1. The inversion of configuration in an S_N2 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. Using these data, show how there is net retention of configuration.

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 S_N2 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 S_N2 displacement. Alternatively, chloride can effect direct S_N2 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 S_N2 inversion to form the reactive, transitory (S)-α-lactone. The strain of the α-lactone allows hydroxide to effect a second S_N2 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.

a) Mercaptans (thiols) are stronger acids than water. See pKa table. Hydroxide ion converts ethanethiol to its thiolate (mercaptide). The anion is a very good nucleophile. Inversion of configuration occurs. Cl and S are both top priority. Thus, (S)- becomes (R)-.
b) Excess methanol as a solvent. Conditions for an S_N1 reaction of the tertiary chloride but not the primary chloride. The ionization produces a planar carbocation. Addition of solvent provides the racemic ether.
c) The dibromide has a two-fold axis of rotation. Both centers are of the same configuration, ---namely, S. Sulfide (charge -2) does an intermolecular S_N2 displacement to give an intermediate that has the new C-S site of the opposite, R-configuration. Now, like 2a, a second S_N2 displacement occurs intramolecularly to give the tetrahydrothiophene A. Both centers have the R-configuration. A is optically active; two-fold axis of symmetry.
d) Protonation of the cyclic ether followed by S_N2 displacement by bromide ion gives a cis-substituted cyclohexyl bromo (primary) alcohol. Similarly, in the presence of of HBr, the primary alcohol is converted into a primary bromide.
e) The asymmetric center is stable under the reaction conditions. It remains intact in the products. A small nucleophile and a primary chloride favors S_N2 displacement (Williamson ether synthesis). A minor amount of E2 product is formed, which is also optically active.
f) Since A and B are constitutional (structural) isomers and the halide is secondary, elimination will dominate. A is the more substituted isomer, which suffers E2 elimination losing HCl in an anti manner. This elimination leads to the Z-isomer shown. [Note that the tertiary hydrogen and the chloride involved in the elimination have a 60^o dihedral angle (H-C-C-Cl). Rotation about the C-C bond must occur to allow anti (180^o) elimination to occur.] Since both asymmetric centers have become sp² hybridized, optical activity is lost in A. The minor E2 product is still optically active. Note how the chiral center in the alkene went from R in the chloride to S in the alkene even though the center wasn't altered.]
g) E2 elimination leads to stereoisomers A and A' having the trisubstituted double bond. B and C are the minor 1,1-disubstituted double bonds. B dominates over C because of the isotope effect in E2 eliminations.
h) The D-C-C-Cl dihedral angle is 0^o. E2 elimination is syn overriding any isotope effect. DCl is lost.
i) A small base will lead to a major Zaitsev product, i.e.; the trisubstituted A having the (Z)-configuration.[In this case, unlike problem f), the hydrogen and chlorine to be eliminated have a 180^o dihedral angle in the conformation shown.]
j) Ionization of the secondary bromide upon heating can occur to capture ethanol in an S_N1 process. The intermediate secondary carbocation can have the vicinal green hydride migrate to give a more stable tertiary carbocation, which captures ethanol in an S_N1 process.