Exam (elaborations) TEST BANK FOR Organic Chemistry 3rd Edition By Jan

Exam (elaborations) TEST BANK FOR Organic Chemistry 3rd Edition By Jan 1. Structure and Bonding: Chapter 1 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 1 2. Acids and Bases: Chapter 2 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 33 3. Introduction to Organic Molecules and Functional Groups: Chapter 3 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 57 4. Alkanes: Chapter 4 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 75 5. Stereochemistry: Chapter 5 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 111 6. Understanding Organic Reactions: Chapter 6 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 139 7. Alkyl Halides and Nucleophilic Substitution: Chapter 7 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 159 8. Alkyl Halides and Elimination Reactions: Chapter 8 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 193 9. Alcohols, Ethers, and Epoxides: Chapter 9 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 223 10. Alkenes: Chapter 10 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 257 11. Alkynes: Chapter 11 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 287 12. Oxidation and Reduction: Chapter 12 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 309 13. Mass Spectrometry and Infrared Spectroscopy: Chapter 13 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 337 14. Nuclear Magnetic Resonance Spectroscopy: Chapter 14 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 351 15. Radical Reactions: Chapter 15 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 373 16. Conjugation, Resonance, and Dienes: Chapter 16 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 397 17. Benzene and Aromatic Compounds: Chapter 17 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 421 18. Electrophilic and Aromatic Substitution: Chapter 18 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 443 19. Carboxylic Acids and the Acidity of the O-H Bond: Chapter 19 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 479 iv 20. Introduction to Carbonyl Chemistry: Chapter 20 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 501 21. Aldehydes and Ketones — Nucleophilic Addition: Chapter 21 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 535 22. Carboxylic Acids and Their Derivatives — Nucleophilic Acyl Substitution: Chapter 22 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 567 23. Substitution Reactions of Carbonyl Compounds at the a Carbon: Chapter 23 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 603 24. Carbonyl Condensation Reactions: Chapter 24 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 631 25. Amines: Chapter 25 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 659 26. Carbon-Carbon Bonding-Forming Reactions in Organic Synthesis: Chapter 26 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 693 27. Carbohydrates: Chapter 27 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 715 28. Amino Acids and Proteins: Chapter 28 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 751 29. Lipids: Chapter 29 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 785 30. Synthetic Polymers: Chapter 30 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 801 v Smith: Study Guide/ Solutions Manual to accompany Organic Chemistry, Third Edition 1. Structure and Bonding Text © The McGraw−Hill Companies, 2011 Structure and Bonding 1–1 CCChhhaaapppttteeerrr111::: SSStttrrruuuccctttuuurrreeeaaannndddBBBooonnndddi iinnnggg


IIImmmpppooorrrtttaaannntttfffaaaccctttsss • The general rule of bonding: Atoms strive to attain a complete outer shell of valence electrons (Section 1.2). H “wants” 2 electrons. Second-row elements “want” 8 electrons. H C N O X X = F, Cl, Br, I nonbonded electron pair Usual number of bonds in neutral atoms Number of nonbonded electron pairs 14 3 2 1 00 1 2 3 The sum (# of bonds + # of lone pairs) = 4 for all elements except H. • Formal charge (FC) is the difference between the number of valence electrons of an atom and the number of electrons it “owns” (Section 1.3C). See Sample Problem 1.4 for a stepwise example. formal charge = number of valence electrons – number of electrons an atom "owns" C C C • C shares 8 electrons. • C "owns" 4 electrons. • FC = 0 • Each C shares 6 electrons. • Each C "owns" 3 electrons. • FC = +1 • C shares 6 electrons. • C has 2 unshared electrons. • C "owns" 5 electrons. • FC =
