Alkenes

In a nutshell

Alkenes are unsaturated hydrocarbons with at least one C=C double bond. That one bond decides everything about them: their flat shape, their stereoisomers, and the way they react.

This subtopic covers the σ and π bonds in C=C, E/Z isomerism, the addition reactions (with their electrophilic addition mechanism and Markownikoff's rule), addition polymers, and what to do with waste plastic.

Assumed knowledge: Basic concepts of organic chemistry (naming, curly arrows, heterolytic fission), Alkanes (σ-bonds), Bonding and structure (electron pair repulsion).

Core content

Properties of alkenes: the C=C bond

Alkenes are unsaturated: they contain a carbon-carbon multiple bond, here a C=C double bond. The simplest alkenes have the general formula C_nH_2n.

The C=C double bond is made of two different bonds:

• a σ-bond (sigma): overlap of orbitals directly between the bonding atoms.
• a π-bond (pi): sideways overlap of adjacent p-orbitals above and below the bonding C atoms.

Because the π-bond only exists while the two p-orbitals line up side by side, the carbons cannot twist about the C=C. This is restricted rotation about the double bond.

Compare a C-C single bond (only a σ-bond), which has free rotation.

You do not need hybridisation for OCR. The orbital overlap description above is what earns the marks.

Properties of alkenes: trigonal planar shape

Each carbon in the C=C is trigonal planar, with bond angles of 120°.

The electron pair repulsion argument, in mark order:

1. Each C=C carbon has three regions of electron density around it (two single bonds and the double bond, which counts as one region).
2. These regions repel each other equally and move as far apart as possible.
3. So the arrangement is trigonal planar, bond angle 120°.

Stereoisomerism in alkenes: E/Z and cis-trans isomers

Stereoisomers are compounds with the same structural formula but a different arrangement (of atoms) in space.

E/Z isomerism is one type of stereoisomerism. An alkene shows it only when both conditions hold:

1. There is restricted rotation about the C=C double bond.
2. Each carbon of the C=C has two different groups attached.

cis-trans isomerism is a special case of E/Z isomerism in which two of the substituent groups attached to each carbon atom of the C=C are the same (one group on the left carbon matches one group on the right carbon).

• cis: the two matching groups are on the same side.
• trans: the two matching groups are on opposite sides.

For but-2-ene the matching groups are the CH₃ groups (and the H atoms), so Z-but-2-ene is also cis-but-2-ene and E-but-2-ene is trans-but-2-ene.

Stereoisomerism in alkenes: the CIP priority rules

To label an isomer E or Z, use the Cahn-Ingold-Prelog (CIP) priority rules on each carbon of the C=C separately:

1. Look at the two atoms directly attached to that carbon. The atom with the higher atomic number has higher priority.
2. If those two atoms are the same, move one atom further out. List the atoms attached to each, highest atomic number first, and compare the lists. The first point of difference decides.

Then compare the two higher-priority groups:

Rule 2 in action: in 1-chloro-2-methylbut-1-ene, C2 carries CH₃ and CH₂CH₃. Both attach through a carbon, so look further out:

• CH₃: that carbon is attached to (H, H, H).
• CH₂CH₃: that carbon is attached to (C, H, H).
• First point of difference: C beats H, so CH₂CH₃ has priority.

Watch out: E = trans and Z = cis is only reliably true when there is an H on each carbon of the C=C. When there is not, work out E/Z from the CIP rules every time (see Worked example 1).

You will not be asked to assign priorities to groups that contain double or triple bonds.

Still don't get it? · deciding E or Z

Picture a short, rigid plank welded at both ends so it cannot twist. At each end there are two hooks, one on the top edge and one on the bottom edge. On each end you hang one heavy bag and one light bag.

If both heavy bags hang on the top edge, that is one arrangement. If one heavy bag is on top and the other is underneath, that is a different arrangement. Because the plank cannot twist, you can never turn one into the other without unhooking a bag.

Now the chemistry, one step at a time:

1. The plank is the C=C. It cannot twist because rotating would pull the p-orbitals out of line and break the π-bond.
2. The hooks are the two groups on each carbon. If one end has two identical bags (say two H atoms), swapping them changes nothing, so there are no isomers at all.
3. "Heavy" means higher atomic number of the atom directly attached (rule 1). If those atoms tie, look at what they are attached to (rule 2).
4. Find the heavy bag at the left end. Then find the heavy bag at the right end.
5. Same side: Z. Opposite sides: E.

