北京化工大学：《有机化学》课程教学资源（习题与答案）Chapter 19 Carboxylic Acids

1559Tch19344-35710/17/0513:51Page344 EQA 19 Carboxylic Acids s.You will dis my wnen you study nu oxy group Outline of the Chapter 19-1 Naming the Carboxylic Acids 192,193rgce 19-4 Acidity and Basicity of Carboxylic Acids An analysis of resonance and inductive effects. 19.5, 197 Reactivity of the Carboxy Group:Addition and Elimination 19-8 Alkanoyl Halides and Anhydrides 199 Esters 19-10 Amides Conversion of acids into their most important derivatives. 19-11 Nucleophilic Attack at a Carboxylate Group Reactions of acids with hydride. 19.12 -Bromination cial value in synthesis. 19-13 Carboxylic Acids in Nature 344

19 Carboxylic Acids The carboxylic acids and their derivatives contain a carbonyl group as one primary source of reactivity and therefore have a lot in common with aldehydes and ketones. However, two major features make carboxylic acids different from aldehydes and ketones. First, the HO proton is made much more acidic by the neighbor￾ing CPO function. Second, the HO may behave as a leaving group under the right conditions. You will dis￾cover the importance of this ability when you study nucleophilic additions to the carboxy group. Outline of the Chapter 19-1 Naming the Carboxylic Acids 19-2, 19-3 Physical Properties of Carboxylic Acids Very strongly hydrogen-bonding molecules; spectroscopy. 19-4 Acidity and Basicity of Carboxylic Acids An analysis of resonance and inductive effects. 19-5, 19-6 Preparation of Carboxylic Acids Reviewing some reactions you’ve seen before. 19-7 Reactivity of the Carboxy Group: Addition and Elimination A new, general mechanism sequence. 19-8 Alkanoyl Halides and Anhydrides 19-9 Esters 19-10 Amides Conversion of acids into their most important derivatives. 19-11 Nucleophilic Attack at a Carboxylate Group Reactions of acids with hydride. 19-12
-Bromination A reaction of special value in synthesis. 19-13 Carboxylic Acids in Nature 344 1559T_ch19_344-357 10/17/05 13:51 Page 344

1559T_ch19_344-35710/17/0513:51Pa9e345 ⊕ EQA Keys to the Chapter·345 Keys to the Chapter 19-1 through 19-3.Nomenclature and Physical Properties of Carboxylic Acids The one characteristic that is unique to carboxylic acids is the strong tendency to form hydrogen-bon nP米本 a process res e-stabilizing ability in the structures of the (D)th :0 RCOH =H++RCO: ←一RC=O: ROn一H++RO: Both negatively charged coniugate bases have the same negative atom (O).but the carboxylate ion is stabi lized by inductive effects and resonance.Because the carboxylate is more stable.the carboxylic acid is a stronger acid than the alcoho Carboxylic acids vs.ketones or aldehydes RCOH=r+RC6:一RC=: 0 :6 RCCH=H*+RCCH2←一RC=CH Neutral Negatively charged Both negatively charged conjugate bases are stabilized by inductive effects of acarbonyl carbon.Both e es and explananons 19-6.Preparation of Carboxylic Acids sions (mainl time to do a quick review and then begin organizing these synthetic methods into appropriate categories