1 Definition: Examples: • Curved arrow notation shows the movement of an electron pair. The tail of the arrow always begins at an electron pair, either in a bond or a lone pair. The head points to where the electron pair “moves” (Section 1.5). H C O N H H C O N H A B Move an electron pair to O. Use this electron pair to form a double bond. • Electrostatic potential plots are color-coded maps of electron density, indicating electron rich and electron deficient regions (Section 1.11). Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition 1 Smith: Study Guide/ Solutions Manual to accompany Organic Chemistry, Third Edition 1. Structure and Bonding Text © The McGraw−Hill Companies, 2011 Chapter 1–2 TTThhheee iiimmmpppooorrrtttaaannnccceeeooofffLLLeeewwwi iisssssstttrrruuuccctttuuurrreeesss(((SSSeeeccct tti iiooonnnsss111...333,,, 111...444))) A properly drawn Lewis structure shows the number of bonds and lone pairs present around each atom in a molecule. In a valid Lewis structure, each H has two electrons, and each second-row element has no more than eight. This is the first step needed to determine many properties of a molecule. Lewis structure Geometry Hybridization Types of bonds (Section 1.6) (Section 1.8) (Sections 1.3, 1.9) [linear, trigonal planar, or tetrahedral] [sp, sp2, or sp3] [single, double, or triple] RRReeesssooonnnaaannnccceee(((SSSeeeccct tti iiooonnn111. ..555))) The basic principles: • Resonance occurs when a compound cannot be represented by a single Lewis structure. • Two resonance structures differ only in the position of nonbonded electrons and
bonds. • The resonance hybrid is the only accurate representation for a resonance-stabilized compound. A hybrid is more stable than any single resonance structure because electron density is delocalized. CH3CH2 C O O CH3CH2 C O O resonance structures CH3CH2 C O O   delocalized
bonds delocalized charges hybrid The difference between resonance structures and isomers: • Two isomers differ in the arrangement of both atoms and electrons. • Resonance structures differ only in the arrangement of electrons. CH3 C O O CH3 CH3CH2 C O O H CH3CH2 C O O H isomers resonance structures GGGeeeooommmeeet ttrrryyyaaannndddhhhyyybbbrrri iidddi iizzzaaat ttiiiooonnn The number of groups around an atom determines both its geometry (Section 1.6) and hybridization (Section 1.8). Number of groups Geometry Bond angle (o ) Hybridization Examples 2 linear 180 sp BeH2, HCCH 3 trigonal planar 120 sp 2 BF3, CH2=CH2 4 tetrahedral 109.5 sp 3 CH4, NH3, H2O 2 Study Guide/Solutions Manual to accompany: Organic Chemistry, 3rd Edition Smith: Study Guide/ Solutions Manual to accompany Organic Chemistry, Third Edition 1. Structure and Bonding Text © The McGraw−Hill Companies, 2011 Structure and Bonding 1–3


DDDrrraaawwwiiinnngggooorrrgggaaannniiicccmmmooollleeecccuuul lleeesss(((SSSeeeccctttiiiooonnn111...777))) • Shorthand methods are used to abbreviate the structure of organic molecules. CH3 C CH3 H C H H C CH3 CH3 CH3 (CH3)2CHCH2C(CH3)3 skeletal structure isooctane condensed structure = = • A carbon bonded to four atoms is tetrahedral in shape. The best way to represent a tetrahedron is to draw two bonds in the plane, one in front, and one behind. H C H H H H C H H H H C H H H H C H H H Four equivalent drawings for CH4 Each drawing has two solid lines, one wedge, and one dashed line.


