CHAPTER 18 ENOLS AND ENOLATES SOLUTIONS TO TEXT PROBLEMS 18.1 (b) There are no H9251-hydrogen atoms in 2,2-dimethylpropanal, because the H9251-carbon atom bears three methyl groups. (c) All three protons of the methyl group, as well as the two benzylic protons, are H9251 hydrogens. (d) Cyclohexanone has four equivalent H9251 hydrogens. O H H H H Cyclohexanone (the hydrogens indicated are the H9251 hydrogens) Benzyl methyl ketone O C 6 H 5 CH 2 CCH 3 Five H9251 hydrogens H 3 C O H CH 3 CH 3 CC 2,2-Dimethylpropanal H9251 470 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website ENOLS AND ENOLATES 471 18.2 As shown in the general equation and the examples, halogen substitution is specific for the H9251-carbon atom. The ketone 2-butanone has two nonequivalent H9251 carbons, and so substitution is possible at both positions. Both 1-chloro-2-butanone and 3-chloro-2-butanone are formed in the reaction. 18.3 The carbon–carbon double bond of the enol always involves the original carbonyl carbon and the H9251-carbon atom. 2-Butanone can form two different enols, each of which yields a different H9251-chloro ketone. 18.4 Chlorine attacks the carbon–carbon double bond of each enol. 18.5 (b) Acetophenone can enolize only in the direction of the methyl group. (c) Enolization of 2-methylcyclohexanone can take place in two different directions. CH 3 OH 2-Methylcyclohex-1-enol (enol form) CH 3 O 2-Methylcyclohexanone CH 3 OH 6-Methylcyclohex-1-enol (enol form) Acetophenone CCH 3 O Enol form of acetophenone CCH 2 OH OH CH 3 C H11001 Cl H11002 CHCH 3 H11001 Cl OH CH 3 C CHCH 3 Cl Cl OH ClCH 2 H11001 Cl H11002 CCH 2 CH 3 H11001 OH H 2 C CCH 2 CH 3 Cl Cl CH 3 CCH 2 CH 3 O 2-Butanone CHCH 3 CH 3 C OH 2-Buten-2-ol (enol) CH 3 CCHCH 3 O Cl 3-Chloro-2-butanone slow Cl 2 fast CH 3 CCH 2 CH 3 O 2-Butanone CCH 2 CH 3 H 2 C OH 1-Buten-2-ol (enol) ClCH 2 CCH 2 CH 3 O 1-Chloro-2-butanone slow Cl 2 fast H11001H11001 H H11001 CH 3 CCHCH 3 O Cl 3-Chloro-2-butanone ClCH 2 CCH 2 CH 3 O 1-Chloro-2-butanone Cl 2 Chlorine CH 3 CCH 2 CH 3 O 2-Butanone Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 472 ENOLS AND ENOLATES 18.6 (b) Enolization of the central methylene group can involve either of the two carbonyl groups. 18.7 (b) Removal of a proton from 1-phenyl-1,3-butanedione occurs on the methylene group between the carbonyls. The three most stable resonance forms of this anion are (c) Deprotonation at C-2 of this H9252-dicarbonyl compound yields the carbanion shown. The three most stable resonance forms of the anion are: 18.8 Each of the five H9251 hydrogens has been replaced by deuterium by base-catalyzed enolization. Only the OCH 3 hydrogens and the hydrogens on the aromatic ring are observed in the 1 H NMR spectrum at H9254 3.9 ppm and H9254 6.7–6.9 ppm, respectively. 18.9 H9251-Chlorination of (R)-sec-butyl phenyl ketone in acetic acid proceeds via the enol. The enol is achi- ral and yields equal amounts of (R)- and (S)-2-chloro-2-methyl-1-phenyl-1-butanone. The product is chiral. It is formed as a racemic mixture, however, and this mixture is not optically active. CC C 6 H 5 HO CH 3 CH 2 CH 3 Enol C 6 H 5 C O CCH 2 CH 3 CH 3 Cl 2-Chloro-2-methyl-1- phenyl-1-butanone (50% R; 50% S) C O C 6 H 5 C H CH 2 CH 3 CH 3 (R)-sec-Butyl phenyl ketone acetic acid Cl 2 K 2 CO 3 CH 2 CCH 3 5D 2 O CH 3 O CH 3 O O H11001H11001 5DOHCD 2 CCD 3 CH 3 O CH 3 O O O CH O H11002 O CH O H11002 O O CH H11002 H 2 O O H CH O O H11001 H11001 CH O H11002 HO H11002 O C 6 H 5 CCH O C 6 H 5 CCHCCH 3 H11002 O O CCH 3 H11002 C 6 H 5 C O H11002 O CHCCH 3 O HO H11002 H 2 OH11001H11001C 6 H 5 CCH 2 CCH 3 C 6 H 5 CCHCCH 3 H11002 O O O O CCH 3 C 6 H 5 CCH HO Enol form O O C 6 H 5 CCH 2 CCH 3 1-Phenyl-1,3-butanedione OOH CHCCH 3 C 6 H 5 C Enol form Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 18.10 (b) Approaching this problem mechanistically in the same way as part (a), write the structure of the enolate ion from 2-methylbutanal. This enolate adds to the carbonyl group of the aldehyde. A proton transfer from solvent yields the product of aldol addition. (c) The aldol addition product of 3-methylbutanal can be identified through the same mechanis- tic approach. O (CH 3 ) 2 CHCH 2 CH CHCH(CH 3 ) 2 H11002 HC O 3-Hydroxy-2-isopropyl-5-methylhexanal OH (CH 3 ) 2 CHCH 2 CH CHCH(CH 3 ) 2 H11001 OH H11002 HC O 3-Methylbutanal Enolate of 3-methylbutanal H11001 O (CH 3 ) 2 CHCH 2 CH CHCH(CH 3 ) 2 H11002 HC O H 2 O H11001 H 2 OH11001 HO H11002 O (CH 3 ) 2 CHCH 2 CH 3-Methylbutanal O (CH 3 ) 2 CHCHCH Enolate of 3-methylbutanal H11002 HO H11002 HO CH 3 CH 2 CHCH CH 3 CCH 2 CH 3 CH 3 HC O H 2 OH11001H11001 O CH 3 CH 2 CHCH CH 3 CCH 2 CH 3 CH 3 HC O H11002 2-Ethyl-3-hydroxy-2,4- dimethylhexanal O CH 3 CH 2 CHCH CH 3 2-Methylbutanal Enolate of 2-methylbutanal CCH 2 CH 3 CH 3 HC O H11002 O CH 3 CH 2 CHCH CH 3 CCH 2 CH 3 CH 3 HC O H11002 HO H11002 H11001 O CH 3 CH 2 CHCH CH 3 2-Methylbutanal Enolate of 2-methylbutanal O CH 3 CH 2 CCH CH 3 O H11002 CH 3 CH 2 CCH CH 3 H11002 ENOLS AND ENOLATES 473 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 18.11 Dehydration of the aldol addition product involves loss of a proton from the H9251-carbon atom and hydroxide from the H9252-carbon atom. (b) The product of aldol addition of 2-methylbutanal has no H9251 hydrogens. It cannot dehydrate to an aldol condensation product. (c) Aldol condensation is possible with 3-methylbutanal. 18.12 The carbon skeleton of 2-ethyl-1-hexanol is the same as that of the aldol condensation product derived from butanal. Hydrogenation of this compound under conditions in which both the carbon–carbon double bond and the carbonyl group are reduced gives 2-ethyl-1-hexanol. 