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. 1.13 a) Hydrogen has one electron. In the ground state, it is contained in the 1s orbital. An electron in the 2s orbital would have to be an excited state of the hydrogen atom. b) Carbon has six electrons. Its electron configuration is 1s2 2s2 2p2 . Therefore, the valence electrons are in 2s and 2p orbitals. The 1s orbital is a core orbital. 1.14 a) The degenerate sets of atomic orbitals are indicated below. Copyright © 2018 Nelson Education Limited 1-15 b) The 3p orbitals are lower in energy than the 3d. c) The 3s orbital should be spherical, as are the 1s and 2s orbitals, but larger. d) The 3p orbitals should be similar in shape to the 2p orbitals, but larger. e) The ground-state electron configuration of silicon: f) The ground-state electron configuration of silicon is 1s2 2s2 2p6 3s2 3s2 3p2 . g) Silicon has four valence electrons (3s2 3p2 ), the same as carbon. The Lewis dot diagram of silicon would therefore be 1.15 a) b) c) d) e) f) g) 1-16 Copyright © 2018 Nelson Education Limited h) i) An intermediate structure is useful in determining the final structure. j) l) k) 1.16 Bond pairs are shown as dashes (–), as indicated. 1.17 a) A molecule of formula C2H5N having no formal charge on any atom: b) A cation of formula C2H8N + : Copyright © 2018 Nelson Education Limited 1-17 c) An anion of formula C2F3O − having a C=O bond: d) Two neutral molecules of formula C2H3N, both having a C–N triple bond: 1.18 a) (CH3)2CHCH2NH2 b) HO(CH2)2CH=C(CH2CH3)2 c) Cl2CHCH2CONHCH3 d) NH(CH2CN)2 1.19 a) b) c) d) e) f) 1-18 Copyright © 2018 Nelson Education Limited 1.20 a) c) b) d) 1.21 Electronegativities are provided in Figure 1.6. a) The N–H bond would be more polar due to the larger electronegativity difference between N(3.0) and H(2.2) than between B(2.0) and H(2.2). The distinction between them is that the N–H bond is polarized toward the nitrogen, while the B–H bond is polarized toward the hydrogen. b) i) ii) iii) iv) c) i) iii) ii) iv) Copyright © 2018 Nelson Education Limited 1-19 1.22 a) c) b) d) 1.23 a) c) b) d) 1.24 a) b) 1-20 Copyright © 2018 Nelson Education Limited c) d) e) 1.25 In each diagram, the hybridization, electron pair geometry, and bond geometry for the nonhydrogen atoms are indicated. a) b) c) Copyright © 2018 Nelson Education Limited 1-21 d) e) 1.26 a) b) c) d) 1-22 Copyright © 2018 Nelson Education Limited 1.27 a) b) c) d) e) f) g) 1.28 a) b) c) d) e) f) Copyright © 2018 Nelson Education Limited 1-23 g) h) i) j) 1.29 a) b) c) d) e) f) 1.30 a) (CH3)CCHCH2CH2CH(CH3)CHO b) C6H5COOH c) (CH3)2CHCHCHCOCHBrCH2OCHO d) CH3CH2OCH2CH2N(CH3) COCH(CH3)2 e) CH3CH2CH2CH2CN f) CH3CH2CCCHCHCH2OH 1-24 Copyright © 2018 Nelson Education Limited 1.31 a) c) b) d) 1.32 a) b) Note the use of the bracket drawn above this condensed structure to indicate the ring in the molecule. The bracket denotes the connection between carbon atoms A and B, which are actually beside each other in the molecule but are located at opposite ends of its condensed structure. All six carbon atoms in the ring are, therefore, in the bracket. c) d) CH3CH2CON(CH3)2 1.33 a) b) c) d) [CH3OCCHCHCHBrCHCH] A B Copyright © 2018 Nelson Education Limited 1-25 1.34 a) c) b) d) 1.35 a) The geometry at each non-H atom in the above molecule appears below. Carbons are numbered in the diagram for clarity. b) Hybridizations for each non-H atom: Bond descriptions: Bond 1 – C sp3 - C sp2 -bond Bond 2 – C sp2 - C sp2 -bond + C p- C p -bond Bond 3 – C sp3 - C sp3 -bond Bond 4 – C sp3 - C sp2 -bond Bond 5 – C sp2 - C sp -bond Bond 6 – O sp2 - H s -bond 1-26 Copyright © 2018 Nelson Education Limited 1.36 a) Lewis structure of formamide with filled valence atomic orbitals and formal charge: b) Formamide resonance forms: 1.37 a) Boron has three valence electrons. b) Lewis structure and geometry of BH3: c) Hybridization of a boron atom: Boron has an incomplete octet, since there are only three shared bond pairs with hydrogen atoms. 1.38 A bonding  molecular orbital has electron density between the bonding atoms (when occupied) and strengthens the bond between them. An anti-bonding  molecular orbital has reduced electron density between the atoms (when occupied) and weakens the bond between them. Copyright © 2018 Nelson Education Limited 1-27 MCAT Style Problems 1.39 Answer: (a). There is an undrawn H atom implied on the carbon atom in this representation. 1.40 Answer: (d). In order to have complete octets on all the CH groups, one of them needs a lone pair, giving the carbon atom a formal negative charge. Challenge Problem 1.41 Assigning the valence electrons as single bonds and lone pairs would lead to the following intermediate Lewis structure: Forming bonds with the lone pairs to reduce the formal charges would lead to the resonance forms for the molecule, as shown below. This shows that the two oxygen atoms and the central carbon atom have negative charges in one of the resonance forms. They will be the most likely sites to act as electron donors in reactions. Chapter 2 Anatomy of an Organic Molecule CHECKPOINT PROBLEMS Practice Problem 2.1 Functional groups are shown in the following molecules. a) b) c) d) H2N OH O carboxylic acid group amine group aromatic ring 2-2 Copyright © 2018 Nelson Education Limited Integrate the Skill 2.2 There are two groups based on the type of bonds involved ( or ) in each of the functional groups in the molecules above. The following groups are all formed using only  bonds: The following groups have both  and  bonds. The  bonds are noted; all the rest are  bonds. Practice Problem 2.3 a) The molecule has four alcohol groups. These can act as hydrogen bond donors and hydrogen bond acceptors. The oxygen in the ketone group can also act as a hydrogen bond acceptor. b) The molecule has three types of functional groups. The amine groups with no attached hydrogen atoms can act as hydrogen bond acceptors. The alcohol groups will be hydrogen bond donors and hydrogen bonds acceptors. The amide groups will act as hydrogen bond donors and hydrogen bond acceptors. Copyright © 2018 Nelson Education Limited 2-3 Integrate the Skill 2.4 Functional groups and hybridizations for atoms in each group are shown. Practice Problem 2.5 The two molecules are drawn below, with hydrophobic and hydrophilic regions indicated. The alcohol-containing molecule would be more soluble in water, since it has an extra hydrogen bond site (donor and acceptor). The long hydrocarbon chain in the other molecule would interact favourably with long-chain hydrocarbon solvents like hexanes and make it more soluble in hexanes. Integrate the Skill 2.6 Lewis structure of ethyl acetate—CH3CO2CH2CH3—with bond dipoles and hydrophobic and hydrophilic regions indicated: 2-4 Copyright © 2018 Nelson Education Limited Practice Problem 2.7 a) Five-carbon main chain: b) Seven-carbon main chain: c) Ten-carbon main chain: d) The largest group of carbons are in the six-membered ring (in bold) with a four-carbon substituent. e) The main chain has nine carbons (in bold) with three sidechains. The sidechains are given in alphabetical order and the main chain is numbered to minimize the numbers on the substituents. f) The main chain has seven carbons (in bold) with two methyl sidechains. 