Keys to the Chapter • 345 Keys to the Chapter 19-1 through 19-3. Nomenclature and Physical Properties of Carboxylic Acids The one characteristic that is unique to carboxylic acids is the strong tendency to form hydrogen-bonded dimers, a process resulting in much higher melting and boiling points relative to comparable compounds (Table 19-2). Notice also the very low field position of the COOH proton in the NMR. 19-4. Acidity and Basicity of Carboxylic Acids Recall that acid and base strengths are determined by the charge-stabilizing ability in the structures of the acid-base conjugate pair. There are three points to consider in evaluating the stability of a charged species: (1) the electronegativity of the charged atom(s), (2) the size of the charged atom(s), (3) stabilization of charge by either inductive effects or resonance. For carboxylic acids, the text sets up two comparisons. Both negatively charged conjugate bases have the same negative atom (O), but the carboxylate ion is stabi￾lized by inductive effects and resonance. Because the carboxylate is more stable, the carboxylic acid is a stronger acid than the alcohol. Both negatively charged conjugate bases are stabilized by inductive effects of a carbonyl carbon. Both are also stabilized by resonance, but the carboxylate ion distributes its negative charge between two elec￾tronegative oxygens, whereas the enolate distributes its charge between a carbon and only one electronegative oxygen. The carboxylate ion is more stable, so the carboxylic acid is more acidic. Similar analyses can be applied to base strength as well. See chapter-end Problems 25 and 41 and their an￾swers for examples and explanations. 19-6. Preparation of Carboxylic Acids Even though the most obvious syntheses of carboxylic acids are simple functional group interconversions (mainly oxidations), there are quite a few that involve breaking or forming carbon–carbon bonds. You might take the time to do a quick review and then begin organizing these synthetic methods into appropriate categories. O RCOH H Neutral Negatively charged O RCO  O  RC O Carboxylic acids vs. ketones or aldehydes O RCCH3 RCCH2  H CH2 O O  RC O RCOH H ROH Neutral Negatively charged H O RCO  O  RO  RC O Carboxylic acids vs. alcohols 1559T_ch19_344-357 10/17/05 13:51 Page 345

1559r.ch19_344-35710/17/0513:51Page346 EQA 346.chapter 19 CARBOXYUC ACIDS 19-7.Reactivity of the Carboxy Group:Addition and Elimination down some of the simply to have the practice of writing them down.Here are two key points to keep in mind. 1.Carbox potential leav m.It is not the sam as either an SN2 or an SN 2.Feaction,me edo not apply to sphy cleophiles ir the presence of an acid catalyst.Strong bases will deprotonate the acid faster than nucleophilic addition can take place nbase and deprotonati rily pow such as LiAlH. reaction().then function that where.peeause the intermediate is relativelyhig energy compared with the carbonyl product.ch has a veryonr nerset nd.So.even e in the enm onstep and y la ith er).The high energy of the strained three-membered ring ether made possible.To Energetically 1.+OH一N-+H 2 (Strain is relieved) Favorable 3.Nu- R +HO- (Stab s bond (From Nu-+R-C-OH) Note how acid catalysis can be beneficial to all these reactions:Adding a proton to each of these oxygens eaving group 19-8,19-9,and 19-10.Alkanoyl Halides,Anhydrides,Esters,and Amides In these sections te previo secton wil be applied note in each case.To understand these reactions,pick outa few whose mechanismsar not given in detail in the text and try to write them out.step-by-step.The practice will prove to be worthwhile 19-11.Nucleophilic Attack at a Carboxylate Group Strongly basic nucleophiles irreversibly deprotonate

