701 CHAPTER 18 ENOLS AND ENOLATES I n the preceding chapter you learned that nucleophilic addition to the carbonyl group is one of the fundamental reaction types of organic chemistry. In addition to its own reactivity, a carbonyl group can affect the chemical properties of aldehydes and ketones in other ways. Aldehydes and ketones are in equilibrium with their enol isomers. In this chapter you’ll see a number of processes in which the enol, rather than the alde- hyde or a ketone, is the reactive species. There is also an important group of reactions in which the carbonyl group acts as a powerful electron-withdrawing substituent, increasing the acidity of protons on the adjacent carbons. This proton is far more acidic than a hydrogen in an alkane. R 2 CCRH11032 H O Aldehyde or ketone R 2 CHCRH11032 O Enol R 2 CCRH11032 OH Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website As an electron-withdrawing group on a carbon–carbon double bond, a carbonyl group renders the double bond susceptible to nucleophilic attack: The presence of a carbonyl group in a molecule makes possible a number of chem- ical reactions that are of great synthetic and mechanistic importance. This chapter is com- plementary to the preceding one; the two chapters taken together demonstrate the extra- ordinary range of chemical reactions available to aldehydes and ketones. 18.1 THE H9251-CARBON ATOM AND ITS HYDROGENS It is convenient to use the Greek letters H9251, H9252, H9253, and so forth, to locate the carbons in a molecule in relation to the carbonyl group. The carbon atom adjacent to the carbonyl is the H9251-carbon atom, the next one down the chain is the H9252 carbon, and so on. Butanal, for example, has an H9251 carbon, a H9252 carbon, and a H9253 carbon. Hydrogens take the same Greek letter as the carbon atom to which they are attached. A hydrogen connected to the H9251-carbon atom is an H9251 hydrogen. Butanal has two H9251 protons, two H9252 protons, and three H9253 protons. No Greek letter is assigned to the hydro- gen attached directly to the carbonyl group of an aldehyde. PROBLEM 18.1 How many H9251 hydrogens are there in each of the following? (a) 3,3-Dimethyl-2-butanone (c) Benzyl methyl ketone (b) 2,2-Dimethylpropanal (d) Cyclohexanone SAMPLE SOLUTION (a) This ketone has two different H9251 carbons, but only one of them has hydrogen substituents. There are three equivalent H9251 hydrogens. The other nine hydrogens are attached to H9252-carbon atoms. Other than nucleophilic addition to the carbonyl group, the most important reac- tions of aldehydes and ketones involve substitution of an H9251 hydrogen. A particularly well studied example is halogenation of aldehydes and ketones. 3,3-Dimethyl-2-butanone CH 3 ±C±C±CH 3 H9251H9251 H9252 H9252 CH 3 H9252 CH 3 O XW W Carbonyl group is reference point; no Greek letter assigned to it. O CH 3 CH 2 CH 2 CH H9253H9252H9251 Normally, carbon–carbon double bonds are attacked by electrophiles; a carbon–carbon double bond that is conjugated to a carbonyl group is attacked by nucleophiles. O R 2 C CHCRH11032 702 CHAPTER EIGHTEEN Enols and Enolates Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 18.2 H9251 HALOGENATION OF ALDEHYDES AND KETONES Aldehydes and ketones react with halogens by substitution of one of the H9251 hydrogens: The reaction is regiospecific for substitution of an H9251 hydrogen. None of the hydrogens farther removed from the carbonyl group are affected. Nor is the hydrogen directly attached to the carbonyl group in aldehydes affected. Only the H9251 hydrogen is replaced. PROBLEM 18.2 Chlorination of 2-butanone yields two isomeric products, each having the molecular formula C 4 H 7 ClO. Identify these two compounds. H9251 Halogenation of aldehydes and ketones can be carried out in a variety of sol- vents (water and chloroform are shown in the examples, but acetic acid and diethyl ether are also often used). The reaction is catalyzed by acids. Since one of the reaction prod- ucts, the hydrogen halide, is an acid and therefore a catalyst for the reaction, the process is said to be autocatalytic. Free radicals are not involved, and the reactions occur at room temperature in the absence of initiators. Mechanistically, acid-catalyzed haloge- nation of aldehydes and ketones is much different from free-radical halogenation of alkanes. Although both processes lead to the replacement of a hydrogen by a halogen, they do so by completely different pathways. 18.3 MECHANISM OF H9251 HALOGENATION OF ALDEHYDES AND KETONES In one of the earliest mechanistic investigations in organic chemistry, Arthur Lapworth discovered in 1904 that the rates of chlorination and bromination of acetone were the same. Later he found that iodination of acetone proceeded at the same rate as chlorination O Cyclohexanone H11001 Cl 2 Chlorine H 2 O O Cl 2-Chlorocyclohexanone (61–66%) H11001 Hydrogen chloride HCl Aldehyde or ketone R 2 CCRH11032 H O R 2 CCRH11032 X O H9251-Halo aldehyde or ketone Halogen X 2 Hydrogen halide HXH11001H11001 H H11001 18.3 Mechanism of H9251 Halogenation of Aldehydes and Ketones 703 HBr Hydrogen bromide CH O H Cyclohexanecarbaldehyde H11001 Br 2 Bromine CH O Br 1-Bromocyclohexanecarbaldehyde (80%) CHCl 3 H11001 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website and bromination. Moreover, the rates of all three halogenation reactions, although first- order in acetone, are independent of the halogen concentration. Thus, the halogen does not participate in the reaction until after the rate-determining step. These kinetic obser- vations, coupled with the fact that substitution occurs exclusively at the H9251-carbon atom, led Lapworth to propose that the rate-determining step is the conversion of acetone to a more reactive form, its enol isomer: Once formed, this enol reacts rapidly with the halogen to form an H9251-halo ketone: PROBLEM 18.3 Write the structures of the enol forms of 2-butanone that react with chlorine to give 1-chloro-2-butanone and 3-chloro-2-butanone. Both parts of the Lapworth mechanism, enol formation and enol halogenation, are new to us. Let’s examine them in reverse order. We can understand enol halogenation by analogy to halogen addition to alkenes. An enol is a very reactive kind of alkene. Its carbon–carbon double bond bears an electron-releasing hydroxyl group, which activates it toward attack by electrophiles. The hydroxyl group stabilizes the carbocation by delocalization of one of the unshared electron pairs of oxygen: Participation by the oxygen lone pairs is responsible for the rapid attack on the carbon–carbon double bond of an enol by bromine. We can represent this participation explicitly: Less stable resonance form; 6 electrons on positively charged carbon. CH 3 CH 2 Br H11001 C O More stable resonance form; all atoms (except hydrogen) have octets of electrons. CH 3 CH 2 BrC H11001 O HH H11001H11001Br H11002 Bromide ion CH 3 CH 2 Br H11001 C OH Stabilized carbocation very fast Br Br Bromine CH 3 CCH 2 OH Propen-2-ol (enol form of acetone) H9251-Halo derivative of acetone CH 3 CCH 2 X O Halogen X 2 Hydrogen halide HXH11001H11001 Propen-2-ol (enol form of acetone) CH 3 CCH 2 OH fast Acetone CH 3 CCH 3 O Propen-2-ol (enol form of acetone) CH 3 CCH 2 OH slow 704 CHAPTER EIGHTEEN Enols and Enolates The graphic that opened this chapter is an electrostatic potential map of the enol of acetone. Lapworth was far ahead of his time in understanding how organic reactions occur. For an account of Lapworth’s contributions to mechanistic organic chemistry, see the November 1972 issue of the Journal of Chemical Educa- tion, pp. 750–752. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Writing the bromine addition step in this way emphasizes the increased nucleophilicity of the enol double bond and identifies the source of that increased nucleophilicity as the enolic oxygen. PROBLEM 18.4 Represent the reaction of chlorine with each of the enol forms of 2-butanone (see Problem 18.3) according to the curved arrow formalism just described. The cationic intermediate is simply the protonated form (conjugate acid) of the H9251-halo ketone. Deprotonation of the cationic intermediate gives the products. Having now seen how an enol, once formed, reacts with a halogen, let us consider the process of enolization itself. 