Alkenes: structure, electrophilic addition and Markovnikov's rule

The double bond: one σ bond and one π bond

An alkene reacts at its carbon–carbon double bond, and the rest of the molecule behaves much like an alkane. Cyclohexene, C₆H₁₀, and cyclohexane, C₆H₁₂, are both rings of six carbon atoms and differ by two hydrogen atoms and one double bond. Shaken with orange bromine water, cyclohexene turns it colourless within seconds; cyclohexane leaves it orange.

Describe the drawing

Two stoppered test tubes, each a third full and shaken with bromine water. Left, cyclohexane, drawn above as a six-membered ring with no double bond: two layers, the upper organic layer deep orange and the lower aqueous layer pale orange, so the colour stays. Right, cyclohexene, drawn as a ring with one double bond: both layers colourless, because bromine has added across the carbon-carbon double bond.

An alkene is a hydrocarbon with at least one C=C double bond. Alkenes with one double bond and no ring have the general formula CₙH₂ₙ: ethene, C₂H₄; propene, C₃H₆; but-1-ene and but-2-ene, both C₄H₈. They are unsaturated. Each carbon of the double bond could hold more atoms than it does, and in every reaction here the double bond opens and each of its two carbons gains a new single bond.

The two bonds in C=C are not the same. The first is a σ bond, made by the direct overlap of an orbital from each carbon, with its electron density on the line between the two nuclei. The second is a π bond, made by the sideways overlap of a p orbital on each carbon. Its electron density lies in two lobes, one above the plane of the molecule and one below it, and none of it lies on the line between the nuclei.

Each carbon of the double bond has three regions of electron density around it, so the arrangement is trigonal planar, with bond angles of about 120°. The two carbons and the four atoms joined to them lie in one plane. The p orbitals overlap only while they stay parallel, so one end cannot twist relative to the other without breaking the π bond. That restricted rotation is why but-2-ene exists as E and Z isomers.

Two facts make the π bond the reactive part of an alkene. Its electrons are exposed, above and below the plane and further from the nuclei than the σ pair, so the double bond is a region of high electron density. And it is the weaker of the two bonds. The mean bond enthalpy of C=C is 612 kJ mol⁻¹ against 347 kJ mol⁻¹ for C–C, so the π bond adds only about 265 kJ mol⁻¹. Breaking it and making two new σ bonds in its place releases energy overall.

A region of high electron density attracts a species that is short of electrons. That species is an electrophile, an electron pair acceptor: a positive ion such as H⁺, or the δ+ end of a polar molecule such as H–Br. Alkanes have only σ bonds, with their electrons held between the nuclei, and they do not react with electrophiles.

Alkene

A hydrocarbon containing at least one carbon–carbon double bond; with one double bond and no ring, the general formula is CₙH₂ₙ.

Unsaturated

Containing at least one carbon–carbon multiple bond, as the C=C bond of an alkene, so that further atoms can be added to the molecule.

π bond

A bond formed by the sideways overlap of p orbitals, with its electron density above and below the line between the two nuclei.

Electrophile

An electron pair acceptor, attracted to a region of high electron density.

Bromine water is the test that follows from this. An alkene decolourises it at room temperature, orange to colourless; an alkane leaves it orange. In bromine water, water molecules compete with bromide ions in the second step of the mechanism, so the main product with cyclohexene is 2-bromocyclohexan-1-ol rather than 1,2-dibromocyclohexane. The colour change is the same either way. Write the result as 'decolourised' or 'orange to colourless'; 'goes clear' does not describe a colour.

Hydrogen bromide and propene

Hydrogen bromide is a polar molecule. Bromine is more electronegative than hydrogen, so the bonding pair sits nearer the bromine, and the molecule is Hδ+–Brδ−. When it meets propene, the hydrogen end is drawn towards the π electrons. The reaction that follows is an electrophilic addition: the double bond attacks the electrophile, and two groups end up added across it with nothing lost.

