Baeyer–Villiger oxidation

Treat acetone, CH₃–CO–CH₃, with the reagent mCPBA and you get methyl acetate, CH₃–CO–O–CH₃. One oxygen atom has slipped into a carbon–carbon bond, between the carbonyl carbon and one of its methyl groups. The ketone has become an ester.

This is the Baeyer–Villiger oxidation. A peroxyacid turns a ketone into an ester by inserting an oxygen between the carbonyl carbon and one of the two groups on it:

R–CO–R′  +  mCPBA  →  R–CO–O–R′

It is unlike the ketone reactions you have met so far. In nucleophilic addition and everything built on it, a new group joins the carbonyl carbon and its C–C bonds stay as they were. Here one of those C–C bonds is cut, and an oxygen fills the gap.

It counts as an oxidation because the carbonyl carbon gains a bond to oxygen: it has two in the ketone (the C=O) and three in the ester (see Oxidation levels).

The usual reagent is mCPBA, short for meta-chloroperoxybenzoic acid. It is the reagent that turns an alkene into an epoxide, and as you saw in Oxidative cleavage & dihydroxylation, it is a peroxyacid: a carboxylic acid with one extra oxygen, so that the acid group reads C(=O)–O–O–H instead of C(=O)–O–H.

Two parts of mCPBA do the work. The oxygen at the O–H end is the one that ends up in the product. The O–O bond beside it is weak, and when it breaks, the rest of the reagent leaves as a carboxylate ion, ArCO₂⁻. Resonance spreads that negative charge over two oxygens, so the carboxylate is a good leaving group.

The mechanism: an addition, then a shift

Follow one ketone through. Butanone, CH₃–CO–CH₂CH₃, reacts with mCPBA to give ethyl acetate, CH₃–CO–O–CH₂CH₃. The oxygen goes in on the ethyl side. The mechanism shows how it gets there; the next section explains why it picks that side.

Step 1. An acid in the flask protonates the carbonyl oxygen. That acid is often the peroxyacid itself, or the carboxylic acid it produces. A positive oxygen pulls electrons away from the carbonyl carbon, which makes that carbon a better electrophile.

Step 2. The peroxyacid adds to the carbonyl carbon through its O–H oxygen, the terminal oxygen, not the oxygen bonded to the reagent’s own C=O. A lone pair on that oxygen forms the new C–O bond. At the same time the C=O π electrons move onto the positive oxygen, which becomes an ordinary OH.

Step 3. The oxygen that just added now has three bonds and a positive charge. It loses its proton to a base in the flask.

The result is the Criegee intermediate. Its central carbon carries four single bonds: to the OH, to the two groups of the ketone, and to the O–O–C(=O)Ar group from the reagent. It has no charge.

Step 4. The Criegee intermediate falls apart in one step, with three pairs of electrons moving at once:

• A lone pair on the OH oxygen moves in to make a new C=O bond.
• The ethyl group moves off the central carbon onto the nearer oxygen of the O–O bond. It takes its bonding pair with it.
• The O–O bond breaks, and its electrons leave with the carboxylate.

The ethyl moves from one atom to the atom right next to it, a move called a 1,2-shift. You met the same move in Carbocations, where a hydride or a methyl shifts onto a neighboring positive carbon.

There the group moves toward an empty p orbital. Here it moves toward the empty σ* orbital of the O–O bond. S_N2 used the same kind of orbital, the σ* of the bond to the leaving group. Filling it is what breaks the O–O bond. In both kinds of shift, the migrating group keeps its configuration (more on this below).

Step 5. The new ester comes out with a proton on its C=O oxygen. The carboxylate takes that proton, and the products are ethyl acetate and the carboxylic acid ArCO₂H.

Which group migrates

Butanone has a methyl and an ethyl on its carbonyl carbon, and the ethyl moved. Acetophenone has a methyl and a phenyl (a benzene ring attached through one of its carbons). It gives phenyl acetate. Methyl benzoate, which would need the methyl to move, does not form.

To read a product, find the inserted oxygen. The group on its far side, away from the C=O, is the group that moved.

A ketone has two groups and only one of them moves, so most Baeyer–Villiger questions come down to which. The groups fall into a fixed order, called migratory aptitude:

H > tertiary > cyclohexyl ≈ secondary ≈ benzyl ≈ aryl > primary > methyl

To place a group in that order, look at the carbon that attaches it to the C=O and count the carbons bonded to it, leaving out the carbonyl carbon itself. Ethyl’s CH₂ has one other carbon, so ethyl is primary. Isopropyl’s CH has two, so it is secondary; so is cyclohexyl. tert-Butyl’s central carbon has three, so it is tertiary. Methyl has none.

Aryl means a benzene ring attached directly, like the phenyl in acetophenone. Benzyl means a CH₂ that carries a benzene ring. The count misleads for benzyl. A benzyl CH₂ has only one other carbon, so the count says primary, but the ring next to it helps hold positive charge, and benzyl ranks with secondary. Aryl ranks with secondary too, for a reason given below.

