(a) (i) Classify the following as aromatic, nonaromatic or antiaromatic : (1) Azulene (2) Pyridine (3) Sydnone (4)…

(a)(i)

A compound is aromatic if it is cyclic, planar and fully conjugated with (4n + 2) π electrons. It is antiaromatic if it meets the same conditions but has 4n π electrons. It is nonaromatic if it cannot keep a planar, fully conjugated ring.

1. Azulene: aromatic.

• It has 10 π electrons on the periphery (4n + 2, n = 2) and is planar.
• Its dipolar form, a 6π cyclopentadienide fused to a 6π tropylium, explains its dipole moment (≈ 1.0 D) and blue colour.

2. Pyridine: aromatic.

• It has 6 π electrons. N contributes one p electron, and its lone pair lies in an sp² orbital in the ring plane, outside the π system.

3. Sydnone: aromatic (mesoionic).

• It has 6 π electrons delocalised over the planar five-membered O–N–N–C–C ring.
• It can be written only with separated charges, but it is a stable 6π heteroaromatic that undergoes electrophilic substitution at C-4.

4. Cyclooctatetraene: nonaromatic.

• It has 8 π electrons (4n), but it adopts a tub shape with alternating single and double bonds.
• It escapes antiaromaticity by bending and reacts as a polyene.

5. Cyclopentadienyl cation: antiaromatic.

• It has 4 π electrons (4n, n = 1) in a planar, fully conjugated ring.
• It is extremely unstable and very hard to generate, even from good leaving groups.

(a)(ii)

The compound is diphenylcyclopropenone. Its carbonyl is strongly polarised, and the polar resonance form C=O ↔ ⁺C–O⁻ turns the three-membered ring into a cyclopropenium ion. That ion has 2 π electrons (4n + 2 with n = 0) and is aromatic. The two phenyl groups delocalise the positive charge further.

As a result:

• The carbonyl carbon carries little effective positive character, because the charge is spread over an aromatic cation.
• Addition of a nucleophile at C=O would make that carbon sp³ and destroy the aromatic 2π system, so the tetrahedral adduct is strongly disfavoured.
• The inertness is electronic, not steric. Ordinary ketones, including strained cyclopropanone, add nucleophiles readily.

Physical evidence for the dipolar aromatic structure:

• an unusually large dipole moment (about 5 D, against about 3 D for benzophenone);
• a low C=O stretching frequency;
• protonation by acids on the oxygen, giving a stable hydroxycyclopropenium ion instead of an addition product.

Diphenylcyclopropenone therefore does not behave as an ordinary ketone.

(b)

The intermediate is benzyne (1,2-didehydrobenzene). The reaction is an elimination–addition substitution.

1. Elimination.

• NH₂⁻ removes the proton ortho to Br, the ring hydrogen made most acidic by the –I effect of Br.
• The aryl anion then expels Br⁻, giving benzyne.
• Its "third bond" is a weak in-plane overlap of two sp² orbitals, outside the aromatic sextet, so it is highly reactive.

2. Addition.

• NH₂⁻ adds to either end of the strained triple bond.
• The resulting aryl anion takes a proton from NH₃ to give aniline.

Experimental proof: ¹⁴C labelling (J. D. Roberts, 1953)

• Chlorobenzene labelled with ¹⁴C only at C-1 (the carbon carrying the halogen) was treated with KNH₂ in liquid NH₃.
• Degradation of the aniline showed the label almost equally divided between C-1 (the carbon carrying NH₂) and C-2 (the carbon next to it).
• Direct displacement (S_NAr) would put NH₂ only on the labelled carbon.
• A roughly 1 : 1 distribution requires a symmetrical intermediate in which C-1 and C-2 have become equivalent. That intermediate is benzyne.

Supporting evidence: haloarenes with no ortho hydrogen, such as 2-bromo-1,3-dimethylbenzene, do not react, and benzyne can be trapped as a Diels–Alder adduct with furan.

(c)

Reagent: thionyl chloride without base, for example neat or in ether or dioxane: ROH + SOCl₂ → RCl + SO₂ + HCl.

Why retention is needed.

• In both structures, Et, CH₃ and H occupy the same positions; only OH has been replaced by Cl.
• The CIP priorities run in the same order: OH > Et > CH₃ > H, and Cl > Et > CH₃ > H.
• So the descriptor (S) → (S) means the spatial arrangement is retained.
• An S_N2 displacement would invert the centre and give (R)-2-butyl chloride, and an S_N1 route would racemise it.
• The reaction must therefore follow the S_Ni (internal nucleophilic substitution) pathway.

