Q5(b). Two epoxides of opposite face from one cyclopentene

Introduction. The substrate, cis-4-tert-butyl-3-methylcyclopentene, has both alkyl groups on the same face (wedged), so one face of the C=C is hindered. The two reagent sequences put the epoxide oxygen on opposite faces.

Reading of the question: the NBS step is read as bromohydrin formation (NBS in the aqueous medium), followed by ring closure with NaOH. NBS in CCl₄ would instead brominate the allylic position.

m-CPBA: epoxide anti to the substituents

The peracid transfers its terminal oxygen in one concerted step (the “butterfly” transition state). It approaches from the less hindered face, the face opposite to the methyl and tert-butyl groups. Both new C–O bonds form on that face (syn addition).

NBS / H₂O, then NaOH: epoxide syn to the substituents

1. Br⁺ from NBS also adds from the less hindered face: the bromonium ion lies anti to the alkyl groups.
2. Water opens it from the opposite face (anti addition), so the OH is on the same face as the alkyl groups and Br is trans to OH: a trans-bromohydrin.
3. NaOH forms the alkoxide, which displaces bromide intramolecularly (SN2, backside, the O and Br are anti-periplanar). The oxygen is already on the hindered face, so the ring closes there.

Scheme: The two routes give the two diastereomeric epoxides (each racemic).

Conclusion. m-CPBA gives the epoxide trans to the methyl and tert-butyl groups (direct attack on the open face). NBS/H₂O then NaOH gives the cis epoxide: bromonium on the open face, water and then oxygen on the hindered face.

Application. The pair of methods is used to make either epoxide of a steroid or terpene alkene; the bromohydrin route is also how propylene oxide was made industrially (chlorohydrin process).