BBBooonnnddd llleeennngggt tthhh • Bond length decreases across a row and increases down a column of the periodic table (Section 1.6A). C H > N H > O H H F < H Cl < H Br Increasing bond length Increasing bond length • Bond length decreases as the number of electrons between two nuclei increases (Section 1.10A). < < Increasing bond length CH3 CH3 CH2 CH2 H C C H • Bond length increases as the percent s-character decreases (Section 1.10B). Csp3 Csp H 2 Csp H H Increasing bond length • Bond length and bond strength are inversely related. Shorter bonds are stronger bonds (Section 1.10). C C C C C C Increasing bond strength shortest C–C bond strongest bond longest C–C bond weakest bond Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition 3 Smith: Study Guide/ Solutions Manual to accompany Organic Chemistry, Third Edition 1. Structure and Bonding Text © The McGraw−Hill Companies, 2011 Chapter 1–4 • Sigma ( ) bonds are generally stronger than
bonds (Section 1.9). C C C C C C 1 strong m bond 1 stronger m bond 1 weaker / bond 1 stronger m bond 2 weaker / bonds EEEllleeeccctttrrrooonnneeegggaaatttiiivvviiitttyyyaaannndddpppooolllaaarrriiitttyyy(((SSSeeeccctttiiiooonnnsss111...,,, 111...))) • Electronegativity increases across a row and decreases down a column of the periodic table. • A polar bond results when two atoms of different electronegativity are bonded together. Whenever C or H is bonded to N, O, or any halogen, the bond is polar. • A polar molecule has either one polar bond, or two or more bond dipoles that reinforce. DDDrrraaawwwiiinnngggLLLeeewwwiiisssssstttrrruuuccctttuuurrreeesss::: AAAssshhhooorrrtttcccuuuttt Chapter 1 devotes a great deal of time to drawing valid Lewis structures. For molecules with many bonds, it may take quite awhile to find acceptable Lewis structures by using trial-and-error to place electrons. Fortunately, a shortcut can be used to figure out how many bonds are present in a molecule. Shortcut on drawing Lewis structures—Determining the number of bonds: [1] Count up the number of valence electrons. [2] Calculate how many electrons are needed if there were no bonds between atoms and every atom has a filled shell of valence electrons; i.e., hydrogen gets two electrons, and second-row elements get eight. [3] Subtract the number obtained in Step [2] from the sum obtained in Step [1]. This difference tells how many electrons must be shared to give every H two electrons and every second-row element eight. Since there are two electrons per bond, dividing this difference by two tells how many bonds are needed. To draw the Lewis structure: [1] Arrange the atoms as usual. [2] Count up the number of valence electrons. [3] Use the shortcut to determine how many bonds are present. [4] Draw in the two-electron bonds to all the H’s first. Then, draw the remaining bonds between other atoms making sure that no second-row element gets more than eight electrons and that you use the total number of bonds determined previously. [5] Finally, place unshared electron pairs on all atoms that do not have an octet of electrons, and calculate formal charge. You should have now used all the valence electrons determined in the first step. Example: Draw all valid Lewis structures for CH3NCO using the shortcut procedure. [1] Arrange the atoms. HCN H H C O • In this case the arrangement of atoms is implied by the way the structure is drawn. 4 Study Guide/Solutions Manual to accompany: Organic Chemistry, 3rd Edition Smith: Study Guide/ Solutions Manual to accompany Organic Chemistry, Third Edition 1. Structure and Bonding Text © The McGraw−Hill Companies, 2011 Structure and Bonding 1–5 [2] Count up the number of valence electrons. = = = = 3 electrons 8 electrons 5 electrons + 6 electrons 22 electrons total 1 electron per H 4 electrons per C 5 electrons per N 6 electrons per O 3H's 2C's 1N 1O x x x x [3] Use the shortcut to figure out how many bonds are needed. • Number of electrons needed if there were no bonds: 3 H's 4 second-row elements x x 2 electrons per H 8 electrons per element = = 6 electrons + 32 electrons 38 electrons needed if there were no bonds • Number of electrons that must be shared: 38 electrons – 22 electrons 16 electrons must be shared • Since every bond takes two electrons, 16/2 = 8 bonds are needed. [4] Draw all possible Lewis structures. • Draw the bonds to the H’s first (three bonds). Then add five more bonds. Arrange