HT3ans-06.cwk - Answers to Hour Examination #3, Chemistry...

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Answers to Hour Examination #3, Chemistry 302X, 2006 1. (a). That S N 2 shown just can t happen with the required inversion. The “back” of that methyl group is not reachable by the putative nucleophile, the pi system of the enolate. (b). Although displacement of the methyl can t happen, addition to the ester carbonyl can. So, one mechanism begins with the addition of the ketone enolate to the ester carbonyl to give A , then, perhaps, the symmetrical intermediate B . O CH 3 O OEt OEt O O CH 3 B O O CH 3 1 A Intermediate B can undergo addition of ethoxide to either carbonyl (right-hand reverse arrow above) to regenerate A and go on to products shown below. Even better (probably) would be the direct opening of the initial adduct A to give an enolate that protonates to product. OEt O O CH 3 OEt O O CH 3 CH 3 O EtOOC A
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Alternatively (and better) we could do the other archetypal reaction of carbonyl groups and add alkoxide to the ketone C=O. A reverse Claisen, followed by a forward Claisen, leads to enolate C . A final protonation gives the product. RO RO O COOEt CH 3 O COOEt CH 3 H O COOEt CH 3 H 3 O + C OR COOEt CH 3 O O COOEt CH 3 OR O COOEt CH 3 O CH 3 OR O OR O COOEt CH 3 OR protonate deprotonate now salt formation is possible: (–) (–) (c). Now, how to tell them apart? Label one (either) carbonyl group. Here are the three possibilities, starting with labelled ketone: Direct opening of A gives product labelled as in D : OEt O O CH 3 OEt O O CH 3 CH 3 O EtOOC A D If the symmetrical intermediate is formed, there must be 50% of 4 produced with the other 50% in the ester carbonyl.
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The reverse Claisen puts the label completely in the final ester product.: RO
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HT3ans-06.cwk - Answers to Hour Examination #3, Chemistry...

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