Take one alkene, 2-methylbut-2-ene, and three reagents. mCPBA adds a single oxygen and makes a three-membered ring. OsO₄ adds two OH groups. Ozone cuts the molecule in two at the double bond.
All three reactions oxidize the alkene: its two carbons gain bonds to oxygen. (Oxidation & Reduction later gives a way to count them.) What sets the reactions apart is how far they go, so choosing a reagent means choosing how much of the alkene survives.
Epoxidation: one oxygen, ring intact
mCPBA turns cis-but-2-ene into an epoxide, a three-membered ring made of the two alkene carbons and one oxygen. The oxygen bridges the two former alkene carbons.
mCPBA is short for meta-chloroperoxybenzoic acid. It is a peroxyacid: a carboxylic acid with a second oxygen in its OH group, so C(=O)–O–H becomes C(=O)–O–O–H. Written short, a carboxylic acid is RCO₂H and a peroxyacid is RCO₃H. The new O–O bond is weak.
The alkene attacks the outer oxygen, the one that carries the H. That oxygen is easy to attack because the O–O bond is weak, and the group it breaks away from is stable: its electrons end up spread over both oxygens of a carboxyl group by resonance.
Everything happens in one step. The pi bond attacks the outer oxygen, the O–O bond breaks, and the proton on that oxygen moves across to the carbonyl oxygen. All of this runs through a single cyclic transition state. It is called the butterfly transition state, because the peroxyacid sits over the alkene at right angles to it, like a pair of wings. The byproduct is the ordinary carboxylic acid.
Two consequences follow, and exams test both. First, the oxygen reaches both carbons from the same face of the alkene, so the addition is syn. Second, no intermediate forms, so no bond has a chance to rotate, and the alkene's geometry carries straight into the ring.
So cis-but-2-ene gives the cis epoxide. It has a mirror plane, so it is a single achiral meso compound. trans-But-2-ene gives the trans epoxide, which is chiral. mCPBA attacks either face of the flat alkene equally often, so the two enantiomers form 50:50: a racemate. Each alkene geometry gives its own stereoisomer, which makes the reaction stereospecific, a term first used for bromine in Addition reactions.
Epoxides are strained, and that makes them reactive. Alcohols, Ethers & Related Chemistry covers their reactions. The one reaction this section needs is ring opening by water, in the next part.
The two diols, and how to choose between them
A 1,2-diol, also called a glycol, has OH groups on two neighboring carbons. Cyclohexene can be turned into cyclohexane-1,2-diol in two ways. One gives the cis diol and the other the trans diol, and telling these two apart is the most examined point in the section.
The syn route: OsO₄. Osmium tetroxide adds to the alkene in one step, as mCPBA does. Two of its oxygens bond to the two carbons at the same time, from the same face, and close a five-membered ring called an osmate ester. Sodium hydrogen sulfite (NaHSO₃) in water then cuts the osmium off, and each oxygen keeps its bond to carbon as an OH. Both OH groups end up on the face the OsO₄ came from: a syn addition, giving a syn diol. On a ring, the syn diol is the cis diol.
OsO₄ is expensive and toxic, so it is usually used in small, catalytic amounts together with NMO (N-methylmorpholine N-oxide). NMO is a co-oxidant: it turns the spent osmium back into OsO₄, so each osmium atom can react many times. The two reagents go into the flask together, and this page writes them as OsO₄, NMO.
Cold, dilute, basic KMnO₄ gives the same syn diol through a similar cyclic ester of manganese. It is cheaper than OsO₄ but less clean, because it can go on to oxidize the product further.
The anti route: mCPBA, then aqueous acid. mCPBA first puts one oxygen on one face, as the epoxide. The acid then adds H⁺ to the epoxide oxygen, which turns it into a better leaving group. Water attacks one ring carbon from the face opposite the C–O bond that breaks, the same backside attack an SN2 reaction makes (Substitution & Elimination). Loss of H⁺ then leaves an OH. The two oxygens end up on opposite faces, so the addition is anti. The result is an anti diol, which on a ring is the trans diol.
