Baixe Carey - Organic Chemistry - sgchapt04 e outras Manuais, Projetos, Pesquisas em PDF para Química, somente na Docsity! Back| CHAPTER 4 ALCOHOILS AND ALKYL HALIDES SOLUTIONS TO TEXT PROBLEMS 4.1 There are four C,H, alkyl groups, and so there are four C,H,CI alkyl chlorides. Each may be named by both the functional class and substitutive methods. The functional class name uses the name of the alkyl group followed by the halide as a second word. The substitutive name modifies the name of the corresponding alkane to show the location of the halogen atom. Functional class name Substitutive name CH,CH,CH,CH,CI n-Butyl chloride 1-Chlorobutane (Butyl chloride) CHHCHCH, sec-Butyl chloride 2-Chlorobutane a (1-Methylpropyl chloride) A Isobutyl chloride 1-Chloro-2-methylpropane cH (2-Methylpropyl chloride) 3 Lo CH.CCH, | tert-Butyl chloride 2-Chloro-2-methylpropane cal (1,1-Dimethylethyl chloride) 4.2 Alcohols may also be named using both the functional class and substitutive methods, as in the previous problem. 67 Forward] MainMenu| TOC] StudyGuideTOC| StudentOLC| | MHHE Website 68 ALCOHOLS AND ALKYL HALIDES Functional class name Substitutive name CH,CH,CH,CH,OH | n-Butyl alcohol 1-Butanol (Butyl alcohol) CHCHCH,CH, sec-Butyl alcohol 2-Butanol 0H (1-MethyIpropyl alcohol) CHQHCH:OH Isobutyl alcohol 2-Methyl-1-propanol cH, (2-MethyIpropyl alcohol) cH, CH.CCH, tert-Butyl alcohol 2-Methyl-2-propanol oH (1,1-Dimethylethyl alcohol) 4.3 Alcohols are classified as primary, secondary, or tertiary according to the number of carbon substituents attached to the carbon that bears the hydroxyl group. t t CHCHCH—G—OH CH—€—CHCH; H oH Primary alcohol Secondary alcohol (one alkyl group bonded to—CH,OH) (two alkyl groups bonded to SCHOH) H cH, (CH 0H CH;—C—OH H CH, Primary alcohol Tertiary alcohol (one alkyl group bonded to—CH,OH) (three alkyl groups bonded to SCOH) 4.4 Dipole moment is the product of charge and distance. Although the electron distribution in the carbon-chlorine bond is more polarized than that in the carbon-bromine bond, this effect is counterbalanced by the longer carbon-bromine bond distance. u=e-d: Dipole moment ) Distance Charge +— +— cH—CI CH;—Br Methyl chloride Methyl bromide (greater value of e) (greater value of d) n19D n18D 4.5 AI the hydrogens in dimethyl ether (CH,OCH.) are bonded to carbon; therefore, intermolecular hydrogen bonding between dimethyl ether molecules does not take place, and its boiling point is lower than that of ethanol (CH,CH,OH), where hydrogen bonding involving the —OH group is important. 4.6 Ammonia is a base and abstracts (accepts) a proton from the acid (proton donor) hydrogen chloride. un HCé *NH, + Base Acid Conjugate — Conjugate acid base Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website Back| Forward ALCOHOLS AND ALKYL HALIDES 71 4.16 (b) Writing the equations for carbon-carbon bond cleavage in propane and in 2-methylpropane, we see that a primary ethyl radical is produced by a cleavage of propane whereas a secondary isopropyl radical is produced by cleavage of 2-methyIpropane. cHcH2CCH, CHÉH, + CH, Propane Ethy radical Methyl radical CH A . cHíCHcH, CHÉHCH, + CH, 2-Methylpropane Isopropy radical Methyl radical A secondary radical is more stable than a primary one, and so carbon-carbon bond cleavage of 2-methylpropane requires less energy than carbon-carbon bond cleavage of propane. (c) | Carbon-carbon bond cleavage of 2,2-dimethylpropane gives a tertiary radical. qu cH, crçem, cH—cí + CH, da, cH, 2,2-Dimethylpropane ter-Butyl radical Methyl radical As noted in part (b), a secondary