Alkenes

In a nutshell

Alkenes are hydrocarbons with a carbon-carbon double bond. That double bond is packed with electrons, so it attracts electrophiles and alkenes take part in addition reactions.

This subtopic covers three things: why the C=C is reactive, the electrophilic addition mechanism (with HBr, H₂SO₄ and Br₂), and addition polymers such as PVC. Mechanism marks and repeating-unit marks turn up on almost every organic paper.

Assumed knowledge: Introduction to organic chemistry, Bonding.

Core content

Structure, bonding and reactivity: why alkenes attract electrophiles

Alkenes are unsaturated hydrocarbons: they contain at least one C=C double bond. A non-cyclic alkene with one C=C has the general formula C_nH_2n.

The C=C double bond is a centre of high electron density: four electrons sit between the two carbon atoms.

That electron-rich region attracts an electrophile, an electron pair acceptor. The electrophile takes a pair of electrons from the C=C, and the double bond becomes a single bond.

The atoms around each C=C carbon are planar, with bond angles of about 120°.

The reaction type is electrophilic addition:

• electrophilic because the attacking species is an electron pair acceptor
• addition because two molecules react to form one molecule.

You need both words when you name the mechanism.

Addition reactions: hydrogen bromide (HBr)

Reagent and conditions: HBr, at room temperature.

Example: ethene + HBr → bromoethane

CH₂=CH₂ + HBr → CH₃CH₂Br

H-Br is polar: bromine is more electronegative than hydrogen, so the H is δ⁺ and the Br is δ^-. The δ⁺ H is the electrophile.

Each curly arrow in the mechanism is a mark. In order:

1. Curly arrow from the C=C double bond to the H of H-Br.
2. Curly arrow from the H-Br bond to the Br (the bond breaks and both electrons go to Br, making Br^-).
3. The carbocation intermediate, with the + on the carbon that did not gain the H.
4. Curly arrow from a lone pair on Br^- to the positive carbon.

A carbocation is an ion with a positive charge on a carbon atom. It is an intermediate, not a product.

Propene is unsymmetrical, so the H can add to either end. Figure 2 shows the route to the major product. The full story is in "Major and minor products" below.

Addition reactions: concentrated sulfuric acid (H₂SO₄)

Reagent and conditions: concentrated H₂SO₄, at room temperature.

Product: an alkyl hydrogensulfate. With ethene it is ethyl hydrogensulfate:

CH₂=CH₂ + H₂SO₄ → CH₃CH₂OSO₂OH

The mechanism is the same shape as HBr, with one trap: you must draw sulfuric acid correctly. Write it as H-O-SO₂-OH, with the H bonded to an oxygen, never to sulfur.

The four mark points:

1. Curly arrow from the C=C to an H on H₂SO₄.
2. Curly arrow from that O-H bond to the O (the O-H bond breaks).
3. The carbocation intermediate.
4. Curly arrow from a lone pair on the negatively charged oxygen of the hydrogensulfate ion to the positive carbon.

The hydrogensulfate ion must carry its negative charge and the lone pair the arrow starts from. Write it as ^-:OSO₂OH (or draw it out with the minus on an O). Writing just "HSO₄^-" does not earn the final arrow mark.

The mechanism you are asked to draw stops at the alkyl hydrogensulfate. Do not add water or draw an alcohol unless the question asks for it. (Making alcohols from alkenes is covered in Alcohols.)

Addition reactions: bromine (Br₂)

Reagent and conditions: bromine (pure, or dissolved in an organic solvent), at room temperature.

Example: ethene + bromine → 1,2-dibromoethane

CH₂=CH₂ + Br₂ → CH₂BrCH₂Br

Br-Br is non-polar: both atoms are identical. So why does an electrophile appear?

1. The C=C is a centre of high electron density.
2. As Br₂ approaches, this electron density repels the electrons in the Br-Br bond.
3. This induces a dipole in Br₂: the nearer Br becomes δ⁺, the further Br becomes δ^-.
4. The δ⁺ Br now acts as the electrophile.

Still don't get it? · how a non-polar bromine molecule becomes an electrophile

Picture a see-saw with two identical children on it, perfectly balanced. Nothing about either child makes it tip. Now someone walks up and pushes hard on one end: the see-saw tips. The tilt came from outside, not from the children.