18.4 ENOLIZATION AND ENOL CONTENT Enols are related to an aldehyde or a ketone by a proton-transfer equilibrium known as keto–enol tautomerism. (Tautomerism refers to an interconversion between two struc- tures that differ by the placement of an atom or a group.) The mechanism of enolization involves two separate proton-transfer steps rather than a one-step process in which a proton jumps from carbon to oxygen. It is relatively slow in neutral media. The rate of enolization is catalyzed by acids as shown by the mechanism in Figure 18.1. In aqueous acid, a hydronium ion transfers a proton to the carbonyl oxygen in step 1, and a water molecule acts as a Br?nsted base to remove a proton from the H9251-carbon atom in step 2. The second step is slower than the first. The first step involves proton transfer between oxygens, and the second is a proton transfer from carbon to oxygen. You have had earlier experience with enols in their role as intermediates in the hydration of alkynes (Section 9.12). The mechanism of enolization of aldehydes and ketones is precisely the reverse of the mechanism by which an enol is converted to a carbonyl compound. The amount of enol present at equilibrium, the enol content, is quite small for sim- ple aldehydes and ketones. The equilibrium constants for enolization, as shown by the following examples, are much less than 1. Keto form RCH 2 CRH11032 O Enol form RCH CRH11032 OH tautomerism H11001 Cationic intermediate Br H11002 O H11001 CH 3 CH 2 BrC H CH 3 CCH 2 Br O Bromoacetone H Br Hydrogen bromide H11001 Br H11002 CH 3 CH 2 BrC Br Br CH 3 CCH 2 OH H11001 OH 18.4 Enolization and Enol Content 705 The keto and enol forms are constitutional isomers. Using older terminology they are referred to as tautomers of each other. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website In these and numerous other simple cases, the keto form is more stable than the enol by some 45–60 kJ/mol (11–14 kcal/mol). The chief reason for this difference is that a car- bon–oxygen double bond is stronger than a carbon–carbon double bond. With unsymmetrical ketones, enolization may occur in either of two directions: The ketone is by far the most abundant species present at equilibrium. Both enols are also present, but in very small concentrations. 2-Butanone (keto form) CH 3 CCH 2 CH 3 O 2-Buten-2-ol (enol form) CH 3 CCHCH 3 OH 1-Buten-2-ol (enol form) CH 2 CCH 2 CH 3 OH Acetaldehyde (keto form) CH 3 CH O Vinyl alcohol (enol form) CH 2 CHOH K H11015 3 H11003 10 H110027 Acetone (keto form) CH 3 CCH 3 O Propen-2-ol (enol form) CH 2 CCH 3 OH K H11015 6 H11003 10 H110029 706 CHAPTER EIGHTEEN Enols and Enolates Overall reaction: Step 1: A proton is transferred from the acid catalyst to the carbonyl oxygen. RCH 2 CRH11032 Aldehyde or ketone Aldehyde or ketone Enol fast O X RCH 2 CRH11032 H11001 ORCH 2 CRH11032 H11001 H ± O H 3 O H11001 RCH ? CRH11032 OH W Enol RCH ? CRH11032 H11001 H ± O O ± H W H11001 H11001 Hydronium ion Conjugate acid of carbonyl compound Water Step 2: A water molecule acts as a Br?nsted base to remove a proton from the H9251 carbon atom of the protonated aldehyde or ketone. Hydronium ion H11001 O ± H X O X H H ± ± H H ± ± H H ± ± RCH ± CRH11032 H11001 O Conjugate acid of carbonyl compound Water H11001 O ± H X H H ± ± BNA BNA slow BNA W H FIGURE 18.1 Mechanism of acid-catalyzed enolization of an aldehyde or ketone in aqueous solution. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website PROBLEM 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 H9251-carbon atom. The ketone 2,4-dimethyl-3-pentanone gives a single enol, since the two H9251 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 H9252-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 H9252-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 H9252-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 CH 2 CCH 2 CCH 3 OH O 2,4-Pentanedione (20%) (keto form) CH 3 CCH 2 CCH 3 O O 4-Hydroxy-3-penten-2-one (80%) (enol form) CH 3 CCHCCH 3 OH O K H11005 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) (CH 3 ) 2 CHCCH(CH 3 ) 2 O 2,4-Dimethyl-2-penten-3-ol (enol form) (CH 3 ) 2 C CCH(CH 3 ) 2 OH 18.5 Stabilized Enols 707 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website does not have its carbon–carbon double bond conjugated with the carbonyl group, is not as stable, and is present in negligible amounts at equilibrium. PROBLEM 18.6 Write structural formulas corresponding to (a) The two most stable enol forms of (b) The two most stable enol forms of 1-phenyl-1,3-butanedione SAMPLE SOLUTION (a) Enolization of this 1,3-dicarbonyl compound can involve either of the two carbonyl groups: Both enols have their carbon–carbon double bonds conjugated to a carbonyl group and can form an intramolecular hydrogen bond. They are of comparable stability. 18.6 BASE-CATALYZED ENOLIZATION: ENOLATE ANIONS The proton-transfer equilibrium that interconverts a carbonyl compound and its enol can be catalyzed by bases as well as by acids. Figure 18.3 illustrates the roles of hydroxide ion and water in a base-catalyzed enolization. As in acid-catalyzed enolization, protons are transferred sequentially rather than in a single step. First (step 1), the base abstracts CH O H O CH 3 C C H CH O H O CH 3 C C H CH 3 CCH 2 CH O O CH 3 CCH 2 CH O X O X 708 CHAPTER EIGHTEEN Enols and Enolates (a) C CH 3 C O H C O CH 3 H O --- H separation in intramolecular hydrogen bond is 166 pm 124 pm 103 pm 133 pm 134 pm 141 pm (b) FIGURE 18.2 (a) A molecular model and (b) bond distances in the enol form of 2,4-pentanedione. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website a proton from the H9251-carbon atom to yield an anion. This anion is a resonance-stabilized species. Its negative charge is shared by the H9251-carbon atom and the carbonyl oxygen. Protonation of this anion can occur either at the H9251 carbon or at oxygen. Protonation of the H9251 carbon simply returns the anion to the starting aldehyde or ketone. Protonation of oxygen, as shown in step 2 of Figure 18.3, produces the enol. The key intermediate in this process, the conjugate base of the carbonyl compound, is referred to as an enolate ion, since it is the conjugate base of an enol. The term “eno- late” is more descriptive of the electron distribution in this intermediate in that oxygen bears a greater share of the negative charge than does the H9251-carbon atom. The slow step in base-catalyzed enolization is formation of the enolate ion. The second step, proton transfer from water to the enolate oxygen, is very fast, as are almost all proton transfers from one oxygen atom to another. CRH11032RCH O H11002 RCH CRH11032 O H11002 Electron delocalization in conjugate base of ketone 18.6 Base-Catalyzed Enolization: Enolate Anions 709 H11002 Overall reaction: Step 1: A proton is abstracted by hydroxide ion from the H9251 carbon atom of the carbonyl compound. RCH 2 CRH11032 Aldehyde or ketone Aldehyde or ketone Enol slow O X RCH ± CRH11032 H11001 ORCH ± CRH11032 H11001 O HO H11002 RCH ? CRH11032 OH W Enol RCH ? CRH11032 H11001 O O ± H W H11002 H11002 H11002 Hydroxide ion Conjugate base of carbonyl compound Water Step 2: A water molecule acts as a Br?nsted acid to transfer a proton to the oxygen of the enolate ion. Hydroxide ion O X O X H ± H H ± ± H ± RCH ? CRH11032 H11001 O Conjugate base of carbonyl compound Water O W H H ± ± BNA BNA fast BNA W H FIGURE 18.3 Mechanism of the base-catalyzed enoliza- tion of an aldehyde or ketone in aqueous solution. Examine the enolate of ace- tone on Learning By Model- ing. How is the negative charge distributed between oxygen and the H9251 carbon? Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Our experience to this point has been that C±H bonds are not very acidic. Com- pared with most hydrocarbons, however, aldehydes and ketones have relatively acidic protons on their H9251-carbon atoms. Equilibrium constants for enolate formation from sim- ple aldehydes and ketones are in the 10 H1100216 to 10 H1100220 range (pK a H11005 16–20). Delocalization of the negative charge onto the electronegative oxygen is responsi- ble for the enhanced acidity of aldehydes and ketones. With K a ’s in the 10 H1100216 to 10 H1100220 range, aldehydes and ketones are about as acidic as water and alcohols. Thus, hydrox- ide ion and alkoxide ions are sufficiently strong bases to produce solutions containing significant concentrations of enolate ions at equilibrium. H9252-Diketones, such as 2,4-pentanedione, are even more acidic: In the presence of bases such as hydroxide, methoxide, and ethoxide, these H9252-diketones are converted completely to their enolate ions. Notice that it is the methylene group flanked by the two carbonyl groups that is deprotonated. Both carbonyl groups partici- pate in stabilizing the enolate by delocalizing its negative charge. PROBLEM 18.7 Write the structure of the enolate ion derived from each of the following H9252-dicarbonyl compounds. Give the three most stable resonance forms of each enolate. (a) 2-Methyl-1,3-cyclopentanedione (b) 1-Phenyl-1,3-butanedione (c) SAMPLE SOLUTION (a) First identify the proton that is removed by the base. It is on the carbon between the two carbonyl groups. CH O O H 3 C C C H C CH 3 O O H11002 H 3 C C H11002 C H C CH 3 O O H 3 C C O C H C CH 3 H11002 O CHCCH 3 O CH 3 C O H11002 H H11001 K a H11005 10 H110029 (pK a H11005 9) H11001CH 3 CCH 2 CCH 3 OO CH(CH 3 ) 2 C O H11002 H H11001 K a H11005 3 H11003 10 H1100216 (pK a H11005 15.5) H11001(CH 3 ) 2 CHCH O 2-Methylpropanal CH 2 C 6 H 5 C O H11002 H H11001 K a H11005 1.6 H11003 10 H1100216 (pK a H11005 15.8) H11001C 6 H 5 CCH 3 O Acetophenone 710 CHAPTER EIGHTEEN Enols and Enolates Learning By Modeling contains molecular models of the enolates of acetone and 2,4- pentanedione. Compare the two with respect to the distribution of negative charge. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website The three most stable resonance forms of this anion are Enolate ions of H9252-dicarbonyl compounds are useful intermediates in organic syn- thesis. We shall see some examples of how they are employed in this way later in the chapter. 18.7 THE HALOFORM REACTION Rapid halogenation of the H9251-carbon atom takes place when an enolate ion is generated in the presence of chlorine, bromine, or iodine. As in the acid-catalyzed halogenation of aldehydes and ketones, the reaction rate is inde- pendent of the concentration of the halogen; chlorination, bromination, and iodination all occur at the same rate. Formation of the enolate is rate-determining, and, once formed, the enolate ion reacts rapidly with the halogen. Unlike its acid-catalyzed counterpart, H9251 halogenation in base cannot normally be limited to monohalogenation. Methyl ketones, for example, undergo a novel polyhalo- genation and cleavage on treatment with a halogen in aqueous base. This is called the haloform reaction because the trihalomethane produced is chloroform, bromoform, or iodoform, depending, of course, on the halogen used. The mechanism of the haloform reaction begins with H9251 halogenation via the eno- late. The electron-attracting effect of an H9251 halogen increases the acidity of the protons on the carbon to which it is bonded, making each subsequent halogenation at that car- bon faster than the preceding one. Methyl ketone RCCH 3 O Carboxylate ion RCO H11002 O Halogen 3X 2 Trihalomethane CHX 3 Water 3H 2 O Halide ion 3X H11002 Hydroxide ion 4HO H11002 H11001H11001 H11001 H11001H11001 Aldehyde or ketone R 2 CHCRH11032 O H9251-Halo aldehyde or ketone R 2 CCRH11032 O X Enolate R 2 C CRH11032 O H11002 HO H11002 , slow X 2 , fast OO H11002 O CH 3 H11002 H11002 O CH 3 O CH 3 O H OH O O CH 3 H11002 H11001OH H11002 O CH 3 H O 18.7 The Haloform Reaction 711 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website The trihalomethyl ketone so formed then undergoes nucleophilic addition of hydroxide ion to its carbonyl group, triggering its dissociation. The three electron-withdrawing halogen substituents stabilize the negative charge of the trihalomethide ion ( H11002 :CX 3 ), permitting it to act as a leaving group in the carbon–carbon bond cleavage step. The haloform reaction is sometimes used for the preparation of carboxylic acids from methyl ketones. The methyl ketone shown in the example can enolize in only one direction and typifies the kind of reactant that can be converted to a carboxylic acid in synthetically accept- able yield by the haloform reaction. When C-3 of a methyl ketone bears enolizable hydro- gens, as in , the first halogenation step is not very regioselective and the isolated yield of CH 3 CH 2 CO 2 H is only about 50%. The haloform reaction, using iodine, was once used as an analytical test in which the formation of a yellow precipitate of iodoform was taken as evidence that a substance was a methyl ketone. This application has been superseded by spectroscopic methods of structure determination. Interest in the haloform reaction has returned with the realiza- tion that chloroform and bromoform occur naturally and are biosynthesized by an anal- ogous process. (See the boxed essay “The Haloform Reaction and the Biosynthesis of Trihalomethanes.”) CH 3 CH 2 CCH 3 O X 3,3-Dimethyl-2-butanone (CH 3 ) 3 CCCH 3 O 2,2-Dimethylpropanoic acid (71–74%) (CH 3 ) 3 CCOH O Tribromomethane (bromoform) CHBr 3 H11001 1. Br 2 , NaOH, H 2 O 2. H H11001 O RCCX 3 Trihalomethyl ketone HO H11002 O RC H11002 CX 3 OH H11001 O RC OH HO H11002 H11002 O RC O Carboxylate ion H 2 O CX 3 H11002 Trihalomethane HCX 3 (RCCX 3 ) O X RCCH 3 O RCCH 2 X O RCCHX 2 O RCCX 3 O (slowest halogenation step) X 2 , HO H11002 X 2 , HO H11002 X 2 , HO H11002 (fastest halogenation step) 712 CHAPTER EIGHTEEN Enols and Enolates Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 18.8 SOME CHEMICAL AND STEREOCHEMICAL CONSEQUENCES OF ENOLIZATION A number of novel reactions involving the H9251-carbon atom of aldehydes and ketones involve enol and enolate anion intermediates. Substitution of deuterium for hydrogen at the H9251-carbon atom of an aldehyde or a ketone is a convenient way to introduce an isotopic label into a molecule and is readily carried out by treating the carbonyl compound with deuterium oxide (D 2 O) and base. 18.8 Some Chemical and Stereochemical Consequences of Enolization 713 THE HALOFORM REACTION AND THE BIOSYNTHESIS OF TRIHALOMETHANES U ntil scientists started looking specifically for them, it was widely believed that naturally oc- curring organohalogen compounds were rare. We now know that more than 2000 such compounds occur naturally, with the oceans being a particularly rich source. * Over 50 organohalogen compounds, including CHBr 3 , CHBrClI, BrCH 2 CH 2 I, CH 2 I 2 , Br 2 CHCH?O, I 2 CHCO 2 H, and (Cl 3 C) 2 C?O, have been found in a single species of Hawaiian red seaweed, for example. It is not surprising that organisms living in the oceans have adapted to their halide-rich environment by incorporating chlorine, bromine, and iodine into their metabolic processes. Chloromethane (CH 3 Cl), bro- momethane (CH 3 Br), and iodomethane (CH 3 l) are all produced by marine algae and kelp, but land-based plants and fungi also contribute their share to the more than 5 million tons of the methyl halides formed each year by living systems. The ice plant, which grows in arid regions throughout the world and is cultivated as a ground cover along coastal highways in California, biosynthesizes CH 3 Cl by a process in which nucleophilic attack by chloride ion (Cl H11002 ) on the methyl group of S- adenosylmethionine is the key step (Section 16.17). Interestingly, the trihalomethanes chloroform (CHCl 3 ), bromoform (CHBr 3 ), and iodoform (CHl 3 ) are biosynthesized by an entirely different process, one that is equivalent to the haloform reaction (Section 18.7) and begins with the formation of an H9251-halo ke- tone. Unlike the biosynthesis of methyl halides, which requires attack by a halide nucleophile (X H11002 ), H9251 halogenation of a ketone requires attack by an elec- trophilic form of the halogen. For chlorination, the electrophilic form of the halogen is generated by oxidation of Cl H11002 in the presence of the enzyme chloroperoxidase. Thus, the overall equation for the enzyme-catalyzed chlorination of a methyl ketone may be written as Further chlorination of the chloromethyl ketone gives the corresponding trichloromethyl ketone, which then undergoes hydrolysis to form chloroform. Purification of drinking water, by adding Cl 2 to kill bacteria, is a source of electrophilic chlorine and contributes a nonenzymatic pathway for H9251 chlorina- tion and subsequent chloroform formation. Al- though some of the odor associated with tap water may be due to chloroform, more of it probably results from chlorination of algae-produced organic com- pounds. Chloromethyl ketone ClCH 2 CR O X chloro- peroxidase Cl H11002 , O 2 chloro- peroxidase Cl H11002 , O 2 Carboxylate RCO 2 H11002 Chloroform Cl 3 CH H11001 Dichloromethyl ketone Cl 2 CHCR O X Trichloromethyl ketone Cl 3 CCR O X H 2 O HO H11002 Chloromethyl ketone ClCH 2 CR O X Hydroxide HO H11002 H11001 Methyl ketone CH 3 CR O X Chloride Cl H11002 H11001H11001 Oxygen O 2 1 2 chloroperoxidase * The November 1994 edition of the Journal of Chemical Education contains as its cover story the article “Natural Organohalogens. Many More Than You Think!” Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Only the H9251 hydrogens are replaced by deuterium in this reaction. The key intermediate is the enolate ion formed by proton abstraction from the H9251-carbon atom of cyclopen- tanone. Transfer of deuterium from the solvent D 2 O to the enolate gives cyclopentanone containing a deuterium atom in place of one of the hydrogens at the H9251 carbon. Formation of the enolate Deuterium transfer to the enolate In excess D 2 O the process continues until all four H9251 protons are eventually replaced by deuterium. PROBLEM 18.8 After the compound shown was heated in D 2 O containing K 2 CO 3 at 70°C the only signals that could be found in its 1 H NMR spectrum were at H9254 3.9 ppm (6H) and H9254 6.7–6.9 ppm (3H). What happened? If the H9251-carbon atom of an aldehyde or a ketone is a stereogenic center, its stereo- chemical integrity is lost on enolization. Enolization of optically active sec-butyl phenyl ketone leads to its racemization by way of the achiral enol form. Each act of proton abstraction from the H9251-carbon atom converts a chiral molecule to an achiral enol or enolate anion. Careful kinetic studies have established that the rate of loss CH 3 O CH 2 CCH 3 CH 3 O O H11001H11001 H H H O H11002 Enolate of cyclopentanone O D H H H Cyclopentanone-2-d 1 H11002 ODOD D H11001 H11001 HOD H H H O H11002 Enolate of cyclopentanone O H H H H Cyclopentanone H11002 OD 4D 2 O 4DOH O Cyclopentanone H11001 KOD reflux O D D D D Cyclopentanone-2,2,5,5-d 4 H11001 714 CHAPTER EIGHTEEN Enols and Enolates HO H11002 , H 2 O, or H 3 O H11001 HO H11002 , H 2 O, or H 3 O H11001 C 6 H 5 C O C H CH 2 CH 3 CH 3 (R)-sec-Butyl phenyl ketone CC 6 H 5 C HO CH 3 CH 2 CH 3 Enol form [achiral; may be converted to either (R)- or (S)- sec-butyl phenyl ketone] C 6 H 5 C O C H CH 2 CH 3 CH 3 (S)-sec-Butyl phenyl ketone Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website of optical activity of sec-butyl phenyl ketone is equal to its rate of hydrogen–deuterium exchange, its rate of bromination, and its rate of iodination. In each case, the rate- determining step is conversion of the starting ketone to the enol or enolate anion. PROBLEM 18.9 Is the product from the H9251 chlorination of (R)-sec-butyl phenyl ketone with Cl 2 in acetic acid chiral? Is it optically active? 18.9 THE ALDOL CONDENSATION As noted earlier, an aldehyde is partially converted to its enolate anion by bases such as hydroxide ion and alkoxide ions. In a solution that contains both an aldehyde and its enolate ion, the enolate undergoes nucleophilic addition to the carbonyl group. This addition is analogous to the addition reactions of other nucleophilic reagents to aldehydes and ketones described in Chapter 17. The alkoxide formed in the nucleophilic addition step then abstracts a proton from the solvent (usually water or ethanol) to yield the product of aldol addition. This product is known as an aldol because it contains both an aldehyde function and a hydroxyl group (ald H11001 ol H11005 aldol). An important feature of aldol addition is that carbon–carbon bond formation occurs between the H9251-carbon atom of one aldehyde and the carbonyl group of another. This is because carbanion (enolate) generation can involve proton abstraction only from the H9251-carbon atom. The overall transformation can be represented schematically, as shown in Figure 18.4. H 2 O RCH O H RCH 2 CH H11002 C O RCH 2 CH CHCH R O H11002 O RCH 2 CH CHCH R OH O Product of aldol addition Aldehyde RCH 2 CH O Hydroxide HO H11002 Water H 2 OH11001H11001 Enolate RCH CH O H11002 18.9 The Aldol Condensation 715 RCH 2 CH H11001 CH 2 CH base RCH 2 CH±CHCH One of these protons is removed by base to form an enolate This is the carbon–carbon bond that is formed in the reaction Carbonyl group to which enolate adds O X O X OH W O X W R W R FIGURE 18.4 The reactive sites in aldol addition are the carbonyl group of one aldehyde molecule and the H9251-carbon atom of another. Some of the earliest studies of the aldol reaction were carried out by Aleksander Borodin. Though a physician by training and a chemist by profession, Borodin is re- membered as the composer of some of the most familiar works in Russian music. See pp. 326–327 in the April 1987 issue of the Journal of Chem- ical Education for a biographical sketch of Borodin. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Aldol addition occurs readily with aldehydes: PROBLEM 18.10 Write the structure of the aldol addition product of (a) (c) (b) SAMPLE SOLUTION (a) A good way to correctly identify the aldol addition prod- uct of any aldehyde is to work through the process mechanistically. Remember that the first step is enolate formation and that this must involve proton abstrac- tion from the H9251 carbon. Now use the negatively charged H9251 carbon of the enolate to form a new car- bon–carbon bond to the carbonyl group. Proton transfer from the solvent com- pletes the process. H 2 O CH 3 CH 2 CH 2 CH 2 CH O Pentanal CHCH H11002 O CH 2 CH 2 CH 3 Enolate of pentanal CH 3 CH 2 CH 2 CH 2 CHCHCH O H11002 O CH 2 CH 2 CH 3 CH 3 CH 2 CH 2 CH 2 CHCHCH OH O CH 2 CH 2 CH 3 3-Hydroxy-2-propylheptanal (aldol addition product of pentanal) H11001 CH 3 CH 2 CH 2 CH 2 CH O Pentanal HO H11002 Hydroxide H11001 Enolate of pentanal CH 3 CH 2 CH 2 CHCH H11002 O CH 3 CH 2 CH 2 CH CH O H11002 2-Methylbutanal, CH 3 CH 2 CHCH CH 3 O X W 3-Methylbutanal, (CH 3 ) 2 CHCH 2 CH O X Pentanal, CH 3 CH 2 CH 2 CH 2 CH O X Acetaldehyde 2CH 3 CH O 3-Hydroxybutanal (50%) (acetaldol) CH 3 CHCH 2 CH O OH NaOH, H 2 O 4–5°C Butanal 2CH 3 CH 2 CH 2 CH O 2-Ethyl-3-hydroxyhexanal (75%) CH 3 CH 2 CH 2 CHCHCH O HO CH 2 CH 3 KOH, H 2 O 6–8°C 716 CHAPTER EIGHTEEN Enols and Enolates Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website The H9252-hydroxy aldehyde products of aldol addition undergo dehydration on heat- ing, to yield H9251,H9252-unsaturated aldehydes: Conjugation of the newly formed double bond with the carbonyl group stabilizes the H9251,H9252-unsaturated aldehyde, provides the driving force for the dehydration, and controls its regioselectivity. Dehydration can be effected by heating the aldol with acid or base. Normally, if the H9251,H9252-unsaturated aldehyde is the desired product, all that is done is to carry out the base-catalyzed aldol addition reaction at elevated temperature. Under these conditions, once the aldol addition product is formed, it rapidly loses water to form the H9251,H9252-unsaturated aldehyde. Reactions in which two molecules of an aldehyde combine to form an H9251,H9252- unsaturated aldehyde and a molecule of water are called aldol condensations. PROBLEM 18.11 Write the structure of the aldol condensation product of each of the aldehydes in Problem 18.10. One of these aldehydes can undergo aldol addition, but not aldol condensation. Which one? Why? SAMPLE SOLUTION (a) Dehydration of the product of aldol addition of pen- tanal introduces the double bond between C-2 and C-3 to give an H9251,H9252-unsaturated aldehyde. The point was made earlier (Section 5.9) that alcohols require acid catalysis in order to undergo dehydration to alkenes. Thus, it may seem strange that aldol addition products can be dehydrated in base. This is another example of the way in which the enhanced acidity of protons at the H9251-carbon atom affects the reactions of carbonyl com- pounds. Elimination may take place in a concerted E2 fashion or it may be stepwise and proceed through an enolate ion. Product of aldol condensation of pentanal (2-propyl-2- heptenal) CH 3 CH 2 CH 2 CH 2 CH CCH O CH 2 CH 2 CH 3 H11002H 2 O Product of aldol addition of pentanal (3-hydroxy-2- propylheptanal) CH 3 CH 2 CH 2 CH 2 CHCHCH OH O CH 2 CH 2 CH 3 heat H11001 H9252-Hydroxy aldehyde RCH 2 CHCHCH OOH R O R RCH 2 CH CCH H9251,H9252-Unsaturated aldehyde H 2 O Water 18.9 The Aldol Condensation 717 Recall from Section 15.7 that a condensation is a reaction in which two molecules com- bine to give a product along with some small (usually in- organic) molecule such as water. Butanal 2CH 3 CH 2 CH 2 CH O 2-Ethyl-3-hydroxyhexanal (not isolated; dehydrates under reaction conditions) CH 3 CH 2 CH 2 CHCHCH OOH CH 2 CH 3 via NaOH, H 2 O 80–100°C O CH 2 CH 3 CH 3 CH 2 CH 2 CH CCH 2-Ethyl-2-hexenal (86%) Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website As with other reversible nucleophilic addition reactions, the equilibria for aldol additions are less favorable for ketones than for aldehydes. For example, only 2% of the aldol addition product of acetone is present at equilibrium. The situation is similar for other ketones. Special procedures for aldol addition and self- condensation of ketones have been