Beckmann rearrangement: push the electrons yourself
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Advanced Organic Chemistry · Free interactive demo
Hot concentrated sulfuric acid turns cyclohexanone oxime into ε-caprolactam, the monomer of nylon 6. The acid protonates the oxime OH; a ring C–C bond then shifts onto nitrogen as water leaves, so nitrogen ends up inside a seven-membered ring as a nitrilium ion. Water captures that cation, and two proton transfers turn the imidic acid into the lactam.
Draw all 15 arrows across six reaction steps above, with one question along the way. Each arrow is checked as you draw it, and Hint reveals the next move a little at a time.
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The Beckmann rearrangement mechanism, step by step
The Beckmann rearrangement converts a ketoxime into an amide by moving a carbon group from carbon to nitrogen. It is stereospecific: the group anti to the leaving group on nitrogen migrates, because its σ bond lines up with the σ* orbital of the breaking N–O bond. Cyclohexanone oxime is symmetric, so either ring bond gives the same lactam; in an unsymmetrical oxime the geometry decides the product. The migration and the loss of water happen together, so no free nitrenium ion forms.
The 8 steps of the demo, in the exercise's own words. Try drawing each one above before you read it.
Acid protonates the oxime hydroxyl
Hydroxide is a poor leaving group, so sulfuric acid first protonates the oxime oxygen. Water attached to nitrogen is now an excellent leaving group.
A ring bond migrates to nitrogen as water leaves
The ring C–C bond anti-periplanar to the N–O bond shifts onto nitrogen at the same moment that water departs. The nitrogen lone pair forms a third bond to the former oxime carbon, so the cation is a nitrilium ion and nitrogen is now inside a seven-membered ring.
Which bond moves?
The Beckmann rearrangement is stereospecific: the group anti (trans) to the leaving group on nitrogen migrates, because its σ bond lines up with the σ* orbital of the N–O bond. In cyclohexanone oxime both ring bonds give the same lactam, but in an unsymmetrical oxime the geometry decides the product.
Water captures the nitrilium ion
The water released in the migration adds to the electrophilic nitrilium carbon. One π pair moves onto nitrogen, leaving an oxonium ion on a C=N carbon.
Hydrogen sulfate removes a proton
Hydrogen sulfate deprotonates the oxonium ion, giving the neutral imidic acid, the tautomer of the lactam.
Acid protonates the imine nitrogen
Tautomerisation starts with protonation of the basic C=N nitrogen by sulfuric acid, giving a cation that is also an O-protonated amide.
Loss of the O–H proton gives the lactam
Hydrogen sulfate removes the O–H proton. The oxygen lone pair forms the C=O bond and the C=N π pair moves onto nitrogen, giving the amide.
ε-Caprolactam
Nitrogen has been inserted into the ring next to the former oxime carbon, turning the six-membered ketone skeleton into a seven-membered cyclic amide, the monomer of nylon 6.
How the mechanism is drawn
H2SO4 (conc.), 100–120 °C; then dilute with water and neutralise with aqueous NH3