What are dienes?
Dienes (alkadienes) are hydrocarbons with two C=C double bonds. Their general formula is CnH2n−2. For example, C4H6 is butadiene and C5H8 is pentadiene or isoprene. We name them with the ending -diene and number the positions: buta-1,3-diene.
Three kinds: cumulated, conjugated, isolated
- Cumulated: C=C=C. The two double bonds share one carbon (propadiene).
- Conjugated: C=C–C=C. One single bond lies between the double bonds (buta-1,3-diene). This is the most important type.
- Isolated: C=C–C–C=C. Two or more single bonds separate them (penta-1,4-diene). It behaves like two separate alkenes.
Why conjugated dienes are special
In a conjugated diene all four carbons are flat and each has a p orbital. The pi electrons are not stuck in two places; they spread over the whole C1 to C4 stretch (delocalisation). This makes the molecule more stable than a diene with isolated bonds. The middle C–C bond is shorter than a normal single bond because it has a little double-bond character.
Industrial preparation of butadiene: heating butane or butene with a catalyst removes hydrogen. A classic lab-scale route (Lebedev) passes ethanol vapour over a ZnO/Al2O3 catalyst at about 400 °C.
1,2- and 1,4-addition
With HBr, the first step puts H+ on a terminal carbon (C1). The positive charge is then shared between C2 and C4 (an allylic cation), so Br− can attach at either place:
- 1,2-addition: CH2=CH–CH=CH2 + HBr → CH3–CHBr–CH=CH2 (3-bromobut-1-ene). The remaining double bond is at C3=C4.
- 1,4-addition: → CH3–CH=CH–CH2Br (1-bromobut-2-ene). The double bond moves to the middle (C2=C3).
At low temperature the 1,2-product forms faster and dominates. At higher temperature the more stable 1,4-product dominates. With excess reagent, both double bonds react (for example 2 mol Br2 per mol diene).
Rubber
Dienes can polymerise: thousands of units join end to end. Isoprene (2-methylbuta-1,3-diene) forms natural rubber, and butadiene forms polybutadiene (synthetic rubber). Each unit joins through a 1,4-link and a double bond stays in the chain, which lets the chain coil and stretch. Raw rubber is soft and sticky, so it is heated with a little sulfur (vulcanisation). Sulfur bridges join the chains, so the rubber becomes strong and springy and goes back to its shape.
Try it
In the 3D, press the 1,2 and 1,4 buttons and say where the Br and the double bond are in each product. Then lengthen the rubber chain and count the units. At home, stretch a rubber band: the long chains uncoil. Let go, and they coil back.
Key formulas and definitions
- Alkadiene: CnH2n−2
- Conjugated: C=C–C=C; isolated: C=C–(C)n–C=C, n ≥ 1 carbon between; cumulated: C=C=C
- 1,2-addition: H on C1, Br on C2; 1,4-addition: H on C1, Br on C4, double bond moves to C2=C3
- Isoprene: CH2=C(CH3)–CH=CH2 → natural rubber (1,4-polyisoprene)
Worked examples
1. Write the molecular formula of a diene with 6 carbon atoms.
CnH2n−2 with n = 6 gives C6H10.
2. Classify: (a) CH2=C=CH2 (b) CH2=CH–CH=CH2 (c) CH2=CH–CH2–CH=CH2.
(a) cumulated, (b) conjugated, (c) isolated. Look between the double bonds: in (a) there is no gap, in (b) there is just one plain single bond, and in (c) there is a CH2 carbon in between.
3. Give the 1,2- and 1,4-products of buta-1,3-diene with HBr.
1,2: CH3–CHBr–CH=CH2 (3-bromobut-1-ene). 1,4: CH3–CH=CH–CH2Br (1-bromobut-2-ene).
4. How many moles of Br2 fully saturate 0.5 mol of buta-1,3-diene?
Each C=C takes one Br2, so one diene takes 2 Br2. 0.5 mol needs 1.0 mol Br2.
5. What mass of HBr (M = 81 g/mol) adds to 5.4 g of C4H6 (M = 54 g/mol) in a 1:1 addition?
Moles C4H6 = 5.4/54 = 0.1 mol. HBr needed = 0.1 mol × 81 = 8.1 g.
Common mistakes
- Calling every diene conjugated. Count the single bonds between the double bonds first.
- Writing the 1,4-product with the double bond left at C3=C4. In 1,4-addition the double bond shifts to C2=C3.
- Forgetting the formula CnH2n−2 and writing CnH2n (that is an alkene).
- Thinking vulcanisation breaks the rubber. It adds sulfur bridges that make it stronger.