18.13 (b) The only enolate that can be formed from tert-butyl methyl ketone arises by proton abstrac- tion from the methyl group. H11001 HO H11002 O (CH 3 ) 3 CCCH 3 tert-Butyl methyl ketone Enolate of tert-butyl methyl ketone O (CH 3 ) 3 CCCH 2 H11002 CH 3 CH 2 CH 2 CH O Butanal 2-Ethyl-1-hexanol CH 3 CH 2 CH 2 CH 2 CHCH 2 OH CH 2 CH 3 CH 3 CH 2 CH 2 CH O CCH CH 2 CH 3 2-Ethyl-2-hexenal H 2 , NiNaOH, H 2 O heat HO H11002 2(CH 3 ) 2 CHCH 2 CH O 3-Methylbutanal (CH 3 ) 2 CHCH 2 CHCHCH(CH 3 ) 2 HO HC O Aldol addition product (CH 3 ) 2 CHCH 2 CH CCH(CH 3 ) 2 HC O 2-Isopropyl-5-methyl-2-hexenal H11002H 2 O 2CH 3 CH 2 CHCH CH 3 O HO H11002 2-Methylbutanal CH 3 CH 2 CHCH CH 3 HO CCH 2 CH 3 HC O CH 3 H9251 (No protons on H9251-carbon atom) H11001 H 2 O H11001 HO H11002 R 2 C CHCH O CHCH H O OH H11002 R 2 C OH heat 474 ENOLS AND ENOLATES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website This enolate adds to the carbonyl group of benzaldehyde to give the mixed aldol addition product, which then dehydrates under the reaction conditions. (c) The enolate of cyclohexanone adds to benzaldehyde. Dehydration of the mixed aldol addition product takes place under the reaction conditions to give the following mixed aldol condensa- tion product. 18.14 Mesityl oxide is an H9251,H9252-unsaturated ketone. Traces of acids or bases can catalyze its isomerization so that some of the less stable H9252,H9253-unsaturated isomer is present. 18.15 The relationship between the molecular formula of acrolein (C 3 H 4 O) and the product (C 3 H 5 N 3 O) corresponds to the addition of HN 3 to acrolein. Because propanal (CH 3 CH 2 CH?O) does not react under these conditions, the carbon-carbon, not the carbon-oxygen, double bond of acrolein is the re- active site. Conjugate addition is the reaction that occurs. N 3 CH 2 CH 2 CH OO H 2 C CHCH NaN 3 acetic acid Acrolein 3-Azidopropanal O CHCCH 3 C H 3 C H 3 C Mesityl oxide; 4-methyl- 3-penten-2-one (more stable) O CH 2 CCH 3 C H 3 C H 2 C 4-Methyl-4-penten-2-one (less stable) HO H11002 H11001 O Cyclohexanone O C 6 H 5 CH Benzaldehyde H11001 H 2 O O CHC 6 H 5 Benzylidenecyclohexanone O H11002 O C 6 H 5 CHCH 2 CC(CH 3 ) 3 H11001 Enolate of tert-butyl methyl ketone CH 2 CC(CH 3 ) 3 H11002 O Benzaldehyde OH O C 6 H 5 CHCH 2 CC(CH 3 ) 3 Product of mixed aldol addition O C 6 H 5 CH H 2 O O C 6 H 5 CH CHCC(CH 3 ) 3 4,4-Dimethyl-1-phenyl-1-penten-3-one (product of mixed aldol condensation) H11002H 2 O ENOLS AND ENOLATES 475 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 18.16 The enolate of dibenzyl ketone adds to methyl vinyl ketone in the conjugate addition step. via The intramolecular aldol condensation that gives the observed product is 18.17 A second solution to the synthesis of 4-methyl-2-octanone by conjugate addition of a lithium dialkylcuprate reagent to an H9251,H9252-unsaturated ketone is revealed by the disconnection shown: According to this disconnection, the methyl group is derived from lithium dimethylcuprate. 18.18 (a) In addition to the double bond of the carbonyl group, there must be a double bond elsewhere in the molecule in order to satisfy the molecular formula C 4 H 6 O (the problem states that the CH 3 CH 2 CH 2 CH 2 CH CHCCH 3 O O CH 3 CH 2 CH 2 CH 2 CHCH 2 CCH 3 H11001 LiCu(CH 3 ) 2 3-Octen-2-one Lithium dimethylcuprate 4-Methyl-2-octanone CH 3 CH 3 CH 2 CH 2 CH 2 CH CHCCH 3 OO CH 3 CH 2 CH 2 CH 2 CHCH 2 CCH 3 Disconnect this bond. CH 3 CH 3 H9252 H9251 H11002 1,3-Diphenyl-2,6-heptanedione CC 6 H 5 CH 2 CHC 6 H 5 O O CH 2 CH 2 CH 3 C H 3 C C 6 H 5 C CHC 6 H 5 C O HO CH 2 CH 2 CH 3-Methyl-2,6-diphenyl-2- cyclohexen-1-one NaOCH 3 CH 3 OH H11002H 2 O C 6 H 5 C 6 H 5 O H 3 C C 6 H 5 CH 2 CCHC 6 H 5 H 2 C O O H11002 CHCCH 3 C 6 H 5 CH 2 CCHC 6 H 5 H11002 O H 2 C O CHCCH 3 Methyl vinyl ketone 1,3-Diphenyl-2,6-heptanedioneDibenzyl ketone NaOCH 3 CH 3 OH C 6 H 5 CH 2 CCH 2 C 6 H 5 H11001 CHCCH 3 H 2 C O O C 6 H 5 CH 2 CCHC 6 H 5 CH 2 CH 2 CCH 3 O O 476 ENOLS AND ENOLATES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website compounds are noncyclic). There are a total of five isomers: (b) The E and Z isomers of 2-butenal are stereoisomers. (c) None of the C 4 H 6 O aldehydes and ketones is chiral. (d) The H9251,H9252-unsaturated aldehydes are (E)- and ; and . There is one H9251,H9252-unsaturated ketone in the group: . (e) The E and Z isomers of 2-butenal are formed by the aldol condensation of acetaldehyde. 18.19 The main flavor component of the hazelnut has the structure shown. 18.20 The characteristic reaction of an alcohol on being heated with KHSO 4 is acid-catalyzed dehydration. Secondary alcohols dehydrate faster than primary alcohols, and so a reasonable first step is The product of this dehydration is an enol, which tautomerizes to an aldehyde. The aldehyde then undergoes dehydration to form acrolein. KHSO 4 heat (H11002H 2 O) HOCH 2 CH CHOH Propene-1,3-diol 3-Hydroxypropanal HOCH 2 CH 2 CH O Acrolein H 2 C CHCH O KHSO 4 heat HOCH 2 CHCH 2 OH OH 1,2,3-Propanetriol HOCH 2 CH CHOH Propene-1,3-diol CC H H 3 C H CC O H CH 3 CH 2 CH 3 (2E,5S)-5-Methyl-2-hepten-4-one H 2 C O CHCCH 3 H 2 C CH 3 CCHOCH 3 CH CHCHO(Z) - O H 2 C CH 3 CCH 2-Methylpropenal O H 2 C CHCCH 3 3-Buten-2-one (methyl vinyl ketone) O O H 2 C CHCH 2 CH 3-Butenal CC H 3 C H H CH (E)-2-Butenal CC H 3 C H H CH (Z)-2-Butenal O ENOLS AND ENOLATES 477 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 18.21 (a) 2-Methylpropanal has the greater enol content. Although the enol content of 2-methylpropanal is quite small, the compound is nevertheless capable of enolization, whereas the other compound, 2,2-dimethylpropanal, cannot enolize— it has no H9251 