有机化学结构与功能5th答案sg_ch19

1、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 m

2、ade 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 19-1Naming the Carboxylic Acids 19-2,

3、 19-3Physical Properties of Carboxylic Acids Very strongly hydrogen-bonding molecules; spectroscopy. 19-4Acidity and Basicity of Carboxylic Acids An analysis of resonance and inductive effects. 19-5, 19-6Preparation of Carboxylic Acids Reviewing some reactions youve seen before. 19-7Reactivity of th

4、e Carboxy Group: Addition and Elimination A new, general mechanism sequence. 19-8Alkanoyl Halides and Anhydrides 19-9Esters 19-10Amides Conversion of acids into their most important derivatives. 19-11Nucleophilic Attack at a Carboxylate Group Reactions of acids with hydride. 19-12?-Bromination A rea

5、ction of special value in synthesis. 19-13Carboxylic Acids in Nature 344 1559T_ch19_344-357 10/17/05 13:51 Page 344 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

6、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

7、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

8、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

9、 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 e

10、nolate 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 explanation

11、s. 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 carboncarbon bonds. You might take the time to do a quick review and then begin

12、organizing these synthetic methods into appropriate categories. O RCOHH? NeutralNegatively charged O RCO ? O ? RCO Carboxylic acids vs. ketones or aldehydes O RCCH3RCCH2?CH2H? OO ? RC O RCOHH? ROH NeutralNegatively charged H? O RCO ? O ? RO ? RCO Carboxylic acids vs. alcohols 1559T_ch19_344-357 10/1

13、7/05 13:51 Page 345 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 ow

14、n, 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 overal

15、l. This is a two-step, additionelimination, mechanism. It is not the same as either an SN2 or an SN1 reaction, mechanistically: SN1 and SN2 reactions do not apply to sp2hybridized carbons. 2. For carboxylic acids, this new type of substitution process occurs best with weakly basic nucleophiles in th

16、e 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

17、 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 her

18、e, because the tetrahedral intermediate is relatively high in energy compared with the carbonyl product, which has a very strong, very stable carbonoxygen 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 simil

19、ar 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: Ad

20、ding 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 t

21、he 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

22、 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- tionelimination

23、 reactions on carboxylate anions are hard to do because (1) the addition is hard to do and (2) HO?NuR R ?C O NuOHCO bond Favorable (Stable C is regenerated) O? O?Nu?Nu?CC O CFavorable (Strain is relieved) C HO?Nu?NuOH?CCUnfavorable Energetically 346 Chapter 19 CARBOXYLIC ACIDS 1559T_ch19_344-357 10/

24、17/05 13:51 Page 346 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, LiAlH4is capable of ad- dition to RCOO?, and the prod

25、uct of addition can go on to eliminate, formally, an aluminum oxide anion leav- ing group. The product of an acid ? LiAlH4is 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 reacti

26、on. 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-benzen

27、edicarboxylic 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 (249C) because of its hydrogen-bonded dimer formation. The alcohol, which can

28、also hydrogen bond, is next (205C), the polar aldehyde third (178C), and the nearly nonpolar hydrocarbon last (115C). 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 w

29、ritten;(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 HCH3 O B CH3CCOOH 1559T_ch19_344-357 10/1

30、7/05 13:51 Page 347 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 cm?1). (b) For B, Hsat? 12 ? 2 ? 14; degree of unsaturation ? ? (14 ? 2 10) ? 2 ? bonds or rings. From the 13C NM

31、R (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 1H 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 togeth

32、er to give cyclohexene! (c) For G, Hsat? 16 ? 2 ? 18; degrees of unsaturation ? ? (18 ? 2 14) ? 2 ? bonds or rings. IR: ? ? 1742 cm?1is 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. T

33、he 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 cm?1) via hydroborationoxidation; ? H 3.4(d) CH2OH E ? CH2 (?C C? 1649 cm?1 and 888 cm?1) via Wittig reaction:?C CH2? D(?C O? 1715 cm?1); ? O ? ? 208.5(13C) Then Cv

