Add HBr to buta-1,3-diene, CH₂=CH–CH=CH₂, using one molecule of HBr for each molecule of diene, so that only one of the two double bonds reacts. You get two products, 3-bromobut-1-ene and 1-bromobut-2-ene, and both come from the same carbocation. Which one you isolate depends on the temperature.
It builds on two earlier ideas: the allylic cation from Conjugated systems, whose charge sits on both ends of a three-carbon system, and the rule from Markovnikov / anti-Markovnikov that the proton adds where it leaves the more stable cation.
Where the proton goes
Number the diene C1 to C4 from one end. The first step is the one you know from any HBr addition to an alkene: a π bond takes the proton from HBr, and the H–Br bond breaks, releasing bromide. The proton can bond to an end carbon, C1, or to an inner carbon, C2. The molecule is symmetric, so C4 and C3 give the same results as C1 and C2.
Put the proton on C1 and the positive charge lands on C2. C2 sits right beside the C3=C4 double bond, so its empty p orbital overlaps with that π bond. This is the allylic cation from the opening: its charge is shared with the far end of that double bond.
A proton on C2 would instead leave the charge on C1. C2 now has four single bonds, so it is sp³ and has no p orbital. Nothing links C1 to the C3=C4 π bond, so C1 is a primary cation with no resonance at all, about the least stable cation this reaction could make.
The barriers to the two cations differ by tens of kJ/mol (several kcal/mol or more), so essentially all of the reaction goes through C1. This is the Markovnikov reasoning applied to a diene: the proton goes where it leaves the more stable cation, and here that cation is allylic.
One cation, two ends
The cation from protonating C1 has two resonance forms. In the first, the charge is on C2 and the double bond is still C3=C4. In the second, the π electrons of C3=C4 have shifted toward C2: the double bond is now C2=C3 and the charge is on C4. These are two drawings of one cation, not two cations that turn into each other. Its charge is shared between C2 and C4 and never sits on C3.
So bromide has two positive carbons it can bond to. It uses one of its lone pairs to bond to C2 or to C4, and each choice gives a different compound.
The two products
Bromide at C2 gives 3-bromobut-1-ene. The H and the Br sit on neighboring carbons, C1 and C2, and the C3=C4 double bond survives at the end of the chain. This is 1,2-addition.
Bromide at C4 gives 1-bromobut-2-ene. The H is on C1 and the Br is on C4, at opposite ends of the original diene, and the double bond now sits in the middle of the chain, between C2 and C3. This is 1,4-addition, also called conjugate addition.
Other routes to the same cation
Nothing in this argument depends on HBr. Any electrophile that adds to C1 of a conjugated diene leaves an allylic cation behind, and the nucleophile then has the same two ends to choose from.
Br₂ is one example. Addition reactions showed that Br₂ adds to an alkene through a bromonium ion, a three-membered ring that holds a positive bromine. With buta-1,3-diene, the ring forms across C1 and C2. It then opens at the C2–Br bond, because breaking that bond puts the charge on C2, next to C3=C4, where it is allylic. Breaking the C1–Br bond instead would leave a primary cation with no resonance.
Bromide then bonds to C2 or to C4. 1,2-addition gives 3,4-dibromobut-1-ene, and 1,4-addition gives 1,4-dibromobut-2-ene. Cold conditions favor the 1,2-product and warm conditions the 1,4-product, just as with HBr; the next sections show why.
The same allylic cation can also form from the other direction, by ionization. Put 3-bromobut-1-ene, the 1,2-product from HBr, in water. Its C–Br bond breaks, bromide leaves, and the allylic cation forms again. Water then bonds to C2 or to C4 and loses a proton, giving but-3-en-2-ol or but-2-en-1-ol. SN1 called this an allylic rearrangement: the new group can end up on a different carbon from the one the leaving group left, and the double bond shifts with it.
The temperature decides the ratio
Run the HBr addition at about −80 °C and the mixture is roughly 80% 1,2-product and 20% 1,4-product. Run it at about 40 °C and the ratio flips to roughly 15:85 in favor of the 1,4-product. The diene, the reagent and the cation are the same at both temperatures.
A second experiment shows what changed. Take the pure 1,2-product, warm it to 40 °C with a trace of HBr, and it turns into the same 15:85 mixture. At that temperature the two products interconvert: bromide leaves again by the ionization shown above, the allylic cation re-forms, and bromide bonds to either end. So the product that forms first is not necessarily the one you end up with.
Two separate comparisons are at work. One asks which product forms faster, and the other asks which product is more stable. For buta-1,3-diene they give different answers, and the next two sections take them one at a time.
