Buta-1,3-diene and penta-1,4-diene both have two C=C double bonds. Yet buta-1,3-diene is measurably more stable, and it reacts in ways that penta-1,4-diene does not. Since the count is the same, the difference must lie in how the double bonds are arranged. In buta-1,3-diene one single bond joins the two double bonds; in penta-1,4-diene a CH₂ group sits between them.
A molecule with two C=C double bonds is a diene. This section sorts dienes by the arrangement of their double bonds, measures what the best arrangement is worth, and then builds the orbital picture that the later sections of this chapter use.
Isolated, conjugated, cumulated
Start with the three smallest cases. In penta-1,4-diene, CH₂=CH–CH₂–CH=CH₂, the middle carbon has four single bonds. In buta-1,3-diene, CH₂=CH–CH=CH₂, a single bond links the two double bonds directly. In propa-1,2-diene, CH₂=C=CH₂, the two double bonds share their middle carbon. Propa-1,2-diene is usually called allene.
What separates the three is whether the p orbitals of one double bond can overlap with those of the other. Drawing Molecules & Moving Electrons set out the test in its resonance section: electrons spread only along an unbroken row of parallel p orbitals, and an sp³ atom, which has no p orbital to offer, ends the row. Applied to dienes, the test gives three classes.
- Isolated: at least one sp³ carbon separates the double bonds, so the row of p orbitals breaks there. Penta-1,4-diene is the standard example. Each of its double bonds behaves as though the other were not there.
- Conjugated: the double bonds alternate with single bonds. Every carbon in the run is sp², so the p orbitals form one continuous row. Buta-1,3-diene is the simplest case.
- Cumulated: the double bonds share a carbon, as in allene. The shared carbon is sp, and like an alkyne carbon it has two leftover p orbitals at right angles to each other. It uses one for each π bond, so the two π bonds are perpendicular and cannot overlap at all.
Conjugation also reaches past double bonds. Take the allyl cation, CH₂=CH–CH₂⁺. Its end carbon is a carbocation: it has three bonds and an empty p orbital. That empty orbital lies parallel to the p orbitals of the C=C beside it, so the row runs over all three carbons.
The allyl radical and the allyl anion work the same way. In the radical, CH₂=CH–CH₂•, the end p orbital holds one electron; in the anion, CH₂=CH–CH₂⁻, it holds a pair.
In a benzylic cation, radical or anion, the carbon that carries the charge or the odd electron is bonded to a benzene ring, and its p orbital joins the ring's p orbitals in the same way.
This is why allylic and benzylic cations, radicals and anions are much more stable than ordinary ones. It is the delocalization you drew as resonance; conjugation is the orbital picture of it.
A lone pair on a neighboring atom can join the row too. In methoxyethene, CH₂=CH–O–CH₃, the oxygen is bonded directly to an alkene carbon. One of its lone pairs sits in a p orbital parallel to the π bond, so the row runs over C1, C2 and the oxygen.
- One sp³ carbon is enough to isolate. The row only has to break once. Reading the rule as "two or more sp³ carbons" makes penta-1,4-diene look conjugated, which it is not.
- Count p orbitals, not double bonds. Methoxyethene has only one C=C and is still conjugated. The same count explains enamines in Carbonyl Chemistry and amides in Carboxylic Acids & Derivatives.
Classify the double bonds of hepta-1,3,6-triene, CH₂=CH–CH=CH–CH₂–CH=CH₂.
Number the chain and mark every sp² carbon. The double bonds are C1=C2, C3=C4 and C6=C7. That makes C1, C2, C3, C4, C6 and C7 sp². C5 is the only sp³ carbon in the molecule.
Walk the chain and look for a break. From C1 to C4 the carbons are all sp², and the bonds alternate: C1=C2, then the single bond C2–C3, then C3=C4. Those two double bonds are conjugated with each other. At C5, which is sp³, the run stops.
Classify the third. C6=C7 lies on the far side of that sp³ carbon, so it is isolated from the other two. It behaves like an ordinary terminal alkene.
