Heat buta-1,3-diene with ethene under pressure, and the two molecules join into cyclohexene, a six-membered ring, in a single step. There is no intermediate and no catalyst, and the starting materials decide the stereochemistry of the product. This is the Diels–Alder reaction. It is one of the most useful reactions in organic chemistry, and almost everything about it follows from one picture: six electrons moving round a ring.
What it is
In that example, buta-1,3-diene is the diene. It is a conjugated diene: two double bonds joined by one single bond, holding four π electrons in all, as in Conjugated systems. Ethene is the dienophile ("diene-loving"), the two-π-electron partner that the diene reacts with. A four-π-electron piece and a two-π-electron piece join to make a ring, so the reaction is called a [4+2] cycloaddition.
The dienophile can also be an alkyne. Ethyne and buta-1,3-diene give cyclohexa-1,4-diene. Only one of the alkyne's two π bonds takes part, so the other survives as a second double bond in the ring.
Three curved arrows describe the whole reaction. Number the diene's carbons C1 to C4. The C1=C2 π electrons form a new σ bond from C1 to one end of the dienophile. The dienophile's π electrons form a new σ bond to C4. The C3=C4 π electrons shift to become a new π bond between C2 and C3, where the diene had a single bond. The three arrows run head to tail, so the six electrons travel round a closed loop.
Counting bonds shows why the reaction is downhill. Three π bonds go in, and two σ bonds and one π bond come out. A σ bond is stronger than a π bond, so trading two π bonds for two σ bonds releases energy.
A reaction in which all the bonds change together, as electrons move round a closed loop, is called pericyclic. That is the word in this chapter's title.
Pericyclic reactions come in three classes. Cycloadditions, like this one, join two π systems into a ring. Electrocyclic reactions close a single conjugated chain into a ring by bonding its two ends together. Sigmatropic shifts move a σ bond to a new position along a π system; the Cope and Claisen rearrangements are examples (not the Claisen condensation met later in the course). This section covers cycloadditions. The other two classes close this chapter, in Electrocyclic & sigmatropic reactions.
Cycloadditions are not new to the course. The ozone and OsO₄ additions in Oxidative cleavage & dihydroxylation were cycloadditions too.
The diene must be able to reach s-cis
Look at the diene in the figure above. C1 and C4 both have to reach the dienophile at the same moment, so they must sit on the same side of the diene. That is true only in the s-cis conformation, which Conjugated systems introduced. A diene that cannot reach s-cis does not react, however good its electronics. This is the most common reason a Diels–Alder fails.
Cyclopentadiene shows the fast extreme. Its ring holds the diene s-cis all the time, so it never has to rotate into position. It is so reactive that two of its molecules combine at room temperature, one acting as the diene and the other as the dienophile. Cyclopentadiene is therefore sold and stored as this dimer, and heated to split it back just before use. The last section of this page shows how that splitting works.
Two rings can lock a diene the other way. In the fused-ring diene in the figure below, the two rings share the diene's central single bond, one double bond sits in each ring, and the rings hold the two ends of the diene pointing away from each other, s-trans. That diene cannot react at all, however electron-rich it is.
An open-chain diene can rotate. Buta-1,3-diene spends most of its time s-trans, the more stable conformation, but rotation about its central single bond is fast, so it reaches s-cis often enough to react well.
Crowding can still slow an open-chain diene. In the s-cis conformation, each end carbon has one position pointing outward, away from the other end, and one pointing inward, toward it. The two inward groups nearly touch. (Z)-Penta-1,3-diene has its methyl in an inward position, where it clashes with the inward hydrogen on the other end. So it reacts far more slowly than the (E) isomer, whose methyl points outward. Any bulky group in an inward position has the same effect.
Electronics: electron-rich diene, electron-poor dienophile
Ethene is a poor dienophile. With buta-1,3-diene it needs high temperature and pressure. Propenal, CH₂=CH–CHO, has an aldehyde group on its C=C and reacts with the same diene far more easily. Maleic anhydride, a five-membered ring with a C=O attached to each carbon of its C=C, is faster still. Good dienophiles share one feature: a group attached directly to the C=C that pulls electron density out of it.
