1559Tch20_358-37210/18/054:24Pag0358 EQA 20 Carboxylic Acid Derivatives the carbonyl carbon itself.Therefore.it is the ahydrogens in most ed by strong bases Carboxylic acid:R-CH2-C-OH+-Most acidie Nucleophiles may add Carboxylic acid derivatives:R-CH2-( Nucleophiles may add L=halide (alkanoyl halide) L=OCR'(anhydride) =OR' L=NR2'(amide) nheamsaocaedwihreactiosof bortant new pointsin derivative into any of the others.for synthetic purposes. Outline of the Chapter 20-2 Alkanoyl Halides 358
20 Carboxylic Acid Derivatives Derivatives of carboxylic acids share two properties in common with carboxylic acids themselves: (1) They have a carbonyl group that is susceptible to nucleophilic addition, and (2) they have a potential leaving group attached to the carbonyl carbon. In addition, they may possess acidic hydrogens to the carbonyl group. However, they lack an acidic OOH group on the carbonyl carbon itself. Therefore, it is the -hydrogens in most carboxylic acid derivatives that are the most readily removed by strong bases. Carboxylic acid: Carboxylic acid derivatives: L halide (alkanoyl halide) O B L OCR (anhydride) L OR (ester) L NR2 (amide) The mechanisms associated with reactions of these compounds have already been introduced (Section 19-7), so this chapter mainly presents additional examples of reactions that follow familiar patterns. The most important new points involve (1) the relative reactivities of these derivatives and (2) how to convert each kind of derivative into any of the others, for synthetic purposes. Outline of the Chapter 20-1 Relative Reactivities, Structures, and Spectra of Carboxylic Acid Derivatives Comparisons of physical properties. 20-2 Alkanoyl Halides R C L O Now most acidic Potential leaving group Nucleophiles may add CH2 R C OH O Less acidic Potential leaving group Most acidic Nucleophiles may add CH2 358 1559T_ch20_358-372 10/18/05 4:24 Page 358
1559T_ch20_358-37210/18/054:24Page359 ⊕ EQA Keys to the Chapter·359 20-3 Carboxylic Anhydrides 20-4,20-5 Esters 20-6,20-7 Amides 20-8 Alkanenitriles Details of the chemistry of the four major carboxylic acid derivatives,plus nitriles,which are relatives of amides. Keys to the Chapter 20-1.Relative Reactivities,Structures,and Spectra of Carboxylic Acid Derivatives In this section.dif physical properties of th four main carbo ns of these com arbon.addition to theca rbon,and protonanon or the 00 each derivative.Then.refer back to this text section from time to time as you go through the chapter to see how the reaction details presented in the upcoming sections reflect the general principles outlined here. 20-2.Akg l Halides y (acyD halidesa er useful in synthesis because()they a asly madefom carborylic secto-)and( hey e easily turned hilic add and the st le p halide) 20-3.Carboxylic Anhydrides red and they nyl groups can add a nucleophile;the ther falls off as part of the carboxylate ion leaving group dride es a are mo use the tw of Cyclic 20-4 and 20-5. Este nien nds for synthetic pun oses because their intermediate reactivity makes them easy to prepare and store for later use mpounds by reactions at both their anter 23) 20-6 and 20-7.Amides reduction reactions that may form either amines or al ofmann rearrangement
