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11.4 Nucleophilic Substitution Reactions of Epoxides 495 11_BRCLoudon_pgs4-4.qxd 11/26/08 8:56 AM Page 495 11.4 NUCLEOPHILIC SUBSTITUTION REACTIONS OF EPOXIDES 495 (d) When tert-butyl methyl ether is heated with sulfuric acid, methanol and 2-methylpropene distill from the solution. (e) Tert-butyl methyl ether cleaves much faster in HBr than its sulfur analog, tert-butyl methyl sulfide. (Hint: See Sec. 8.7.) (f) When enantiomerically pure (S)-2-methoxybutane is treated with HBr, the products are enantiomerically pure (S)-2-butanol and methyl bromide. 11.16 What products are formed when each of the following ethers reacts with concentrated aque- ous HI? (a) diisopropyl ether (b) 2-ethoxy-2,3-dimethylbutane 11.4 NUCLEOPHILIC SUBSTITUTION REACTIONS OF EPOXIDES A. Ring-Opening Reactions under Basic Conditions Epoxides readily undergo reactions in which the epoxide ring is opened by nucleophiles. O OH Na C2H5O L (CH ) CCH C H OH | _ (CH ) C CH OC H (11.29) 3 2 ) ) 2 2 5 5 h, 80 °C 3 2 2 2 5 L + L L 2,2-dimethyloxirane ethanol 1-ethoxy-2-methyl-2-propanol (isobutylene oxide) (solvent) (83% yield) A reaction of this type is an SN2 reaction in which the epoxide oxygen serves as the leaving group. In this reaction, though, the leaving group does not depart as a separate entity, but rather remains within the same product molecule. leaving group an SN2 reaction 2 2 O O OH 2 _ L 3 3 332 L H OC2H5 3 2 (CH3)2C) CH) 2 (CH3)2C CH2 OC2H5 L 2 (CH3)2C CH2 OC2H5 _ OC2H5 L L .. L 2 L L 2 + 3 2 _OC2H5 2 2 (11.30) 2 nucleophile Because an epoxide is a type of ether, the ring opening of epoxides is an ether cleavage. Re- call that ordinary ethers do not undergo cleavage in base (Eq. 11.23, p. 492). Epoxides, how- ever, are opened readily by basic reagents. Epoxides are so reactive because they, like their carbon analogs, the cyclopropanes, possess significant angle strain (Sec. 7.5B). Because of this strain, the bonds of an epoxide are weaker than those of an ordinary ether, and are thus more easily broken. The opening of an epoxide relieves the strain of the three-membered ring just as the snapping of a twig relieves the strain of its bending. In an unsymmetrical epoxide, two ring-opening products could be formed corresponding to the reaction of the nucleophile at the two different carbons of the ring. As Eq. 11.30 illus- trates, nucleophiles typically react with unsymmetrical epoxides at the carbon with fewer alkyl substituents. This regioselectivity is expected from the effect of alkyl substitution on the 11_BRCLoudon_pgs4-4.qxd 11/26/08 8:56 AM Page 496 496 CHAPTER 11 • THE CHEMISTRY OF ETHERS, EPOXIDES, GLYCOLS, AND SULFIDES rates of SN2 reactions (Sec. 9.4D). Because alkyl substitution retards the SN2 reaction, the re- action of a nucleophile at the unsubstituted carbon is faster and leads to the observed product. Like other SN2 reactions, the ring opening of epoxides by bases involves backside substitu- tion of the nucleophile on the epoxide carbon. When this carbon is a stereocenter, inversion of configuration occurs, as illustrated by Study Problem 11.2. Study Problem 11.2 What is the stereochemistry of the 2,3-butanediol formed when meso-2,3-dimethyloxirane reacts with aqueous sodium hydroxide? Solution First draw the structure of the epoxide. The meso stereoisomer of 2,3-dimethyloxirane has an internal plane of symmetry, and its two asymmetric carbons have opposite configurations. S O R H C C) C) CH 3 L 3 H H meso-2,3-dimethyloxirane Because the two different carbons of the epoxide ring are enantiotopic (Sec. 10.8A), the hydroxide ion reacts at either one at the same rate. Backside substitution on each carbon should occur with inversion of configuration. 1 O OH O CH3 CH _ 3 3 % 3 3 332 % H H2O C %% C %C C %C C H OH H C CH H C H C _ 3 L 3 3 L 3 L + H H H OH H OH OH 3 1 3 1 _ inversion 3 12 of configuration The product shown is the 2S,3S stereoisomer. Reaction at the other carbon gives the 2R,3R stereoisomer. (Verify this point!) Because the starting materials are achiral, the two enantiomers of the product must be formed in equal amounts (Sec. 7.8A). Hence, the product of the reaction is racemic 2,3-butanediol. (This predicted result is in fact observed in the laboratory.) Although the examples in this section have involved hydroxide and alkoxides as nucle- ophiles, the pattern of reactivity is the same with any nucleophile: The nucleophile reacts at the carbon with no alkyl substituents and opens the epoxide to form an alkoxide, which