(a) (i) Complete the following reaction by showing stepwise reaction mechanism for the formation of products : trans-but-2-ene…

(a)(i) Epoxidation of trans-but-2-ene and ring opening by OH⁻

The printed reagent is perbenzoic acid, PhCO₃H.

Step 1, epoxidation (Prilezhaev). The peracid transfers its terminal oxygen to the C=C in one concerted, cyclic "butterfly" transition state. The alkene π electrons attack the electrophilic oxygen, the O–O bond breaks, and the peracid H moves to the carbonyl oxygen of the departing benzoic acid. Both new C–O bonds form on the same face (syn), so the alkene geometry is kept. trans-But-2-ene gives only trans-2,3-dimethyloxirane, as the racemate (2R,3R) + (2S,3S). The by-product benzoic acid is taken up by the base as benzoate.

Step 2, ring opening by OH⁻. Hydroxide attacks one ring carbon from the back side (SN2), anti to the C–O bond, with inversion at that carbon. Water then protonates the alkoxide. Anti opening of a trans-epoxide gives the meso diol, so the products are meso-butane-2,3-diol, (2R,3S), and benzoate (PhCOO⁻). Both epoxide enantiomers give the same meso diol, because the two ring carbons are equivalent.

The overall result is anti-dihydroxylation: the trans-alkene gives the meso diol, whereas cis-but-2-ene would give (±)-butane-2,3-diol. If OH⁻ only neutralises the acid, the isolated product is (±)-trans-2,3-dimethyloxirane.

(a)(ii) Loss of optical activity of the imine A

A is the benzophenone imine of 1-phenylethylamine, Ph(Me)CH–N=CPh₂. Its only stereocentre is the CH carbon. That hydrogen is unusually acidic, because it is benzylic and α to the C=N. Removing it gives an anion delocalised over Ph–C–N=C–Ph₂, a planar 2-azaallyl anion.

EtO⁻ removes this proton, the stereocentre becomes sp² and planar, and the chirality is lost. Reprotonation by EtOH can then happen in two ways:

• At the original carbon, equally from either face, giving the racemate.
• At the other terminus, giving the tautomeric imine Ph(Me)C=N–CHPh₂, which is achiral.

Either way the net rotation falls to zero. The cause is a base-catalysed 1,3-proton shift (imine–imine tautomerism) through a planar carbanion. No C–N bond is broken.

(b)(i) Dibromocyclopentanes from cyclopentane and excess Br₂

At high temperature, radical bromination can put the second Br on the same carbon, on C-2 or on C-3. C-4 and C-5 repeat C-3 and C-2. So there are three constitutional isomers, which become five compounds once stereochemistry is counted:

• 1,1-Dibromocyclopentane: achiral, one compound.
• cis-1,2-Dibromocyclopentane: meso. A mirror plane bisects the C1–C2 bond and passes through C4, so it is optically inactive.
• trans-1,2-Dibromocyclopentane: chiral, a pair of enantiomers (1R,2R) and (1S,2S).
• cis-1,3-Dibromocyclopentane: meso. A mirror plane passes through C2 and the midpoint of C4–C5.
• trans-1,3-Dibromocyclopentane: chiral, (1R,3R) and (1S,3S).

That gives 5 distinct dibromocyclopentanes (7 stereoisomers if each enantiomer is counted separately). Radical bromination forms the chiral ones as racemates, so five products are expected.

(b)(ii) p-Nitrotoluene to m-nitrotoluene

The nitro group cannot be moved directly, so it is removed and a new one is introduced meta to CH₃ under the control of a temporary directing group.

1. Sn/HCl (or H₂, Pd–C) reduces the NO₂: p-toluidine.
2. Ac₂O protects the amine: 4-methylacetanilide.
3. HNO₃/H₂SO₄ nitrates ortho to NHAc. NHAc is the stronger o/p director, and its para position is blocked by CH₃, so NO₂ enters C-3, which is meta to CH₃: 4-acetamido-3-nitrotoluene.
4. Acid hydrolysis frees the amine: 4-methyl-2-nitroaniline.
5. NaNO₂/HCl at 0–5 °C, then H₃PO₂, replaces the diazonium group by H (deamination).

The product is m-nitrotoluene (3-nitrotoluene).

(c) Phenol to coumarin

Route 1, Reimer–Tiemann then Perkin (the standard route to unsubstituted coumarin).

1. Phenol to salicylaldehyde. Phenol is treated with CHCl₃ and aqueous NaOH at 60–70 °C, then acidified. OH⁻ deprotonates CHCl₃, and ⁻CCl₃ loses Cl⁻ to give dichlorocarbene, :CCl₂ (α-elimination). The electron-rich ortho carbon of the phenoxide attacks the carbene. Tautomerisation restores the aromatic ring and gives an o-(dichloromethyl)phenoxide. Hydrolysis of the CHCl₂ group gives the aldehyde. Ortho attack is favoured because the carbene approaches the phenoxide close to the O⁻/Na⁺ ion pair.
2. Salicylaldehyde to coumarin. Salicylaldehyde is heated with acetic anhydride and sodium acetate at about 180 °C. Acetate removes an α-H from Ac₂O, and the anhydride enolate adds to the CHO, an aldol-type step. Acyl transfer and loss of acetate (or water) give the conjugated side chain. The phenolic oxygen then attacks the side-chain carbonyl, closing the six-membered lactone. Elimination fixes the C3=C4 double bond of coumarin (2H-chromen-2-one). Ring closure is possible because the phenol O and the side chain are held cis on the ring.

Route 2, Pechmann condensation (one pot). Phenol is heated with malic acid in concentrated H₂SO₄. The acid dehydrates and decarbonylates malic acid to formylacetic acid, OHC–CH₂–COOH. The protonated aldehyde attacks the ring ortho to OH (electrophilic substitution). The phenolic OH then lactonises onto the COOH, and dehydration gives coumarin. With ethyl acetoacetate in place of malic acid, the same Pechmann sequence gives 4-methylcoumarin, not coumarin, and simple phenol reacts sluggishly. Activated phenols such as resorcinol work far better.