radical is produced on carbon-carbon bond cleavage of 2-methylpropane. We therefore expect a lower carbon-carbon bond dissociation energy for 2,2-dimethylpropane than for 2-methylpropane, since a tertiary radical is more stable than a secondary one. 4.17 First write the equation for the overall reaction. CHCL + CC — cHa, + HcI Chloromethane — Chlorine Dichloromethane Hydrogen chloride The initiation step is dissociation of chlorine to two chlorine atoms. Chiorine 2 Chlorine atoms A chlorine atom abstracts a hydrogen atom from chloromethane in the first propagation step. H 7) H AO 8 —— cid. + d d Chloromethane Chlorine atom Chloromethyl radical Hydrogen chloride Chloromethyl radical reacts with Cl, in the next propagation step. Chloromethyl radical Chlorine Dichloromethane Chiorine atom MainMenu) TOC] StudyGuideTOC| Student OLC| | MHHE Website 72 Back| ALCOHOLS AND ALKYL HALIDES 4.18 Writing the structural formula for ethyl chloride reveals that there are two nonequivalent sets of hydrogen atoms, in either of which a hydrogen is capable of being replaced by chlorine. CH,CH,CI CH,CHCL, + CICH;CH,CI — o, Tighr or hear Ethyl chloride 1,1-Dichloroethane 1.2-Dichloroethane The two dichlorides are 1,1-dichloroethane and 1,2-dichloroethane. 4.19 Propane has six primary hydrogens and two secondary. In the chlorination of propane, the relative proportions of hydrogen atom removal are given by the product of the statistical distribution and the relative rate per hydrogen. Given that a secondary hydrogen is abstracted 3.9 times faster than a pri- mary one, we write the expression for the amount of chlorination at the primary relative to that at the secondary position as: Number of primary hydrogens X rate of abstraction of primary hydrogen 6x1 0.77 Number of secondary hydrogens X rate of abstraction of a secondary hydrogen 2xX3.9 1.00 Thus, the percentage of propyl chloride formed is 0.77/1.77, or 43%, and that of isopropyl chloride is 57%. (The amounts actually observed are propy1 45%, isopropyl 55%.) 4.20 (b) Im contrast with free-radical chlorination, alkane bromination is a highly selective process. The major organic product will be the alkyl bromide formed by substitution of a tertiary hydrogen with a bromine. De Br, DX CHCH). tm C(CH;), A [ Tertiary hydrogen Br L-Isopropyl-1- 1-(1-Bromo- |-methylethy])- methyleyclopentane I-methyleyclopentane (c) — Asin part (b), bromination results in substitution of a tertiary hydrogen. cH, CH, cH, CH, CHiÇCH.CHCH; —a— cuscnçer, cH, cH, Br 2,2,4-Trimethylpentane 2-Bromo-2,4,4-trimethylpentane 4.21 (a) Cyclobutanol has a hydroxyl group attached to a four-membered ring. Son Cyclobutanol (b) | sec-Butyl alcohol is the functional class name for 2-butanol. CH,CHCH,CH, du sec-Butyl alcohol (c) The hydroxyl group is at C-3 of an unbranched seven-carbon chain in 3-heptanol. CH,CH,CHCH.CH,CH,CH, | 0H 3-Heptanol Forward] MainMenu| TOC] StudyGuideTOC| StudentOLC| | MHHE Website ALCOHOLS AND ALKYL HALIDES 73 (d) A chlorine at C-2 is on the opposite side of the ring from the C-1 hydroxyl group in trans- 2-chlorocyclopentanol. Note that it is not necessary to assign a number to the carbon that bears the hydroxyl group; naming the compound as a derivative of cyclopentanol automatically requires the hydroxyl group to be located at C-1. e ai trans-2-Chloroeyclopentanol (e) This compound is an alcohol in which the longest continuous chain that incorporates the hydroxyl function has eight carbons. It bears chlorine substituents at C-2 and C-6 and methyl and hydroxyl groups at C-4. cH, CHICHCH CH CHCH CH; cl 0H 