Pinacol rearrangement: practice and mechanism
The 1,2-diol gives a carbocation, then a methyl group moves to the neighbouring carbon. The new cation is stabilised by the lone pair on O, and a ketone (pinacolone) forms.
Example
One OH group is protonated and leaves as water, giving a tertiary carbocation. A methyl group migrates from the neighbouring carbon (1,2-shift). The new cation is stabilised by the remaining OH group, and a ketone (pinacolone) forms.
Mechanism step by step
- 1
The acid protonates an OH group and water leaves: a tertiary carbocation forms.
- 2
A methyl group from the neighbouring carbon moves to the cation with its electron pair (1,2-shift).
- 3
The new cation is on the carbon with the OH group and is stabilised by the O (oxocarbenium ion). Losing the proton gives the ketone.
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
- No double loss of water to a diene. After the first water is lost the cation rearranges, because a C=O bond (oxocarbenium ion) is much more stable.
- Epoxides form from alkenes with peracid. In the pinacol rearrangement an alkyl group migrates and the result has a C=O group.
- Cleaving the C–C bond is diol cleavage with NaIO₄. With acid the carbon skeleton stays together; only a methyl group migrates.
Often confused with: Acid-catalysed dehydration (E1)
Frequently asked questions
Which group migrates when the groups differ?
The cation forms where it is most stable. Then the group with the highest migratory aptitude moves, roughly aryl > H > alkyl.
Why does no alkene form?
The O-stabilised cation (C=OH⁺) is so much more stable that rearrangement is faster than elimination.