The frontier orbitals from Conjugated systems explain why. The diene supplies the electrons, so its HOMO matters: the highest-energy orbital that holds electrons. The dienophile accepts them, so its LUMO matters: the lowest-energy orbital with room for them. The closer these two orbitals are in energy, the more strongly they interact, and the faster the reaction goes.
Substituents move those two orbitals. An electron-donating group pushes electron density into a π system; on the diene, it raises the HOMO. An OCH₃ group is a strong donor, and an alkyl group such as CH₃ is a weak one. An electron-withdrawing group pulls electron density out; on the dienophile, it lowers the LUMO. A carbonyl group (as in an aldehyde, ketone, ester or anhydride), a nitrile (C≡N) and a nitro group (NO₂) all withdraw. Either change narrows the gap. Ethene has no withdrawing group, so its LUMO stays high, and it reacts slowly.
Regiochemistry: which way round the partners join
Now give both partners a substituent. When 1-methoxybuta-1,3-diene reacts with propenal, the two molecules can line up in two ways, and each way puts the OCH₃ group (often written OMe, where Me means methyl) and the CHO in different places on the new ring. One product strongly dominates. When the atoms can bond in more than one way, the question of which way wins is called regiochemistry, the same question Markovnikov's rule answers for additions to alkenes.
On this page, a diene's four carbons are always numbered C1 to C4 along the diene, from the end nearer its substituent. These numbers are not the locants in the compound's name. In penta-1,3-diene, for example, the methyl-bearing carbon is C1 of the diene, although its name calls it C4.
Resonance shows where each partner is polarized. In 1-methoxybuta-1,3-diene, a lone pair on the oxygen can push into the π system. Follow the arrows in the top row of the figure below: the negative charge they push along ends up on C4. That resonance form is only a minor contributor, but it means C4 carries a small negative charge, δ−, in the real molecule.
Propenal is polarized the other way. Its C=O pulls electrons toward the oxygen, and the small positive charge, δ+, appears on the CH₂ carbon at the far end of the C=C, not on the carbon that carries the CHO.
The two atoms that attract each other bond together: C4 of the diene, δ−, bonds to the CH₂ carbon of propenal, δ+. The second new bond then has only one place to go, from C1 to the carbon that carries the CHO. Lining up the other way would join the δ− end of the diene to the carbon of propenal that is not δ+, so that orientation is slower. The OCH₃ on C1 and the CHO end up on neighboring ring carbons.
Move the methoxy group to C2 and the answer changes. In 2-methoxybuta-1,3-diene, the oxygen lone pair pushes the negative charge onto C1 instead (bottom row of the figure above). So C1 bonds to the CH₂ carbon of propenal, C4 bonds to the carbon carrying the CHO, and the two groups end up across the ring from each other.
The same reasoning works for any donor and any withdrawing group:
- A donor on C1 of the diene makes C4 the δ− end.
- A donor on C2 of the diene makes C1 the δ− end.
- A withdrawing group on the dienophile makes the other alkene carbon, the one it is not attached to, the δ+ end.
Joining δ− to δ+ gives one of two patterns. Chemists name them with the words ortho and para, borrowed from benzene rings (see Naming rings and unsaturation):
- A 1-substituted diene with a monosubstituted dienophile gives the "ortho" product, with the two groups on neighboring ring carbons, 1,2.
- A 2-substituted diene with a monosubstituted dienophile gives the "para" product, with the two groups 1,4 across the ring.
Take (E)-penta-1,3-diene, CH₂=CH–CH=CH–CH₃, with propenal, CH₂=CH–CHO. Numbered along the diene, the methyl sits on C1, so this is a 1-substituted diene. A methyl is only a weak donor, but it has the same effect as OCH₃, just smaller.
