Organic Chemistry

The Favorskii Rearrangement

Shrink a ring by one carbon through a strained little cyclopropanone

The Favorskii rearrangement converts an α-halo ketone into a carboxylic acid, ester, or amide by base-induced ring contraction through a cyclopropanone intermediate. It shrinks a six-membered ring to a five-membered ester and is the classic proof that a symmetric cyclopropanone sits at the mechanistic heart.

  • First reported1894 (A. E. Favorskii)
  • Substrateα-halo ketone (Cl, Br)
  • Key intermediateCyclopropanone
  • Base / nucleophileRO⁻, HO⁻, R₂N⁻
  • ProductEster, acid, or amide
  • Ring changeContracts by one carbon

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What the Favorskii rearrangement does

Take a ketone that carries a halogen on the carbon next to the carbonyl — an α-halo ketone — and expose it to an alkoxide or hydroxide base. Instead of the simple substitution or elimination you might expect, the whole carbon skeleton reorganizes: the ketone disappears and a carboxylic acid derivative (an ester, acid, or amide) appears, with the carbon framework knitted together in a new way. When the α-halo ketone is cyclic, the ring loses one carbon from the loop and gains it as an exocyclic acyl group. A six-membered ring becomes a five-membered ring bearing a –COOR arm.

The engine behind all of this is a fleeting, highly strained three-membered ring ketone called a cyclopropanone. It forms, it gets attacked by a nucleophile, and it springs open — and that spring-open step is where the rearrangement happens.

    2-chlorocyclohexanone  ──NaOMe / MeOH──→  methyl cyclopentanecarboxylate  +  NaCl

         O                                            COOMe
         ‖                                             |
      ⌌──C──⌍                                        ⌌─CH─⌍
     (6-membered ring, C–Cl at α)      →         (5-membered ring)

The mechanism, arrow by arrow

The canonical mechanism has three moving parts. Watch which carbon the base deprotonates — it is not the one holding the halide.

  1. Deprotonate the α′-carbon. The base removes a proton from the carbon on the opposite side of the carbonyl from the halide — the α′-carbon. This works because that C–H is acidic (α to a C=O, pKa ≈ 20), and its conjugate base is a resonance-stabilized enolate. The lone pair now sits on the α′-carbon.
  2. Close the cyclopropanone. That α′-carbanion swings across the carbonyl and attacks the C–X carbon from the back side, in an intramolecular SN2 displacement. The halide leaves with the bonding electrons; a new C–C bond forms; and a three-membered cyclopropanone ring is born. This is the committed, rate-relevant step.
  3. Open the ring with a nucleophile. The cyclopropanone carbonyl is fantastically electrophilic (ring strain + the carbonyl). The nucleophile — methoxide, hydroxide, or an amide ion — adds to the carbonyl carbon to give a tetrahedral alkoxide. One of the two strained C–C ring bonds then breaks heterolytically, kicking the electrons onto whichever carbon gives the more stabilized carbanion. Protonation of that carbanion delivers the ester / acid / amide.
  Step 1  base pulls H off the α′-carbon:
            X–Cα–C(=O)–Cα′H     →     X–Cα–C(=O)–Cα′(⁻)   (enolate)

  Step 2  α′ carbanion displaces X⁻ (intramolecular SN2):
            X–Cα–C(=O)–Cα′(⁻)   →     [ cyclopropanone ]  +  X⁻
                                          Cα — Cα′
                                            \ /
                                             C=O

  Step 3  Nu⁻ opens the strained ring:
            cyclopropanone + Nu⁻ → Nu–C(–O⁻)(Cα)(Cα′)  →  bond cleaves →  Nu–C(=O)–CHR–R'
            (protonation)                                      the ester / acid / amide

The electron bookkeeping to keep straight: step 1 is an acid–base event (make the nucleophile), step 2 is a C–C bond-forming event with C–X bond cleavage (make the strained ring), and step 3 is addition–elimination on a carbonyl (open the strained ring). Nothing exotic — just three familiar moves stacked to give a non-obvious skeleton.