Give the whole answer. Hepta-1,3,6-triene contains a conjugated diene and an isolated alkene. Questions often ask how many π electrons are in conjugation. Here the answer is four, the two π bonds from C1 to C4, not six.
Conjugation is worth about 15 kJ/mol (3.6 kcal/mol)
The heat of hydrogenation, which you used in Alkenes & Alkynes to rank alkenes, is the heat an alkene gives out when H₂ adds across its double bond. When two compounds end at the same kind of product, here an alkane, the one that gives out less heat started lower in energy, and so it was more stable. The same measurement puts a number on the delocalization energy: the extra stability a molecule gains because its π electrons spread over more atoms.
You are given three heats of hydrogenation: but-1-ene −127, penta-1,4-diene −254 and buta-1,3-diene −239 kJ/mol (about −30, −61 and −57 kcal/mol). Find how much conjugation stabilizes buta-1,3-diene.
Build the reference first. Ask how much heat buta-1,3-diene would give out if its two double bonds ignored each other. One ordinary C=C gives out 127 kJ/mol, so two independent ones should give out 2 × 127 = 254. Penta-1,4-diene, whose double bonds really are independent, measures almost exactly that. So the reference is backed by a measurement, not just arithmetic.
Subtract the measurement from the reference. 254 − 239 = 15 kJ/mol (3.6 kcal/mol).
Decide which molecule the 15 kJ/mol belongs to. Buta-1,3-diene gave out less heat, so it started 15 kJ/mol lower. The 15 kJ/mol stabilizes the starting diene, not the product.
The trap: comparing buta-1,3-diene with one but-1-ene instead of two. The reference must contain as many π bonds as the compound you are testing. Otherwise the subtraction measures the second double bond, not the conjugation.
So the delocalization energy of buta-1,3-diene is about 15 kJ/mol (about 3.6 kcal/mol). Only the conjugated arrangement earns it. Allene sits at the other end. Hydrogenating it gives out about 298 kJ/mol (71 kcal/mol), about 44 kJ/mol (10.5 kcal/mol) more than two isolated C=C (254), so it starts higher still. From most to least stable, the measured order is conjugated, isolated, cumulated.
Bond lengths give a second, independent measurement. In buta-1,3-diene, the C2–C3 bond is formally a single bond, yet it is only about 1.47 Å long. An ordinary C–C single bond is 1.54 Å, and the C=C bonds in the same molecule are 1.34 Å. The usual reading is that C2–C3 has partial double-bond character: the π electrons spread over all four carbons, so some of them sit between C2 and C3.
s-cis and s-trans: a conformation, not a configuration
The C2–C3 bond of buta-1,3-diene is a single bond, so the two halves of the molecule can rotate about it, just as the two halves of butane rotate about their central bond. Two flat arrangements matter. Their names, s-trans and s-cis, carry an s for single, the bond the rotation happens about.
- s-trans: the two double bonds point in opposite directions across C2–C3. This is the more stable conformation, and most butadiene molecules are in it at any moment.
- s-cis: both double bonds point the same way, which brings C1 and C4 close together.
The s-cis form costs about 12 kJ/mol (roughly 3 kcal/mol), nearly as much as the whole delocalization energy. The cost comes from one clash. C1 and C4 each carry a hydrogen that points inward in the flat s-cis form, and those two hydrogens run into each other. Butadiene eases the clash by twisting a little out of plane, so its real minor conformer is skewed rather than flat. "s-cis" names the idealized flat form.
Do not confuse s-cis and s-trans with cis and trans across a double bond. Cis and trans are configurations: turning one into the other means breaking a π bond, so they are two different compounds. s-cis and s-trans are conformations of one compound, and they interconvert freely by rotation at room temperature. The prefix s is the only thing that marks the difference in writing.
A ring can remove the choice. In cyclopentadiene, the ring holds both double bonds on the same side of the central single bond, so the diene is locked s-cis. When two fused rings share the central bond and each ring holds one of the double bonds, the double bonds sit on opposite sides of it, and the diene is locked s-trans.
