Solution Manual for Organic Chemistry Mechanistic Patterns Canadian 1st Edition Ogilvie Ackroyd Br

Chapter 1 Carbon and Its Compounds CHECKPOINT PROBLEMS Practice Problem 1.1 a) S — 1s2 2s2 2p6 3s2 3p4 b) Cl — 1s2 2s2 2p6 3s2 3p5 c) Na+ — 1s2 2s2 2p6 1-2 Copyright © 2018 Nelson Education Limited Practice Problem 1.2 a) Count valence electrons. Build a basic bonding framework and account for electrons used. Add remaining electrons and check for formal charges. The molecule has a lone pair on the nitrogen. All other electrons are bonding electrons. b) Count valence electrons. Build a basic bonding framework and account for electrons used. Copyright © 2018 Nelson Education Limited 1-3 Add remaining electrons and check for formal charges. There are formal charges on the S and O atoms. They can be removed by making an additional bond between O and S. The molecule has a lone pair on the sulfur and two lone pairs on oxygen. All other electrons are bonding electrons. c) Count valence electrons. Build a basic bonding framework and account for electrons used. 3 carbons (group 4) 5 hydrogens (group 1) 1 nitrogen (group 5) 3 X 4 = 12 5 X 1 = 5 1 X 5 = 5 22 valence e - CH3CH2CN (nitrogen is connected to one carbon only) 8 bonds = 16 bonding e22 valence e- – 16 bonding e- = 6 non-bonded e H C C C - H H H H N 1-4 Copyright © 2018 Nelson Education Limited Add remaining electrons and check for formal charges. The formal charges on C and N show the carbon needs more electrons and the N has too many. Forming two more bonds between C and N alleviates this problem. The molecule has a lone pair on the nitrogen. All other electrons are bonding electrons. d) Count valence electrons. Build a basic bonding framework and account for electrons used. Add remaining electrons and check for formal charges. The oxygen atom has three lone pairs and a positive charge. All other electrons are bonding electrons. Copyright © 2018 Nelson Education Limited 1-5 e) Count valence electrons. Build a basic bonding framework and account for electrons used. Add remaining electrons and check for formal charges. The ion has a formal positive charge on the nitrogen. All electrons are bonding electrons. f) Count valence electrons. Build a basic bonding framework and account for electrons used. 1-6 Copyright © 2018 Nelson Education Limited Add remaining electrons and check for formal charges. There are formal charges on the sulfur (+1) and two of the oxygens (-1). These can be reduced by forming a double bond between sulfur and either of the oxygen atoms carrying a formal charge. The ion has seven lone pairs on oxygen atoms and one lone pair on sulfur. All other electrons are bonding electrons. One oxygen has a formal negative charge. g) Count valence electrons. Build a basic bonding framework and account for electrons used. Add remaining electrons (on oxygen first) and check for formal charges. Copyright © 2018 Nelson Education Limited 1-7 There are formal charges on the sulfur (+3) and two of the oxygens (-1). These can be reduced by forming double bonds between sulfur and both charged oxygen atoms. This expands the octet of the sulfur but, since it is a third-row element, this is allowed. The final ion has six lone pairs on oxygen atoms. The sulfur has a formal positive charge. All other electrons are bonding electrons. Practice Problem 1.3 a) b) c) S O H O O H S O OR O O valence e- = 9 (9H) + 12 (3C) + 7 (1Cl) = 28 e - 28 valence e- – 18 bonding e- = 6 non-bonded eC C C O H H H H H H Lewis Structure C C C O H H H H H H reduce formal charges H3C C CH3 O dd+ dipole 1-8 Copyright © 2018 Nelson Education Limited d) Practice Problem 1.4 In the solutions, “BG” is used as an abbreviation for “bond group” and “LP” is used as an abbreviation for “lone pair.” a) b) c) H O C C C C C H H H H H H H H O H H–O bond dipole C–O bond dipole C=O bond dipole d+ dd+ dd+ Copyright © 2018 Nelson Education Limited 1-9 Integrate the Skill 1.5 All carbons have four bonds and so will not have lone pairs. Lone pairs are added to the nitrogen atoms according to the formal charges indicated. “BP” refers to shared pairs of electrons in bonds between atoms where each atom formally has one of the electrons. The geometry of the atoms can then be established for all of the atoms. Practice Problem 1.6 a) b) C C N C C N H H H H H H H formal charge of –1 = 6 valence e - = 2 (2BP) + 4 (2LP) formal charge of 0 = 5 valence e - = 3 (3BP) + 2 (1LP) C C C O C H H H H H H 3 BG trigonal planar 4 BG tetrahedral 2 BG + 2 LP bent C C C O C H H H H H H sp2 sp3 1-10 Copyright © 2018 Nelson Education Limited Integrate the Skill 1.7 a) Accounting for all of the electrons leaves four non-bonded electrons to add as lone pairs. These are added to the oxygen first (most electronegative), leaving the carbon with a formal positive charge. b) Electron geometry: c) Hybridization: d) The charged carbon is not saturated, so a second bond to the oxygen can be formed. This moves the formal charge to the oxygen atom. H C H H O C H H H C H H O C H H Copyright © 2018 Nelson Education Limited 1-11 This leads to the following geometries for the new structure. The corresponding hybridizations would then be Practice Problem 1.8 a) b) c) d) H C H H O C H H tetrahedral sp3 trigonal planar sp2 trigonal planar sp2 H3C NH2 O NH2 H3C NH2 O H3C O 1-12 Copyright © 2018 Nelson Education Limited Integrate the Skill 1.9 Drawing the basic structure leaves a formal positive charge on the carbon atom and lone pairs on each nitrogen atom. Three more resonance forms can be produced by forming a double bond between each of the nitrogen atoms and the central carbon atom. This leaves the formal charge on a nitrogen atom for each of these new forms. Practice Problem 1.10 a) i) ii) b) i) ii) c) i) ii) d) i) ii) H2N NH2 NH2 H2N NH2 NH2 H2N NH2 NH2 H2N NH2 NH2 Copyright © 2018 Nelson Education Limited 1-13 Integrate the Skill 1.11 The formal charge can be on the carbon or the oxygen. Both are acceptable Lewis structures. 1-14 Copyright © 2018 Nelson Education Limited PROBLEMS 1.12 a) b) The ground-state configuration is 1s2 2s2 2p1 . c) Boron has four valence orbitals