The exam version: "E/Z isomers exist because there is restricted rotation about the C=C and each carbon of the C=C has two different groups attached. Using CIP rules, the higher-priority groups are on the same side, so it is the Z isomer."

Stereoisomerism in alkenes: finding the possible stereoisomers

Given a structural formula, work through these checks:

1. Find every C=C. No C=C, no E/Z to consider.
2. For each C=C, check both carbons. If either carbon carries two identical groups (=CH₂, or two CH₃ groups), that C=C gives no E/Z isomers.
3. If both carbons carry two different groups, there are two stereoisomers, E and Z.
4. Draw both with the C=C at the centre and 120° bonds, then label each using CIP priorities.

Addition reactions of alkenes: why alkenes are reactive

Alkenes are much more reactive than alkanes because of the relatively low bond enthalpy of the π-bond.

The π-bond is weaker than the σ-bond, so it breaks more easily, while the σ-bond stays intact. The two carbons are then free to bond to new atoms.

The C=C is less than twice as strong as C-C, so the π part (about 612 - 347 = 265 kJ mol^-1) is weaker than a σ-bond. You do not need to recall these values; they show why the π-bond is the one that breaks.

In an addition reaction, two reactants join to form one product. The π-bond breaks and each C=C carbon gains a new atom or group.

Addition reactions of alkenes: the four reactions you need

Product names for the examples:

• with H₂: ethane
• with Br₂: 1,2-dibromoethane (one Br on each carbon of the old C=C)
• with HBr: bromoethane
• with steam: ethanol

The reaction with steam is reversible, which is why it is written with ⇌. With an unsymmetrical alkene, steam and hydrogen halides can each give two possible products (see Markownikoff's rule below).

PAG7: qualitative analysis of organic functional groups (the bromine test)

Bromine is the test for unsaturation (a C=C) in a carbon chain.

Method:

1. Put about 1 cm³ of the test compound in a test tube.
2. Add a few drops of bromine water and shake.
3. Record the colour change.

The colour disappears because the Br₂ is used up as it adds across the C=C to form a colourless dihaloalkane.

Bromine water is harmful and an irritant: wear eye protection and gloves.

Addition reactions of alkenes: electrophiles and electrophilic addition

An electrophile is an electron pair acceptor.

The C=C is a region of high electron density (the π electrons sit above and below the bond, exposed). Electrophiles are attracted to it and accept a pair of electrons from it.

The reactions of alkenes with halogens and hydrogen halides go by electrophilic addition. The mechanism uses heterolytic fission: when a bond breaks, one bonding atom receives both electrons from the bonded pair.

HBr (a polar electrophile). Br is more electronegative than H, so the molecule is H^δ+-Br^δ-. The H^δ+ is the end that accepts the electron pair.

1. Curly arrow from the C=C (the π-bond) to the H^δ+.
2. Curly arrow from the H-Br bond to the Br: heterolytic fission makes Br^-.
3. This leaves a carbocation: a positively charged carbon intermediate.
4. Curly arrow from a lone pair on Br^- to the C⁺, forming the C-Br bond.

Br₂ (a non-polar molecule). Br₂ has no permanent dipole. As it approaches, the high electron density of the C=C repels the electrons in the Br-Br bond. This induces a dipole: the nearer Br becomes δ+.

The steps then match HBr: arrow from the C=C to Br^δ+, arrow from the Br-Br bond to the far Br, a carbocation and Br^- form, and the Br^- lone pair attacks the C⁺.

For the halogen mechanism, OCR accepts either a carbocation or a halonium ion (bridged) intermediate. The carbocation route shown is the one to learn, because it matches the HBr mechanism.

Curly arrow rules that carry marks:

• A curly arrow shows the movement of an electron pair.
• It must start from a bond, a lone pair or a negative charge, and point to where the pair ends up.
• Show all relevant dipoles (δ+ and δ-), the + charge on the carbocation, and the lone pair and - charge on Br^-.

Addition reactions of alkenes: Markownikoff's rule

When H-X adds to an unsymmetrical alkene, two products are possible. Markownikoff's rule predicts the major product: the H atom adds to the C=C carbon that already has more H atoms, so X ends up on the carbon with fewer H atoms.

The reason, in mark-scheme terms: the major product forms via the more stable carbocation intermediate.

Carbocations are classified by how many alkyl groups (carbons) are attached to the positive carbon:

Stability order: tertiary > secondary > primary. You only need the order. OCR does not ask you to explain why.