In the first step, the pair of electrons in the π bond moves to form a new bond between one carbon of the double bond and the hydrogen atom. Hydrogen can hold only one pair, so at the same moment the H–Br bond breaks heterolytically. Both its electrons go to bromine, which leaves as a bromide ion, Br⁻. The other carbon of the old double bond has lost its share of the π pair. It now has only three bonds and a positive charge: it is a carbocation.

In the second step, the bromide ion uses one of its lone pairs to form a bond to the positive carbon. The charges cancel and the product is a bromoalkane. Say that the H–Br bond breaks heterolytically; 'the bond breaks' alone does not say where the electrons go, and that is the point being tested.

CH₃CH=CH₂(g) + HBr(g) → CH₃CHBrCH₃(l)2-Bromopropane is the major product. 1-Bromopropane forms by the same two steps through a different carbocation, as the minor product.

Describe the drawing

Three stages of the electrophilic addition of hydrogen bromide to propene. First, propene with its double bond and a hydrogen bromide molecule above it, hydrogen delta plus and bromine delta minus; one curly arrow runs from the double bond to the hydrogen and a second from the H–Br bond to the bromine. Second, the secondary carbocation with the positive charge on the middle carbon and the new hydrogen on the end carbon, and a bromide ion whose lone pair sends a third curly arrow to the positive carbon. Third, the major product 2-bromopropane. A grey note records the minor route through the primary carbocation to 1-bromopropane.

Describe the drawing

The electrophilic addition of hydrogen bromide to propene in three stages. First propene, with its C=C double bond, sits below an H–Br molecule marked H δ+ and Br δ−; curly arrow 1 runs from the double bond to the hydrogen and arrow 2 from the H–Br bond onto the bromine. Second, the secondary carbocation, with the positive charge on the middle carbon and the new hydrogen on the end carbon, and a bromide ion whose lone pair sends arrow 3 to the positive carbon. Third, the product 2-bromopropane.

Every curly arrow shows a pair of electrons moving, so it starts where the pair is now and ends where the pair goes. The tail sits on a bond or on a lone pair, never on an atom's symbol or on a positive charge. The head points at the atom that gains the pair, or between two atoms where a new bond will form.

AQA's mark schemes give a mark for each of four things: the arrow from the double bond to the hydrogen; the arrow from the H–Br bond to the bromine; the structure of the carbocation, with the + on the right carbon; and the arrow from a lone pair on the bromide ion to that positive carbon. The δ+ and δ− on H–Br are worth drawing too, because they show why the first arrow goes to the hydrogen.

Electrophilic addition

A reaction in which the double bond of an alkene attacks an electrophile, and two groups add across the double bond to give one saturated product.

Carbocation

An ion in which a carbon atom has only three bonds and carries a positive charge.

Heterolytic fission

The breaking of a covalent bond in which both electrons of the bonding pair go to the same atom.

Bromine: a dipole made on arrival

Bromine is harder to explain. A Br₂ molecule is two identical atoms sharing a pair of electrons equally, so it has no permanent dipole, yet liquid bromine decolourises a gas jar of ethene. The reason is what the double bond does to the bromine molecule as it approaches.

As a Br₂ molecule comes close to the C=C bond, the high electron density of the π bond repels the electrons in the Br–Br bond and pushes them towards the far bromine atom. The near bromine becomes δ+ and the far one δ−. That induced dipole lasts only while the molecule is close to the double bond, but it is enough: the δ+ bromine is now an electrophile, and the mechanism runs as it did with H–Br.

The π pair forms a bond to the near bromine atom. The Br–Br bond breaks heterolytically and its pair leaves with the far bromine as Br⁻. The carbocation has a bromine on one carbon and the + on the other. A lone pair on the bromide ion then forms a bond to that positive carbon, and the product has a bromine on each carbon of the old double bond.