hydrogens. (b) Benzophenone has no H9251 hydrogens; it cannot form an enol. Dibenzyl ketone enolizes slightly to form a small amount of enol. (c) Here we are comparing a simple ketone, dibenzyl ketone, with a H9252-diketone. The H9252-diketone enolizes to a much greater extent than the simple ketone because its enol form is stabilized by conjugation of the double bond with the remaining carbonyl group and by intramolecular hy- drogen bonding. (d) The enol content of cyclohexanone is quite small, whereas the enol form of 2,4-cyclo- hexadienone is the aromatic compound phenol, and therefore enolization is essentially complete. O Keto form OH Enol form (aromatic; much more stable) 1,3-Diphenyl-1,3- propanedione C 6 H 5 CCH 2 CC 6 H 5 O O Enol form O C 6 H 5 C CH CC 6 H 5 O H Dibenzyl ketone C 6 H 5 CH 2 CCH 2 C 6 H 5 O Enol form C 6 H 5 CH CCH 2 C 6 H 5 OH C O (Enolization is impossible.) C H CH 3 C CH 3 CH 3 O (Enolization is impossible.) 2-Methylpropanal (CH 3 ) 2 CHCH O Enol form (CH 3 ) 2 CCH OH 478 ENOLS AND ENOLATES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (e) A small amount of enol is in equilibrium with cyclopentanone. Cyclopentadienone does not form a stable enol. Enolization would lead to a highly strained allene-type compound. ( f ) The H9252-diketone is more extensively enolized. The double bond of the enol form of 1,4-cyclohexanedione is not conjugated with the car- bonyl group. Its enol content is expected to be similar to that of cyclohexanone. 18.22 (a) Chlorination of 3-phenylpropanal under conditions of acid catalysis occurs via the enol form and yields the H9251-chloro derivative. (b) Aldehydes undergo aldol addition on treatment with base. HC O C 6 H 5 CH 2 CH 2 CHCHCH 2 C 6 H 5 OH 2-Benzyl-3-hydroxy-5-phenylpentanal O 2C 6 H 5 CH 2 CH 2 CH 3-Phenylpropanal NaOH ethanol, 10H11034C Cl 2 H11001 3-Phenylpropanal C 6 H 5 CH 2 CH 2 CH O HClH11001 2-Chloro-3- phenylpropanal C 6 H 5 CH 2 CHCH O Cl acetic acid Enol form (not particularly stable; double bond and carbonyl group not conjugated) OH O 1,4-Cyclohexanedione O O O O 1,3-Cyclohexanedione OH O Enol form (double bond conjugated with carbonyl group) H O OH (Not stable; highly strained) O Cyclopentanone OH Enol form ENOLS AND ENOLATES 479 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (c) Dehydration of the aldol addition product occurs when the reaction is carried out at elevated temperature. (d) Lithium aluminum hydride reduces the aldehyde function to the corresponding primary alcohol. (e) A characteristic reaction of H9251,H9252-unsaturated carbonyl compounds is their tendency to undergo conjugate addition on treatment with weakly basic nucleophiles. 18.23 (a) Ketones undergo H9251 halogenation by way of their enol form. (b) The combination of C 6 H 5 CH 2 SH and NaOH yields C 6 H 5 CH 2 S H11002 (as its sodium salt), which is a weakly basic nucleophile and adds to H9251,H9252-unsaturated ketones by conjugate addition. (c) Bromination occurs at the carbon atom that is H9251 to the carbonyl group. Br 2 diethyl ether C 6 H 5 O C 6 H 5 2,2-Diphenylcyclo- pentanone C 6 H 5 O Br C 6 H 5 2-Bromo-5,5- diphenylcyclopentanone (76%) H 3 C C(CH 3 ) 2 O 2-Isopropylidene-5- methylcyclohexanone C 6 H 5 CH 2 SH NaOH, H 2 O 2-(1-Benzylthio-1-methylethyl)-5- methylcyclohexanone (89–90%) H 3 C C(CH 3 ) 2 SCH 2 C 6 H 5 H O CCH 2 CH 3 Cl O 1-(o-Chlorophenyl)- 1-propanone CCHCH 3 Cl Cl O 2-Chloro-1-(o-chlorophenyl)- 1-propanone Cl 2 CH 2 Cl 2 HC O C 6 H 5 CH 2 CH 2 CH CCH 2 C 6 H 5 2-Benzyl-5-phenyl-2-pentenal HC CN O C 6 H 5 CH 2 CH 2 CHCHCH 2 C 6 H 5 2-Benzyl-3-cyano-5-phenylpentanal NaCN H H11001 CH 2 OH C 6 H 5 CH 2 CH 2 CH CCH 2 C 6 H 5 2-Benzyl-5-phenyl-2-penten-1-ol HC O C 6 H 5 CH 2 CH 2 CH CCH 2 C 6 H 5 2-Benzyl-5-phenyl-2-pentenal 1. LiAlH 4 2. H 2 O HC O C 6 H 5 CH 2 CH 2 CH CCH 2 C 6 H 5 2-Benzyl-5-phenyl-2-pentenal O 2C 6 H 5 CH 2 CH 2 CH 3-Phenylpropanal NaOH ethanol, 70H11034C 480 ENOLS AND ENOLATES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (d) The reaction is a mixed aldol condensation. The enolate of 2,2-diphenylcyclohexanone reacts with p-chlorobenzaldehyde. Elimination of the aldol addition product occurs readily to yield the H9251,H9252-unsaturated ketone as the isolated product. (e) The aldehyde given as the starting material is called furfural and is based on a furan unit as an aromatic ring. Furfural cannot form an enolate. It reacts with the enolate of acetone in a manner much as benzaldehyde would. ( f ) Lithium dialkylcuprates transfer an alkyl group to the H9252-carbon atom of H9251,H9252-unsaturated ketones. A mixture of stereoisomers was obtained in 67% yield in this reaction. (g) Two nonequivalent H9251-carbon atoms occur in the starting ketone. Although enolate formation is possible at either position, only reaction at the methylene carbon leads to an intermediate that can undergo dehydration. O O CHC 6 H 5 OH HO H11002 C 6 H 5 CHO H11002H 2 O O CHC 6 H 5 Observed product (75% yield) LiCu(CH 3 ) 2 H11001 CH 3 CH 3 O H 3 C 1. diethyl ether 2. H 2 O 2,4,4-Trimethyl-2- cyclohexenone CH 3 CH 3 CH 3 O H 3 C 2,3,4,4-Tetramethyl- cyclohexanone H11001 O NaOH water H11002H 2 O CH O O Furfural CH 3 CCH 3 Acetone CHCH 2 CCH 3 OH O O (Not isolated) CH CHCCH 3 O O 4-Furyl-3-buten-2-one (60–66%) H11001 CHCl C 6 H 5 C 6 H 5 O KOH ethanol H11002H 2 O CH O Cl p-Chlorobenzaldehyde C 6 H 5 C 6 H 5 O 2,2-Diphenylcyclo- hexanone CH OH Cl C 6 H 5 C 6 H 5 O (Not isolated) 2-(p-Chlorobenzylidene)-6,6- diphenylcyclohexanone (84%) ENOLS AND ENOLATES 481 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Reaction at the other H9251 position gives an intermediate that cannot dehydrate. (h) H9252-Diketones readily undergo alkylation by primary halides at the most acidic position, on the carbon between the carbonyls. 