1559T_ch23_409-42210/19/0519:50Pa9e411 ⊕ EQA Solutions to Problems.411 Note that the ne 3-ketoeste ondensation.as a result.RCo.CH CH en cone crossed cond he be a dis on will additions can be followed by Robinson annulations.thereby resulting in six-membered rings. 23-4.Alkanoyl(Acyl)Anion Equivalents roxy keto made'? Because th f 0 of a nucleophilic alkanoyl anion contradicts everything you've leamed s to ne (1.3-dithiane)anion.Anothe way is to expose an aldehyde to thiaz lum salts.Again,deprotona on of th nginal carb nyl carbon may nthe usual way.The result is anew carbon-carbon bond between car. All you've done carbonyl groups into new functions that ae able tosuppor negative charges.Then.after finishing with them. ng for to ca be (hard t me ke)alk low along one of the two Solutions to Problems 24.Claisen condensations.(a).(b).(c)involve two identical molecules:(d).(e)are intramolecular cm(()(h)() 0 (a)CH.CHCH2C -CHCOCH2CH CH;CH2 P 0 (b)CHCH(CH)CH C-CHCOCH-CH; CH CHCH (e)Unfavorable equilibrium.Claisen product is not stable:no reaction is observed
Solutions to Problems • 411 Note that the necessary 3-ketoester comes from a crossed Claisen condensation. As a result, RCO2CH2CH3 must be an ester that does not do Claisen condensations with itself (i.e., R must not have an -CH2 group). Otherwise the crossed condensation will be a disaster. Section 23-3 reinforces the fact that anions of -dicarbonyl compounds are still enolates. Thus, you will find that they do 1,4-additions to , -unsaturated carbonyl compounds (Michael additions). Furthermore, these additions can be followed by Robinson annulations, thereby resulting in six-membered rings. 23-4. Alkanoyl (Acyl) Anion Equivalents How are -hydroxy ketones made? Because they contain alcohol groups, you might consider methods first introduced in Chapter 8 for the synthesis of alcohols: addition reactions of organometallic (carbanionic) reagents to aldehydes and ketones. However, if you tried to apply this approach, you would encounter a O B problem: The necessary carbanion would have the structure RC), with a nucleophilic carbonyl carbon. Such a species is called an alkanoyl (acyl) anion and is not readily prepared. Indeed, ever since Chapter 8, you have had drilled into your brain the fact that carbonyl carbons are electrophilic, not nucleophilic. The whole idea of a nucleophilic alkanoyl anion contradicts everything you’ve learned. So, given all that, can anything be done to get around this limitation? Yes: Carbonyl carbons can be chemically modified to become nucleophilic. One way is to convert the carbonyl group of an aldehyde into a thioacetal and then deprotonate it with a strong base, forming a 1,3-dithiacyclohexane (1,3-dithiane) anion. Another way is to expose an aldehyde to thiazolium salts. Again, deprotonation of the original carbonyl carbon may follow, giving a nucleophilic anion. Either way, you have reversed the normal polarity of this carbon atom. Once such an anion (an alkanoyl anion equivalent) has been formed, it can add to the normal, electrophilic carbonyl group of another molecule in the usual way. The result is a new carbon–carbon bond between carbons that both started out with the same polarity (electrophilic). This is important. Application of polarity reversal in organic chemistry isn’t magical. All you’ve done is turn carbonyl groups into new functions that are able to support negative charges. Then, after finishing with them, you’ve changed them back to carbonyl groups again. Still, this material can be troublesome to learn. You might try the following for study purposes: Write down several aldehydes and their (hard to make) alkanoyl anions. Next, follow along one of the two general reaction sequences shown in the text section, drawing the corresponding alkanoyl anion equivalent, adding it to another carbonyl group, and regenerating the original carbonyl carbon. The practice will be good for you. Solutions to Problems 24. Claisen condensations. (a), (b), (c) involve two identical molecules; (d), (e) are intramolecular examples; (f), (g), (h), (i) are mixed condensations. Make your new carbon–carbon bond (boldface) between the carbonyl carbon of one ester and the -carbon of another. (a) (b) (c) Unfavorable equilibrium. Claisen product is not stable; no reaction is observed. CHCOCH2CH3 C6H5CHCH3 O O C6H5CH(CH3)CH2C CHCOCH2CH3 CH3CH2 CH3CH2CH2C O O 1559T_ch23_409-422 10/19/05 19:50 Page 411
1559rch23409-42210/19/0519:50Page412 EQA 412 chapter 23 ESTER ENOLATES AND THE CLAISEN CONDENSATON This oth OCH.CH D HC-CHCOCH-CH (C.H.C- OCH.CH CH:CH2 O OCH.CH COCH.CH COCH,CH OCH-CH 25.The second ester.(CH)CHCOCH.should be nt in stabl product from Claisen from the first ester.Side reaction (condensation of first ester with itself): 0 2CH.CH.CO.CH,CH.CH.CCHCO.CH CH 26.Analyze as you did for Problem 24.Claisen"means 1.NaOCH,CH CH,CH,OH.2.HH.O. CHCO.CH.CH,-CH.Co.CH.CH C.H.CCHCO.CH.CH,C.H.CO.CH.CH,+C.H.CH.CO.CH.CH c COzCHCH [Problem 24(e)!]
