(a) Ionization of 3,4-dichloro-1,2,3,4-tetramethylcyclobutene in SbF₅-SO₂ at –75 °C produces a reaction intermediate of its own kind. Predict the intermediate and comment on its stability and aromaticity. (10 marks)

(b) Predict the structures of X a

Question

(a) Ionization of 3,4-dichloro-1,2,3,4-tetramethylcyclobutene in SbF₅-SO₂ at –75 °C produces a reaction intermediate of its own kind. Predict the intermediate and comment on its stability and aromaticity. (10 marks)

(b) Predict the structures of X and Y, and also mention the major product: [figure: single enantiomer of 2-(4-nitrophenyl)oxirane, MeO⁻/MeOH → X + Y] (10 marks)

(c) Out of the following list of reactants along with the reagents, identify the reaction which is not a condensation reaction. Propose a suitable mechanism for the products formed in this reaction: (10 marks)
(i) Acetaldehyde is reacted with dilute KOH
(ii) Benzaldehyde is reacted with acetic anhydride in presence of sodium acetate
(iii) Benzaldehyde is reacted with malonic ester in presence of a base
(iv) Benzaldehyde is reacted with concentrated KOH

(d) Write down the structure(s) of the product(s) obtained in the following reactions. Provide suitable justification and propose the mechanisms: (10 marks)
(i) C₆H₅–C(C₆H₅)(OH)–C(CH₃)(OH)–CH₃ —H⁺, Δ→
(ii) C₆H₅–C(CH₃)(OH)–C(CH₃)(OH)–C₆H₅ —H⁺, Δ→

(e) Predict the product(s) formed on heating the cyclopentadiene and provide a suitable justification to your answer. (10 marks)

UPSC Answer Check · Accepted Answer

(a) Tetramethylcyclobutadiene dication

Both C–Cl bonds ionise. The intermediate is the 1,2,3,4-tetramethylcyclobutadiene dication.

Four contiguous p orbitals, 2 π electrons (the original double bond; the two cationic carbons contribute empty p orbitals). Hückel 4n+2 with n=0: aromatic. Neutral cyclobutadiene is the 4π antiaromatic contrast.

Four equivalent resonance forms spread the two positive charges. That closed-shell 2π configuration is why the ion survives at −75 °C in this superacid (Olah, 1969, characterised by NMR).

(b) Epoxide opening by methoxide: X, Y and the major product

The substrate is the single enantiomer of 4-nitrostyrene oxide drawn in the paper. MeO⁻ in MeOH is a strong nucleophile under basic conditions, so the ring opens by SN2 attack on a neutral epoxide, with backside approach and inversion at the carbon attacked.
X (major) = ArCH(OH)CH₂OMe, 2-methoxy-1-(4-nitrophenyl)ethanol, from attack at the unhindered CH₂. No bond to the stereocentre is broken, so its configuration is retained.
Y (minor) = ArCH(OMe)CH₂OH, 2-methoxy-2-(4-nitrophenyl)ethanol, from attack at the benzylic carbon with inversion of configuration.

Steric control decides the regiochemistry in base. The p-NO₂ group withdraws electron density and destabilises any positive charge at the benzylic carbon, so the SN1-like benzylic bias seen under acid catalysis is weakened further, and attack at CH₂ dominates. Mechanism: MeO⁻ attacks CH₂ anti to the C–O bond, the C–O bond breaks to give an alkoxide, and MeOH protonates it, regenerating MeO⁻.

(c) The non-condensation reaction: Cannizzaro

(i) Aldol (acetaldehyde + dil. KOH).
(ii) Perkin (PhCHO + Ac₂O/NaOAc).
(iii) Knoevenagel (PhCHO + malonic ester / base).

Not a condensation: (iv) PhCHO + conc. KOH = Cannizzaro.

Mechanism. OH⁻ adds to one aldehyde → PhCH(OH)O⁻. That tetrahedral anion transfers hydride to a second PhCHO. The donor is oxidised to benzoate; the acceptor is reduced to PhCH₂O⁻. Proton inventory gives benzyl alcohol + potassium benzoate. No small molecule is eliminated.

(d) Pinacol rearrangement of the two diols

Both substrates are vicinal diols that undergo the pinacol rearrangement with H⁺ and heat. The mechanism is: protonation of the OH whose loss gives the more stable carbocation, loss of water, a 1,2-shift from the neighbouring carbinol carbon to the cation, and loss of H⁺ from the resulting oxocarbenium ion to give the ketone.

(i) Ph₂C(OH)–C(OH)Me₂ gives 3,3-diphenylbutan-2-one, Ph₂C(Me)COCH₃. The OH on the diphenyl carbon leaves because Ph₂C⁺ is a highly stabilised benzhydryl cation. Only methyl groups are available on the neighbouring carbon, so a methyl shift occurs and the remaining C–OH becomes the C=O.

(ii) Ph(Me)C(OH)–C(OH)(Me)Ph also gives 3,3-diphenylbutan-2-one. The diol is symmetrical, so either OH leaves to give the benzylic cation PhC⁺(Me). The neighbouring carbon carries Ph and Me, and phenyl migrates in preference to methyl (higher migratory aptitude, through a bridged phenonium-type transition state). The product is Ph₂C(Me)COCH₃. A methyl shift would give the minor ketone 2-methyl-1,2-diphenylpropan-1-one, PhC(Me)₂COPh.

Both diols therefore converge on the same ketone, one by Me migration to the more stable Ph₂C⁺ and the other by Ph migration to PhC⁺(Me).

(e) Heating cyclopentadiene

Thermal [4+2] Diels–Alder dimerisation → endo-dicyclopentadiene (tricyclo[5.2.1.0²,6]deca-3,8-diene). One molecule is the diene, one the dienophile. Suprafacial–suprafacial HOMO–LUMO match (Woodward–Hoffmann). Endo is kinetic (secondary orbital interactions). The dimer is what is bottled; cracking at about 170 °C returns the monomer.

(a) Ionization of 3,4-dichloro-1,2,3,4-tetramethylcyclobutene in SbF₅-SO₂ at –75 °C produces a reaction intermediate of its own…

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(a) Tetramethylcyclobutadiene dication

(b) Epoxide opening by methoxide: X, Y and the major product

(c) The non-condensation reaction: Cannizzaro

(d) Pinacol rearrangement of the two diols

(e) Heating cyclopentadiene

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