Put cyclohexene and benzene in one flask with hydrogen gas and palladium on carbon, at room temperature. Cyclohexene takes up one H2 and becomes cyclohexane. Benzene does not react. Raise the pressure and the temperature far enough and benzene does react, but it goes all the way to cyclohexane. Any cyclohexadiene (a six-membered ring with two C=C bonds) or cyclohexene formed along the way is reduced faster than benzene itself, so none of it survives.
Now dissolve sodium in liquid ammonia, add a little ethanol, and add the benzene. This time the ring is reduced only partway, and the product is cyclohexa-1,4-diene (often written 1,4-cyclohexadiene): a six-membered ring with its two C=C bonds on opposite sides of the ring. This is the Birch reduction. It works by a different mechanism, the dissolving-metal reduction that turns an alkyne into a trans alkene in Alkenes & Alkynes.
That mechanism also explains why it stops halfway. Sodium in ammonia cannot reduce an isolated C=C; that is what stops the alkyne reduction at the alkene. Once the ring is two separate C=C bonds, nothing is left that it can reduce.
The conditions
The metal is sodium or lithium, and the solvent is cold liquid ammonia, as in the alkyne reduction. When a sodium atom dissolves, it gives up its outer electron and becomes Na+. That electron does not join any one molecule. Ammonia molecules cluster around it and hold it in solution, and an electron held this way is called a solvated electron. Solvated electrons color the solution deep blue, and they are the reducing agent.
The third ingredient is an alcohol, usually ethanol or tert-butyl alcohol, and it supplies the protons. In the alkyne reduction, ammonia supplied the protons. With benzene, ammonia is too slow. Benzene takes on an electron reluctantly and gives it back easily, so a proton must arrive before the electron leaves. Ammonia, with a pKa of 38, is too weak an acid to manage that. An alcohol, with a pKa of about 16 to 18, is strong enough.
The alcohol cannot be much more acidic than that. Any acid reacts with sodium directly, turning the metal into Na+ and hydrogen gas. A strong acid would do this so fast that no electrons would be left for the ring. Ethanol and tert-butyl alcohol react with sodium only slowly in cold ammonia, so most of the electrons still reach the benzene.
Electron, proton, electron, proton
The steps run in the same order as for the alkyne. Call the carbon that takes the first proton C1:
- Sodium gives benzene one electron. The result is a radical anion: the ring now carries one unpaired electron and one negative charge.
- C1 takes a proton from the alcohol. That leaves a cyclohexadienyl radical: C1 is now an sp³ CH2, and the other five carbons form one delocalized system, a pentadienyl unit. Resonance spreads the unpaired electron over C2, C4 and C6.
- A second sodium atom gives the radical one more electron. The result is a cyclohexadienyl anion: the same sp³ CH2, but the pentadienyl unit now holds a pair of electrons instead of one. Resonance spreads the negative charge over C2, C4 and C6 (the next section shows how).
- C4, straight across the ring from C1, takes a second proton from the alcohol, and cyclohexa-1,4-diene is finished.
C1 and C4 are labels for following the mechanism. The product's name numbers the ring again, starting from a double-bond carbon, so in cyclohexa-1,4-diene the C=C bonds start at C1 and C4 and the CH2 groups are C3 and C6. The worked example below draws a product numbered this way.
Why the double bonds end up 1,4
Look again at the cyclohexadienyl anion, with C1 as its sp³ carbon. Resonance puts the negative charge on three carbons: C2, C4 and C6. C2 and C6 sit next to C1, at the two ends of the delocalized system. C4 sits in the middle, straight across the ring from C1.
Where the second proton lands decides the product. A proton on C4 leaves the two C=C bonds apart, as cyclohexa-1,4-diene. A proton on C2 or C6 would leave them joined by a single bond, as cyclohexa-1,3-diene. That is a conjugated diene, whose p orbitals overlap across all four carbons, and it is the more stable of the two. Yet the reaction gives the 1,4-diene.
So stability does not decide the outcome. The alcohol protonates the middle carbon faster than either end. The 1,4-diene does not turn into the 1,3-diene under these conditions. So the product that forms faster is the one you isolate. That is kinetic control, as in Conjugation & Pericyclic Reactions.
Substituents decide where the double bonds end up
Take two examples. Anisole (methoxybenzene) gives 1-methoxycyclohexa-1,4-diene. The carbon carrying OCH3 is still on a double bond, and the two sp³ carbons are C3 and C6. Benzoic acid gives cyclohexa-2,5-diene-1-carboxylic acid. Here the carbon carrying COOH has become sp³, with a hydrogen of its own, and the other sp³ carbon is C4, straight across from it.
