VKC Notes · Reaction mechanisms · Civil Services P-II · IFoS P-II · Geo-Scientist P-III

2.6 Molecular Rearrangements

★★★ Asked in every paper of all three exams, 2021–2025: 34 parts, 265 marks (IFoS 14 parts / 120 m · Civil Services 9 / 85 m · Geo-Scientist 11 / 60 m).

Syllabus lines. Civil Services and IFoS: “Pinacol-pinacolone, Hofmann, Beckmann, Baeyer–Villiger, Favorskii, Fries, Claisen, Cope, Stevens and Wagner–Meerwein rearrangements”. Geo-Scientist: “carbenes and nitrenes and their rearrangements (Reimer–Tiemann, Hofmann, Curtius, Lossen and Schmidt)” and “benzil–benzilic acid rearrangement”. Claisen and Cope are [3,3] shifts: see the pericyclic notes.

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IFoS 2024 Q3(b) = asked there; tap to open the solved answerSN = line kept in the short notes

2.6.1 The common idea: a 1,2-shift to an electron-deficient atom ★★★

Almost every rearrangement in the syllabus is the same event. An atom becomes electron-deficient (a carbocation, a nitrogen losing a leaving group, an oxygen losing a carboxylate), and a group on the neighbouring atom swings across to it.SN

Fig. 2.6.1 One pattern, three acceptor atoms.

Draw it in 60 s: write R–C–X three times with X = C⁺, N, O → one curved arrow from the C–R bond to X → product C–X–R → name the reactions under each row.

The group migrates with its bonding pair (a nucleophilic or anionotropic 1,2-shift). It never leaves completely, so a chiral migrating group keeps its configuration.SN
The migrating group and the leaving group are usually anti-periplanar, the same geometric demand as in E2. In Beckmann this is absolute: only the group anti to the OH moves.
Driving force: a more stable cation (2° → 3°, alkyl → benzylic), a C=O or C=N bond formed, or ring strain released (cyclobutyl → cyclopentyl).SN

Subtle point · why nucleophilic shifts dominate IFoS 2023 Q1(f) 5m

The migration passes through a three-centre transition state. With 2 electrons (shift to a cation) it is a bonding, Hückel-aromatic arrangement and is easy. With 4 electrons (shift to a carbanion, electrophilic shift) the arrangement is antibonding; with 3 (radical) it is in between. So 1,2-shifts to carbanions are rare, to radicals uncommon, to cations everywhere.SN The Stevens rearrangement looks like a shift to a carbanion, and that is exactly why it goes through a radical pair instead (2.6.9).

Fig. 2.6.2 Migratory aptitude, highest first.

Draw it in 60 s: vertical arrow labelled “migrates” → list H, 3° alkyl, 2° and cyclohexyl, aryl, 1° alkyl, CH₃ from top → put H in red.

Aptitude reflects how well the group carries positive charge in the transition state: tertiary alkyls and electron-rich aryls migrate best, methyl worst. p-MeO-C₆H₄ > C₆H₅ > p-NO₂-C₆H₄.SN

2.6.2 Wagner–Meerwein and other carbocation shifts ★★★

Whenever a carbocation forms (S_N1, E1, addition of HX to an alkene, dehydration), a 1,2-H or 1,2-alkyl shift will occur if it gives a more stable cation. Expect it with neopentyl, 2° next to 3°, and small rings next to C⁺.SN
Neopentyl: (CH₃)₃C–CH₂–X ionises only slowly, and the 1° cation (or the departing system) takes a methyl shift to the 3° tert-amyl cation. Products: 2-methylbut-2-ene and tert-amyl ethers or alcohols, not neopentyl ones. See the daily answer of 20 Oct 2026.
Bicyclic terpenes: isoborneol or camphene hydrochloride in acid → camphene, by an alkyl (ring bond) shift. Asked as IFoS 2022 Q2(c) 10m. A methyl shift in the same family gives santene CSE 2024 Q4(c)(i).
Ring expansion: a C⁺ next to a cyclobutane takes a ring bond and becomes a cyclopentyl cation: cyclobutane carries about 110 kJ mol⁻¹ of ring strain, cyclopentane only about 26.SN

Fig. 2.6.3 Wagner–Meerwein ring expansion: 1-methyl-1-vinylcyclobutane + HCl.