4 3 2 1 butylcyclohexane Copyright © 2018 Nelson Education Limited 2-5 g) Integrate the Skill 2.8 a) The IUPAC name for this compound is isooctane. b) Octane is a straight-chain molecule. It can efficiently pack with other octane molecules to establish dispersion attractions. Isooctane is a branched molecule and, therefore, has a shorter carbon chain to effectively interact with other isooctane molecules. The branches will also inhibit effective packing of the molecule. So, octane will have greater dispersion attractions and a higher boiling point. Practice Problem 2.9 a) b) c) The double bonds are in the ring, so it is named as a cyclohexadiene, shown in bold. d) The molecule is a redrawn in a planar projection for clarity. 1 2 3 4 6 5 1-methyl-4-(1-methylethyl)cyclohex-1-ene 2-6 Copyright © 2018 Nelson Education Limited Integrate the Skill 2.10 First, draw the main chain, numbering the atoms in the chain. This helps with substituents. Now, add the substituents. Practice Problem 2.11 a) The only functional group is an aldehyde, so the suffix is -al. Numbering is from the functional group end. b) The highest priority group is the carboxylic acid, with seven carbons in the longest chain. The alcohol group is named as a substituent, as it is not the highest priority functional group. c) The ketone is the highest priority group. The substituents are named alphabetically. d) The highest priority group is the alcohol. The locations of the double bond and the alcohol must be specified. Copyright © 2018 Nelson Education Limited 2-7 Integrate the Skill 2.12 The systematic name of the molecule is based on an eight-carbon main chain (in bold). The highest priority functional group is the alcohol. Numbering starts at the alcohol end and leads to the following name. There are many possible results from searches. Wikipedia is one location with a large amount of chemical information that is generally properly referenced. Searching for “3,7-dimethyloct-6- en-1-ol” leads to the following page: name of this molecule is citronellol, reminiscent of citronella oil, of which citronellol is a component. It is often used to ward off biting insects like mosquitos and blackflies. Practice Problem 2.13 The root is cyclohexa-1,3-diene, so the framework upon which to build the molecule is 1-(4-methylheptan-2-yl) is a seven-carbon group attached by the second carbon in the chain. This group also has a methyl attached on the fourth carbon. This is connected to carbon 1 in the ring. “5,5-difluoro” refers to two fluorine atoms, which are both attached to carbon 5 in the ring. F F 5,5-difluoro-1-(1,4-dimethylhexanyl)cyclohexa-1,3-diene 2-8 Copyright © 2018 Nelson Education Limited PROBLEMS 2.14 a) e) b) f) c) g) d) h) 2.15 For unsaturated molecules, the site of unsaturation is indicated in the structures shown. a) e) b) f) c) g) d) h) Copyright © 2018 Nelson Education Limited 2-9 2.16 Line structures and functional groups are shown. a) c) b) d) 2.17 The electron density is concentrated on the fluorine atoms (outer regions; red in the textbook) since fluorine is highly electronegative and attracts electron density away from the carbon atom (central region; blue in the textbook). In methane, the charge density is evenly distributed, since carbon and hydrogen have similar electronegativities. 2.18 The non-polar region of the molecule is the long chain on the left. In the electrostatic potential map in the textbook, it is a light blue-green. The negative parts of the molecule are red in the electrostatic potential map in the textbook. The positive regions are dark blue in the electrostatic potential map in the textbook. 2-10 Copyright © 2018 Nelson Education Limited 2.19 The intermolecular forces would involve hydrogen bonding from the carboxylic acid groups. This would be similar for both molecules. The only difference would be in the dispersion forces between the hydrocarbon chains. Line drawings of both would look like: The alkane region in 2-ethylhexanoic acid is seven carbons long, while it is only five in hexanoic acid. This would result in more attractions in 2-ethylhexanoic acid and, therefore, 2- ethylhexanoic acid should have a higher boiling point. 2.20 Molecules ranked in decreasing order of their expected boiling points. a) Since all the molecules are of similar size, the types of forces determine the order of expected boiling points. H-bonding is the strongest (highest BP), followed by dipole interactions, with dispersion forces being the weakest (lowest BP). The straight-chain alkane will have a higher boiling point, since the chains have more effective contact with neighbouring molecules and a higher net force of attraction. b) The highest BP would be for the sodium salt, since there are full charges and electrostatic forces of attraction. The carboxylic acid group is more polar than the alcohol, so the Hbonding would be stronger in the acid. O HO highest BP lowest BP H-bonding dipole interactions dispersion forces Copyright © 2018 Nelson Education Limited 2-11 c) The ammonium salt has isolated charges and electrostatic interactions. The molecules with hydrogen-substituted N atoms will have H-bonding. It will be stronger for the nitrogen with two attached hydrogens. The last molecule has no H-bond donors, so only dipole interactions are possible. 