Alkenes

H2 Chemistry (9476), JC1

By the end of this topic, you can answer

1

Why does but-2-ene exist as two different compounds, while but-1-ene does not?

2

Why does trans-but-2-ene melt at a higher temperature than cis-but-2-ene?

3

How can you turn ethanol into ethene?

4

When HBr adds to propene, why does the bromine mostly end up on the middle carbon?

5

How does bromine water show that a hydrocarbon has a C=C bond, and how does hot KMnO4 show where that bond is?

Recall: what you know already

1How do you test for C=C at O levels?

Bromine water: orange to colourless.[1]

2From Introduction to Organic Chemistry: what is an electrophile?

An electron pair acceptor.[1]

3From Introduction to Organic Chemistry: which carbocation is more stable, 1° or 2°?

2°: two electron-donating alkyl groups disperse the positive charge.[1]
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

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

Structure, Isomerism and Naming

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.
1

Structure, Isomerism and Naming

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

Your turnWhich shows cis–trans isomerism: (a) but-1-ene, (b) pent-2-ene, (c) 2-methylbut-2-ene?

Only (b). In (a), C1 has two H. In (c), C2 has two CH₃.[1]
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.

  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

Physical Properties

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
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
3

Preparation of Alkenes

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
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

Electrophilic Addition

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

  • 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.
4

Electrophilic Addition

4.2Markovnikov's rule: propene + HBr

H 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
4

Electrophilic Addition

Your turnGive the major product of but-1-ene with HBr, and explain why it is major.

2-bromobutane, CH₃CHBrCH₂CH₃[1]
H adds to C1 (more H atoms), giving a 2° carbocation[1]
stabilised by two electron-donating alkyl groups which disperse the positive charge more than in the 1° carbocation.[1]

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

Quick checkIn step 1 of Br₂ + ethene, the Br–Br bond breaks

Ahomolytically
Bheterolytically
Cin the fast step
Dby UV light
NoteB: Brδ− takes both bonding electrons, forming Br⁻.
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⁻.

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

Oxidative Cleavage and Tests

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.

  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
6

Oxidative Cleavage and Tests

Your turnAn alkene, C₅H₁₀, gives propanone and ethanoic acid with hot acidified KMnO₄. Deduce its structure.

Propanone comes from the =C(CH₃)₂ end of the alkene, and ethanoic acid comes from the =CHCH₃ end.[1]
(CH₃)₂C=CHCH₃, 2-methylbut-2-ene[1]

Exit ticket

1Why can but-2-ene show cis–trans isomerism?

Restricted rotation about C=C, and each C of C=C has two different groups.[1]

2Reagents and conditions to make propan-2-ol from propene?

Steam, H₃PO₄ catalyst, 300 °C, 60 atm.[1]

3Name the mechanism of propene + HBr.

Electrophilic addition.[1]

4What does =CH₂ give with hot acidified KMnO₄?

CO₂ and H₂O.[1]