Amide coupling through an O-acylisourea: the mechanism, step by step
The reagent turns the acid into something an amine will attack, and then it is an ordinary acyl substitution: attack, tetrahedral intermediate, collapse, proton transfer.
The pathway runs in 4 steps. What sits between them is a real structure: it is what the reaction passes through, and on a bad day it is what you isolate instead of the product.
Below is the sequence as it is taught. On the platform it is drawn on whatever structures you supply, with the curly arrows resolved to the actual atoms involved rather than to a general case with R groups on it.
The steps
The acid adds across the carbodiimide
The acid adds to the middle carbon of the reagent. What results has a very good leaving group attached where the hydroxyl used to be.
Addition of the carboxylic acid across the carbodiimide to give the O-acylisourea. This species is reactive and short-lived, and it can also rearrange to an unreactive N-acylurea, which is the side reaction an additive is there to prevent.
The amine attacks the activated carbonyl
Now that the carbon is easy to attack, the nitrogen shares its pair of electrons with it, and the electrons of the double bond move up onto the oxygen.
Addition of the amine lone pair into the π* of the activated ester. The carbon goes from three groups to four, the oxygen takes a negative charge and the nitrogen a positive one.
The carbonyl comes back and the activating group goes
The oxygen pushes its electrons back down to remake the double bond, and that forces the activating group off the other side.
The alkoxide collapses: the oxygen lone pair reforms the π bond and the bond to the activating group breaks, the pair leaving with it. That group is a far better leaving group than the hydroxyl it replaced, which is the whole point of the exercise.
A base takes the proton
The nitrogen is still carrying a positive charge and a spare hydrogen. The base in the flask takes it, and the amide is left.
Proton transfer to the tertiary amine base that is present in every one of these reactions for exactly this purpose.
What this drawing assumes
- Drawn without the additive. HOBt or HOAt is usually present and converts this intermediate to an active ester first, which suppresses the rearrangement below; the bond-forming steps that follow are the same either way.
- The base is not drawn.
Other pathways this class runs by
- An acid chloride or an anhydride reaches the same tetrahedral intermediate without any of this, which is the older way of making the same bond.
- With a hindered acid or a weak amine the activated species can be attacked by a second molecule of acid instead, giving the anhydride.
Run it on your own structures
Enter the reactants and the product you expect. If the transformation is one the engine can perform and this pathway reaches that product from those structures, it is drawn on them — with the arrows on the right atoms and the intermediates you would actually pass through. If it does not reach the product, you are told that instead of being shown a drawing that does not apply.
Other mechanisms
Nucleophilic acyl substitution
The nucleophile adds to the carbonyl, the carbon holds four groups for a moment, and the leaving group is pushed out as the double bond comes back.
Bimolecular nucleophilic substitution
One step. The nucleophile comes in on the opposite side from the leaving group, and the carbon turns inside out as the exchange happens.
Imine formation
The amine adds to the carbonyl, the resulting alcohol-amine loses water, and a carbon–nitrogen double bond is left.
Diels-Alder cycloaddition
One step. Six electrons move round a ring at once, two new single bonds form at the ends and the double bond ends up in the middle.
Bimolecular elimination
One step. The base takes a hydrogen from one carbon while the leaving group departs from the next, and a double bond forms between them.
Unimolecular nucleophilic substitution
The leaving group goes first, on its own, leaving a flat carbon with a positive charge. Whatever is around then attacks it from either side.