1,2- vs 1,4-addition — Conjugation & Pericyclic Reactions

Add one equivalent of HBr to buta-1,3-diene and you get two products, not one. Neither is a mistake, both come from the same intermediate, and which one dominates depends on the temperature — which is the most surprising thing in this chapter and the reason the next section exists.

One protonation, one intermediate, two ends

The mechanism opens exactly as an ordinary alkene addition does. The π system is nucleophilic, HBr supplies the proton, and the proton adds to a terminal carbon, C1. What is left is a carbocation — but not an ordinary one.

Protonating C1 puts the positive charge on C2, which is next door to the remaining C3=C4 double bond. The empty p orbital and that π bond are conjugated, so the cation is allylic and delocalized:

CH₃–C⁺H–CH=CH₂ ↔ CH₃–CH=CH–C⁺H₂

The two resonance forms put the positive charge on C2 and on C4, and on nothing in between. So when bromide arrives, there are two electrophilic carbons to choose from, and the two choices give two different compounds:

Bromide attacks C2 — the 1,2-addition product, 3-bromobut-1-ene. The new C–Br and the H that was added sit on adjacent carbons, and the double bond that survives is terminal.
Bromide attacks C4 — the 1,4-addition product, 1-bromobut-2-ene. The H went on C1 and the Br on C4, at opposite ends of the original diene, and the double bond has moved into the middle.

"1,4-addition" is also called conjugate addition, and the numbers refer to the positions of the original diene, not to the product's own numbering. The double bond appearing somewhere it was not drawn is not a rearrangement — it is the other resonance form of the intermediate being captured.

The ratio depends on the temperature, and that is the real finding

Run the reaction cold, at around −80 °C, and the 1,2-product dominates, roughly 80 to 20. Run the same reaction warm, at around 40 °C, and the ratio inverts to roughly 15 to 85 in favor of the 1,4-product.

Nothing about the mechanism changed. Same diene, same reagent, same intermediate. What changed is which of two competing considerations is allowed to decide the outcome — and that is the subject of the next section.

A decisive experiment: take the pure 1,2-product, warm it to 40 °C with a trace of HBr, and you get the same 15:85 mixture. The products are interconverting, which means the first-formed product is not necessarily the final one. Hold that fact — it is the entire mechanism of thermodynamic control.

Why the 1,2-product forms faster

One protonation, one cation, and then a choice. Bromide can land on C2 or on C4 — the two carbons the resonance forms put the charge on — and the two landings give compounds that differ in where the bromine sits and where the surviving double bond ended up.Note what is not different between the two products: the bromine came from the same bromide and the hydrogen went to C1 in both cases. The double bond looks as though it moved, and it did not; the cation never had a double bond in one fixed place to begin with. The temperature rows are the finding — the mechanism is identical at both, which is exactly why the next section is about conditions rather than about arrows.

Look again at the two resonance forms of the allylic cation. They are not equally good descriptions of it. The form with the charge on C2 is a secondary carbocation; the form with the charge on C4 is primary. The secondary form is the major contributor, so the real intermediate carries more positive charge on C2 than on C4.

Bromide attacks the more electrophilic carbon faster, and that is C2. The 1,2-product is therefore the one that forms first, and at a temperature low enough that nothing can go back, it is the one you isolate.

Why the 1,4-product is more stable

Now compare the products rather than the transition states. The 1,2-product, 3-bromobut-1-ene, has a monosubstituted terminal double bond. The 1,4-product, 1-bromobut-2-ene, has a disubstituted internal one. Alkene stability rises with substitution, so the 1,4-product sits lower in energy.

Given enough thermal energy for bromide to leave again and re-form the allylic cation, the system samples both products repeatedly and drains into the more stable one. That is why warming inverts the ratio, and why the inversion also happens when you warm the pure 1,2-product on its own.

Worked example — predicting both products

Add HBr to 2-methylbuta-1,3-diene (isoprene), CH₂=C(CH₃)–CH=CH₂.

Protonate at a terminus. Adding H⁺ to C1 gives a cation delocalized over C2 and C4. The C2 form is tertiary and the C4 form is primary, so the charge sits overwhelmingly on C2 and this is much the better protonation.

Capture at each end. Bromide at C2 gives the 1,2-product, 3-bromo-3-methylbut-1-ene, keeping the terminal alkene. Attack at the far end instead moves the double bond inward, giving 1-bromo-3-methylbut-2-ene.

Assign kinetic and thermodynamic. C2 carries far more positive charge, so the 1,2-product forms faster and dominates cold. But its alkene is monosubstituted and terminal, while the 1,4-product's is trisubstituted — a large stability gap, so warming and allowing the bromide to leave again drains the mixture into the 1,4-product.

Pick your test diene with care. Penta-1,3-diene looks like a good example and is not: protonating it at C1 gives a symmetric allylic cation with a methyl at each end, so capturing at C2 and at C4 give the same compound and there is no pair of products to compare.

What carries forward

The pattern — one delocalized intermediate, two places to capture it, and a product ratio that depends on conditions rather than on the mechanism — is not confined to dienes. You will meet it again in enolate chemistry, where the same molecule gives a kinetic enolate at low temperature with a bulky base and a thermodynamic enolate under equilibrating conditions. It is the same idea with a different intermediate, and the section after this one is the general statement of it.