Bimolecular elimination: the mechanism, step by step
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.
This class has one step and no intermediate. Everything happens at once, and that is not a simplification — it is what the class means. Bonds break and form in the same motion, through a single transition state that is never isolated.
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 base takes a hydrogen as the leaving group goes
A base pulls a hydrogen off the carbon next to the one carrying the leaving group. Those electrons have nowhere to go but into a second bond between the two carbons, and that pushes the leaving group off.
Concerted anti-periplanar elimination: the C-H σ electrons become the π bond while the C-leaving-group σ bond breaks heterolytically, through one transition state.
What this drawing assumes
- Drawn as one step, which is what bimolecular means here: the C-H and C-leaving-group bonds break together.
- The hydrogen taken has to be on the opposite side from the leaving group, which the drawing does not show but which decides the geometry of the alkene.
Other pathways this class runs by
- A carbon that supports a cation well can ionise first and lose the proton afterwards, which is the stepwise route and gives the same alkene by a different path.
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.
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.
Organometallic addition to a carbonyl
The carbon attached to the metal is nucleophilic. It adds to the carbonyl, and the alkoxide that results is protonated when the reaction is worked up.