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1、CHAPTER 18 ENOLS AND ENOLATES SOLUTIONS TO TEXT PROBLEMS 18.1(b)There are no ?-hydrogen atoms in 2,2-dimethylpropanal, because the ?-carbon atom bears three methyl groups. (c)All three protons of the methyl group, as well as the two benzylic protons, are ? hydrogens. (d)Cyclohexanone has four equiva
2、lent ? hydrogens. O H H H H Cyclohexanone (the hydrogens indicated are the ? hydrogens) Benzyl methyl ketone O C6H5CH2CCH3 Five ? hydrogens H3C O H CH3 CH3 CC 2,2-Dimethylpropanal ? 470 BackForwardMain MenuTOCStudy Guide TOCStudent OLCMHHE Website ENOLS AND ENOLATES471 18.2 As shown in the general e
3、quation and the examples, halogen substitution is specifi c for the ?-carbon atom. The ketone 2-butanone has two nonequivalent ? carbons, and so substitution is possible at both positions. Both 1-chloro-2-butanone and 3-chloro-2-butanone are formed in the reaction. 18.3The carboncarbon double bond o
4、f the enol always involves the original carbonyl carbon and the ?-carbon atom. 2-Butanone can form two different enols, each of which yields a different ?-chloro ketone. 18.4Chlorine attacks the carboncarbon double bond of each enol. 18.5(b)Acetophenone can enolize only in the direction of the methy
5、l group. (c)Enolization of 2-methylcyclohexanone can take place in two different directions. CH3 OH 2-Methylcyclohex-1-enol (enol form) CH3 O 2-Methylcyclohexanone CH3 OH 6-Methylcyclohex-1-enol (enol form) Acetophenone CCH3 O Enol form of acetophenone CCH2 OH OH CH3C? Cl?CHCH3 ? Cl OH CH3CCHCH3 ClC
6、l OH ClCH2? Cl?CCH2CH3 ? OH H2CCCH2CH3 ClCl CH3CCH2CH3 O 2-Butanone CHCH3CH3C OH 2-Buten-2-ol (enol) CH3CCHCH3 O Cl 3-Chloro-2-butanone slowCl2 fast CH3CCH2CH3 O 2-Butanone CCH2CH3H2C OH 1-Buten-2-ol (enol) ClCH2CCH2CH3 O 1-Chloro-2-butanone slowCl2 fast ? H? CH3CCHCH3 O Cl 3-Chloro-2-butanone ClCH2
7、CCH2CH3 O 1-Chloro-2-butanone Cl2 Chlorine CH3CCH2CH3 O 2-Butanone BackForwardMain MenuTOCStudy Guide TOCStudent OLCMHHE Website 472ENOLS 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-butaned
8、ione 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 ?-dicarbonyl compound yields the carbanion shown. The three most stable resonance forms of the anion are: 18.8 Each of the fi ve ? hydrogens has been repl
9、aced by deuterium by base-catalyzed enolization. Only the OCH3hydrogens and the hydrogens on the aromatic ring are observed in the 1H NMR spectrum at ? 3.9 ppm and ? 6.76.9 ppm, respectively. 18.9?-Chlorination of (R)-sec-butyl phenyl ketone in acetic acid proceeds via the enol. The enol is achi- ra
10、l 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 C6H5 HO CH3 CH2CH3 Enol C6H5C O CCH2CH3 CH3 Cl 2-Chloro-2-methyl-1- phenyl-1-butanone (50% R; 50% S) C O C
11、6H5C H CH2CH3 CH3 (R)-sec-Butyl phenyl ketone acetic acidCl2 K2CO3 CH2CCH35D2O CH3O CH3O O ?5DOHCD2CCD3 CH3O CH3O O O CH O ? O CH O ? O O CH ? H2O O H CH O O ? CH O ? HO ? O C6H5CCH O C6H5CCHCCH3 ? OO CCH3 ? C6H5C O?O CHCCH3 O HO?H2O?C6H5CCH2CCH3C6H5CCHCCH3 ? OOO O CCH3C6H5CCH HO Enol form OO C6H5CC
12、H2CCH3 1-Phenyl-1,3-butanedione OOH CHCCH3C6H5C Enol form BackForwardMain MenuTOCStudy Guide TOCStudent OLCMHHE 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 o
13、f the aldehyde. Aproton transfer from solvent yields the product of aldol addition. (c) The aldol addition product of 3-methylbutanal can be identifi ed through the same mechanis- tic approach. O (CH3)2CHCH2CHCHCH(CH3)2 ? HCO 3-Hydroxy-2-isopropyl-5-methylhexanal OH (CH3)2CHCH2CHCHCH(CH3)2? OH? HCO
14、3-MethylbutanalEnolate of 3-methylbutanal ? O (CH3)2CHCH2CHCHCH(CH3)2 ? HCO H2O ? H2O? HO? O (CH3)2CHCH2CH 3-Methylbutanal O (CH3)2CHCHCH Enolate of 3-methylbutanal ? HO? HO CH3CH2CHCH CH3 CCH2CH3 CH3 HCO H2O? O CH3CH2CHCH CH3 CCH2CH3 CH3 HCO ? 2-Ethyl-3-hydroxy-2,4- dimethylhexanal O CH3CH2CHCH CH3
15、 2-MethylbutanalEnolate of 2-methylbutanal CCH2CH3 CH3 HCO ? O CH3CH2CHCH CH3 CCH2CH3 CH3 HCO ? HO? O CH3CH2CHCH CH3 2-MethylbutanalEnolate of 2-methylbutanal O CH3CH2CCH CH3 O? CH3CH2CCH CH3 ? ENOLS AND ENOLATES473 BackForwardMain MenuTOCStudy Guide TOCStudent OLCMHHE Website 18.11Dehydration of th
16、e aldol addition product involves loss of a proton from the ?-carbon atom and hydroxide from the ?-carbon atom. (b)The product of aldol addition of 2-methylbutanal has no ? hydrogens. It cannot dehydrate to an aldol condensation product. (c)Aldol condensation is possible with 3-methylbutanal. 18.12The 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 carbonca