Now rebuild it step by step.

1. In Br-Br the two atoms are identical, so the shared pair of electrons sits exactly in the middle. No end is positive or negative. This is the balanced see-saw.
2. The C=C double bond holds four electrons squeezed between two carbons. That is a dense patch of negative charge.
3. Electrons repel electrons. As the Br₂ molecule gets close end-on, the C=C pushes the Br-Br electrons away, towards the far bromine. This is the push on one end.
4. Now the far Br has slightly more than its share of electrons (δ^-) and the near Br has slightly less (δ⁺). The molecule has been made polar by its neighbour.
5. An atom that is short of electrons will accept a pair. So the near, δ⁺ Br accepts the electron pair from the C=C, and the Br-Br bond breaks with both electrons going to the far Br, which leaves as Br^-.

Exam wording: "the C=C is a centre of high electron density; it induces a dipole in the bromine molecule (Br^δ+-Br^δ-); the δ⁺ bromine is attacked by the double bond". Do not say bromine is polar because of electronegativity: it is not, the dipole is induced.

The mark points:

1. Curly arrow from the C=C to the δ⁺ Br of Br-Br.
2. Curly arrow from the Br-Br bond to the other Br.
3. The carbocation, which now carries a Br on the neighbouring carbon.
4. Curly arrow from a lone pair on Br^- to the positive carbon.

Addition reactions: testing for unsaturation with bromine water

This is the test for a C=C double bond.

Two wording rules:

• Say colourless, never "clear". Clear means see-through, and coloured solutions are clear too, so "goes clear" earns nothing.
• Give the reagent in full: "bromine water" or Br₂, never "Br". No correct reagent means no marks for the observations either.

Practical skill: you may carry out this test on organic compounds and record your observations. It distinguishes functional group isomers such as hex-1-ene and cyclohexane (both C₆H₁₂), see Worked example 3.

Addition reactions: major and minor products

An unsymmetrical alkene (such as propene) can give two different products with HBr or H₂SO₄, because the H can add to either carbon of the C=C. Each route goes through a different carbocation.

Carbocations are classified by how many alkyl groups are bonded to the positive carbon:

Stability order: tertiary > secondary > primary.

The explanation examiners want has three links:

1. Name the carbocation each product forms via: e.g. 2-bromopropane forms via a secondary carbocation; 1-bromopropane forms via a primary carbocation.
2. Compare stability: the secondary carbocation is more stable than the primary.
3. Say why: alkyl groups are electron releasing (a positive inductive effect). A secondary carbocation has more alkyl groups pushing electron density towards the C⁺, which spreads out and reduces the positive charge.

The major product is the one formed via the more stable carbocation. The minor product forms via the less stable one.

Still don't get it? · why the more stable carbocation gives the major product

Imagine carrying a heavy box. On your own it is a struggle; with friends taking some of the weight, it is easy. The more friends helping, the more comfortably the box gets carried, so if there are two routes to take a box home, the one with more helpers is the one that usually gets used.

Now rebuild it step by step.

1. Propene's C=C has a different number of groups on each carbon, so the H from HBr can land on either one.
2. Whichever carbon does not get the H is left with the positive charge. So there are two possible carbocations.
3. Count the alkyl groups on each positive carbon. H on the end carbon leaves the + on the middle carbon, which has two CH₃ groups (secondary). H on the middle carbon leaves the + on the end carbon, with only one alkyl group (primary).
4. Alkyl groups push electron density towards the positive carbon. This spreads the positive charge out, which makes the ion more stable. These are the friends taking the weight.
5. Two alkyl groups do this better than one, so the secondary carbocation is more stable and forms more readily.
6. Most molecules therefore go through the secondary carbocation, and Br^- adds there. That product, 2-bromopropane, is the major product.

Exam wording: "2-bromopropane is formed via a secondary carbocation; the secondary carbocation is more stable than the primary carbocation (which gives 1-bromopropane) because it has more electron-releasing alkyl groups / a greater positive inductive effect." Always compare the carbocations, never the stability of the final products.

If both possible carbocations are of the same class (for example both secondary, as in pent-2-ene), neither is favoured much, and you get similar amounts of the two products.

