SE - Draw structures corresponding to the following...Ch. 15.SE - Prob. 20APCh. 15.SE - Prob. 21APCh. 15.SE - Draw and name all possible aromatic compounds with...Ch. 15.SE - Propose structures for aromatic hydrocarbons that...Ch. 15.SE - Look at the three resonance structures of...Ch. 15.SE - Prob. 25APCh. 15.SE - Prob. 26APCh. 15.SE - Look at the five resonance structures for...Ch. 15.SE - Prob. 28APCh. 15.SE - 3-Chlorocyclopropene, on treatment with AgBF4,...Ch. 15.SE - Prob. 30APCh. 15.SE - Prob. 31APCh. 15.SE - Prob. 32APCh. 15.SE - Which would you expect to be most stable,...Ch. 15.SE - How might you convert 1, 3, 5, 7-cyclononatetraene...Ch. 15.SE - Calicene, like azulene (Problem 15-17), has an...Ch. 15.SE - Pentalene is a most elusive molecule that has been...Ch. 15.SE - Prob. 37APCh. 15.SE - Prob. 38APCh. 15.SE - Compound A, C8H10, yields three substitution...Ch. 15.SE - Prob. 40APCh. 15.SE - Propose structures for compounds that fit the...Ch. 15.SE - Prob. 42APCh. 15.SE - Prob. 43APCh. 15.SE - N-Phenylsydnone, so-named because it was first...Ch. 15.SE - Prob. 45APCh. 15.SE - Prob. 46APCh. 15.SE - Prob. 47APCh. 15.SE - Propose a structure for a molecule C14H12 that has...Ch. 15.SE - The proton NMR spectrum for a compound with...Ch. 15.SE - The proton NMR spectrum of a compound with formula...Ch. 15.SE - Aromatic substitution reactions occur by addition...Ch. 15.SE - Prob. 52APCh. 15.SE - Consider the aromatic anions below and their...Ch. 15.SE - After the reaction below, the chemical shift of Ha...Ch. 15.SE - Prob. 55APCh. 15.SE - Azo dyes are the major source of artificial color...
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- 16-17 Propylamine (bp 48°C), ethylmethylamine (bp 37°C), and trimethylamine (bp 3°C) are constitutional isomers with the molecular formula C3HgN. Account for the fact that trimethylamine has the lowest boiling point of the three and propylamine has the highest.arrow_forward13-43 (Chemical Connections 13E) Which features of Allura Red and Sunset Yellow make them water-soluble?arrow_forward16-6 Answer true or false. te/7-Butylamine is a 3° amine. In an aromatic amine, one or more of the groups bonded to nitrogen is an aromatic ring. In a heterocyclic amine, the amine nitrogen is one of the atoms of a ring. The Lewis structures of both NH4~ and CH4show the same number (eight) of valence electrons, and the VSEPR model predicts tetrahedral geometry for each. There are four constitutional isomers with the molecular formula CgH^N.arrow_forward
- 13-57 Benzene, as we have seen in this chapter, is the simplest aromatic compound. Pyidine is an analog of benzene in which a CH group is replaced by a nitrogenarrow_forward13-20 Three products with the molecular formula CgH^BrCl form when bromobenzene is treated with chlorine, Cl2, in the presence of FeCl3 as a catalyst. Name and draw a structural formula for each product.arrow_forward17-27 Pentane, 1-butanol, and butanal all have approximately the same molecular weights but different boiling points. Arrange them in order of increasing boiling point. Explain the basis for your ranking.arrow_forward
- 13-29 Show that if you add Steps 2a and 2b of the radical- chain mechanism for the autoxidation of a fatty acid hydrocarbon chain, you arrive at the following net equation: H I —CH2CH=CH—CH— + 0—0 • » Section of a fatty acid Oxygen hydrocarbon chain O—O—H I —CH2CH=CH—CH— A hydroperoxidearrow_forward10-22 Suppose you forget to take into account the presence of the unshared pair of electrons on nitrogen in the molecule NH3. What would you then predict for the H—N—H bond angles and the geometry (bond angles and shape) of ammonia?arrow_forward13-44 (Chemical Connections 13E) What color would you get if you mixed Allura Red and Sunset Yellow? (Hint: Remember the color wheel.)arrow_forward
- 13-56 Four alternatives to the structure of benzene proposed by Kekule are shown in the text. In this problem, we ask you to propose additional alternatives for compounds with the molecular formula C6Hg. As you think about possible alternatives keep in mind the building blocks available to you. Also keep in mind that carbon must be tetravalent, that is, each carbon must have no more and no fewer than four bonds. Three- four-, five-, and six-membered rings Carbon—carbon single, double, and triple bonds The possibility for cis /trans isomerism for carbon-carbon double bondsarrow_forward13-10 Account for the fact that the six-membered ring in benzene is planar but the six-membered ring in cyclohexane is not.arrow_forward14-30 Show how to distinguish between cyclohexanol and cyclohexene by a simple chemical test. Tell what you would do, what you would expect to see, and how you would interpret your observation.arrow_forward
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