Unimolecular elimination: the mechanism, step by step
Acid turns the hydroxyl into a good leaving group, it leaves on its own, and the cation then loses a proton to give the alkene.
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
Acid protonates the hydroxyl
The acid gives a hydrogen to the oxygen. A hydroxyl is a poor leaving group; water is a good one.
Protonation of the alcohol to the oxonium. This is what converts hydroxide, which does not leave, into water, which does.
Water leaves on its own
Water breaks away by itself, taking both electrons and leaving a flat, positively charged carbon.
Rate-determining heterolysis giving the carbocation. Nothing attacks and nothing is removed in this step.
A base takes a proton from the next carbon
A hydrogen on the carbon next door is removed, and its electrons become the second bond between the two carbons.
Loss of the β proton to any base present, giving the alkene. Which neighbour loses the proton decides which alkene, and the more substituted one is usually favoured.
What this drawing assumes
- Drawn with the carbon holding the charge on its own. Where a neighbouring group can shift across, the alkene ends up somewhere else, and this does not show that.
- Losing the leaving group is the slow step, which is what unimolecular means here.
Other pathways this class runs by
- A primary alcohol cannot support the cation and goes by a different route, with the leaving group and the proton departing together.
- The cation can also be captured by whatever is around, which is substitution rather than elimination.
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.