The symmetric intermediate and Loftfield's label

Here is the subtle part. In the parent case where the two ring carbons flanking the cyclopropanone carbonyl are equivalent, the cyclopropanone is symmetric — the nucleophile cannot tell the two C–C bonds apart, so it opens either one with equal probability. That symmetry is the reaction's fingerprint, and it was proven with isotopes.

In 1950 Robert Loftfield ran the Favorskii of 2-chlorocyclohexanone with a 14C label on the α-carbon (C-2, the carbon bearing the chlorine), then degraded the cyclopentanecarboxylic acid product carbon by carbon to locate the label. A naive "direct displacement" mechanism predicts the label ends up entirely on one specific carbon. What Loftfield actually found was that the 14C came out scrambled roughly 50:50 between the ring carbon bearing the carboxyl group and the ring carbon next to it. Only a symmetric intermediate — the cyclopropanone, with its two equivalent ring bonds — can scramble the label that way, because a nucleophile opens either equivalent C–C bond with equal probability. That single experiment cemented the cyclopropanone mechanism.

  * = ¹⁴C label starts on the α-carbon (C-2, the one that bore the Cl)

  symmetric cyclopropanone:      Ca*— Cb        (Ca, Cb = the two α-carbons,
                                    \  /           equivalent by symmetry;
                                     C=O           C = carbonyl carbon)

  Nu⁻ opens the C–Cb bond → Ca* becomes the ring carbon bearing COOH
  Nu⁻ opens the C–Ca bond → Cb becomes the ring carbon bearing COOH, Ca* the neighbor
  Because the two bonds are identical, both open ~equally
  → the label ends up ~1:1 on the ring carbon bearing COOH vs the ring carbon next to it
  → observed ~1:1 scrambling — the smoking gun for a symmetric intermediate

Reagents, base, and conditions

The Favorskii is forgiving but the base does double duty, so you choose it for the product you want:

  • Ester product: sodium or potassium alkoxide in the parent alcohol. NaOMe/MeOH gives the methyl ester; NaOEt/EtOH the ethyl ester. This is the most common laboratory version.
  • Carboxylic acid product: aqueous or alcoholic NaOH/KOH. The nucleophile is hydroxide, so ring-opening gives the carboxylate; acidify on workup to get the free acid.
  • Amide product: a secondary amine with a base, or the amide anion R₂N⁻, opens the cyclopropanone to the amide.
  • Loading and temperature: typically 1–3 equivalents of base, 0 °C to reflux of the alcohol solvent, minutes to a few hours.
  • Halide: α-chloro and α-bromo ketones are standard. The rate roughly follows the ease of the intramolecular displacement, but very good leaving groups (α-iodo) can lose out to plain SN2 substitution or elimination side reactions.

Two practical cautions. First, strong, non-nucleophilic bases that only deprotonate (no good nucleophile around) can stall the reaction at the cyclopropanone or divert it to enolate chemistry. Second, α,α′-dihalo ketones under Favorskii conditions are the standard entry to isolable cyclopropanones and to α,β-unsaturated products — a related but distinct reaction manifold.

Regioselectivity: which bond opens?

When the cyclopropanone is not symmetric — the two flanking carbons carry different substituents — the nucleophile still adds to the carbonyl, but the ring then opens to give the more stabilized carbanion. In practice that means the C–C bond breaks so as to place the developing negative charge on the more substituted or more anion-stabilizing carbon (benzylic, or bearing electron-withdrawing groups). The result is regioselective: you get predominantly the ester or acid whose α-carbon is the better carbanion.

This is why acyclic Favorskii reactions of unsymmetrical α-halo ketones can be steered. For a phenyl-substituted case, ring-opening favors the bond that puts charge on the benzylic carbon, funneling to a single major ester. Stereochemistry follows from the concerted, inversion-at-carbon SN2 closure in step 2 and the geometry of the cyclopropanone: the reaction is generally stereospecific, and the relative configuration of substituents on the two ring carbons is set during cyclization.

