2,2-Dimethyloxirane 2026-10-20
An unsymmetrical epoxide with two methyl substituents at one ring carbon. Acidic opening can favour nucleophile attachment to that more substituted carbon.
Bromohydrin 2026-10-20
A halohydrin with bromine on the carbon adjacent to OH. Its relative stereochemistry can select between properly aligned epoxide closure and semipinacol rearrangement.
Cyclohexene oxide 2026-10-20
Electrophilic oxygen transfer 2026-10-20
Epoxidation 2026-10-20
Epoxide ring opening 2026-10-20
Ethylene oxide 2026-10-20
The simplest epoxide, a three-membered ring. Opening by a carbon nucleophile, followed by protonation, adds a two-carbon primary-alcohol unit.
Halohydrin 2026-10-20
Meta-chloroperoxybenzoic acid 2026-10-20
Past exam of the chemistry B course of the University of Cambridge 2021 Paper 1 B3 c Solution 2026-10-20
Number the nitrogen N1, the phenyl-bearing carbon C2, and then continue through C3–C6 around the piperidine. Call the solid-wedge epoxide face up; the phenyl group is down in both starting structures. The bulky phenyl group favors a pseudoequatorial orientation.
epoxide opening follows the Fürst-Plattner preference for a chair-like, trans-diaxial opening trajectory. Attack at the two epoxide carbons is not equivalent: the favored route evolves toward an ordinary chair, whereas the alternative demands a less favorable twist/boat-like distortion. Such conformational control of epoxy-2-arylpiperidine opening is supported by synthetic studies.[ref]
1,3-Allylic Strain as a Strategic Diversification Element for Constructing Libraries of Substituted 2-Arylpiperidines by Thomas C. Coombs and others. Scheme 1, chair-like trans-diaxial epoxide opening; applied here with the free-NH conformer
For the first epoxide, azide attacks C5 from down, breaking the C5–O bond. The residual C4–O bond is retained, so work-up gives C5–N3 down and C4–OH up, with the original C2–Ph down. For the second, down-face epoxide, azide attacks C4 from up, leaving C4–N3 up and C5–OH down. These assignments are shown in the middle row of the original stereochemical map.
In each mechanism an azide terminal nitrogen attacks the epoxide carbon, the C–O bonding pair goes to oxygen, and protonation gives the alcohol. Each attacked center inverts, and the azide and OH are trans. The phenyl-bearing center is unchanged. Reversing the epoxide face changes which carbon has the favored axial attack trajectory; a steric rule based only on distance from phenyl misses that conformational effect.
Past exam of the chemistry B course of the University of Cambridge 2021 Paper 1 B3 d ii Solution 2026-10-20
sulfide first opens one epoxide by backside attack, giving a carbon-bound thiolate and an alkoxide. Using the same numbering, take this first attack at C2: the C2–S bond is down and C3–O remains up.
The attached sulfur still bears a nucleophilic lone pair and negative charge. It attacks C5 of the second epoxide intramolecularly, breaking C5–O and closing a bridge between C2 and C5. Work-up protonates the oxygen anions at C3 and C6. Thus B is 7-thiabicycloheptane-2,5-diol, with both OH groups on the face opposite the sulfur bridge.
Both C–S bonds are formed by inversion at their attacked epoxide carbons. Because the original epoxides are cis, both new C–S bonds point to the same opposite face and one sulfur chemical atom can connect them. The 1,4 bridge gives two five-membered sulfur-containing circuits, unlike the more strained closure between nearby 1,3 positions.
The sulfur therefore replaces two separate external nucleophiles: two epoxides are opened but only one sulfur chemical atom is incorporated. Adding that sulfur and two work-up protons gives , consistent with B.
Past exam of the chemistry B course of the University of Cambridge 2021 Paper 1 B3 d i Solution 2026-10-20
Number the cyclohexane consecutively, putting the top methylene group at C1 and the two original epoxides at C2/C3 and C5/C6. Both epoxide oxygens are up.
A is a diazidocyclohexanediol, with each epoxide replaced by one azide and one alcohol. A standard successive chair-favored opening gives N3 down at C2 and C5, OH up at C3 and C6. This is the 1,4-diazido/1,4-diol connectivity drawn in the bottom row of the original map.
For each opening, azide attacks from the face opposite oxygen, breaking one C–O bond with inversion at the attacked carbon. protonation of the two resulting alkoxides gives the two OH groups. After the first opening, the preferred conformation of the opened portion must be used for the second trans-diaxial opening; simply imposing one unchanged chair on both steps can select the wrong regioisomer.
