Alkenes

CIE A-Level Chemistry· 9701 Unit 13 Learning Outcome 2· 20 min read

1. Structure and Bonding of Alkenes★★☆☆☆⏱ 5 min

📘 Definition

Alkene

General formula C $_{n}$ H $_{2 n}$ (monounsaturated alkenes)

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

Example:

Ethene (C $_{2}$ H $_{4}$ ), propene (C $_{3}$ H $_{6}$ )

The carbon-carbon double bond is made of one strong sigma ( $σ$ ) bond from head-on orbital overlap, plus one weaker pi ( $π$ ) 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.

2. E/Z Stereoisomerism in Alkenes★★★☆☆⏱ 6 min

📘 Definition

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.

📐 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.

3. Electrophilic Addition Reactions★★★★☆⏱ 7 min

🚫 No Calculator

📘 Definition

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
$H^{\ ext{δ}+}-Br^{\text{δ}-}$
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

✓ Quick check

Test your understanding of Markovnikov's rule

What is the major product when HBr adds to propene (CH $_{3}$ CH=CH $_{2}$ )?
- 1-bromopropane
- 2-bromopropane
- propanal
- propane
Reveal answer
2-bromopropane —
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.

4. Addition Polymerisation of Alkenes★★☆☆☆⏱ 4 min

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: $- [C H_{2} - C H C l] -$ , 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.

5. Common Pitfalls

Wrong move:

Claiming all C-C bonds in alkenes have restricted rotation

Why:

Only the C=C double bond has restricted rotation; single C-C bonds in alkenes rotate freely

Correct move:

Only the C=C double bond has restricted rotation due to the pi bond, which causes E/Z isomerism

Wrong move:

Drawing curly arrows starting from the electrophile in electrophilic addition mechanisms

Why:

The C=C pi bond of the alkene is the electron-rich source that attacks the electrophile

Correct move:

Start the curly arrow from the C=C double bond, pointing at the electrophilic atom

Wrong move:

Writing the subscript n inside the brackets when drawing a repeating unit

Why:

The n indicates the number of repeating units, which goes outside the brackets

Correct move:

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 move:

Assigning E/Z based on molecular mass of groups instead of atomic number

Why:

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 move:

Compare the atomic number of the first atom attached to each double bond carbon to assign priority

6. Quick Reference Cheatsheet

Topic	Key CIE Exam Fact
General formula (mono-alkene)	C $_{n}$ H $_{2 n}$
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

7. Frequently Asked

What is the difference between E/Z and cis-trans isomerism?

Cis-trans is a subset of E/Z isomerism that only works when each carbon of the double bond has one identical group attached. E/Z works for all cases, and is the required system in CIE exams.

Why are alkenes more reactive than alkanes?

Alkenes have a weaker, exposed pi bond with high electron density, which is easily attacked by electrophiles. Alkanes only have strong sigma bonds with no exposed electron density, so they are less reactive.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 12
E/Z isomerism identification
2023 · 22
Electrophilic addition mechanism
2021 · 31
Addition polymer repeating unit drawing

Going deeper

Official SyllabusCIE 9701 A-Level Chemistry SyllabusCheck for latest syllabus updates

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.