(a) (i) Consider the following reaction : furan + (CH₃CO)₂O →(BF₃, CH₃COOH) X. Name the product X and outline the mechanism…

4. (a) (i) Furan + (CH₃CO)₂O / BF₃ → X

• X = 2-acetylfuran (1-(furan-2-yl)ethan-1-one), formed by Friedel–Crafts acylation, an electrophilic aromatic substitution.
• Step 1 (fast, pre-equilibrium): BF₃ coordinates to a carbonyl oxygen of acetic anhydride. This polarises the anhydride and releases the acylium ion CH₃–C≡O⁺, or delivers it as the Ac₂O·BF₃ complex, together with CH₃CO₂BF₃⁻.

Step 2 (slow, rate-determining): the π system of furan attacks the electrophile at C-2, giving a σ-complex (Wheland intermediate).

• Aromaticity is lost and a cation forms, so this step has the highest barrier.
• The cation is stabilised by three resonance forms: + on C-3, + on C-5, and an oxonium ion with + on O and every atom having an octet.
• Step 3 (fast): acetate (or CH₃CO₂BF₃⁻) removes H⁺ from C-2, restoring the aromatic sextet and giving 2-acetylfuran.
• Why C-2 and not C-3: attack at C-3 gives a σ-complex with only two resonance forms, because the charge cannot reach C-5. Its transition state is therefore higher in energy.
• Why BF₃: furan is π-excessive (the O lone pair is part of the sextet), so it reacts under mild conditions. Strong Lewis acids such as AlCl₃ cause ring opening and polymerisation, which is why BF₃ is used.

(a) (ii) Products and mechanisms

(1) (CH₃)₃CBr + C₂H₅OH, 60 °C: SN1/E1

• Slow ionisation gives the planar tertiary carbocation (CH₃)₃C⁺. The rate law is rate = k[RBr].
• Ethanol, a weak base and weak nucleophile, then either removes a β-H, giving 2-methylpropene (E1, major on heating), or bonds to C⁺, giving tert-butyl ethyl ether (SN1, minor).
• Heat favours elimination because it produces more particles (ΔS > 0).
• SN2 is impossible at a crowded tertiary carbon.

(2) CH₃CH=CHCl + NaNH₂: E2

• The very strong base NH₂⁻ removes the vinylic H on C-2 while Cl⁻ leaves from C-1 in one concerted step, giving propyne, CH₃C≡CH.
• The Z-isomer, which has H anti to Cl, eliminates fastest.
• SN1 and SN2 are not possible: the C–Cl bond is on an sp² carbon, has partial double-bond character, backside attack is blocked, and a vinyl cation would be very unstable.
• Because propyne (pKa ≈ 25) is more acidic than NH₃ (≈ 38), excess NaNH₂ converts it to sodium propynide. Aqueous work-up then gives propyne.

(b) (i) Claisen rearrangement: the crossover test

The reaction: heating allyl phenyl ether (about 200 °C) gives 2-allylphenol.

• It proceeds by a [3,3]-sigmatropic shift through a six-membered chair transition state to 6-allylcyclohexa-2,4-dienone.
• This dienone tautomerises to the aromatic phenol.

Design of the experiment: heat an equimolar mixture of two ethers that are labelled in both halves, for example:

• C₆H₅–O–CH₂CH=CH₂ (unlabelled), and
• p-CH₃C₆H₄–O–CD₂CD=CD₂ (ring methyl plus a d₅-allyl group).

Predictions:

• If the reaction is intramolecular, each allyl group stays on its own ring, giving only 2-allylphenol and 4-methyl-2-(allyl-d₅)phenol.
• If it were intermolecular, for example via free allyl radicals, cations or ion pairs, the fragments would mix. The crossover products 2-(allyl-d₅)phenol and 4-methyl-2-allylphenol would also form, giving four phenols in total.
• Analysis: mass spectrometry (the m/z of each phenol) or ²H NMR. Only the two non-crossover products are found, which confirms the rearrangement is intramolecular.

Supporting evidence:

• With PhO–CH₂CH=¹⁴CH₂, the ¹⁴C (the γ-carbon) ends up bonded to the ring. The allyl group is inverted, exactly as a cyclic [3,3] transition state requires.
• The reaction follows first-order kinetics and has a large negative ΔS‡ (ordered cyclic transition state).

(b) (ii) Energy profile: C₆H₆ + Cl₂/FeCl₃ → C₆H₅Cl + HCl

The profile has two humps with one valley.