CH₂=CH₂(g) + Br₂(l) → CH₂BrCH₂Br(l)1,2-Dibromoethane, the only product: ethene is symmetrical, so either carbon can take the first bromine.

Describe the drawing

Three stages of the electrophilic addition of bromine to ethene. First, ethene with a bromine molecule pointing end-on at one carbon from above; the nearer bromine is marked delta plus and the far one delta minus, both induced by the double bond. A curly arrow runs from the double bond to the near bromine and another from the Br–Br bond to the far bromine. Second, an open carbocation with bromine on one carbon and the positive charge on the other, and a bromide ion whose lone pair sends a third arrow to the positive carbon. Third, 1,2-dibromoethane.

Two errors appear in drawings of this mechanism. One is to call Br₂ polar in its own right: the dipole is induced, by the π electrons repelling the Br–Br pair, and has to be described that way. The other is to start the last arrow on the carbocation. The electrons that form the new C–Br bond come from the bromide ion, so the arrow starts on its lone pair and ends on the positive carbon.

Induced dipole

A temporary separation of charge in a molecule caused by a nearby region of charge; the π electrons of an alkene repel the Br–Br bonding pair, making the nearer bromine δ+.

Which carbocation forms: Markovnikov's rule

Ethene and but-2-ene are symmetrical, so it does not matter which carbon the electrophile bonds to: the product is the same either way. Propene is not symmetrical. If the hydrogen from H–Br bonds to the end carbon, the CH₂, the positive charge is left on the middle carbon, which is joined to two other carbons: a secondary carbocation, CH₃C⁺HCH₃. If the hydrogen bonds to the middle carbon, the charge is left on the end carbon, joined to one other carbon: a primary carbocation, CH₃CH₂C⁺H₂.

Both routes happen. The secondary carbocation leads to 2-bromopropane and the primary one to 1-bromopropane, and the mixture is mostly 2-bromopropane. It is the major product because the secondary carbocation is more stable than the primary one, so it forms faster.

Describe the drawing

Three carbocations side by side, each with the positive carbon in the centre: primary with one alkyl group and two hydrogens, secondary with two methyl groups and one hydrogen, tertiary with three methyl groups. Small solid arrowheads on each carbon-carbon bond point towards the positive carbon, showing electron release, and a long arrow beneath reads stability increases from left to right. An inset above shows H⁺ adding to C1 of propene to give the secondary ion by the major route, or to C2 to give the primary ion by the minor route.

The reason is the inductive effect. An alkyl group such as CH₃ releases electron density along the σ bond that joins it to the positive carbon. That partly offsets the positive charge and spreads it over more of the ion, and a charge that is spread out is more stable than one concentrated on a single atom. A tertiary carbocation has three alkyl groups doing this, a secondary two and a primary one, so the order of stability is tertiary > secondary > primary.

Stability matters because of rate. The step that makes the more stable carbocation has the lower activation energy, so at the same temperature more collisions succeed by that route and more of the product comes through it. The stability of the carbocation is the reason to give, not the stability of the product: 1-bromopropane and 2-bromopropane are close in energy.

Markovnikov's rule is the shortcut. When H–X adds to an unsymmetrical alkene, the hydrogen goes to the carbon of the double bond that already has more hydrogen atoms. It gives the major product correctly in every case at this level, because putting the hydrogen there leaves the positive charge on the carbon with more alkyl groups. The rule names the answer and the carbocation explains it, and an Explain question asks for the carbocation.

Inductive effect

The release of electron density along a σ bond by an alkyl group, which stabilises a positive charge on the carbon it is joined to.

Markovnikov's rule

When a hydrogen halide adds to an unsymmetrical alkene, the hydrogen bonds to the carbon of the double bond that already carries more hydrogen atoms.