(d) (e) (f) (g) (h) (i) O B 25. The second ester, (CH3)2CHCOCH3, should be present in excess because (1) it does not form a stable product from Claisen condensation with itself and (2) it will be able to preferentially react with enolate ions from the first ester. Side reaction (condensation of first ester with itself): 26. Analyze as you did for Problem 24. “Claisen” means 1. NaOCH2CH3, CH3CH2OH, 2. H, H2O. (a) (b) (c) [Problem 24(e)!] Claisen CO2CH2CH3 CO2CH2CH3 CO2CH2CH3 CH3 CH3 O Claisen C6H5C CHCO2CH2CH3 C6H5CO2CH2CH3 C6H5CH2CO2CH2CH3 C6H5 O Claisen CH2C CHCO2CH2CH3 CH2CO2CH2CH3 O 2 2 CH3CH2CO2CH3 CH3CH2CCHCO2CH3 CH3 NaOCH3, CH3OH O OCH2CH3 OCH2CH3 O O O O C C COCH2CH3 COCH2CH3 O O O O CHCOCH2CH3 CH3CH2 C6H5C O O HC CHCOCH2CH3 C6H5 O O O O This other possible product is not stable and will not be isolated. CH3 COCH2CH3 O O C CH3 OCH2CH3 O O C OCH2CH3 412 • Chapter 23 ESTER ENOLATES AND THE CLAISEN CONDENSATION 1559T_ch23_409-422 10/19/05 19:50 Page 412
1559T_ch23_409-42210/19/0519:50Pa9e413 EQA Soutionso Problems413 ()HCCCH.CO.CH.CH,HcCO.CH.CH,+CH.CO.CH.CH, 0 CaeCCHtCllCo.CHC+cHcC (Ketone ester version) 0 CH.CH.OCCH.COCH.CH,CH.CH.OCOCH.CH.+CH.CO.CH.CH (Carbonate ester version 00 (,co.CH.CH,+CH. (Ester ketone) 00 27.HC-CH-CH HCO-CH-CH:CH:CH? Not likely to work because aldol condensation of 2 CHCHO would be a major competing process 0 O R 28.Analysis:CH,CCH-R CH;C- -CO.CH.CH,CH.CCH.CO.CH.CH3 R The solvent for each reaction in this and the next problem can be ethanol (a)R--CHCH(CH)2 R'-H.1.NaOCH,CH.2.(CH)CHCH,Br.3.NaOH,H.O,4.H" R=-CHCHR CH. CH.CO.CH,.NaOCH.CHs.2.BrCH.CO.CH.CH.3.NaOCH.CHs.4.CH,CH,Br. 29.General pattern CH-COOR 、CO2CH,CH3 CO2CH2CH, CH2 R CO2CH-CH; CO.CH-CH ()1.NaOCH.CH,2.CH,CH.CH.CHl.3.NaOCH,CH. 》-CHBr(completes necessary alkylations).5.NaOH.H2O(hydrolyzes esters).6.H.HaO.A(decarboxylation): 茶器
Solutions to Problems • 413 (d) (e) (f) (g) 27. Not likely to work because aldol condensation of 2 CH3CHO would be a major competing process. 28. The solvent for each reaction in this and the next problem can be ethanol. (a) R OCH2CH(CH3)2, R H. 1. NaOCH2CH3, 2. (CH3)2CHCH2Br, 3. NaOH, H2O, 4. H, H2O, ; (b) R R OCH2CH2CH2O. 1. 2 NaOCH2CH3, 2. BrCH2CH2CH2Br, 3. NaOH, H2O, 4. H, H2O, ; (c) R OCH2C6H5, R OCH2CHPCH2. 1. NaOCH2CH3, 2. C6H5CH2Br, 3. NaOCH2CH3, 4. CH2PCHCH2Br, 5. NaOH, H2O, 6. H, H2O, ; (d) R OCH2CH3, R OCH2CO2CH2CH3. 