2.6-Dichloro-4-methyl-4-octanol (f) The hydroxyl group is at C-1 in trans-4-tert-buty lcyclohexanol; the tert-butyl group is at C-4. The structures of the compound can be represented as shown at the left; the structure at the right depicts it in its most stable conformation. Ni OH O (CH.C H trans-4-tert-Butyleyclohexanol (CH)€ (g) The cyclopropyl group is on the same carbon as the hydroxyl group in 1-cyclopropylethanol. Dçuom CH, 1-Cyelopropylethanol (h) The cyclopropyl group and the hydroxyl group are on adjacent carbons in 2-cyclopro- pylethanol. D-ca,cn,oH 2-Cyelopropylethanol 4.22 (a) This compound has a five-carbon chain that bears a methyl substituent and a bromine. The numbering scheme that gives the lower number to the substituent closest to the end of the chain is chosen. Bromine is therefore at C-1, and methyl is a substituent at C-4. CHSCHCHCHsCH;Br cH, 1-Bromo-4-methylpentane Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website 76 ALCOHOLS AND ALKYL HALIDES In the other isomers methyl and hydroxyl will be in a 1,2, 1,3, or 1,4 relationship and can be cis or trans in each. We can write the preferred conformation by recognizing that the methyl group will always be equatorial and the hydroxy1 either equatorial or axial. 0H 0H SOJA [28 LT ne cH, trans-2-Methyleyclohexanol cis-3-Methyleyclohexanol trans-4-Methyleyclohexanol oH OH oH [Tea / 7 S H LI : CH, 3 cis-2-Methyleyclohexanol trans-3-Methyleyclohexanol cis-4-Methyleyclohexanol 4.25 The assumption is incorrect for the 3-methylcyclohexanols. cis-3-Methylcyclohexanol is more stable than trans-3-methylcyclohexanol because the methyl group and the hydroxyl group are both equa- torial in the cis isomer, whereas one substituent must be axial in the trans. oH [rm oa CH, CH, cis-3-Methyleyclohexanol more trans-3-Methyleyclohexanol less stable: smaller heat of combustion stable; larger heat of combustion 4.26 (a) The most stable conformation will be the one with all the substituents equatorial. HC OH LL entao, The hydroxyl group is trans to the isopropyl group and cis to the methyl group. (b) All three substituents need not always be equatorial; instead, one or two of them may be axial. Since neomenthol is the second most stable stereoisomer, we choose the structure with one axial substituent. Furthermore, we choose the structure with the smallest substituent (the hydroxyl group) as the axial one. Neomenthol is shown as follows: oH ur CH(CH;), 4.27 | Inall these reactions the negatively charged atom abstracts a proton from an acid. (a) HI + HO == I + H,O Hydrogen iodide: acid Hydroxide ion: lodide ion: Water: conjugate acid (stronger acid, K, = 10º) base conjugate base (weaker acid, K,= 10719) | f (b) CHCHO | + | CcHCOH = cH,CHOH | + CHCO Ethoxide ion: Acetic acid: acid Ethanol: conjugate acid Acetate ion: base (stronger acid, K, = 105) (weaker acid, K, = 10-19) conjugate base Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website ALCOHOLS AND ALKYL HALIDES 77 (c) HF + HN == F- + H;N Hydrogen fluoride: acid Amide ion: Fluoride ion: Ammonia: conjugate acid (stronger acid, K,= 10) base conjugate base (weaker acid, K, = 10) (d) CH;CO” + HCI = cH,COH + cr Acetate ion: Acetic acid: conjugate acid Chloride ion: base (weaker acid, K, = 10) conjugate base (e) (CH)CO” + H,O == (CH).COH | + | HO” tert-Butoxide ion: Water: acid tert-Butyl alcohol: Hydroxide ion: base (stronger acid, K, = 10719) conjugate acid conjugate base (weaker acid, K, = 1018) (9) (CH)CHOH + HN ==> (CHycHo + H;N Isopropyl alcohol: acid Amide ion: Isopropoxide ion: Ammonia: conjugate acid (stronger acid, K, = 