The orbital picture: one molecular orbital per p orbital
Start with ethene. Its two p orbitals combine into two molecular orbitals, orbitals that spread over both carbons instead of belonging to one. In the lower one the two lobes are in phase and reinforce each other between the carbons: this is the bonding π orbital. In the upper one they are out of phase and cancel between the carbons, leaving a node, a plane where the orbital is zero. That is the antibonding orbital, written π*.
Ethene's two π electrons fill π, and π* stays empty. It is the same in-phase and out-of-phase overlap that Foundations used for two hydrogen atoms.
Buta-1,3-diene has four p orbitals, so it has four π molecular orbitals, labeled ψ₁ to ψ₄ (ψ, the Greek letter psi, is the usual symbol for an orbital). In ψ₁ every lobe is in phase and there is no node between the carbons. Each orbital above it has one more node, so ψ₄ has three. The four π electrons fill the lowest two, ψ₁ and ψ₂, two to each.
That makes ψ₂ the highest occupied molecular orbital, or HOMO, and ψ₃ the lowest unoccupied one, or LUMO. Together they are the frontier orbitals, and they are usually the only two that matter in a reaction. The HOMO holds the electrons the molecule gives up most easily, and the LUMO is the lowest-energy place for arriving electrons to go.
The pattern holds for any length. Overlap n parallel p orbitals and you get n π molecular orbitals. The lowest has no node between the carbons, and each one above it adds one, up to n − 1 in the highest. In a neutral chain with an even number of sp² carbons, each carbon brings one π electron, so the n electrons fill the lowest n/2 orbitals and the top half stay empty. An odd-numbered or charged system, such as the allyl case below, has the same kind of ladder but a different electron count, so count its electrons before you fill anything.
As the chain grows, the ladder changes shape in two ways. It stretches a little, because the bottom orbital sinks and the top one rises. It also gains orbitals, and it gains them much faster than it gains height, so they crowd together. The result is that the HOMO rises, the LUMO falls, and the HOMO–LUMO gap narrows. UV-Vis spectroscopy, later in this chapter, measures that gap: the longer the conjugated system, the longer the wavelength of light it absorbs.
Look again at ψ₂ and ψ₃ of butadiene. In both, the lobes on C1 and C4 are larger than those on C2 and C3. So the end carbons are where butadiene most readily donates electrons from its HOMO and accepts them into its LUMO.
Why the ends of the system are where the chemistry happens
Draw the resonance forms of the allyl cation. In the first, the positive charge is on C3. A curved arrow moves the π electrons of C1=C2 into the C2–C3 bond. The double bond is now C2=C3, and C1 has lost its share of those electrons, so the positive charge sits on C1. The charge appears on C1 in one form and on C3 in the other, and on C2 in neither.
The allyl radical and the allyl anion follow the same pattern. Their resonance forms put the odd electron, or the negative charge, on C1 and C3, and never on C2.
The orbital picture gives the same answer. Three p orbitals make three molecular orbitals. In the middle one, ψ₂, the node passes straight through C2, so C2 has no lobe in it at all. ψ₂ is also the orbital whose filling changes across the three species: it is empty in the cation, holds one electron in the radical and holds two in the anion. Whatever ψ₂ holds, or lacks, therefore belongs to C1 and C3. The resonance forms and the orbitals describe the same distribution.
Compare this with butadiene's ψ₂. There the node falls between C2 and C3, so every carbon still carries a lobe. In the allyl ψ₂ the node falls on an atom, and that atom drops out.
This sets up the next section. When H⁺ adds to a conjugated diene, the intermediate is an allylic cation. That cation has two positive ends, so the bromide or other nucleophile that follows has two places to attack.
What carries forward
Three ideas from this section return later. The extra stability of conjugated systems explains the allylic bromination previewed in How Reactions Happen. It also explains why an elimination that can form a conjugated diene favors that product. And it is the starting point for the special stability of benzene in Aromatic Chemistry. The s-cis conformation returns in the Diels–Alder section later in this chapter. And the orbital ladder, whose gap narrows as the chain lengthens, is what UV-Vis spectroscopy measures.