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Solid state peptide synthesis


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#1 Sciencegeeknm

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Posted 18 December 2016 - 01:08 PM


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.
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#2 hypervalent_iodine

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Posted 19 December 2016 - 06:02 AM

Your picture isn't attached
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#3 Sciencegeeknm

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Posted 19 December 2016 - 06:34 PM

Sorry, I keep forgetting to attach file.

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#4 hypervalent_iodine

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Posted 19 December 2016 - 06:45 PM

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.
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#5 Sciencegeeknm

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Posted 20 December 2016 - 07:18 PM

Thanks for your reply it really helped.
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#6 hypervalent_iodine

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Posted 20 December 2016 - 07:32 PM

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.
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#7 Sciencegeeknm

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Posted 21 December 2016 - 10:12 PM

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?

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#8 hypervalent_iodine

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Posted 22 December 2016 - 04:01 AM

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.
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#9 Sciencegeeknm

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Posted 22 December 2016 - 07:41 AM

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?

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#10 hypervalent_iodine

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Posted 22 December 2016 - 07:58 AM

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.

image.gif

From http://www.chemgaped...rbox.vscml.html
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