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22.3 A-HALOGENATION of CARBONYL COMPOUNDS 1057 22_BRCLoudon_pgs4-4.qxd 11/26/08 12:27 PM Page 1057 22.3 a-HALOGENATION OF CARBONYL COMPOUNDS 1057 enol introduces hydrogen from solvent at the a-carbon; this fact accounts for the observed isotope exchange. This carbon of an enol, like that of an enolate ion, is not asymmetric. The absence of chirality in the enol accounts for the racemization observed in acid. PROBLEM 22.11 Give a curved-arrow mechanism for (a) the racemization shown in Eq. 22.19; (b) the deu- terium exchange shown in Eq. 22.18. 22.3 a-HALOGENATION OF CARBONYL COMPOUNDS A. Acid-Catalyzed a-Halogenation This section begins a survey of reactions that involve enols and enolate ions as reactive intermediates. Halogenation of an aldehyde or ketone in acidic solution usually results in the replacement of one a-hydrogen by halogen. O O S S 25 °C Br2 Br C CH3 HOAc BrC CH2 Br HBr (22.20) + LLL LLLL + p-bromoacetophenone 1-(4-bromophenyl)-2-bromoethanone (69–72% yield) Cl ) H3O| A O Cl2 A O HCl (22.21) + + cyclohexanone 2-chlorocyclohexanone (61–66% yield) Enols are reactive intermediates in these reactions. O HO S H3O C "CH | $C A C $ (22.22a) L L ) aldehyde" or enol ) ketone Like other “alkenes,” enols react with halogens; but unlike ordinary alkenes, enols add only one halogen atom. After addition of the first halogen to the double bond, the resulting carbocation intermediate loses a proton instead of adding the second halogen. (Addition of the second halo- gen would form a tetrahedral addition intermediate which, in this case, is relatively unstable.) 1 1 1 1 1 OH O H |O H 1 O 3 3 L SL S 3 "C A C $ "CC" CC" Br C "C HBr _ 1 L 1 L | LL L L LL L 3 1 3 LLL+ 1 3 ) "Br "Br "Br Br Br 3 1 3 3 1 3 3 1 3 (22.22b) 3 1 L 1 3 22_BRCLoudon_pgs4-4.qxd 11/26/08 12:27 PM Page 1058 1058 CHAPTER 22 • THE CHEMISTRY OF ENOLATE IONS, ENOLS, AND a,b-UNSATURATED CARBONYL COMPOUNDS Acid-catalyzed halogenation provides a particularly instructive case study that shows the importance of the rate law in determining the mechanism of a reaction. Under the usual reac- tion conditions, the rate law for acid-catalyzed halogenation is rate k[ketone][H3O|] (22.23) = This rate law implies that, even though the reaction is a halogenation, the rate is independent of the halogen concentration. Thus, halogens cannot be involved in the transition state for the rate-limiting step of the reaction (Sec. 9.3B). From this observation and others, it was deduced that enol formation (Eq. 22.22a) is the rate-limiting process in the acid-catalyzed halogenation of aldehydes and ketones. Because the halogen is not involved in enol formation, it does not ap- pear in the rate law at the concentrations of halogen ordinarily used. O OH O S S H3O Br | A 2 (22.24) C" CH "C C $ (fast) C"C Br HBr L L L LLL + " rate-limiting ) " process Enol formation is described in this equation as the rate-limiting process. This process consists of two elementary steps, as shown in Eq. 22.17b. The rate-limiting step of acid-catalyzed enolization is the second step, removal of the a-proton. The same step, therefore, is also the rate-limiting step of a- halogenation. Because only one halogen is introduced at a given a-carbon in acidic solution, it follows that introduction of a second halogen is much slower than introduction of the first. The slower halogenation is probably a consequence of the stability of the carbocation intermediate that is formed by reaction of the halogen with the halogenated enol. This carbocation is destabilized by the electron-attracting polar effect of two halogens: 1 1 Br Br 1 HO 3 1 L 1 3 HO Br 1 3 3 A | (22.25) $C C $ $C "C Br Br _ LL 3 1 3 Br ) ) carbocation) " intermediate halogenated enol is destabilized by the polar effect of two bromines If the rate-limiting transition state resembles this carbocation, then the transition state should have very high energy and the rate should be small. PROBLEMS 22.12 (a) Sketch a reaction-free energy diagram for acid-catalyzed enol formation using the mech- anism in Eq. 22.17b as your guide. Assume that the second step, proton removal from the a-carbon, is rate-limiting. (b) Incorporating the results of part (a), sketch a reaction-free energy diagram for the acid- catalyzed halogenation of an aldehyde or ketone. 