11_BRCLoudon_pgs4-4.qxd 11/26/08 8:56 AM Page 488 488 CHAPTER 11 • THE CHEMISTRY OF ETHERS, EPOXIDES, GLYCOLS, AND SULFIDES 11.2 SYNTHESIS OF EPOXIDES A. Oxidation of Alkenes with Peroxycarboxylic Acids One of the best laboratory preparations of epoxides involves the direct oxidation of alkenes with peroxycarboxylic acids. O O S S ) C O ) C O O H 25 °C OH CH (CH ) CHA CH % % CH (CH ) CH CH % % 3 2 5 2 LL benzene 3 2 5 2 + i L + i 1-octene 2-hexyloxirane (81% yield) "Cl "Cl meta-chloroperoxybenzoic meta-chlorobenzoic acid (mCPBA) acid a peroxycarboxylic acid (11.15) The use of alkenes as starting materials for epoxide synthesis is one reason that certain epox- ides are named traditionally as oxidation products of the corresponding alkenes (Sec. 8.1C). The oxidizing agent in Eq. 11.15, meta-chloroperoxybenzoic acid (abbreviated mCPBA), is an example of a peroxycarboxylic acid, which is a carboxylic acid that contains an O O H(hydroperoxy)groupinsteadofan OH (hydroxy) group. L L L L O O hydroperoxy group S S RC OH or RCO2H R C O OH or RCO3H LLa carboxylic acid LLa peroxycarboxylicL acid (The terms peroxyacid or peracid are sometimes used instead of peroxycarboxylic acid. These are actually more general terms that refer not only to peroxycarboxylic acids, but also to any acid containing an O O H group instead of an OH group.) Many peroxycar- boxylic acids are unstable,L but theyL canL be formed just prior toL use by mixing a carboxylic acid and hydrogen peroxide. In principle, any one of several peroxycarboxylic acids can be used for the epoxidation of alkenes. The peroxyacid used in Eq. 11.15, mCPBA, has been popular because it is a crystalline solid that can be shipped commercially and stored in the laboratory. However, mCPBA, like most other peroxides, will detonate if it is not handled carefully. A less hazardous peroxycarboxylic acid that has essentially the same reactivity is the magnesium salt of monoperoxyphthalic acid, abbreviated MMPP. O S C % % O O H % % Mg2 O LL | i _ SC O 2 magnesium monoperoxyphthalate (MMPP) 11_BRCLoudon_pgs4-4.qxd 11/26/08 8:56 AM Page 489 11.2 SYNTHESIS OF EPOXIDES 489 To understand why epoxidation occurs so readily, we’ll take the same approach that we used in Chapter 5 to understand other electrophilic additions. That is, let’s think of this reac- tion initially as a stepwise electrophilic addition. First, in an electron-pair displacement reac- tion, the p electrons of the double bond act as a nucleophile towards the terminal oxygen of the peroxycarboxylic acid. A carboxylate ion, which is a relatively weak base, serves as the leaving group. electrophile leaving group O .. H .. H .. .. .. O O.. C R´ O O .. .. RCH CHR RCH CHR + O.. C R´ (11.16a) a carbocation a carboxylate ion p electrons act as (a weak base) the nucleophile The OH group is positioned to react as a nucleophile with the carbocation. This reaction is a LewisL acid–base association, much like the formation of a bromonium ion in bromine addi- tion (Eq. 5.11, p. 182). This process closes the three-membered ring and forms the conjugate acid of the epoxide. nucleophile .. H .. H O .. O RCH CHR RCH CHR (11.16b) conjugate acid of electrophile an epoxide Like the conjugate acids of other ethers (Sec. 8.7), the conjugate acid of the epoxide has a neg- ative pKa value. This very acidic proton is removed in a Brønsted acid–base reaction by the carboxylate ion to form the epoxide and a carboxylic acid. You can convince yourself using the method described in Sec. 3.4E that the equilibrium for this final step is highly favorable. O .. .. O.. C R´ H .. .. .. O O O .. RCH CHR RCH CHR + H O.. C R´ (11.16c) pKa < –2 a carboxylic acid pKa ≈ 4–5 Although this stepwise mechanism is useful in analyzing why the reaction occurs, the ac- tual mechanism is concerted. That is, it occurs in a single step, and there is no carbocation in- termediate. We can represent this concerted process as follows: 11_BRCLoudon_pgs4-4.qxd 11/26/08 8:56 AM Page 490 490 CHAPTER 11 • THE CHEMISTRY OF ETHERS, EPOXIDES, GLYCOLS, AND SULFIDES O R´ C # % O R´ H C H "O % %% .. .. S O % % O O carboxylic acid RCH CHR RCH CHR (11.17) epoxide We know this process is concerted because carbocation rearrangements do not occur. In addi- tion, the reaction is a stereospecific syn-addition. That is, the reaction takes place with com- plete retention of