Structure of the double bond
An alkene has one C=C double bond. General formula: CₙH₂ₙ (ethene C₂H₄, propene C₃H₆). Alkenes are unsaturated: they can take up more atoms. Old name: olefins ("oil makers"), because ethene + chlorine gives an oily liquid.
- Each C of the double bond is sp². Its three σ bonds lie in one flat plane at 120°.
- The leftover p orbitals on the two carbons overlap sideways. This makes the π bond: a cloud above and below the plane.
- C=C bond length 134 pm (shorter than C–C 154 pm). Bond energy 681 kJ/mol (more than C–C 348, but less than two C–C).
- The π electrons are loosely held and easy to reach, so alkenes attract electrophiles (electron-lovers) and do addition reactions.
Naming
Pick the longest chain that holds the C=C. Number from the end nearer the double bond. Change -ane to -ene and give its position: CH₃–CH=CH–CH₃ is but-2-ene. CH₂=C(CH₃)–CH₃ is 2-methylprop-1-ene.
Isomerism in alkenes
Structural isomerism
From C₄H₈ on, alkenes show chain and position isomers: but-1-ene (CH₂=CH–CH₂–CH₃), but-2-ene (CH₃–CH=CH–CH₃) and 2-methylpropene ((CH₃)₂C=CH₂).
Geometrical (cis–trans) isomerism
The π bond stops rotation about C=C. If each carbon of the double bond carries two different groups, the groups can be fixed on the same side or on opposite sides.
- cis: similar groups on the same side (cis-but-2-ene).
- trans: similar groups on opposite sides (trans-but-2-ene).
- If one carbon has two identical groups (like CH₂=CH–CH₃), there is no cis–trans isomer.
How cis and trans differ
- Polarity: in the trans form, bond dipoles point opposite ways and cancel. cis-but-2-ene is slightly polar, trans is almost non-polar. cis-1,2-dichloroethene has a dipole moment of 1.9 D; the trans form has 0 D.
- Boiling point: cis is usually higher (it is polar).
- Melting point: trans is usually higher (it is more symmetrical and packs better in a solid).
Preparation of alkenes
- From alkynes (partial reduction): H₂ with Lindlar's catalyst (Pd on BaSO₄/CaCO₃, poisoned with quinoline or sulphur) stops at the alkene and gives the cis alkene. Sodium in liquid ammonia gives the trans alkene.
- From alkyl halides (dehydrohalogenation): heat with alcoholic KOH. H from the β-carbon and X from the α-carbon leave: CH₃CH₂Br + KOH(alc) → CH₂=CH₂ + KBr + H₂O. This is β-elimination. Ease: I > Br > Cl; 3° > 2° > 1°.
- From vicinal dihalides (dehalogenation): two X on neighbouring carbons are removed by zinc dust: BrCH₂–CH₂Br + Zn → CH₂=CH₂ + ZnBr₂.
- From alcohols (acidic dehydration): heat with concentrated H₂SO₄ at 443 K: CH₃CH₂OH → CH₂=CH₂ + H₂O. Also a β-elimination.
Physical properties
C2–C4 are gases, C5–C17 liquids, higher ones solids. Ethene is colourless with a faint sweet smell. Alkenes do not mix with water but dissolve in non-polar solvents. Boiling point rises with size and falls with branching, just like alkanes.
Addition reactions of alkenes
1. Hydrogen
CH₂=CH₂ + H₂ → CH₃–CH₃ (Ni, Pt or Pd).
2. Halogens
CH₂=CH₂ + Br₂ → BrCH₂–CH₂Br (1,2-dibromoethane). Red-brown bromine water (or Br₂ in CCl₄) is decolourised. This is the lab test for a C=C. Reactivity: Cl₂ > Br₂ > I₂.
3. Hydrogen halides and the Markovnikov rule
Reactivity: HI > HBr > HCl. For a symmetric alkene there is only one product. For an unsymmetric one like propene, two products are possible.
Markovnikov rule: the negative part of the reagent (Br) goes to the carbon of the double bond that has fewer hydrogens. CH₃–CH=CH₂ + HBr → CH₃–CHBr–CH₃ (2-bromopropane, main product).
Why: H⁺ adds first and a carbocation forms. If H⁺ joins the CH₂ end, the + charge sits on the middle carbon (2° carbocation). If it joins the middle, the + sits on the end (1°). A 2° carbocation is more stable (more alkyl groups push electrons towards it), so that path wins. Br⁻ then joins the + carbon.