34、ia oxymercurationdemercuration;? 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 Solutions to Problems 349 equivalent OCH2Os (? ? 1.7). The splitting between the upfield signals suggests that the set

35、s of CH2s are connected, so a reasonable answer is (d) (e) How about (f) 27. (a) (CH3)2CH2CH2COCl (alkanoyl chloride) OO BB (b) (CH3)2CHCH2COOOCCH3(mixed anhydride) (c) (d) (e) (f) O O O (cyclic anhydride) CONH2 (carboxylic amide)CH3O COO?NH4 (ammonium salt)CH3O CO2CH2CH3 (ethyl ester) H2SO4, ? OH L

36、iAlH4 (CH3CH2)2O 1. O3, CH2Cl2 2. Zn, H?, H2O CH2O LiAlH4 (CH3CH2)2O 1. PBr3, (CH3CH2)2O 2. K?OC(CH3)3 CH2OHCO2H 1. PBr3, (CH3CH2)2O 2. Mg OH 1. CO2, (CH3CH2)2O 2. H?, H2O MgBr CO2H 1. O3, CH2Cl2 2. Zn, H?, H2ONa2Cr2O7, H2SO4, H2O CHO CHO H?, CH3OH CO2H G CO2H CH2C O OCH2 CH2 CH3 CH2COCH3 O 1559T_ch

37、19_344-357 10/17/05 13:51 Page 349 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

38、 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, H2S

39、O4, 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

40、, O This is the product. CH3CH2C 18O CH2CH3 18OCH 2CH3 OHCCH3CH2 CH2CH3 H OH 18O ? ?H? ?H? OH2CCH3CH2 OH ?H?, H2O CH3CH2OHC O CH3CH2OHC OH H? ? CH3CH218OH 1. Mg, (CH3CH2)2O 2. CO2 3. H?, H2O (CH3)3CClproduct 1. NaCN, DMSO 2. KOH, H2O, ? 3. H?, H2O, ?Cl2, H2O CH3CHCH2CH3CHCH2Clproduct OH (O.K. to sta

41、rt here, actually) Either 1. Mg, (CH3CH2)2O 2. CO2 3. H?, H2O or 1. NaCN, DMF 2. KOH, H2O, ? 3. H?, H2O, ? CH3(CH2)5Brproduct O ? O OO O O H ? Cl Cl O O HO 350 Chapter 19 CARBOXYLIC ACIDS 1559T_ch19_344-357 10/17/05 13:51 Page 350 Solutions to Problems 351 (b) Alternatively, intermediate 2 could pro

42、tonate 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! OOO O BBBB 32. (a) CH3CH2CCl;(b) CH3CH2CBr;(c) CH3CH2COCCH2CH3; O B (d) CH3CH2COCH(CH3)2;(e) (f)(g) CH

43、3CH2CH2OH; Br A (h) CH3CHCO2H 33. (a)(b) COBr;COCl; CH3CH2CNHCH2; O CH3CH2CO?H3NCH2; ? O 18OH 2 RROR? ? C 18O H OH2 ROR? ? C HO OR? (!)C ?H?, H2O 18O ?H?, H2O R 18OH CROR ? C O H H OH 18O ROR? OHOH C H218O ROR? ? C O H? ROR? ? C H? ?H? 18O H H Intermediate 2 1559T_ch19_344-357 10/17/05 13:51 Page 35

44、1 (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 c

45、arbanion: 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 pKain the mid-teens. Proton transfer between this anion and the carboxylic acid gives the final products

46、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. ? cleavage ?C6H13 ? CO2H m /z = 45

47、 CH3CH2CH2CCH2CO2H H CH3 ? ? O H H2C H2C CH3 CH OH McLafferty rearrangement ?H2CCH2 CH3CHC(OH)2 m /z = 74 ? C3 C4 cleavage ?CH3CH2CH2 CH3CH2CH2CCH2CO2H H CH3 34 m /z = 87 ? CH3 CH2C(OH)2C OC Br3C R HO ?C O CBr3 RCO2H HO R ? ?CBr3 ? RCO2? HCBr3 CO2H Br CH2OH;CNH O CH2; CO? O CH2NH3? (a salt); CCOCH(C