2.21 Molecules ranked in the expected order of increasing solubility in water: The two most soluble molecules have two hydrogen bonding groups on them. The smaller of the two will be most soluble, since it has a smaller hydrophobic region. The two least-soluble molecules have one hydrogen bonding group, but the least-soluble one has a much larger hydrophobic region due to the extra methyl substituents. N H HO N H HO HO HO least water-soluble most water-soluble most H-bonding smallest hydrophobic region least H-bonding largest hydrophobic region 2-12 Copyright © 2018 Nelson Education Limited 2.22 a) All of the –OH groups are hydrogen-bonding donors and acceptors. The ring carbons will be the hydrophobic region, but they are all close to polar groups. This will make this molecule very soluble in water and not very soluble in non-polar organic solvents. b) The majority of this molecule is hydrophobic and it should be soluble in organic solvents. The hydrophilic region is small in comparison, so only limited solubility in water would be expected. c) The amide and carboxylate groups are all hydrophilic. The carbon backbone is hydrophobic. This molecule has a formal charge (carboxylate group) and hydrogen bond acceptors (amide and carboxylate). So, it should have good solubility in water and polar organic solvents. It would have limited solubility in non-polar organic solvents. d) The amine (–NH2) and –SO2NH2 groups are hydrophilic hydrogen bond donors and acceptors. The aromatic ring would be the hydrophobic region. It should have good solubility in water and polar organic solvents. It would have limited solubility in non-polar organic solvents. Copyright © 2018 Nelson Education Limited 2-13 2.23 Carbons in the main chain are numbered in some diagrams to assist in naming. a) c) b) d) 2.24 Carbons in the main chain are numbered in some diagrams to assist in naming. a) c) b) Line structure makes naming easier. d) 2-14 Copyright © 2018 Nelson Education Limited 2.25 Carbons in the main chain are numbered in some diagrams to assist in naming. a) This might also be referred to as “4-(n-butyl)-3-methyldecane.” b) Converting condensed structure to a line structure makes naming easier. c) d) Copyright © 2018 Nelson Education Limited 2-15 2.26 Carbons in the main chain are numbered in some diagrams to assist in naming. a) c) b) d) 2.27 Parts (a) through (h) refer to this molecule. Answers to (a), (b), (c), and (e) appear in the diagram below. d) Root name: hex-5-en-3-one f) Substituents: C1 hydroxy, C4 methyl, C5 methyl g) 1-hydroxy-4,5-dimethyl h) 1-hydroxy-4,5-dimethylhex-5-en-3-one 2-16 Copyright © 2018 Nelson Education Limited 2.28 Carbons in the main chain are numbered in some diagrams to assist in naming. a) c) b) 2.29 Carbons in the main chain are numbered in some diagrams to assist in naming. a) d) b) e) c) Copyright © 2018 Nelson Education Limited 2-17 2.30 a) First, account for the atoms needed for substituents. The three groups would require seven carbon atoms, sixteen hydrogen atoms, and the chlorine atom. This leaves C6H9 for the ring. This could be a cyclohexane ring, a cyclopentane ring with a –CH2– sidechain, a cyclobutane with a –(CH2)2– sidechain, or a cyclopropane with a –(CH2)3– sidechain. One of the substituents needs to be on the sidechain of the smaller rings to meet the criteria presented. Other substituents must be on the ring in all four cases. There are many possible molecules, depending on where the substituents are attached. Some examples are shown below. b) There are no unsaturations (rings or double bonds), based on the molecular formula. So, all bonds will have to be single bonds. The oxygen atoms will have lone pairs, which will provide the hydrogen bond acceptor property. Alcohols cannot be used, since that would make the molecule a hydrogen bond donor. So, the oxygen atoms need to be bonded to two carbon atoms, which leads to the only possible molecule: an acetal. 2-18 Copyright © 2018 Nelson Education Limited c) The required groups will consume four C, eight H, and four O atoms. This leaves two C and two H for the backbone. There is more than one possible answer, depending on where the groups are attached to the two-carbon framework. One possible molecule is shown. 2.31 In order for a substituent to not be a functional group, it would have to be an alkyl group. An ethyl group would be an example of this. A functional group is not a substituent if it is the highest priority group in the molecule and, therefore, the basis for the name of the molecule. The carboxylic acid group in 3-chlorobutanoic acid would be an example of this. 2.32 Carbons in the main chain are numbered in some diagrams to assist in naming. a) c) b) Copyright © 2018 Nelson Education Limited 2-19 2.33 2.34 a) b) 2.35 a) i) 2-methylpropan-1-al prefix: 2-methyl root: propan suffix: 1-al ii) 2,3-dichlorocyclopent-1-ene prefix: 2,3-dichloro root: cyclopent suffix: 1-ene iii) heptan-3-ol prefix: none root: heptan suffix: 3-ol iv) 5,6-diethyl-7-hydroxyoct-1-yn-3-one prefix: 5,6-diethyl-7-hydroxy root: oct-1-yn suffix: 3-one b) i) aldehyde (-al) ii) alkene (-ene) iii) alcohol (-ol) iv) ketone (-one) c) i) prop = 3 carbon chain C=O double bond at C1 ii) pent = 5 carbon chain C=C double bond at C1 iii) hept = 7 carbon chain no multiple bonds iv) oct = 8 carbon chain C≡C triple bond at C1 C=O double bond at C3 d) i) methyl group at C2 ii) chloro group at C2 chloro group at C3 iii) no substituents iv) ethyl group at C5 ethyl group at C6 hydroxyl group at C7 e) i) iii) ii) iv) 2-20 Copyright © 2018 Nelson Education Limited 2.36 a) c) b) d) 2.37 a) c) b) d) Copyright © 2018 Nelson Education Limited 2-21 2.38 a) d) b) e) c) MCAT Style Problems 2.39 Answer: (b). For (a), 2-oxo-propylbenzene, there is no aromatic ring. Answer (c), 1-cyclohexylacetaldehyde, is a ketone, not an aldehyde. For (d), 2-oxo-propylcyclohexane, the highest-priority group is the ketone, so the suffix should be -one. O 1-cyclohexylpropan-2-one 2-22 Copyright © 2018 Nelson Education Limited 2.40 Place the following molecules in order of increasing solubility in water. Answer: (d); C < D < B < A. The charged molecule (A) would be most soluble, due to electrostatic interactions with the water. B would be next most soluble, with the carboxylic acid being both a hydrogen bond donor and acceptor. C and D are only hydrogen bond acceptors and would be the least soluble. C would be less soluble than D, since it has a larger hydrophobic region. 