Keys to the Chapter • 359 20-3 Carboxylic Anhydrides 20-4, 20-5 Esters 20-6, 20-7 Amides 20-8 Alkanenitriles Details of the chemistry of the four major carboxylic acid derivatives, plus nitriles, which are relatives of amides. Keys to the Chapter 20-1. Relative Reactivities, Structures, and Spectra of Carboxylic Acid Derivatives In this section, differences in the physical properties of the four main carboxylic acid derivatives are highlighted. The principal factor determining these differences is the electronegativity of the atom attached to the carbonyl carbon. These differences are reflected in a consistent way throughout all the reactions of these compounds: deprotonation of the -carbon, addition to the carbonyl carbon, and protonation of the carbonyl oxygen. Make sure you understand the conceptual arguments here before moving ahead to the specific sections on each derivative. Then, refer back to this text section from time to time as you go through the chapter to see how the reaction details presented in the upcoming sections reflect the general principles outlined here. 20-2. Alkanoyl Halides Alkanoyl (acyl) halides are very useful in synthesis because (1) they are easily made from carboxylic acids (Section 19-8) and (2) they are easily turned into every major type of carbonyl compound, including aldehydes, ketones, and all the other carboxylic acid derivatives. Of all the carboxylic acid derivatives, alkanoyl halides possess the most reactive carbonyl group toward nucleophilic addition and the best leaving group (halide). 20-3. Carboxylic Anhydrides Carboxylic acid anhydrides are used in synthesis less often than are alkanoyl halides. They are generally not as easily prepared, and they are less reactive. A major drawback to using an anhydride in synthesis is that only one of its carbonyl groups can add a nucleophile; the other falls off as part of the carboxylate ion leaving group (see “Nucleophilic Addition–Elimination of Anhydrides”). Cyclic anhydrides are more useful because the two ends still stay stuck together by a carbon chain after an addition takes place (see “Nucleophilic Ring Opening of Cyclic Anhydrides”). 20-4 and 20-5. Esters Esters are the most prevalent carboxylic acid derivatives in nature. They also are convenient compounds for synthetic purposes because their intermediate reactivity makes them easy to prepare and store for later use. Alkanoyl halides, in contrast, are so reactive toward water that it takes some care to prevent them from hydrolyzing to some extent upon extended storage. Esters are easily prepared and readily converted into many types of compounds by reactions at both their carbonyl and -carbons (Chapter 23). 20-6 and 20-7. Amides Amides are much less reactive than esters toward nucleophilic acyl substitution. They have greater resonance stabilization by the nitrogen lone pair, and they have a poorer leaving group (NH3 in acid, and NH2 in base). Amides do show some special reactions: the possibility of deprotonation of the nitrogen in 1° or 2° amides, reduction reactions that may form either amines or aldehydes, and a new process, the Hofmann rearrangement, 1559T_ch20_358-372 10/18/05 4:24 Page 359