Sulfuric acid, water and steam

Ethene bubbled through cold concentrated sulfuric acid is absorbed. The acid is H–O–SO₂–O–H, and each O–H bond is polar, with the hydrogen δ+ and the oxygen δ−. The δ+ hydrogen is the electrophile. The π pair forms a bond to it, the O–H bond breaks with its pair going to the oxygen, and a carbocation and a hydrogensulfate ion, HSO₄⁻, are left. A lone pair on the negatively charged oxygen of the hydrogensulfate ion then forms a bond to the positive carbon. The product is ethyl hydrogensulfate, CH₃CH₂OSO₂OH.

CH₂=CH₂(g) + H₂SO₄(l) → CH₃CH₂OSO₂OH(l)

Describe the drawing

Electrophilic addition of concentrated sulfuric acid to ethene in three stages. First, ethene below a sulfuric acid molecule whose lower O–H points at the double bond, the hydrogen delta plus and the oxygen delta minus; one curly arrow runs from the double bond to that hydrogen and another from the O–H bond to the oxygen. Second, the carbocation with the positive charge on the carbon not bonded to the new hydrogen, and a hydrogensulfate ion whose negative oxygen sends a curly arrow from its lone pair to the positive carbon. Third, ethyl hydrogensulfate. Below, a plain equation shows warm water hydrolysing it to ethanol and sulfuric acid.

Adding water and warming hydrolyses the ethyl hydrogensulfate to ethanol and gives the sulfuric acid back. Over the two stages the acid is a catalyst, and the overall change is the addition of water across the double bond.

CH₃CH₂OSO₂OH(l) + H₂O(l) → CH₃CH₂OH(aq) + H₂SO₄(aq)

Industry adds water to ethene directly, as steam over a phosphoric acid catalyst at about 300 °C and 60 to 70 atm, the acid supplying the H⁺ electrophile. The mechanism is the same three arrows: the π pair to H⁺, a lone pair on water to the carbocation, and the O–H bond of the protonated alcohol back to the oxygen as H⁺ returns to the catalyst.

CH₂=CH₂(g) + H₂O(g) ⇌ CH₃CH₂OH(g)ΔH = −45 kJ mol⁻¹

Other boards: hybridisation, hydrogenation and manganate(VII)

Cambridge describes the bonds of the double bond through hybridisation. Each carbon of the double bond mixes its 2s orbital with two of its 2p orbitals to make three sp² hybrid orbitals, which lie in one plane at 120° and form the three σ bonds. The third 2p orbital is left unhybridised, at right angles to that plane, and the sideways overlap of two of them is the π bond. A carbon with four single bonds is sp³, at 109.5°; a carbon in a triple bond is sp, at 180°, with two π bonds.

Hydrogen also adds across the double bond, at about 150 °C over a nickel catalyst, giving the alkane. OCR, Edexcel and Cambridge teach it with the other additions, and it is how vegetable oils are hardened into spreads. AQA uses it only in the enthalpies of hydrogenation that are the evidence for the structure of benzene.

CH₂=CH₂(g) + H₂(g) → CH₃CH₃(g)Nickel catalyst, about 150 °C.

Edexcel and Cambridge also set the oxidation of alkenes by acidified potassium manganate(VII). Shaken cold and dilute with an alkene, the purple solution turns colourless and an OH group adds to each carbon of the old double bond, giving a diol: ethene gives ethane-1,2-diol. The colour change is a second test for C=C.

CH₂=CH₂(g) + [O] + H₂O(l) → HOCH₂CH₂OH(aq)Cold, dilute, acidified KMnO₄: purple to colourless.

Cambridge adds the hot, concentrated reagent, which breaks the double bond completely. Each carbon of the old C=C becomes a C=O. A carbon carrying two alkyl groups ends as a ketone, one carrying a hydrogen ends as a carboxylic acid, and a CH₂ at the end of a chain is oxidised to carbon dioxide and water. 2-Methylbut-2-ene, (CH₃)₂C=CHCH₃, gives propanone and ethanoic acid, and the two fragments show where the double bond was.

Diol

A compound with two OH groups; cold dilute acidified manganate(VII) adds one to each carbon of a C=C bond.