22.13 Explain why: (a) the rate of iodination of optically active 1-phenyl-2-methyl-1-butanone in acetic acid/HNO3 is identical to its rate of racemization under the same conditions. (b) the rates of bromination and iodination of acetophenone are identical at a given acid con- centration. 22_BRCLoudon_pgs4-4.qxd 11/26/08 12:27 PM Page 1059 22.3 a-HALOGENATION OF CARBONYL COMPOUNDS 1059 B. Halogenation of Aldehydes and Ketones in Base: The Haloform Reaction Halogenation of aldehydes and ketones with a-hydrogens also occurs in base. In this reaction, all a-hydrogens are substituted by halogen. a-hydrogens O O S S NaOH, 0 °C 3NaOH (CH ) CCCH 3Br (CH ) CCCBr 3Na Br 3H O (22.26a) 3 3 3 2 H O/dioxane 3 3 3 | _ 2 + L L + 2 L L + + no a-hydrogens When the aldehyde or ketone starting material is either acetaldehyde or a methyl ketone (as in Eq. 22.26a), the product of halogenation is a trihalo carbonyl compound, which is not stable under the reaction conditions. This compound reacts further to give, after acidification of the re- action mixture, a carboxylic acid and a haloform. (Recall from Sec. 8.1A that a haloform is a trihalomethane—that is, a compound of the form HCX3, where X halogen.) = O O S S _OH H3O| (CH3)3CCCBr3 (CH3)3CCOH HCBr3 (22.26b) L L (71–74%L yield)L + bromoform The conversion of acetaldehyde or a methyl ketone into a carboxylic acid and a haloform by halogen in base, followed by acidification, as exemplified by Eq. 22.26a–b, is called the halo- form reaction. Notice that, in a haloform reaction, a carbon–carbon bond is broken. The mechanism of the haloform reaction involves the formation of an enolate ion as a re- active intermediate. O O S S _ (22.27a) RCCH3 OH_ RCCH2 H2O L L + LenolateL ion2 + The enolate ion reacts as a nucleophile with halogen to give an a-halo carbonyl compound. 1 1 1 O 1 O S S _ (22.27b) RCCH2 Br Br RCCH2Br Br _ L L 2 + 3 1 L 1 3 L L 1 3 + 3 1 3 However, halogenation does not stop here, because the enolate ion of the a-halo ketone is formed even more rapidly than the enolate ion of the starting ketone. The reason is that the polar effect of the halogen stabilizes the enolate ion and, by Hammond’s postulate, the transi- tion state for enolate-ion formation. Consequently, a second bromination occurs. OH OH O S S S Br Br RCC" Br RCC" Br L RC CHBr2 Br (22.27c) 1 _ LLL LLL LL + 1 _ "H 2 H2O _ OH + 3 3 1 The dihalo carbonyl compound brominates again even more rapidly. (Why?) O O S Br S RC CHBr 2 RC CBr Br H O (22.27d) 2 OH 3 _ 2 LL _ LL + + 22_BRCLoudon_pgs4-4.qxd 11/26/08 12:27 PM Page 1060 1060 CHAPTER 22 • THE CHEMISTRY OF ENOLATE IONS, ENOLS, AND a,b-UNSATURATED CARBONYL COMPOUNDS A carbon–carbon bond is broken when the trihalo carbonyl compound undergoes a nucle- ophilic acyl substitution reaction. O O _ O O S3 3 3 2 3 S3 3 S3 3 R C CBr3 R "C CBr3 R C OH _ CBr3 R C O _ H CBr3 LL LL L LL2 a+ trihalomethyl3 L L 2 3 + L OH OH _ " 2 anion 2 (22.27e) 3 2 3 2 The leaving2 group in this reaction is a trihalomethyl anion. Usually, carbanions are too basic to serve as leaving groups; but trihalomethyl anions are much less basic than ordinary carban- ions. (Why?) However, the basicity of trihalomethyl anions, although low enough for them to act as leaving groups, is high enough for them to react irreversibly with the carboxylic acid by- product, as shown in the last part of Eq. 22.27e. This acid–base reaction drives the overall haloform reaction to completion. (This situation is analogous to saponification, which is also driven to completion by ionization of the carboxylic acid product; Sec. 21.7A.) The carboxylic acid itself can be isolated by acidifying the reaction mixture, as shown in Eq. 22.26b. Occasionally, the haloform reaction can be used to prepare carboxylic acids from readily available methyl ketones. This reaction can also be used as a qualitative test for methyl ketones, called the iodoform test. In the iodoform test, a compound of unknown structure is mixed with alkaline I2. A yellow precipitate of iodoform (HCI3) is taken as evidence for a methyl ketone (or acetaldehyde, the “methyl aldehyde”). The iodoform test is specific for methyl ketones because only by replacement of three hydrogens with halogen does the carbon become a good enough leaving group for the nucleophilic acyl substitution reaction shown in Eq. 22.27e to occur. Alcohols of the form shown in Eq.
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