Mechanism

1. Alkyl chlorosulfite formation. The OH oxygen attacks sulfur, Cl⁻ is lost, and deprotonation (with HCl released) gives the alkyl chlorosulfite, (S)-CH₃CH(Et)–O–S(=O)–Cl. No bond to the stereocentre has been broken, so its configuration is unchanged.
2. Ion pair. The C–O bond ionises to form an intimate (tight) ion pair, R⁺ ⁻O–S(=O)–Cl, and the anion collapses by losing SO₂.
3. Front-side delivery. The Cl is delivered from the same face that the oxygen has just left, before the carbocation can rotate or be attacked from the back. The result is (S)-2-butyl chloride, with retention.

If pyridine is added, it captures HCl as pyridinium chloride. Free Cl⁻ then attacks the chlorosulfite from the back (S_N2), and the product is (R)-2-butyl chloride, with inversion.

(d)

1. (i) Camphor → camphor lactone (O inserted between the C=O and the quaternary bridgehead carbon).

• Reaction: Baeyer–Villiger oxidation.
• Reagent: a peroxy acid: m-CPBA, CH₃CO₃H or CF₃CO₃H. Baeyer and Villiger originally used Caro's acid, H₂SO₅.
• Note: the more substituted carbon migrates, with retention.

2. (ii) Ar–CO–CH₃ → Ar–CH₂CH₃, with the secondary alcohol untouched.

• Reaction: Wolff–Kishner reduction (Huang-Minlon modification).
• Reagent: NH₂NH₂, KOH, diethylene glycol, heated to about 200 °C.
• Note: the basic conditions spare the alcohol. Clemmensen conditions (Zn–Hg, conc. HCl) would attack it (dehydration or substitution).

3. (iii) Dimethyl adipate → methyl 2-oxocyclopentanecarboxylate.

• Reaction: Dieckmann condensation (intramolecular Claisen).
• Reagent: NaOMe in MeOH or toluene, then dilute H₃O⁺.

4. (iv) CH₃CH₂N⁺Me₃ → CH₂=CH₂ + NMe₃.

• Reaction: Hofmann elimination (E2, anti-periplanar; gives the least substituted alkene).
• Reagent: moist Ag₂O, which gives the quaternary ammonium hydroxide, then heat.

5. (v) 2 PhCHO → PhCOO⁻ + PhCH₂OH.

• Reaction: Cannizzaro reaction.
• Reagent: concentrated (about 50%) NaOH or KOH.
• Note: the aldehyde has no α-H, so a hydride transfers from the tetrahedral alkoxide to a second aldehyde molecule.

(e)

(i) Modes of ring closure and opening (Woodward–Hoffmann rules for 6π systems)

A 6π electrocyclic reaction is 4n + 2 (n = 1). Thermal reactions use the HOMO ψ₃ and are disrotatory; photochemical reactions use ψ₄* and are conrotatory.

• Step I: (Z,Z,Z)-cyclonona-1,3,5-triene → bicyclo[4.3.0]nona-2,4-diene, thermal (100 °C): disrotatory closure.
• Step II: bicyclic diene → M, photochemical (hν): conrotatory ring opening.
• Step III: M → bicyclic diene, thermal (25 °C): disrotatory closure.

(ii) Structure of M

• Disrotatory closure of the all-cis triene in step I gives the cis-fused bicycle.
• Photochemical conrotatory opening of that cis-fused diene turns one terminus "in" and the other "out", so one of the terminal C=C bonds becomes trans.
• M is (1Z,3Z,5E)-cyclonona-1,3,5-triene (cis,cis,trans-cyclononatriene).
• The strained trans double bond in the nine-membered ring explains why M closes again thermally at only 25 °C, whereas step I needs 100 °C.

(iii) The indicated ring-fusion hydrogens

• In the step I product they are cis (cis-bicyclo[4.3.0]nona-2,4-diene). A (Z,Z,Z) triene with both terminal substituents "outside" closes disrotatorily to put the two H atoms on the same face.
• In the final product of step III they are trans (trans-bicyclo[4.3.0]nona-2,4-diene). Disrotatory closure of the (Z,Z,E) triene M, which has one terminal substituent inside and one outside, gives the trans ring fusion.