2.41 Answer: (d). a) c) b) e) Copyright © 2018 Nelson Education Limited 2-23 2.42 Answer: (d). The map has to have negatively polarized areas (red) for the fluorine and oxygen atoms. The maps in (a) and (d) are the only ones showing this. This rules out (b) and (c). The hydrogen attached to the oxygen atom would be positively polarized (dark blue). Map (d) is the only one with this feature, so (d) is the correct map. Challenge Problems 2.43 a) b) The electrostatic map shows that the carbon nearest the oxygen and the one on the double bond furthest from the oxygen are positively polarized, while the middle one is more neutral. The one closest to the oxygen is electron deficient, due to the electronegative oxygen drawing electron density away from it. The second resonance form (shown below) shows a formal charge on the other electron deficient carbon, explaining its lack of electron density. 2-24 Copyright © 2018 Nelson Education Limited 2.44 a) b) 2.45 The main chain is numbered and the various substituents are identified in the figure below. The isopropyl groups need to be expanded into line-drawing form to correctly name the molecule. The systematic name would then be: 5-(1-chloroethyl)-9,10-dimethyl-8-(1,2-dimethyl-1- propyl)undec-2-en-4,6-dione. Chapter 3 Molecules in Motion: Conformations by Rotations CHECKPOINT PROBLEMS Practice Problem 3.1 a) This is an eclipsed conformation: b) This is a staggered conformation: c) This is an eclipsed conformation: d) This is a staggered conformation: Integrate the Skill 3.2 a) Newman projections of staggered and eclipsed 2,2,2-trifluoroethan-1-ol: 3-2 Copyright © 2018 Nelson Education Limited b) Line bond drawing of a Newman projection: Practice Problem 3.3 a) Newman projections for the staggered and eclipsed conformations of butane: b) i) 2-methylpentane (along the C2–C3 bond with C2 in front): ii) 1-bromopentane (along the C1–C2 bond with C1 in front): iii) N,N,O-trimethylhydroxylamine (along the O-N bond with the O in front): Copyright © 2018 Nelson Education Limited 3-3 Integrate the Skill 3.4 a) Use the energy diagram in Figure 3.3. i) Energy barrier for gauche → anti is (5 kcal/mol – 0.6 kcal/mol) = 4.4 kcal/mol. ii) Energy barrier for anti → gauche is (5 kcal/mol – 0 kcal/mol) = 5 kcal/mol. iii) The highest energy barrier for gauche (60°) to gauche (300°) is anti → gauche (300°). The energy barrier for this is 5 kcal/mol. b) The only thing that would change are the x-axis labels. The relation between the H–H torsion angle and the Cl–Cl torsion angle is shown below. This would give the following graph: H H Cl H H Cl H Cl H H H H H H H Cl Cl H H Cl H Cl H Cl gauche Cl`s 60° anti H`s 180° gauche Cl`s 300° gauche H`s 60° anti Cl`s 180° gauche H`s 300° 0° 60° 120° 180° 0 Strain energy (kcal/mol) Cl–C–C–Cl torsion angle 5 8.8 0.6 240° 300° 360° 0° 180° 300° 360° 60° 120° H–C–C–H torsion angle 120° 240° 3-4 Copyright © 2018 Nelson Education Limited Practice Problem 3.5 a) The three-membered ring would have more angle strain. b) The biggest difference would be the amide having more angle strain than the alcohol. This is a result of the carbonyl carbon being sp2 -hybridized (120° angle preferred), while the carbon with the alcohol group is sp3 -hybridized (109° angle preferred). c) The cyclopropene would have more angle strain due to hybridization. d) The alkene would have more angle strain due to hybridization. Integrate the Skill 3.6 a) If cyclohexane were planar, there would be 12 eclipsed H–H interactions. b) Planar cyclohexane would have angle strain, since planar cyclohexane would require 120° bond angles, while the carbon atoms would prefer tetrahedral (109°) angles. H H H H H H H H H H H H 12 H–H interactions 6 top ones indicated all C—C bonds eclipsed Copyright © 2018 Nelson Education Limited 3-5 Practice Problem 3.7 a) i) 1,1,3,3-tetramethylcyclohexane ii) 4-bromo-1,1-dimethylcyclohexane (Br in an equatorial position) iii) b) i) The C–F bond should have a break in it to indicate that it’s behind the ring C–C bond. ii) Adjacent axial positions need to point in opposite directions (one up, one down). Assuming the ethyl groups are cis, the correct structure is shown. iii) The main bonds in the cyclohexane ring should not be horizontal. F F F H3C Ph C–F bond should be in back H3C Ph F correct representation 3-6 Copyright © 2018 Nelson Education Limited Integrate the Skill 3.8 a) Newman projection of a chair structure with hydrogen atoms: b) Two differences between an envelope conformation of cyclopentane and a chair conformation of cyclohexane: Practice Problem 3.9 a) Equatorial substituents are more stable than axial ones. b) Equatorial substituents are more stable than axial ones. c) No difference in stability. d) Equatorial substituents are more stable than axial ones. Copyright © 2018 Nelson Education Limited 3-7 Integrate the Skill 3.10 a) Newman projections for the two possible chair conformations of methylcyclohexane: b) The ethyl group is bulkier than the ethynyl group at the carbon attached to the ring. So, torsional interactions would be larger for the ethyl group. The energy difference between axial and equatorial positions would be larger for ethylhexane, making equatorial preferred. The fact that ethynylcyclohexane has almost equal amounts of axial and equatorial conformers means there is very little difference in energy between them. Practice Problem 3.11 a) Only one substituent can be equatorial. The largest group is the bromine atom, so the preferred conformation will have it in an equatorial position. b) Only one substituent can be equatorial. The largest group is the chlorine atom, so the preferred conformation will have it in an equatorial position. 