For propene + HBr:

• H adds to the CH₂ end: secondary carbocation → 2-bromopropane, the major product.
• H adds to the middle carbon: primary carbocation → 1-bromopropane, the minor product.

Still don't get it? · why one product is major

Imagine leaving a crowded hall by one of two exits. One leads to a wide, stable staircase; the other to a wobbly rope ladder. Both get you out, but most people take the staircase, so far more people end up at its bottom.

Now the chemistry, one step at a time:

1. The first curly arrow puts the H^δ+ onto one of the two C=C carbons. Whichever carbon does not get the H becomes C⁺.
2. In propene, putting H on the end CH₂ leaves C⁺ on the middle carbon, which has two carbons attached: a secondary carbocation.
3. Putting H on the middle carbon leaves C⁺ on the end carbon, which has one carbon attached: a primary carbocation.
4. The secondary carbocation is more stable (the staircase), so most molecules go that way.
5. Br^- then bonds to wherever the + charge is. Most molecules had the + on the middle carbon, so the main product has Br on the middle carbon: 2-bromopropane.

The exam version: "The major product, 2-bromopropane, is formed via the more stable carbocation. The secondary carbocation is more stable than the primary carbocation."

Polymers from alkenes: addition polymerisation

In addition polymerisation, many alkene molecules (monomers) join together. The π-bond in each C=C opens and the carbons link into a long saturated chain with no other product.

Rules for drawing the repeat unit:

1. Take the two carbons of the C=C and draw them joined by a single bond.
2. Keep every other group exactly where it was, hanging off the chain (side groups are never part of the main chain).
3. Add a trailing bond at each end that passes through the brackets.
4. In an equation, put n in front of the monomer and n after the bracket.

To go backwards from a polymer section to the monomer:

1. The main chain of an addition polymer has a repeating two-carbon pattern. Box one two-carbon unit.
2. Put a C=C back between those two carbons.
3. Remove the trailing bonds.

Waste polymers and alternatives: processing waste polymers

Addition polymers are saturated and have strong, non-polar C-C and C-H bonds, so they are unreactive and do not break down in landfill. Cheap oil-derived plastics therefore build up as long-lasting waste.

The spec names three ways of processing waste polymers, each with a sustainability benefit:

Sorting waste plastics and remoulding them (recycling) is another way of re-using waste and saving resources.

Waste polymers and alternatives: biodegradable and photodegradable polymers

Chemists are developing polymers that break down instead of persisting:

• Biodegradable polymers are broken down by microorganisms. Many are made from plant-based materials such as starch.
• Photodegradable polymers contain bonds that absorb light energy and break, so the polymer breaks down when exposed to light.

Benefits to the environment:

• Less persistent plastic waste in landfill and the environment.
• Plant-based polymers come from renewable sources, reducing dependency on finite resources (crude oil).

A limitation worth knowing: a photodegradable polymer buried in landfill gets no light, so it will not break down there.

Worked examples

1. Assigning E/Z and cis-trans to 2-bromobut-2-ene

Task: 2-bromobut-2-ene is CH₃CBr=CHCH₃. Decide whether it has stereoisomers, then label the isomer drawn below as E or Z, and as cis or trans.

1. Check each carbon of the C=C. C2 carries Br and CH₃ (different). C3 carries H and CH₃ (different). So E/Z isomers exist.
2. Priority on C2: Br (atomic number 35) beats C (atomic number 6). Br has priority.
3. Priority on C3: C (6) beats H (1). CH₃ has priority.
4. Compare positions: in the drawing, Br on C2 and CH₃ on C3 are on the same side, so this is the Z isomer: Z-2-bromobut-2-ene.
5. cis or trans? One group on each carbon is the same (CH₃), so cis-trans naming applies. The two CH₃ groups are on opposite sides, so it is also trans-2-bromobut-2-ene.

Where marks go: this is the trap. Here Z = trans, because there is no H on C2. Students who assume "Z means cis" lose the mark.

2. Predicting the major product with Markownikoff's rule

Task: 2-methylpropene, (CH₃)₂C=CH₂, reacts with HBr. Predict the major product and explain your choice in terms of carbocation stability.

