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Topic: peptide chains  (Read 6987 times)

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Offline rleung

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peptide chains
« on: April 29, 2006, 03:19:06 PM »
Hi,

I am having a little trouble with peptide bonds.   My teacher showed us to always protect the C-terminal of the amino acid being added (the one on the right), but if we are making a polypeptide, how do we reactivate the C-terminal with DCC?  Since he showed us to use the Fischer Esterification to turn C-terminal of the amino acid on the right into an ester, we would need to hydrolyze that bond to turn it back into a carboxylic acid so we can reactivate it with DCC for addition of the new amino acid, but if we hydrolyze it, we will also hydrolyze everything else in the peptide chain thus far, wouldn't we?  Thanks so much for answering my last-minute questions.

Ryan

Offline Albert

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Re: peptide chains
« Reply #1 on: April 30, 2006, 06:26:57 AM »
It depends on the technique you're using. However, according to what you wrote, I think you should protect a C-terminal with benzyl alcohol (forming an ester). The carboxyl function can be deprotected with catalytic hydrogenation.

Offline rleung

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Re: peptide chains
« Reply #2 on: April 30, 2006, 09:28:03 AM »
So would that mean that after the addition of each amino acid, you would have to remove the protecting groups on both the N-terminus and the C-terminus, activate the C-terminus with DCC, and re-protect the N-terminus?  Thanks.

Ryan

Offline Albert

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Re: peptide chains
« Reply #3 on: April 30, 2006, 02:52:36 PM »
First of all, you protect only the N-terminal of an amino acid.
Then you add DCC for activating the carboxyl group.
After that, you add the second amino acid with its C-terminal already protected with benzyl alcohol.
That's it.  :)

Shall I show you what I mean with the help of a picture?

Offline rleung

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Re: peptide chains
« Reply #4 on: April 30, 2006, 04:07:58 PM »
Thank you :)

I am slightly confused on one thing tho.  I think I will show you with the below picture.  After the last product I have there, say, if I want to add another alanine to the chain.....wouldn't I need to reactivate the C-terminus of the second alanine in the chain (the carbon that is bonded to the benzyl alcohol protecting grp all the way on the right).  In order to reactivate it with, say, DCC, wouldn't I have to de-protect the C-terminus from its protecting grp by turning it back into a carboxylic acid before I react it with DCC (according to what I know, DCC only reacts with carboxylic acids)?  But if I use, say, H2/Pd to deprotect the C-terminus, then that would mean I would also de-protect the N-terminus (the CBZ protecting grp), so I would need to re-protect the N-terminus again (all the way on the left) if I want to add another alanine to the chain, right???

From your previous post, I am sort of guessing you mean that the benzyl alcohol protecting grp that is already in place at the end of the synthesis in the picture is already a sufficient leaving group for the next alanine to react with the carbonyl carbon of the C-terminus.  Is that correct?

Hehe, sorry for the confusing post.  Hopefully I make some sense in all this rambling...

Ryan
« Last Edit: April 30, 2006, 04:11:10 PM by rleung »

Offline Albert

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Re: peptide chains
« Reply #5 on: May 01, 2006, 06:33:44 AM »
In fact, you use di-tert-butyl carbonate as protecting group for the N-terminal, because the product won't be altered by the hydrogenation.
Deprotection procedure is carried out with a dilute acid.

Offline rleung

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Re: peptide chains
« Reply #6 on: May 01, 2006, 02:08:04 PM »
Ahh...it's all becoming clear now...thank you SO much :)

odedarad

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Re: peptide chains
« Reply #7 on: May 01, 2006, 04:17:45 PM »
Do not forget, however, that as you try to extend your dipeptide to a tripeptide by activating the C-terminal Alanine with DCC, you will probably cause some racemization...

Offline movies

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Re: peptide chains
« Reply #8 on: May 02, 2006, 02:16:31 PM »
Just a note, the CBZ group is drawn wrong above.  It should look like this:

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