48、H3)2; O CCOCCH2CH3; OO 352 Chapter 19 CARBOXYLIC ACIDS 1559T_ch19_344-357 11/9/05 15:35 Page 352 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;(

49、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

50、 carbon atoms of a cyclic bromonium ion, the initial product of Br2addition 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 m

51、ore highly substituted carbon atom (which is more positively polarizedsee 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) CH 3CH2CH(CH3)CH2CH2COOH Br2, cat. P p

52、roduct 1. K2CO3, H2O, ? 2. H?, H2O Then CH2CO2H 1. Br2, cat. P 2. KCN, DMSO CHCO2Hproduct 1. HO?, H2O 2. H?, H2O CN CH3CH2CHCOOHCH3CH2CHCOOHproductNH3 ?NH3 ? ?Br? SN2 ?H? Br O exchange CH3CH2CHBrCBr. via CH3CH2CH2CBrCH3CH2CH? O BrBrCBrC OH CH3CH2CH2COOHCH3CH2CHCOOH Br2, cat. P Br O O O Br Br O OO OH

53、 Aq. base a ab or or b Br ? O ? Br ? 1559T_ch19_344-357 10/17/05 13:51 Page 353 (d) (e) BrCH2COOH ? (CH3CH2)2NH n product (f) 40. No tricks here! The mechanism is almost exactly as shown in Section 19-12, with only minor differences. Because the method starts with an alkanoyl halide, step 1 is unnec

54、essary. Step 2 (enolization) occurs as written. The third step uses Cl2, generated in low concentration from NCS, or Br2, formed in the same way from NBS, or I2. The fourth step in the mechanism doesnt apply because only alkanoyl halides are present, not carboxylic acids. OOOO BBBB 41. Acidity: CH3C

55、OH ? CH3CNH2? CH3CCH3. The most acidic hydrogens in CH3CNH2are on the nitrogen. Acidity order is determined by electronegativity. Two protonation possibilities: on N, giving O B CH3CN GH 3, and on O, giving Resonance stabilization causes protonation on O to be favored. 42. See Problem 34 in Chapter

56、17. 43. (a) CH3CH2 ?Cl? Elimination CO Cl OH Tetrahedral intermediate CH3CH2 ?H? CO ClH ? CH3CH2 H? CO Cl ? O? O CH3CH2OH Addition C? Cl O HOCH2CH2CH2CH2OHHOCH2CH2CH2CH CrO3 Initially O CrO3 OOHO Hemiacetal form O ? OH CH3CNH2 ? OH .CH3CNH2 CH3CH2COOHproduct 1. Br2, cat. P 2. (C6H5)3P, (CH3CH2)2O CH

57、3COOHBrCH2COOH Br2, cat. P product 1. Excess KSH, CH3CH2OH 2. I2 (Section 910) 354 Chapter 19 CARBOXYLIC ACIDS 1559T_ch19_344-357 10/17/05 13:51 Page 354 Solutions to Problems 355 (b) 44. 45. (a) Haloalkanes are generally not very soluble in water (too much polarity difference). The reaction mixture

58、 is heterogeneous, which prevents good mixing of reactants. Water also forms hydrogen bonds to the nucleophile, which doesnt help. (b) Acetic acid is a better solvent for the haloalkane, so the system is homogeneous, allowing for better mixing of reacting molecules. (c) Sodium dodecanoate is a soap,

59、 and it dissolves in water to make micelles. The less polar interior regions of the micelles form good solvents for molecules of low polarity like haloalkanes. The iodobutane dissolves in the micelles and is therefore in close proximity to the nucleophilic carboxylate groups, allowing the SN2 reaction to proceed. A lactone (?CL ? H H O O O? 1770 cm?1) CH2COOH ?J O O H H CH2COOH OH(via ketone);K ? H H CH2CHC(CH3)2 H ?O H H (Alkylation on less hindered bottom face) CH2CHC(CH3)2 I ? O O H H O CH3COH ?NH3 NH