Exam (elaborations) TEST BANK FOR Organic Chemistry 4th Edition By Francis A. Carey, Robert C. Atkins (Study Guide and Solutions Manual)

CHAPTER 1 CHEMICAL BONDING 1 CHAPTER 2 ALKANES 25 CHAPTER 3 CONFORMATIONS OF ALKANES AND CYCLOALKANES 46 CHAPTER 4 ALCOHOLS AND ALKYL HALIDES 67 CHAPTER 5 STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS 90 CHAPTER 6 REACTIONS OF ALKENES: ADDITION REACTIONS 124 CHAPTER 7 STEREOCHEMISTRY 156 CHAPTER 8 NUCLEOPHILIC SUBSTITUTION 184 CHAPTER 9 ALKYNES 209 CHAPTER 10 CONJUGATION IN ALKADIENES AND ALLYLIC SYSTEMS 230 CHAPTER 11 ARENES AND AROMATICITY 253 CHAPTER 12 REACTIONS OF ARENES: ELECTROPHILIC AROMATIC SUBSTITUTION 279 iii Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website iv CONTENTS CHAPTER 13 SPECTROSCOPY 320 CHAPTER 14 ORGANOMETALLIC COMPOUNDS 342 CHAPTER 15 ALCOHOLS, DIOLS, AND THIOLS 364 CHAPTER 16 ETHERS, EPOXIDES, AND SULFIDES 401 CHAPTER 17 ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP 426 CHAPTER 18 ENOLS AND ENOLATES 470 CHAPTER 19 CARBOXYLIC ACIDS 502 CHAPTER 20 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 536 CHAPTER 21 ESTER ENOLATES 576 CHAPTER 22 AMINES 604 CHAPTER 23 ARYL HALIDES 656 CHAPTER 24 PHENOLS 676 CHAPTER 25 CARBOHYDRATES 701 CHAPTER 26 LIPIDS 731 CHAPTER 27 AMINO ACIDS, PEPTIDES, AND PROTEINS. NUCLEIC ACIDS 752 APPENDIX A ANSWERS TO THE SELF-TESTS 775 APPENDIX B TABLES 821 B-1 Bond Dissociation Energies of Some Representative Compounds 821 B-2 Acid Dissociation Constants 822 B-3 Chemical Shifts of Representative Types of Protons 822 B-4 Chemical Shifts of Representative Carbons 823 B-5 Infrared Absorption Frequencies of Some Common Structural Units 823 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 1 CHAPTER 1 CHEMICAL BONDING SOLUTIONS TO TEXT PROBLEMS 1.1 The element carbon has atomic number 6, and so it has a total of six electrons. Two of these electrons are in the 1s level. The four electrons in the 2s and 2p levels (the valence shell) are the valence electrons. Carbon has four valence electrons. 1.2 Electron configurations of elements are derived by applying the following principles: (a) The number of electrons in a neutral atom is equal to its atomic number Z. (b) The maximum number of electrons in any orbital is 2. (c) Electrons are added to orbitals in order of increasing energy, filling the 1s orbital before any electrons occupy the 2s level. The 2s orbital is filled before any of the 2p orbitals, and the 3s orbital is filled before any of the 3p orbitals. (d) All the 2p orbitals (2px, 2py, 2pz) are of equal energy, and each is singly occupied before any is doubly occupied. The same holds for the 3p orbitals. With this as background, the electron configuration of the third-row elements is derived as follows [2p6  2px 22py 22pz 2]: Na (Z  11) 1s22s22p63s1 Mg (Z  12) 1s22s22p63s2 Al (Z  13) 1s22s22p63s23px 1 Si (Z  14) 1s22s22p63s23px 13py 1 P (Z  15) 1s22s22p63s23px 13py 13pz 1 S (Z  16) 1s22s22p63s23px 23py 13pz 1 Cl (Z  17) 1s22s22p63s23px 23py 23pz 1 Ar (Z  18) 1s22s22p63s23px 23py 23pz 2 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 2 CHEMICAL BONDING 1.3 The electron configurations of the designated ions are: Number of Electrons Electron Configuration Ion Z in Ion of Ion (b) He 2 1 1s1 (c) H 1 2 1s2 (d) O 8 9 1s22s22px 22py 22pz 1 (e) F 9 10 1s22s22p6 ( f) Ca2 20 18 1s22s22p63s23p6 Those with a noble gas configuration are H, F, and Ca2. 1.4 A positively charged ion is formed when an electron is removed from a neutral atom. The equation representing the ionization of carbon and the electron configurations of the neutral atom and the ion is: A negatively charged carbon is formed when an electron is added to a carbon atom. The additional electron enters the 2pz orbital. Neither C nor C has a noble gas electron configuration. 1.5 Hydrogen has one valence electron, and fluorine has seven. The covalent bond in hydrogen fluoride arises by sharing the single electron of hydrogen with the unpaired electron of fluorine. 