In the epoxide of cyclohexene the two ring carbons are alike. Water attacks each of them equally often, and the two choices give mirror images, so the trans diol forms as a racemate.
Which diastereomer, exactly — and is it chiral?
A double bond in a small ring is always cis, so on a ring only syn or anti matters. An open-chain alkene can be cis or trans, and the product depends on both choices. But-2-ene shows all four combinations.
| Alkene | OsO₄ (syn) | mCPBA then H₃O⁺ (anti) |
|---|---|---|
| cis-But-2-ene | meso-butane-2,3-diol (achiral) | racemic (±)-butane-2,3-diol |
| trans-But-2-ene | racemic (±)-butane-2,3-diol | meso-butane-2,3-diol (achiral) |
Start with cis-but-2-ene and OsO₄, top row in the middle. Both OH groups point toward you, and the two CH₃ groups stay on the upper side, where the alkene had them. A mirror plane runs between the two carbons, so the diol is achiral even though it has two stereocenters. That is the meso case from Meso compounds.
Now switch to anti addition on the same alkene. One OH points toward you and the other away, so the mirror plane is gone and the product is chiral. The epoxide from cis-but-2-ene is meso, just like the epoxide of cyclohexene, and water opens it at C2 or at C3 equally often. The two choices give the two enantiomers in equal amounts: a racemate.
The second row of the table is the first row with its two answers swapped. Changing either the alkene geometry or the reagent flips the answer, and changing both flips it back. So trans-but-2-ene with anti addition gives the meso diol again. In the figure it does not look symmetric, but give C3 a half turn about the C2–C3 bond and it becomes the meso drawing in the top row.
Ozonolysis: cutting the double bond in half
Treat 2-methylbut-2-ene with ozone (O₃) and then dimethyl sulfide, and the molecule comes apart into propanone and ethanal. Each carbon of the old double bond has become a C=O. This reaction is ozonolysis: ozone breaks both the sigma and the pi bond of the C=C.
Ozone alone does not give the carbonyl compounds. It first adds across the double bond in one step, giving an unstable ring called the molozonide. That ring splits in two, and the pieces join up the other way round as the ozonide, in which the two old alkene carbons are linked only through oxygen.
The ozonide is what the flask contains when the ozone is switched off, and it has no C=O yet. A second reagent, the workup, takes it apart. That is why ozonolysis is always written with two reagents, and why the workup, not the ozone, decides whether each end stops as an aldehyde or goes on to an acid.
The two usual workups differ in what they do to a new C=O carbon that carries an H.
- Reductive workup: dimethyl sulfide (Me₂S), or zinc in acetic acid. Me₂S removes the extra oxygen from the ozonide and becomes dimethyl sulfoxide. The fragments stop as carbonyl compounds: an alkene carbon that carried two carbon groups becomes a ketone, and one that carried an H becomes an aldehyde.
- Oxidative workup: hydrogen peroxide (H₂O₂). The peroxide stays in the flask as an oxidant, so any fragment that would have been an aldehyde is oxidized on to a carboxylic acid. A ketone has no H on its carbonyl carbon to lose, so it is unaffected.
Ozonolysis also shows where the double bond was in an unknown alkene. Identify the two carbonyl fragments, then join them back together at their carbonyl carbons. A question that gives you the fragments and asks for the alkene wants exactly this, as the next example shows.
An unknown alkene, C₆H₁₂, is ozonolyzed with Me₂S workup. The products are propanal and propanone (acetone).
Find the two carbonyl carbons. Propanal is CH₃CH₂CHO, so its carbonyl carbon carried one hydrogen and an ethyl group. Propanone is (CH₃)₂CO, so its carbonyl carbon carried two methyl groups.
Rebuild the double bond. Remove both oxygens and join those two carbons with a C=C: CH₃CH₂–CH=C(CH₃)₂, which is 2-methylpent-2-ene. Check the formula: C₆H₁₂. ✓
The tell in the question is that one fragment is an aldehyde and one a ketone. So one alkene carbon bore a hydrogen and the other did not, and the alkene was trisubstituted.