Draw it in 60 s: alkene + H⁺ (arrow from C=C to H⁺) → 2° cation beside the four-ring → arrow from the ring C–C bond to C⁺ → five-ring 3° cation → Cl⁻ adds.

Solved PYQ Geo 2022 Q6(d) 4m

Write the mechanism: 1-methyl-1-vinylcyclobutane + HCl.

H⁺ adds to the terminal CH₂ (Markovnikov): a 2° cation, CH₃–C⁺H–, next to the ring carbon C1.
A C1–C2 ring bond shifts to the cation carbon: the four-membered ring becomes five-membered and the charge moves to C1, now a 3° carbon bearing CH₃.
Cl⁻ adds: 1-chloro-1,2-dimethylcyclopentane.

Two reasons to write: 2° → 3° cation, and ring strain relieved (cyclobutane → cyclopentane).

Also asked: IFoS 2025 Q2(b) 10m dehydration of 2-cyclohexylpropan-2-ol with BF₃: direct alkenes plus a ring-expansion channel.

2.6.3 Pinacol–pinacolone and its relatives ★★★ ring expansion asked again and again

Fig. 2.6.4 Pinacol rearrangement of Ph₂C(OH)–C(OH)Me₂ in cold concentrated H₂SO₄.

Draw it in 60 s: diol → H⁺ on the OH next to the Ph₂C → arrow from C–OH₂⁺ bond to O → Ph₂C⁺ → arrow from the C–Me bond to C⁺ → OH lone pair to the C–O bond → −H⁺ → ketone.

Which OH leaves? The one that leaves the more stable cation (Ph₂C⁺ before Me₂C⁺).SN
Which group moves? From the other carbon, by migratory aptitude, but conformation can override it (anti-periplanar group moves). The cation left behind is stabilised by OH (an oxocarbenium ion), which is the driving force.
Ring expansion: when the cation sits next to a ring, a ring bond migrates. 1-(hydroxydiphenylmethyl)cyclopentanol → 2,2-diphenylcyclohexanone.SN

Solved PYQ CSE 2021 Q3(a)(i) 10m

1-(Hydroxydiphenylmethyl)cyclopentanol with conc. H₂SO₄: product and mechanism.

The OH on the CPh₂ carbon leaves (Ph₂C⁺ is benzylic, doubly stabilised).
From the ring carbon, a ring C–C bond migrates to Ph₂C⁺: the five-membered ring takes in that carbon and becomes six-membered.
The charge lands on the old carbinol carbon, stabilised by its OH; loss of H⁺ gives the ketone: 2,2-diphenylcyclohexanone.

The same move with a cyclopentyl–cyclopentyl diol gives spiro[4.5]decan-6-one CSE 2024 Q4(c)(i).

Subtle point · conditions can change the product IFoS 2025 Q1(f) 5m

Me₂C(OH)–C(OH)Ph₂ in cold concentrated H₂SO₄ ionises at the CPh₂ end and a methyl moves: MeCOC(Me)Ph₂. In acetic acid with a little H₂SO₄ the product is PhCOCMe₂Ph, from phenyl migration. The answer earns marks only if it explains both: first the more stable cation, then the better migrating group.

Relatives worth one line each

Semipinacol / Tiffeneau–Demjanov: a β-amino alcohol + HNO₂ makes a diazonium leaving group. 1-(aminomethyl)cyclopentanol → cyclohexanone: one-carbon ring expansion IFoS 2023 Q4(e).SN
Epoxide → carbonyl (Meinwald): a Lewis acid (BF₃) opens the epoxide to the more stable cation and a hydride or alkyl shifts IFoS 2023 Q8(a) 10m.

Also asked: CSE 2022 Q1(d) 10m two phenyl-substituted diols.