Addition polymers: forming them and drawing the repeating unit

Addition polymers are formed from alkenes and substituted alkenes (alkenes with other atoms or groups attached to the C=C). Many monomer molecules join in a long chain. The C=C in each monomer opens, and no other product is formed.

The repeating unit is the smallest section of the chain that repeats. For a monomer with one C=C, it is the two carbons that were the C=C, now joined by a single bond, with everything else hanging off them.

How to draw the repeating unit from a monomer:

1. Draw the monomer with the C=C horizontal and the other four groups above and below it.
2. Change C=C to a single C-C bond.
3. Keep every group in the same place, above and below the two carbons. Side groups (CH₃, Cl) go off the chain, never into it.
4. Add a trailing bond to each end, poking out through the brackets.
5. For the polymer, put square brackets round the unit and an n outside. For "draw the repeating unit", leave the n off.

Working backwards uses the same idea in reverse:

• From a section of chain to the repeating unit: find the shortest two-carbon section that repeats, and draw it with trailing bonds.
• From the polymer to the monomer: take one repeating unit, remove the trailing bonds, and turn the C-C back into C=C.

Practical skill: you may make poly(phenylethene) from phenylethene in the lab. It forms in exactly the same way as the polymers above.

Addition polymers: naming them

The IUPAC name is poly followed by the monomer's name in brackets: poly(ethene), poly(propene), poly(chloroethene), poly(but-2-ene).

The name comes from the monomer, not from the structure of the chain. Poly(propene) is named after propene, even though its chain contains no C=C.

Addition polymers: why they are unreactive

Addition polymers are unreactive, for two reasons:

1. The chain is saturated: the C=C has gone, so there is no electron-rich centre to attract electrophiles. A polyalkene behaves like a very long alkane.
2. The chain is made of strong C-C and C-H bonds, which are non-polar. There is no δ⁺ carbon to attract a nucleophile.

So a polyalkene does not decolourise bromine water.

Addition polymers: intermolecular forces

The forces between polyalkene chains are van der Waals forces (induced dipole-dipole forces) between the molecules.

• The chains are very long, with many electrons and a large surface area in contact, so the van der Waals forces between them add up to a lot.
• That is why polyalkenes are solids, while their small monomers (ethene, propene) are gases at room temperature.
• In poly(chloroethene), the polar C-Cl bonds also give permanent dipole-dipole forces between chains.

When a polymer softens or melts, it is these intermolecular forces that are overcome. No covalent bonds break.

Addition polymers: PVC and plasticisers

Poly(chloroethene), commonly called PVC, is rigid, strong, waterproof and an electrical insulator.

Rigid (unplasticised) PVC: the chains pack closely, so the intermolecular forces are strong and the chains cannot slide past each other. Typical uses:

• window frames
• drainpipes and guttering.

Plasticised PVC: a plasticiser is a small molecule added to the polymer.

1. Plasticiser molecules get between the polymer chains and push them apart.
2. This weakens the intermolecular forces between the chains.
3. The chains can slide past each other more easily, so the PVC becomes flexible.

Typical uses of plasticised PVC:

• insulation on electrical cables (it must bend)
• waterproof clothing.

Addition polymers: how our understanding developed

Knowledge of polymers has built up over time. Poly(ethene) was first made by accident in the 1930s, when ethene was heated under high pressure.

Since then, chemists have learned how chain length, the forces between chains and additives such as plasticisers control a polymer's properties. They now choose monomers, conditions, catalysts and additives to design polymers for particular uses.

Worked examples

Worked example 1: predicting and explaining the major product

Task: 2-methylpropene, CH₂=C(CH₃)₂, reacts with HBr. Name the two possible products, say which is major, and explain why.

1. Find the two routes. The H can add to C1 (the CH₂ end) or to C2 (the carbon carrying the two CH₃ groups).
2. H adds to C1: the + is left on C2, giving (CH₃)₃C⁺. C2 now has three alkyl groups, so this is a tertiary carbocation. Br^- adds to it to give 2-bromo-2-methylpropane, (CH₃)₃CBr.
3. H adds to C2: the + is left on C1, giving (CH₃)₂CHCH₂⁺. C1 has one alkyl group, so this is a primary carbocation. Br^- adds to give 1-bromo-2-methylpropane, (CH₃)₂CHCH₂Br.
4. Model answer:
   • 2-bromo-2-methylpropane is the major product; it forms via a tertiary carbocation.
   • 1-bromo-2-methylpropane is the minor product; it forms via a primary carbocation.
   • The tertiary carbocation is more stable than the primary carbocation, because it has more electron-releasing alkyl groups (a greater positive inductive effect) attached to the C⁺.