19-7. Reactivity of the Carboxy Group: Addition and Elimination The mechanistic discussion at the start of this section is central to both the chemistry of carboxylic acids and the chemistry of their derivatives. Read it very carefully, and copy down some of the schemes on your own, simply to have the practice of writing them down. Here are two key points to keep in mind. 1. Carboxylic acids and their derivatives contain potential leaving groups attached to their carbonyl carbon. After a nucleophile adds, the leaving group may leave, giving a net substitution reaction overall. This is a two-step, addition–elimination, mechanism. It is not the same as either an SN2 or an SN1 reaction, mechanistically: SN1 and SN2 reactions do not apply to sp2 hybridized carbons. 2. For carboxylic acids, this new type of substitution process occurs best with weakly basic nucleophiles in the presence of an acid catalyst. Strong bases will deprotonate the acid faster than nucleophilic addition can take place. Therefore, if the nucleophile is a strong base and deprotonation is essentially irreversible, nucleophilic addition will be very difficult and will occur only with extraordinarily powerful reagents, such as LiAlH4. In some of these reactions you will see some unexpected species act as leaving groups, including strong bases such as HO and RO. Although these were generally far too basic to be leaving groups in ordinary SN2 reactions (Chapter 6), they can function that way here, because the tetrahedral intermediate is relatively high in energy compared with the carbonyl product, which has a very strong, very stable carbon–oxygen double bond. So, even HO and RO can leave in the elimination step and still result in an energetically favorable process overall. You saw a similar effect in the nucleophilic ring-opening of oxacyclopropanes with bases (Chap￾ter 9). The high energy of the strained three-membered ring ether made expulsion of an alkoxide possible. To summarize, 1. 2. 3. O B (From Nu ROCOOH) Note how acid catalysis can be beneficial to all these reactions: Adding a proton to each of these oxygens before it leaves should help by converting it from a strongly basic alkoxide (i.e., bad) leaving group to a neu￾tral alcohol or water (weakly basic, good leaving group). 19-8, 19-9, and 19-10. Alkanoyl Halides, Anhydrides, Esters, and Amides In these sections the general mechanisms of the previous section will be applied in illustrating the transfor￾mations of carboxylic acids into their four most important derivatives. For later synthetic applications, carefully note the reagents involved in each case. To understand these reactions, pick out a few whose mechanisms are not given in detail in the text and try to write them out, step-by-step. The practice will prove to be worthwhile. 19-11. Nucleophilic Attack at a Carboxylate Group Strongly basic nucleophiles irreversibly deprotonate carboxylic acids, forming carboxylate anions. Addi￾tion–elimination reactions on carboxylate anions are hard to do because (1) the addition is hard to do and (2) HO Nu R R C O Nu OH C O bond Favorable (Stable C is regenerated) O O Nu C C Nu O C Favorable (Strain is relieved) C HO Nu C OH Nu C Unfavorable Energetically 346 • Chapter 19 CARBOXYLIC ACIDS 1559T_ch19_344-357 10/17/05 13:51 Page 346

1559T_ch19_344-35710/17/0513:51Pa9e347 ⊕ EQA Solutions to Problems.347 nother due to electro an派o，或itcul toe可 bad leaving groups Nevertheless.LiAH s capable ofad cohol. te.formally.n aluminum oxide anion leav 19-12.a-Bromination A useful method of -substitution in carboxylic acids is via the -bromo derivative.synthesized using the Hell- Volhard-Zelinsky reaction. Solutions to Problems 23.(a)2-Chloro-4-methylpentanoic acid:(b)2-ethyl-3-butenoic acid (e)E-2-bromo-3.4-dimethyl-2-pentenoic acid (d)cyclopentylacetic acid: (e)trans-2-hydroxycyclohexanccarboxylic acid: (f)E-2-chlorobutenedioic acid: (g)2.4-dihydroxy-6-methylbenzoic acid: (h)1.2-benzenedicarboxylic acid:(i) COOH (k)O H CH.CCOOH CH CH3- -H COOH COOH CO2H CO2H COOH CH COOH CH2OH CHO 24. > The order is the same for both boiling point and solubility in water.The acid has the most hydrogen- bonding capability and will have the highest boiling point (249C)because of its hydrogen-bonde d th quemnater becase they canbyo 25.(a)Order is as written:(b)order is reverse of that given: (c)CHCH,CHCICO,H>CH CHCICH CO,H>CICH CH,CH,CO,H: (d)order is as written:(e)2.4-dinitro>4-nitro>unsubstituted>4-methoxybenzoic