18.24 (a) Conversion of 3-pentanone to 2-bromo-3-pentanone is best accomplished by acid-catalyzed bromination via the enol. Bromine in acetic acid is the customary reagent for this transformation. (b) Once 2-bromo-3-pentanone has been prepared, its dehydrohalogenation by base converts it to the desired H9251,H9252-unsaturated ketone 1-penten-3-one. Potassium tert-butoxide is a good base for bringing about elimination reactions of secondary alkyl halides; suitable solvents include tert-butyl alcohol and dimethyl sulfoxide. (c) Reduction of the carbonyl group of 1-penten-3-one converts it to the desired alcohol. Catalytic hydrogenation would not be suitable for this reaction because reduction of the dou- ble bond would accompany carbonyl reduction. 1-Penten-3-one H 2 C CHCCH 2 CH 3 O 1-Penten-3-ol H 2 C CHCHCH 2 CH 3 OH 1. LiAlH 4 , diethyl ether 2. H 2 O KOC(CH 3 ) 3 CH 3 CHCCH 2 CH 3 O Br 2-Bromo-3-pentanone CHCCH 2 CH 3 H 2 C O 1-Penten-3-one Br 2 acetic acid CH 3 CH 2 CCH 2 CH 3 O 3-Pentanone CH 3 CHCCH 2 CH 3 O Br 2-Bromo-3-pentanone H11001 KOH O O 1,3-Cyclo- hexanedione H 2 C CHCH 2 Br Allyl bromide CH 2 CH 2 CH O O 2-Allyl-1,3-cyclohexanedione (75%) O O C 6 H 5 CH OH (Cannot dehydrate; reverts to starting materials) HO H11002 , C 6 H 5 CHO 482 ENOLS AND ENOLATES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (d) Conversion of 3-pentanone to 3-hexanone requires addition of a methyl group to the H9252-carbon atom. The best way to add an alkyl group to the H9252 carbon of a ketone is via conjugate addition of a dialkylcuprate reagent to an H9251,H9252-unsaturated ketone. (e) The compound to be prepared is the mixed aldol condensation product of 3-pentanone and benzaldehyde. The desired reaction sequence is 18.25 (a) The first step is an H9251 halogenation of a ketone. This is customarily accomplished under condi- tions of acid catalysis. In the second step the carbonyl group of the H9251-bromo ketone is reduced to a secondary alcohol. As actually carried out, sodium borohydride in water was used to achieve this transformation. (CH 3 ) 3 CCCH 2 Br O 1-Bromo-3,3-dimethyl-2-butanone (CH 3 ) 3 CCHCH 2 Br OH 1-Bromo-3,3-dimethyl-2-butanol (54%) NaBH 4 H 2 O (CH 3 ) 3 CCCH 3 O 3,3-Dimethyl-2-butanone (CH 3 ) 3 CCCH 2 Br O 1-Bromo-3,3-dimethyl-2-butanone (58%) Br 2 H H11001 CH 3 CCCH 2 CH 3 O C 6 H 5 CH 2-Methyl-1-phenyl- 1-penten-3-one CH 3 CHCCH 2 CH 3 O C 6 H 5 CHOH Aldol addition product (not isolated; dehydration occurs under conditions of its formation) CH 3 CH 2 CCH 2 CH 3 O 3-Pentanone CH 3 CHCCH 2 CH 3 O Enolate of 3-pentanone HO H11002 H11002H 2 OC 6 H 5 CH H11002 O C 6 H 5 CH O H11001CH 3 CHCCH 2 CH 3 CH 3 CCCH 2 CH 3 O C 6 H 5 CH 2-Methyl-1-phenyl- 1-penten-3-one O H11002 1-Penten-3-one [prepared as described in part (b)] H 2 C CHCCH 2 CH 3 O 3-Hexanone O CH 3 CH 2 CH 2 CCH 2 CH 3 1. LiCu(CH 3 ) 2 2. H 2 O CH 3 CH 2 CH 2 CCH 2 CH 3 O O H 2 CCH 3 H11001 CHCCH 2 CH 3 H11002 H9252 ENOLS AND ENOLATES 483 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website The third step is conversion of a vicinal bromohydrin to an epoxide in aqueous base. (b) The overall yield is the product of the yields of the individual steps. Yield H11005 100(0.58 H11003 0.54 H11003 0.68) H11005 21% 18.26 The product is a sulfide (thioether). Retrosynthetic analysis reveals a pathway that begins with ben- zene and acetic anhydride. The desired synthesis can be accomplished with the following series of reactions: The synthesis is completed by reacting bromomethyl phenyl ketone with 1-propanethiolate anion. 18.27 All these problems begin in the same way, with exchange of all the H9251 protons for deuterium (Section 18.8). NaOD D 2 O Cyclopentanone O H HH H Cyclopentanone-2,2,5,5-d 4 O D DD D KOH CH 3 CH 2 CH 2 SH CCH 2 SCH 2 CH 2 CH 3 O CCH 2 Br O CH 3 CH 2 CH 2 S H11002 1-Propanethiol 1-Propanethiolate Phenyl (1-thiopropyl)- methyl ketone AlCl 3 Br 2 H H11001 H11001 Benzene CH 3 COCCH 3 O O Acetic anhydride CCH 3 O Acetophenone CCH 2 Br O Bromomethyl phenyl ketone CCH 2 SCH 2 CH 2 CH 3 O CCH 2 Br H11001 O SCH 2 CHCH 2 CH 3 H11002 CCH 3 O H11001 OO CH 3 COCCH 3 (CH 3 ) 3 CCHCH 2 Br OH 1-Bromo-3,3-dimethyl-2-butanol KOH H 2 O 2-tert-Butyloxirane (68%) (CH 3 ) 3 CC CH 2 H O 484 ENOLS AND ENOLATES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Once the tetradeuterated cyclopentanone has been prepared, functional group transformations are employed to convert it to the desired products. (a) Reduction of the carbonyl group can be achieved by using any of the customary reagents. (b) Acid-catalyzed dehydration of the alcohol prepared in part (a) yields the desired alkene. (c) Catalytic hydrogenation of the alkene in part (b) yields cyclopentane-1,1,3-d 3 . (d) Carbonyl reduction of the tetradeuterated ketone under Wolff–Kishner conditions furnishes the desired product. Alternatively, Clemmensen reduction conditions (Zn, HCl) could be used. 18.28 (a) Hydroformylation converts alkenes to aldehydes having one more carbon atom by reaction with carbon monoxide and hydrogen in the presence of a cobalt octacarbonyl catalyst. (b) Aldol condensation of acetaldehyde to 2-butenal, followed by catalytic hydrogenation of the carbon–carbon double bond, gives butanal. 