Buchwald-Hartwig amination (catalytic cycle): the mechanism, step by step
The metal inserts into the aryl-halide bond, the amine takes the halide's place on the metal and loses its proton to the base, and the two groups join as the metal lets go.
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 metal inserts into the aryl-halide bond
The metal pushes itself into the bond between the ring and the halide, ending up holding both.
Oxidative addition: the metal is formally oxidised by two and now carries the aryl group and the halide. Usually the slow step, which is why an iodide reacts more readily than a chloride.
The amine takes the halide's place on the metal
The amine reaches the metal and binds to it, the halide comes off, and the base takes the amine's proton. The metal is now holding both pieces that are to be joined.
Ligand exchange at the metal followed by deprotonation of the coordinated amine. The base is doing real work here: without it the nitrogen keeps its proton and the cycle stops.
The two groups are joined and the metal lets go
The ring and the nitrogen join to each other and come off together. The metal is back exactly as it started.
Reductive elimination forming the carbon-nitrogen bond. The metal is reduced by two and returns to where it began, which is what makes the sequence catalytic.
What this drawing assumes
- The metal is drawn as a bare atom, and may be palladium, nickel or copper: the cycle is the same and the ligand, which is not drawn, is what decides whether it turns over. The ligand is what makes this reaction work at all — it is the difference between a coupling and a flask of starting material — and the drawing does not show it.
- The amine binding and losing its proton are drawn as one step. They are separable, and which order they happen in depends on the base.
Other pathways this class runs by
- An activated ring does not need the metal: a nitro group opposite the halide is enough for the amine to attack directly, which is a different pathway and a different template.
- With a weak base the amine can bind and fail to lose its proton, which stalls the cycle at that point rather than turning over.
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.