1. Two possible carbocations.
   • H adds to the CH₂ carbon (which becomes CH₃): C⁺ is on the central carbon, now bonded to three CH₃ groups. This is (CH₃)₃C⁺, a tertiary carbocation.
   • H adds to the central carbon: C⁺ is on the end carbon, bonded to only one carbon. This is (CH₃)₂CHCH₂⁺, a primary carbocation.
2. Compare stability. Tertiary is more stable than primary.
3. Major product (Br bonds to the tertiary C⁺): 2-bromo-2-methylpropane, (CH₃)₃CBr.
4. Minor product: 1-bromo-2-methylpropane, (CH₃)₂CHCH₂Br.

Model statement: "The major product is 2-bromo-2-methylpropane because it forms via the more stable carbocation: the tertiary carbocation is more stable than the primary carbocation."

Both product equations balance: C₄H₈ + HBr → C₄H₉Br.

3. Finding the monomer from a polymer section

Task: a section of a polymer chain is -CH₂-CHCl-CH₂-CHCl-CH₂-CHCl-. Give the repeat unit and the monomer.

1. Spot the repeating two-carbon pattern: -CH₂-CHCl- appears three times.
2. Repeat unit: -[CH₂-CHCl]- with a trailing bond through each bracket.
3. Put the C=C back between those two carbons: CH₂=CHCl, chloroethene.
4. The polymer is poly(chloroethene), PVC.

Where marks go: the repeat unit has two carbons in the chain, not four; and the Cl stays as a side group, it is never drawn in the chain.

4. Using hydrogenation to count C=C bonds

Task: 0.205 g of a hydrocarbon Y, C₆H₁₀, reacts completely with 120 cm³ of hydrogen at RTP, using a nickel catalyst. Each H₂ adds across one C=C. How many C=C bonds does one molecule of Y contain? (Molar gas volume = 24.0 dm³ mol^-1 at RTP, given on the data sheet.)

Step 1: molar mass of Y.

$M(\text{C}_6\text{H}_{10}) = (6 \times 12.0) + (10 \times 1.0) = 82.0\ \text{g mol}^{-1}$

Step 2: moles of Y.

$n(\text{Y}) = \frac{m}{M} = \frac{0.205\ \text{g}}{82.0\ \text{g mol}^{-1}} = 2.50 \times 10^{-3}\ \text{mol}$

Step 3: convert the gas volume to dm³, then to moles.

$120\ \text{cm}^3 = \frac{120}{1000}\ \text{dm}^3 = 0.120\ \text{dm}^3$

$n(\text{H}_2) = \frac{V}{24.0} = \frac{0.120\ \text{dm}^3}{24.0\ \text{dm}^3\ \text{mol}^{-1}} = 5.00 \times 10^{-3}\ \text{mol}$

Step 4: mole ratio.

$\frac{n(\text{H}_2)}{n(\text{Y})} = \frac{5.00 \times 10^{-3}\ \text{mol}}{2.50 \times 10^{-3}\ \text{mol}} = 2$

Answer: each molecule of Y contains 2 C=C bonds. The equation is C₆H₁₀ + 2H₂ → C₆H₁₄.

Where marks are dropped: dividing 120 by 24.0 without converting cm³ to dm³ (giving 5.00 mol, a factor of 1000 out); and dividing the moles the wrong way round, giving a ratio of 0.5.

Common exam mistakes

• Bromine water "goes clear". Clear is not a colour: a coloured solution can be clear. The mark needs orange to colourless (or "decolourises").
• Describing the π-bond as just "overlap of p-orbitals". The mark needs sideways overlap, above and below the bonding carbons. Likewise the σ-bond needs directly between the bonding atoms.
• Giving 109.5° around a C=C carbon, or saying there are "four bonds". The carbon has three regions of electron density, so it is trigonal planar, 120°.
• Leaving one condition out of the E/Z explanation. Both are needed: restricted rotation about the C=C and two different groups on each carbon of the C=C.
• Claiming but-1-ene or propene has E/Z isomers. One carbon carries two H atoms, so it cannot.
• Assuming Z always means cis. That only holds with an H on each C=C carbon (see Worked example 1).
• Choosing CIP priority by the size or mass of the whole group rather than the atomic number of the attached atom, then the first point of difference.
• Curly arrow starting from a carbon atom instead of from the C=C bond; or the arrow to C⁺ starting from the Br^- symbol rather than its lone pair.
• Missing labels in the mechanism: no δ+/δ- on H-Br or the induced dipole on Br₂; no + on the carbocation; no lone pair on Br^-; or drawing Br₂ with full charges as though it splits into ions before it reaches the alkene.
• Saying Br₂ is polar because of electronegativity. The two atoms are identical; the dipole is induced by the electron-rich C=C.
• Defining an electrophile as "something attracted to negative charge". The definition is an electron pair acceptor.
• Explaining Markownikoff's rule only as "H goes to the carbon with more H" when asked in terms of the mechanism. The mark needs via the more stable carbocation and secondary (or tertiary) more stable than primary.
• Hydration slips: writing "water" instead of steam, or leaving out the H₃PO₄ acid catalyst.
• Repeat unit errors: drawing the side group (CH₃, Cl) in the main chain; leaving the C=C in the repeat unit; missing the trailing bonds; or writing n inside the brackets.
• Vague sustainability answers such as "better for the environment". Name the benefit: less landfill, energy used to generate electricity, less crude oil (a finite resource) used, or toxic HCl removed.

