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Asymmetric Epoxidation Exemplified Facial Descriptors: Re and Si	K. Barry Sharpless Co-Nobelist 2001 Lecture: Video Text Asymmetric Epoxidation and Dihydroxylation

 

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Reading Assignment:

On the Epoxidation of Alkenes

Sharpless Asymmetric Dihydroxylation

Enrichment Assignment:

Professor Sharpless visits Chem 125

Diethyl ether (ether) may well be the first organic compound prepared that does not appear in Nature. For a chemical history of ether click here. A different Powerpoint version is here.
Theory of Etherification, A. W. Williamson, Quarterly J. Chem. Soc., 1852, 4, 106. [See page 106 in the .pdf file.]

"The following experiments were made with the view of obtaining new alcohols, by substituting carburetted hydrogen for hydrogen in a known alcohol. Iodide of potassium was readily formed on the application of a gentle heat, and the desired substitution was effected; but, contrary to expectation, the compound thus formed had none of the properties of an alcohol -- it was nothing else than common ether, C⁴H¹⁰O." On Etherification, A. W. Williamson, Quarterly J. Chem. Soc., 1852, 4, 229.

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1. Each of the following reactions is missing a reactant, reagent and/or product. Provide the missing items along with an explanation. Answer any questions.

a) The diol is optically active (S, S) and trans. Since OsO4 adds syn to a double bond, A cannot be cis-cyclooctene, which would provide the cis-diol. A is the enantiomer of trans-cyclooctene shown. Its enantiomer, A', would afford the R,R-enantiomer. The reaction of OsO4 with the double bond of trans-cyclooctene can only react with the external face because the internal face is hindered by the methylene chain. b) Isobutylene must be used twice and a carbon-carbon bond must be formed. The epoxide is the electrophile and the cuprate reagent is the nucleophile. Cuprates may be generated from Grignard reagents (or organolithium reagents). The problem with magnesium cations, and Grignard reactions with highly substituted epoxides, is that they can effect rearrangement of epoxides. In this instance isobutyraldehyde can form and act as an electrophile toward the magnesium reagents. Attack occurs in an SN2 fashion at the less substituted carbon of the epoxide to afford the diol. Acid catalyzed E1 dehydration gives the more substituted double bond. c) Direct epoxidation or halohydrin formation followed by base will give racemic epoxide. Moreover an direct mode of epoxidation will not provide the correct stereochemistry. There must be one inversion of stereochemistry. The epoxide can be formed from the tosylate via internal displacement by the alkoxide of the alcohol. The requisite diol arises by Sharpless asymmetric dihydroxylation. The optically active diol is formed from the achiral alkene with AD-mix-α. [Go here for a review.] d) cis-1,2-Cycloheptanediol is formed in situ. Cleavage to the dialdehyde occurs. Two equivalents of periodic acid are required. Four electron oxidation. e) There is no indication that the product is optically active; it is racemic. Peracid gives the epoxide. Alternatively, halohydrin formation (X = Cl, Br, I) gives two racemates. The mixture, when treated with base, gives the epoxide. f) Since the aldehyde is opt. act. and a single product, reconstituting the alkene means that both sides of the double bond are substituted identically. Permanganate adds cis to the (Z)-double bond. Syn addition of two hydroxyl groups to either face of the alkene gives a single diol. Both A and B are optically active. Periodic acid will effect cleavage of the diol B. Why is (E)-A excluded?

 

2. Identify each of the structures below. Pay attention to stereochemistry. Pay attention to the equilibrium arrows in the equilibria among B, C and D. A good place to start is the formation of M. n-Propyl magnesium bromide adds to ethyl formate to eliminate ethoxide forming n-butyraldehyde. A second equivalent of Grignard adds to the aldehyde to form the secondary alcohol, 4-heptanol (M). Consequently, E must be 4-heptanone. E adds methyl magnesium bromide to form tertiary alcohol A, 4-methyl-4-heptanol. E1 acid-catalyzed dehydration of A affords three possible alkenes under equilibrating conditions. The arrows show that the most stable alkene is C, (E)-4-methyl-3-heptene. B is the (Z)-stereoisomer of A given that B and C give the same products, G and H, upon double bond cleavage. Additionally, the equilibrium arrows show that C is slightly more stable than B and very much more stable than D, which is only disubstituted. Vigorous permanganate oxidation of D gives A and F, which is CO2. Carbon dioxide is the result of over oxidation of formaldehyde via formic acid. G is propionaldehyde while H is 2-pentanone because of a lack of reaction with dichromate. Compound I is propionic acid. Syn dihydroxylation gives racemic diol J.

Derivatization of the secondary alcohol of J as a tosylate followed by base treatment gives epoxide K with a single SN2 inversion at the secondary site. Direct (zero inversions) peracid epoxidation of alkene gives epoxide K as does bromohydrin formation (L) followed by base treatment, a two inversion process.

 

3. Toprol (4; Lopressor; Metoprolol) is a β-blocker for the treatment of hypertension. It is a racemic drug sold as its monosuccinate salt. A seasoned chemist wants to prepare Toprol as its (R)-enantiomer using the synthesis of the racemate. She follows the steps shown on the right. a) Why is Toprol a racemate? When prepared from racemic 1, 4 is a racemate because the carbon bearing the OH group has both enatiomers present. b) Draw a structure of the (±)-succinate salt. c) What is the role of aq. NaOH? Check here. The strong base NaOH forms the anion of the phenol 2. d) What does the sequence of reactions and their chirality tell you about the mechanism of the double SN2 reaction sequence? Write mechanisms for the two reactions. There are 2 sites for SN2 reactions to occur; on the C-Cl bond and the less substituted end of the epoxide ring. If Path A is followed, the phenoxide anion displaces chloride on (S)-1 to give epoxide (S)-3. Subsequent use of isopropylamine as a nucleophile would lead to (S)-4, the wrong enantiomer of toprol. On the other hand, if Path B is followed, phenoxide opens the epoxide first to yield a chlorohydrin anion which closes to the epoxide which is now (R)-3. Opening of this epoxide with the amine gives the correct enantiomer of toprol, (R)-4. The epoxide, and not the C-Cl entity, is the better electrophile.

f) How can she prepare (S)-4 without resorting to the use of (R)-1? Use the amine as the first nucleophile and then employ the phenoxide. Use the amine as the first nucleophile and then employ the phenoxide.