hellen Posted May 20, 2009 Posted May 20, 2009 (edited) I don't know how to do this:embarass:, can anyone explain a bit, please? Thanks a lot for the help! The transformation shown below can be accomplished in TWO reaction steps. From the following list, select the two reagents (or sets of reagents) which would most effectively bring about this conversion. Reagents: 1. Conc H2SO4/heat 2. CN- 3. KMnO4 4. (i) CH3MgI (ii) H3O+ 5. PBr3 6. (i) BH3 (ii) H2O2 7. CH3CH2O-/CH3CH2OH 8. H2O/H+ 9. OH-/H2O 10. (i) O3 (ii) Zn 11. NaBH4 12. HCl 13. KI 14. Na 15. CH3CH2Br Please enter the reagent number (from the above list) for each of the two steps in the following synthesis: Step 1: Step 2: Edited May 20, 2009 by hellen
hermanntrude Posted May 20, 2009 Posted May 20, 2009 OK have you any idea what might be the intermediate? have you any thoughts on this at all?
UC Posted May 20, 2009 Posted May 20, 2009 Well, what classification does the product you want to make fall under? There is a named reaction that generates that kind of structure. If you are doing this homework, I suspect you've heard of it before, and it uses 2 reagents on that list.
hellen Posted May 21, 2009 Author Posted May 21, 2009 I don't know how to do this:embarass:, can anyone explain a bit, please? Thanks a lot for the help! The transformation shown below can be accomplished in TWO reaction steps. From the following list, select the two reagents (or sets of reagents) which would most effectively bring about this conversion. Reagents: 1. Conc H2SO4/heat 2. CN- 3. KMnO4 4. (i) CH3MgI (ii) H3O+ 5. PBr3 6. (i) BH3 (ii) H2O2 7. CH3CH2O-/CH3CH2OH 8. H2O/H+ 9. OH-/H2O 10. (i) O3 (ii) Zn 11. NaBH4 12. HCl 13. KI 14. Na 15. CH3CH2Br Please enter the reagent number (from the above list) for each of the two steps in the following synthesis: Step 1: Step 2: Ok, so I think it's Step1: 4, Step2: 15 Yes, anyone know can just tell me I am on the right track or not, thank you.
hermanntrude Posted May 21, 2009 Posted May 21, 2009 as far as i can tell, you got step 2 right. What would the intermediate be if you used #4 on the starting material? By my estimation it'd just be the starting material again. try answering UC's questions... the answers will be helpful
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