Suzuki coupling (catalytic cycle): the mechanism, step by step
The metal inserts into the aryl-halide bond, the boron hands over its carbon, and the two groups on the metal join and leave together.
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 boron hands its carbon to the metal
The boron gives up its organic group to the metal and takes the halide in exchange. Now the metal is holding both of the pieces that are to be joined.
Transmetalation. The base is needed to make the boron electron-rich enough to give up its carbon, which is why the reaction does nothing without one.
The two groups are joined and the metal lets go
The two organic groups on the metal join to each other and come off together. The metal is back exactly as it started.
Reductive elimination: the metal is reduced by two as the new carbon-carbon bond forms, returning it to where it began. This is what makes the sequence catalytic.
What this drawing assumes
- The metal is drawn as a bare atom; in the flask it carries ligands throughout, and which ligands they are decides how readily each step happens.
- The handover from boron is drawn as a simple exchange. It needs the base, which makes the boron give up its carbon, and that is why a Suzuki run without one does nothing.
Other pathways this class runs by
- Whether the base acts on the boron or on the metal is still argued about, and the drawing does not take a side.
- The same cycle with tin or zinc in place of boron is the Stille and the Negishi chemistry; only the middle step differs.
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.