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

Exam (elaborations) TEST BANK FOR Organic Chemistry 4th Edition By Francis A. Carey, Robert C. Atkins (Study Guide and Solutions Manual) CONTENTS Preface v To the Student vii 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 hydrogens contribute 26 valence electrons. The structure shown has 13 covalent bonds, and so all the valence electrons are accounted for. The molecule has no unshared electron pairs. (c) The number of valence electrons in ClCH2CH2Cl is 26 (2Cl  14; 4H  4; 2C  8). The constitution at the left below shows seven covalent bonds accounting for 14 electrons. The remaining 12 electrons are divided equally between the two chlorines as unshared electron pairs. The octet rule is satisfied for both carbon and chlorine in the structure at the right below. H H C H H Cl C Cl H H C H H Cl C Cl H H H C H H C H H H C H H C H H H B H  H H  B  H  H  H H H N H  H H  N  H  H  H H N H H H H B H H Ammonium ion Borohydride ion 4 CHEMICAL BONDING Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website CHEMICAL BONDING 5 (d) This compound has the same molecular formula as the compound in part (c), but a different structure. It, too, has 26 valence electrons, and again only chlorine has unshared pairs. (e) The constitution of CH3NHCH2CH3 is shown (below left). There are 26 valence electrons, and 24 of them are accounted for by the covalent bonds in the structural formula. The remaining two electrons complete the octet of nitrogen as an unshared pair (below right). ( f ) Oxygen has two unshared pairs in (CH3)2CHCH?O. 1.13 (b) This compound has a four-carbon chain to which are appended two other carbons. (c) The carbon skeleton is the same as that of the compound in part (b), but one of the terminal carbons bears an OH group in place of one of its hydrogens. (d) The compound is a six-membered ring that bears a @C(CH3)3 substituent. 1.14 The problem specifies that nitrogen and both oxygens of carbamic acid are bonded to carbon and one of the carbon–oxygen bonds is a double bond. Since a neutral carbon is associated with four which may be rewritten as H C C C C C C H H C H H H H H H H H CH3 CH3 CH3 is equivalent to C(CH3)3 CH2OH CH3CHCH(CH3)2 H CH3 H CH C 3 HO H C CH3 H C is equivalent to which may be rewritten as HO CH3 CH3 H CH3 H C CHC 3 (CH3)2CHCH(CH3)2 is equivalent to which may be rewritten as H H H C H H C H H C H C O H H H C H H H C H H H N C H H H C H H H C H H N C H H H C H H C Cl Cl Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website bonds, a neutral nitrogen three (plus one unshared electron pair), and a neutral oxygen two (plus two unshared electron pairs), this gives the Lewis structure shown. 1.15 (b) There are three constitutional isomers of C3H8O: (c) Four isomers of C4H10O have @OH groups: Three isomers have C@O@C units: 1.16 (b) Move electrons from the negatively charged oxygen, as shown by the curved arrows. The resonance interaction shown for bicarbonate ion is more important than an alternative one involving delocalization of lone-pair electrons in the OH group. (c) All three oxygens are equivalent in carbonate ion. Either negatively charged oxygen can serve as the donor atom. O O O C O O C   O   O O C O O C  O    O O O O C H O O O C H     Not equivalent to original structure; not as stable because of charge separation O O O C H O O O C H   Equivalent to original structure CH3OCHCH3 CH3 CH3CH2OCH2CH3 CH3OCH2CH2CH3 CH3CH2CH2CH2OH OH CH3CHCH2CH3 CH3CHCH2OH CH3 CH3COH CH3 CH3 CH3CH2OCH3 OH CH3CHCHCH 3 3CH2CH2OH Carbamic acid H H N C O H O 6 CHEMICAL BONDING Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website CHEMICAL BONDING 7 (d) Resonance in borate ion is exactly analogous to that in carbonate. and 1.17 There are four B@H bonds in BH4 . The four electron pairs surround boron in a tetrahedral orientation. The H@B@H angles are 109.5°. 1.18 (b) Nitrogen in ammonium ion is surrounded by 8 electrons in four covalent bonds. These four bonds are directed toward the corners of a tetrahedron. (c) Double bonds are treated as a single unit when deducing the shape of a molecule using the VSEPR model. Thus azide ion is linear. (d) Since the double bond in carbonate ion is treated as if it were a single unit, the three sets of electrons are arranged in a trigonal planar arrangement around carbon. 1.19 (b) Water is a bent molecule, and so the individual O@H bond dipole moments do not cancel. Water has a dipole moment. (c) Methane, CH4, is perfectly tetrahedral, and so the individual (small) C@H bond dipole moments cancel. Methane has no dipole moment. (d) Methyl chloride has a dipole moment. C Cl H H C Cl H H Directions of bond dipole moments in CH3Cl Direction of molecular dipole moment H H O H H O H H Individual OH bond moments in water Direction of net dipole moment C O O O The OCO angle is 120º.   N N  N The NNN angle is 180°. Each HNH angle is 109.5º.  N H H H H O O OB B O  O O      O O O B B O O  O    Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (e) Oxygen is more electronegative than carbon and attracts electrons from it. Formaldehyde has a dipole moment. ( f ) Nitrogen is more electronegative than carbon. Hydrogen cyanide has a dipole moment. 1.20 The orbital diagram for sp3-hybridized nitrogen is the same as for sp3-hybridized carbon, except nitrogen has one more electron. The unshared electron pair in ammonia (•• NH3) occupies an sp3-hybridized orbital of nitrogen. Each N@H bond corresponds to overlap of a half-filled sp3 hybrid orbital of nitrogen and a 1s orbital of hydrogen. 