ORGANIC CUME 2/29/ (4 pts) Formal refers to a synthesis of an important precursor to the target product, in this case: (S,S)-Reboxetine.

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1 GAIC CUME 2/29/08 1. (4 pts) Formal refers to a synthesis of an important precursor to the target product, in this case (S,S)-eboxetine. (4 pts) o, it describes the actual synthes of (S,S)-eboxetine. See the discussion on page 708, second column. 2. (10 pts) It is important to consider the fact that an inversion of stereochemistry is involved in this reaction. This indicates that no carbocations are involved. Also, the authors point out that the a is used to saponify the actual product to obtain the β-amino alcohol (thus it is not involved in the rearrangement. A reasonable mechanism would be ( C) rotate F 3 C 1 2 a 1 2

2 3. (8 pts) Based on the stereochemical outcome that arises due to formation of aziridine in the mechanism, there would be no way to retain the configuration at the #1 carbon. If racemization occurred in the rearrangement (the only way to obtain the configuration), the authors would not be able to synthesize (S,S)-eboxetine in a stereodefined fashion. 4. (8 pts) The key to the Mitsunobu reaction is the conversion of the group into a good leaving group (one that can be displaced by a weak nucleophile) under very mild conditions. In this case the nucleophile is a phenol. The reaction involves activation of the alcohol by converting it to an alkoxyphosphonium ion which then undergoes and S 2 inversion. The DIAD (diisopropylazodicarboxylate) activates both the triphenylphosphine (by allowing formation of the phosphonium species) and the phenol (by forming the phenolate nucleophiles) ipr 2 C C 2 ipr Ph 3 P Et ipr 2 C PPh 3 C 2 ipr 8 Et 9 Ph 3 P Boc Et 5. (10 pts) Bn = benzyl Bz = benzoyl Boc = butoxycarbonyl (normally, tert-butyl) DIAD = diisopropylazodicarboxylate DMAP = dimethylaminopyridine (normally, para) DMF = dimethylformamide ed-al = bis(2-methoxyethoxy)aluminum hydride (a salt) 6. (5 pts) Since one enantiomer must be in excess in the amount stated, the ratio of enantiomers in 99.5 to 0.5. The 0.5 of the minor isomer negates 0.5 of the other isomer, leaving 99% excess of the major isomer.

3 7. (5 pts) The structure of the product is actually not easy to figure out.. If it is generated from racemization of carbon #2, it is A (this would presumably occur during the formation of 3 via ring opening of the aziridine or, perhaps, by racemization of the ring opened product. Bn 2 A (5 pts) Bn 2 (5 pts) The answer is dependent on your answer to part (a) BUT the relationship is almost certainly diastereomeric and not enantiomeric! 8. (7 pts) This is an interesting reaction sequence. The important thing to note is that the stereochemistry at carbon #2 was not inverted. Thus the oxygen at carbon #2 eventually displaces the tosylate. The a is used to generate the oxy anion at carbon #2 either directly or by displacement of a Ts group. easonable Bn Bn Ts

4 (3 pts) Compound 10 has no free, polar groups and would be quite easy to separate from polar byproducts by chromatography. 9. (8 pts) The reaction of B 3 involves reduction of the carbonyl group in both the benzamide and the benzoate. The reduced benzamide ends up getting reduced further (either through loss of a oxyboron species or through an intramolecular hydride migration). For the benzoate, the reaction results in the boron acetal which, after hydrolysis, decomposes to the alcohol and benzaldehyde. B loss of B- B- 2 Bn B The fact that the authors report only 24 when there should be 25 complicates the interpretation a bit. Also, the splitting patterns are a bit odd. The M resonances would be expected to be (a) (5 pts) 1 -M assignments 4.62 δ 3.61 and 3.56 δ δ But this one probably has too δ diastereotopic

5 (b) (5 pts) 13 C M assignments δ 140 δ 78 δ 58.9 δ δ 79 δ 52.5 δ (c) (2 pts) Absorption at 3387 cm-1 is due to stretching motions 11. (3pts) (a) 5 ppm (for a typical organic cpd (MW = 200) that would be about amu) (3 pts) (b) 0.3 (0.4 also reasonable)