2CH 3 CH CH 3 CH 2 CH 2 CH O H 2 Ni NaOH heat Acetaldehyde O CH 3 CH O CHCH 2-Butenal Butanal H11001 CO H11001 H 2 CH 3 CH CH 2 CH 3 CH 2 CH 2 CH O Co 2 (CO) 8 Propene Carbon monoxide Hydrogen Butanal H 2 NNH 2 KOH, diethylene glycol, heat Cyclopentanone-2,2,5,5-d 4 O Cyclopentane-1,1,3,3-d 4 D DD D D DD D H11001 H 2 Hydrogen Cyclopentane-1,1,3-d 3 Pt Cyclopentene-1,3,3-d 3 H D D D HH H DD D H 2 SO 4 heat Cyclopentene-1,3,3-d 3 Cyclopentanol-2,2,5,5-d 4 H D HOH D DD D D D NaBH 4 or LiAlH 4 or H 2 , Pt Cyclopentanone-2,2,5,5-d 4 O Cyclopentanol-2,2,5,5-d 4 HOH D DD D D DD D ENOLS AND ENOLATES 485 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 18.29 (a) The first conversion is the H9251 halogenation of an aldehyde. As described in Section 18.2, this particular conversion has been achieved in 80% yield simply by treatment with bromine in chloroform. Dehydrohalogenation of this compound can be accomplished under E2 conditions by treat- ment with base. Sodium methoxide in methanol would be appropriate, for example, although almost any alkoxide could be employed to dehydrohalogenate this tertiary bromide. As the reaction was actually carried out, the bromide was heated with the weak base N,N-diethylaniline to effect dehydrobromination in 71% yield. (b) Cleavage of vicinal diols to carbonyl compounds can be achieved by using periodic acid (HIO 4 ) (Section 15.12). The conversion of this dialdehyde to cyclopentene-1-carbaldehyde is an intramolecular aldol condensation and is achieved by treatment with potassium hydroxide. As the reaction was actually carried out, cyclopentene-1-carbaldehyde was obtained in 58% yield from trans-1,2-cyclohexanediol by this method. (c) The first transformation requires an oxidative cleavage of a carbon–carbon double bond. Ozonolysis followed by hydrolysis in the presence of zinc is indicated. 3-Isopropyl-6- oxoheptanal 1. O 3 2. H 2 O, Zn 4-Isopropyl-1- methylcyclohexene CH 3 CH(CH 3 ) 2 CH 3 CH(CH 3 ) 2 O CH 2 CH O CH O CH O OH CH O H11002 H11002H 2 O CH O Cyclopentene-1- carbaldehyde HO H11002 CH O CH O HIO 4 O 1,6-Hexanedial OH OH trans-1,2-Cyclohexanediol CH CH O Cyclohexene-1- carbaldehyde NaOCH 3 CH 3 OH 1-Bromocyclohexane- carbaldehyde Br CH O CH O Cyclohexane- carbaldehyde Br 2 CHCl 3 CH O 1-Bromocyclohexane- carbaldehyde Br CH O 486 ENOLS AND ENOLATES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Cyclization of the resulting keto aldehyde is an intramolecular aldol condensation. Base is required. (d) The first step in this synthesis is the hydration of the alkene function to an alcohol. Notice that this hydration must take place with a regioselectivity opposite to that of Markovnikov’s rule and therefore requires a hydroboration–oxidation sequence. Conversion of the secondary alcohol function to a carbonyl group can be achieved with any of a number of oxidizing agents. Cyclization of the dione to the final product is a base-catalyzed intramolecular aldol conden- sation and was accomplished in 71% yield by treatment of the dione with a 2% solution of sodium hydroxide in aqueous ethanol. 18.30 Intramolecular aldol condensations occur best when a five- or six-membered ring is formed. Carbon–carbon bond formation therefore involves the aldehyde and the methyl group attached to the ketone carbonyl. KOH O OCH 2 C H 2 C H 3 C C CH CH 3 H11002 H9015 CH 3 CH 3 OH CH 3 CCH 2 CCH O O CH 3 CH 3 2,2-Dimethyl-4- oxopentanal H9015 CH 3 CH 3 4,4-Dimethyl-2- cyclopentenone (63%) H11002H 2 O O CCH 2 CH 2 C CH 2 (CH 3 ) 2 CH O H11002 CCH 2 CH 2 C H 2 C O (CH 3 ) 2 CH HO O (CH 3 )CH HO H11002 O H11002 O (CH 3 ) 2 CHCCH 2 CH 2 CCH 2 OO (CH 3 ) 2 CHCCH 2 CH 2 CCH 3 H 2 CrO 4 5-Hydroxy-6-methyl-2-heptanone O (CH 3 ) 2 CHCHCH 2 CH 2 CCH 3 OH 6-Methyl-2,5-heptanedione OO (CH 3 ) 2 CHCCH 2 CH 2 CCH 3 6-Methyl-5-hepten-2-one (CH 3 ) 2 C CHCH 2 CH 2 CCH 3 O 1. B 2 H 6 , THF 2. H 2 O 2 , HO H11002 5-Hydroxy-6-methyl-2-heptanone O (CH 3 ) 2 CHCHCH 2 CH 2 CCH 3 OH HO H11002 CH 3 H H CH(CH 3 ) 2 O CH 2 CH O CH 3 H CH(CH 3 ) 2 O CH 2 CH O H11002 CH 3 C O H OH CH(CH 3 ) 2 H11002H 2 O CH 3 C O CH(CH 3 ) 2 ENOLS AND ENOLATES 487 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 18.31 (a) By realizing that the primary alcohol function of the target molecule can be introduced by re- duction of an aldehyde, it can be seen that the required carbon skeleton is the same as that of the aldol addition product of 2-methylpropanal. The synthetic sequence is The starting aldehyde is prepared by oxidation of 2-methyl-1-propanol. (b) Retrosynthetic analysis of the desired product shows that the carbon skeleton can be con- structed by a mixed aldol condensation between benzaldehyde and propanal. The reaction scheme therefore becomes Reduction of the aldehyde to the corresponding primary alcohol gives the desired compound. LiAlH 4 , then H 2 O or NaBH 4 , CH 3 OH CH 3 CCHC 6 H 5 CH O 2-Methyl-3-phenyl-2- propenal CH 3 CCH 2 OHC 6 H 5 CH 2-Methyl-3-phenyl-2- propen-1-ol H11001 HO H11002 CH 3 CCHC 6 H 5 CH O 2-Methyl-3-phenyl-2- propenal CH 3 CH 2 CH O Propanal C 6 H 5 CH O Benzaldehyde CH 3 CCH 2 OHC 6 H 5 CH CH 3 CCHC 6 H 5 CH O C 6 H 5 CH H11001 CH 3 CH 2 CH O O PCC CH 2 Cl 2 O (CH 3 ) 2 CHCH 2-Methylpropanal (CH 3 ) 2 CHCH 2 OH 2-Methyl-1-propanol NaBH 4 CH 3 OH NaOH ethanol O (CH 3 ) 2 CHCH 2-Methylpropanal CH 3 CH 3 HO (CH 3 ) 2 CHCHCC H O 3-Hydroxy-2,2,4- trimethylpentanal CH 3 CH 3 HO (CH 3 ) 2 CHCHCCH 2 OH 2,2,4-Trimethyl-1,3-pentanediol O 2(CH 3 ) 2 CHCH CH 3 CH 3 HO (CH 3 ) 2 CHCHCCH 2 OH CH 3 CH 3 HO (CH 3 ) 2 CHCH CC H O 488 ENOLS AND ENOLATES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website The starting materials for the mixed aldol condensation—benzaldehyde and propanal—are prepared by oxidation of benzyl alcohol and 1-propanol, respectively. (c) The cyclohexene ring in this case can be assembled by a Diels–Alder reaction. 