Sciencegeeknm

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About Sciencegeeknm

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  • Birthday 07/14/72

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    Leicester, England
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    Chemistry
  1. Sorry, I didn't mean how does the carboxylate remove the proton from nitrogen, I was referring instead to the Tert-butyl bicarbonate molecule in the by- products box. I was wondering why the oxygen attached to the tert-butyl group deprotanates the OH and starts the decarboxylation reaction?
  2. Ok, now I understand. The first amino acid is joined to the linkage unit by way of subsitution reaction so no activation needed. I have attached a screenshot showing the mechanism showing how the boc group is added. step 3 says that the tert butyl bicarbonate breaks down into CO2 and tert butanol. What causes the ester oxygen to grab the hydrogen? Would this reaction require heat?
  3. Thanks for your reply it really helped.
  4. Sorry, I keep forgetting to attach file.
  5. In the attached photo, step one has the amino acid N terminus blocked and the C terminus activated by DCC. In step 2 the amino acid does not show the DCC unit. Would it not be needed in order to attach the amino acid to the linkage molecule? Also, what reaction takes place when joining the C terminus to the benzyl chloride linkage unit. I can see that the Cl is substituted for the oxygen so would this go via SN2 or SN1? Thanks for your help. Nick.
  6. Ah ok. Stabilisation through resonance. So would it ever be SN2?
  7. Interesting that you say SN1 as the teacher says this would go by SN2 alkylation. If it went by SN1 then you get a primary carbocation?
  8. Sorry, here it is. Here it is.
  9. Hi, The following reaction(see photo attached) was talked about on a lecture I watched on YouTube. There was no mechanism mentioned so I tried to work it out myself. Can you confirm if I'm on the right track? Firstly the Na H (in excess) would de protonate all of the OH groups on the glucose molecule. Then the now O- groups would react by SN2 pathway on the benzyl chloride molecules. Now all the former O- groups would be joined to benzyl CH2 units? Thanks. Nick.
  10. Hi, I'm just learning the Aldol Condensation mechanism and in one video it's says under basic catalytic conditions the final product is usually the Aldol. Only under acidic conditions will the Aldol react further to form the a,b unsaturated compound. But on another video (by a different person) she says that by adding heat you can do the condensation under basic conditions as a molecule of OH- will deprotanate the alpha proton forming a pi bond between the alpha and beta carbons and kicking off the OH group. The first video must be wrong in saying that under basic conditions you stop at the Aldol product? I'm not planning to do this reaction just learning it out of interest. Any advice would great as the two videos seem to be telling me different things? I have attached screenshots from the two videos.
  11. Thanks for your suggestion of using the Adol reaction. On the attached file I have drawn out the reaction which I think is correct? You said to substitute the OH for Br. Would you use H-Br and would the reaction follow the Sn1 pathway with water as the leaving group? Many thanks.
  12. Hi, Sorry it is heptanone I have drawn. The target molecule is one I have just randomly drawn. I do not intend to make it. I study organic chemistry as a hobby/interest and watch online lectures. The lectures are now covering synthesis so I have tried to do one myself. I learnt that the starting material should only be 4 or 5 carbons and possibly an alcohol? Im trying to do a retro synthesis here but probably not very well. How would you go about making this target molecule? Or have I drawn something which can't be made? Thanks Nick.
  13. Hello, I am a complete beginner to synthesis and wondered if someone knowledgeable would mind checking these steps to make 5-Bromo 3-hexanone When you make the two molecules how is the carbon carbon bond formed to join them together and make the target molecule? Thanks.
  14. Thank you for your answer it really helped. I'm glad I found this forum.
  15. Hello, This is the first reaction in the synthesis of C60. As you can see from the attached picture the starting molecule is benzene with a chlorine attached and a bromine para to it. After a grignard reaction with Mg and MeCHO(Acetaldehyde) you get a chlorine attached and a methyl and alcohol para to it. Can someone please confirm that my thinking below is correct on how I think the initial carbonyl is reduced to an alcohol. 1. The bromine which was kicked of by the Grignard reagent now carries a negative charge and attacks the carbon double bonded to the oxygen. The 2 elections from one of the double bonds goes onto the oxygen 2. The oxygen now attacks the adjacent hydrogen and is protonated. The bond which was attached to it collapses and kicks of the bromine? I might be completely wrong! Many thanks. Nick.