1.6 We are told that C2H6 has a carbon–carbon bond. There are a total of 14 valence electrons distributed as shown. Each carbon is surrounded by eight electrons. 1.7 (b) Each carbon contributes four valence electrons, and each fluorine contributes seven. Thus, C2F4 has 36 valence electrons. The octet rule is satisfied for carbon only if the two carbons are attached by a double bond and there are two fluorines on each carbon. The pattern of connections shown (below left) accounts for 12 electrons. The remaining 24 electrons are divided equally (six each) among the four fluorines. The complete Lewis structure is shown at right below. (c) Since the problem states that the atoms in C3H3N are connected in the order CCCN and all hydrogens are bonded to carbon, the order of attachments can only be as shown (below left) so as to have four bonds to each carbon. Three carbons contribute 12 valence electrons, three hydrogens contribute 3, and nitrogen contributes 5, for a total of 20 valence electrons. The nine C F F C F F C F F C F F H C H H H H H Thus, we combine two C H C to write the Lewis structure of ethane and six Combine H and F to give the Lewis structure for hydrogen fluoride H F C 1s22s22px 1py 12pz 1 C 1s22s22px 12py 1  e C 1s22s22px 12py 1 C 1s22s22px 1  e Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website CHEMICAL BONDING 3 bonds indicated in the partial structure account for 18 electrons. Since the octet rule is satisfied for carbon, add the remaining two electrons as an unshared pair on nitrogen (below right). 1.8 The degree of positive or negative character at carbon depends on the difference in electronegativity between the carbon and the atoms to which it is attached. From Table 1.2, we find the electronegativity values for the atoms contained in the molecules given in the problem are: Li 1.0 H 2.1 C 2.5 Cl 3.0 Thus, carbon is more electronegative than hydrogen and lithium, but less electronegative than chlorine. When bonded to carbon, hydrogen and lithium bear a partial positive charge, and carbon bears a partial negative charge. Conversely, when chlorine is bonded to carbon, it bears a partial negative charge, and carbon becomes partially positive. In this group of compounds, lithium is the least electronegative element, chlorine the most electronegative. 1.9 (b) The formal charges in sulfuric acid are calculated as follows: Valence Electrons in Neutral Atom Electron Count Formal Charge Hydrogen: 1  1 2 (2) 1 0 Oxygen (of OH): 6  1 2 (4)  4 6 0 Oxygen: 6  1 2 (2)  6  7 1 Sulfur: 6  1 2 (8)  0  4 2 (c) The formal charges in nitrous acid are calculated as follows: Valence Electrons in Neutral Atom Electron Count Formal Charge Hydrogen: 1  1 2 (2) 1 0 Oxygen (of OH): 6  1 2 (4)  4 6 0 Oxygen: 6  1 2 (4)  4 6 0 Nitrogen: 5  1 2 (6)  2 5 0 H O N O S2 O O H O O H   H H H C Li H H H C H H H H C Cl Methyllithium; most negative character at carbon Chloromethane; most positive character at carbon N C H H C H C N C C H H H C Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 1.10 The electron counts of nitrogen in ammonium ion and boron in borohydride ion are both 4 (one half of 8 electrons in covalent bonds). Since a neutral nitrogen has 5 electrons in its valence shell, an electron count of 4 gives it a formal charge of 1. A neutral boron has 3 valence electrons, and so an electron count of 4 in borohydride ion corresponds to a formal charge of 1. 1.11 As shown in the text in Table 1.2, nitrogen is more electronegative than hydrogen and will draw the electrons in N@H bonds toward itself. Nitrogen with a formal charge of 1 is even more electronegative than a neutral nitrogen. Boron (electronegativity  2.0) is, on the other hand, slightly less electronegative than hydrogen (electronegativity  2.1). Boron with a formal charge of 1 is less electronegative than a neutral boron. The electron density in the B@H bonds of BH4  is therefore drawn toward hydrogen and away from boron. 1.12 (b) The compound (CH3)3CH has a central carbon to which are attached three CH3 groups and a hydrogen. Four carbons and 10 hy