Why the 1,2-product forms faster
Compare the two resonance forms of the cation as carbocations. When the charge is on C2, the charged carbon is bonded to two other carbons, C1 and C3, so that form is a secondary cation. When the charge is on C4, the charged carbon is bonded only to C3, so that form is primary.
The secondary form is the more stable of the two, so it is the major contributor: it describes more of the real cation. The real cation therefore carries more of its positive charge on C2 than on C4. Bromide bonds faster to the carbon with more positive charge, so the 1,2-product forms faster. Bromide very rarely leaves again at −80 °C, so whatever forms first stays, and the 1,2-product dominates.
Why the 1,4-product is more stable
Now compare the products themselves. In 3-bromobut-1-ene, the 1,2-product, the double bond is at the end of the chain, and only one carbon is bonded to it: it is monosubstituted. In 1-bromobut-2-ene, the 1,4-product, the double bond is inside the chain and has a carbon bonded to each end: it is disubstituted. Alkene structure showed that a more substituted alkene is more stable, so the 1,4-product is lower in energy.
At 40 °C both products keep re-forming the allylic cation, so the mixture settles at a ratio set by how stable each product is, which favors the 1,4-product.
So the 1,2-product wins on speed and the 1,4-product wins on stability, and the temperature decides which of the two comparisons sets the ratio. The next section, Kinetic vs thermodynamic control, names these two outcomes and turns this result into a general rule.
Unsymmetrical dienes: choose the end first
Buta-1,3-diene is symmetric, so protonating C1 and protonating C4 give the same cation. A diene with a substituent has two different ends, and you must decide which one takes the proton before you can predict any products. Isoprene shows how.
Add HBr to 2-methylbuta-1,3-diene (isoprene), CH₂=C(CH₃)–CH=CH₂. Number the diene so that the methyl sits on C2.
Try C1. The charge lands on C2. C2 is bonded to three carbons: the new CH₃ at C1, its own methyl, and C3. So this resonance form is a tertiary cation. The other form puts the charge on C4, which is primary. The better form is tertiary.
Try C4. The charge lands on C3, which is bonded to C2 and C4: a secondary cation. Its partner form, with the charge on C1, is only primary. The better form is secondary.
Pick the end. Tertiary beats secondary, so the proton goes to C1, and the reaction runs through the cation shared between C2 and C4.
Bond bromide to each end. Bromide at C2 gives the 1,2-product, 3-bromo-3-methylbut-1-ene, which keeps a monosubstituted double bond at the end of the chain. Bromide at C4 gives the 1,4-product, 1-bromo-3-methylbut-2-ene. Its double bond, C2=C3, carries three carbons, so it is trisubstituted.
Decide which wins cold and which wins warm. The tertiary form is the major contributor, so C2 carries far more of the charge. Bromide bonds there faster, so the 1,2-product forms faster and is the main product cold. The 1,4-product has a trisubstituted double bond against a monosubstituted one, so it is the more stable product and wins when the mixture is warmed.
The general rule, then: try both ends of the diene, compare the best resonance form of each cation, and put the proton on the end that gives the more stable one. That cation is the one both products come from.
Add HBr to 2-methylpenta-1,3-diene, CH₂=C(CH₃)–CH=CH–CH₃. Both C1 and C4 are ends of the conjugated system, so the first decision is which one takes the proton.
Try C1. The charge lands on C2, which carries the new CH₃, its own methyl and C3, so that form is tertiary. The other form puts the charge on C4, which is bonded to C3 and C5: secondary. The better form is tertiary.
Try C4. Now C3 holds the charge. It has two carbon neighbors, so this form is secondary, and the alternative form puts the charge on C1, which is primary. The better form is secondary.
Pick the end. Tertiary beats secondary, so the proton goes to C1, and the cation shares its charge between the tertiary C2 and the secondary C4, with more of it on C2.
Bond bromide to each end. Bromide at C2 gives the 1,2-product, 4-bromo-4-methylpent-2-ene, whose double bond is inside the chain and disubstituted. Bromide at C4 gives the 1,4-product, 4-bromo-2-methylpent-2-ene, whose double bond is trisubstituted.
Decide which wins cold and which wins warm. Bromide bonds to the tertiary carbon faster, so the 1,2-product forms faster and is the one you isolate cold. On stability the order reverses, because a trisubstituted double bond beats a disubstituted one, so warming drains the mixture into the 1,4-product.
What carries forward
The pattern on this page is one delocalized intermediate, two places where a nucleophile can bond to it, and a product ratio set by the conditions rather than by a change of mechanism. It is not limited to dienes. Enolate Chemistry meets the same choice when a ketone can lose a proton from either side of its C=O. The name conjugate addition returns in Cuprates and conjugate addition and again in Enolate Chemistry, for nucleophiles that add to the far end of a C=C–C=O system.