Conjugate addition: the mechanism, step by step
A base makes the enolate, and it adds to the far end of the enone rather than to the carbonyl. The charge travels through the double bond to the oxygen and then comes back as the product is protonated.
The pathway runs in 3 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 base makes the enolate
A base takes a hydrogen from the carbon sitting between two C=O groups. Those hydrogens come off easily because two oxygens can share the charge left behind.
Deprotonation of the active methylene. Two flanking carbonyls make this position far more acidic than an ordinary one, which is why a mild alkoxide is enough.
The enolate adds to the far end of the enone
The nucleophile attacks the far carbon of the C=C rather than the C=O next to it. The electrons run along the chain and end up on the oxygen.
1,4-addition. The electrons of the alkene shift into the carbonyl and the charge arrives on the oxygen, which is what makes attacking the far carbon worthwhile rather than attacking the carbonyl directly.
The enolate takes a proton back at carbon
The negatively charged oxygen pushes its electrons back to remake the C=O group, and the carbon picks up a hydrogen. The product is left.
Protonation at carbon rather than at oxygen, which is where the enolate ends up after tautomerising. The carbonyl is restored and the addition is complete.
What this drawing assumes
- The enolate is drawn as the carbanion. The charge is shared with both flanking carbonyls, and that sharing is what makes the hydrogen acidic enough for a mild base to remove.
- Adding at the far end is drawn as the only outcome. Adding straight to the carbonyl is a real competing reaction; which one wins depends on the nucleophile, and this drawing does not decide it.
Other pathways this class runs by
- A hard, reactive nucleophile adds to the carbonyl carbon instead, giving the alcohol rather than this product. Organolithiums do that; stabilised enolates and cuprates do what is drawn here.
- Under acid the enol rather than the enolate adds, and the enone is protonated on oxygen first.
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.