Sciencegeeknm Posted December 18, 2016 Posted December 18, 2016 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.
Sciencegeeknm Posted December 19, 2016 Author Posted December 19, 2016 Sorry, I keep forgetting to attach file.
hypervalent_iodine Posted December 19, 2016 Posted December 19, 2016 Sorry, I keep forgetting to attach file. The DCC isn't required in the second rection, because it isn't making a peptide bond. It's just being bound to the resin bead. Carboxylates tend not to be great nucleophiles, but under certain conditions you can convince them to undergo substitution. In this case it's SN1, because the chloro is benzylic.
Sciencegeeknm Posted December 20, 2016 Author Posted December 20, 2016 Thanks for your reply it really helped.
hypervalent_iodine Posted December 20, 2016 Posted December 20, 2016 No problems. Just to elaborate a little, the DCC (as with any activating agent) is used to derivatise the carboxylic acid (as an ester, anhydride, uronium, etc.), thereby making the carbonyl carbon more electrophilic and susceptible to nucleophilic attack by the unprotected amine on the terminal end of the growing chain. When binding the first amino acid to the resin, you need the amine to be at the terminal end for deprotection and then addition of the second amino acid, and for the nucleophilic oxygen to be the reactive centre on the carboxyl group. It makes no sense to hide the latter by way of an activating group, because then it can't react.
Sciencegeeknm Posted December 21, 2016 Author Posted December 21, 2016 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?
hypervalent_iodine Posted December 22, 2016 Posted December 22, 2016 It's no longer an ester oxygen, it's a carboxylate, which is basic. The ammonium nitrogen is acidic, so abstraction of a proton / hydrogen from it by the carboxylate is just an acid base reaction. The reaction generally either done at ambient temperature, or where the addition of the amine is performed drop wise to a solution of the Boc, and then allowed to warm to room temp.
Sciencegeeknm Posted December 22, 2016 Author Posted December 22, 2016 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?
hypervalent_iodine Posted December 22, 2016 Posted December 22, 2016 Oh, sorry. It's possibly oversimplified. I don't think the geometry really favours intramolecular hydrogen abstraction, but overall you are essentially seeing the acidic proton of the acid moving to the nucleophilic oxygen (on the right). This drives the formation of another C=O (on the left), and the breaking of the other C-O bond. This isn't the exact same sort of molecule, but is illustrative of an analogous mechanism. From http://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/12/oc/vlu_organik/c_acid/reaktionen_organoli_carbons.vlu/Page/vsc/en/ch/12/oc/c_acid/decarbox/decarbox.vscml.html
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