1559Tch20359-37210/18/054:24Page360 360.Chapter 20 CARBOXYUC ACID DERIVATIVES CH(CH2+NH2←一sae HAOCH.CINO 0 CH.(CH2)CNH Br.NaOH.H.O CH,CH.NH,一-oe This Hofmann re nt has a mechanism that is related cond otually to s The nitrile functional g up is,formally,a dehydrated primary amide.In fact.a little-used nitrile synthesis in- volves amide dehydration: ⊕ R-C-NH:O R-C-N In this arboxvlic acids and their derivatives The n by the nitrile may be conve d to carboxyli trophilie arbonyl earbon.Thisisa good wayto use substitution to introducea carbonyl car bon into a molecule. R-CH2-CONH R一CH-CHO R一CHh-COR Solutions to Problems 28.(a)-Methylbutanoyl iodide: (b)1-methylcyclopentanecarbonyl chloride (e)2.2.2-trifluoroacetic anhydride: (d)benzoic propanoic anhydride (e)ethyl 2.2-dimethylpropanoate: (f)N-phenylacetamide (g)CH-CH-CH-COCH CH-CH h)CHCH-COCH-CH-CH-CH Φ
which forms an amine with one carbon less than the original amide. Consider the following two amine syntheses. This Hofmann rearrangement has a mechanism that is related conceptually to some you’ve seen before but has some very new and unusual details. It merits a careful look, especially steps 4 and 5. 20-8. Alkanenitriles The nitrile functional group is, formally, a dehydrated primary amide. In fact, a little-used nitrile synthesis involves amide dehydration: In this sense, therefore, it is related to carboxylic acids and their derivatives. The importance of nitriles in synthesis derives from the fact that the nitrile carbon may be introduced into a molecule as the nucleophilic cyanide ion in an SN2 reaction. Then, by the reactions in this text section, the nitrile may be converted to carboxylic acid derivatives, ketones, or aldehydes, in which the originally nucleophilic cyanide carbon has become an electrophilic carbonyl carbon. This is a good way to use a nucleophilic substitution to introduce a carbonyl carbon into a molecule. Solutions to Problems 28. (a) 3-Methylbutanoyl iodide; (b) 1-methylcyclopentanecarbonyl chloride; (c) 2,2,2-trifluoroacetic anhydride; (d) benzoic propanoic anhydride; (e) ethyl 2,2-dimethylpropanoate; (f) N-phenylacetamide O O B B (g) CH3CH2CH2COCH2CH2CH3 (h) CH3CH2COCH2CH2CH2CH3 CN SN2 Various reactions CH2 R CH2 CN R R R CH2 COR R CH2 CHO R CH2 CONH2 CH2 CO2H R X CH2 CO2R P2O5 (H2O) RCN O R C NH2 CH3(CH2)nCNH2 1. LiAlH4, (CH3CH2)2O 2. H, H2O Br2, NaOH, H2O O CH3(CH2)n1NH2 Same number of carbons as starting amide CH3(CH2)nNH2 One carbon less than starting amide 360 • Chapter 20 CARBOXYLIC ACID DERIVATIVES 1559T_ch20_358-372 10/18/05 4:24 Page 360
1559T_ch20_358-37210/18/054:24Pa9e361 EQA Soutionso Problems361 re given (Section 18-1)that resonance delocalization of the negative charge of an enolate from the acarbon into the carbony group is partly responsible for the acidity of the a-hydrogen.Resonance donation of an electron pair from a-carbon.The net effect is de 30.(a)Acetyl chloride (Clis bigger than F.and the bonds to it are longer): (b)CH(COCH)(H'sto ketones are more acidic than H's to esters) r on N is shared ir so it and an amide is similar to that between an anhydride and an ester.) d)(Ccucu-9-CH-H resonance reduces electron donation from oxygen toward the carbonyl carbon.So the resonance 0- NHCCH NH.C (e)(CHa)CCH
