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14.2 Alkenes

Syllabus
9701–2028–2029
Topic
14.2
Level
AS

Learning objectives

14.2.1Preparation of alkenes• Recall preparation of alkenes: (a) elimination of HX from a halogenoalkane by ethanolic NaOH and heat; (b) dehydration of an alcohol, by using a heated catalyst (e.g. Al2O3) or a conc. acid (e.g. conc. H2SO4); (c) cracking of a longer chain alkane14.2.2Reactions of alkenes• Describe reactions of alkenes:- (a) the electrophilic addition of- (i) hydrogen in a hydrogenation reaction, H2(g) and Pt/Ni catalyst and heat- (ii) steam, H2O(g) and H3PO4 catalyst- (iii) a hydrogen halide, HX(g), at room temp.- (iv) a halogen, X2- (b) the oxidation by cold dil. acidified KMnO4 to form the diol- (c) the oxidation by hot conc. acidified KMnO4 leading to the cleavage of the carbon–carbon double bond and the identities of the subsequent products to determine the position of alkene linkages in larger molecules- (d) addition polymerisation exemplified by the reactions of ethene and propene14.2.3Of aq. bromine to show the presence of a C=C• Describe the use of aq. bromine to show the presence of a C=C bond14.2.4Mechanism: electrophilic addition in alkenes,• Describe the mechanism of electrophilic addition in alkenes, using bromine/ethene and hydrogen bromide/propene as examples14.2.5Inductive effects of alkyl groups• Describe/explain: the inductive effects of alkyl groups on the stability of primary, secondary and tertiary cations formed during electrophilic addition (this should be used to explain Markovnikov addition)

Prepare alkenes by elimination, dehydration or cracking

Starting material Change Reagent and conditions Example
halogenoalkane eliminate HX NaOH in ethanol, heat CH₃CH₂Br → CH₂=CH₂
alcohol eliminate H₂O (dehydration) heated Al₂O₃ catalyst, or concentrated H₂SO₄ and heat CH₃CH₂OH → CH₂=CH₂ + H₂O
longer-chain alkane crack C–C bonds heat, with Al₂O₃ C₁₀H₂₂ → C₈H₁₈ + C₂H₄

The first two routes form C=C by removing atoms from neighbouring carbon atoms; cracking redistributes the atoms of a longer alkane into smaller hydrocarbons. In every route, check that the proposed equation conserves C, H and any halogen or oxygen atoms.

Solvent is part of the reagent choice: hot ethanolic NaOH favours elimination from a halogenoalkane, whereas aqueous NaOH is associated with nucleophilic substitution. Do not interchange these conditions.

The C=C bond supports addition, oxidation and polymerisation

Alkene reaction Reagent and conditions Product change
hydrogenation H₂(g), Pt or Ni catalyst, heat alkane
hydration H₂O(g), H₃PO₄ catalyst alcohol
hydrogen-halide addition HX(g), room temperature halogenoalkane
halogen addition X₂ 1,2-dihalogenoalkane
mild oxidation cold, dilute, acidified KMnO₄ two –OH groups replace C=C, forming a diol
oxidative cleavage hot, concentrated, acidified KMnO₄ C=C breaks and each alkene carbon becomes an oxidised product
Groups originally attached to one C of C=C Final fragment from that carbon
two carbon groups, no H ketone
one carbon group and one H carboxylic acid
two H (terminal =CH₂) CO₂ and H₂O

Apply the cleavage rule separately to the two C=C carbon atoms. The carbon skeletons retained in the products reveal which two atoms were joined by the original double bond; conversely, joining the carbonyl or carboxyl carbon atoms reconstructs the alkene position.

nCHX2=CHX2→[−CHX2−CHX2X−]Xnn\ce{CH2=CH2 -> [-CH2-CH2-]_{n}}

nCHX2=CHCHX3→[−CHX2−CH(CHX3)X−]Xnn\ce{CH2=CHCH3 -> [-CH2-CH(CH3)-]_{n}}

In addition polymerisation, many alkene monomers open their π bonds and form new C–C σ bonds. No small molecule is eliminated, and the propene CH₃ side group remains attached to every other backbone carbon in the repeat unit.

Aqueous bromine tests for a reactive C=C bond

Shake the sample with aqueous bromine. An alkene changes the reagent from orange to colourless at room temperature because Br₂ adds across C=C; this observable decolourisation is the positive result.

CHX2=CHX2+BrX2→CHX2BrCHX2Br\ce{CH2=CH2 + Br2 -> CH2BrCH2Br}

An alkane does not rapidly decolourise aqueous bromine under these conditions: alkane bromination requires ultraviolet light for free-radical substitution. Keep the reagent, conditions and observation together when interpreting the test.

A positive result shows that bromine has reacted under the test conditions; it does not count how many C=C bonds are present. Aqueous bromine is not a colourless bromide solution.

Electrophilic addition follows electron-pair movement from the π bond

The alkene π bond is an electron-pair donor. A full curly arrow therefore starts at the π bond and points to the electrophilic atom; a second arrow moves the reagent's bonding pair to the atom that leaves as an anion.

With ethene and Br₂, the nearby π electrons induce Brδ+–Brδ−. Draw π → Brδ+ and Br–Br → Brδ−, giving a positively charged carbon intermediate and Br⁻; then draw the Br⁻ lone pair to the positive carbon to form 1,2-dibromoethane.

With propene and HBr, Hδ+ is the electrophile. Draw π → H and H–Br → Br⁻. Proton attachment can place the positive charge on either alkene carbon; the favoured route forms the more stable secondary carbocation, which Br⁻ attacks to give mainly 2-bromopropane.

Every curly-arrow tail must begin at a bond or lone pair, and every head must show where that electron pair goes. Do not start an arrow at H⁺ or a positive carbon, and do not use radical fish-hook arrows for these mechanisms.

Alkyl-group induction favours the more stable carbocation pathway

Carbocation carbon bonded to Class Relative stability
three alkyl groups tertiary, 3° highest
two alkyl groups secondary, 2° intermediate
one alkyl group primary, 1° lowest

An alkyl group has a positive inductive effect: electron density in its σ bonds is displaced towards the electron-deficient, positively charged carbon. More attached alkyl groups spread and reduce the intensity of the positive charge, lowering the carbocation's energy.

When HX adds to an unsymmetrical alkene, compare the carbocations that could form after protonation. The lower-energy pathway is favoured, so H bonds in the direction that leaves the more substituted carbocation; X then bonds to that carbon. This gives the Markovnikov major product.

CHX3CH=CHX2+HBr→CHX3CHBrCHX3(major)\ce{CH3CH=CH2 + HBr -> CH3CHBrCH3}\quad\text{(major)}

The stability order predicts a favoured pathway and major product, not an absolutely exclusive product. The positive charge belongs to a reaction intermediate and is absent from the final halogenoalkane.