18.5 Stabilized enols PROBLEM 18.5 Write structural formulas corresponding to (a) The enol form of 2, 4-dimethyl-3-pentanone b)The enol form of acetophe (c) The two enol forms of 2-methylcyclohexanone SAMPLE SOLUTION(a) Remember that enolization involves the a- carbon atom The ketone 2, 4-dimethyl-3-pentanone gives a single enol, since the two a carbons are equivalent. (CH3)2 CHCCH(CH3)2 (CH3)2C-CCH(CH3)2 2, 4-Dimethyl-3-pentanone 2, 4-Dimethyl-2-penten-3-ol (keto form) (enol form) It is important to recognize that an enol is a real substance, capable of indepen dent existence. An enol is not a resonance form of a carbonyl compound; the two are constitutional isomers of each other 18.5 STABILIZED ENOLS Certain structural features can make the keto-enol equilibrium more favorable by stabi lizing the enol form. Enolization of 2, 4-cyclohexadienone is one such example K is too large to measure. (keto form, not one Phenol (enol form, aromatic The enol is phenol, and the stabilization gained by forming an aromatic ring is more than enough to overcome the normal preference for the keto form. A 1, 3 arrangement of two carbonyl groups(compounds called B-diketones) leads to a situation in which the keto and enol forms are of comparable stability CH3CCHCCH3 F CH3C=CHCCH3 K= 4 2, 4-Pentanedione(20%o 4-Hydroxy-3-penten-2-one(80%c The two most important structural features that stabilize the enol of a B-dicarbonyl com pound are(1)conjugation of its double bond with the remaining carbonyl group and (2) the presence of a strong intramolecular hydrogen bond between the enolic hydroxyl group and the carbonyl oxygen(Figure 18.2) involved in enolization. The alternative enoy p flanked by the two carbonyls that is In B-diketones it is the methylene grou CH=CCH,CCH 4-Hydroxy-4-penten-2- Back Forward Main MenuToc Study Guide ToC Student o MHHE WebsitePROBLEM 18.5 Write structural formulas corresponding to (a) The enol form of 2,4-dimethyl-3-pentanone (b) The enol form of acetophenone (c) The two enol forms of 2-methylcyclohexanone SAMPLE SOLUTION (a) Remember that enolization involves the -carbon atom. The ketone 2,4-dimethyl-3-pentanone gives a single enol, since the two carbons are equivalent. It is important to recognize that an enol is a real substance, capable of indepen￾dent existence. An enol is not a resonance form of a carbonyl compound; the two are constitutional isomers of each other. 18.5 STABILIZED ENOLS Certain structural features can make the keto–enol equilibrium more favorable by stabi￾lizing the enol form. Enolization of 2,4-cyclohexadienone is one such example: The enol is phenol, and the stabilization gained by forming an aromatic ring is more than enough to overcome the normal preference for the keto form. A 1,3 arrangement of two carbonyl groups (compounds called -diketones) leads to a situation in which the keto and enol forms are of comparable stability. The two most important structural features that stabilize the enol of a -dicarbonyl com￾pound are (1) conjugation of its double bond with the remaining carbonyl group and (2) the presence of a strong intramolecular hydrogen bond between the enolic hydroxyl group and the carbonyl oxygen (Figure 18.2). In -diketones it is the methylene group flanked by the two carbonyls that is involved in enolization. The alternative enol 4-Hydroxy-4-penten-2-one CH2 CCH2CCH3 OH O 2,4-Pentanedione (20%) (keto form) CH3CCH2CCH3 O O 4-Hydroxy-3-penten-2-one (80%) (enol form) CH3C CHCCH3 OH O K 4 K is too large to measure. O 2,4-Cyclohexadienone (keto form, not aromatic) OH Phenol (enol form, aromatic) 2,4-Dimethyl-3-pentanone (keto form) (CH3)2CHCCH(CH3)2 O 2,4-Dimethyl-2-penten-3-ol (enol form) (CH3)2C CCH(CH3)2 OH 18.5 Stabilized Enols 707 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website