3-8 Copyright © 2018 Nelson Education Limited c) The favoured conformation has all but one group in an equatorial position, which should be the lowest energy. d) The favoured conformation has the two largest substituents in equatorial positions. Integrate the Skill 3.12 In both of the chair conformers of cis-1,4-di-tert-butylcyclohexane, at least one of the very bulky tertiary butyl groups is in an axial position. The methyl groups would be in very close proximity to the adjacent axial hydrogens. A twist-boat conformation allows the axial t-butyl group to move out of a pure axial position and, thus, can reduce its steric interactions with axial hydrogen atoms. Copyright © 2018 Nelson Education Limited 3-9 PROBLEMS 3.13 a) The conformations of a molecule are different forms produced by rotations around single bonds. b) A conformer of a molecule is generated by single bond rotations. An isomer of a molecule requires breaking bonds to produce it. 3.14 a) Torsional strain is caused by the repulsions between bonding electrons attaching groups to two adjacent atoms. Conformers have higher strain when the rotational orientation around the single bond between the adjacent atoms brings these groups closer together in space. b) Steric strain is produced when groups that are not directly bonded to each other are close enough that some of their electron density occupies the same region of space and repel each other. Torsional strain is when the rotational orientation around a single bond produces repulsions between bond electrons. Steric strain arises when the overall geometry of the molecule forces groups to be close enough that their electron clouds overlap. It does not require that the groups be adjacent, just that they be close enough to significantly repel each other. c) The steric strain in ethane is small, since the only groups that can interact are hydrogen atoms. They are small and their electron clouds do not get close enough to interact with each other in all conformations. 3.15 Lewis structures by definition do not depict geometry, just connectivity. So, there would be no way to show different conformations. 3.16 a) b) 3-10 Copyright © 2018 Nelson Education Limited c) d) 3.17 a) b) c) d) 3.18 a) b) Copyright © 2018 Nelson Education Limited 3-11 c) d) e) 3.19 a) b) The graph would resemble that of ethane. All eclipsed conformations have the same interactions and would have the same strain energy. c) All eclipsed conformers have the same energy. All staggered conformers have the same energy. d) All barriers are the same height. e) The barriers would be lower for CH3NH2, since the bulkier methyl group has been replaced with a smaller hydrogen atom. 3-12 Copyright © 2018 Nelson Education Limited 3.20 a) The highest energy conformer of 1-chloropropane has two eclipsed H–H interactions and an eclipsed methyl-Cl interation. The lowest energy conformation has the two largest groups as far apart as possible and no eclipsed interations. b) Defining the torsion angle with the Cl–C–C–C group, the highest energy conformation has a torsion angle of 0° and the lowest energy conformation has a torsion angle of 180°. 3.21 a) The staggered conformations at 60° and 300° have equivalent interactions and are lower in energy than the 180° conformation. The 0° conformation is the lowest energy eclipsed conformation (methyl-hydrogen interaction), while the 120° and 240° conformers are equivalent and higher in energy (methyl–methyl interaction). b) 0° 60° 120° 180° Strain energy C–C–C–H torsion angle 240° 300° 360° Copyright © 2018 Nelson Education Limited 3-13 c) The staggered conformations at 60° and 300° have equivalent interactions and are lower in energy than the 180° conformation. The main difference is that the front methyl has staggered interactions, with an H and a CH3 at 60° and 300°, while at 180° the front methyl has interactions with two CH3 groups. d) The 0° conformation is the only eclipsed conformation with the lowest energy. The other two eclipsed conformations have methyl–methyl interactions. 3.22 It is not possible to have isomers in a simple alkane. Any conformation of the alkane can be changed into any other by rotations around single bonds, so the cis and trans forms are the same molecule in different conformations. 3.23 a) The equatorial form of cis-1,3-dimethylcyclobutane would be more stable, since it maximizes the distance of the larger methyl groups from other groups in the molecule. As well, in the axial form, the two methyl groups exhibit both torsional and steric strain. b) Both conformers of trans-1,3-dimethylcyclobutane would have equivalent interactions. They are exactly the same in different orientations. 3.24 Favourably: The phrasing is true. Inversion takes place due to rotation about the C–C bonds of the cyclohexane ring. Unfavourably: While rotation does bring about inversion, the rotation is limited in its range of motion, unlike the rotations in acyclic molecules in which the degree of rotation is unrestricted. 3-14 Copyright © 2018 Nelson Education Limited 3.25 a) The steric interactions are between a hydrogen and a methyl group in the same geometry in both cases. Therefore, the strain should be the same: ~0.9 kcal/mol. Hydrogen atoms are too small to create any significant steric strain through H–H interactions, so the total steric strain should be ~0.9 kcal/mol for the gauche conformation of butane. b) There are two hydrogen atoms in an axial geometry that would interact with the axial methyl group in the conformer of methylcyclohexane drawn. So, this conformer would have 2 × 0.9, or 1.8 kcal/mol, of steric strain from H–methyl interactions. 3.26 a) b) Both chlorine atoms in molecule A are axial. Both equatorial chlorine atoms in molecule B are equatorial. c) As shown below, trans-1,3-dichlorocyclohexane always has one axial and one equatorial chlorine in its chair forms. 3.27 a) b) c) OH CH3 OH CH3 cis axial OH trans axial OH HO CH3 HO CH3 cis equatorial OH trans equatorial OH Copyright © 2018 Nelson Education Limited 3-15 d) In part (b), the difference in the axial OH conformers would be trans, having 1.8 kcal/mol more strain arising from the CH3 group being axial. In part (c), the difference in the equatorial OH conformers would be cis, having 1.8 kcal/mol more strain arising from the CH3 group being axial. The cis form would be most stable, having an axial OH and equatorial CH3. The trans form would be most stable, having equatorial OH and CH3. 3.28 a) b) c) 3.29 3.30 The 1,1 dimethylcyclopentane cannot have cis/trans isomers, since there are no two carbon atoms in the ring with two different substituents. 