Favorskii vs neighboring rearrangements

FavorskiiSemibenzilic (quasi-Favorskii)Ramberg-Bäcklund
Substrateα-halo ketone with an α′-Hα-halo ketone with NO α′-Hα-halo sulfone
Base roleDeprotonate α′, then act as NuNu adds directly to C=ODeprotonate α, then close ring
Strained intermediateCyclopropanone (all C)None — a 1,2-shift insteadEpisulfone (C–C–SO₂)
Key stepIntramolecular SN2 closure1,2-alkyl migration to C–XSO₂ extrusion (retro-cheletropic)
ProductEster / acid / amideEster / acid (ring-contracted)Alkene (C=C)
Carbon countPreservedPreservedLost SO₂, forms new C=C
Isotope scramblingYes (symmetric cyclopropanone)NoN/A
Ring effectContracts ring by one CContracts ring by one CSame-size ring or acyclic alkene

Worked example: 2-chlorocyclohexanone → methyl cyclopentanecarboxylate

The textbook case, and a genuine ring contraction of a six-membered ring to a five-membered one.

    2-chlorocyclohexanone  +  NaOMe (1.5 eq)  ──MeOH, 0 → 25 °C, 1–2 h──→
                                                methyl cyclopentanecarboxylate  +  NaCl
  • Deprotonation. Methoxide removes an H from C-6 (the α′-carbon, across the ring from the C-2 chloride). Enolate forms; the carbanion character is on C-6.
  • Ring closure. C-6 attacks C-2 from behind the departing chloride. The new C-6–C-2 bond fuses a cyclopropanone onto the residual ring, giving bicyclo[3.1.0]hexan-6-one — a five-membered ring cis-fused to a cyclopropanone.
  • Ring-opening. Methoxide adds to the cyclopropanone carbonyl. The external C–C bond of the three-membered ring breaks, splitting off the carbonyl carbon as the exocyclic ester carbon and leaving a plain cyclopentane ring. Protonation gives methyl cyclopentanecarboxylate.
  • Net result. The ring shrank from six carbons to five, and the former carbonyl carbon became the ester carbon dangling off the ring. Isolated yields for well-behaved substrates run 50–80%.

Run the same substrate with aqueous NaOH instead of NaOMe and you get cyclopentanecarboxylic acid (after acidification) rather than its methyl ester — same skeleton, different nucleophile trapping the cyclopropanone.

Limitations and side reactions

  • No α′-hydrogen → no classical Favorskii. If the carbon opposite the halide has no removable proton, the cyclopropanone can't form. The reaction switches to the semibenzilic (quasi-Favorskii) route, or fails.
  • Competing substitution and elimination. The very C–X bond the mechanism relies on is also a target for direct SN2 by the base, and the α′-enolate can eliminate to an enone. Milder, less nucleophilic bases and controlled temperature bias toward the rearrangement.
  • Over-strained closures. Cyclopropanone formation across a large or geometrically awkward ring can be slow; entropy and ring geometry both matter for step 2.
  • Sensitive nucleophiles. Because the base is also the nucleophile that opens the cyclopropanone, functional groups that react with strong alkoxide/hydroxide (esters elsewhere in the molecule, base-labile protecting groups) can be casualties.
  • α,α′-dihalo ketones behave differently. They lead toward isolable cyclopropanones or, on double dehydrohalogenation, to α,β-unsaturated products — plan around this if a second halide is present.

Who discovered it, and when

The reaction is named for Alexei Yevgrafovich Favorskii (1860–1945), a foundational figure of Russian organic chemistry and a student of Aleksandr Butlerov's school in St. Petersburg. Favorskii first reported the rearrangement of α-halo ketones with base in 1894, and he and his students developed the chemistry over the following two decades. Favorskii's broader legacy includes pioneering acetylene chemistry (the Favorskii and Favorskii–Babayan reactions of terminal alkynes with ketones).

The mechanism itself was contested for decades — proposals ranged from direct displacement to various concerted shifts. It was Robert B. Loftfield's 1950 14C-labeling study at MIT that decisively established the symmetric cyclopropanone intermediate, turning a named curiosity into a textbook exemplar of how isotope labels reveal hidden symmetry in a mechanism. The non-enolizable variant was clarified separately and is often called the semibenzilic or quasi-Favorskii rearrangement, by analogy to the benzilic acid rearrangement it resembles.