Polarize the diene. With the donor on C1, the δ− end is C4, the CH₂ at the far end.
Polarize the dienophile. The δ+ end of propenal is its CH₂ carbon, the one not carrying the CHO.
Join those two. The diene's CH₂ end bonds to the dienophile's CH₂. That leaves the diene's methyl-bearing carbon to bond to the carbon carrying the CHO. The methyl and the aldehyde are now on neighboring ring carbons: 2-methylcyclohex-3-ene-1-carbaldehyde, the "ortho" product. The stereochemistry section below finishes this answer.
Now move the methyl to C2. That diene is isoprene, 2-methylbuta-1,3-diene. With the donor on C2, the δ− end is C1, so C1 bonds to the dienophile's CH₂ and C4 bonds to the carbon carrying the CHO. Count round the ring and the methyl and the aldehyde are 1,4 apart: 4-methylcyclohex-3-ene-1-carbaldehyde, the "para" product.
Stereochemistry, and why it is predictable
Because every bond forms in one step, nothing in either partner can rotate while the reaction happens. The geometry of both starting materials is copied into the product, so the reaction is stereospecific, as defined in Addition reactions. Three rules describe the stereochemistry of the product, and they are the main reason chemists use this reaction so often.
First, the dienophile keeps its geometry. In dimethyl maleate, the two ester groups (CO₂Me) are cis across the C=C. With buta-1,3-diene it gives a cyclohexene whose two esters are cis on the ring, on neighboring carbons and on the same face. Dimethyl fumarate, the trans isomer, gives the ring with the esters trans. Each dienophile gives one diastereomer. This works because both new bonds form on the same face of the dienophile, which is called suprafacial addition.
Second, the diene keeps its geometry. Draw (2E,4E)-hexa-2,4-diene s-cis, and both of its methyls sit in outward positions. With ethene it gives 3,6-dimethylcyclohexene with the two methyls cis on the ring. In (2E,4Z)-hexa-2,4-diene, one methyl is outward and the other inward, and the methyls come out trans. The addition is suprafacial on the diene too, and the rule is simple: two outward groups end up cis, and an outward group and an inward group end up trans. An E double bond at the end of the diene puts its substituent outward; a Z double bond puts it inward.
Third, the endo product is favored. Cyclopentadiene and maleic anhydride can stack in two orientations as they meet. In the endo orientation, the dienophile's substituent, here the anhydride ring, is tucked under the diene. In the exo orientation, it points away. Both orientations form the same two bonds but give different products.
The endo orientation is more crowded, yet it reacts faster. The usual explanation is a secondary orbital interaction. Tucked under the diene, the π system of the substituent's C=O overlaps with the π system the diene is using, and that extra overlap lowers the energy of the transition state without ever forming a bond. Other effects, such as electrostatic attraction, are also thought to help, and chemists still debate how much each one contributes.
So endo is the kinetic product, the one that forms faster (see Kinetic vs thermodynamic control). A Lewis acid catalyst (see the box above) often strengthens the endo preference.
Endo and exo are usually taught with this reaction, because the two orientations give clearly different compounds.
Skeleton. Cyclopentadiene is locked s-cis, with a CH₂ joining its C1 and C4. C1 and C4 bond to the two carbons of the anhydride's C=C and close a new six-membered ring. The CH₂ still links C1 and C4, so the product has two rings sharing atoms: it is bicyclic. It is a norbornene, the norbornane skeleton from Alkene structure with one C=C in it. The old CH₂ becomes a one-carbon bridge, and the new C=C sits in a two-carbon bridge.
Dienophile geometry. The anhydride holds its two C=O groups cis across the C=C, and the addition is suprafacial, so they stay cis: on neighboring carbons, on the same face.
Endo or exo. Endo wins. In the product, endo means the anhydride points to the face opposite the CH₂ bridge: down in the figure below. The two H's on those same carbons take the exo positions, pointing out sideways, on the CH₂-bridge side of the ring.
