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8 Introduction to Alkyl Halides, Alcohols, Ethers, Thiols, and Sulfides Solutions to In-Text Problems ·········································································································· 8.1 (b) Hexyl iodide is a primary alkyl halide. 8.2 (b ) (d) Tert-butyl chloride is a tertiary alkyl halide. (d) Cl I CH2 I methylene iodide chlorocyclopropane 8.3 (b) (Z)-3-Chloro-2-pentene (f) 1-Bromo-2,2-dimethylpropane (d) Trichloromethane (h) 1,3-Dibromocyclobutane 8.4 (b) (f) (d) (CH3)2CHCH2OH isobutyl alcohol OH CH3CHCHCH2CH3 OH CH3 3-methyl-2-pentanol 8.5 44 (b) 1-Butanol (f) 1,2,3-Propanetriol 2-cyclohexenol (d) (E)-1-Chloro-3-methyl-3-penten-2-ol (i) 2-Methyl-2-propanethiol Instructor’s Solutions to In-Text Problems 8.6 (b) (d) O Chapter 8 45 (f) (CH3)3C S CH(CH3)2 O CH CH2 tert-butyl isopropyl sulfide dicyclohexyl ether phenyl vinyl ether 8.7 (b) 2-methoxy-2-methylpropane 8.10 By the logic used in the solution to Problem 8.9, the cis stereoisomer has the higher dipole moment and hence, the higher boiling point. Cl Cl C C H Cl H C H cis-1,2-dichloroethylene (bp = 60.3°) C Cl H trans-1,2-dichloroethylene (bp = 47.4°) 8.11 (b) A chlorine contributes about the same molecular mass (35 units) as an ethyl group (29 units), and alkyl chlorides have about the same boiling points as alkanes of the same molecular mass. Hence, chloromethane has about the same boiling point as propane, which has a lower boiling point than the fivecarbon alkene 1-pentene. The alcohol has the highest boiling point because it has about the same molecular mass as 1-pentene, but can donate and accept hydrogen bonds. Consequently, it has the highest boiling point of all. The order of increasing boiling points is, therefore, chloromethane (–42°) < 1-pentene (30°) < 1-butanol (118°). 8.12 (b) Hydrogen fluoride is an excellent hydrogen-bond donor, and the fluorine is an excellent hydrogenbond acceptor. (d) N-methylacetamide can serve as both a hydrogen-bond donor and a hydrogen-bond acceptor. (See structure below.) (f) The ethylammonium ion can donate its N— H hydrogens to hydrogen bonds, but it cannot accept hydrogen bonds because it has no unshared electron pairs. accepts hydrogen bonds O CH3 C can be donated to a hydrogen bond H N CH3 CH3CH2 H accepts hydrogen bonds N-methylacetamide H + N H can be donated to a hydrogen bond ethylammonium ion 8.13 (b) 2,2,2-Trifluoroethanol is a polar, protic, donor solvent. (d) 2,2,4-Trimethylpentane (a major component of gasoline) is an apolar, aprotic, nondonor solvent. 8.15 Hexane should be least soluble in ethanol for the same reasons that ethanol is not very soluble in hexane. 8.17 (b) Na+ (CH3)2CH—O– sodium isopropoxide 8.18 (b) Calcium methoxide (d) Li+ CH3O– lithium methanolate 46 Chapter 8 8.19 (b) HCF2CH2CH2OH Introduction to Alkyl Halides, Alcohols, Ethers, Thiols, and Sulfides < < HCF2CH2OH C CF3CF2CH2OH A B Compound B is most acidic because it has the most fluorines with the smallest separation from the oxygen; compound A is more acidic than compound C because the fluorines are closer to the oxygen in compound A. (d) CH3CH2CH2CH2OH < CH3OCH2CH2OH A B Compound B is more acidic than compound A because of the electron-withdrawing polar effect of the oxygen in the methoxy group. 8.20 (b) CH3 + Mg I (d) (CH3)2CH 8.21 ether CH3 MgI hexane Cl + 2 Li (CH3)2CH Li + LiCl (b) (CH3)3C—Li + LiCl tert-butyllithium Early printings of the text erroneously posed (CH3 )3 C — Br as the starting material; the byproduct in this case is LiBr. The starting material was changed to (CH 3)3C — Cl because the bromide actually gives little or no tert-butyllithium, but instead gives mostly the coupling product (CH3)3C — C(CH 3)3 (2,2,3,3-tetramethylbutane). ! 8.22 (b) – HO H (CH3)2CHCH2 MgCl HO + ClMg + (CH3)3CH isobutane 8.23 (b) (CH3)2CH—D and CH3CH2CH2—D 8.25 The initiation step is the dissociation of Br2 into bromine atoms; this step is analogous to the dissociation of chlorine shown in text Eq. 8.29. The propagation steps are as follows: Br Br H Br CH2CH3 Br H + CH2CH3 Br + CH2CH3 Br CH2CH3 bromoethane 8.27 In the free-radical bromination of ethane, shown in the solution to Problem 8.25, the ethyl radical is formed in the first propagation step. Recombination of two ethyl radicals gives butane. CH3CH2 CH2CH3 CH3CH2 CH2CH3 butane Instructor’s Solutions to Additional Problems Chapter 8 47 Solutions to Additional Problems ·········································································································· 8.29 (a–d) The alcohols with the formula C4H9OH CH3CH2CH2CH2OH 1-butanol an achiral primary alcohol OH CH3CHCH2OH CH3CHCH2CH3 2-butanol a chiral secondary alcohol CH3 CH3COH CH3 2-methyl-1-propanol an achiral primary alcohol CH3 2-methyl-2-propanol an achiral tertiary alcohol 8.30 (b) The systematic name of methoxyflurane is 2,2-dichloro-1,1-difluoro-1-methoxyethane. 