Solutions to Problems • 361 (i) (j) CH3 A (k) CH3CH2CH2CH2CHCN (l) 29. The guiding principle is that the strength of an acid is related to the ability of its conjugate base to accommodate the negative charge that remains after loss of a proton (Section 2-2). We are given (Section 20-1) that the order of acidity of -hydrogens in carboxylic acid derivatives increases in the series carboxamide (weakest acid) ester anhydride alkanoyl halide (strongest acid). (a) We know (Section 18-1) that resonance delocalization of the negative charge of an enolate from the -carbon into the carbonyl group is partly responsible for the acidity of the -hydrogen. Resonance donation of an electron pair from O B L in RCL to the carbonyl group essentially competes with possible donation from the deprotonated -carbon. The net effect is destabilization of the anionic enolate. Therefore, as the resonance donation ability of L increases, the acidity of the -hydrogen decreases. (b) Inductively, as the electronegativity of L increases, stabilization of the negative charge of the enolate function should increase accordingly, as is observed. 30. (a) Acetyl chloride (Cl is bigger than F, and the bonds to it are longer); (b) CH2(COCH3)2 (H’s to ketones are more acidic than H’s to esters); (c) imide (The lone pair on N is shared in resonance by two carbonyl groups, so it does not reduce their electrophilicity as much as the N in an amide. Note that the relationship between an imide and an amide is similar to that between an anhydride and an ester.) (d) resonance reduces electron donation from oxygen toward the carbonyl carbon. So the resonance O A form CH3CP G OOCHPCH2 is reduced in importance, strengthening the COO double bond and raising the carbonyl stretching frequency, actually to about 1760 cm1 .) 31. (a) (b) CH3(CH2)4C O (CH3)3CCH O (c) NHCCH3 O NH3Cl Ethenyl acetate (CH3C CH O O CH2 CH3C CH O O CH2 CN O CN(CH3)2 O COCH2CH2Cl 1559T_ch20_358-372 10/18/05 4:24 Page 361
1559r.ch20.358-37210/18/054:24Pag8362 EQA 362.chapter 20 CARBOXYUC ACID DERIVATIVES CH CH 32. 0 cg+CH.CH.CH一c-气g H八aa 0: →HC OCH-CH-CHs H 33.Assume aqueous acid work-up for both this and the next problem. (+CH..CNH+CH.COH 0 00 34.(a)(CH3CHOCCH.CH2COH (b)HOCCH-CH2CNH2 CO OCH O+CH.OH
362 • Chapter 20 CARBOXYLIC ACID DERIVATIVES (d) (e) 32. 33. Assume aqueous acid work-up for both this and the next problem. O O OO B B BB (a) CH3COCH(CH3)2 CH3COH (b) CH3CNH2 CH3COH (c) (d) 2 CH3CH2OH OO OO BB BB 34. (a) (CH3)2CHOCCH2CH2COH (b) HOCCH2CH2CNH2 (c) (d) HOCH2CH2CH2CH2OH 35. OH CH3O O O O CH3O O O H O O O CH 3OH O OCH3 OCH3 O O O O H H O HOCCH2CH2C O OH C OH CH3COH CH3 O CH2CH2CH3 H3C O OCH2CH2CH3 C O H3C C O H HOCH2CH2CH3 H3C Cl C O H CH2CH2CH3 H3C C Cl O O CH3 CH3 CH O CH2OCCH2CH2COCH2 etc. O O 1559T_ch20_358-372 10/18/05 4:24 Page 362
1559T_ch20_358-37210/18/054:24Page363 EQA Soutionso Problems363 36.(a)CH.CH.CH.CH.CO.H (b)CH,CH,CH CH,COCH CH,CH(CH,) 0 OH (e)CH,CH,CH.CHCN(CHCH)(d)CH,CH,CH,CH,CCH, 0 (e)CH CH2CH2CH-CH2OH (f)CH CH2CH2CH2CH OH OH 0 37.(a)CH,CHCH2CH2CO2H (b)CH,CHCH,CH,COCH2CH,CH(CH3)2 OH OH OH (e)CH,CHCH,CH,CN(CH,CH2(d)CH,CHCH,CH.CCH, OH (e)CH CHCH CH,CHOH OH (f)CH,which forms the cyclic hemiaceta H,C、0oH =0 38.(a (b) (c)y CH:CH2 CH3 0 (d)NH e (0 B N-CH 0: COH CH.OH 39.(.C,(CH)..CH :0-H 0 CHCHCCHCH CH).CHE H 入CH,CH3 H-Q-CHs