Solutions to Problems • 347 the elimination is hard to do. More specifically, (1) it is difficult to add one anion to another due to electro￾static repulsion and (2) oxide anions are extremely bad leaving groups. Nevertheless, LiAlH4 is capable of ad￾dition to RCOO, and the product of addition can go on to eliminate, formally, an aluminum oxide anion leav￾ing group. The product of an acid LiAlH4 is a primary alcohol. 19-12.
-Bromination A useful method of
-substitution in carboxylic acids is via the
-bromo derivative, synthesized using the Hell￾Volhard-Zelinsky reaction. Solutions to Problems 23. (a) 2-Chloro-4-methylpentanoic acid; (b) 2-ethyl-3-butenoic acid (c) E-2-bromo-3,4-dimethyl-2-pentenoic acid; (d) cyclopentylacetic acid; (e) trans-2-hydroxycyclohexanecarboxylic acid; (f) E-2-chlorobutenedioic acid; (g) 2,4-dihydroxy-6-methylbenzoic acid; (h) 1,2-benzenedicarboxylic acid; (i) H2NCH2CH2CH2COOH (j) (k) (l) (m) (n) 24. The order is the same for both boiling point and solubility in water. The acid has the most hydrogen￾bonding capability and will have the highest boiling point (249°C) because of its hydrogen-bonded dimer formation. The alcohol, which can also hydrogen bond, is next (205°C), the polar aldehyde third (178°C), and the nearly nonpolar hydrocarbon last (115°C). Solubilities follow similar considerations, except that the acid and alcohol are quite similar in water solubility because they can both hydrogen bond with H2O. 25. (a) Order is as written; (b) order is reverse of that given; (c) CH3CH2CHClCO2H  CH3CHClCH2CO2H  ClCH2CH2CH2CO2H; (d) order is as written; (e) 2,4-dinitro  4-nitro  unsubstituted  4-methoxybenzoic COOH  CH2OH  CHO  CH3 CO2H CO2H C COOH H C CH3 COOH CH O H COOH COOH CH3 CH3 H O B CH3CCOOH 1559T_ch19_344-357 10/17/05 13:51 Page 347

1559r.ch19_344-35710/17/0513:51Page348 348.chapter 19 CARBOXYUC ACIDS 26.(a)H=14+2-16:degrees of unsaturation-2 bonds or rings are present. Compare Figure 19-3.This is a carboxylic acid (=1704 and 3040 cm). (b)For B.H14 degrce of unsaturation bonds or rings.From the alarea 4 Hcach)allne igmal Se the picetbe -C- 一C 16 20 which can be simply put together to give cyclohexene! OH Then C via oxymereution-demercurstion: ⊕ (c=c =1649 cm-1 and ic=cH:=888 cm-1)via Wittig reaction: F= 34 (o=3328 cm-)via hydroboration-oxidation: A=[ (e)For G,H=16+2=18:degrees of unsaturation==2 bonds or rings 1R:=1742cm Is C -O.very possibly an ester because of the high value and the large number with pieces such as 2 CH3-0-(8=3.7).2-CH2-C-(?)(=2.4).and two more Φ

26. (a) Hsat  14 2  16; degrees of unsaturation   (16  2  12)  2  bonds or rings are present. Compare Figure 19-3. This is a carboxylic acid ( ˜  1704 and 3040 cm1 ). (b) For B, Hsat  12 2  14; degree of unsaturation   (14  2  10)  2  bonds or rings. From the 13C NMR (three signals), the molecule would seem to have twofold symmetry, with two sets of equivalent pairs of alkyl carbons and a pair of equivalent alkene carbons. The 1 H NMR shows two equal-area upfield signals (4 H each) and a 2 H alkene signal. So the pieces seem to be which can be simply put together to give cyclohexene! (c) For G, Hsat  16 2  18; degrees of unsaturation   (18  2  14)  2  bonds or rings. IR: ˜  1742 cm1 is CPO, very possibly an ester because of the high value and the large number of oxygens in the formula. NMR: Only three signals, with integrations of 4, 4, and 6 H. The molecule must be symmetrical, O B with pieces such as 2 CH3OOO (  3.7), 2 OCH2OCO(?) (  2.4), and two more A  CO2H F ( O H  3328 cm1  ) via hydroboration–oxidation; H 3.4(d) CH2OH ~ E  CH2 ( C C  1649 cm1 and 888 cm1 C CH ) via Wittig reaction: ~ ~ 2  D ( C O  1715 cm1  ); O  208.5(13C) ~ Then C  via oxymercuration–demercuration; OH  69.5(13C) , CH2 Equiv. CH2 CH CH Equiv. CH2 1.6 CH2 2.0 Equiv. 5.7 348 • Chapter 19 CARBOXYLIC ACIDS 1559T_ch19_344-357 10/17/05 13:51 Page 348