developed, but are rarely used. Aldol condensations of dicarbonyl compounds—even diketones—occur intra- molecularly when five- or six-membered rings are possible. Aldol condensations are one of the fundamental carbon–carbon bond-forming processes of synthetic organic chemistry. Furthermore, since the products of these aldol condensa- tions contain functional groups capable of subsequent modification, access to a host of useful materials is gained. To illustrate how aldol condensation may be coupled to functional group modifi- cation, consider the synthesis of 2-ethyl-1,3-hexanediol, a compound used as an insect repellent. This 1,3-diol is prepared by reduction of the aldol addition product of butanal: Butanal CH 3 CH 2 CH 2 CH O 2-Ethyl-3-hydroxyhexanal CH 3 CH 2 CH 2 CHCHCH OOH CH 2 CH 3 2-Ethyl-1,3-hexanediol CH 3 CH 2 CH 2 CHCHCH 2 OH OH CH 2 CH 3 aldol addition H 2 Ni O O 1,6-Cyclodecanedione O Bicyclo[5.3.0]dec- 1(7)-en-2-one (96%) O OH Not isolated; dehydrates under reaction conditions Na 2 CO 3 , H 2 O reflux Acetone 2CH 3 CCH 3 O 2% 98% 4-Hydroxy-4-methyl-2-pentanone CH 3 CCH 2 CCH 3 OOH CH 3 H11001 H9252-Hydroxy aldehyde RCH 2 CHCHCH OOH R HOHH11001 HO H11002 fast Enolate ion of H9252-hydroxy aldehyde OH RCH 2 CHC CH R O H11002 Enolate ion of H9252-hydroxy aldehyde OH RCH 2 CH CH O C R H11002 slow H11001 O R RCH 2 CH CCH H9251,H9252-Unsaturated aldehyde HO H11002 718 CHAPTER EIGHTEEN Enols and Enolates Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website PROBLEM 18.12 Outline a synthesis of 2-ethyl-1-hexanol from butanal. The carbon–carbon bond-forming potential of the aldol condensation has been extended beyond the self-condensations described in this section to cases in which two different carbonyl compounds react in what are called mixed aldol condensations. 18.10 MIXED ALDOL CONDENSATIONS Mixed aldol condensations can be effective only if we limit the number of reaction pos- sibilities. It would not be useful, for example, to treat a solution of acetaldehyde and propanal with base. A mixture of four aldol addition products forms under these condi- tions. Two of the products are those of self-addition: Two are the products of mixed addition: The mixed aldol condensations that are the most synthetically useful are those in which: 1. Only one of the reactants can form an enolate; or 2. One of the reactants is more reactive toward nucleophilic addition than the other. Formaldehyde, for example, cannot form an enolate but can react with the enolate of an aldehyde or ketone that can. Indeed, formaldehyde is so reactive toward nucleophilic addition that it suppresses the self-condensation of the other component by reacting rapidly with any enolate present. Aromatic aldehydes cannot form enolates, and a large number of mixed aldol con- densations have been carried out in which an aromatic aldehyde reacts with an enolate. Formaldehyde HCH O H11001 3-Methylbutanal (CH 3 ) 2 CHCH 2 CH O 2-Hydroxymethyl-3- methylbutanal (52%) (CH 3 ) 2 CHCHCH O CH 2 OH K 2 CO 3 water–ether 3-Hydroxypentanal (from addition of enolate of acetaldehyde to propanal) CH 3 CH 2 CHCH 2 CH O OH 3-Hydroxy-2-methylbutanal (from addition of enolate of propanal to acetaldehyde) CH 3 CHCHCH OOH CH 3 3-Hydroxybutanal (from addition of enolate of acetaldehyde to acetaldehyde) CH 3 CHCH 2 CH O OH 3-Hydroxy-2-methylpentanal (from addition of enolate of propanal to propanal) CH 3 CH 2 CHCHCH OHO CH 3 18.10 Mixed Aldol Condensations 719 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Recall that ketones do not readily undergo self-condensation. Thus, in the preceding example, the enolate of acetone reacts preferentially with the aromatic aldehyde and gives the mixed aldol condensation product in good yield. Mixed aldol condensations using aromatic aldehydes always involve dehydration of the product of mixed addition and yield a product in which the double bond is conjugated to both the aromatic ring and the carbonyl group. PROBLEM 18.13 Give the structure of the mixed aldol condensation product of benzaldehyde with (a) (b) (c) Cyclohexanone SAMPLE SOLUTION (a) The enolate of acetophenone reacts with benzaldehyde to yield the product of mixed addition. Dehydration of the intermediate occurs, giving the H9251,H9252-unsaturated ketone. As actually carried out, the mixed aldol condensation product, 1,3-diphenyl-2- propen-1-one, has been isolated in 85% yield on treating benzaldehyde with ace- tophenone in an aqueous ethanol solution of sodium hydroxide at 15–30°C. 18.11 EFFECTS OF CONJUGATION IN H9251,H9252-UNSATURATED ALDEHYDES AND KETONES Aldol condensation offers an effective route to H9251,H9252-unsaturated aldehydes and ketones. These compounds have some interesting properties that result from conjugation of the carbon–carbon double bond with the carbonyl group. As shown in Figure 18.5, the H9266 systems of the carbon–carbon and carbon–oxygen double bonds overlap to form an extended H9266 system that permits increased electron delocalization. This electron delocalization stabilizes a conjugated system. Under conditions cho- sen to bring about their interconversion, the equilibrium between a H9252,H9253-unsaturated ketone and an H9251,H9252-unsaturated analog favors the conjugated isomer. H11002H 2 O C 6 H 5 CH O Benzaldehyde H11001 CH 2 CC 6 H 5 H11002 O Enolate of acetophenone C 6 H 5 CHCH 2 CC 6 H 5 O OH 1,3-Diphenyl-2-propen-1-one O C 6 H 5 CH CHCC 6 H 5 tert-Butyl methyl ketone, (CH 3 ) 3 CCCH 3 O X Acetophenone, C 6 H 5 CCH 3 O X NaOH, H 2 O 30°C CH 3 OCH O p-Methoxybenzaldehyde H11001 CH 3 CCH 3 O Acetone CHCCH 3 O CH 3 OCH 4-p-Methoxyphenyl-3- buten-2-one (83%) 720 CHAPTER EIGHTEEN Enols and Enolates Mixed aldol condensations in which a ketone reacts with an aromatic aldehyde are known as Claisen–Schmidt condensations. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website PROBLEM 18.14 Commercial mesityl oxide, , is often contam- inated with about 10% of an isomer having the same carbon skeleton. What is a likely structure for this compound? In resonance terms, electron delocalization in H9251,H9252-unsaturated carbonyl compounds is represented by contributions from three principal resonance structures: The carbonyl group withdraws H9266 electron density from the double bond, and both the carbonyl carbon and the H9252 carbon are positively polarized. Their greater degree of charge separation makes the dipole moments of H9251,H9252-unsaturated carbonyl compounds signifi- cantly larger than those of comparable aldehydes and ketones. The diminished H9266 electron density in the double bond makes H9251,H9252-unsaturated alde- hydes and ketones less reactive than alkenes toward electrophilic addition. Electrophilic reagents—bromine and peroxy acids, for example—react more slowly with the car- bon–carbon double bond of H9251,H9252-unsaturated carbonyl compounds than with simple alkenes. On the other hand, the polarization of electron density in H9251,H9252-unsaturated carbonyl compounds makes their H9252-carbon atoms rather electrophilic. Some chemical conse- quences of this enhanced electrophilicity are described in the following section. O H9254H11002 HH9254H11001 Butanal H9262 H11005 2.7 D H O H9254H11002 H9254H11001 H9254H11001 trans-2-Butenal H9262 H11005 3.7 D C C C O H9252 H9251 Most stable structure C C C H11001 O H11002 H9252 H9251 H11001 C C C O H11002 H9252 H9251 (CH 3 ) 2 C?CHCCH 3 O X CHCH 2 CCH 3 O CH 3 CH 4-Hexen-2-one (17%) (H9252,H9253-unsaturated ketone) CHCCH 3 O CH 3 CH 2 CH 3-Hexen-2-one (83%) (H9251,H9252-unsaturated ketone) K H11005 4.8 25°C 18.11 Effects of Conjugation in H9251,H9252-Unsaturated Aldehydes and Ketones 721 FIGURE 18.5 Acrolein (H 2 C?CHCH?O) is a pla- nar molecule. Oxygen and each carbon are sp 2 - hybridized, and each con- tributes one electron to a conjugated H9266 electron sys- tem analogous to that of 1,3- butadiene. Figure 3.17 (page 107) shows how the composition of an equilibrium mixture of two components varies according to the free-energy difference between them. For the equi- librium shown in the accom- panying equation, H9004G° H11005H110024 kJ/mol (H110021 kcal/mol). Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 18.12 CONJUGATE ADDITION TO H9251,H9252-UNSATURATED CARBONYL COMPOUNDS H9251,H9252-Unsaturated carbonyl compounds contain two electrophilic sites: the carbonyl car- bon and the carbon atom that is H9252 to it. Nucleophiles such as organolithium and Grig- nard reagents and lithium aluminum hydride tend to react by nucleophilic addition to the carbonyl group, as shown in the following example: This is called direct addition, or 1,2 addition. (The “1” and “2” do not refer to IUPAC locants but are used in a manner analogous to that employed in Section 10.10 to distin- guish between direct and conjugate addition to conjugated dienes.) With certain other nucleophiles, addition takes place at the carbon–carbon double bond rather than at the carbonyl group. Such reactions proceed via enol intermediates and are described as conjugate addition, or 1,4-addition, reactions. The nucleophilic portion of the reagent (Y in HY) becomes bonded to the H9252 carbon. For reactions carried out under conditions in which the attacking species is the anion , an enolate ion precedes the enol. Ordinarily, nucleophilic addition to the carbon–carbon double bond of an alkene is very rare. It occurs with H9251,H9252-unsaturated carbonyl compounds because the carbanion that results is an enolate, which is more stable than a simple alkyl anion. Conjugate addition is most often observed when the nucleophile is weakly basic. The nucleophiles in the two examples that follow are and , respectively. Both are much weaker bases than acetylide ion, which was the nucleophile used in the example illustrating direct addition. C 6 H 5 CH 2 S H11002 CPN H11002 (Y H11002 ) C O C C H11002 Y Enolate ion formed by nucleophilic addition of : Y H11002 to H9252 carbon H11002 YC O C C H11002 Y O CC C Y H11002 YC O HC C Isolated product of 1,4-addition pathway fastHY Y HO CC C Enol formed by 1,4 addition C O 1 2 34 C C H9251,H9252-Unsaturated aldehyde or ketone CHCH O CH 3 CH 2-Butenal CMgBrHC Ethynylmagnesium bromide H11001 1. THF 2. H 3 O H11001 4-Hexen-1-yn-3-ol (84%) CHCHC OH CH 3 CH CH 722 CHAPTER EIGHTEEN Enols and Enolates Hydrogen cyanide and al- kanethiols have K a values in the 10 H110029 –10 H1100210 range (pK a H11005 9–10), and K a for acetylene is 10 H1100226 (pK a H11005 26). Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website One explanation for these observations is presented in Figure 18.6. Nucleophilic addition to H9251,H9252-unsaturated aldehydes and ketones may be governed either by kinetic control or by thermodynamic control (Section 10.10). 1,2 Addition is faster than 1,4 addi- tion and, under conditions in which the 1,2- and 1,4-addition products do not equilibrate, is the predominant pathway. Kinetic control operates with strongly basic nucleophiles to give the 1,2-addition product. A weakly basic nucleophile, however, goes on and off the carbonyl carbon readily and permits the 1,2-addition product to equilibrate with the more slowly formed, but more stable, 1,4-addition product. Thermodynamic control is observed with weakly basic nucleophiles. The product of 1,4 addition, which retains the carbon–oxygen double bond, is more stable than the product of 1,2 addition, which retains the carbon–carbon double bond. In general, carbon–oxygen double bonds are more stable than carbon–carbon double bonds because the greater electronegativity of oxygen permits the H9266 electrons to be bound more strongly. PROBLEM 18.15 Acrolein (CH 2 ?CHCH?O) reacts with sodium azide (NaN 3 ) in aqueous acetic acid to form a compound, C 3 H 5 N 3 O in 71% yield. Propanal (CH 3 CH 2 CH?O), when subjected to the same reaction conditions, is recovered unchanged. Suggest a structure for the product formed from acrolein, and offer an explanation for the difference in reactivity between acrolein and propanal. CHCC 6 H 5 O C 6 H 5 CH 1,3-Diphenyl-2-propen-1-one 4-Oxo-2,4-diphenylbutanenitrile (93–96%) C 6 H 5 CHCH 2 CC 6 H 5 O CN KCN ethanol– acetic acid C 6 H 5 CH 2 SH HO H11002 , H 2 O O CH 3 3-Methyl-2-cyclohexenone CH 3 SCH 2 C 6 H 5 O 3-Benzylthio-3-methylcyclohexanone (58%) 18.12 Conjugate Addition to H9251,H9252-Unsaturated Carbonyl Compounds 723 CC C O H11001 H Y fast 1,2-addition fast keto–enol isomerism CC C YHO HO Less stable product slow 1,4-addition CC C Y CC C O Y H More stable product FIGURE 18.6 Nucleophilic addition to H9251,H9252-unsaturated aldehydes and ketones may take place either in a 1,2- or 1,4 manner. Direct addition (1,2) occurs faster than con- jugate addition (1,4) but gives a less stable product. The product of 1,4 addition retains the carbon–oxygen double bond, which is, in general, stronger than a car- bon–carbon double bond. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 18.13 ADDITION OF CARBANIONS TO H9251,H9252-UNSATURATED KETONES: THE MICHAEL REACTION A synthetically useful reaction known as the Michael reaction, or Michael addition, involves nucleophilic addition of carbanions to H9251,H9252-unsaturated ketones. The most common types of carbanions used are enolate ions derived from H9252-diketones. These enolates are weak bases (Section 18.6) and react with H9251,H9252-unsaturated ketones by conjugate addition. The product of Michael addition has the necessary functionality to undergo an intramolecular aldol condensation: The synthesis of cyclohexenone derivatives by Michael addition followed by intramolec- ular aldol condensation is called the Robinson annulation, after Sir Robert Robinson, who popularized its use. By annulation we mean the building of a ring onto some start- ing molecule. (The alternative spelling “annelation” is also often used.) PROBLEM 18.16 Both the conjugate addition step and the intramolecular aldol condensation step can be carried out in one synthetic operation without isolat- ing any of the intermediates along the way. For example, consider the reaction Write structural formulas corresponding to the intermediates formed in the con- jugate addition step and in the aldol addition step. 18.14 CONJUGATE ADDITION OF ORGANOCOPPER REAGENTS TO H9251,H9252-UNSATURATED CARBONYL COMPOUNDS The preparation and some synthetic applications of lithium dialkylcuprates were described earlier (Section 14.11). The most prominent feature of these reagents is their capacity to undergo conjugate addition to H9251,H9252-unsaturated aldehydes and ketones. H11001C 6 H 5 CH 2 CCH 2 C 6 H 5 O Dibenzyl ketone CH 2 O CHCCH 3 Methyl vinyl ketone NaOCH 3 CH 3 OH C 6 H 5 CH 3 O C 6 H 5 3-Methyl-2,6-diphenyl-2- cyclohexen-1-one (55%) KOH methanol CH 3 O O 2-Methyl-1,3- cyclohexanedione H11001 CH 2 O CHCCH 3 Methyl vinyl ketone O CH 3 O CH 2 CH 2 CCH 3 O 2-Methyl-2-(3H11032-oxobutyl)- 1,3-cyclohexanedione (85%) 724 CHAPTER EIGHTEEN Enols and Enolates Arthur Michael, for whom the reaction is named, was an American chemist whose career spanned the period between the 1870s and the 1930s. He was independently wealthy and did much of his research in his own private laboratory. O CH 3 O CH 2 CH 2 CCH 3 O 2-Methyl-2-(3H11032-oxobutyl)- 1,3-cyclohexanedione NaOH heat H11002H 2 O CH 3 OH O O Intramolecular aldol addition product; not isolated CH 3 O O H9004 4 -9-Methyloctalin-3,8-dione Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website PROBLEM 18.17 Outline two ways in which 4-methyl-2-octanone can be pre- pared by conjugate addition of an organocuprate to an H9251,H9252-unsaturated ketone. SAMPLE SOLUTION Mentally disconnect one of the bonds to the H9252 carbon so as to identify the group that comes from the lithium dialkylcuprate. According to this disconnection, the butyl group is derived from lithium dibutyl- cuprate. A suitable preparation is Now see if you can identify the second possibility. Like other carbon–carbon bond-forming reactions, organocuprate addition to enones is a powerful tool in organic synthesis. 18.15 ALKYLATION OF ENOLATE ANIONS Since enolate anions are sources of nucleophilic carbon, one potential use in organic syn- thesis is their reaction with alkyl halides to give H9251-alkyl derivatives of aldehydes and ketones: O R 2 CHCRH11032 H9001ldehyde or ketone CRH11032 O H11002 R 2 C Enolate anion base RH11033X S N 2 H9251-Alkyl derivative of an aldehyde or a ketone O CRH11032 RH11033 R 2 C CH 3 CH 2 CH 2 CH 2 CHCH 2 CCH 3 O CH 3 4-Methyl-2-octanone H11001 3-Penten-2-one CH 3 CH O CHCCH 3 Lithium dibutylcuprate LiCu(CH 2 CH 2 CH 2 CH 3 ) 2 1. diethyl ether 2. H 2 O Disconnect this bond CH 3 CH 2 CH 2 CH 2 O CH 3 CHCH 2 CCH 3 4-Methyl-2-octanone CH 3 CH 2 CH 2 CH 2 H11002 H11001 CH 3 CH O CHCCH 3 CHCRH11032 O R 2 C H9251,H9252-Unsaturated aldehyde or ketone LiCuRH11033 2 Lithium dialkylcuprate H11001 Aldehyde or ketone alkylated at the H9252 position R 2 CCH 2 CRH11032 O RH11033 1. diethyl ether 2. H 2 O LiCu(CH 3 ) 2 Lithium dimethylcuprate H11001 1. diethyl ether 2. H 2 O O CH 3 3-Methyl-2- cyclohexenone 3,3-Dimethylcyclohexanone (98%) O CH 3 CH 3 18.15 Alkylation of Enolate Anions 725 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Alkylation occurs by an S N 2 mechanism in which the enolate ion acts as a nucleophile toward the alkyl halide. In practice, this reaction is difficult to carry out with simple aldehydes and ketones because aldol condensation competes with alkylation. Furthermore, it is not always pos- sible to limit the reaction to the introduction of a single alkyl group. The most success- ful alkylation procedures use H9252-diketones as starting materials. Because they are rela- tively acidic, H9252-diketones can be converted quantitatively to their enolate ions by weak bases and do not self-condense. Ideally, the alkyl halide should be a methyl or primary alkyl halide. 