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Topic 5: Alkenes

Syllabus
2017
Topic
—
Level
AS

Learning objectives

5.1The general formula of alkenes and understand that alkenes and cycloalkenes are hydrocarbons which are unsaturated (haveKnow the general formula of alkenes and understand that alkenes and cycloalkenes are hydrocarbons which are unsaturated (have a carbon-carbon double bond which consists of a σ bond and a π bond)5.2Geometric isomerism in terms of restricted rotation around a C=C double bond and the nature of the substituents onBe able to explain geometric isomerism in terms of restricted rotation around a C=C double bond and the nature of the substituents on the carbon atoms5.3The E–Z naming system for geometric isomers and why it is necessary to use this when the cis- and trans- naming systemUnderstand the E–Z naming system for geometric isomers and why it is necessary to use this when the cis- and trans- naming system breaks down5.4The reactions of alkenesBe able to describe the reactions of alkenes, limited to: i the addition of hydrogen, using a nickel catalyst, to form an alkane ii the addition of halogens to produce a di-substituted halogenoalkane iii the addition of hydrogen halides to produce mono-substituted halogenoalkanes iv the addition of steam, in the presence of an acid catalyst, to produce alcohols v oxidation of the double bond by acidified potassium manganate(VII) to produce a diol5.5The qualitative test for a C=C double bond using bromine or bromine waterKnow the qualitative test for a C=C double bond using bromine or bromine water5.6The mechanism (including diagrams), giving evidence where possible, of: i the electrophilic addition of bromine and hydrogenBe able to describe the mechanism (including diagrams), giving evidence where possible, of: i the electrophilic addition of bromine and hydrogen bromide to ethene ii the electrophilic addition of hydrogen bromide to propene Use of the curly arrow notation is expected – the curly arrows should start from either a bond or from a lone pair of electrons. Knowledge of the relative stability of primary, secondary and tertiary carbocation intermediates is expected.5.7The addition polymerisation of alkenes and draw the repeat unit given the monomer, and vice versaBe able to describe the addition polymerisation of alkenes and draw the repeat unit given the monomer, and vice versa5.8How chemists limit the problems caused by polymer disposal by: i developing biodegradable polymers ii removing toxic wasteUnderstand how chemists limit the problems caused by polymer disposal by: i developing biodegradable polymers ii removing toxic waste gases produced by the incineration of polymers

An alkene double bond contains one sigma and one pi bond

An alkene is an unsaturated hydrocarbon containing a carbon–carbon double bond. An acyclic alkene with one C=C bond has general formula Cn_nH2n_{2n}; a cycloalkene also contains a ring, so a monocyclic species with one C=C has two fewer hydrogens, Cn_nH2n−2_{2n-2}.

Component Orbital overlap Electron-density location Consequence
σ\sigma bond head-on overlap along the internuclear axis directly between the carbon nuclei strong framework bond
π\pi bond sideways overlap of parallel p orbitals two regions above and below the C–C axis prevents free rotation and is exposed to electrophilic attack

The C=C consists of one σ\sigma bond and one π\pi bond. Addition reactions break the weaker π\pi component and form two new σ\sigma bonds, converting the two carbon centres from double-bonded to single-bonded.

A double bond is not two identical bonds. It contains one σ\sigma and one π\pi bond with different overlap and electron-density geometry.

Restricted C=C rotation creates geometric isomers

A π\pi bond requires parallel p orbitals. Rotation around C=C would destroy their sideways overlap, so rotation is restricted unless the π\pi bond is broken.

Geometric isomerism is possible only when each carbon of the C=C is bonded to two different substituents. The restricted arrangement then locks two different spatial patterns that cannot interconvert by simple rotation.

Alkene Groups on each double-bond carbon Geometric isomerism?
but-2-ene H/CH3_3 on both carbons yes
but-1-ene first carbon has H/H no
2-methylpropene one carbon has CH3_3/CH3_3 no

Restricted rotation is necessary but not sufficient. If either double-bond carbon has two identical substituents, swapping sides does not create a distinct isomer.

E/Z names compare the higher-priority groups

On each carbon of the C=C, rank the two attached atoms by atomic number: higher atomic number gives higher priority. If the directly attached atoms tie, compare the next set of attached atoms at the first point of difference.

Position of the two higher-priority groups Descriptor Memory aid
same side of C=C Z zusammen, together
opposite sides of C=C E entgegen, opposite

Draw the alkene with the C=C fixed, assign priority separately at its left and right carbon, then compare only the two higher-priority substituents. Place E or Z in parentheses before the complete IUPAC name.

Cis/trans works only when a suitable identical group occurs on both double-bond carbons. E/Z remains unambiguous when all four substituents differ, so it is the general naming system.

Do not choose the visually largest group. E/Z priority is determined by atomic number and the first point of difference, not mass of the whole substituent.