3.31 The gauche conformers have the two large chlorine atoms close to each other. By increasing the distance between them (60° → 67°, 300° → 293°), the Cl–Cl repulsions can be reduced. The H–H interactions should not increase significantly due to the small size of the hydrogen. NH2 NH2 CH3 CH3 NH2 NH2 cis trans 3-16 Copyright © 2018 Nelson Education Limited 3.32 3.33 3.34 a) b) c) The first, part (a), would be most stable, with all of the large substituents located in equatorial positions. d) 3.35 a) b) c) The conformer in (a) would be lower energy. The very large tertiary butyl group creates the most steric strain and would be most stable in an equatorial position to reduce the strain. d) Copyright © 2018 Nelson Education Limited 3-17 3.36 a) b) Since all the bonds are in staggered conformations, the major source of strain is from 1,3 diaxial interactions between the methyl groups and axial hydrogen atoms. c) The main source of strain would be the interactions of the staggered methyl groups. In the 1,4 compound, these are all with hydrogen atoms. In the 1,2 compound, the methyl groups interact with each other in a gauche butane interaction. The 1,2 compound would be expected to have more total strain energy. d) The less stable (axial methyls) conformers of each would have approximately the same strain energy from 1,3 diaxial interactions. The 1,2 compound would have more strain energy than the 1,4 in the more stable (equatorial methyls) form than the methyl–methyl interactions. So, the overall difference would be less for the 1,2 compound. e) The atypical behaviour would have to be due to interactions in the equatorial form being larger than anticipated. Due to the electronegativity difference between bromine and carbon, the C–Br bond would be polar, giving the bromine atoms a slight negative charge. In the equatorial form, these partial charges are close to each other and would produce a large repulsion. In the axial form, the bromines are far apart and would not interact with each other. 3-18 Copyright © 2018 Nelson Education Limited 3.37 Taking the number 1 carbon as reference, the five isomers can be described by the relative orientation of the 2,4,5 methyl groups (cis or trans). The five unique isomers are shown below. 3.38 a) b) The major source of strain in these molecules would be the 1,3 diaxial interactions between the methyl and axial hydrogens or the chlorine and axial hydrogens. c) The strain from 1,3 interactions can be estimated using Table 3.1. Comparing a chlorine (A = 0.6) and a methyl (A = 1.6), the methyl would be expected to have a larger steric interaction with axial hydrogens. Therefore, the conformer with less strain would be the one with an equatorial methyl and be favoured in the equilibrium. 3.39 a) b) Since there are different 1,3 diaxial interactions (=O,H or CH3,H), the strain is most likely different in each conformer. Copyright © 2018 Nelson Education Limited 3-19 3.40 The ring atoms in both are sp3 hybridized. The 109° angle is closer to the 90° found in the fourmembered ring than the 60° in the three-membered ring. The overlaps are more effective in azetane and are, therefore, more stable. 3.41 3.42 a) The bond angles are not accurate (they should all be sp3 or 109°). The ring is not flat but a chair form. There are two different types of substituent (axial/equatorial) on the ring. b) 3.43 There are no chair or half-chair conformations in the inversion process. 3-20 Copyright © 2018 Nelson Education Limited 3.44 The chair cis form can have both bulky t-butyl groups in equatorial positions. In the trans chair form, one is forced into an axial position. Adopting a twist boat form must relieve some of the strain for the trans isomer. 3.45 Cyclopropene would have more strain. The hybrid orbitals forming the -bonds (60° bond angles) are sp2 and 120° apart on the alkene carbons. This would lead to poorer overlap than in cyclopropane where all of the -bond formation is with sp3 orbitals with 109° angles. 3.46 a) b) The eight hydrogens in the boat form indicated will all be eclipsed and create torsional strain. 3.47 a) b) t-Bu t-Bu trans cis t-Bu t-Bu Copyright © 2018 Nelson Education Limited 3-21 3.48 a) b) c) The trans form has no significant steric strain. The cis form has CH2 groups, which would interact with the axial hydrogen atoms and cause steric strain. The trans form would be more stable. d) Cis decalin can undergo ring flip. It exchanges one equatorial ring connection to an axial position and one axial connection to an equatorial position. Trans decalin would convert both the equatorial carbon connections to axial ones, which would force one ring to be highly strained. 3.49 a) In cyclopropane, two of the bonds on each carbon are strained. In tetrahedrane, three of them are strained. So, the total strain energy would be expected to be 4 × cyclopropane + strain from third bond + loss of stability from one unstrained C–H bond. b) Although the total strain in cubane is more, each carbon atom would undergo less strain than in tetrahedrane. The bond angles are 90°, which is closer to the preferred 109° for sp3 hybridization. So, the C–C bonds should be stronger and the molecule more stable as a result. trans form cis form H H H H trans cis 3-22 Copyright © 2018 Nelson Education Limited 3.50 a) The axial bromines interact with axial hydrogens in the form on the left. The equatorial bromine atoms are polarized and will repel each other due to like charges in the form on the right. Also, the large size of the bromine atom would add to the steric strain in the equatorial orientation. b) The equatorial form should be more polar, since the C–Br dipoles are not cancelling each other out. So, the axial form should be favoured in non-polar solvents, while the equatorial form would be favoured in polar solvents. MCAT Style Problems 3.51 Answer: (d). Only statement (iii) is true. i) False. This is steric strain. ii) False. Torsional strain exists due to eclipsed C–H bonds. 3.52 The most likely answer is (b). Both methyl groups will be axial in one form and both equatorial in the other form. The 1,3 diaxial steric Me–Me interaction will be very destabilizing. There is little or no strain in the equatorial version. a) b) c) d) Molecule (d) only has one 1,3 diaxial interaction and one of the groups is a hydrogen. The molecules in (a) and (c) have two groups: one axial and one equatorial in chair conformations. So, the difference between their chair forms should be less than it is for (b) or (d). Copyright © 2018 Nelson Education Limited 3-23 3.53 Answer: (d). None of the statements is true. i) False. The half-chair conformation is the highest energy. ii) False. Conformers are configurations of the same isomer. iii) False. Conformational changes cannot change molecules into different isomers. 