Key definitions

• Unsaturated: the presence of multiple carbon-carbon bonds (for alkenes, a C=C double bond).
• σ-bond: overlap of orbitals directly between the bonding atoms.
• π-bond: sideways overlap of adjacent p-orbitals above and below the bonding C atoms.
• Stereoisomers: compounds with the same structural formula but with a different arrangement (of atoms) in space.
• E/Z isomerism: a type of stereoisomerism caused by restricted rotation about a C=C double bond, where each carbon of the C=C has two different groups attached.
• cis-trans isomerism: a special case of E/Z isomerism in which two of the substituent groups attached to each carbon atom of the C=C are the same.
• Electrophile: an electron pair acceptor.
• Heterolytic fission: bond breaking in which one bonding atom receives both electrons from the bonded pair.
• Carbocation: an intermediate with a positive charge on a carbon atom.
• Markownikoff's rule: in the addition of H-X to an unsymmetrical alkene, the major product is formed via the more stable carbocation (H adds to the C=C carbon with more H atoms already).
• Addition polymerisation: the joining of many alkene (monomer) molecules, by opening their C=C bonds, to form a long saturated chain as the only product.

Specification

• I can describe an alkene as an unsaturated hydrocarbon whose C=C is one σ-bond (overlap directly between the bonding atoms) and one π-bond (sideways overlap of adjacent p-orbitals above and below the bonding C atoms), and explain restricted rotation. (4.1.3 a)
• I can explain the trigonal planar shape and 120° bond angle around each C=C carbon using electron pair repulsion. (4.1.3 b)
• I can explain the terms stereoisomers, E/Z isomerism and cis-trans isomerism. (4.1.3 c i)
• I can use the CIP priority rules to identify the E and Z stereoisomers, and know when E = trans and Z = cis is safe. (4.1.3 c ii)
• I can work out the possible E/Z or cis-trans stereoisomers of a molecule from its structural formula. (4.1.3 d)
• I can explain the reactivity of alkenes using the relatively low bond enthalpy of the π-bond. (4.1.3 e)
• I can give the reagents, conditions, products and equations for addition of H₂ (Ni), halogens, hydrogen halides and steam (H₃PO₄), and use bromine to test for a C=C. (4.1.3 f, PAG7)
• I can define an electrophile as an electron pair acceptor. (4.1.3 g)
• I can draw the electrophilic addition mechanism with curly arrows, dipoles, the carbocation and the halide lone pair, using heterolytic fission. (4.1.3 h)
• I can use Markownikoff's rule to predict the major product of H-X adding to an unsymmetrical alkene, using primary, secondary and tertiary carbocation stability. (4.1.3 i)
• I can draw the repeat unit of an addition polymer from its monomer, and identify the monomer from a polymer section. (4.1.3 j)
• I can describe the sustainability benefits of combustion for energy, use as an organic feedstock, and removal of toxic products such as HCl from burning PVC. (4.1.3 k)
• I can describe the environmental benefits of biodegradable and photodegradable polymers. (4.1.3 l)

Related notes

• Basic concepts of organic chemistry: naming alkenes, structural isomers, curly arrows and heterolytic fission
• Alkanes: σ-bonds, free rotation and the low reactivity of saturated hydrocarbons
• Bonding and structure: electron pair repulsion and shapes
• Amount of substance: moles and molar gas volume for Worked example 4
• Alcohols: elimination of water from alcohols to make alkenes (the reverse direction)
• Haloalkanes: what the products of HBr addition do next
• Organic synthesis: using alkene reactions in multi-step routes
• Chemical equilibrium: le Chatelier's principle, which applies to the reversible hydration of ethene