them between the C’s, N, and O, making sure that no atom gets more than eight electrons. There are three possible arrangements of bonds; i.e., there are three resonance structures. • Add additional electron pairs to give each atom an octet and check that all 22 electrons are used. HCN H H C O HCN H H C O HCN H H C O HCN H H C O HCN H H C O HCN H H C O HCN H H C O All bonds drawn in. Three arrangements possible. Electron pairs drawn in. Every atom has an octet. Bonds to H's added. • Calculate the formal charge on each atom. HCN H H HCN C O H H C O HCN H H C O +1 –1 –1 +1 • You can evaluate the Lewis structures you have drawn. The middle structure is the best resonance structure, since it has no charged atoms. Note: This method works for compounds that contain second-row elements in which every element gets an octet of electrons. It does NOT necessarily work for compounds with an atom that does not have an octet (such as BF3), or compounds that have elements located in the third row and later in the periodic table. Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition 5 Smith: Study Guide/ Solutions Manual to accompany Organic Chemistry, Third Edition 1. Structure and Bonding Text © The McGraw−Hill Companies, 2011 Chapter 1–6 CCChhhaaapppttteeerrr111::: AAAnnnssswwweeerrrsss tttoooPPPrrrooobbbllleeemmmsss 1.1 The mass number is the number of protons and neutrons. The atomic number is the number of protons and is the same for all isotopes. Nitrogen-14 Nitrogen-13 a. number of protons = atomic number for N = 7 7 7 b. number of neutrons = mass number – atomic number 7 6 c. number of electrons = number of protons 7 7 d. The group number is the same for all isotopes. 5A 5A 1.2 The atomic number is the number of protons. The total number of electrons in the neutral atom is equal to the number of protons. The number of valence electrons is equal to the group number for second-row elements. The group number is located above each column in the periodic table. [1] 31P [2] 19F [3] 2H a. atomic number 15 9 1 b. total number of e– 15 9 1 c. valence e– 5 7 1 d. group number 5A 7A 1A 15 9 1 1.3 Ionic bonds form when an element on the far left side of the periodic table transfers an electron to an element on the far right side of the periodic table. Covalent bonds result when two atoms share electrons. a. F F b. Li+ Br c. H C C H H N H H d. Na+ Both N–H bonds are covalent. All C–H and C–C bonds are covalent. covalent ionic ionic H H H 1.4 a. Ionic bonding is observed in NaF since Na is in group 1A and has only one valence electron, and F is in group 7A and has seven valence electrons. When F gains one electron from Na, they form an ionic bond. b. Covalent bonding is observed in CFCl3 since carbon is a nonmetal in the middle of the periodic table and does not readily transfer electrons. 1.5 Atoms with one, two, three, or four valence electrons form one, two, three, or four bonds, respectively. Atoms with five or more valence electrons form [8 – (number of valence electrons)] bonds. a. O 8
6 valence e
= 2 bonds b. Al 3 valence e
= 3 bonds c. Br 8
7 valence e
= 1 bond d. Si 4 valence e
= 4 bonds 6 Study Guide/Solutions Manual to accompany: Organic Chemistry, 3rd Edition Smith: Study Guide/ Solutions Manual to accompany Organic Chemistry, Third Edition 1. Structure and Bonding Text © The McGraw−Hill Companies, 2011 Structure and Bonding 1–7 1.6 [1] Arrange the atoms with the H’s on the periphery. [2] Count the valence electrons. [3] Arrange the electrons around the atoms. Give the H’s 2 electrons first, and then fill the octets of the other atoms. [4] Assign formal charges (Section 1.3C). Count valence e
. 2C x 4 e
= 8 6H x 1 e
= 6 total e
= 14 a. HCCH H H H H HCCH H H H H All 14 e
used. All second-row elements have an octet. [1] [2] [3] Count valence e
. 1C x 4 e
= 4 5H x 1 e
= 5 1N x 5 e
= 5 total e
= 14 b. HCNH HH H HCNH H H H 12 e
used. N needs 2 more electrons for an octet. [1] [2] [3] HCNH H H H Count valence e
. 1C x 4 e
= 4 3H x 1 e
= 3 negative charge = 1 total e
= 8 c. H C H H H C H H 6 e