3.54 Answer: (a). The two chair conformations are equal in energy. There is one axial and one equatorial group in each chair conformation. Challenge Problems 3.55 a) Allowing the oppositely charged functional groups to get as close as possible would be a strong stabilizing influence. It is likely that GABA, which has no restricted rotation, would adopt a conformation that maximizes this interaction. b) This is a different conformation. The ring prevents the charged functional groups from forming the conformation where they are eclipsed. 3-24 Copyright © 2018 Nelson Education Limited c) The two functional groups are trans. This is a similar conformation to (b). 3.56 The major product would arise from the more stable conformation in the reactive step. Ph H3C H O CH3 Ph H3C H H3C O Ph CH3 R O BH4 BH4 Me–Me interaction less favourable conformation major pathway Ph H3C H H HO CH3 Ph H HO CH3 CH3 Ph CH3 CH3 OH Chapter 4 Stereochemistry: Three-Dimensional Structure in Molecules CHECKPOINT PROBLEMS Practice Problem 4.1 a) There are no chirality centres in this molecule, so it cannot be a chiral molecule. b) There is one chirality centre. Rotating the mirror image molecule so the main chain is lined up like the original shows that this mirror image is not superimposable. This is a chiral molecule. c) This molecule has two chirality centres. Looking at the mirror image, if it is rotated 180° on the plane of the page, the original orientation of the molecule results. 4-2 Copyright © 2018 Nelson Education Limited d) This molecule is chiral. First, drawing the mirror image and then rotating it to put it in the same orientation as the original molecule shows it to be non-superposable. In the final orientation of the mirror image, it is clear that each of the chiral centres has the opposite configuration of the original molecule. Integrate the Skill 4.2 a) The first step is to draw the molecule and determine whether there are chirality centres. Looking at the drawing below, there are no asymmetric carbon centres in this molecule and it is, therefore, not chiral. b) This molecule has one chirality centre. The molecule is, therefore, chiral. c) There is one asymmetric centre in this molecule. This molecule is chiral. Practice Problem 4.3 a) The screw is chiral. The threads follow a right-handed helix along the length. The mirror image would have the threads in a left-handed helix. Most common screws are right-handed. b) The mug is achiral. The mug may appear chiral, but, the mug is rotated so that its handle faces toward the user, there is a vertical mirror plane going through the middle of the handle, which splits the cup into two mirror images. Since it has this internal mirror plane, it cannot be chiral. Copyright © 2018 Nelson Education Limited 4-3 c) The doghouse is achiral. A vertical mirror plane running along the peak of the roof splits it into a left half and a right half, which are mirror images. The presence of this mirror plane means that the doghouse is not chiral. Aside: If the name of the dog were on the front, in most cases the doghouse would be chiral. Try to come up with a name where this would not be true. d) If the molecule is achiral, the two chiral carbons will be mirror images of each other. Rotating about a C–C bond so the presence or absence of symmetry is easier to assess, it is clear that there is no mirror plane in the molecule that reflects the left half into the right half and the right half into the left half. This molecule is chiral. e) There are two chirality centres, so an achiral compound is possible. Redrawing the molecule reveals the internal mirror plane of symmetry that reflects the top half into the bottom half and vice versa. This molecule is achiral (meso). f) There are no chirality centres in this molecule. There is a mirror plane running through the middle of the molecule, bisecting the bromine and chlorine atoms. This molecule is achiral. Integrate the Skill 4.4 a) To see the symmetry, a boat form is necessary. 4-4 Copyright © 2018 Nelson Education Limited b) Rotation about the C–C bond is required until the same groups on the front carbon and back carbon are eclipsed. The symmetry is most easily seen by then rotating the molecule by 90°, as shown. c) Rotating around the centre C–C bond by 180° to eclipse equivalent groups shows the symmetry in this molecule. Practice Problem 4.5 a) There are two sp3 carbons with four different substituents. b) There are two chiral sp3 carbon atoms, as shown. c) The five carbon atoms indicated are chiral sp3 centres. d) There are two chirality centres: one an sp3 carbon and the other a sulfoxide group. Et Ph H Et H Ph rotate front atom 180° Et Ph H Et Ph H * chirality centres S O NH2 O OH H * * Copyright © 2018 Nelson Education Limited 4-5 Integrate the Skill 4.6 The constitutional isomers contain no rings or double bonds, since the molecular formula has n C atoms and 2n + 2 H atoms (see textbook Section 2.2; n = 5 in this case), which is the ratio found in an alkane. Formation of a ring or double bond would require removing two hydrogen atoms. So, the molecules will be ethers or alcohols and contain no rings. Drawing each basic carbon framework then adding in the oxygen in all possible unique sites gives the 14 isomers. Chirality centres are indicated where they exist. Pentane backbone Methylbutane backbone Dimethylpropane backbone Practice Problem 4.7 a) 4-6 Copyright © 2018 Nelson Education Limited b) c) d) Integrate the Skill 4.8 a) Drawing the molecule without stereochemistry, it can be seen that the chirality centre will be of R configuration if the undrawn H is pointing away. This gives the final structure with the stereochemistry indicated. Copyright © 2018 Nelson Education Limited 4-7 b) After assigning the priorities on the chiral carbon, the lowest priority group is the methyl group, so it should be pointing away to assign the stereochemistry. If the determined stereochemistry was incorrect, switching any two groups would give rise to the correct R stereochemistry. Practice Problem 4.9 a) Phantom atoms are added to the double bond, and the two lowest priorities can be assigned at the first attachment. At the second carbon, the assignments can be completed. This is an S configuration. b) There are three chirality centres in this molecule. The first centre has only carbon atoms at the first comparison, but a difference in attached atoms allows the first priority to be assigned. 