Problem Set 5, Solution Set

Chapter 6, Alkyl Halides: Substitution and Elimination

Due: Monday, October 12, 2009

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A Walden Cycle

Paul Walden (1863-1957) here also   xxxxxxxxxxxxxxxxx

1. The inversion of configuration in an SN2 reaction is often called a Walden inversion, named after its discoverer, Paul Walden. In the cycle shown above, the overall conversion of one enantiomer of malic acid to the other one must require an inversion of configuration. Similarly, the same is true of the chloro acids. More generally, each interconversion of enantiomers must require an odd number of inversions. The PCl5 reaction requires a single inversion which means that the Ag2O reaction involves an even number of inversions of configuration, namely two in this instance. (-)-Malic acid is of the (S)-configuration. a) Show how malic acid, like any alcohol, might react with PCl5 and then undergo inversion to form a chloride. Remember that phosphoric acid is a strong acid and its conjugate base and analogs thereof are also good leaving groups. b) Silver oxide is an anhydrous form of AgOH. The carboxylic acid group closest to the hydroxyl group plays a role in the process. The reaction medium is mildly alkaline. c) Draw these four enantiomers as Fischer projections with the CO2H closest to the OH or Cl in the topmost position. (-)-Malic acid is of the (S)-configuration.

a) An electron pair on the hydroxyl group of (S)-malic acid does the equivalent of an SN2 displacement of chloride on the phosphorus atom of PCl5. The proton at the positive site may be removed by chloride at this point and then chloride from dissociated HCl can effect an SN2 displacement. Alternatively, chloride can effect direct SN2 displacement with inversion of configuration (1st inversion) to form (R)-chlorosuccinic acid, POCl3 and HCl. [Note: It is also likely that the carboxyl groups are converted to acyl chlorides during the reaction. Aqueous workup would rapidly reform carboxyl groups].

b) (R)-Chlorosuccinic acid under alkaline conditions is converted to its dicarboxylate salt. Ag+1 may or may not complex with the chlorine atom at this point to enhance chloride as a leaving group. AgOH is not critical. The reaction works using NaHCO3 as a base. The proximate carboxylate acts as a nucleophile with SN2 inversion to form the reactive, transitory (R)-a-lactone. The strain of the a-lactone allows hydroxide to effect a second SN2 displacement to form (R)-malic acid upon acidification. This step has an even number of inversions -- net retention. The overall process of (R)- to (S)-malic acid has an odd number of inversions -- net inversion.

c)

2.

a) This is a primary bromide that is chiral by virtue of the H, D, Br and n-butyl group attached to carbon 1. RSH: pKa = 10; H2O: pKa = 15.7. LINK TO PKA RSH is a weaker acid than water, so the equilibrium favors RS- + H2O. Mercaptide ion effects an SN2 inversion to form the (S)-sulfide. b) Chloride is exchanged for bromide by an SN2 reaction. One can't see the inversion at the primary site (proved in 2a) but it is seen at the secondary site. cis-Dichloride to trans-dibromide. c) An SN2 transition state apNote that there is approximates a trigonal bipyramid. The apices are the nucleophile and the leaving group; the trigonal part is CH2, CH2 and H at ~120o. The C-C-C bond angle for cyclohexane is 111o; cyclobutane, 88o. It is easier for cyclohexane to undergo the displacement as long as elimination to cyclohexene is not a factor. Note that there is a limited amount of nucleophile. d) 1-Bromo-2-methylpentane is more hindered toward SN2 displacement than 1-bromo-3-methylpentane. Competition for limited nucleophile favors 1-bromo-3-methylpentane.

3. Show how you would convert (S)-2-octanol into (S)-2-octanethiol. [Hint: how do you make a secondary hydroxyl group a good leaving group for an S_N2 reaction?] a goo

4. (2S,5S)-5-Bromo-2-hexanol (A) is expected to form optically inactive B (C₆H₁₂O) upon exposure to aqueous NaOH. An optically active stereoisomer of A, namely, C also forms optically in active B under the same conditions. Two other optically active stereoisomers of A, namely D and E as a racemic mixture, form optically inactive F, a diastereoisomer of B. What are the structures of A-F? The structures D and E are not distinguishable. Explain and illustrate with mechanisms. Why are B and F both optically inactive?