4. Peroxide effect (anti-Markovnikov, Kharash effect)
With an organic peroxide (like benzoyl peroxide), HBr adds the other way: CH₃–CH=CH₂ + HBr → CH₃–CH₂–CH₂Br (1-bromopropane). It goes by free radicals: the peroxide makes Br•, which joins the CH₂ end to leave the more stable 2° radical. HCl and HI do not show this effect (the H–Cl bond is too strong; the I• radical joins back to I₂).
5. Water (acid-catalysed hydration)
With a few drops of conc. H₂SO₄, water adds by the Markovnikov rule: (CH₃)₂C=CH₂ + H₂O → (CH₃)₃C–OH. Cold conc. H₂SO₄ itself adds to give alkyl hydrogen sulphates.
6. Oxidation
- Baeyer's reagent (cold, dilute, alkaline KMnO₄): purple colour fades and a glycol forms. CH₂=CH₂ + H₂O + [O] → HOCH₂–CH₂OH. Another test for C=C.
- Acidic KMnO₄ or K₂Cr₂O₇ (strong): breaks the C=C and gives ketones and/or acids. CH₃CH=CHCH₃ → 2CH₃COOH.
7. Ozonolysis
Ozone adds to C=C to form an ozonide. Zn + water then cuts it into two carbonyl compounds. Rule of thumb: cut the C=C and put =O on each end. CH₃–CH=CH₂ → CH₃CHO + HCHO. By looking at the pieces we can find where the double bond was.
8. Polymerisation
Under high pressure and heat with a catalyst, many ethene molecules join into one long chain: n CH₂=CH₂ → –(CH₂–CH₂)ₙ– (polythene). Propene gives polypropene.
Key formulas and definitions
- Alkene: CₙH₂ₙ
- CH₃CH₂OH → CH₂=CH₂ + H₂O (conc. H₂SO₄, 443 K)
- R–CH₂CH₂X + KOH(alc) → R–CH=CH₂ + KX + H₂O
- X–CH₂–CH₂–X + Zn → CH₂=CH₂ + ZnX₂
- Alkyne + H₂ (Lindlar) → cis-alkene
- CH₃CH=CH₂ + HBr → CH₃CHBrCH₃ (Markovnikov)
- CH₃CH=CH₂ + HBr (peroxide) → CH₃CH₂CH₂Br
- Ozonolysis: R₂C=CR'₂ → R₂C=O + O=CR'₂
Worked examples
1. Which of these show cis–trans isomerism: (a) propene, (b) but-2-ene, (c) 2-methylpropene?
Check each C of the double bond for two different groups. (a) CH₂=CH–CH₃: first C has H and H (same) → no. (b) CH₃–CH=CH–CH₃: each C has H and CH₃ → yes. (c) (CH₃)₂C=CH₂: both ends have identical pairs → no. Only but-2-ene.
2. Write the main product when 2-bromobutane is heated with alcoholic KOH.
β-H can be taken from C1 or C3. Taking from C3 gives but-2-ene (CH₃CH=CHCH₃, more substituted); from C1 gives but-1-ene. The more substituted alkene is the main product (Saytzeff rule): but-2-ene.
3. Give the product of HBr with propene (a) without and (b) with benzoyl peroxide.
(a) Markovnikov: H to CH₂, Br to the middle C → CH₃CHBrCH₃ (2-bromopropane). (b) Peroxide effect: Br to the end C → CH₃CH₂CH₂Br (1-bromopropane).
4. An alkene on ozonolysis gives ethanal (CH₃CHO) and propanone (CH₃COCH₃). Find the alkene.
Remove the two =O and join the carbons with a double bond: CH₃CH= and =C(CH₃)₂ → CH₃–CH=C(CH₃)–CH₃. It is 2-methylbut-2-ene.
5. An alkene C₄H₈ on ozonolysis gives only one product, ethanal. Name it.
Only one product means both halves are the same: CH₃CH= + =CHCH₃ → CH₃CH=CHCH₃. It is but-2-ene.
6. How many moles of H₂ react with 5.6 g of ethene? What mass of ethane forms?
Molar mass of C₂H₄ = 28 g/mol, so 5.6 g = 0.2 mol. One C=C takes one H₂, so 0.2 mol H₂. Ethane C₂H₆ = 30 g/mol → 0.2 × 30 = 6.0 g.
Common mistakes
- Saying every alkene has cis and trans forms. Each double-bond carbon must carry two different groups.
- Applying the peroxide effect to HCl or HI. It works only for HBr.
- Using aqueous KOH to make an alkene from R–X. Aqueous KOH gives an alcohol; you need alcoholic KOH.
- In ozonolysis, forgetting that each carbon of the old C=C gets an =O and keeps all its other groups.