Aldol condensation: the mechanism, step by step
A base makes the enolate, the enolate attacks the other carbonyl, and the alcohol that results loses water to leave a conjugated enone.
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 base makes the enolate
A base takes a hydrogen from the carbon next to a C=O group. That carbon is left with a pair of electrons and is now the nucleophile.
Deprotonation at the α carbon. The anion is stabilised by sharing the charge with the carbonyl oxygen, which is what makes those hydrogens acidic at all.
The enolate attacks the other carbonyl
The carbon with the spare electrons attacks the carbon of the other C=O group, and those double bond electrons move onto its oxygen.
Aldol addition: the enolate carbon adds into the second carbonyl, giving an alkoxide.
The alkoxide takes a proton
The negatively charged oxygen picks up a hydrogen and becomes an alcohol. This is the aldol product.
Protonation of the alkoxide gives the β-hydroxy carbonyl compound, which is where the reaction stops if it is not pushed further.
Water leaves and the double bond forms
A base takes a hydrogen from the carbon between the alcohol and the C=O group, and water is pushed out. The double bond that forms is next to the C=O, which is what makes it worth doing.
E1cb elimination: deprotonation at the α carbon and loss of hydroxide, giving the conjugated enone. Conjugation is the reason this step happens at all.
What this drawing assumes
- The enolate is drawn as the carbanion. The charge is really shared with the oxygen, and that sharing is why the proton comes off at all.
- The last step is drawn as losing water in one go. It goes through the enolate again, and it happens because the product is conjugated.
Other pathways this class runs by
- Stopping at the alcohol is possible and is the plain aldol reaction; whether it goes on to lose water depends on the temperature and on whether the alkene would be conjugated.
- Under acid the enol rather than the enolate does the attacking, and the carbonyl is protonated 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.