Where marks are lost: comparing the stability of the two products instead of the two carbocations; writing "it follows Markovnikov's rule" with no carbocation argument; calling the carbocation in step 3 "secondary" (count the carbons bonded to the C⁺: there is only one).

Worked example 2: model answer for "why does bromine react with ethene?"

Task: Explain why bromine reacts with ethene even though bromine molecules are non-polar.

1. The C=C double bond is a centre of high electron density.
2. It repels the electrons in the Br-Br bond and induces a dipole in the bromine molecule, making the nearer Br δ⁺.
3. The δ⁺ bromine acts as an electrophile (electron pair acceptor) and is attacked by the C=C, in electrophilic addition.

Where marks are lost: writing just "ethene has a C=C" (the mark needs electron density or electron rich); saying bromine is polar because of electronegativity; saying Br₂ first splits into Br⁺ and Br^- before it reaches the alkene.

Worked example 3: telling an alkene from a functional group isomer

Task: Hex-1-ene and cyclohexane are both C₆H₁₂. Give a test to tell them apart.

1. Reagent: shake each with bromine water.
2. Hex-1-ene: the bromine water turns from orange to colourless, because bromine reacts with the C=C.
3. Cyclohexane: the bromine water stays orange (no reaction), because cyclohexane is saturated.

Where marks are lost: "goes clear" instead of colourless; writing "Br" as the reagent; giving an observation for only one of the two compounds; describing burning the compounds instead of a test-tube test.

Worked example 4: repeating unit from a monomer, and monomer from a polymer

Part (a): draw the repeating unit of poly(but-2-ene). The monomer is but-2-ene, CH₃CH=CHCH₃.

1. Draw but-2-ene with the C=C horizontal: each C=C carbon has one H and one CH₃.
2. Open the double bond to a single bond and keep the groups where they were.
3. Add a trailing bond to each end.
4. The repeating unit is -CH(CH₃)-CH(CH₃)-, with the two CH₃ groups off the chain, not in it.

Part (b): a section of a polymer chain is -CH₂-C(CH₃)₂-CH₂-C(CH₃)₂-CH₂-C(CH₃)₂-. Identify the monomer.

1. Find the two-carbon unit that repeats: -CH₂-C(CH₃)₂-.
2. Turn its C-C bond back into C=C: CH₂=C(CH₃)₂.
3. The monomer is 2-methylpropene, so the polymer is poly(2-methylpropene).

Where marks are lost: drawing more than one repeating unit; leaving a C=C in the polymer; putting the CH₃ groups into the main chain (making a four-carbon backbone); missing the trailing bonds.

Worked example 5: model answer on polymer intermolecular forces

Task: Propene is a gas at room temperature but poly(propene) is a solid. Explain why.

1. Poly(propene) molecules are much longer, with many more electrons and a larger surface area.
2. So the van der Waals forces between the molecules are much stronger in poly(propene).
3. More energy is needed to overcome these intermolecular forces, so poly(propene) has a much higher melting point.

Where marks are lost: saying covalent bonds break when the polymer melts; writing "forces between atoms" or "van der Waals bonds"; mentioning hydrogen bonding (there are no H-bonds in a polyalkene); describing only one substance with no comparison.