Answer. The endo adduct, cis-bicyclo[2.2.1]hept-5-ene-2,3-dicarboxylic anhydride, formed almost exclusively at room temperature.
The trap. "Endo is favored" does not mean "endo is more stable". The exo adduct is usually the more stable one. Endo wins on rate, and it stays the product only because the reaction does not reverse at room temperature. Heat it hard enough for long enough, and it starts to reverse, so the exo adduct can take over.
The endo rule also links the diene's substituents to the dienophile's. In the endo orientation, the dienophile's substituent ends up cis to the diene's outward groups on the new ring. The end-on view in the figure after the next example shows why. Seen end-on, the endo substituent points in under the diene, while the diene's inward groups point the opposite way. Groups that point opposite ways end up on opposite faces, so the endo substituent is trans to the inward groups, and therefore cis to the outward ones.
Go back to (E)-penta-1,3-diene and propenal. The regiochemistry gave 2-methylcyclohex-3-ene-1-carbaldehyde. Ring carbons 1 and 2 are neighboring stereocenters, so there is one more question: are the methyl and the aldehyde cis or trans?
Place the methyl. The diene is E, so in the s-cis conformation its methyl points outward.
Place the aldehyde. Endo wins, so the CHO is tucked under the diene as the bonds form.
Combine them. An endo substituent ends up cis to the diene's outward groups. So the methyl and the aldehyde are cis, and the full answer is cis-2-methylcyclohex-3-ene-1-carbaldehyde.
Put the three rules together and, in practice, one diene and one dienophile give one main diastereomer.
Running it backwards
Heat can run a Diels–Alder in reverse. The retro-Diels–Alder splits a cyclohexene of the right kind back into a diene and a dienophile, and it breaks the same two σ bonds that the forward reaction made. It runs well only when something pulls it forward, for example when one piece escapes as it forms. Cracking cyclopentadiene's dimer is exactly this: heating splits the dimer, and the cyclopentadiene distills off as it forms.
The same cut is a planning tool. Chemists often design a synthesis backwards: they start from the molecule they want and ask which simpler pieces could make it. The Synthesis & Retrosynthesis chapter develops that method. When the target contains a cyclohexene, a retro-Diels–Alder cut is one of the most useful first moves, because one reaction then builds the ring and places the substituents.
The target is 4-methylcyclohex-3-ene-1-carbaldehyde. Which diene and dienophile make it?
Find the double bond. It runs from C3 to C4, and C4 carries the methyl.
Place the cuts. Step one carbon out from each end of the double bond, to C2 and to C5. Those were the diene's end carbons. Cut the bond from each of them to the next atom round the ring: C2–C1 and C5–C6. The four atoms C2, C3, C4 and C5 become the diene, and C6 and C1 become the dienophile.
Restore the π bonds. Each diene end that lost a σ bond gets its double bond back, and the old C3=C4 bond becomes single. That gives CH₂=C(CH₃)–CH=CH₂, isoprene. The dienophile gets its π bond back: CH₂=CH–CHO, propenal.
Check the regiochemistry forwards. A 2-substituted diene with a monosubstituted dienophile gives the "para" product, and the methyl and the aldehyde in the target are indeed 1,4. So the plan works, and one reaction builds the ring with both substituents in place.
The method works for any cyclohexene. Find the ring's double bond. The two ring carbons one step out from its ends were the diene's C1 and C4, so cut the bond from each of them that leads away from the double bond. What falls out is a four-carbon diene and a two-carbon dienophile.
What carries forward
Three ideas carry forward. First, the Diels–Alder is the first reaction in this course whose outcome is read from frontier orbitals. Hydroboration was concerted too, but was not explained that way. That reasoning, together with suprafacial addition and stereospecificity, is the template for pericyclic chemistry generally. Second, the reaction is a standard way to make a six-membered ring, which makes it a fixture of planning syntheses backwards. Third, the endo rule is a clear reminder that "more stable" and "forms faster" are separate questions, which is the lesson of Kinetic vs thermodynamic control.