Solutions to Problems • 363 O B 36. (a) CH3CH2CH2CH2CO2H (b) CH3CH2CH2CH2COCH2CH2CH(CH3)2 O OH B A (c) CH3CH2CH2CH2CN(CH2CH3)2 (d) CH3CH2CH2CH2CCH3 A CH3 O B (e) CH3CH2CH2CH2CH2OH (f) CH3CH2CH2CH2CH OH OH O A AB 37. (a) CH3CHCH2CH2CO2H (b) CH3CHCH2CH2COCH2CH2CH(CH3)2 OH O OH OH AB AA (c) CH3CHCH2CH2CN(CH2CH3)2 (d) CH3CHCH2CH2CCH3 A CH3 OH A (e) CH3CHCH2CH2CH2OH OH O A B (f) CH3CHCH2CH2CH, which forms the cyclic hemiacetal 38. (a) (b) (c) (d) (e) (f) 39. H, CH3CH2OH (CH3)2CHC (CH3)2CHCOCH3 CH2CH3 OCH3 O O O H H H H (CH3)2CHCOCH2CH3 H CH3 O O H H (CH3)2CHCOCH2CH3 OH (CH3)2CHCOCH2CH3 O CH3OH O N CH3 O CH3 NH O NH O O CH3CH2 O O H3C O O H3C O OH 1559T_ch20_358-372 10/18/05 4:24 Page 363
1559T_ch20_358-37211/09/0518:53Pa9e365 EQA Solutions to Problems.365 42.Amide the following options: CON(CH-CH32 Method 1:Direct reaction between an amine and an acid first gives a salt:strong heating converts the salt osofHsof water.Method 2:Converts to alkanoyl chloride;reactionwthm The starting cyclic amide(a lactam)p we can proce ⊕ 0-CH,2-0 -0-CH OCH (See Problem40) 44.CH COCH3:NH CH CH 0 CH;C COCH CH;C NH2 +CH,OH NH. entin CH.CH.CH.CH.CH NCH acid.(e) a)and (f).whe ntanal:N.N-dimethylpentanamide
Solutions to Problems • 365 42. Amides may be made directly from the acids, or indirectly via alkanoyl halides, anhydrides (less efficiently, because only one of the two alkanoyl groups is converted into an amide), or esters. Thus we have the following options: Method 1: Direct reaction between an amine and an acid first gives a salt; strong heating converts the salt to the amide with loss of water. Method 2: Converts to alkanoyl chloride; reaction with amine gives amide with loss of HCl. Often done in the presence of excess amine to consume the HCl. Method 3: Esterification with any inexpensive alcohol (ethanol chosen here), then reaction with the amine under mild heating. 43. Dimethyl sulfate is a highly reactive SN2 substrate with respect to attack by a nucleophile on a methyl group (similar to CH3I in reactivity). The process displaces the methyl sulfate anion, a good leaving group. The starting cyclic amide (a lactam) possesses two possible nucleophilic atoms, nitrogen and oxygen. Reaction of the oxygen with an electrophile gives a resonance-stabilized intermediate; reaction of the nitrogen does not. Therefore, we can proceed as follows: 44. 45. Pentanamide gives (a) pentanoic acid, (e) pentanamine, and (f) pentanal; N,N-dimethylpentanamide gives the same products for (a) and (f), whereas LiAlH4 reduction in this case gives N,N-dimethylpentanamine, CH3CH2CH2CH2CH2N(CH3)2. OCH3 CH3CNH2 CH3OH H O NH2 CH3C O CH3COCH3 OCH3 O NH3 NH3 CH3C O (See Problem 40) OCH3 H N OCH3 N OCH3 H N H N H O O O H3C O S CH O 3 SN2 O O O S CH O 3 CO2H 1. (CH3CH2)2NH 2. strong heating 1. CH3CH2OH, H 2. (CH3CH2)2NH, 1. SOCl2 2. (CH3CH2)2NH H3C CON(CH2CH3)2 H3C 1559T_ch20_358-372 11/09/05 18:53 Page 365
1559Tch20359-37210/18/054:24Page366 366.chapter 20 CARBOXYUC ACID DERIVATIVES 0: 46.、人 in,+ C-NH2 C-OH :OH LOCCecCM (b)N=C一 NH )N=C入-C=N:+H* NEC- 入c+H N=C一 H一N=C Then continue as in Problem 46 and repeat at the other nitrile group
366 • Chapter 20 CARBOXYLIC ACID DERIVATIVES 46. 47. (a) (CH3CH2CH2CH2)2CuLi; then H, H2O; (b), (d) LiAlH4, (CH3CH2)2O; then H, H2O; (c) LiAl[OC(CH3)3]H; then H, H2O; (e), (f) CH3CH2CH2MgBr, (CH3CH2)2O; then H, H2O 48. (a) (b) (c) Then continue as in Problem 46 and repeat at the other nitrile group. N C NH2 O N C NH2 H O N C NH H H O H N C NH H O N C C NH H2O N C C N H N C C N H, H2O HO O O OH N C C N H2, PtO2 H2N NH2 N H N H n O O H C NH3 O OH OH C O H H H H C NH2 O O H H C NH2 O O H NH2 C H2O O H NH2 C O 1559T_ch20_358-372 10/18/05 4:24 Page 366