One case breaks the two-fragment habit. When the double bond is in a ring, cutting it does not make two molecules, because the rest of the ring still joins the two ends. Cyclohexene with O₃ and then Me₂S gives a single molecule, hexanedial, OHC–(CH₂)₄–CHO, with a CHO group at each end.
So a single product with two C=O groups is the signature of a ring alkene. A single product with one C=O group, formed as two moles per mole of alkene, points instead to a symmetric open-chain alkene, whose two halves are identical.
The substrate is 2-methylbut-2-ene, (CH₃)₂C=CH–CH₃. Before choosing a reagent, label the two alkene carbons. The left one carries two methyl groups and no hydrogen; the right one carries one methyl group and one hydrogen. Every answer below comes from that labeling.
(a) mCPBA. One oxygen goes across the double bond and nothing is cut. The product is the epoxide, 2,2,3-trimethyloxirane (oxirane is the systematic name for the epoxide ring). No OH appears.
(b) OsO₄, then NaHSO₃. Two OH groups add to the same face, giving 2-methylbutane-2,3-diol, the syn diol. This reaction creates only one stereocenter, C3. C2 carries two methyl groups, so it is not a stereocenter. With one stereocenter, syn has no visible consequence here, and the product is simply racemic. To see syn and anti give different products, pick a substrate that gains two stereocenters.
(c) mCPBA, then H₃O⁺. Backside opening of the epoxide gives the anti diol with the same constitution. For the reason given in (b), it is the same compound as the product of (b).
(d) O₃, then Me₂S. Cut the double bond and give each carbon a C=O. The carbon with two methyl groups has no hydrogen, so it can only become a ketone: propanone. The other carbon has a hydrogen, so it becomes an aldehyde: ethanal. Products: propanone + ethanal, as drawn in the workup figure above.
(e) O₃, then H₂O₂. Same cut, oxidative workup. The ketone is unchanged, because its carbonyl carbon has no hydrogen to lose. The aldehyde is oxidized one step further, to the acid. Products: propanone + ethanoic acid.
(f) Hot, concentrated KMnO₄. Same outcome as (e): propanone + ethanoic acid. Cold, dilute and basic KMnO₄ would instead give the syn diol from (b).
The method comes down to two questions. Does the reagent cut the double bond or add across it? If it cuts, does each carbon carry a hydrogen? The summary table below answers the first question for each reagent. You answer the second from the substrate.
The whole section as one decision
| Reagent | What happens to the C=C | Product |
|---|---|---|
| mCPBA | One O added, ring kept | Epoxide (stereospecific) |
| OsO₄, NMO | Two OH, same face | syn diol |
| mCPBA then H₃O⁺ | Two OH, opposite faces | anti diol |
| O₃ then Me₂S | Cleaved | Aldehydes and/or ketones |
| O₃ then H₂O₂ | Cleaved | Carboxylic acids and/or ketones |
| Hot, concentrated KMnO₄ | Cleaved | Carboxylic acids and/or ketones |
| Cold, dilute, basic KMnO₄ | Two OH, same face | syn diol |
KMnO₄ appears twice on purpose, because its conditions do all the work. Cold, dilute and basic, it stops at the syn diol. Hot and concentrated, it cuts the double bond and gives the oxidative result, the same products as ozone followed by H₂O₂ (except at a =CH₂ end; see the trap above); some questions write it in place of that pair. A question that gives KMnO₄ without conditions is incomplete, and one that gives them is telling you which row to read.
What carries forward
The syn and anti diols are a stereochemical choice you make by picking the reagent. Choices of that kind are what synthesis questions are built around, and Catalytic hydrogenation adds another one, for alkynes.
Ozonolysis run backwards is a way to work out structure. Given the fragments, you find the parent alkene. That makes it a structure-finding tool, like the methods of IR & Mass Spectrometry, the next chapter.