Sonogashira coupling (catalytic cycle): the mechanism, step by step
The metal inserts into the aryl-halide bond, the alkyne loses its hydrogen and is carried across, and the two groups join and leave together.
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. Usually the slow step, which is why an iodide reacts more readily than a chloride.
The base takes the alkyne's hydrogen
The hydrogen on the end of the alkyne is removed by the amine. That hydrogen is unusually easy to take off for a carbon-hydrogen bond.
Deprotonation of the terminal alkyne, in practice assisted by copper. The sp carbon holds the electrons far better than an sp3 one would, which is what makes an amine base sufficient.
The alkyne is carried to the metal
The deprotonated alkyne is handed to the palladium, and the halide comes off. The metal is now holding both pieces.
Transmetalation from copper to palladium, with loss of the halide. The copper acetylide is the species that does this, and it is why a copper salt is usually present.
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 copper is not drawn. Its job is to take the deprotonated alkyne and hand it to the palladium, and the two cycles are shown here as one.
- The amine is both base and solvent in most of these reactions, so the proton it takes is not shown going anywhere in particular.
Other pathways this class runs by
- The reaction runs without copper under the right conditions, by a route in which the alkyne binds the palladium directly before losing its proton.
- An aryl chloride needs a far more active catalyst than this drawing implies, and often does not react at all.
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.