1077) base conjugate base (weaker acid, K,= 10) [5 + HO, = HF + HSO, Fluoride ion: Sulfuric acid: acid Hydrogen fluoride: Hydrogen base (stronger acid, K, = 10º) conjugate acid sulfate ion: (weaker acid, K,= 10") conjugate base 4.28 (a) The proton-transfer transition state can represent the following reaction, or its reverse: OC NARA > (CH, CH— HCBr (CH), CH—O-—-H--Br” (CH).CH—OH + Br Base Acid (stronger acid) Conjugate acid Conjugate K = 10º (weaker acid) base K= 10" When the reaction proceeds as drawn, the stronger acid (hydrogen bromide) is on the left, the weaker acid (isopropyl alcohol) is on the right, and the equilibrium lies to the right. (b) Hydroxide is a strong base; methyloxonium ion is a strong acid. H H CN a/ K=1 n va + H-oS — HOM + O CH, CH, Hydroxideion | Methyloxonium Water Methanol (base) ion (acid) (conjugate acid) (conjugate base) 4.29 (a) This problem reviews the relationship between logarithms and exponential numbers. We need to determine K,, given pK,. The equation that relates the two is PK, = logo K, Therefore K,= 10" = 1048 =33x 10! Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website 78 ALCOHOLS AND ALKYL HALIDES (b) As described in part (a), K, = 10-P&, therefore K, for vitamin C is given by the expression: K, = 107 =67x 105 (c) Similarly, K, = 1.8 X 107 for formic acid (pK, 3.75). (d) K,=6.5 X 1072 for oxalic acid (pK, 1.19). In ranking the acids in order of decreasing acidity, remember that the larger the equilibrium constant K,, the stronger the acid; and the lower the pK, value, the stronger the acid. Acid K, PK, Oxalic (strongest) 6.5 x 107? 1.19 Aspirin 3.3 x 10* 348 Formic acid 1.8 x 1074 3.75 Vitamin C (weakest) 6.7 x 10 417 4.30 Because the pK, of CH;SH (11) is smaller than that of CH;OH (16), CH;SH is the stronger acid of the two. Its conjugate base (as in KSCH,) is therefore weaker than the conjugate base of CH;OH (as in KOCH;). 4.31 This problem illustrates the reactions of a primary alcohol with the reagents described in the chapter. (a) CH;CH,CH,CH,0H + NaNH, ——» CH;CH.CH,CH,O Na” + NH, Sodium butoxide HBr (b) | CH.CH,;CH,CH,0H e CH,CH,CH,CH,Br 1-Bromobutane NaBy, H;SO, (0) CH,CH,CH,CH,0H CH,CH.CH,CH,Br 1-Bromobutane (d) CH;CH,CH,CH,OH ———+> CH,CH,CH,CH;Br 1-Bromobutane SOCL, (e) CH,CH;CH,CH;OH —É, cH,CH,CH,CH,CI 1-Chlorobutane 4.32 (a) This reaction was used to convert the primary alcohol to the corresponding bromide in 60% yield. PBr, CHCHOH dn” CH,CH;Br (b) Thionyl chloride treatment of this secondary alcohol gave the chloro derivative in 59% yield. cH, ? cH, 7 COCH,CH, soci, COCH,CH, — Soh, pyridine 0H ci Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website Back| ALCOHOLS AND ALKYL HALIDES 81 (b) The positively charged carbon in the carbocation intermediate is sp”-hybridized, and is planar. H ue /27 Carbocation intermediate (c) | Bromide ion attacks the carbocation from both above and below, giving rise to two stereo- isomers, cis- and trans-1-bromo-4-methylcyclohexane. Br ue Z27 ne Per cis-1-Bromo-4-methyleyclohexane trans-1-Bromo-4-methyleyclohexane 4.37 | Examine the equations to ascertain which bonds are made and which are broken. Then use the bond dissociation energies in Table 4.3 to calculate AHº for each reaction. (a) (CH)CH—OH | + HF (CH)CH—F + H—OH 385 kJ/mol 568 kJ/mol 439 kJ/mol 497 K/mol (92 keal/mol) (136 kcal/mol) (105 kcal/mol) (119 kcalmol) Bond breaking: 953 kJ/mol (228 kcal/mol) Bond making: 936 kJ/mol (224 kcal/mol) AHº = energy cost of breaking bonds — energy given off in making bonds = 953 kJ/mol — 936 kJ/mol (228 kcal/mol — 224 kcal/mol) = +17 