Past exam of the chemistry B course of the University of Cambridge 2022 Paper 1 B3 a Solution 2026-10-20
Number the OH-bearing carbon C1 and the adjacent brominated carbon C2. In the trans bromohydrin, base generates an alkoxide. A chair conformation can place O at C1 and Br at C2 trans-diaxial, aligning the oxygen nucleophile for backside attack at C2. Intramolecular SN2 closure expels bromide and inverts C2: the retained C1–O bond and the newly formed C2–O bond are both on the original OH face, giving the epoxide shown. ring inversion can provide the reactive chair even if it is not the dominant ground-state conformer.
The cis bromohydrin cannot place OH and Br trans-diaxial: one is axial when the other is equatorial. In the chair with Br axial, the hydrogen on C1 can instead be antiperiplanar to C2–Br. An alkoxide-assisted semipinacol pathway moves the C1–H bonding pair to C2 while the oxygen lone pair forms C1=O and bromide departs. hydride migration gives the ketone without changing the carbon skeleton. The hydrogen is transferred to C2 rather than expelled as a free hydride. The original phenyl-bearing stereocentre is not involved in either bond-reorganization sequence. Configuration controls which properly aligned displacement or migration can occur; a flat ring drawing alone conceals that distinction.
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 1 B3 b Solution 2026-10-20
In both cases the alkene undergoes concerted oxygen transfer from metachloroperbenzoic acid. The alkene pi bond forms one C–O bond to the terminal peroxide oxygen; electron redistribution forms the second C–O bond, breaks O–O, and transfers the peracid proton to its carbonyl oxygen as the carboxylic acid product forms. This cyclic epoxidation pathway delivers both C–O bonds from one face, without a freely rotating carbon intermediate.
For 1-methylcyclohexene, the two alkene faces are enantiotopic. Attack from opposite faces gives the two mirror-related epoxides, and an achiral peracid in an achiral environment cannot distinguish their Gibbs energies of activation. The product is racemic, 1:1.
For the cis-fused bicyclic alkene, the junction configuration and its puckered ring conformation make approach from the two faces sterically different. The folded saturated ring shields the concave face; oxygen transfer preferentially approaches the more exposed convex face, directed away from that ring. The alternative attack places the incoming peracid closer to the folded skeleton. The 95:5 ratio represents facial diastereoselectivity, while concerted syn oxygen delivery is retained for either product. It should not be explained by free rotation or by one epoxide subsequently changing configuration. For scale only, if this ratio were measured under irreversible kinetic conditions at 298 K, it would correspond to difference between the competing Gibbs energies of activation; the exam does not specify that temperature.
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 1 B3 d Solution 2026-10-20
With triethylamine present, proton transfer from thiophenol produces a more nucleophilic thiolate population. sulfur attacks the anomeric epoxide carbon C1 from the face opposite its epoxide C–O bond. Displacement by backside attack opens the epoxide, breaks C1–O, and leaves the former epoxide oxygen attached to C2. protonation of that alkoxide by thiophenol or triethylammonium gives G: SPh at C1 on the front face and OH at C2 on the back face, a trans relationship. This is inversion at C1 with retention of the C2 C–O bond; there is no need to invert every stereocentre. Rapid protonation also suppresses competing silyl transfer by the alkoxide.
Without added base, neutral thiophenol is less nucleophilic. acid-assisted cleavage at the anomeric epoxide bond can instead be accompanied by donation from the ring oxygen, giving an oxocarbenium ion with charge shared with the ring oxygen. thiophenol capture from the face that puts SPh back on the original epoxide face, followed by deprotonation, gives H with SPh and C2–OH both on the back face, the net-retention pathway. An ion-pair or highly asynchronous opening can provide this outcome; the drawing does not establish a long-lived free carbocation.
The distinction is configuration, not a mandatory fixed chair. For the closely matching trisilylated substrate, the primary study obtained both anomers with thiophenol alone, and established an unusual inverted chair for the alpha thioglycoside by NMR and crystallography. Thus H should be explained as an accessible product, not asserted to be the sole product or necessarily to have axial SPh in its preferred conformation.[ref]
Reaction of thiophenol with glucal epoxides: X-ray structure of 3,4,6-tri-O-tert-butyldimethylsilyl-1-S-phenyl-1-thio-α-D-glucopyranoside by Caroline Walford and others. 1997, Schemes 1–2 and Figure 1; neutral reaction mixture and configuration/conformation assignment
Past exam of the chemistry B course of the University of Cambridge 2024 Paper 1 B4 a Solution 2026-10-20
The alkene undergoes concerted oxygen transfer from the terminal peroxide oxygen of mCPBA. In the cyclic electron-flow description, the alkene pi bond begins one C–O bond, the peroxide oxygen supplies the second, O–O breaks, and proton transfer/carboxyl rearrangement gives m-chlorobenzoic acid. Both C–O bonds are installed from the same alkene face, so the alkene's relative substituent geometry is retained in the epoxide. There is no freely rotating carbocation intermediate.