• Reactants: C₆H₆ + Cl–Cl···FeCl₃. FeCl₃ polarises Cl₂ into Cl^δ+···Cl–FeCl₃^δ−.
• TS1 (highest point): the ring C–Cl bond is forming and the Cl–Cl bond is breaking. Positive charge builds up in the ring and the π-sextet is being disrupted.
• Intermediate: the σ-complex (arenium ion) with an sp³ C bearing H and Cl, its + charge spread over the other five carbons, paired with FeCl₄⁻. It is a true energy minimum.
• TS2 (lower): FeCl₄⁻ removes the H⁺ (C–H bond breaking) as aromaticity returns.
• Products: C₆H₅Cl + HCl + FeCl₃. The overall reaction is exothermic.

The rate-determining step is step 1 (Eₐ1 ≫ Eₐ2), because aromaticity is lost there.

• By the Hammond postulate, TS1 resembles the σ-complex.
• There is no primary kinetic isotope effect (k_H/k_D ≈ 1), which confirms that C–H breaking is not rate-determining.

(c) (i) Major products

(A) 1,1′-Bi(cyclopentyl)-1,1′-diol + conc. H₂SO₄: pinacol rearrangement with ring expansion → spiro[4.5]decan-6-one

• One OH is protonated and lost as water, giving a tertiary carbocation.
• A C–C bond of the other (carbinol-bearing) cyclopentane ring migrates to the cation. This enlarges that ring from 5 to 6 members, relieving strain.
• The positive charge moves onto the carbon bearing OH, becoming an oxocarbenium ion. Loss of H⁺ gives the cyclohexanone.
• The spiro centre joins the cyclopentane and the new cyclohexanone.
• Simple dehydration to a diene is not observed under these conditions.

(B) 3-Methylbutan-2-one: NH₂OH, then Al₂O₃/heat: Beckmann rearrangement → N-isopropylacetamide, CH₃CONH–CH(CH₃)₂

• The ketone forms mainly the E-oxime, in which the OH lies anti to the bulkier isopropyl group.
• When the OH (activated by the Lewis acid) leaves, the group anti to it migrates to N in a concerted step, giving the nitrilium ion CH₃–C≡N⁺–CH(CH₃)₂.
• Water adds and tautomerisation gives the amide.
• The minor Z-oxime gives N-methylisobutyramide, (CH₃)₂CH–CO–NHCH₃.

(C) Camphenilol (3,3-dimethylbicyclo[2.2.1]heptan-2-ol) + H⁺ → santene (2,3-dimethylbicyclo[2.2.1]hept-2-ene)

• Protonation and loss of water give the secondary 2-norbornyl cation next to the gem-dimethyl carbon.
• A 1,2-methyl shift (a Wagner–Meerwein shift) from C-3 to C-2 converts it into the more stable tertiary cation at C-3.
• Loss of the C-2 proton gives the tetrasubstituted C2=C3 alkene. It is not at a bridgehead, so Bredt's rule is satisfied.
• Dehydrating camphenilol (for example with KHSO₄) is a known preparation of santene.

(c) (ii) PhCOCF₂Cl + Ac₂O / AcONa, heat: Perkin condensation

• Product: (E)-4-chloro-4,4-difluoro-3-phenylbut-2-enoic acid, Ph(ClF₂C)C=CH–COOH.
• Why this ketone reacts: ordinary ketones fail in the Perkin reaction. The strongly electron-withdrawing CF₂Cl group, however, makes this carbonyl as electrophilic as an aldehyde's. The ketone has no α-H, so it cannot self-condense.

Steps:

• Enolate formation: AcO⁻ (from AcONa) removes an α-H from acetic anhydride, giving ⁻CH₂–CO–O–COCH₃.
• Aldol-type addition: the enolate adds to the ketone carbonyl, giving a tertiary alkoxide.
• Acetylation: intramolecular acyl transfer (or reaction with Ac₂O) converts the alkoxide to an acetate, leaving a mixed anhydride.
• Elimination: E1cB loss of AcOH, through the enolate of the mixed anhydride, forms the conjugated C=C.
• Hydrolysis: on aqueous work-up, the mixed anhydride hydrolyses to the free carboxylic acid.
• Geometry: Perkin condensations of 2,2,2-trifluoroacetophenone give 3-phenyl-4,4,4-trifluorobut-2-enoic acids with high E-selectivity (CF₃ trans to CO₂H). By analogy, the E-acid (CF₂Cl trans to COOH) is expected to predominate here, possibly with some Z.
• Not a Darzens reaction (there is no α-halo-ester enolate and no epoxide forms), and not a Favorskii reaction.