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184 Chapter 9 FURTHER REACTIONS OF ALCOHOLS AND THE CHEMISTRY OF ETHERS O excess NH₃ S CH₃ O (R)-1-Deuterio-1-pentyl tosylate (S)-1-Deuterio-1-pentanamine The product should be 100% optically pure (S), because SN2 reactions proceed with 100% inversion. But it is not: It is 70% (S) and 30% (R). How? One possibility is that some of the molecules reacted by the mechanism. (In fact, in a related series of experiments carried out in the 1950s, this was the conclusion reached by the chemists involved.) But this turns out to be wrong: Displacements at simple primary carbons do not pro- ceed by the mechanism, because simple primary carbocations are too unstable to form. How else can we ex- plain the appearance of products with the same (R) configuration as was present in the starting material? Back in Section 6-6 you were shown that two successive displacements ("double inversion") at the same carbon atom would result in net retention of the original configuration. Could something of this sort be happening in the example given for this problem? Using the hint, let us consider the implications of the fact that chloride ion is produced in the course of tosylate formation. Suppose some of this chloride ion, a nucle- ophile, were to react with some of the tosylate formed in the first step. This process would be an dis- placement, at with inversion, producing an (S) chloroalkane: CH₃ (R)-1-Deuterio-1-pentyl tosylate (S)-1-Chloro-1-deuteriopentane Then we would have a mixture containing both (R) tosylate and (S) chloroalkane going into the second reaction, the SN2 with ammonia. This second reaction would convert the (R) tosylate present into (S) amine, and any (S) chloroalkane into (R) amine. This explains how an unequal mixture of amine enantiomers came out of the reaction: About 30% of the first-formed tosylate reacted with chloride before the reaction with ammonia was carried out, giving a second inversion and therefore overall net retention of configuration at for the corresponding 30% of the final product. H O D 64. Product: CH₃ C C H Note the inversion at the carbon that formerly contained the bromine. Make a model if necessary. CH₂ CH₂ CH₂ The reaction is first order, because the nucleophile and the haloalkane functional groups are in the same molecule. The mechanism is identical to that of the familiar SN2 reaction, but the "2" here is not applicable because both reacting components are in the same molecule. 65. We'll go through the retrosynthetic analyses in detail, followed by the syntheses. As usual, the solvent for all organometallic preparations and reactions is (CH₃CH₂)₂O. (a) Retrosynthetic analysis