1559T_ch19_344-35710/17/0513:51Pa9e349 ⊕ EQA Solutions to Problems.349 CH;-0-C-CH2-CH2 CH3-0-C-CH2-CHz ◇2品q0 .MOO入oH CHO CO2H .G ○o 2KOC -0 CHCH-O 27.(a)(CHa)CHaCHCOCI(alkanoyl chloride) 00 (b)(CH3)2CHCH2C-O-CCH3(mixed anhydride) >COONHa (ammonium salt) (e)CH,0- -CONH2 (carboxylic amide) (cyclic anhydride)

Solutions to Problems • 349 equivalent OCH2O’s (  1.7). The splitting between the upfield signals suggests that the sets of CH2’s are connected, so a reasonable answer is (d) (e) How about (f) 27. (a) (CH3)2CH2CH2COCl (alkanoyl chloride) O O B B (b) (CH3)2CHCH2COOOCCH3 (mixed anhydride) (c) (d) (e) (f) O O O (cyclic anhydride) CH3O CONH2 (carboxylic amide) COO CH3O NH4 (ammonium salt) CO2CH2CH3 (ethyl ester) H2SO4, OH LiAlH4 (CH3CH2)2O 1. O3, CH2Cl2 2. Zn, H, H2O CH2 O LiAlH4 (CH3CH2)2O 1. PBr3, (CH3CH2)2O 2. KOC(CH3)3 CO2H CH2OH 1. PBr3, (CH3CH2)2O 2. Mg OH 1. CO2, (CH3CH2)2O 2. H, H2O MgBr CO2H 1. O3, CH2Cl2 2. Zn, H, H2O Na2Cr2O7, H2SO4, H2O CHO CHO H, CH3OH CO2H G CO2H C CH2 O O CH2 CH2 CH3 CH3 O C CH2 O 1559T_ch19_344-357 10/17/05 13:51 Page 349

1559r.eh19.344-35710/17/0513:51Page350 350.chapter 19 CARBOXYUC ACIDS may reactwith it intramolecuyto form the cyclicahydride.For xampe: 0 H 6. -a (e)1.SOC(makes alkanoyl chloride).2.Add one more mole of starting acid.: (f)(CH3)2CHOH,H';(g)CHsCOOH.A 30.(a)CH:(CH2)sBr 盛器 →product OH bCH影CH=CH2aH -CH CHCHCI product y) 1.Mg.(CH.CH2)O product 31.(a)Acid catalysis is understood for esterification.So. CH.CH-e CH.CH GG :OH 6m,-o HCH.CH rche