18.16 SUMMARY Section 18.1 Greek letters are commonly used to identify various carbons in alde- hydes and ketones. Using the carbonyl group as a reference, the adjacent carbon is designated H9251, the next one H9252, and so on as one moves down the chain. Attached groups take the same Greek letter as the carbon to which they are connected. Sections Because aldehydes and ketones exist in equilibrium with their corre- 18.2–18.15 sponding enol isomers, they can express a variety of different kinds of chemical reactivity. Reactions that proceed via enol or enolate intermediates are summarized in Table 18.1. PROBLEMS 18.18 (a) Write structural formulas or build molecular models for all the noncyclic aldehydes and ketones of molecular formula C 4 H 6 O. (b) Are any of these compounds stereoisomeric? (c) Are any of these compounds chiral? (d) Which of these are H9251,H9252-unsaturated aldehydes or H9251,H9252-unsaturated ketones? (e) Which of these can be prepared by a simple (i.e., not mixed) aldol condensation? 18.19 The main flavor component of the hazelnut is (2E,5S)-5-methyl-2-hepten-4-one. Write a structural formula or build a molecular model showing its stereochemistry. CRH11032 OH R 2 C Carbonyl group is electrophilic; nucleophilic reagents add to carbonyl carbon. H9251 carbon atom of enol is nucleophilic; it attacks electrophilic reagents. H9251 proton is relatively acidic; it can be removed by strong bases. O CRH11032 H R 2 C K 2 CO 3 2,4-Pentanedione CH 3 CCH 2 CCH 3 O O 3-Methyl-2,4-pentanedione (75–77%) CH 3 CH 3 CCHCCH 3 O O Iodomethane CH 3 IH11001 726 CHAPTER EIGHTEEN Enols and Enolates Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Problems 727 TABLE 18.1 Reactions of Aldehydes and Ketones That Involve Enol or Enolate Ion Intermediates Reaction (section) and comments H9251 Halogenation (Sections 18.2 and 18.3) Halogens react with aldehydes and ketones by substi- tution; an H9251 hydrogen is replaced by a halogen. Reaction occurs by electrophilic attack of the halo- gen on the carbon–carbon dou- ble bond of the enol form of the aldehyde or ketone. An acid catalyst increases the rate of enolization, which is the rate- determining step. Enolate ion formation (Section 18.6) An H9251 proton of an aldehyde or a ketone is more acidic than most other protons bound to car- bon. Aldehydes and ketones are weak acids, with K a ’s in the range 10 H1100216 to 10 H1100220 (pK a 16–20). Their enhanced acidity is due to the electron-withdrawing effect of the carbonyl group and the reso- nance stabilization of the enolate anion. Haloform reaction (Section 18.7) Methyl ketones are cleaved on reaction with excess halogen in the presence of base. The prod- ucts are a trihalomethane (halo- form) and a carboxylate salt. Enolization (Sections 18.4 through 18.6) Aldehydes and ketones exist in equilibrium with their enol forms. The rate at which equilibrium is achieved is increased by acidic or basic cata- lysts. The enol content of simple aldehydes and ketones is quite small; H9252-diketones, however, are extensively enolized. (Continued) General equation and typical example Aldehyde or ketone R 2 CH±CRH11032 O X R 2 C?CRH11032 W OH Enol K H11005 1 H11003 10 H110028 O Cyclopentanone OH Cyclopenten-1-ol K Aldehyde or ketone R 2 CHCRH11032 O X X 2 Halogen HX Hydrogen halide H11001H11001R 2 CCRH11032 O X W X H9251-Halo aldehyde or ketone H11001 Br 2 Bromine H11001 Hydrogen bromide HBr p-Bromophenacyl bromide (69–72%) CCH 2 BrBr O X acetic acid CCH 3 Br O X p-Bromoacetophenone Aldehyde or ketone R 2 CHCRH11032 O X HO H11002 Hydroxide ion H 2 O Water H11001H11001 Enolate anion R 2 C?CRH11032 W O H11002 3-Pentanone CH 3 CH 2 CCH 2 CH 3 O X HO H11002 Hydroxide ion H 2 O Water H11001H11001 Enolate anion of 3-pentanone CH 3 CH?CCH 2 CH 3 W O H11002 Methyl ketone RCCH 3 O X 3X 2 Halogen HCX 3 Trihalomethane (haloform) H11001H11001RCO H11002 O X Carboxylate ion HO H11002 4,4-Dimethyl-2-pentanone (CH 3 ) 3 CCH 2 CCH 3 O X Bromoform CHBr 3 H11001(CH 3 ) 3 CCH 2 CO 2 H 3,3-Dimethylbutanoic acid (89%) 1. Br 2 , NaOH 2. H H11001 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 728 CHAPTER EIGHTEEN Enols and Enolates TABLE 18.1 Reactions of Aldehydes and Ketones That Involve Enol or Enolate Ion Intermediates (Continued) Reaction (section) and comments Aldol condensation (Section 18.9) A reaction of great synthetic val- ue for carbon–carbon bond for- mation. Nucleophilic addition of an enolate ion to a carbonyl group, followed by dehydration of the H9252-hydroxy aldehyde, yields an H9251,H9252-unsaturated aldehyde. Conjugate addition to H9251,H9252-unsat- urated carbonyl compounds (Sec- tions 18.11 through 18.14) The H9252- carbon atom of an H9251,H9252-unsaturat- ed carbonyl compound is electro- philic; nucleophiles, especially weakly basic ones, yield the prod- ucts of conjugate addition to H9251,H9252- unsaturated aldehydes and ketones. Robinson annulation (Section 18.13) A combination of conju- gate addition of an enolate anion to an H9251,H9252-unsaturated ketone with subsequent intramolecular aldol condensation. Claisen-Schmidt reaction (Section 18.10) A mixed aldol condensa- tion in which an aromatic alde- hyde reacts with an enolizable aldehyde or ketone. (Continued) General equation and typical example Aldehyde 2RCH 2 CRH11032 O X H 2 O Water H11001RCH 2 C?CCRH11032 RH11032 O X W R W H9251,H9252-Unsaturated aldehyde HO H11002 Aromatic aldehyde ArCH O X Aldehyde or ketone RCH 2 CRH11032 O X H 2 O Water H11001H11001 ArCH?CCRH11032 O X R W H9251,H9252-Unsaturated carbonyl compound HO H11002 Benzaldehyde C 6 H 5 CH O X 3,3-Dimethyl-2- butanone (CH 3 ) 3 CCCH 3 O X H11001 C 6 H 5 CH?CHCC(CH 3 ) 3 O X 4,4-Dimethyl-1-phenyl- 1-penten-3-one (88–93%) NaOH ethanol– water Octanal CH 3 (CH 2 ) 6 CH O X CH 3 (CH 2 ) 6 CH?C(CH 2 ) 5 CH 3 HC?O W 2-Hexyl-2-decenal (79%) NaOCH 2 CH 3 CH 3 CH 2 OH H9251,H9252-Unsaturated aldehyde or ketone R 2 C?CHCRH11032 O X H11001 R 2 CCH 2 CRH11032 O X Y W Product of conjugate addition Nucleophile HY 4-Methyl-3-penten-2-one (mesityl oxide) (CH 3 ) 2 C?CHCCH 3 O X (CH 3 ) 2 CCH 2 CCH 3 O X NH 2 W 4-Amino-4-methyl-2- pentanone (63–70%) NH 3 H 2 O H11001 CH 3 O 2-Methylcyclohexanone CH 2 ?CHCCH 3 O X Methyl vinyl ketone 1. NaOCH 2 CH 3 , CH 3 CH 2 OH 2. KOH, heat 6-Methylbicyclo[4.4.0]- 1-decen-3-one (46%) CH 3 O Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 18.20 The simplest H9251,H9252-unsaturated aldehyde acrolein is prepared by heating glycerol with an acid catalyst. Suggest a mechanism for this reaction. 18.21 In each of the following pairs of compounds, choose the one that has the greater enol con- tent, and write the structure of its enol form: (a) (b) orC 6 H 5 CC 6 H 5 O C 6 H 5 CH 2 CCH 2 C 6 H 5 O or(CH 3 ) 3 CCH O (CH 3 ) 2 CHCH O HOCH 2 CHCH 2 OH OH KHSO 4 heat CH 2 CHCH O H11001 H 2 O Problems 729 TABLE 18.1 Reactions of Aldehydes and Ketones That Involve Enol or Enolate Ion Intermediates (Continued) Reaction (section) and comments Conjugate addition of organocop- per compounds (Section 18.14) The principal synthetic application of lithium dialkylcuprate reagents is their reaction with H9251,H9252- unsaturated carbonyl compounds. Alkyl-ation of the H9252 carbon occurs. H9251 Alkylation of aldehydes and ketones (Section 18.15) Alkylation of simple aldehydes and ketones via their enolates is difficult. H9252- Diketones can be converted quan- titatively to their enolate anions, which react efficiently with pri- mary alkyl halides. General equation and typical example H9251,H9252-Unsaturated aldehyde or ketone R 2 C?CHCRH11032 O X H11001 R 2 C±CH 2 CRH11032 O X RH11033 W H9252-Alkyl aldehyde or ketone Lithium dialkylcuprate RH11033 2 CuLi 1. diethyl ether 2. H 2 O H9252-Diketone RCCH 2 CR O X O X RCCHCR O X O X CH 2 RH11032 W H9251-Alkyl-H9252-diketone RH11032CH 2 X, HO H11002 1. LiCu(CH 3 ) 2 2. H 2 O O CH 3 6-Methylcyclohept- 2-enone O CH 3 CH 3 3,6-Dimethylcycloheptanone (85%) H11001 KOCH 2 CH 3 ethanol CH 