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Topic: CDI (carbonyldiimidazole) usage in peptide synthesis  (Read 4971 times)

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Offline Rutherford

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CDI (carbonyldiimidazole) usage in peptide synthesis
« on: February 15, 2014, 10:25:43 AM »
How to solve the attached problem? How does CDI activate the carbonyl group?

Also: Explain why the C=O stretching frequency in 1,1’-carbonyldiimidazole is 100 cm–1 higher than that of 1,1’-carbonyldipyrrolidine.

I thought that the participation in conjugation would reduce the second bond character of the C=O bond, and move it lower, not higher. Where am I wrong at?

Offline discodermolide

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Re: CDI (carbonyldiimidazole) usage in peptide synthesis
« Reply #1 on: February 15, 2014, 10:30:10 AM »
The a acid OH attacks the CDI carbonyl and eliminates imidazole to produce a sort of mixed anhydride.
See if you can draw the structure and find out what happens next.
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Offline Rutherford

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Re: CDI (carbonyldiimidazole) usage in peptide synthesis
« Reply #2 on: February 15, 2014, 10:34:09 AM »
That should be C right? What are A and B? A protonated CDI (only one nitrogen), B deprotonated amino acid?

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Re: CDI (carbonyldiimidazole) usage in peptide synthesis
« Reply #3 on: February 15, 2014, 10:47:02 AM »
This is C I think.
O=C(OC(N1C=CN=C1)=O)C(N)C
plus imidazole.
God knows what A is.
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Offline Rutherford

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Re: CDI (carbonyldiimidazole) usage in peptide synthesis
« Reply #4 on: February 15, 2014, 10:52:48 AM »
I got this. But can't figure out F :(.

Offline discodermolide

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Re: CDI (carbonyldiimidazole) usage in peptide synthesis
« Reply #5 on: February 15, 2014, 10:59:46 AM »
Well the mixed anhydride is attacked by imidazole at the carbonyl to eliminate CO and another imidazole. This gives G!
The other intermediates are ??
So F is probably the ionic intermediate?.
Never seen something like this before, strange scheme.

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Offline Rutherford

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Re: CDI (carbonyldiimidazole) usage in peptide synthesis
« Reply #6 on: February 15, 2014, 11:07:37 AM »
How does the imidazole attack D? The lone pair on 1-nitrogen is contributing to the aromaticity.

Offline discodermolide

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Re: CDI (carbonyldiimidazole) usage in peptide synthesis
« Reply #7 on: February 15, 2014, 11:13:40 AM »
A nitrogen of imidazole is nucleophillic.
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Offline Rutherford

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Re: CDI (carbonyldiimidazole) usage in peptide synthesis
« Reply #8 on: February 15, 2014, 11:20:41 AM »
Okay, I see. There would be also more logic if the other N atom in CDI was protonated.

At the end activated Ala is produced, and the amino group of Gly can attack it and form Ala-Gly dipeptide right?

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Re: CDI (carbonyldiimidazole) usage in peptide synthesis
« Reply #9 on: February 15, 2014, 11:23:20 AM »
Yes that is the whole idea, peptide coupling.
Logic, who applies logic? You can ask where do the protons come from. What starts the whole process off? Protonation of a carbonyl by an amino acid? You could go on for years with this and similar mechanisms.
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Offline Rutherford

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Re: CDI (carbonyldiimidazole) usage in peptide synthesis
« Reply #10 on: February 15, 2014, 11:27:31 AM »
Good. Two more questions: How does the imidazole ring activate the carbonyl group? Why the C=O stretching frequency in 1,1’-carbonyldiimidazole is 100 cm–1 higher than that of 1,1’-carbonyldipyrrolidine.

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Re: CDI (carbonyldiimidazole) usage in peptide synthesis
« Reply #11 on: February 15, 2014, 11:30:34 AM »
Well the imidazole is a good leaving group, and presumably it is electron withdrawing giving the C=O a partial positive charge, the partial negative charge on the imidazole is resonance stabilised into the ring. Draw the resonance structures out.
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Offline Rutherford

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Re: CDI (carbonyldiimidazole) usage in peptide synthesis
« Reply #12 on: February 15, 2014, 11:34:58 AM »
I get a positive charge on imidazole  ???.

Offline discodermolide

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Re: CDI (carbonyldiimidazole) usage in peptide synthesis
« Reply #13 on: February 15, 2014, 11:41:59 AM »
Move the lone pair into the ring.
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Offline Rutherford

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Re: CDI (carbonyldiimidazole) usage in peptide synthesis
« Reply #14 on: February 15, 2014, 11:45:33 AM »
Okay, thanks. But why is the C=O stretching frequency in 1,1’-carbonyldiimidazole is 100 cm–1 higher than that of 1,1’-carbonyldipyrrolidine? Higher position in IR means stronger bond, but the C=O bond has a single bond character in 1,1’-carbonyldiimidazole, so it should be lower in IR.

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