28. Recall (Section 19-8) that carboxylic acids may react with alkanoyl halides to produce anhydrides. Upon conversion of one carboxylic acid function of a 1,4- or 1,5-diacid to a halide, the other acid function may react with it intramolecularly to form the cyclic anhydride. For example: Are you surprised that I used the oxygen of the carbonyl group to form the ring instead of the O of the carboxy OOH group? Which oxygen is more basic (and therefore more nucleophilic)? Check out Section 19-4 in the textbook. 29. (a) Na2Cr2O7, H2O, H2SO4; (b) 1. NaCN, H2O, H2SO4, 2. H, H2O, ; (c) 1. Mg, (CH3CH2)2O, 2. CO2, 3. H, H2O; (d) 1. NaCN, DMSO, 2. KOH, H2O, , 3. H, H2O, ; (e) 1. SOCl2 (makes alkanoyl chloride), 2. Add one more mole of starting acid, ; (f) (CH3)2CHOH, H; (g) CH3COOH, 30. (a) (b) (c) 31. (a) Acid catalysis is understood for esterification. So, O This is the product. CH3CH2C 18O CH2CH3 18 OCH2CH3 CH3CH2 C OH H CH2CH3 OH 18O H H CH3CH2 C OH2 O H H, H2O CH3CH2 C OH O CH3CH2 C OH OH H CH3CH2 18OH 1. Mg, (CH3CH2)2O 2. CO2 3. H, H2O (CH3)3CCl product 1. NaCN, DMSO 2. KOH, H2O, 3. H Cl2, H2O , H2O, CH3CH CH2 CH3CHCH2Cl product OH (O.K. to start here, actually) Either 1. Mg, (CH3CH2)2O 2. CO2 3. H, H2O or 1. NaCN, DMF 2. KOH, H2O, 3. H, H2O, CH3(CH2)5Br product O  O O O O O H Cl Cl O O HO 350 • Chapter 19 CARBOXYLIC ACIDS 1559T_ch19_344-357 10/17/05 13:51 Page 350

1559T_ch19_344-35710/17/0513:51Pa9e351 ⊕ EQA Solutions to Problems351 OH -H Intermediate :0-H 0: R-6RR-C6 Altematively.intermediate 2 could protonate on the unlabeled OH group instead of the OR' group: H OH 0: R- C-R'RC-ORR-C-OR() 62H ⊕ Ester containingin the carbonyl oxygen.Now if you follow the hydrolysis mechanism written above,the product will be 1*0 R-C-1%OH! 0 0 99 32.(a)CH.CHaCCl:(b)CHaCHaCBr:(e)CHCH.COCCH2CH 0 (d)CH,CH2COCH(CH3):(e)CH3CH.CO- (CH.CH,CNHCH (g)CH,CH2CH2OH Br (h)CH;CHCO2H 33.()-coct:

Solutions to Problems • 351 (b) Alternatively, intermediate 2 could protonate on the unlabeled OH group instead of the OR group: Ester containing 18O in the carbonyl oxygen. Now if you follow the hydrolysis mechanism written above, the product will be 18O B ROCO18OH! O O OO B B BB 32. (a) CH3CH2CCl; (b) CH3CH2CBr; (c) CH3CH2COCCH2CH3; O B (d) CH3CH2COCH(CH3)2 ; (e) (f) (g) CH3CH2CH2OH; Br A (h) CH3CHCO2H 33. (a) (b) COCl; COBr; CH3CH2CNHCH2 ; O CH3CH2CO H3NCH2 ; O 18 OH2 R OR R C 18 O H OH2 R OR C HO C OR (!) H, H2O 18 O H, H2O R 18 R OR C OH C O H H O H 18 O R OR OH OH C H2 18O R OR C O H R OR C H H 18 O H H Intermediate 2 1559T_ch19_344-357 10/17/05 13:51 Page 351

15597.eh19_344-35711/9/0515:35Page352 EQA 352.chapter 19 CARBOXYUC ACIDS oO是o是aawO是。eas oa 34.The only sensible way to start is by addition of the main nucleophile present (hydroxide)to the car- bonyl carbon of the ketone.The resulting tetrahedral intermediate can be taken in the direction of the prod- ct by eliminating a novel leaving group,a tribromomethyl carbanion R R: -RCO2H+:CBr Br:C HO CCBrs 一RCO2+HCBS 35.Here are the processes that give rise to the labeled peaks. CH3 CHCH-CH2-CH2-C-CO.H H mk=87 H,CC人oH 「cH,cH=CoH2 mk=74 CH;CH-CH2-CH2-C-CO:H [CO,H] H mk=45

