Carbocation Rearrangements: Predicting the Unexpected Product
When the product has a skeleton different from the starting material, a carbocation rearranged. Knowing when to expect it is most of the skill.
The stability ladder
with allylic and benzylic cations stabilised further by resonance. A cation will rearrange whenever a single migration produces a cation higher on this ladder, and the migration happens essentially as fast as the cation forms.
The two migrations
Hydride shift
A hydrogen on an adjacent carbon migrates with its bonding pair to the cationic centre. The positive charge moves to the carbon the hydrogen left.
It is important that the hydrogen takes its electrons with it — describing it as a proton moving is wrong and gives the opposite charge outcome.
Alkyl shift
An alkyl group migrates the same way, again with its bonding pair. Methyl shifts are the most commonly encountered.
Ring expansion
Where a cation sits next to a strained ring, a carbon–carbon bond of the ring can migrate to the cationic centre. The ring becomes one atom larger and the strain is relieved.
This is a favourite in exam questions because the product skeleton is obviously different from the starting material, which signals rearrangement immediately. A cyclobutane adjacent to a cation expanding to a cyclopentane is the standard case.
When to expect rearrangement
- Does the mechanism generate a carbocation at all? SN1, E1, acid-catalysed additions and dehydrations all do.
- Is the cation secondary or primary? Tertiary cations usually have nothing better to rearrange to.
- Is there a hydrogen or alkyl group on an adjacent carbon whose migration would give a more stable cation?
- Would migration relieve ring strain?
If the answer to one and any of two to four is yes, expect the rearranged product to dominate.
Reactions where this matters most
| Reaction | Typical consequence |
|---|---|
| SN1 substitution | Nucleophile ends up on a different carbon than expected |
| E1 elimination | Alkene appears in a different position |
| Acid-catalysed hydration | Alcohol forms at a rearranged position |
| Friedel–Crafts alkylation | A straight-chain halide gives a branched product |
| Dehydration of alcohols | Skeletal rearrangement in the alkene |
The Friedel–Crafts case is the most frequently examined. Attempting to attach a straight-chain alkyl group gives the rearranged branched product instead, which is why Friedel–Crafts acylation followed by reduction is the standard workaround — the acylium ion does not rearrange, being resonance stabilised.
Reactions that avoid it
Mechanisms that never form a free carbocation cannot rearrange. SN2 substitution, E2 elimination and hydroboration all proceed without one, which is precisely why they give predictable, unrearranged products.
Where a question requires an unrearranged product from a substrate prone to rearrangement, choosing a mechanism that avoids the cation is the intended answer.
Frequently asked questions
Why does the migrating group take its electrons?
Because it is migrating to an electron-deficient centre. Moving without electrons would leave the origin carbon with a positive charge as well, which is not what happens.
Can a rearrangement give a less stable cation?
Not productively. Rearrangements are reversible in principle, but the equilibrium favours the more stable cation, so that is what is observed.
How do I know from a product that rearrangement occurred?
The carbon skeleton differs from the starting material, or the functional group sits on a carbon that could not have borne it without migration.
Why does the acylium ion not rearrange?
Because it is stabilised by resonance involving the carbonyl oxygen, so it is already low in energy and there is nothing more stable to rearrange to.
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