3 4 H3C H add phantom atoms H3C CH3 3 4 H3C H C C tied higher priority at first C atom: C, C, H lower priority at first C atom: H, H, H H3C CH3 3 4 H3C H C C lower priority at second C atom: H, H, H higher priority at second C atom: C, H, H 1 2 S 4-8 Copyright © 2018 Nelson Education Limited At the second pair of C atoms to be compared, there is a difference and the groups can be fully assigned priorities. This centre has an R configuration. The second centre is simpler to assign. The O atom is top priority and the attached carbon atoms are second and third as indicated. This is an R configuration. For the third centre, it is useful to rotate the lowest priority H atom to the back. This is an S configuration. Copyright © 2018 Nelson Education Limited 4-9 c) There are four chirality centres in this sugar molecule, as shown. Centre 1: S Configuration Centre 2: S Configuration Centre 3: S Configuration 4-10 Copyright © 2018 Nelson Education Limited Centre 4: R Configuration d) There is one chirality centre in this molecule. It is of S configuration. Integrate the Skill 4.10 a) Copyright © 2018 Nelson Education Limited 4-11 b) H3C CH3 H3C CH3 CH3 C C C C C C Higher Priority Pair Lower Priority Pair lower priority: C atom with H, H, H higher priority: C atom with C, C, H 4 3 1 2 H3C CH3 H3C CH3 CH3 C C C C C C lower priority: C atom with H, H, H higher priority: C atom with C, H, H 4-12 Copyright © 2018 Nelson Education Limited Practice Problem 4.11 a) b) Copyright © 2018 Nelson Education Limited 4-13 c) d) Integrate the Skill 4.12 Enantiomers of compounds in Question 4.11. a) 4-14 Copyright © 2018 Nelson Education Limited b) c) d) Copyright © 2018 Nelson Education Limited 4-15 Practice Problem 4.13 a) The pair must be enantiomers, since 2-butanol has only one chirality centre. To confirm the enantiomeric relationship, the quickest approach is to assign the stereochemistry of the structure and compare it to (R)-2-butanol. b) The first step would be to redraw the Newman projection and assign configurations. Next, assign configurations to the other molecule. Since the configuration at only one chirality centre has changed, these must be diastereomers. c) There are many centres to consider here, so redrawing the chair into the first form will allow a direct comparison of each chiral centre without assigning an absolute configuration to each. First, add missing H atoms to the ring, then assign up (solid wedge) or down (dashed wedge) to each position for the non-hydrogen substituent. 4-16 Copyright © 2018 Nelson Education Limited Now, compare the two molecules directly. The configuration differs in only one position, so they must be diastereomers. d) These are best compared to see whether they are mirror images of each other. First, reorient B by rotation. Comparing the two molecules side-by-side, it is clear that they are mirror images. One more rotation of B and comparison with A shows that these mirror images are not superposable and are, therefore, enantiomers: Copyright © 2018 Nelson Education Limited 4-17 Integrate the Skill 4.14 a) These molecules are identical even though they are drawn as mirror images. b) In A and B, the carbons at the front and at the back of the Newman projection are both chirality centres. To compare, first reorient B into a mirror image of A. Putting both molecules side by side with a mirror plane shows that the groups on the back carbon are reflected and are therefore of opposite configuration. The groups on the front carbon are not reflected and, therefore, are of the same configuration. A and B are diastereomers. 4-18 Copyright © 2018 Nelson Education Limited c) These are constitutional isomers, since their –OH groups are positioned at different carbons around the ring. One is 1,3,5 substituted and the other is 1,2,4. Practice Problem 4.15 a) Draw 2,3,5,6-tetrahydroxy-2-methylheptanoic acid and determine the chirality centres. There are 4 here for 16 possible stereoisomers. b) Draw 1,3,5-trimethylcyclopent-1-ene and determine the chirality centres. There are two here for four possible stereoisomers. c) There are eight possible stereoisomers for this molecule. d) There are 7 chiral centres and 128 possible stereoisomers. Copyright © 2018 Nelson Education Limited 4-19 Integrate the Skill 4.16 R configurations for molecules in Practice Problem 4.15. a) b) c) d) Practice Problem 4.17 a) This molecule is achiral. There are no carbon atoms with four different substituents. b) The molecule has a mirror plane of symmetry that reflects the two centres into each other, so this is a meso compound (achiral). COOH OH OH OH OH * * * * OH COOH HO H H OH all R HO H centres 4-20 Copyright © 2018 Nelson Education Limited c) This molecule has a mirror plane, so this is a meso compound (achiral). d) There are two chirality centres. Redrawing the ring shows there is no plane of symmetry between the two chiral centres, so this is not a meso compound. This is a chiral molecule. e) There are two chirality centres in this molecule. Redrawing the molecule in a different representation shows there is a mirror plane, so this is a meso compound (achiral). f) There are two chirality centres. As there is no mirror plane, this is not a meso compound. It is a chiral molecule. Copyright © 2018 Nelson Education Limited 4-21 Integrate the Skill 4.18 Upon first inspection, the (2R,4S) and (2S,4R) stereoisomers of pentane-2,4-diol appear to be enantiomers. However, due to the symmetry in the molecule, these two are actually the same molecule. Therefore, there are only three stereoisomers of pentane-2,4-diol. Practice Problem 4.19 a) The ring structure means there can only be one isomer. So, the correct name is 4-propylcyclohex-1-ene. b) The first double bond is symmetric at one end, so no E/Z can be assigned. The correct name is (4E,6E)-2-methylocta-2,4,6-triene. c) E/Z should be used instead of cis/trans in naming. This is (E)-3-methylhex-3-ene. trans-3-methylhex-3-ene higher priority (E) 4-22 Copyright © 2018 Nelson Education Limited d) This is an E arrangement. The correct name would be (E)-(4-methylhexa-1,3-dien-3-yl)benzene. Integrate the Skill 4.20 a) There are three chirality centres, so 23 , or eight, stereoisomers are possible. b) There is one asymmetric double bond, so there would be two stereoisomers of this compound. c) There are seven stereochemical sites (five chirality centres, two asymmetric double bonds), so there are 27 , or 128, possible stereoisomers. Copyright © 2018 Nelson Education Limited 4-23 Practice Problem 4.21 a) First, calculate the concentration of the compound in g/mL.