Problem_Set_6_solutions

Problem_Set_6_solutions - O OR O CH 2 PPh 3 3...

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Chem 358 Spring 2007 Problem Set #6 - Answers Text Problems – See posted solutions. Chapter 19 : 19 .37, 40, 43, 44, 46, 47, 49, 53, 55, 56, 64 Additional problems There is always more than one approach to making a particular compound. What follows are just examples of the possible ways of making the given compounds. These are based on what we have done so far. As we learn more and more reactions you will be able to effect the same synthesis in fewer steps. You may want to explore on your own alternate methods for each. 1. (a) Retrosynthesis: OH O BrMg + Br HO BrMg H H O + Br Forward synthesis: Mg BrMg Br H H O 1. 2. H 3 O + HO HBr or PBr 3 Br Mg BrMg O 2. H 3 O + OH 1. (b) Retrosynthesis OH OH MgBr OH O + Br OH OH OH OH O Forward synthesis OH O HO OH H 3 O + OH O O Br O O PBr 3 Mg MgBr O O O 1. 2. H 3 O + OH O NaBH 4 OH OH (c) Retrosynthesis H 3 C O H 3 C OH H O H O O H O OH CH 3 MgBr +
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Forward synthesis H O H OH HO OH H OH O O H Br O O H O H 3 C OH H 3 C O HCl (trace) PBr 3 1. Mg 2. H 3 O + 1. CH 3 MgBr 2. H 3 O + CrO 3 pyridine Alternate approach: H O H OH HO OH H OH O O O O O H O H 3 C OH H 3 C O 1. CH 3 MgBr 2. H 3 O + CrO 3 pyridine PCC 1. H 2 NNH 2 /KOH triglyme 2. H 3 O + HCl (trace) 2. CH CHCH 2 CH 2 CH 3 CH Ph 3 P CHCH 2 CH 2 CH 3 CH PPh 3 O CHCH 2 CH 2 CH 3 O +
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Unformatted text preview: O + + OR O CH 2 PPh 3 + 3. Cyclopropanone has more bond angle strain than does the corresponding hydrate. The carbonyl carbon is sp 2 hybridized and must have 120° angle. However, in the cyclopropanone ring the angle is 60° leading to a bond angle compression of 120°- 60° = 60°. In the hydrate the carbon is sp 3 hybridized and must have the tetrahedral angle of 109°. The strain is therefore 109°-60° = 49°, which is less than in the corresponding starting ketone. O OH HO H 2 O sp 2 C sp 3 C 4. 4-Hydroxybutanal reacts first to form the cyclic hemiacetal, which being an alcohol, gets oxidized to the carbonyl group. HO O O H O O OH O O PCC 5. .. H O BF 3 H O BF 3 CH 3 —S—H H O BF 3 S CH 3 H H O BF 3 S CH 3 + H + H O BF 3 S CH 3 H H S CH 3 CH 3 —S—H H S S CH 3 CH 3 H H S S CH 3 CH 3 H 2 O + BF 3 + HO—BF 3 + HO—BF 3...
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