In both products the two sp³ carbons sit straight across the ring from each other, as they do in benzene's own product. What changes is which two carbons they are. OCH3 pushes electron density into the ring, so it is an electron-donating group. COOH pulls electron density out, so it is an electron-withdrawing group. The two kinds give opposite results:
| Ring bears | The substituted carbon ends up | Product |
|---|---|---|
| Electron-donating group (OCH3, CH3) | On a double bond | 1-substituted cyclohexa-1,4-diene |
| Electron-withdrawing group (COOH, COOR) | On an sp³ carbon | 1-substituted cyclohexa-2,5-diene |
The reason is where the negative charge can best sit. A withdrawing group stabilizes negative charge on the carbon it is attached to. The anion places its charge there, so that carbon takes the last proton and ends up sp³. A donating group destabilizes negative charge on its own carbon. The charge sits elsewhere, so that carbon never takes a proton, and it stays on a double bond.
Benzoic acid has one extra step. Its O–H proton is far more acidic than any C–H, so sodium removes it first (giving H2), and the ring being reduced is really the benzoate ion. A carboxylate still withdraws electron density from the ring, so the result is the same. Adding acid at the end of the reaction puts the proton back on the COO−.
Aldehydes and ketones are missing from the table on purpose. Sodium in ammonia reduces their C=O group itself, so an aryl ketone such as acetophenone is not a clean case of ring reduction.
Esters need gentler conditions: tert-butyl alcohol at about −78 °C. With sodium and ethanol, sodium can reduce the ester C=O instead.
Step 1. Sort the substituents by type. OCH3 is a strong donor, and CH3 is a weak one. They are para to each other, so there is no conflict: both keep the negative charge away from their own carbon.
Step 2. Ask where the charge sits. It avoids C1, which carries the OCH3, and C4, which carries the CH3. So the protons go to carbons that carry neither group.
Step 3. Place the sp³ carbons. They sit straight across the ring from each other, so here they are C3 and C6, both unsubstituted.
Step 4. Draw the product. 1-Methoxy-4-methylcyclohexa-1,4-diene has double bonds C1=C2 and C4=C5 and sp³ carbons at C3 and C6. Both substituted carbons are still on double bonds.
Step 5. Check it. Donors keep their carbons unsaturated, and the two double bonds are not conjugated. Both hold.
Why you would want a cyclohexa-1,4-diene
A cyclohexa-1,4-diene has two separate C=C bonds, and each one reacts as an ordinary alkene. It can be epoxidized, hydrated or hydrogenated with the reactions of earlier chapters, often at just one of the two double bonds.
Anisole gives the most useful case. Its product, 1-methoxycyclohexa-1,4-diene, has an OCH3 group sitting directly on a C=C. That arrangement is an enol ether (see Enolate Chemistry). It is the oxygen version of an enamine, which has a nitrogen on a C=C (see Carbonyl Chemistry).
Mild aqueous acid hydrolyzes an enol ether to a ketone and methanol, much as it hydrolyzes an enamine to a ketone and an amine. The first step puts a proton on C2, the C=C carbon that does not carry the oxygen, just as an enamine takes its first proton on carbon when it hydrolyzes. From the anisole product, the ketone is cyclohex-3-en-1-one.
Be precise about that ketone. Label the carbon next to the C=O as α, the next one β and the one after that γ. The C=C of cyclohex-3-en-1-one sits between the β and γ carbons, so it is a β,γ-unsaturated ketone, and its C=C is not conjugated with the C=O. Stronger acid or base moves the double bond one carbon closer. That gives cyclohex-2-en-1-one, an α,β-unsaturated ketone.
The same hydrolysis turns the worked example's product, 1-methoxy-4-methylcyclohexa-1,4-diene, into 4-methylcyclohex-3-en-1-one, whose C=C is likewise not conjugated with the C=O.
Only the conjugated isomer is a Michael acceptor, the enone of the Michael and Robinson reactions in Enolate Chemistry. So Birch reduction, hydrolysis and isomerization make a cyclohexenone from an aromatic ring, rather than from the condensation of two carbonyl compounds.
Naphthalene: one ring at a time
Naphthalene is two benzene rings fused along one shared C–C bond. It takes on an electron far more easily than benzene does, so its first ring reduces readily. With just enough metal, only that ring is reduced, giving 1,4-dihydronaphthalene. What is left is an ordinary benzene ring, which is reduced more slowly. More metal and alcohol reduce it too. Run that way from the start, the reaction gives 1,4,5,8-tetrahydronaphthalene in one step.
What carries forward
Sodium or lithium in liquid ammonia with an alcohol turns a benzene ring into a cyclohexa-1,4-diene by adding electrons and protons in turn. A carbon carrying a donating group stays on a double bond, and a carbon carrying a withdrawing group ends up sp³. Both rules come from asking where the negative charge sits. It is the one reaction in this course that stops partway when reducing a benzene ring. Hydrogenation, by contrast, reduces the ring all the way to a cyclohexane.