KJ/mol (+4 kcal/mol) The reaction of isopropyl alcohol with hydrogen fluoride is endothermic. (b) (CH)CH—OH + H—a (CH)CH—CI + H—OH 385 kJ/'mol 431 K/mol 339 kJ/mol 497 KJ/mol (92 kcal/mol) (103 kcal/mol) (81 kcal/mol) (119 kcal/mol) Bond breaking: 816 kJ/mol (195 kcal/mol) Bond making: 836 kJ/mol (200 kcal/mol) AHº = energy cost of breaking bonds — energy given off in making bonds 816 kJ/mol — 836 kJ/mol (195 kcal/mol — 200 kcal/mol) = —20 kJ/mol (—5 kcal/mol) The reaction of isopropyl alcohol with hydrogen chloride is exothermic. (o) CHÇHCH; + H—CI —— CHCHCH; + HH H a 397 KJ/mol 431 KJ/mol 339 kJ/mol 435 KJ/mol (95 kcal/mol) (103 kcal/mol) (81 kcal/mol) (104 kcal/mol) Bond breaking: 828 kJ/mol (198 kcal/mol) Bond making: 774 kJ/mol (185 kcal/mol) AHº = energy cost of breaking bonds — energy given off in making bonds = 828 kJ/mol — 774 kJ/mol (198 kcal/mol — 185 kcal/mol) +54 kJ/mol (+13 kcal/mol) The reaction of propane with hydrogen chloride is endothermic. 4.38 Im the statement of the problem you are told that the starting material is 2,2-dimethyIpropane, that the reaction is one of fluorination, meaning that F, is a reactant, and that the product is (CF;),C. You Forward] MainMenu| TOC] StudyGuideTOC| StudentOLC| | MHHE Website 82 Back| ALCOHOLS AND ALKYL HALIDES need to complete the equation by realizing that HF is also formed in the fluorination of alkanes. The balanced equation is therefore: (CH) + I2F, (CF) + 12HF 4.39 The reaction is free-radical chlorination, and substitution occurs at all possible positions that bear a replaceable hydrogen. Write the structure of the starting material, and identify the nonequivalent hydrogens. q Ci —E—cH; F H 1.2-Dichloro-1, | -difluoropropane The problem states that one of the products is 1,2,3-trichloro-1,1-difluoropropane. This compound arises by substitution of one of the methyl hydrogens by chlorine. We are told that the other product is an isomer of 1,2,3-trichloro-1,1-difluoropropane; therefore, it must be formed by replacement of the hydrogen at C-2. Li Li area oq F H F cl 1,23-Trichloro-1,1-difluoropropane 1.2,2-Trichloro-l,I-difluoropropane 4.40 Free-radical chlorination leads to substitution at each carbon that bears a hydrogen. This problem es- sentially requires you to recognize structures that possess various numbers of nonequivalent hydro- gens. The easiest way to determine the number of constitutional isomers that can be formed by chlo- rination of a particular compound is to replace one hydrogen with chlorine and assign an IUPAC name to the product. Continue by replacing one hydrogen on each carbon in the compound, and compare names to identify duplicates. (a) 2,2-DimethyIpropane is the C;H,, isomer that gives a single monochloride, since all the hydro- gens are equivalent. cH, cH, I ci | CH,CCH, e CH,CCH,CI cH, cH, 2,2-Dimethylpropane 1-Chloro-2,2-dimethylpropane (b) The CH, isomer that has three nonequivalent sets of hydrogens is pentane. It yields three isomeric monochlorides on free-radical chlorination. CICH,CH,CH.CH,CH, 1-Chloropentane cl, CH,CH,CH,CH,CH, — CH;CHCH,CH,CH, 2-Chloropentane CHCH.CHCH.CH; cl 3-Chloropentane Forward] MainMenu| TOC] StudyGuideTOC| StudentOLC| | MHHE Website Back| 441 ALCOHOLS AND ALKYL HALIDES 83 (c) | 2-Methylbutane forms four different monochlorides. CICH,CHCH,CH, du, 1-Chloro-2-methylbutane CH.CHCH,CH, —S (CH;CCH,CH, du, d 2-Chiloro-2-methylbuta 2-Methylbutane Oro-d-methyIbutane (CH CHÇHCH; a 2-Chloro-3-methylbutane (CH;)CHCH,CH,CI 1-Chloro-3-methylbutane (d) | For only two dichlorides to be formed, the starting alkane must have a structure that is rather symmetrical; that is, one in which most (or all) of the hydrogens are equivalent. 