Fluoride removal of a silyl group: the mechanism, step by step
Fluoride attacks the silicon, which is where it would rather be than anywhere else, and the alcohol is released.
The pathway runs in 2 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
Fluoride attacks the silicon
Fluoride goes straight for the silicon. The bond it makes there is the strongest bond silicon forms, and making it is what pulls the group off the oxygen.
Nucleophilic attack at silicon with displacement of the alkoxide. The silicon-fluorine bond is stronger than almost any other bond silicon makes, and that difference is the whole driving force.
The alkoxide takes a proton
The negatively charged oxygen picks up a hydrogen and the alcohol is back.
Protonation from the solvent or on work-up.
What this drawing assumes
- Drawn as a single attack. Silicon can hold five groups for a moment, and the true path goes through that; the drawing does not show it.
- The alcohol is shown taking its proton back from the solvent or on work-up.
Other pathways this class runs by
- Acid also removes these groups, by protonating the oxygen instead, and the more hindered silyl groups need it.
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.