1,3-Butadiene is one of the given starting materials; the H9251,H9252-unsaturated ketone is the mixed aldol condensation product of 4-methylbenzaldehyde and acetophenone. The complete synthetic sequence is H9251,H9252-Unsaturated ketones are good dienophiles in Diels–Alder reactions. 4-Methylbenzaldehyde H 3 CCHH11001 O O Acetophenone CH 3 C NaOH ethanol H 3 C CHCCH O CHCHH 2 CCH 2 trans-4-Benzoyl-5- (4-methylphenyl)cyclohexene CH 3 CC 6 H 5 O H 3 CCH 2 OH 4-Methylbenzyl alcohol 4-Methylbenzaldehyde H 3 CCH O PDC CH 2 Cl 2 H11001H 3 C CHCC 6 H 5 CH H 3 CCH OO CH 3 C O CH 3 H11001 CC 6 H 5 CC 6 H 5 H H O O H 3 C CH 3 CH 2 CH 2 OH 1-Propanol CH 3 CH 2 CH O Propanal PCC CH 2 Cl 2 C 6 H 5 CH 2 OH Benzyl alcohol C 6 H 5 CH O Benzaldehyde PCC CH 2 Cl 2 ENOLS AND ENOLATES 489 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 18.32 It is the carbon atom flanked by two carbonyl groups that is involved in the enolization of terreic acid. Of these two structures, enol A, with its double bond conjugated to two carbonyl groups, is more sta- ble than enol B, in which the double bond is conjugated to only one carbonyl. 18.33 (a) Recall that aldehydes and ketones are in equilibrium with their hydrates in aqueous solution (Section 17.6). Thus, the principal substance present when (C 6 H 5 ) 2 CHCH?O is dissolved in aqueous acid is (C 6 H 5 ) 2 CHCH(OH) 2 (81%). (b) The problem states that the major species present in aqueous base is not (C 6 H 5 ) 2 CHCH?O, its enol, or its hydrate. The most reasonable species is the enolate ion: 18.34 (a) At first glance this transformation seems to be an internal oxidation–reduction reaction. An aldehyde function is reduced to a primary alcohol, and a secondary alcohol is oxidized to a ketone. Once one realizes that enolization can occur, however, a simpler explanation, involving only proton-transfer reactions, emerges. The enol form of compound A is an enediol; it is at the same time the enol form of compound B. The enediol can revert to compound A or to compound B. C 6 H 5 CCH 2 OH O C 6 H 5 CHCH O OH OH C 6 H 5 C CHOH HO H11002 , H 2 O (Compound B) (Compound A) C 6 H 5 CHCH O OH Compound A Enol form of compound A OH C 6 H 5 C CHOH HO H11002 , H 2 O HO H11002 H 2 O C 6 H 5 CHCH O OH Compound A C 6 H 5 CCH 2 OH O Compound B (C 6 H 5 ) 2 CCHO H11002 (C 6 H 5 ) 2 CCHO(C 6 H 5 ) 2 CHCH O H11002H11002H H11001 Terreic acid Enol A Enol B O O O CH 3 H O O O OH CH 3 O O and O O CH 3 OH 490 ENOLS AND ENOLATES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website At equilibrium, compound B predominates because it is more stable than A. A ketone carbonyl is more stabilized than an aldehyde, and the carbonyl in B is conjugated with the benzene ring. (b) The isolated product is the double hemiacetal formed between two molecules of compound A. 18.35 (a) The only stereogenic center in piperitone is adjacent to a carbonyl group. Base-catalyzed enol- ization causes this carbon to lose its stereochemical integrity. Both the enolate and enol of piperitone are achiral and can revert only to a racemic mixture of piperitones. (b) The enol formed from menthone can revert to either menthone or isomenthone. Only the stereochemistry at the H9251-carbon atom is affected by enolization. The other stereogenic center in menthone (the one bearing the methyl group) is not affected. 18.36 In all parts of this problem the bonding change that takes place is described by the general equation (a) The compound given is nitrosoethane. Nitrosoalkanes are less stable than their oxime isomers formed by proton transfer. CH 3 CH N H O Nitrosoethane (less stable) CH 3 CH N OH Acetaldehyde oxime (more stable) HX N Z XNZH O CH(CH 3 ) 2 CH 3 H Menthone OH CH(CH 3 ) 2 CH 3 Enol form H O CH(CH 3 ) 2 CH 3 Isomenthone H H11001 H H11001 CH 3 CH 2 O H11002 CH 3 CH 2 OH O CH(CH 3 ) 2 CH 3 H (H11002)-Piperitone O H11002 CH(CH 3 ) 2 CH 3 Enolate of piperitone OH CH(CH 3 ) 2 CH 3 Enol of piperitone HC C C O H O O CH C 6 H 5 OHC 6 H 5 H HO C 6 H 5 CH C O H O C 6 H 5 H H O C CH C 6 H 5 C 6 H 5 OH HO O O Compound C ENOLS AND ENOLATES 491 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (b) You may recognize this compound as an enamine. It is slightly different, however, from the enamines we discussed earlier (Section 17.11) in that nitrogen bears a hydrogen substituent. Stable enamines are compounds of the type where neither R group is hydrogen; both R’s must be alkyl or aryl. Enamines that bear a hy- drogen substituent are converted to imines in a proton-transfer equilibrium. (c) The compound given is known as a nitronic acid; its more stable tautomeric form is a nitroalkane. (d) The six-membered ring is aromatic in the tautomeric form derived from the compound given. (e) This compound is called isourea. Urea has a carbon–oxygen double bond and is more stable. 18.37 (a) This reaction is an intramolecular alkylation of a ketone. Although alkylation of a ketone with a separate alkyl halide molecule is usually difficult, intramolecular alkylation reactions can be carried out effectively. The enolate formed by proton abstraction from the H9251-carbon atom carries out a nucleophilic attack on the carbon that bears the leaving group. O H11002 CH 2 CH 2 CH 2 Br CH 2 O O O H11002 CH 2 CH 2 CH 2 CH 2 Br CH 2 CH 2 CH 2 CH 2 Br KOC(CH 3 ) 3 Isourea (less stable) HN C NH 2 OH Urea (more stable) H 2 N C NH 2 O NH 2 N N N NH NH Nitronic acid Nitroalkane CH 3 CH N H11001 OH O H11002 CH 3 CH 2 N H11001 O O H11002 (CH 3 ) 2 C CH NCH 3 H Enamine (less stable) (CH 3 ) 2 CH CH NCH 3 Imine (more stable) CC N R R 492 ENOLS AND ENOLATES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (b) The starting material, known as citral, is converted to the two products by a reversal of an aldol condensation. The first step is conjugate addition