The neighboring free OH hydrogen-bonds to the approaching peracid and favors a cyclic, organized transition-state geometry. oxygen is delivered from the OH-directed face, giving the indicated syn relationship between epoxide oxygen and the allylic hydroxyl. In the supplied projection, the epoxide C–O bonds and the original C–OH bond are on the rear/hashed face. Steric interactions still influence the preferred hydrogen-bonded conformer, but “attack from the least hindered face” alone misses the positive directing role of OH. The distant benzyl ether is not the directing free hydroxyl.
The classical experimental comparison of allylic alcohols and protected derivatives established this directing principle; it supports the mechanistic interpretation without itself measuring the selectivity of this acyclic substrate.[ref]
Past exam of the chemistry B course of the University of Cambridge 2025 Paper 1 B2 c Solution 2026-10-20
Base deprotonates the sulfonamide. Its nitrogen attacks one epoxide carbon from the back, breaking the C–O bond at that carbon and producing an adjacent alkoxide. The nitrogen and oxygen attachments on the cyclopentane are therefore trans. The alkoxide then attacks the aromatic carbon bearing the fluorine ortho to the sulfonyl group. Intramolecular nucleophilic aromatic substitution expels fluoride and closes a seven-membered heterocycle with the path . The para fluorine remains.
This tandem epoxide-opening/ring-closing pathway is required by the formula of F: simple epoxide addition would retain both fluorines, whereas F contains only one. In the second stage, pyrrolidine replaces the remaining para fluorine by another nucleophilic aromatic substitution, producing G. The mass balance is F plus pyrrolidine minus HF, as required by the given formula of G.
Past exam of the chemistry B course of the University of Cambridge 2025 Paper 1 B3 d Solution 2026-10-20
The products are the two mirror-image 2,3-epoxy-2,3-diphenylindan-1-ones, shown in the bottom row of the stereoisomer diagram. The original carbonyl remains, and an additional oxygen bridges the former double bond, giving the stated formula.
In the Weitz-Scheffer epoxidation, hydroperoxide anion adds conjugatively to the enone. The resulting enolate closes onto the attached peroxide oxygen with O–O bond cleavage and loss of hydroxide. Attack on the two faces of the achiral planar substrate gives the two enantiomers in equal amounts. This is a racemic epoxide pair, rather than a one-to-one mixture of different diastereomeric epoxide connectivities.
Let . One standard route is:where the last two carbons are joined through oxygen to form the epoxide shown in the product drawing.
Use rigorously dry conditions for Grignard formation/addition; quench only after the aldehyde addition. Dehydration gives the styrene, and peracid oxidation transfers oxygen across its double bond. The methoxy substituent is compatible with this sequence. No asymmetric reagent is specified, so the target epoxide is racemic.
The oxygen substituent changes position through a halohydrin–epoxide sequence. First, hydroxide attacks the acetate carbonyl of , giving a tetrahedral intermediate. Collapse expels the substrate alkoxide, and proton transfer gives plus acetate. This is saponification, breaking the acyl–oxygen bond rather than replacing bromide directly with hydroxide.
A small equilibrium population of undergoes intramolecular SN2 reaction at the adjacent primary carbon: the O lone pair attacks and its C–Br pair leaves with bromide. The result is 2,2-Dimethyloxirane. Addition of HBr protonates the ring oxygen. bromide then attacks preferentially at the more substituted carbon, where the protonated ring has greater carbocation character, breaking that C–O bond and leaving oxygen attached to the former primary carbon:The middle formula denotes a three-membered epoxide, not an open-chain ether. Acidic opening of this highly unsymmetrical epoxide has the opposite regioselectivity to a simple basic nucleophilic opening at its less substituted carbon.
Finally, the primary alcohol oxygen attacks acetyl chloride. The tetrahedral intermediate collapses with chloride departure, and triethylamine removes the O–H proton and captures HCl. The product is . oxygen is carried through the epoxide to the other carbon; it is not necessary to invent a methyl migration or a carbon-skeleton rearrangement.