2,2-Di- methylpropane satisfies this requirement. 1º ii ii CH,;CCH —., CH,CCHCI, + CICH,CCH,CI o Tight a 2 ado cH, cH, cH, 2,2-Dimethylpropane 1,1-Dichloro-2,2- 1,3-Dichloro-2,2- dimethylpropane dimethylpropane (a) -Heptane has five methylene groups, which on chlorination together contribute 85% of the total monochlorinated product. CH,(CH,.CH, CH;(CH,),CH,CI + (2-chloro + 3-chloro + 4-chloro) 15% 85% Since the problem specifies that attack at each methylene group is equally probable, the five methylene groups each give rise to 85/5, or 17%, of the monochloride product. Since C-2 and C-6 of heptane are equivalent, we calculate that 2-chloroheptane will con- stitute 34% of the monochloride fraction. Similarly, C-3 and C-5 are equivalent, and so there should be 34% 3-chloroheptane. The remainder, 17%, is 4-chloroheptane. These predictions are very close to the observed proportions. Calculated, % Observed, % 2-Chloro 34 35 3-Chloro 34 34 4-Chloro 17 16 (b) There are a total of 20 methylene hydrogens in dodecane, CH(CH,) )CH,. The 19% 2-chlorododecane that is formed arises by substitution of any of the four equivalent methyl- ene hydrogens at C-2 and C-11. The total amount of substitution of methylene hydrogens must therefore be: Px 19% = 95% Forward] MainMenu| TOC] StudyGuideTOC| StudentOLC| | MHHE Website 86 ALCOHOLS AND ALKYL HALIDES SELF-TEST PART A A-1. Give the correct substitutive IUPAC name for each of the following compounds: 7” cH,0H (a) (b) CH;CH,CHCH,CHCH, “Br CH.CH, A-2. Draw the structures of the following substances: (a) | 2-Chloro-1-iodo-2-methylheptane (b) | cis-3-Isopropylcyclohexanol A-3. Give both a functional class and a substitutive TUPAC name for each of the following compounds: oH cl (a) AA O) ps A-4. What are the structures of the conjugate acid and the conjugate base of CH;OH? A-5. Supply the missing component for each of the following reactions: SOCL, (a) CH;CHCHOH — > 9 Br db ? HBr CH,CHC(CH;), A-6. (a) Write the products of the acid-base reaction that follows, and identify the stronger acid and base and the conjugate of each. Will the equilibrium lie to the left (K < 1) or to the right (K > 1)? The approximate pK, of NH, is 36; that of CH,CH,OH is 16. CH,CH,O” + NH, (b) | Draw a representation of the transition state of the elementary step of the reaction in part (a). A-7. (a) How many different free radicals can possibly be produced in the reaction between chlorine atoms and 2,4-dimethylpentane? (b) Write their structures. (c) | Which is the most stable? Which is the least stable? A-8. Write a balanced chemical equation for the reaction of chlorine with the pentane isomer that gives only one product on monochlorination. Write the propagation steps for the light-initiated reaction of bromine with methylcyclo- hexane. A-10. Using the data in Table B-1 of this Study Guide, calculate the heat of reaction (AH?) for the light-initiated reaction of bromine (Br,) with 2-methylpropane to give 2-bromo-2- methylpropane and hydrogen bromide. A-M. (a) Write out each of the elementary steps in the reaction of tert-butyl alcohol with hydrogen bromide. Use curved arrows to show electron movement in each step. (b) | Draw the structure of the transition state representing the unimolecular dissociation of the alkyloxonium ion in the preceding reaction. Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website ALCOHOLS AND ALKYL HALIDES 87 (c) | How does the mechanism of the reaction between 1-butanol and hydrogen bromide differ from the reaction in part (a)? A-12. (Choose the correct response for each part.) Which species or compound: (a) Reacts faster with sodium bromide and sulfuric acid? 2-methyl-3-pentanol or 3-methyl-3-pentanol (b) Isa stronger base? KOoC(CH), or HOC(CH), (c) - Reacts more vigorously with cyclohexane? Fluorine or iodine (d) | Has an odd number of electrons? Ethoxide ion or ethyl radical (e) - Undergoes bond cleavage in the initiation step in the reaction by which methane is converted to chloromethane? CH, or Ch PART B B-1. A certain alcohol has the functional class TUPAC name 1-ethyl-3-methylbutyl alcohol. What is its substitutive name? (a) —1-Ethyl-3-methyl-1-butanol (d) | 2-Methyl-4-hexanol (b) | 2-Methyl-1-hexanol (e) | 5-Methyl-3-hexanol (c) | 3-Methyl-1-hexanol B-2. Rank the following substances in order of increasing boiling point (lowest — highest): CH,CH;CH.CH,OH (CH),CHOCH, (CH,),;COH (CH, 1 2 3 4 (a) 1<3<2<4 (cd) 4<2<3<1 (0) 4<3<2<1 (b) 2<4<3<1 (d) 2<3<1<4 B-3. Which one of the following reacts with HBr at the fastest rate? cH, oH oH ” = CH; (4) ue 477 B-4. What is the decreasing stability order (most stable > least stable) of the following carbo- cations? A A AA A Dn ON + Õ 1 2 3 4 5 (a) 3>2>1>4>5 () 3>2=5>1=4 b) 1=4>2=5>3 (d) 3>1=4>2=5 Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website 88 ALCOHOLS AND ALKYL HALIDES B-5. Rank the bond dissociation energies (BDEs) of the bonds indicated with the arrows from smallest to largest. H (a) 1<2<3 (d) 1<3<2 1 [NO (b) 3<2<1 (e) 3<1I<2 mn CH, (0) 2<3<1 EN 3 B-6. What are the chain-propagating steps in the free-radical chlorination of methane? 1. Ch 2Cl- 4. H + CL HCl + Cl- 2. Cr+CH — CHCI+ H: 5. CH, + CL CH,CI + CI- 3 Cr+CH — “CH, + HCl 6. CH, + CH — CH, + “CH, (a) 2,4 b 1,2 () 35 (d) 1,3,5 (e) A combination different from those listed B-7. Which of the following is least able to serve as a nucleophile in a chemical reaction? (a) Br (b) OH” (c) NH, (d) CH B-8. Thiols are alcohol analogs in which the oxygen has been replaced by sulfur (e.g., CH,SH). Given the fact that the S—H bond is less polar than the O—H bond, which of the following statements comparing thiols and alcohols is correct? (a) - Hydrogen bonding forces are weaker in thiols. (b) | Hydrogen bonding forces are stronger in thiols. (c) | Hydrogen bonding forces would be the same. (d) | No comparison can be made without additional information. B-9. Rank the transition states that occur during the following reaction steps in order of increas- ing stability (least — most stable): 1 cH—ÓH, cH;* + HO 2 (CHICÓH, — (CHC! + HO 3. (CH)CH—ÓH, —— (CH),CH* + HO (a) 1<2<3 (b) 2<3<1 () 1<3<2 (d) 2<1<3 B-10. Using the data from Appendix B (Table B-1), calculate the heat of reaction AHº for the following: CH,CH,: + HBr CH,CH, + Br: (a) | +69 kJ/mol (+16.5 kcal/mol) (b) | —69 KJ/mol (— 16.5 kcal/mol) (c) +44 kJ/mol (+10.5 kcal/mol) (d) | —44 kJ/mol (—10.5 kcal/mol) B-11. An alkane with a molecular formula C,H,, reacts with chlorine in the presence of light and heat to give four constitutionally isomeric monochlorides of molecular formula C;H CI. What is the most reasonable structure for the starting alkane? (a) - CH;CH,CH,CH,CH,CH, (d) (CH;);CCH,CH, (b) (CH;)CHCH,CH,CH, (e) (CH),CHCH(CH;), (e) CH;CH(CH,CH;), B-12. The species shown in the box represents of the reaction between isopropyl alcohol and hydrogen bromide. as + (a) — the alkyloxonium ion intermediate (CH;),CH------0H, (b) the transition state of the bimolecular proton transfer step Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website