Topic 11.3 · Alkenes

Alkenes

1

Structure, Isomerism and Naming

1.1Structure and naming

  • Alkenes: unsaturated hydrocarbons with a C=C bond, CnH2n. Each C of C=C is sp², trigonal planar, 120°.
  • Naming: longest chain containing the C=C, number from the end giving C=C the lowest locant, e.g. CH₃CH=CHCH₂CH₃ is pent-2-ene
123451234

Substituents take the numbering fixed by the C=C: 4-methylpent-2-ene, 3,3-dimethylbut-1-ene. Name cis or trans as drawn.

1.2Isomers of C₄H₈

  • Chain and position isomers: structural isomers
  • Cis–trans isomers: due to restricted rotation about C=C (the π bond would have to break). Each C of C=C must have two different groups.
but-1-ene2-methylpropenecis-but-2-enetrans-but-2-ene

Why is trans more stable than cis?

  1. In the cis isomer, the bulky groups are on the same side
  2. → greater steric hindrance (repulsion) between them
2

Physical Properties

Similar to alkanes: non-polar, held by id–id, insoluble in water, less dense than water, b.p. rises with chain length and falls with branching.

2.1cis isomers have a higher b.p.

ClHClHcis: μ > 0ClHHCltrans: μ = 0net dipoledipoles cancel
  1. cis has a net dipole moment (polar), in trans the dipoles cancel
  2. → additional pd–pd between cis molecules → more energy to overcome

2.2cis isomers have a lower m.p.

  1. cis has lower symmetry (U-shape)
  2. → molecules pack less closely in the solid
  3. → weaker IMF in the lattice → less energy to overcome
trans: packs closelycis: gaps, weaker IMF
62Topic 11.3 · Alkenes
3

Preparation of Alkenes

3.1Dehydration of alcohols

  • Elimination of H₂O: excess conc. H₂SO₄ at 170 °C, or pass the vapour over heated Al₂O₃
  • If the OH carbon has different neighbouring C–H groups, more than one alkene can form
OHHHbutan-2-olexcess conc. H₂SO₄170 °Cbut-2-enemajorexcess conc. H₂SO₄170 °Cbut-1-eneminor

3.2Elimination of HX from halogenoalkanes

  • Elimination of HX: KOH in ethanol, heat under reflux
  • Major product: the more substituted alkene (more alkyl groups on the C=C carbons), which is more stable
BrHH2-bromobutaneKOH in ethanolheat under refluxbut-2-enemajorKOH in ethanolheat under refluxbut-1-eneminor
4

Electrophilic Addition

Why do alkenes react with electrophiles?
  1. The C=C has a high electron density, the π electrons lie above and below the plane and are exposed
  2. → attracts electron-deficient species: one weak π bond broken, two σ bonds formed

4.1Mechanism: CH₂=CH₂ + Br₂ in CCl₄, room temperature

CCHHHHBrBrδ+δ−slowCCBrHHHH+Br−carbocationfastCCBrHHBrHH1,2-dibromoethane
Step 1 (slow)The π electrons induce a dipole in Br₂ and attack Brδ+, Br–Br breaks heterolytically → carbocation + Br⁻
Step 2 (fast)The lone pair on Br⁻ is donated to the C⁺ atom.
Answer structureDescribe the mechanism Name it (electrophilic addition), then draw it with reagents and conditions, δ+/δ−, lone pairs, charges and curly arrows. No written explanation is needed.
Topic 11.3 · Alkenes63
4

Electrophilic Addition (continued)

4.2Markovnikov's rule: propene + HBr

propene + HBrH⁺+2° carbocationBr⁻Br2-bromopropanemajorH⁺+1° carbocationBr⁻Br1-bromopropaneminor
The ruleH adds to the C=C carbon with more H atoms, so the more stable 2° carbocation forms → 2-bromopropane is major.

Why is the 2° carbocation more stable?

  1. Alkyl groups are electron-donating
  2. → two alkyl groups disperse the positive charge more than one

Reactivity of HX: HI > HBr > HCl (H–X bond strength decreases).

5

Reactions of Alkenes

ReactionEquation, with reagents and conditions on the arrow (propene)Observation
+ HXCH₃CH=CH₂ + HBr HBr(g), r.t. CH₃CHBrCH₃–
+ X₂CH₃CH=CH₂ + Br₂ Br₂ in CCl₄, r.t. CH₃CHBrCH₂Brorange-red Br₂ decolourises
+ X₂(aq)CH₃CH=CH₂ + Br₂ + H₂O Br₂(aq), r.t. CH₃CH(OH)CH₂Br + HBrorange decolourises
+ H₂OCH₃CH=CH₂ + H₂O steam, H₃PO₄, 300 °C, 60 atm CH₃CH(OH)CH₃–
+ H₂CH₃CH=CH₂ + H₂ Ni, heat (or Pt / Pd, r.t.) CH₃CH₂CH₃–
mild [O]CH₃CH=CH₂ + [O] + H₂O cold KMnO₄, NaOH(aq) CH₃CH(OH)CH₂OHpurple decolourises, brown ppt MnO₂
strong [O]CH₃CH=CH₂ + 5[O] KMnO₄, H₂SO₄(aq), heat CH₃CO₂H + CO₂ + H₂Opurple decolourises

Hydration can also use cold conc. H₂SO₄, then warm water. With Br₂(aq), the bromohydrin is major as H₂O (the solvent) is present in far larger amount than Br⁻.

64Topic 11.3 · Alkenes
6

Oxidative Cleavage and Tests

Hot acidified KMnO₄ breaks C=C. Each half becomes:

Half of C=CProduct
=CH₂ (terminal)CO₂ + H₂O
=CHRcarboxylic acid, RCO₂H
=CRR′ketone, RCOR′

Split C=C, add =O to each C, then replace any –H on that C by –OH.

Test for C=C: add Br₂(aq) or Br₂ in CCl₄ at r.t. → orange decolourises.
Test for a terminal alkene (=CH₂): heat with acidified KMnO₄ → purple KMnO₄ decolourises, and a colourless, odourless gas forms that gives a white ppt with limewater (CO₂ from the =CH₂).

6.1Identify the alkene

Worked exampleHydrocarbon X, C₇H₁₂, is heated with acidified KMnO₄. The only organic products are CO₂, propanone and propanedioic acid, HO₂CCH₂CO₂H. Deduce the structure of X.

123X: CH₂=CHCH₂CH=C(CH₃)₂hot KMnO₄, H⁺CO₂+ H₂O1OOHOOH2O3
  1. Piece 1: CO₂ → from a terminal =CH₂
  2. Piece 3: propanone, (CH₃)₂C=O → from =C(CH₃)₂
  3. Piece 2: HO₂CCH₂CO₂H (two –CO₂H) → from =CH–CH₂–CH=, a C=C at each end
  4. Join the pieces at the C=C: X is CH₂=CHCH₂CH=C(CH₃)₂, 5-methylhexa-1,4-diene (C₇H₁₂ ✓)
Topic 11.3 · Alkenes65