1.21 Silicon lies below carbon in the periodic table, and it is reasonable to assume that both carbon and silicon are sp3-hybridized in H3CSiH3. The C@Si bond and all of the C@H and Si@H bonds are  bonds. The principal quantum number of the carbon orbitals that are hybridized is 2; the principal quantum number for the silicon orbitals is 3. 1.22 (b) Carbon in formaldehyde (H2C?O) is directly bonded to three other atoms (two hydrogens and one oxygen). It is sp2-hybridized. (c) Ketene has two carbons in different hybridization states. One is sp2-hybridized; the other is sp-hybridized. H2C C O Bonded to three atoms: sp2 Bonded to two atoms: sp C H H H H H Si H C(2sp3) Si(3sp3)  bond C(2sp3) H(1s)  bond Si(3sp3) H(1s)  bond Ground electronic state of nitrogen sp3 hybrid state of nitrogen (a) (b) Energy 2s 2p 2sp3 H C N Direction of bond dipole moments in HCN H C N Direction of molecular dipole moment Direction of bond dipole moments in formaldehyde Direction of molecular dipole moment C O H H C O H H 8 CHEMICAL BONDING Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website CHEMICAL BONDING 9 (d) One of the carbons in propene is sp3-hybridized. The carbons of the double bond are sp2-hybridized. (e) The carbons of the CH3 groups in acetone [(CH3)2C?O] are sp3-hybridized. The C?O carbon is sp2-hybridized. ( f ) The carbons in acrylonitrile are hybridized as shown: 1.23 All these species are characterized by the formula •• XY• •, and each atom has an electron count of 5. Electron count X  electron count Y  2  3  5 (a) A neutral nitrogen atom has 5 valence electrons: therefore, each atom is electrically neutral in molecular nitrogen. (b) Nitrogen, as before, is electrically neutral. A neutral carbon has 4 valence electrons, and so carbon in this species, with an electron count of 5, has a unit negative charge. The species is cyanide anion; its net charge is 1. (c) There are two negatively charged carbon atoms in this species. It is a dianion; its net charge is 2. (d) Here again is a species with a neutral nitrogen atom. Oxygen, with an electron count of 5, has 1 less electron in its valence shell than a neutral oxygen atom. Oxygen has a formal charge of 1; the net charge is 1. (e) Carbon has a formal charge of 1; oxygen has a formal charge of 1. Carbon monoxide is a neutral molecule. 1.24 All these species are of the type •• Y• • ?X?Y• • • •. Atom X has an electron count of 4, corresponding to half of the 8 shared electrons in its four covalent bonds. Each atom Y has an electron count of 6; 4 unshared electrons plus half of the 4 electrons in the double bond of each Y to X. (a) Oxygen, with an electron count of 6, and carbon, with an electron count of 4, both correspond to the respective neutral atoms in the number of electrons they “own.” Carbon dioxide is a neutral molecule, and neither carbon nor oxygen has a formal charge in this Lewis structure. (b) The two terminal nitrogens each have an electron count (6) that is one more than a neutral atom and thus each has a formal charge of 1. The central N has an electron count (4) that is one less than a neutral nitrogen; it has a formal charge of 1. The net charge on the species is (1  1  1), or 1. (c) As in part (b), the central nitrogen has a formal charge of 1. As in part (a), each oxygen is electrically neutral. The net charge is 1. 1.25 (a, b) The problem specifies that ionic bonding is present and that the anion is tetrahedral. The cations are the group I metals Na and Li. Both boron and aluminum are group III O N O N N N O C O C O N O C C C N N N Unshared electron pair contributes 2 electrons to electron count of X. Unshared electron pair contributes 2 electrons to electron count of Y. Triple bond contributes half of its 6 electrons, or 3 electrons each, to separate electron counts of X and Y. X Y H CH C N 2C sp2 sp2 sp CH CH2 H3C sp3 sp2 sp2 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website elements, and thus have a formal charge of 1 in the tetrahedral anions BF4  and AlH4  respectively. (c, d) Both of the tetrahedral anions have 32 valence electrons. Sulfur contributes 6 valence electrons and phosphorus 5 to the anions. Each oxygen contributes 6 electrons. The double negative charge in sulfate contributes 2 more, and the triple negative charge in phosphate contributes 3 more. The formal charge on each oxygen in both ions is 1. The formal charge on sulfur in sulfate is 2; the charge on phosphorus is 1. The net charge of sulfate ion is 2; the net charge of phosphate ion is 3. 1.26 (a) Each hydrogen has a formal charge of 0, as is always the case when hydrogen is covalently bonded to one substituent. Oxygen has an electron count of 5. A neutral oxygen atom has 6 valence electrons; therefore, oxygen in this species has a formal charge of 1. The species as a whole has a unit positive charge. It is the hydronium ion, H3O. (b) The electron count of carbon is 5; there are 2 electrons in an unshared pair, and 3 electrons are counted as carbon’s share of the three covalent bonds to hydrogen. An electron count of 5 is one more than that for a neutral carbon atom. The formal charge on carbon is 1, as is the net charge on this species. (c) This species has 1 less electron than that of part (b). None of the atoms bears a formal charge. The species is neutral. (d) The formal charge of carbon in this species is 1. Its only electrons are those in its three covalent bonds to hydrogen, and so its electron count is 3. This corresponds to 1 less electron than in a neutral carbon atom, giving it a unit positive charge. Electron count of carbon  1 1 (6)  4 2 H C H H Unshared electron Electrons shared in covalent bonds H C H H Two electrons “owned” by carbon. One of the electrons in eachC Hbond “belongs” to carbon. H O H H Electron count of oxygen  2  1 (6)  5 2 Unshared pair Covalently bonded electrons S2 O O