1559T_ch20_358-37210/18/054:24Pa9e36 EQA Solutions to Problems.367 9 d by protonatio 个co0cH for B) The a-hydrogens are not removed,so no isomerization is alie only those proton tass thatth In th ⊕ C=00 me
49. In A and B, the most acidic hydrogens are -hydrogens. Deprotonation followed by protonation provides a mechanism for isomerization. In C the most acidic hydrogens are on nitrogen. The -hydrogens are not removed, so no isomerization is observed. 50. The mechanisms are related to amide formation and Hofmann rearrangement, respectively. In the formation of phthalimide, only those proton transfers that occur along the pathway to the product are shown. Then Now what? There’s no hydrogen on the N anymore, so how can you get to an intermediate N-haloamidate that can decompose to the necessary acyl nitrene? Br O C R NBr N-Haloamidate O C R N Acyl nitrene O O NH O O N Na O O NBr H2O NaOH NaBr Br Br C C O O NH2 OH NH2 O OH NH OH2 O Phthalimide NH H2O O O O O O O O O NH3 NH3 O O NH2 O OH O LDA THF H, H2O COOCH3 COOCH3 H COOCH3 COOCH3 H H H A (Similar for B) H COOCH3 COOCH3 Solutions to Problems • 367 1559T_ch20_358-372 10/18/05 4:24 Page 367
1559rch20358-37210/18/054:24Page368 EQA 368.chapter 20 CARBOXYUC ACID DERIVATIVES The answer.The reaction takes place in strong base,souse hydroxide in addition-elimination :6:OH OH NB -Br Now you are in bus in Section 20-7. 51.Self-explanatory. mthyations of amine rogen)Noie athad anon that the Hom rearrangement remoy use of th 54.(a)Bad idea.Residual hexanoyl chloride will be converted to hexanoic acid.which smells like a herd of goats on a hot day. CH.c-o-steroid,C-O-CH+HO-stcroid
The answer: The reaction takes place in strong base, so use hydroxide in addition–elimination. 51. Self-explanatory. 52. From A: 1. SOCl2, 2. (CH3)2NH (makes carboxamide), 3. LiAlH4, (CH3CH2)2O, then H, H2O. From B: 1. SOCl2, 2. NH3, 3. Cl2, NaOH (Hofmann rearrangement, loses CO2 and makes simple amine), 4. 2 CH3I, NaOH (SN2 methylations of amine nitrogen). Notice that B had an extra carbon that the Hofmann rearrangement removed. 53. The dipolar resonance forms that weaken the carbonyl COO bond and reduce its stretching frequency are less and less favorable in small rings because of the increased strain associated with a second sp2 atom. 54. (a) Bad idea. Residual hexanoyl chloride will be converted to hexanoic acid, which smells like a herd of goats on a hot day. (b) Wash out the glassware with an alcohol like ethanol. Reaction with hexanoyl chloride produces the ester ethyl hexanoate, which smells like fresh fruit. Much better. 55. React with H and CH3OH! The methyl ester group at the upper right will remain unchanged because the only nucleophile around for attacking carbonyl groups is methanol. However, the acetate function at the lower right will exchange to give methyl acetate, and the steroid alcohol group will be displaced. H, CH3OH CH3C O O steroid CH3C CH3 O O HO steroid O O Very strained, relatively unimportant resonance form O O (See Exercise 20-20) C C O O OH N C N O O OH C COH NH2 O O O NBr NBr O OH OH O C C O O Now you are in business. Continue exactly as in Section 20-7. OH Br NBr O 368 • Chapter 20 CARBOXYLIC ACID DERIVATIVES 1559T_ch20_358-372 10/18/05 4:24 Page 368