of hydroxide. The product of this conjugate addition is a H9252-hydroxy ketone. It undergoes base-catalyzed cleavage to the observed products. (c) The product is formed by an intramolecular aldol condensation. (d) In this problem stereochemical isomerization involving a proton attached to the H9251-carbon atom of a ketone takes place. Enolization of the ketone yields an intermediate in which the O CH 3 O HCCH 2 CH 2 CHCCH 3 H11001 HO H11002 O CH 3 O HCCH 2 CH 2 CHCCH 2 H11001 H 2 O H11002 HC CH 3 O CH 2 O H11002 C H H11002 O CH 2 CH 3 O CH 3 O H HO HO H11002 (H11002H 2 O) CH 3 O H 2 O OH CHCH 2 CH 2 C(CH 3 ) 2 C CH 3 CH 2 CH O HO H11002 CH 3 CH O O CHCH 2 CH 2 C(CH 3 ) 2 C CH 3 CH 2 CH O H11002 O CHCH 2 CH 2 CCH 3 (CH 3 ) 2 CH11001H 2 CCH O H11002 H 2 O (CH 3 ) 2 C CH 3 OH CHCH 2 CH 2 CCHCH O O H11002 O (CH 3 ) 2 C CH 3 OH CHCH 2 CH 2 CCHCH (CH 3 ) 2 C CH 3 OH CHCH 2 CH 2 CCH 2 CH H 2 O H11002 ENOLS AND ENOLATES 493 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website stereochemical integrity of the H9251 carbon is lost. Reversion to ketone eventually leads to the formation of the more stable stereoisomer at equilibrium. The rate of enolization is increased by heating or by base catalysis. The cis ring fusion in the product is more stable than the trans because there are not enough atoms in the six-membered ring to span trans-1,2 positions in the four-membered ring without excessive strain. (e) Working backward from the product, we can see that the transformation involves two aldol condensations: one intermolecular and the other intramolecular. The first reaction is a mixed aldol condensation between the enolate of dibenzyl ketone and one of the carbonyl groups of the dione. This is followed by an intramolecular aldol condensation. ( f ) This is a fairly difficult problem because it is not obvious at the outset which of the two possible enolates of benzyl ethyl ketone undergoes conjugate addition to the H9251,H9252-unsaturated ketone. A good idea here is to work backward from the final product—in effect, do a ret- rosynthetic analysis. The first step is to recognize that the enone arises by dehydration of a H9252-hydroxy ketone. C 6 H 5 C 6 H 5 CH 3 C 6 H 5 O C 6 H 5 CH 3 C 6 H 5 O C 6 H 5 OH C 6 H 5 C CC 6 H 5 O O C 6 H 5 C CHC 6 H 5 C H11002 H11002H 2 O O C 6 H 5 C 6 H 5 C 6 H 5 OH C 6 H 5 O C 6 H 5 C 6 H 5 C 6 H 5 C 6 H 5 2,3,4,5-Tetraphenyl- cyclopentadienone H11001 O O C 6 H 5 CCCH 2 C 6 H 5 C 6 H 5 CCC 6 H 5 O C 6 H 5 CH 2 CCH 2 C 6 H 5 Dibenzyl ketone C 6 H 5 CCC 6 H 5 OO Benzil O C 6 H 5 C 6 H 5 C 6 H 5 C 6 H 5 C 6 H 5 C CC 6 H 5 O O C 6 H 5 C CHC 6 H 5 C H11002 C 6 H 5 CCC 6 H 5 OO O C 6 H 5 CHCCH 2 C 6 H 5 H11002 O H H Less stable ketone; starting material HO H Enol O H H More stable ketone; preferred at equilibrium CH 3 CH 3 CH 3 CH 3 CH 3 CH 3 494 ENOLS AND ENOLATES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Now, mentally disconnect the bond between the H9251-carbon atom and the carbon that bears the hydroxyl group to reveal the intermediate that undergoes intramolecular aldol condensation. The H9252-hydroxy ketone is the intermediate formed in the intramolecular aldol addition step, and the diketone that leads to it is the intermediate that is formed in the conjugate addition step. The relationship of the starting materials to the intermediates and product is now more evident. 18.38 (a) The reduced C?O stretching frequency of H9251,H9252-unsaturated ketones is consistent with an en- hanced degree of single bond character as compared with simple dialkyl ketones. Resonance is more important in H9251,H9252-unsaturated ketones. Conjugation of the carbonyl group with the carbon–carbon double bond increases opportunities for electron delocalization. (b) Even more single-bond character is indicated in the carbonyl group of cyclopropenone than in that of typical H9251,H9252-unsaturated ketones. The dipolar resonance form contributes substantially to the electron distribution because of the aromatic character of the three-membered ring. Recall that cyclopropenyl cation satisfies the 4n H11001 2 rule for aromaticity (text Section 11.20). (c) The dipolar resonance form is a more important contributor to the electron distribution in diphenylcyclopropenone than in benzophenone. C 6 H 5 C 6 H 5 O C 6 H 5 C 6 H 5 O H11002 O O H11002 H11013O H11002 H11001 Equivalent to an oxyanion-substituted cyclopropenyl cation R 2 CCHCRH11032 O R 2 CCHCRH11032 O H11002 H11001 R 2 CCHCRH11032 O H11002 H11001 RCRH11032 O RCRH11032 O H11002 H11001 CH 3 OH C 6 H 5 C 6 H 5 O C 6 H 5 CH 3 O Intermediate formed in conjugate addition step C 6 H 5 CCH 2 C C 6 H 5 O CH C 6 H 5 CCH 2 CH 3 H11002 O C 6 H 5 C 6 H 5 O C 6 H 5 CH 3 O H11002 C 6 H 5 C 6 H 5 O C 6 H 5 CH 3 O H 5 C 6 O C 6 H 5 CH 3 OH Disconnect this bond. ENOLS AND ENOLATES 495 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website is more pronounced than The dipolar resonance form of diphenylcyclopropenone has aromatic character. Its stability leads to increased charge separation and a larger dipole moment. (d) Decreased electron density at the H9252 carbon atom of an H9251,H9252-unsaturated ketone is responsible for its decreased shielding. The decreased electron density arises from the polarization of its H9266 electrons as represented by a significant contribution of the dipolar resonance form. 18.39 Bromination can occur at either of the two H9251-carbon atoms. The 1 H NMR spectrum of the major product, compoundA, is consistent with the structure of 1-bromo- 3-methyl-2-butanone. The minor product B is identified as 3-bromo-3-methyl-2-butanone on the basis of its NMR spectrum. 