(c) (d) (e) (f) (g) (h) 34. The only sensible way to start is by addition of the main nucleophile present (hydroxide) to the car￾bonyl carbon of the ketone. The resulting tetrahedral intermediate can be taken in the direction of the prod￾uct by eliminating a novel leaving group, a tribromomethyl carbanion: This carbanion is capable of leaving because its negative charge is stabilized by the combined inductive effects of the three halogen atoms. It is still a moderately strong base with a pKa in the mid-teens. Proton transfer between this anion and the carboxylic acid gives the final products of the reaction, the salt of the acid and the haloform, bromoform (CHBr3). The same process occurs with trichloro- and triiodo-substituted methyl ketones, to give chloroform and iodoform, respectively. 35. Here are the processes that give rise to the labeled peaks. α cleavage C6H13 CO2H m/z = 45 CH3CH2CH2 CH2 C CO2H H CH3 O H H2C H2C CH3 CH OH McLafferty rearrangement H2C CH2 CH3CH C(OH)2 m/z = 74 C3 C4 cleavage CH3CH2CH2 CH3CH2CH2 CH2 C CO2H H CH3 4 3 m/z = 87 CH3 CH2 C C(OH)2 C O Br3C R HO C O CBr3 RCO2H HO R CBr3 RCO2 HCBr3 CO2H Br C NH CH2OH; O CH2 ; C O O CH2NH3 (a salt); C C OH(CH3)2; O C C O CH2CH3; O O 352 • Chapter 19 CARBOXYLIC ACIDS 1559T_ch19_344-357 11/9/05 15:35 Page 352

1559T_ch19_344-35710/17/0513:51Pa9e35 ⊕ EQA Solutions to Problems.353 38.As usual,think mechanistically.The first intermediate should be the cyclic bromonium ion: OH Ag.base r Br The bromolactones shown are formed by intramolecular attack by a carboxylate oxyger n on one of the for two reasons.First.attack bya nucleophile on a brominium ion normally takes place at the atom (whi amic edge th hered rings have B 39.(CH,CH.,CH.CHCOOH viaCH.CH,CH.CBr -CH;CH,CH=C-Br Br-CBr_ CH,CH,CHBrCBr Then Br NH, CH,CH.CHCOOH+NHCH,CH.CHCOOH Product CN product (e)CH;CH,CH(CH)CH:CH.COOHP The t0. product

Solutions to Problems • 353 36. 1. LiAlH4, (CH3CH2)2O (makes 1-pentanol); 2. KBr, H2SO4, ; 3. KCN, DMSO (makes hexanenitrile); 4. KOH, H2O, ; 5. H, H2O, 37. (a) SOCl2; (b) H, CH3OH; (c) H, 2-butanol; (d) alkanoyl chloride from (a); (e) CH3NH2 (via the ammonium salt), ; (f) LiAlH4, (CH3CH2)2O; (g) Br2, cat. P 38. As usual, think mechanistically. The first intermediate should be the cyclic bromonium ion: The bromolactones shown are formed by intramolecular attack by a carboxylate oxygen on one of the carbon atoms of a cyclic bromonium ion, the initial product of Br2 addition to the double bond (Section 12-6). Which pathway, a or b, should be favored? The five-membered ring should form preferentially, for two reasons. First, attack by a nucleophile on a brominium ion normally takes place at the more highly substituted carbon atom (which is more positively polarized—see Section 12-6 again). Second, five-membered rings form faster than do sixes (Section 9-6), compensating for the slight thermodynamic edge that six-membered rings have. 39. (a) Then (b) (c) CH3CH2CH(CH3)CH2CH2COOH Br2, cat. P product 1. K2CO3, H2O, 2. H, H2O Then CH2CO2H 1. Br2, cat. P 2. KCN, DMSO CHCO2H product 1. HO, H2O 2. H, H2O CN CH3CH2CHCOOH CH NH3 3CH2CHCOOH product NH3 Br SN2 H Br O exchange CH3CH2CHBrCBr . via CH3CH2CH2CBr CH3CH2CH O C Br Br Br O H CH3CH2CH2COOH CH3CH2CHCOOH Br2, cat. P Br O O O Br Br O O O OH Aq. base a a b or or b Br O  Br 1559T_ch19_344-357 10/17/05 13:51 Page 353