# Alkenes

> CIE A-Level Chemistry · Unit 13: Hydrocarbons
> Source: https://www.owlsprep.com/study/cie-9701-u13-alkenes/

This sub-topic covers the structure, stereoisomerism, key reactions and polymerisation of alkenes, unsaturated hydrocarbons containing carbon-carbon double bonds, a core topic for CIE A-Level Chemistry organic questions.

**Prerequisites:** [Structure and bonding of alkanes](https://www.owlsprep.com/study/cie-9701-u13-alkanes/); Basics of covalent bonding and reaction mechanisms

## Learning objectives

- Describe the structure and bonding of alkenes
- Explain the relative reactivity of alkenes compared to alkanes
- Name and draw alkenes including E/Z stereoisomers
- Recall and draw mechanisms for electrophilic addition reactions of alkenes
- Draw repeating units for addition polymers of alkenes

## Structure and Bonding of Alkenes

**Alkene** — An unsaturated hydrocarbon containing at least one carbon-carbon double covalent bond between two carbon atoms

*Notation:* General formula C$_n$H$_{2n}$ (monounsaturated alkenes)

*Example:* Ethene (C$_2$H$_4$), propene (C$_3$H$_6$)

The carbon-carbon double bond is made of one strong sigma ($\sigma$) bond from head-on orbital overlap, plus one weaker pi ($\pi$) bond from side-on overlap of p-orbitals. The pi bond has exposed electron density above and below the plane of the double bond, making it easily attacked by electrophiles.

**Worked example:** Explain why the average bond enthalpy of C=C (+612 kJ mol⁻¹) is less than twice the average bond enthalpy of C-C (+347 kJ mol⁻¹)

1. A single C-C bond only contains one sigma bond, with bond enthalpy +347 kJ mol⁻¹.
2. A C=C double bond contains one sigma bond and one weaker pi bond, which has a lower bond enthalpy than a sigma bond.
3. Calculate twice the C-C bond enthalpy: $2 \times 347 = 694$ kJ mol⁻¹, which is higher than the measured C=C enthalpy of 612 kJ mol⁻¹.
4. The difference arises from the lower strength of the pi bond, which is more easily broken than a sigma bond.

> **Exam tip:** When asked why alkenes are more reactive than alkanes, always mention the exposed electron density of the pi bond and its lower bond enthalpy for full marks.

## E/Z Stereoisomerism in Alkenes

**E/Z Isomerism** — A form of stereoisomerism that arises because rotation around the C=C double bond is restricted, leading to different spatial arrangements of groups attached to the double bond

*Example:* E-1,2-dichloroethene and Z-1,2-dichloroethene

E/Z isomerism only occurs if *each carbon* in the double bond is bonded to two different groups. Priority is assigned by the Cahn-Ingold-Prelog (CIP) rule: higher atomic number of the atom directly attached = higher priority.

> **mnemonic**
>
> Z = the higher priority groups are on the **Zame (same) side**; E = the higher priority groups are on **opposite Ends**

**Worked example:** Assign the E/Z configuration to CH$_3$CH=C(Cl)CH$_3$

1. List groups on each double bond carbon: Left C: H (Z=1) and CH$_3$ (C, Z=6); Right C: Cl (Z=17) and CH$_3$ (C, Z=6)
2. Assign priorities: Left C: CH$_3$ (higher) > H (lower); Right C: Cl (higher) > CH$_3$ (lower)
3. Check position of higher priority groups: CH$_3$ (left higher) and Cl (right higher) are on opposite sides of the double bond
4. Conclusion: This is the **E isomer**

> **Exam tip:** If one carbon in the double bond has two identical groups attached, E/Z isomerism is not possible — always check this first.

## Electrophilic Addition Reactions

**Electrophilic Addition** — A reaction mechanism where an electron-deficient electrophile attacks the electron-rich pi bond of the alkene, breaking the pi bond and forming two new single bonds

Common tested reactions include hydrogenation (addition of H$_2$), halogenation (addition of Cl$_2$/Br$_2$), hydrohalogenation (addition of HCl/HBr) and hydration (addition of steam). For unsymmetrical alkenes, Markovnikov's rule predicts the major product: the H atom adds to the double bond carbon that already has more H atoms.