1. NaOCH2CH3, 2. BrCH2CO2CH2CH3, 3. NaOCH2CH3, 4. CH3CH2Br, 5. NaOH, H2O, 6. H, H2O, (decarboxylates only the COOH on the -carbon to the ketone), 7. CH3CH2OH, H (converts the other COOH group back to ethyl ester) 29. General pattern (a) 1. NaOCH2CH3, 2. CH3CH2CH2CH2I, 3. NaOCH2CH3, 4. (completes necessary alkylations), 5. NaOH, H2O (hydrolyzes esters), 6. H, H2O, (decarboxylation); (b) 1. NaOCH2CH3, 2. (CH3)2CHCH2I, 3. NaOCH2CH3, 4. CH3I (completes alkylations), 5. NaOH, H2O, 6. H, H2O, ; (c) 1. NaOCH2CH3, 2. BrCH2CO2CH2CH3 [alkylation, makes CH3CH2O2CCH2CH(CO2CH2CH3)2], 3. NaOH, H2O, 4. H, H2O, ; CH2Br CH COOH R R C R R CO2CH2CH3 CH2 CO2CH2CH3 CO2CH2CH3 CO2CH2CH3 Starting compound for each synthesis. Analysis: CH3CCH CH3C C CO2CH2CH3 CH3CCH2CO2CH2CH3 O R R R O O Starting material R for each synthesis. HC CH2CH HCO2CH2CH3 CH3CH ? O O O (Ester ketone) C CH2CCH3 CO2CH2CH3 CH3CCH3 O O O Claisen Claisen CH3CH2OC CH2COCH2CH3 O O (Carbonate ester version) CH3CH2OCOCH2CH3 CH3CO2CH2CH3 O Claisen CH3CC6H5 (Ketone ester version) C6H5C C CH2CC6H5 6H5CO2CH2CH3 O O O Claisen HC C CH2CO2CH2CH3 HCCO2CH2CH3 CH3CO2CH2CH3 O O O 1559T_ch23_409-422 10/19/05 19:50 Page 413
15597.eh23.409-42210/21/0523:33Page414 EQA 414.Chapter 23 ESTER ENOLATES AND THE CLAISEN CONDENSATON CH Br (d)1.2 NaOCH2CH3.2. ,3.NaOH.H,0,4.H,H,0.4 人CH,B 30.a +CH=CHCCH, →product CH.CHOH b NaOCH.CH CO2CH-CH 31.Here is one way that gets the job done 32.(a)R (b)H2N (c),(d)Does not work:An O-H bond is necessary (see the mechanisms above)
(d) 1. 2 NaOCH2CH3, 2. , 3. NaOH, H2O, 4. H, H2O, 30. (a) (b) (c) 31. Here is one way that gets the job done: 32. (a) (b) (c), (d) Does not work: An OOH bond is necessary (see the mechanisms above). O O C H O H H O O C H2N H2N H O H H O O O R C H O H H O O C O R H O H H O O O H C H O H H O O C O H H O H H Cat. NaOCH2CH3 CH3CH2OH (Michael addition) 1. NaOH, H2O 2. H, H2O, CO2 O O O O CH product 3CCH2CO2CH2CH3 CO2CH2CH3 Cat. NaOCH2CH3 CH3CH2OH CH2(CO2CH2CH3)2 product O Cat. NaOCH2CH3 CH3CH2OH (Michael addition) CH2 CHCCH3 O O product O CH2Br , CH2Br 414 • Chapter 23 ESTER ENOLATES AND THE CLAISEN CONDENSATION 1559T_ch23_409-422 10/21/05 23:33 Page 414
1559T_ch23_409-42211/7/0516:42Pa9e415 EQA Solutions to Problems.415 33.(CHCH2OzC)CH- -CH-CH-OH .(CH.CH.O.C.CH +CH.CHCCH (CH,CHOC)CHCH.CHCOCH,COCC CH.CHO..C..CHC 34.Work backward.Note carbon-carbon bonds being formed (arrows). acetoacetic ester. CH ⊕ CH,CH.O.C H,H0. CHs 89c -2 CH. NaOH.HO. CHCOCH. Michael Claisen 广.CH.CH.OH COOH 上08 Re-form ester -C0 CH:
Solutions to Problems • 415 33. The step marked with an asterisk is reversible and, in fact, is an unfavorable equilibrium, because the product (a simple ketone enolate) is a less stable anion than is the doubly stabilized malonate anion. However, the next step, reaction with more malonic ester to make a new malonate anion, drives the equilibrium to product. The reaction is catalytic in base because malonate is regenerated in this last step. 34. Work backward. Note carbon–carbon bonds being formed (arrows). (a) (b) CH3 O CO2CH2CH3 CH3 O COOH Re-form ester H, CH3CH2OH 1. NaOH, H2O CO2 2. H, H2O, 1. NaOH Alkylation 2. CH3I 1. NaOCH2CH3 Claisen 2. H, H2O O O CH3 O O NaOH, H2O, Aldol 1. NaOH Michael 2. CH2 CHCOCH3 O O O O (A Robinson annulation sequence) O CH3 O 2. H, H2O 1. NaOCH2CH3 CH3CCH2CO2CH2CH3 2 CH3CO2CH2CH3 CH3 CH3 O O 2. CH2 CHCOCH3 1. NaOCH2CH3 H, H2O, CH3CH2O2C NaOH, H2O, Aldol CH3 CH3 CH3 O O O How? Consider Michael addition of acetoacetic ester. (CH3CH2O2C)2CHCH2CH2COCH3 CH(CO2CH2CH3)2 (Regenerated, to continue on) (Product) (CH3CH2O2C)2CHCH2CHCOCH3 CH(CO2CH2CH3)2 H CH3CH2OH OCH2CH3 (CH3CH2O2C)2CH H (CH3CH2O2C)2CH CH2 CHCCH3 O * 1559T_ch23_409-422 11/7/05 16:42 Page 415
1559r.ah23_409-42210/19/0520:17Page416 EQA 416.chapter 23 ESTER ENOLATES AND THE CLAISEN CONDENSATON CH O CH.CO.CH.CH) CH:C (c)A sequence identical to that of(b),but substitute two alkylations with BrCH,COCH,for the two Michael additions to CH2=CHCOCH3. 00H 35.(a)(CH3)CH-C-CH-CH(CH3) (b)CoHs-C-CH-CoHs c-ce CoHs H CoHs In the same order as in Problem 35: O OH (a)C.H-C-CH-CH(CH (b)CoHsCCH-CaHs (ame product!) ac-c-m○ dCHC-CH-CH,CH 37.(a)A IR:Ketone and alcohol groups(cannot be an amine because molecular weight is an even OH O NMR:CH-CH CH-C- 1.4d42g220
(c) A sequence identical to that of (b), but substitute two alkylations with BrCH2COCH3 for the two Michael additions to CH2PCHCOCH3. O OH B A 35. (a) (CH3)2CHOCOCHOCH(CH3)2 O OH B A (b) C6H5OCOCHOC6H5 (c) O OH B A (d) C6H5CH2OCOCHOCH2C6H5 36. (a) (b) In the same order as in Problem 35: O OH B A (a) C6H5OCOCHOCH(CH3)2 O OH B A (b) C6H5OCOCHOC6H5 (same product!) (c) O OH B A (d) C6H5OCOCHOCH2C6H5 37. (a) A IR: Ketone and alcohol groups (cannot be an amine because molecular weight is an even number). Structure is CH3CH CH3 (C4H8O2) C OH O CH3 1.4(d) 4.2(q) 2.2(s) 3.7 NMR: CH3 C OH O CH C6H5 C CH O OH S S Li C6H5 S H S C6H5 C O OH CH 1. NaOCH2CH3 Michael 2. CH2 CHCOCH3 CH2(CO2CH2CH3)2 CH3CH2O2C CH CH2 CH2 CH3 C O CO2CH2CH3 416 • Chapter 23 ESTER ENOLATES AND THE CLAISEN CONDENSATION 1559T_ch23_409-422 10/19/05 20:17 Page 416