then reacts in a Brønsted acid–base reaction with a proton source to give an alcohol. Letting Nuc _ be a general nucleophile, we can summarize this pattern of reactivity with Eq. 11.31: 3 nucleophilic proton substitution transfer .. .. .. .. .. .. .. O O OH L L .. .. H Nuc R2C) CH) 2 Nuc R2C CH2 Nuc R2C CH2 Nuc Nuc (11.31) L + L L L L + 11_BRCLoudon_pgs4-4.qxd 11/26/08 8:56 AM Page 497 11.4 NUCLEOPHILIC SUBSTITUTION REACTIONS OF EPOXIDES 497 PROBLEMS 11.17 Predict the products of the following reactions. (a) O C) CH) 2 NH3 H3C C2H5OH L +(excess) (CH3)2CH (b) O C CH Na N H C ) ) 2 + 3- C H OH/H O 3 L + 2 5 2 CH3CH2 sodium azide 11.18 From what epoxide and what nucleophile could each of the following compounds be pre- pared? (Assume each is racemic.) (a) OH (b) OH CH3(CH2)4"CHCH2CN SCH3 B. Ring-Opening Reactions under Acidic Conditions Ring-opening reactions of epoxides, like those of ordinary ethers, can be catalyzed by acids. However, epoxides are much more reactive than ethers under acidic conditions because of their angle strain. Hence, milder conditions can be used for the ring-opening reactions of epoxides than are required for the cleavage of ordinary ethers. For example, very low concen- trations of acid catalysts are required in ring-opening reactions of epoxides. O H2SO4 (trace) (CH3)2C) CH) 2 CH3OH (CH3)2C CH2 OH (11.32) L L LL 2,2-dimethyloxirane+ methanol (isobutylene oxide) (solvent) OCH3 2-methoxy-2-methyl-1-propanol (76% yield) The regioselectivity of the ring-opening reaction is different under acidic and basic condi- tions. The structure of the product in Eq. 11.32 shows that the nucleophile methanol reacts at the more substituted carbon of the epoxide. Contrast this with the result in Eq. 11.31, in which the nucleophile reacts at the less substituted carbon under basic conditions. In general, if one of the carbons of an unsymmetrical epoxide is tertiary, nucleophiles react at this carbon under acidic conditions. Some insight into why different regioselectivities are observed under different conditions comes from mechanistic considerations. The first step in the mechanism of Eq. 11.32, like the first step of ether cleavage, is protonation of the oxygen. protons come from protonated solvent molecule H H |"OCH3 O L 1 OH| HOCH3 (11.33a) 3 3 3 21 (CH3)2C) CH) 2 (CH3)2C) CH) 2 L protonatedL epoxide 11_BRCLoudon_pgs4-4.qxd 11/26/08 8:56 AM Page 498 498 CHAPTER 11 • THE CHEMISTRY OF ETHERS, EPOXIDES, GLYCOLS, AND SULFIDES The structural properties of the protonated epoxide show that it can be expected to behave like a tertiary carbocation. d+ long, weak bond .. OH nearly trigonal H3C C CH2 (11.33b) planar geometry H3C d+ a large amount of positive charge First, calculations show that the tertiary carbon bears about 0.7 of a positive charge. Second, the geometry at the tertiary carbon is nearly trigonal planar. This means that the tertiary car- bon and the groups around it are very nearly flattened into a common plane so that little or no steric hindrance prevents the approach of a nucleophile to this carbon. Finally, the bond be- tween the tertiary carbon and the OH group is unusually long and weak. This means that this bond is more easily broken thanL the other C O bond. In fact, this cation resembles a car- bocation solvated by the leaving group (see Fig.L 9.13, p. 419). The OH leaving group blocks the front side of the carbocation so that the nucleophilic reactionL must occur from the back side with inversion of stereochemistry. In other words, we can think of this reaction as an SN1 reaction with stereochemical inversion. Thus, a solvent molecule reacts with the protonated epoxide at the tertiary carbon, and loss of a proton to solvent gives the product. OH OH OH L 3 3 2 L 3 2 (CH3)2CCH) 2 (CH3)2C CH2 (CH3)2C CH2 L LLL L O CH3 O CH3 (11.33c) | L L 3 1 L HOCH3 3 H 1 2 H L | HOCH3 HOCH3 1 2 + 1 It is a solvent molecule, not the alkoxide conjugate base of the solvent, that reacts with the pro- tonated epoxide. The alkoxide conjugate base cannot exist in acidic solution; nor is it neces- sary, because the protonated epoxide is very reactive and because the nucleophile is also the solvent and is thus present in great excess. When the carbons of an unsymmetrical epoxide are secondary or primary, there is much less carbocation character at either carbon in the protonated epoxide, and acid-catalyzed ring- opening reactions tend to give mixtures of products; the exact compositions of the mixtures vary from case to case.
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