**Worked example:** Draw the mechanism for the reaction of ethene with HBr and name the product

1. The electron-rich pi bond of ethene attacks the partially positive H atom of HBr (the electrophile)
2. $$H^{\text{δ}+}-Br^{\text{δ}-}$$
3. The H-Br bond breaks heterolytically, forming a positively charged carbocation intermediate and a Br⁻ ion
4. The Br⁻ ion attacks the carbocation, forming a new covalent bond
5. Final product name: bromoethane

**Check your understanding**

Test your understanding of Markovnikov's rule

1. What is the major product when HBr adds to propene (CH$_3$CH=CH$_2$)?

   - 1-bromopropane
   - 2-bromopropane
   - propanal
   - propane

   *Why:* Correct! H adds to the CH₂ end (which has more H atoms), so Br adds to the central carbon to form 2-bromopropane as the major product.

> **Exam tip:** Curly arrows must start from the electron source (the C=C pi bond or a lone pair) — starting from the wrong position loses marks in mechanism questions.

*Calculator:* forbidden

## Addition Polymerisation of Alkenes

Alkenes undergo addition polymerisation, where many small alkene monomers join together to form a long polymer chain. The pi bond in each monomer breaks, and new single bonds form between adjacent monomers. No other products are formed in this reaction.

**Worked example:** Draw the repeating unit of the polymer formed from chloroethene (CH₂=CHCl) and name the polymer

1. Break the C=C double bond in the chloroethene monomer
2. Draw the carbon backbone with open bonds extending out from the two carbons that were double bonded
3. Keep all substituents attached to their original carbons
4. Repeating unit: $-[CH_2-CHCl]-$, polymer name: poly(chloroethene) (PVC)
5. $$-[CH_2-CHCl]-_n$$

> **Exam tip:** Always draw the open bonds extending outside the brackets when drawing repeating units — missing this is a common mistake that costs marks.

## Common pitfalls

- **Wrong:** Claiming all C-C bonds in alkenes have restricted rotation
  - Why it fails: Only the C=C double bond has restricted rotation; single C-C bonds in alkenes rotate freely
  - Correct: Only the C=C double bond has restricted rotation due to the pi bond, which causes E/Z isomerism
- **Wrong:** Drawing curly arrows starting from the electrophile in electrophilic addition mechanisms
  - Why it fails: The C=C pi bond of the alkene is the electron-rich source that attacks the electrophile
  - Correct: Start the curly arrow from the C=C double bond, pointing at the electrophilic atom
- **Wrong:** Writing the subscript n inside the brackets when drawing a repeating unit
  - Why it fails: The n indicates the number of repeating units, which goes outside the brackets
  - Correct: Draw the repeating unit inside brackets, with the open bonds extending out of the brackets and n as a subscript outside the right bracket
- **Wrong:** Assigning E/Z based on molecular mass of groups instead of atomic number
  - Why it fails: Priority is assigned based on the atomic number of the atom directly attached to the double bond, not the total mass of the group
  - Correct: Compare the atomic number of the first atom attached to each double bond carbon to assign priority

## Cheatsheet

| Topic | Key CIE Exam Fact |
| --- | --- |
| General formula (mono-alkene) | C$_n$H$_{2n}$ |
| C=C bonding | 1 σ + 1 π bond; restricted rotation |
| E/Z priority rule | Higher atomic number = higher priority; Z = same side, E = opposite |
| Markovnikov's Rule | H adds to the C of C=C with more H atoms |
| Addition products | H$_2$ → alkane; Br$_2$ → dibromoalkane; HBr → bromoalkane; H$_2$O → alcohol |
| Repeating unit rule | Break C=C, extend open bonds out of brackets |

## What's next

Alkenes are a core functional group in organic chemistry, and their electrophilic addition mechanism is the foundation for understanding most other organic reactions tested in CIE A-Level. Mastery of E/Z isomerism and mechanism drawing is critical for high marks in both paper 1 and paper 2 organic sections. The next class of hydrocarbons you will study is arenes (aromatic hydrocarbons), which have different bonding and reactivity patterns compared to alkenes. You can also deepen your understanding of stereoisomerism and organic reaction mechanisms more broadly with the linked resources.

- [Halogen derivatives and alcohols](https://www.owlsprep.com/study/cie-9701-u14-overview/)
- [Halogenoalkanes](https://www.owlsprep.com/study/cie-9701-u14-halogenoalkanes/)
- [Alcohols](https://www.owlsprep.com/study/cie-9701-u14-alcohols/)

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