Where it earns its keep in synthesis

  • Ring contraction on demand. The Favorskii is one of the cleanest ways to shrink a ring by exactly one carbon while installing a carboxylic acid handle — invaluable when a five-membered carbocycle bearing a –COOH is easier to reach from a six-membered ketone than by direct construction.
  • Strained and small rings. It has been used to build cyclobutane- and cyclopropanecarboxylic acids from the corresponding α-halo cyclopentanones and cyclobutanones, giving access to strained carboxylic acids that are otherwise awkward to make.
  • The Grewe route to morphinan-type frameworks and various terpenoid and steroid syntheses have exploited Favorskii ring contractions to reshape a carbocyclic core at a key stage.
  • Cubane synthesis. Philip Eaton's landmark synthesis of cubane (1964) uses a Favorskii ring contraction to forge one of the strained C–C bonds of the cube — a showcase of the reaction building geometry that looks almost impossible.

The through-line: whenever a target needs a carboxylic acid derivative sitting on a ring that is one carbon smaller than the ketone you can easily make, the Favorskii rearrangement is the reaction to reach for.

Frequently asked questions

What is the intermediate in the Favorskii rearrangement?

A cyclopropanone. Base removes the acidic proton on the α′-carbon (the carbon on the far side of the carbonyl, not the one bearing the halide). The resulting enolate carbon attacks the C–X carbon intramolecularly in an SN2-like closure, expelling the halide and forming a strained three-membered cyclopropanone ring. A nucleophile then opens this cyclopropanone to give the rearranged product.

How does the Favorskii rearrangement contract a ring?

Cyclization of the α′-enolate onto the C–X carbon builds a new C–C bond that is one carbon shorter across the ring than the old C(=O)–CαX linkage it replaces. When you start from 2-chlorocyclohexanone, the new bond fuses a bicyclic cyclopropanone onto a five-membered ring; nucleophilic ring-opening then discards the extra carbon as the exocyclic ester or acid group, leaving a cyclopentane ring. Net: a six-membered ring becomes a five-membered ring bearing a carboxylic acid derivative.

What is the evidence for the symmetric cyclopropanone intermediate?

The classic isotope-labeling experiment of Loftfield (1950). He ran the Favorskii rearrangement of 2-chlorocyclohexanone with the α-carbon (C-2, the carbon bearing the chlorine) labeled as ¹⁴C, then degraded the product cyclopentanecarboxylic acid carbon by carbon. If the mechanism were a direct, unsymmetrical displacement, the label would appear at only one specific carbon. Instead the ¹⁴C was scrambled roughly 50:50 between the ring carbon bearing the carboxyl group and the ring carbon next to it — exactly what a symmetric cyclopropanone predicts, because a nucleophile can open either of its two equivalent C–C bonds with equal probability.

What happens when there is no α′-hydrogen (the quasi-Favorskii)?

You cannot form the cyclopropanone, so the reaction diverts to the semibenzilic (quasi-Favorskii) pathway. Here the nucleophile adds directly to the ketone carbonyl to give a tetrahedral alkoxide, and then the C–C bond migrates onto the C–X carbon with displacement of halide — a 1,2-shift analogous to the benzilic acid rearrangement. This still contracts the ring, but it goes through no cyclopropanone and gives no ¹⁴C scrambling.

What base and conditions does the Favorskii need?

An alkoxide or hydroxide, matched to the nucleophile you want in the product. Sodium or potassium methoxide/ethoxide in the parent alcohol gives the methyl or ethyl ester; aqueous NaOH gives the carboxylic acid (as its salt) after acidic workup; an amide (R₂NH plus base) gives the amide. Typical runs use 1–3 equivalents of base in the alcohol solvent at 0 °C to reflux. The α-halide is usually chloride or bromide; iodides sometimes favor simple substitution over rearrangement.

How is the Favorskii different from the Ramberg-Bäcklund reaction?

Both use base to close a strained three-membered ring from an α-halo carbonyl-type substrate, but the ring is different. Favorskii closes a cyclopropanone (all carbon) from an α-halo ketone and ends in a carboxylic acid derivative. Ramberg-Bäcklund closes a three-membered episulfone from an α-halo sulfone, then extrudes SO₂ to give an alkene. Favorskii makes a C–C bond and keeps the carbon count; Ramberg-Bäcklund loses SO₂ and makes a C=C double bond.