1559T_ch23_409-42210/19/0519:50Pa9e417 EQA Solutions to Problems.417 0 CHO fragments.Simplest interpretation:molecule is CH.CCCH (b)Oxidation.Churning cream mixes it with air,thus allowing O to react with ketoalcohol A to 38.Addition to carbonyl 0 CHaCH +/CHCH2O--CHaC-OCH.CHs Deprotonation of a-carbon 0 Deprotonation of the aldehyde carbon leads to a much poorer anion than the enolate:CHC:electron pair in an sporbital,unable to be stabilized by resonance.Given the two favorable processes shown above. deprotonation of the -CH group is simply not competitive 39 CO.CH: CO.CH 290 -CH;OH -H0
Solutions to Problems • 417 B Molecular weight is reduced by 2 units, so formula is probably now C4H6O2. IR: Ketone signal only. NMR: All H’s equivalent. MS: Molecule breaks in half readily, giving m/z 43, O OO B BB C2H3O fragments. Simplest interpretation: CH3OCO, so molecule is CH3CCCH3 (b) Oxidation. Churning cream mixes it with air, thus allowing O2 to react with ketoalcohol A to make diketone B. (c) You can synthesize A by reaction of acetaldehyde with catalytic N-dodecyl-thiazolium salt (Section 23-4). Oxidation gives B. (d) The diketone is conjugated. 38. Addition to carbonyl: Deprotonation of -carbon: O B Deprotonation of the aldehyde carbon leads to a much poorer anion than the enolate: CH3C), electron pair in an sp2 orbital, unable to be stabilized by resonance. Given the two favorable processes shown above, O B deprotonation of the OCH group is simply not competitive. 39. 1. NaOCH3, CH3OH 2. H, H2O H2O O CO2CH3 1. NaOCH3, CH3OH 2. H, H2O CH3OH CO2CH3 O 1. NaOCH3 2. O O O CO2CH3 Br Br 1. Na CH(CO2CH3)2 2. NaOH, H2O 3. H2SO4, H2O, 2 CH3OH 2 CO2 CH3OH, H CO2H CO2H CO2CH3 CO2CH3 CH3CH2O O O HCCH2 HCCH2 Enolate H HOCH2CH3 CH3CH2O O O CH3CH CH 3C H OCH2CH3 1559T_ch23_409-422 10/19/05 19:50 Page 417
15597_ch23_409-42210/19/0520:17Page419 418.Chapter 23 ESTER ENOLATES AND THE CLAISEN CONDENSATION 40.The Kn usm is the same a (a)CH.CCH-CO2CH2CH .HCoCH.CH NaOCH2CH :o:CCH CH.CCHCO.CH.CH, CO.CH2CHs CH CH: OCH:CHs CO.CHCH CO2CH2CH3 (b) arbon double bond: -CH=C(CO2CH2CH3)2 &098 CO:CH2CH CO2CH2CH3 CH2OH CH,B 41.The acetoacetic ester ketone synthesis is only good for methyl ketones. CH,C-CHRR'from CH,C-CH2CO,CH2CH, For other ketones,the appropriate 3-ketoester must be prepared by using a Claisen condensation. 0 0 CHCO2CH.CHs CHa CH.CH.CO.CH.CH
40. The Knoevenagel condensation is a variation of aldol condensation using a -dicarbonyl compound as the source of the enolate reaction partner. Its mechanism is the same as that of the aldol: (a) (b) As in the aldol, remove the elements of water (two hydrogen atoms from the carbon of the malonate ester and the oxygen atom from the aldehyde) and replace them with a carbon–carbon double bond: (c) 41. The acetoacetic ester ketone synthesis is only good for methyl ketones. O O B B CH3COCHRR from CH3COCH2CO2CH2CH3 