8.31 (b) 3-Methyl-1-butanethiol 8.32 (b) The order of boiling points is tert-butyl alcohol < butyl alcohol. Increased branching lowers boiling point. (See text pp. 69–70; the actual boiling points are 108° and 118°, respectively.) (d) The order of boiling points is 1-chloropropane < 1-chlorobutane < 1-pentanol. Even though it has lower molecular mass, the hydrogen-bonding capability of the alcohol causes it to have a higher boiling point than 1-chlorobutane. (The actual boiling points are 77°, 109°, and 137°, respectively). (f) The order of boiling points is cyclobutane < chlorocyclobutane < cyclooctane. Use the facts that the chlorine has about the same mass as an ethyl group, and that alkyl chlorides and alkanes of the same molecular mass have about the same boiling points. Then order the boiling points by molecular mass. (The actual boiling points are 12°, 83°, and 151°, respectively.) 8.33 (b) Use the same reasoning as in part (a). The N— H hydrogens of the first compound, acetamide, can be involved in hydrogen bonds between molecules that involve the oxygen or the nitrogen as an acceptor. The second compound, N,N-dimethylacetamide, has no hydrogens that can be involved in hydrogen bonding. The hydrogen bonding in the liquid state of acetamide is reflected in a higher boiling point. 8.34 (b) Two other chiral ethers with the formula C5H10O and no double bonds are CH3 O CH3 O CH2CH3 (d) Two other chiral alcohols with the formula C4H6O are H 2C H 3C OH OH (g) The stereoisomeric forms of 1,2-butanediol consist of a pair of enantiomers: OH HOCH2 CH CH2CH3 1,2-butanediol 48 Chapter 8 Introduction to Alkyl Halides, Alcohols, Ethers, Thiols, and Sulfides (i) O O OH CH3 CH3 (S)-2-methyloxirane O oxetane cyclopropanol (R)-2-methyloxirane 8.35 (b) deuterium, D2 (d) methane, CH4 (f) propane, CH3CH 2CH 3 8.37 (a) Only the alcohol, 2-methylcyclohexanol, will react with sodium hydride to give a gaseous product (H2). (b) The alcohol 2-(methylthio)ethanol, CH3SCH2CH2OH, because of its considerably greater ability to serve as both a hydrogen-bond acceptor and a hydrogen-bond donor, is the water-soluble compound. 8.39 (b) Isobutane, (CH3)3CH, gives two achiral monochlorination products: isobutyl chloride, (CH3)2CHCH 2Cl, and tert-butyl chloride, (CH3)3C—Cl. 8.40 (b) 2-chloroethanol < 3-chloro-1-propanethiol < 2-chloro-1-propanethiol. Thiols are more acidic than alcohols (element effect). 2-Chloro-1-propanethiol is the more acidic thiol because the electronegative chlorine is closer to the site of negative charge in the conjugate-base thiolate anion. (d) –O— CH2CH2— OH < CH3CH2CH2— OH < CH3O— CH2CH2— OH. 2-Methoxyethanol is most acidic because of the electron-withdrawing polar effect of the oxygen. The anion is least acidic because the negative charge on the oxygen interacts repulsively with a second negative charge formed on ionization of the O— H group: repulsive interaction of two negative charges is a destabilizing effect – O 8.46 CH2CH2 O H ionization – O CH2CH2 – O (a) The greater equilibrium constant for dimerization of ethanol shows that ethanol has the stronger hydrogen bond. (b) Use the same equation as in the solution to Problem 8.45(b): ∆G° = –2.303RT(log K eq) = –5.706(log 0.004) = +13.7 kJ/mol (c) Neglect the volume of added thiol and use the same technique used in solving Problem 8.45(c). In this case, however, the equilibrium constant is very small; hence, x is small compared to 1.0, and the denominator can be taken as 1.0. The concentration of dimer is about 0.004 M, and that of free ethanethiol is about 0.992 M. Thus, hardly any (less than 1%) of the ethanethiol is dimerized. (Substituting these values back into the mass law for dimerization in the solution to Problem 8.45(c) shows that the assumptions made above are quite good.) This calculation shows quantitatively that the conclusion given in part (a) above is valid, namely, that hydrogen bonding in ethanethiol is much weaker than that in ethanol. 8.51 (b) The gauche conformation is stabilized by intramolecular hydrogen bonding: Instructor’s Solutions to Additional Problems Chapter 8 49 H intramolecular hydrogen bond O H OH H H H gauche conformation of ethylene glycol In the anti conformation, the two — OH groups are too far apart for intramolecular hydrogen bonding to occur. 8.54 (b) The reasoning is much the same as that in part (a). Because the C— O bonds are shorter in the ether than the corresponding C— C bonds in butane, the methyl groups are brought closer together in the gauche conformation of the ether than they are in the gauche conformation of butane. Consequently, van der Waals repulsions in the gauche conformation of the ether are somewhat greater than they are in the gauche conformation of butane. The greater energy of the gauche conformation of the ether causes less of this conformation to be present at equilibrium. Thus, butane contains more gauche conformation at equilibrium. ··········································································································