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HT3an-302X-07 - Answers to Hour Examination#3 Chemistry...

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Answers to Hour Examination #3, Chemistry 302X - 2007 1. Start with an analysis of exactly what molecule must have produced the product. The product is a β -hydroxy ketone, so a reverse aldol to give X seems in order: HO O H 3 C O O = O O CH 3 X If we had compound 2 , the product of a different aldol condensation of X , a reverse aldol of 2 to X would solve the problem. O O CH 3 X forward aldol here CH 3 O HO 2 The difficulty, of course, is that we don t have 2 , and 2 can t be made through hydration of starting material. Amy s mechanism involves an unstabilized anion! CH 2 O NaOH CH 3 O HO
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However, we can move that double bond in base, and then we can do a Michael to give 2 . Problem solved. CH 2 O CH 3 O HO 2 NaOH CH 2 O CH 2 O CH 3 O H 2 O CH 3 O HO HO Michael
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2. (a) A classic acetoacetic ester synthesis. From benzene: 1. CH 3 Cl/AlCl 3 2. NBS 3. Mg (now you have PhCH 2 MgBr) 4. CO 2 5. pH = 7 (now you have PhCH 2 COOH) 6. CH 3 OH/acid (now you have PhCH 2 COOCH 3) 7. full eq. methoxide 8. acid/heat (done) (b) A classic Robinson.
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