2.6.4 Benzil–benzilic acid rearrangement ★★ Geo favourite

A 1,2-diketone + OH⁻ → α-hydroxy carboxylate. .SN
Mechanism: OH⁻ adds to one C=O; the alkoxide re-forms C=O and pushes a phenyl onto the neighbouring carbonyl carbon; proton transfer gives the stable carboxylate (irreversible step).SN
Here the group migrates to a carbonyl carbon, not a cation: the anion pushes, the C=O pulls. Rate = k[diketone][OH⁻].
Cyclic 1,2-diones contract: cyclohexane-1,2-dione → 1-hydroxycyclopentane-1-carboxylate Geo 2021 Q8(c). Furil → furilic acid Geo 2025 Q7(b)(ii).SN
With an alkoxide instead of OH⁻ the product is the ester (benzilic ester rearrangement). Alkoxides that can be oxidised (EtO⁻) reduce the diketone instead, so use MeO⁻ or t-BuO⁻.

Fig. 2.6.5 Benzil–benzilic acid rearrangement.

Draw it in 60 s: benzil + HO⁻ (arrow to one C=O carbon, C=O → O) → tetrahedral alkoxide → three arrows: O⁻ to C–O, C–Ph bond to the next C, C=O → O → acid alkoxide → H⁺ transfer → benzilate.

2.6.5 Favorskii rearrangement ★★★ most asked · 9 parts

Fig. 2.6.6 Favorskii ring contraction through a cyclopropanone.

Draw it in 60 s: chloroketone → MeO⁻ takes the α′-H → enolate → arrow from C=C to the C–Cl carbon, C–Cl → Cl → cyclopropanone → MeO⁻ adds → O⁻ re-forms C=O, ring C–C bond → C → five-ring ester → write the opening rule underneath.

Base removes an α′-H (on the side away from Cl): enolate.
The enolate carbon displaces Cl⁻ intramolecularly (S_N2, so inversion at the C–Cl carbon): a cyclopropanone.SN
RO⁻ adds to the strained C=O; the ring opens by breaking the C–C bond that leaves the more stable carbanion, which is protonated.SN
Result: an ester (RO⁻) or acid (OH⁻) or amide (amine) with a rearranged skeleton. α-Halocyclic ketones contract by one carbon.SN

Solved PYQ CSE 2023 Q1(e)(i) 5m

2-Chloro-6-methylcyclohexanone + MeO⁻: major product?

The α′-H at C6 (the CH bearing CH₃) goes; C6 attacks C2: a bicyclic cyclopropanone whose bridging carbons are C2 (H) and C6 (CH₃).
MeO⁻ adds to C=O. Breaking the bond to C2 gives a 2° carbanion; breaking to C6 would give a 3° one. The 2° (more stable) carbanion forms.
Product: methyl 1-methylcyclopentane-1-carboxylate. The ester carbon stays on the methyl-bearing carbon.

Subtle point · two substrates, one product IFoS 2023 Q4(g) 10m CSE 2024 Q8(b)(i) 10m

ClCH₂COCH₂Ph and PhCHClCOCH₃ form the same phenylcyclopropanone, which opens towards the benzylic carbanion: both give methyl 3-phenylpropanoate. This is the classic proof of the symmetrical intermediate. Likewise 2,2-dimethylcyclopropanone + MeO⁻ opens to the CH₂⁻ side and gives methyl pivalate, (CH₃)₃CCOOCH₃ (NMR: 9H and 3H singlets).

No α′-H? The quasi-Favorskii (semibenzilic) path: RO⁻ adds to C=O and the alkyl group migrates with loss of Cl⁻, a benzilic-type shift.
α,α′- or α,β-dihalo ketones give α,β-unsaturated esters: the second halide is lost after ring opening IFoS 2022 Q2(b) IFoS 2025 Q2(c). Pulegone dibromide → methyl pulegenate IFoS 2024 Q1(f).

Also asked: CSE 2021 Q3(a)(ii) 10m with the inversion · CSE 2025 Q3(a)(i) ring contraction.

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