Heck coupling (catalytic cycle): the mechanism, step by step
The metal inserts into the aryl-halide bond, the alkene inserts into the aryl-metal bond, the metal takes back a hydrogen and leaves, and a base strips it so the metal can start again.
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 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. This is usually the slow step, and it is why an iodide reacts more readily than a chloride.
The alkene inserts into the aryl-metal bond
The alkene slots into the bond between the ring and the metal. The ring ends up on one carbon and the metal on the other.
Migratory insertion: the alkene binds to the metal and then the aryl group migrates onto it, leaving the metal on the neighbouring carbon. Both new bonds form on the same face.
The metal takes a hydrogen back and lets go
The metal takes a hydrogen from the carbon next to it. Those electrons become the double bond, and the product falls off the metal.
Beta-hydride elimination. The hydrogen and the metal leave from the same face, and the conformation that arranges that with the two substituents apart is the lower one, which is why the trans alkene is what is isolated.
A base strips the acid and the metal starts again
The base takes the hydrogen and the halide off the metal. The metal is back exactly as it began, ready for the next molecule.
Reductive elimination of the acid, taken up by the base. The metal returns to where it started, which is what makes the whole sequence catalytic: it is not consumed however much substrate is put through it.
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. In the flask it carries ligands throughout, and which ligands they are decides how readily each step happens.
- The metal ends the cycle as it began it. That is what catalytic means here, and it is why the metal is not consumed however much substrate is used.
- The alkene and the metal add to the same face and the hydrogen leaves from the same face, which is what fixes the geometry of the product; this drawing does not show faces.
Other pathways this class runs by
- With a triflate or under cationic conditions the halide leaves the metal first, and the alkene binds to a positively charged metal instead.
- A related cycle with a boron or tin partner transfers the group to the metal rather than inserting an alkene, which is the Suzuki and Stille chemistry.
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.