18.40 Three dibromination products are possible from H9251 halogenation of 2-butanone. The product is 1,3-dibromo-2-butanone, on the basis of its observed 1 H NMR spectrum, which showed two signals at low field. One is a two-proton singlet at H9254 4.6 ppm assignable to CH 2 Br and the other a one-proton quartet at H9254 5.2 ppm assignable to CHBr. 18.41 Solutions to molecular modeling exercises are not provided in this Study Guide and Solutions Man- ual. You should use Learning By Modeling for this exercise. Br 2 CHCCH 2 CH 3 O 1,1-Dibromo-2-butanone BrCH 2 CCHCH 3 O Br 1,3-Dibromo-2-butanone CH 3 C CCH 3 O Br Br 3,3-Dibromo-2-butanone BrCH 2 CCH 3 C O CH 3 H H9254 4.1 ppm (singlet) H9254 3.0 ppm (septet) H9254 1.2 ppm (doublet) Compound A CH 3 C O CH 3 Br H 3 CC H9254 1.9 ppm (singlet) H9254 2.5 ppm (singlet) Compound B H11001 Br 2 CH 3 CC(CH 3 ) 2 Br O 3-Bromo-3-methyl- 2-butanone BrCH 2 CCH(CH 3 ) 2 O 1-Bromo-3-methyl- 2-butanone CH 3 CCH(CH 3 ) 2 O 3-Methyl-2-butanone H 2 CCHCR O H 2 CCHCR O H11002 H11001 C 6 H 5 C 6 H 5 OC C 6 H 5 C 6 H 5 O H11002 C H11001 496 ENOLS AND ENOLATES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website SELF-TEST PART A A-1. Write the correct structure(s) for each of the following: (a) The two enol forms of 2-butanone (b) The enolate ion derived from reaction of 1,3-cyclohexanedione with sodium methoxide (c) The carbonyl form of the following enol A-2. Give the correct structures for compounds A and B in the following reaction schemes: (b) A-3. Write the structures of all the possible aldol addition products that may be obtained by reac- tion of a mixture of propanal and 2-methylpropanal with base. A-4. Using any necessary organic or inorganic reagents, outline a synthesis of 1,3-butanediol from ethanol as the only source of carbons. A-5. Outline a series of reaction steps that will allow the preparation of compound B from 1,3- cyclopentanedione, compound A. A-6. Give the structure of the product formed in each of the following reactions: CH 3 O CH 3 H11001 CH(c) O (H11002H 2 O) NaOH, heat 2CH 3 CH 2 CH 2 CH 2 CH(b) O NaOH 5H11034C CH 3 CH 2 CH 2 CH 2 CH(a) O Br 2 acetic acid ? O O O O O CH 2 CH 3 CH 2 CH 2 CCH 3 A B Propanal CH 3 CH 2 CH O 2-Methylpropanal CH 3 CH 3 CHCH O CH 3 CH 2 CH CHCCH 2 CH 3 LiCu(CH 2 CH 3 ) 2 BH11001 1. ether 2. H 2 O O 2C 6 H 5 CH 2 CH(a)A O 1. HO H11002 2. heat (H11002H 2 O) HO ENOLS AND ENOLATES 497 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website A-7. Write out the mechanism, using curved arrows to show electron movement, of the following aldol addition reaction. A-8. Identify the two starting materials needed to make the following compound by a mixed aldol condensation. PART B B-1. When enolate A is compared with enolate B which of the following statements is true? (a) A is more stable than B. (b) B is more stable than A. (c) A and B have the same stability. (d) No comparison of stability can be made. B-2. Which structure is the most stable? B-3. Which one of the following molecules contains deuterium ( 2 H H11005 D) after reaction with NaOD in D 2 O? (CH 3 ) 3 CCC(CH 3 ) 3 (d) O C 6 H 5 CH 2 CH(b) O C 6 H 5 CC(CH 3 ) 3 (c) O C 6 H 5 CH(a) O OO (b)(a) OH O (d) OH OH (e) OH OH (c) OH O BA O O H11002 O O H11002 CHH11001 C CH 3 C O CH 2 CH 3 H 2 O? NaOH, H 2 O heat HCH O CH 3 CH 2 CHH11001 O NaOH, H 2 O 5°C CH(d) CHC O NaSCH 3 ethanol 498 ENOLS AND ENOLATES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website B-4. Which of the following RX compounds is (are) the best alkylating agent(s) in the reaction shown? (CH 3 ) 3 CBr (CH 3 ) 2 CHCH 2 OH C 6 H 5 Br (CH 3 ) 2 CHCH 2 Br 1234 (a) 1 and 4 (c) 2 and 4 (b) 4 only (d) 1, 3, and 4 B-5. Which of the following pairs of aldehydes gives a single product in a mixed aldol condensation? B-6. What is the principal product of the following reaction? B-7. Which of the following forms an enol to the greatest extent? (a)(c) (b d) CH 3 CCCH 2 CH 3 OO CH 3 CCH 2 CH 2 CCH 3 OO CH 3 CCH 2 CH OO CH 3 CH 2 CH O (d) O O (b) OH OH (c) CH O CH (a) O CH O CH 2 CH O CH 2 CH H 2 O H11001 ? O NaOH ethanol, water, heat CH 3 CH H11001(d) O (CH 3 ) 2 CHCH O C 6 H 5 CH H11001(b) O (CH 3 ) 3 CCH O C 6 H 5 CH H11001(c) O H 2 COC 6 H 5 CH 2 CH H11001(a) O C 6 H 5 CH O O O H11001 RX O R O NaOCH 2 CH 3 ENOLS AND ENOLATES 499 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website B-8. Which of the following species is (are) not intermediates in the aldol condensation of ac- etaldehyde (ethanal) in aqueous base? (a) 1 and 2 (c) 4 only (e) 3 and 4 (b) 3 only (d) 2 and 3 B-9. The compound shown in the box undergoes racemization on reaction with aqueous acid. Which of the following structures best represents the intermediate responsible for this process? (a)(b)(c)(d)(e) B-10. Which one of the following compounds is the best candidate for being prepared by an effi- cient mixed aldol addition reaction? (a)(b)(c) (d)(e) B-11. Which one of the following undergoes 1,4-addition with CH 3 SK (in ethanol)? (a)(b)(c)(d) O CH CH 3 O O O O CH 3 O CH 2 CCH 2 CCH 3 OO CCH 2 CH 2 CH OO CH 2 CCHCH 3 O C CH 3 CH 3 HOCCHCH 3 O CH 2 OH CCH 2 CH OH CH 3 O O O H HO OH OH OH OH H OH OH OH O H O OH CH O H11002 CH 3 CHCH O H11002 4 CH 3 CH H11001 OH 3 O CH 3 CHCH 2 CH O H11002 2 H 2 CCH O H11002 1 500 ENOLS AND ENOLATES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website B-12. Benzalacetone is the mixed aldol condensation product formed between benzaldehyde (C 6 H 5 CH?O) and acetone [(CH 3 ) 2 C?O]. What is its structure? (a)(c) (b d) C 6 H 5 CH 2 CCH CH 2 O C 6 H 5 CCH CHCH 3 O C 6 H 5 CH C(CH 3 ) 2 C 6 H 5 CH CHCCH 3 O ENOLS AND ENOLATES 501 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website