[Rutherford]

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?

[discodermolide]

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.

[Rutherford]

That should be C right? What are A and B? A protonated CDI (only one nitrogen), B deprotonated amino acid?

[discodermolide]

This is C I think.

plus imidazole.
God knows what A is.

[Rutherford]

I got this. But can't figure out F .

[discodermolide]

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.

[Rutherford]

How does the imidazole attack D? The lone pair on 1-nitrogen is contributing to the aromaticity.

[discodermolide]

A nitrogen of imidazole is nucleophillic.

[Rutherford]

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?

[discodermolide]

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.

[Rutherford]

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.

[discodermolide]

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.

[Rutherford]

I get a positive charge on imidazole  .

[discodermolide]

Move the lone pair into the ring.

[Rutherford]

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.