Alkenes undergo five required addition or oxidation reactions

Reagent and conditions Product type Example with ethene
H2_2, nickel catalyst, heat alkane CH2=CH2+H2→CH3CH3\mathrm{CH_2{=}CH_2+H_2 \rightarrow CH_3CH_3}
Cl2_2 or Br2_2 1,2-dihalogenoalkane CH2=CH2+Br2→CH2BrCH2Br\mathrm{CH_2{=}CH_2+Br_2 \rightarrow CH_2BrCH_2Br}
HCl or HBr monohalogenoalkane CH2=CH2+HBr→CH3CH2Br\mathrm{CH_2{=}CH_2+HBr \rightarrow CH_3CH_2Br}
steam, acid catalyst alcohol CH2=CH2+H2O→CH3CH2OH\mathrm{CH_2{=}CH_2+H_2O \rightarrow CH_3CH_2OH}
dilute acidified KMnO4_4 vicinal diol ethene forms HOCH2_2CH2_2OH

For an unsymmetrical alkene, H–X or H–OH can add in two orientations and may form more than one structural product. Product proportions depend on the relative stability of the possible carbocation intermediates.

In the mild oxidation test, purple manganate(VII) solution is decolourised as two –OH groups are added across C=C. In every listed reaction, the C=C π\pi bond is replaced by new σ\sigma bonds.

Hydration with steam makes an alcohol; acidified manganate(VII) makes a diol under the specified mild conditions. Do not treat these as the same addition reagent.

Bromine decolourisation tests for C=C

Shake the sample with bromine or bromine water at room temperature. An alkene rapidly changes the orange bromine colour to colourless because Br2_2 adds across the C=C bond.

For ethene: CH2=CH2+Br2→CH2BrCH2Br\mathrm{CH_2{=}CH_2 + Br_2 \rightarrow CH_2BrCH_2Br}. The product is 1,2-dibromoethane and no C=C remains.

Compare with a blank if the sample itself is coloured, use small quantities and avoid confusing dilution with reaction. Under these conditions, a saturated alkane does not rapidly decolourise bromine without ultraviolet initiation.

The observation detects reactive unsaturation such as C=C; it does not by itself identify the alkene's chain length or double-bond position.

Electrophilic addition follows electron-pair arrows and carbocations

A full curly arrow starts at an electron pair—a bond or lone pair—and points to where that pair forms a new bond. The alkene π\pi bond is electron-rich and attacks an electrophilic, δ+\delta+ atom.

Reaction First step Intermediate and second step
ethene + Br2_2 the π\pi electrons induce Brδ+^{\delta+}–Brδ−^{\delta-}; arrow C=C to Brδ+^{\delta+} and arrow Br–Br to Brδ−^{\delta-} a carbocation and Br−^- form; a Br−^- lone pair attacks C+^+ to give 1,2-dibromoethane
ethene + HBr label Hδ+^{\delta+}–Brδ−^{\delta-}; arrow C=C to H and arrow H–Br to Br ethyl carbocation and Br−^- form; Br−^- attacks C+^+ to give bromoethane
propene + HBr protonation can give a primary or secondary carbocation the more stable secondary carbocation forms more readily; Br−^- attack gives 2-bromopropane as major product

Carbocation stability follows tertiary > secondary > primary because surrounding alkyl groups stabilise the positive centre. This explains major-product proportions; it is evidence for the carbocation pathway.

A non-polar halogen still reacts because the electron-rich π\pi bond distorts its electron cloud and induces a dipole. Product distributions from unsymmetrical alkenes support intermediates of different stability.

Curly arrows must not start from a positive charge or from empty space. Use full two-electron arrows here, not the half-arrows used for free radicals.

Addition polymerisation opens C=C into a repeat unit

In addition polymerisation, many alkene monomers join when their C=C π\pi bonds open. No small molecule is eliminated; the monomer atoms are conserved in the polymer chain.

Direction Procedure
monomer to repeat unit replace C=C by C–C; keep every substituent on its original carbon; place the two-carbon segment in brackets with extension bonds crossing both sides; write subscript nn
repeat unit to monomer identify the two backbone carbons inside one repeat; remove the extension bonds and restore C=C between those carbons; retain all substituents

Propene, CH2_2=CHCH3_3, gives repeat unit [–CH2_2–CH(CH3_3)–]n_n. Chloroethene, CH2_2=CHCl, gives [–CH2_2–CHCl–]n_n.

The repeat bracket must enclose exactly one repeating connectivity unit, show bonds continuing through both bracket edges, and preserve the monomer's atom count and substituent placement.

Do not leave a C=C in an addition-polymer repeat unit, and do not add or remove H atoms or a small-molecule product.

Polymer disposal needs degradable materials and cleaner incineration

Strategy How it limits the problem Limitation to manage
develop biodegradable polymers microorganisms, water or environmental conditions break susceptible links so waste persists for less time degradation requires suitable conditions and products must be acceptably non-toxic
remove toxic incineration gases scrub acidic gases with alkaline material; use filters or other gas-cleaning stages before flue gases are released equipment and reagents are required, and captured residues still need safe disposal

Many addition polymers have strong, chemically unreactive carbon backbones and persist in landfill or the environment. Incineration reduces waste volume and can recover energy, but some polymers or additives can produce harmful gases if emissions are untreated.

A useful disposal decision considers the polymer composition, collection route, actual degradation environment, energy recovery and the emissions and residues from treatment rather than relying on one label.

Biodegradable does not mean that a polymer disappears immediately in every environment. Incineration does not remove pollution unless toxic waste gases are captured or converted before release.