The order has a reason. While the group is moving, its carbon is partly bonded to where it came from and partly bonded to where it is going, and in that transition state it runs short of electrons: it carries a partial positive charge. A group that is good at holding a positive charge makes that transition state easier to reach, so it moves faster. That is why the alkyl groups line up in the same order as carbocation stability: tertiary, then secondary, then primary, then methyl.

Aryl groups seem to break the pattern. A phenyl cation is very unstable, yet phenyl migrates about as well as a secondary group. But the ring never becomes a free phenyl cation. As it moves, its ipso carbon, the ring carbon bonded to the carbonyl carbon, is partly bonded to the carbon it is leaving and to the oxygen it is joining. For a moment the ring bridges the two atoms. The ring’s π electrons then spread the positive charge onto three of its other carbons: the two next to the ipso carbon and the one directly across the ring. A benzylic cation shares its charge with the same three ring carbons (Carbocations).

Hydrogen tops the list, and that gives the reaction a second use. An aldehyde has an H on its carbonyl carbon, and that H migrates ahead of any carbon group. So butanal and mCPBA give butanoic acid, not propyl formate (HCO–O–CH₂CH₂CH₃), which would need the propyl to move. Tollens’ reagent, Jones reagent or KMnO₄ give the same acid by a different mechanism (Oxidizing an aldehyde).

Aptitude is not the only condition. The migrating group must sit anti-periplanar to the O–O bond: seen down the C–O bond, the two point in opposite directions, 180° apart, as in an E2. Only then does the filled orbital of the bond that moves line up with the empty σ* orbital of the O–O bond.

In an open-chain ketone, rotation about the C–O bond can bring either group into that position, so aptitude alone decides. In a rigid ring system one of the groups sometimes cannot get there, and then the lower-ranked group migrates. That is why aptitude is a strong tendency rather than a law.

The most useful case is the methyl ketone. Methyl comes last in the order, so the other group almost always moves, and the product is the acetate ester of that group.

The migrating group keeps its configuration

(R)-3-Methylpentan-2-one has a stereocenter right next to its C=O. That carbon is secondary and the group on the other side is a methyl, so the stereocenter migrates, and the product is (R)-butan-2-yl acetate. The CH₃ that pointed toward you still points toward you, and the H that pointed away still points away.

The bond from the migrating carbon never breaks completely. It slides from the carbonyl carbon to the oxygen in one motion, so the carbon never becomes a free cation that could flatten out. Its other three bonds are never touched. When a reaction leaves a stereocenter’s arrangement unchanged like this, it goes with retention of configuration (see Alcohol reactions).

In this example the R/S label also stays the same. The new O sits where the carbonyl carbon was, and both outrank every other group on the stereocenter. In other molecules the label can change even though the arrangement does not, because the priorities around the carbon change.

Rings become lactones

Cyclohexanone and mCPBA give oxepan-2-one, a seven-membered ring that contains one oxygen. That product is a lactone, a cyclic ester (Esters & amides). Its common name is ε-caprolactone.

Both groups on the carbonyl carbon of a cyclic ketone are ring carbons. Whichever one moves stays in the ring and simply bonds to the new oxygen, so the ring does not open. It grows by one atom, and that atom is the oxygen. Cyclopentanone gives the six-membered lactone oxan-2-one (δ-valerolactone) in the same way.

This makes the Baeyer–Villiger one of the few reliable ways in this course to enlarge a ring by one atom.

When the molecule also has an alkene

6-Methylhept-5-en-2-one has a methyl ketone at one end and a C=C at the other. Three alkyl groups sit on that C=C. Alkyl groups push electron density into the C=C, and the electron-rich π bond attacks the peroxyacid’s oxygen faster. With one equivalent of mCPBA, the C=C becomes an epoxide and the ketone is left alone.

Such an alkene epoxidizes much faster than a ketone undergoes the Baeyer–Villiger, so when both are present the alkene usually reacts first. A planned Baeyer–Villiger on such a molecule therefore needs the steps reordered, or the alkene protected: turned for a while into a group that mCPBA ignores, then turned back.

The balance can tip the other way. An alkene with an electron-withdrawing group on it, such as a C=O attached directly, epoxidizes slowly. A strained ketone, such as cyclobutanone, undergoes the Baeyer–Villiger unusually fast. So “the alkene reacts first” describes the usual case, not every case.

What carries forward

Very little here is new machinery. The reagent is the epoxidation reagent, and the key step is the 1,2-shift from carbocation chemistry. A peroxyacid inserts an oxygen between the carbonyl carbon and the group best able to carry a positive charge, so the migration order follows carbocation stability. Methyl ketones give acetate esters of the other group, cyclic ketones give lactones one atom larger, and the migrating group keeps its configuration. The product is an ester, which you already know how to hydrolyze.

A Baeyer–Villiger can also make esters that Fischer esterification makes badly. tert-Butyl acetate is one. In hot acid, tert-butyl alcohol loses water to give a tertiary cation, which mostly eliminates to an alkene, so that route fails. A tertiary group is the easiest carbon group to migrate, though, so 3,3-dimethylbutan-2-one and mCPBA give tert-butyl acetate directly.