2 C 6 H 5 O O H 2-Benzyl-1,3- cyclohexanedione CH 2 C 6 H 5 O O CH 2 C 6 H 5 2,2-Dibenzyl-1,3- cyclohexanedione (69%) C 6 H 5 CH 2 Cl Benzyl chloride Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (c) (d) (e) (f) 18.22 Give the structure of the expected organic product in the reaction of 3-phenylpropanal with each of the following: (a) Chlorine in acetic acid (b) Sodium hydroxide in ethanol, 10°C (c) Sodium hydroxide in ethanol, 70°C (d) Product of part (c) with lithium aluminum hydride; then H 2 O (e) Product of part (c) with sodium cyanide in acidic ethanol 18.23 Each of the following reactions has been reported in the chemical literature. Write the struc- ture of the product(s) formed in each case. (a) (b) (c) (d) (e) NaOH water O CH O H11001 CH 3 CCH 3 O KOH ethanol Cl CH O H11001 C 6 H 5 C 6 H 5 O Br 2 diethyl ether O C 6 H 5 C 6 H 5 C 6 H 5 CH 2 SH NaOH, H 2 O C(CH 3 ) 2 O CH 3 Cl 2 CH 2 Cl 2 CCH 2 CH 3 Cl O O O or O O O or O O Oor orC 6 H 5 CCH 2 CC 6 H 5 O O C 6 H 5 CH 2 CCH 2 C 6 H 5 O 730 CHAPTER EIGHTEEN Enols and Enolates Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (f) (g) (h) 18.24 Show how each of the following compounds could be prepared from 3-pentanone. In most cases more than one synthetic transformation will be necessary. (a) 2-Bromo-3-pentanone (d) 3-Hexanone (b) 1-Penten-3-one (e) 2-Methyl-1-phenyl-1-penten-3-one (c) 1-Penten-3-ol 18.25 (a) A synthesis that begins with 3,3-dimethyl-2-butanone gives the epoxide shown. Suggest reagents appropriate for each step in the synthesis. (b) The yield for each step as actually carried out in the laboratory is given above each arrow. What is the overall yield for the three-step sequence? 18.26 Using benzene, acetic anhydride, and 1-propanethiol as the source of all the carbon atoms, along with any necessary inorganic reagents, outline a synthesis of the compound shown. 18.27 Show how you could prepare each of the following compounds from cyclopentanone, D 2 O, and any necessary organic or inorganic reagents. (a) (c) (b) (d) D D D D H D D D D D H D D D D D H OH CCH 2 SCH 2 CH 2 CH 3 O 58% 54% 68% (CH 3 ) 3 CCCH 3 O (CH 3 ) 3 CCCH 2 Br O (CH 3 ) 3 CCHCH 2 Br OH (CH 3 ) 3 CC O CH 2 H KOH O O H11001 CH 2 CHCH 2 Br NaOH ethanol–water O H11001 C 6 H 5 CH O 1. diethyl ether 2. H 2 O CH 3 CH 3 H 3 C O H11001 LiCu(CH 3 ) 2 Problems 731 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 18.28 (a) At present, butanal is prepared industrially by hydroformylation of propene (Section 17.4). Write a chemical equation for this industrial synthesis. (b) Before about 1970, the principal industrial preparation of butanal was from acetalde- hyde. Outline a practical synthesis of butanal from acetaldehyde. 18.29 Identify the reagents appropriate for each step in the following syntheses: (a) (b) (c) (d) 18.30 Give the structure of the product derived by intramolecular aldol condensation of the keto aldehyde shown: 18.31 Prepare each of the following compounds from the starting materials given and any neces- sary organic or inorganic reagents: (a) (b) CH 3 C 6 H 5 CH CCH 2 OH from benzyl alcohol and 1-propanol (CH 3 ) 2 CHCHCCH 2 OH from (CH 3 ) 2 CHCH 2 OH CH 3 HO CH 3 KOH, H 2 O C 7 H 10 OCH 3 CCH 2 CCHO CH 3 OCH 3 (CH 3 ) 2 C O CHCH 2 CH 2 CCH 3 (CH O 3 ) 2 CHCHCH 2 CH 2 CCH 3 OH (CH 3 ) 2 CHCCH 2 CH 2 CCH 3 O O (CH 3 ) 2 CH O CH(CH 3 ) 2 CH 3 CH(CH 3 ) 2 CH 3 CCH 2 CH 2 CHCH 2 CH O O O (CH 3 ) 2 CH CCH 3 CH O HCCH 2 CH 2 CH 2 CH 2 CH O O OH OH CH O Br CH O CH O 732 CHAPTER EIGHTEEN Enols and Enolates Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (c) 18.32 Terreic acid is a naturally occurring antibiotic substance. Its actual structure is an enol iso- mer of the structure shown. Write the two most stable enol forms of terreic acid, and choose which of those two is more stable. 18.33 In each of the following, the indicated observations were made before any of the starting material was transformed to aldol addition or condensation products: (a) In aqueous acid, only 17% of (C 6 H 5 ) 2 CHCH?O is present as the aldehyde; 2% of the enol is present. Some other species accounts for 81% of the material. What is it? (b) In aqueous base, 97% of (C 6 H 5 ) 2 CHCH?O is present as a species different from any of those in part (a). What is this species? 18.34 (a) For a long time attempts to prepare compound A were thwarted by its ready isomer- ization to compound B. The isomerization is efficiently catalyzed by traces of base. Write a reasonable mechanism for this isomerization. (b) Another attempt to prepare compound A by hydrolysis of its diethyl acetal gave only the 1,4-dioxane derivative C. How was compound C formed? 18.35 Consider the ketones piperitone, menthone, and isomenthone. O CH 3 CH(CH 3 ) 2 (H11002)-Piperitone O CH 3 CH(CH 3 ) 2 Menthone Isomenthone O CH 3 CH(CH 3 ) 2 C 6 H 5 CHCH(OCH 2 CH 3 ) 2 OH H 2 O H H11001 C 6 H 5 O O C6H5 OH HO Compound C HO H11002 H 2 O Compound A C 6 H 5 CHCH OH O Compound B C 6 H 5 CCH 2 OH O O H O O O CH 3 CC 6 H 5 O CH 3 from acetophenone, 4-methylbenzyl alcohol, and 1,3-butadiene Problems 733 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Suggest reasonable explanations for each of the following observations: (a) Optically active piperitone (H9251 D H1100232°) is converted to racemic piperitone on standing in a solution of sodium ethoxide in ethanol. (b) Menthone is converted to a mixture of menthone and isomenthone on treatment with 90% sulfuric acid. 18.36 Many nitrogen-containing compounds engage in a proton-transfer equilibrium that is anal- ogous to keto–enol tautomerism: HX±N?Z BA X?N±ZH Each of the following compounds is the less stable partner of such a tautomeric pair. Write the structure of the more stable partner for each one. (a) CH 3 CH 2 N?O (d) (b) (CH 3 ) 2 C?CHNHCH 3 (c) (e) 18.37 Outline reasonable mechanisms for each of the following reactions: (a) (b) (c) (d) (e) H11001 KOH ethanol C 6 H 5 C 6 H 5 C 6 H 5 C 6 H 5 O C 6 H 5 CCC 6 H 5 OO C 6 H 5 CH 2 CCH 2 C 6 H 5 O heat or base O H H CH 3 CH 3 O H H CH 3 CH 3 KOH H 2 O, CH 3 OH O CH 3 (40%) HCCH 2 CH 2 CHCCH 3 O O CH 3 H11001 HO H11002 heat CHCH 2 CH 2 C O CH 3 (CH 3 ) 2 C CHCH (96%) CHCH 2 CH 2 CCH 3 O (CH 3 ) 2 C O CH 3 CH KOC(CH 3 ) 3 benzene O CH 2 CH 2 CH 2 CH 2 Br O (76%) HN OH NH 2 CCH 3 CH O H11002 H11001 OH N NH NH N 734 CHAPTER EIGHTEEN Enols and Enolates Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (f) 18.38 Suggest reasonable explanations for each of the following observations: (a) The C?O stretching frequency of H9251,H9252-unsaturated ketones (about 1675 cm H110021 ) is less than that of typical dialkyl ketones (1710–1750 cm H110021 ). (b) The C?O stretching frequency of cyclopropenone (1640 cm H110021 ) is lower than that of typical H9251,H9252-unsaturated ketones (1675 cm H110021 ). (c) The dipole moment of diphenylcyclopropenone (H9262 H11005 5.1 D) is substantially larger than that of benzophenone (H9262 H11005 3.0 D) (d) The H9252 carbon of an H9251,H9252-unsaturated ketone is less shielded than the corresponding car- bon of an alkene. Typical 13 C NMR chemical shift values are 18.39 Bromination of 3-methyl-2-butanone yielded two compounds, each having the molecular formula C 5 H 9 BrO, in a 95:5 ratio. The 1 H NMR spectrum of the major isomer A was character- ized by a doublet at H9254 1.2 ppm (6 protons), a septet at H9254 3.0 ppm (1 proton), and a singlet at H9254 4.1 ppm (2 protons). The 1 H NMR spectrum of the minor isomer B exhibited two singlets, one at H9254 1.9 ppm and the other at H9254 2.5 ppm. The lower field singlet had half the area of the higher field one. Suggest reasonable structures for these two compounds. 18.40 Treatment of 2-butanone (1 mol) with Br 2 (2 mol) in aqueous HBr gave C 4 H 6 Br 2 O. The 1 H NMR spectrum of the product was characterized by signals at H9254 1.9 ppm (doublet, 3 protons), 4.6 ppm (singlet, 2 protons), and 5.2 ppm (quartet, 1 proton). Identify this compound. 18.41 2-Phenylpropanedial [C 6 H 5 CH(CHO) 2 ] exists in the solid state as an enol in which the con- figuration of the double bond is E. In solution (CDCl 3 ), an enol form again predominates but this time the configuration is Z. Make molecular models of these two enols, and suggest an explana- tion for the predominance of the Z enol in solution. (Hint: Think about intermolecular versus intramolecular hydrogen bonding.) (H9254 H11015 129 ppm) CH 2 O CHCR (H9254 H11015 114 ppm) CH 2 CHCH 2 R H11001 NaOCH 3 CH 3 OH O C 6 H 5 CH 3 C 6 H 5 C 6 H 5 C 6 H 5 CH 2 CCH 2 CH 3 O CH 2 O C 6 H 5 CCC 6 H 5 Problems 735 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website