For other ketones, the appropriate 3-ketoester must be prepared by using a Claisen condensation. (a) Claisen (Problem 26) 2CH3CH2CO2CH2CH3 1. NaOH, H2O 2. H, H2O, CH3CH2CCHCO2CH2CH3 CH3 O CH3CH2CCH2CH3 O 1. LiAlH4 2. H, H2O PBr3 CH2OH CH2OH CH2Br CH2Br O CH2(CO2CH2CH3) 2 NaOCH2CH3, CH3CH2 OH CO2CH2CH3 CO2CH2CH3 CH C(CO2CH2CH3)2 OCH2CH3 O CCH3 CO2CH2CH3 O CCH3 CO2CH2CH3 HO H NaOCH2CH3 CH3CCH2CO2CH2CH3 O CH3CCHCO2CH2CH3 O O O O CCH3 CO2CH2CH3 H OCH2CH3 418 • Chapter 23 ESTER ENOLATES AND THE CLAISEN CONDENSATION 1559T_ch23_409-422 10/19/05 20:17 Page 418
1559T_ch23_409-42210/19/0519:50Pa9e419 EQA Soufionso Problems419 CH -C-CO2CH2CH3 ,29a,c4, CH CH2CH2CH2CHs 〉-co,cH,c CH.CH-CH-CH CHa(CH2).CO.CH-CH CO.CH-CH c) CH.CH-CH CH-CH=CHz 上8ca CO2CH2CH; —C0,CH,CH -CH2CO2CH2CH3 0 d 0 -CH.CO-CH-CH CO2CH2CHs CO2CH2CHs CH:CHQ 22《 >CH2B 0 CO2CH2CH3 CO2CH2CH3 CH3CH20 42.Ethanol is the reaction solvent unless specified otherwise. Cyclopentanon HCO.CH.CH,+CH.CO-CH.CH,HCCH.CO.CH.CH. CH,CCH,BrCH.CCH
Solutions to Problems • 419 (b) (c) (d) 42. Ethanol is the reaction solvent unless specified otherwise. Cyclopentanone Br2, CH3CO2H BrCH2CCH3 O CH3CCH3 O Claisen HCO2CH2CH3 CH3CO2CH2CH3 HCCH2CO2CH2CH3 O Double Claisen 1. 2 NaOCH2CH3 2. 2 CH2Br CO2CH2CH3 O O CO2CH2CH3 CO2CH2CH3 CH3CH2O O O CO2CH2CH3 CH3CH2O 1. NaOH, H2O 2. H, H2O, CH2 CH2 O O CH2 CO2CH2CH3 CH2 CO2CH2CH3 O O 2. BrCH2CH CH2 1. NaOCH2CH3 CO2CH2CH3 Dieckmann O CO2CH2CH3 CO2CH2CH3 1. NaOH, H2O 2. H, H2O, CO2CH2CH3 CH2CH CH2 O CH2CH CH2 O Mixed Claisen CHCO2CH2CH3 CH2CH2CH2CH3 C O CO2CH2CH3 CH3(CH2)4CO2CH2CH3 1. NaOCH2CH3 2. CH3I CH3 CO2CH2CH3 CH2CH2CH2CH3 C C O 1. NaOH, H2O 2. H, H2O, CH3 C CHCH2CH2CH2CH3 O 1559T_ch23_409-422 10/19/05 19:50 Page 419
15597ch23_409-42211/7/0516:45Pag0420 EQA 420.chapter 23 ESTER ENOLATES AND THE CLAISEN CONDENSATON CCHCH.CCH CO2CH2CH3 HCCH.CH.CCH:oLo4了》N-C人 Aldol Cyclohexanone CH.CCH,+CH-0CH.CCH.CH.O CHCCH-CH CO.CH-CH: is-
Cyclohexanone 43. CH3O HO H H Dehydration H product CH3O O O H CH3O Variation of Friedel-Crafts HO CH3O O H HO (Enol as nucleophile) CH3O O O OH H HO H2O Acid-catalyzed dehydration H CH3O O O H Acid-catalyzed aldol condensation O CH3 CH3O O O Michael addition O HCCH2CO2CH2CH3 CO2CH2CH3 O HCCHCH2CH2CCH3 1. NaOCH O O 2CH3 2. CH3COCH CH2 Same steps as above From above Michael O CH3CCH3 CH2 O CH3CCH CH2 O CH3CCH2CH2OH O O NaOH, H2O Aldol H, HCCH2CH2CCH3 O O NaOH, H2O, H2, Pd C Aldol HCCH2CO2CH2CH3 CO2CH2CH3 O HCCHCH2CCH3 1. NaOCH O O 2CH3 2. BrCH2COCH3 H, H2O, 420 • Chapter 23 ESTER ENOLATES AND THE CLAISEN CONDENSATION 1559T_ch23_409-422 11/7/05 16:45 Page 420