used. C needs 2 more electrons for an octet. [1] [2] [3] H C H H [The –1 charge on C is explained in Section 1.3C.] Count valence e
. 1C x 4 e
= 4 3H x 1 e
= 3 1Cl x 7 e– = 7 total e
= 14 d. H C H H H C H H 8 e
used. Cl needs 6 more electrons for an octet. [1] [2] [3] H C H H Cl Cl Cl Complete octet. 1.7 Follow the directions from Answer 1.6. a. HCN H C N Count valence e
. 1C x 4 e
= 4 1H x 1 e
= 1 1N x 5 e
= 5 total e
= 10 H C N 4 e
used. H C N Complete N and C octets. b. H2CO H C O H Count valence e
. 1C x 4 e
= 4 2H x 1 e
= 2 1O x 6 e
= 6 total e
= 12 H C O H 6 e
used. H C O H Complete O and C octets. c. HOCH2CO2H HOCC H H Count valence e
. 2C x 4 e
= 8 4H x 1 e
= 4 3O x 6 e
= 18 total e
= 30 16 e
used. Complete octets. O O H HOCC H H O O H HOCC H H O O H Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition 7 Smith: Study Guide/ Solutions Manual to accompany Organic Chemistry, Third Edition 1. Structure and Bonding Text © The McGraw−Hill Companies, 2011 Chapter 1–8 1.8 Formal charge (FC) = number of valence electrons – [number of unshared electrons + 1/2 (number of shared electrons)] 5
[0 + 1/2(8)] = +1 4
[0 + 1/2(8)] = 0 4
[2 + 1/2(6)] =
1 5
[0 + 1/2(8)] = +1 6
[2 + 1/2(6)] = +1 6
[4 + 1/2(4)] = 0 6
[6 + 1/2(2)] =
1 H N CH3 N C H H a. H b. c. O O O + 1.9 a. CH3O HCO H H [1] [2] H C O H H Count valence e [3] [4]
. 1C x 4 e
= 4 3H x 1 e
= 3 1O x 6 e
= 6 total e
= 13 Add 1 for (
) charge = 14 8 e
used. Assign charge. H C O H H H C O H H b. HC2 [1] [2] Count valence e [3] [4]
. 2C x 4 e
= 8 1H x 1 e
= 1 total e
= 9 Add 1 for (
) charge = 10 HCC HCC 4 e
used. HCC HCC Assign charge. c. (CH3NH3) [1] [2] Count valence e [3] [4]
. 1C x 4 e
= 4 6H x 1 e
= 6 1N x 5 e
= 5 total e
= 15 Subtract 1 for (+) charge = 14 HCN H H H H H HCNH H H 14 e
used. H H HCNH H H H H Assign charge. d. (CH3NH)– [1] [2] Count valence e [3] [4]
. 1C x 4 e
= 4 4H x 1 e
= 4 1N x 5 e
= 5 total e
= 13 Add 1 for (
) charge = 14 HCN H H H HCNH 10 e
used. H H HCNH H H Complete octet and assign charge. 1.10 a. C2H4Cl2 (two isomers) Count valence e
. 2C x 4 e
= 8 4H x 1 e
= 4 2Cl x 7 e
= 14 total e
= 26 H C C Cl Cl H H H H C C Cl H H Cl H b. C3H8O (three isomers) Count valence e
. 3C x 4 e
= 12 8H x 1 e
= 8 1O x 6 e
= 6 total e
= 26 H C C C H H H H O H H H H C C C H O H H H H H H C C O H H H H C H H H H 8 Study Guide/Solutions Manual to accompany: Organic Chemistry, 3rd Edition Smith: Study Guide/ Solutions Manual to accompany Organic Chemistry, Third Edition 1. Structure and Bonding Text © The McGraw−Hill Companies, 2011 Structure and Bonding 1–9 c. C3H6 (two isomers) Count valence e
. 3C x 4 e
=12 6H x 1 e
= 6 total e
= 18 H C C C H H H H H C C C H H H H H H 1.11 Two different definitions: • Isomers have the same molecular formula and a different arrangement of atoms. • Resonance structures have the same molecular formula and the same arrangement of atoms. different arrangement of atoms = isomers same arrangement of atoms = resonance structures N at the end N in the middle a. N C O and C N O HO C O b. O and HO C O O 2 lone pairs 3 lone pairs 1.12 Isomers have the same molecular formula and a different arrangement of atoms. Resonance structures have the same molecular formula and the same arrangement of atoms. CH3 C O OH CH3 C O OH H H C O CH2OH A C D 2 lone pairs 3 lone pairs CH3 C O CH3 C OH O OH B A H C O CH2OH D CH3 C O OH B CH3 C O OH A a. b. c. d. same arrangement of atoms = resonance structures different arrangement of atoms = isomers different arrangement of atoms = isomers CH3 bonded to C=O H bonded to C=O different molecular formulas = neither C2H4O2 (C2H5O2) – 1.13 Curved arrow notation shows the movement of an electron pair. The tail begins at an electron pair (a bond or a lone pair) and the head points to where the electron pair moves. CH3 C H C H H C CH2 H a. O H C b. H O CH3 C H C H CH2 The net charge is the same in both resonance structures. The net charge is the same in both resonance structures. Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition 9 Smith: Study Guide/ Solutions Manual to accompany Organic Chemistry, Third Edition 1. Structure and Bonding Text © The McGraw−Hill Companies, 2011 Chapter 1–10 1.14 Compare the resonance structures to see what electrons have “moved.” Use one curved arrow to show the movement of each electron pair. a. CH2 C C CH3 CH2 C C CH3 b. O C O O O C O O H H H H One electron pair moves: one curved arrow. Two electron pairs move: two curved arrows. 1.15 To draw another resonance structure, move electrons only in multiple bonds and lone pairs and keep the number of unpaired electrons constant. a. C C C CH3 b. H H CH3 H CH3 C CH3 Cl c. H C H C H C C C CH3 Cl H H CH3 H CH3 C CH3 Cl H C H C H Cl 1.16 A “better” resonance structure is one that has more bonds and fewer charges. The better structure is the major contributor and all others are minor contributors. To draw the resonance hybrid, use dashed lines for bonds that are in only one resonance structure, and use partial charges when the charge is on different atoms in the resonance structures. a. CH3 C H N H CH3 CH3 C H N H CH3 b. CH2 C CH2 H CH2 C CH2 H one more bond major contributor These two resonance structures are equivalent. They both have one charge and the same number of bonds. They are equal contributors to the hybrid. CH3 C H N H CH3 CH2 C CH2 H 
+ + 
hybrid: hybrid: All atoms have octets. 1.17 Draw a second resonance structure for nitrous acid. H O N O major contributor fewer charges H O N O minor contributor hybrid H O N O
+
– 1.18 All representations have a carbon with two bonds in the plane of the page, one in front of the page (solid wedge) and one behind the page (dashed line). Four possibilities: H C H Cl Cl H C H Cl Cl Cl C Cl H H H C H Cl Cl 10 Study Guide/Solutions Manual to accompany: Organic Chemistry, 3rd Edition Smith: Study Guide/ Solutions Manual to accompany Organic Chemistry, Third Edition 1. Structure and Bonding Text © The McGraw−Hill Companies, 2011 Structure and Bonding 1–11 1.19 To predict the geometry around an atom, count the number of groups (atoms + lone pairs), making sure to draw in any needed lone pairs or hydrogens: 2 groups = linear, 3 groups = trigonal planar, 4 groups = tetrahedral. CH3 C O CH3 CH3 O CH3 CH3 C N a. b. c. d. 3 groups = trigonal planar 3 groups = trigonal planar 4 groups = tetrahedral (or bent molecular shape) N has 2 atoms + 2 lone pairs 4 groups = tetrahedral (or bent molecular shape) 2 groups = linear 2 groups = linear 4 groups = tetrahedral 4 groups = tetrahedral 4 groups = tetrahedral 4 groups = tetrahedral NH2 4 groups = tetrahedral 1.20 To predict the bond angle around an atom, count the number of groups (atoms + lone pairs), making sure to draw in any needed lone pairs or hydrogens: 2 groups = 180°, 3 groups = 120°, 4 groups = 109.5°. 2 groups = 180° 2 groups = 180° CH3 C C Cl CH2 C H Cl CH3 C H Cl H a. b. c. This C has 3 groups, so both angles are 120°. This C has 4 groups, so both angles are 109.5°. 1.21 To predict the geometry around an atom, use the rules in Answer 1.19. 2 groups linear C C C C C C C C H H C C H H C C H H H H C H H C O H C enanthotoxin H H HO H H H C H H CH3 4 groups tetrahedral (or bent molecular shape) 4 groups tetrahedral (or bent molecular shape) 4 groups 3 groups tetrahedral trigonal planar 3 groups trigonal planar Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition 11 Smith: Study Guide/ Solutions Manual to accompany Organic Chemistry, Third Edition 1. Structure and Bonding Text © The McGraw−Hill Companies, 2011 Chapter 1–12 1.22 Reading from left to right, draw the molecule as a Lewis structure. Always check that carbon has four bonds and all heteroatoms have an octet by adding any needed lone pairs. a. H C C C C C C H C C H H H H H H H H H H H H H H H H b. H C C H C C C C H H H H C H H H H H H C H H H H H H c. C C C C C H H H O C C H H H H H H H H H H H H H d. C C C C H O H H H H H H H (CH3)2CHCH(CH2CH3)2 (CH3)3CCH(OH)CH2CH3 (CH3)2CHCHO CH3(CH2)4CH(CH3)2 double bond needed to give C an octet 1.23 Simplify each condensed structure using parentheses. CH3CH2CH2CH2CH2Cl CH3CH2CH2 C CH2CH3 H CH2CH3 HOCH2 C CH2OH H CH2CH2CH2 C CH3 CH2 CH3 C CH3 CH3 CH3 a. b. c. CH3(CH2)4Cl (HOCH2)2CH(CH2)3C(CH3)2CH2C(CH3) CH3 3 (CH2)2CH(CH2CH3)2 1.24 Draw the Lewis structure of lactic acid. CH3CH(OH)CO2H CC O H C O H O H H H H 1.25 In shorthand or skeletal drawings, all line junctions or ends of lines represent carbon atoms. The carbons are all tetravalent. O O O O O O octinoxate (2-ethylhexyl 4-methoxycinnamate) avobenzone a. b. 3 H's 0 H's 0 H's 0 H's 3 H's 1 H 1 H 1 H 1.26 In shorthand or skeletal drawings, all line junctions or ends of lines represent carbon atoms. Convert by writing in all carbons, and then adding hydrogen atoms to make the carbons tetravalent. C C C C CH3 CH3 CH3 C C C C C C Cl CH3 C C C C C O CH3 N C C N CH3 H CH3 a. b. c. d. H H H H H H H H H H H H H H H H H H H H H H H H H HH 12 Study Guide/Solutions Manual to accompany: Organic Chemistry, 3rd Edition Smith: Study Guide/ Solutions Manual to accompany Organic Chemistry, Third Edition 1. Structure and Bonding Text © The McGraw−Hill Companies, 2011 Structure and Bonding 1–13 1.27 A charge on a carbon atom takes the place of one hydrogen atom. A negatively charged C has one lone pair, and a positively charged C has none. a. b. H positive charge no lone pairs no H's needed negative charge one lone pair one H needed c. d. positive charge no lone pairs one H needed negative charge one lone pair one H needed O H H 1.28 Draw each indicated structure. Recall that in the skeletal drawings, a carbon atom is located at the intersection of any two lines and at the end of any line. (CH3)2C CH(CH2)4CH3 C C C C H CH3 H2N H H CH2CH2Cl H H N H O HO O O a. b. c. d. = = = = H2N Cl (CH3)2CH(CH2)2CONHCH3 HO(CH2)2CH=CHCO2CH(CH3)2 1.29 To determine the orbitals used in bonding, count the number of groups (atoms + lone pairs): 4 groups = sp 3 , 3 groups = sp 2 , 2 groups = sp, H atom = 1s (no hybridization). All covalent single bonds are , and all double bonds contain one  and one
bond. H C C C H H H H H H H Each C has 4 groups and is sp3 hybridized. Each H uses a 1s orbital. All single bonds are
bonds. Each C–C bond is Csp3–Csp3. Each C–H bond is Csp3–H1s. Total of 10
bonds. 1.30 [1] Draw a valid Lewis structure for each molecule. [2] Count the number of groups around each atom: 4 groups = sp 3 , 3 groups = sp 2 , 2 groups = sp, H atom = 1s (no hybridization). Note: Be and B (Groups 2A and 3A) do not have enough valence e– to form an octet, and do not form an octet in neutral molecules. H C Be H H H a. [1] Be has 2 bonds. [2] Count groups around each atom: [3] All C–H bonds: Csp3–H1s C–Be bond: Csp3–Besp Be–H bond: Besp–H1s 4 groups sp3 2 groups sp H C Be H H H Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition 13 Smith: Study Guide/ Solutions Manual to accompany Organic Chemistry, Third Edition 1. Structure and Bonding Text © The McGraw−Hill Companies, 2011 Chapter 1–14 CH3 B CH3 CH3 b. [1] B forms 3 bonds. [3] All C–H bonds: Csp3–H1s C–B bonds: Csp3–Bsp2 3 groups sp2 [2] Count groups around each atom: CH3 B CH3 CH3 4 groups sp3 H C O C H H H c. [1] H H [3] All C–H bonds: Csp3–H1s C–O bonds: Csp3–Osp3 [2] Count groups around each atom: H C O C H H H H H 4 groups sp3 4 groups sp3 1.31 To determine the hybridization, count the number of groups around each atom: 4 groups = sp 3 , 3 groups = sp 2 , 2 groups = sp, H atom = 1s (no hybridization). a. CH3 C CH b. N CH3 c. CH2 C CH2 4 groups sp3 2 groups sp 2 groups sp 3 groups sp2 3 groups sp2 3 groups sp2 1.32 All single bonds are . Multiple bonds contain one  bond, and all others are
bonds. a. b. CH3 C N  bond  bond  bond one  + two
bonds All CH bonds are  bonds. c. H C O O CH3  bond  bond one  bond, one
bond  bond one  bond, one
bond CH3 C H O 1.33 Bond length and bond strength are inversely related: longer bonds are weaker bonds. Single bonds are weaker and longer than double bonds, which are weaker and longer than triple bonds. C C bond 1: single bond bond 2: triple bond bond 3: double bond increasing bond strength: 1 < 3 < 2 a. increasing bond length: 2 < 3 < 1 b. N H N CH3 C N bond 1: single bond bond 2: double bond bond 3: triple bond increasing bond strength: 1 < 2 < 3 increasing bond length