Common exam mistakes

• Naming the mechanism with one word. "Addition" or "electrophilic" alone does not score; the mark needs electrophilic addition. Some students write "elimination" or "nucleophilic", confusing it with halogenoalkanes.
• Defining an electrophile as a positive ion or as "having a surplus of electrons". The mark needs electron pair acceptor.
• Defining addition as "one molecule adding on to another". The credited idea is two molecules react to form one molecule.
• Curly arrows that do not start at a bond or a lone pair, or that point the wrong way (for example from Br towards the C=C). The first arrow must start at the C=C and the last at a lone pair on the anion.
• Missing charges. Leaving the + off the carbocation, drawing it as δ⁺, or putting it on the wrong carbon; leaving the - or the lone pair off Br^- or the hydrogensulfate ion.
• Drawing a bond as a line plus two dots. This is penalised; a covalent bond is just a line.
• Drawing Br₂ with full charges, or splitting it into Br⁺ and Br^- before it reaches the alkene. The dipole is induced (δ⁺/δ^-).
• Drawing the primary carbocation as the intermediate in the mechanism for the major product.
• Drawing sulfuric acid wrongly: H bonded to S, or the rest of the molecule written as -SO₄H. The H is on an O, and you need an arrow showing the O-H bond breaking.
• Adding water and drawing an alcohol when the question asks for the product with sulfuric acid. The product is the alkyl hydrogensulfate.
• Quoting "Markovnikov's rule" instead of explaining. Examiners give no credit for the name alone. Explain using the stability of the carbocations.
• Comparing the stability of the products, or saying the products are carbocations. Carbocations are intermediates; the comparison must be between them.
• Stopping at "secondary is more stable". The second mark needs why: secondary has more electron-releasing alkyl groups (positive inductive effect).
• "Goes clear" for the bromine water test. It must be colourless.
• Thinking a polymer still has a C=C. In one series only about half of students got this right; many thought poly(chloroethene) decolourises bromine water. It does not: the chain is saturated.
• Repeating-unit errors: including n when asked for the repeating unit, drawing two units, missing the trailing bonds, or putting the side group in the chain (very common for poly(propene)).
• "Additional polymerisation" or "condensation" for alkenes. It is addition polymerisation.
• Product names: "bromopropane" instead of 1,2-dibromopropane for propene + Br₂.
• Not knowing why plasticisers are added. They push chains apart and weaken intermolecular forces so the PVC is flexible, which is why cable insulation contains one.

Key definitions

• Unsaturated: contains at least one carbon-carbon double bond (C=C).
• Electrophile: an electron pair acceptor.
• Addition reaction: a reaction in which two molecules react to form one molecule.
• Electrophilic addition: an addition reaction started by an electrophile attacking the electron-rich C=C (both words needed when naming the mechanism).
• Carbocation: an ion with a positive charge on a carbon atom; an intermediate in electrophilic addition.
• Primary / secondary / tertiary carbocation: the positive carbon is bonded to one / two / three alkyl groups.
• Stability of carbocations: tertiary > secondary > primary, because alkyl groups are electron releasing (positive inductive effect).
• Major product: the product formed via the more stable carbocation.
• Addition polymer: a polymer formed when many alkene (or substituted alkene) monomers join together, with no other product.
• Repeating unit: the smallest section of a polymer chain that repeats, drawn with trailing bonds.
• Plasticiser: a small molecule added to a polymer that gets between the chains, weakening the intermolecular forces so the chains slide more easily and the polymer is more flexible.

Specification

• I can state that alkenes are unsaturated hydrocarbons.
• I can explain that the C=C double bond is a centre of high electron density, which is why alkenes are attacked by electrophiles.
• I can outline (draw with curly arrows) the electrophilic addition mechanism for an alkene with HBr.
• I can outline the electrophilic addition mechanism for an alkene with concentrated H₂SO₄, drawing the acid and the hydrogensulfate ion correctly.
• I can outline the electrophilic addition mechanism for an alkene with Br₂, and explain how the dipole in Br₂ is induced.
• I can describe the bromine water test for unsaturation and give both observations.
• I can predict the major and minor products of addition to an unsymmetrical alkene.
• I can explain the major and minor products using the relative stabilities of primary, secondary and tertiary carbocation intermediates.
• I can state that addition polymers are formed from alkenes and substituted alkenes.
• I can draw the repeating unit from a monomer structure.
• I can draw the repeating unit from a section of the polymer chain.
• I can draw the monomer from a section of the polymer.
• I can name an addition polymer using IUPAC rules.
• I can explain why addition polymers are unreactive.
• I can explain the nature of the intermolecular forces between polyalkene molecules.
• I can give typical uses of PVC, and explain how a plasticiser modifies its properties.
• I can appreciate that knowledge and understanding of the production and properties of polymers has developed over time.

Related notes

• Introduction to organic chemistry
• Bonding
• Alkanes
• Halogenoalkanes
• Alcohols
• Organic analysis
• Polymers (A-level only: condensation polymers and the disposal of polyalkenes)