alkene stereochemistry. (Recall from Eq. 7.50, p. 311, that a stepwise process would not be stereospecific.) This means that a cis alkene gives a cis-substituted epox- ide and a trans alkene gives a trans-substituted epoxide. Ph H O % % PhCO H %C A C % 3 C C (11.18a) benzene; 25 C Ph ) ) H ° L H Ph H Ph trans-stilbene trans-stilbene oxide (55% yield) Ph Ph O % % PhCO H %C A C % 3 C C (11.18b) benzene; 25 C Ph ) ) Ph ° L H H H H cis-stilbene cis-stilbene oxide (52% yield) The reaction is a syn-addition because, in an anti-addition, the epoxide oxygen would have to bridge opposite faces of the two alkene carbons simultaneously. The double bond in the tran- sition state for anti-addition would thus be significantly twisted and the transition state would be highly strained. As we shall see, the stereospecificity of this reaction is one reason why epoxide formation is a highly valuable synthetic reaction. Because epoxides contain three-membered rings, they, like cyclopropanes (Sec. 7.5B), have significant angle strain. As we’ll see in Sec. 11.4, this strain imparts valuable reactivity to epoxides. It is possible to form such strained compounds so easily because the O O bond of a peroxycarboxylic acid is very weak. In other words, it is the high energy of theL peroxy- carboxylic acid that drives epoxide formation. PROBLEMS 11.11 Give the structure of the alkene that would react with mCPBA to give each of the following epoxides. (a) % (b) H3C O (c) O (d) O O % % %C CH C C C C ` ) ) 2 Ph ) ) CH Ph ) ) H L L 3 L H3C H H H CH3 11.12 Give the product expected when each of the following alkenes is treated with MMPP. (a) trans-3-hexene (b) A CH2 11_BRCLoudon_pgs4-4.qxd 11/26/08 8:56 AM Page 491 11.2 SYNTHESIS OF EPOXIDES 491 B. Cyclization of Halohydrins Epoxides can also be synthesized by the treatment of halohydrins (Sec. 5.2B) with base: OH O 60 °C (11.19) (CH3)2"C CH2 Br Na| OH_ (CH3)2C) CH) 2 Na| Br_ H OH L L + L ++L 1-bromo-2-methyl-2-propanol 2,2-dimethyloxirane (a halohydrin) (81% yield) This reaction is an intramolecular variation of the Williamson ether synthesis (Sec. 11.1A); in this case, the alcohol and the alkyl halide are part of the same molecule. The alkoxide anion, formed reversibly by reaction of the alcohol with NaOH, displaces halide ion from the neigh- boring carbon: 1 1 _ OH H OH 3 1 L 1 1 1 1 1 1 O H O _ O 3 L 3 3 33 (CH ) C (CH ) C (11.20) (CH3)2"C CH2 Br 3 2" CH2 Br 3 2 ) CH) 2 Br _ L L 1 3 L L 1 3 L + 1 33 Like bimolecular SN2 reactions, this reaction takes place by backside substitution of the nu- cleophile—in this case, the oxygen anion—at the halide-bearing carbon (Sec. 9.4C). Such a backside substitution requires that the nucleophilic oxygen and the leaving halide assume an anti relationship in the transition state of the reaction. In most noncyclic halohydrins, this re- lationship can generally be achieved through a simple internal rotation. O O O _ H internal _ H rotation 332 % Br 332 H 33 %C C $C C C %% C Br H C 213 H3C H3C H _ 3 L L L 3321 CH H CH $Br CH H (11.21a) 3 3 3 1 3 3 O_ O O 2 _ 33 corresponding H 33 Br internal H 332 H Newman 213 rotation projections H H Br _ H3C CH3 H3C CH3 H3C CH3 3321 H Br 331 C—O and C—Br bonds C—O and C—Br bonds are gauche; are anti; backside substitution backside substitution is not possible is possible (11.21b) Halohydrins derived from cyclic compounds must be able to assume the required anti rela- tionship through a conformational change if epoxide formation is to succeed. The following cyclohexane derivative, for example, must undergo the chair interconversion before epoxide formation can occur. 11_BRCLoudon_pgs4-4.qxd 11/26/08 8:56 AM Page 492 492 CHAPTER 11 • THE CHEMISTRY OF ETHERS, EPOXIDES, GLYCOLS, AND SULFIDES O _ 2 Br 332 O % (11.22) 2 O1 _ 12 3 33 3 3 Br _ 332 % 1 3 + diequatorial "Br conformation diaxial conformation Even though the diaxial conformation of the halohydrin is less stable than the diequatorial conformation, the two conformations are in rapid equilibrium. As the diaxial conformation re- acts to give epoxide, it is replenished by the rapidly established conformational equilibrium. PROBLEMS 11.13 From models of the transition states for their reactions, predict which of the following two diastereomers of 3-bromo-2-butanol should form an epoxide at the greater rate when treated with base, and explain your reasoning.