What are alkane reactions?

Alkanes are saturated and fairly unreactive (strong, non-polar C-C and C-H bonds). They undergo: - Complete combustion to CO_2 and H_2O in plenty of oxygen. - Incomplete combustion to CO (toxic) or soot when oxygen is limited. - Free-radical substitution with halogens in UV light, by a three-stage mechanism: initiation (homolytic bond fission of X_2 forming radicals), propagation (chain-carrying steps that regenerate a radical), and termination (two radicals combine to end the chain).

What are alkene reactions?

The C=C double bond is electron-rich (a pi bond above and below the sigma bond) and attracts electrophiles, so alkenes undergo electrophilic addition with Br_2 (the test for unsaturation, orange to colourless), HBr, and H_2O (steam with a phosphoric acid catalyst, to make alcohols).

EnglandChemistrySyllabus dot point

How do we name, classify and predict the reactions of the simplest hydrocarbons?

Nomenclature and isomerism, the reactions of alkanes (combustion and free-radical substitution), and the reactions of alkenes (electrophilic addition and addition polymerisation) including Markownikoff's rule.

An Edexcel 9CH0 Topic 6 answer covering organic nomenclature and isomerism, alkane combustion and free-radical substitution, and alkene electrophilic addition and addition polymerisation.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section

What this topic is asking
The answer
Examples in context
Try this

What this topic is asking

Edexcel Topic 6 wants you to name and classify organic compounds, recognise the types of structural and stereo isomerism, and describe and explain the reactions and mechanisms of alkanes (combustion and free-radical substitution) and alkenes (electrophilic addition and addition polymerisation), applying Markownikoff's rule.

The answer

Nomenclature and isomerism

Compounds are named from the longest carbon chain (the stem: meth, eth, prop, but), a suffix for the functional group (-ane, -ene, -ol), and prefixes with locant numbers for substituents, chosen to give the lowest set of numbers.

Alkane reactions

Alkanes are saturated and fairly unreactive (strong, non-polar $\text{C-C}$ and $\text{C-H}$ bonds). They undergo:

Complete combustion to $\text{CO}_2$ and $\text{H}_2\text{O}$ in plenty of oxygen.
Incomplete combustion to $\text{CO}$ (toxic) or soot when oxygen is limited.
Free-radical substitution with halogens in UV light, by a three-stage mechanism: initiation (homolytic bond fission of $\text{X}_2$ forming radicals), propagation (chain-carrying steps that regenerate a radical), and termination (two radicals combine to end the chain).

Alkene reactions

The $\text{C=C}$ double bond is electron-rich (a pi bond above and below the sigma bond) and attracts electrophiles, so alkenes undergo electrophilic addition with $\text{Br}_2$ (the test for unsaturation, orange to colourless), $\text{HBr}$ , and $\text{H}_2\text{O}$ (steam with a phosphoric acid catalyst, to make alcohols).

Alkenes also undergo addition polymerisation, in which many monomers join by opening the $\text{C=C}$ bond, with no other product, for example $n\,\text{CH}_2=\text{CH}_2 \rightarrow [\text{-CH}_2\text{-CH}_2\text{-}]_n$ (poly(ethene)). This has $100\%$ atom economy.

The electrophilic addition mechanism for propene and HBr

Show the mechanism for the addition of $\text{HBr}$ to propene and explain why 2-bromopropane is the major product.

step 1: The double bond attacks the electrophile

The electron-rich $\text{C=C}$ attacks the $\delta+$ hydrogen of the polar $\text{H-Br}$ molecule. A curly arrow goes from the double bond to the H, and another from the $\text{H-Br}$ bond to the Br.

step 2: A carbocation forms

The H adds to the end carbon, leaving a positive charge on the middle carbon (a secondary carbocation) and releasing a bromide ion $\text{Br}^-$ .

step 3: The bromide ion attacks

A curly arrow from a lone pair on $\text{Br}^-$ to the positive carbon forms the $\text{C-Br}$ bond, giving 2-bromopropane.

step 4: Explain the major product

A secondary carbocation is more stable than the primary one that would form if H added to the middle carbon, so the secondary route dominates and 2-bromopropane is the major product, in line with Markownikoff's rule.

Examples in context

Example 1. Cracking crude oil for petrol and plastics. Long-chain alkanes from crude oil are catalytically cracked into shorter alkanes (for fuels) and alkenes such as ethene and propene. The alkenes are then used in addition polymerisation to make poly(ethene) and poly(propene). This connects the alkane and alkene chemistry of Topic 6 to the petrochemical industry, where the reactive $\text{C=C}$ of the cracked alkenes is the gateway to plastics.

Example 2. The bromine water test in the field. Field chemists and students test for unsaturation by shaking a sample with orange bromine water: an alkene decolourises it rapidly (electrophilic addition across the double bond), while an alkane does not react in the dark. This simple, vivid test distinguishes saturated from unsaturated hydrocarbons and is a direct application of the difference in reactivity between the $\text{C-C}$ single bond and the $\text{C=C}$ double bond.

Try this

Q1. Name the three stages of the free-radical substitution mechanism. [3 marks]

Cue. Initiation, propagation, termination.

Q2. Predict the major product of the reaction between propene and hydrogen bromide, and explain your choice. [3 marks]

Cue. 2-bromopropane; Markownikoff's rule, formation of the more stable secondary carbocation.

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20185 marksMethane reacts with chlorine in the presence of ultraviolet light. (a) Name the type of mechanism. (b) Write equations for the initiation step and for the two propagation steps. (c) Give one termination step.

Show worked answer →

Show homolytic fission and the chain-carrying radicals.

(a) Free-radical substitution (1).

(b) Initiation: $\text{Cl}_2 \xrightarrow{\text{UV}} 2\text{Cl}\bullet$ (homolytic fission) (1). Propagation 1: $\text{CH}_4 + \text{Cl}\bullet \rightarrow \text{CH}_3\bullet + \text{HCl}$ (1). Propagation 2: $\text{CH}_3\bullet + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \text{Cl}\bullet$ (1).

(c) Termination (any radical-radical combination), e.g. $\text{CH}_3\bullet + \text{Cl}\bullet \rightarrow \text{CH}_3\text{Cl}$ , or $2\text{Cl}\bullet \rightarrow \text{Cl}_2$ , or $2\text{CH}_3\bullet \rightarrow \text{C}_2\text{H}_6$ (1).

Edexcel 20214 marksBut-1-ene reacts with hydrogen bromide. (a) State and explain, using carbocation stability, which is the major product. (b) Outline the mechanism, naming it and showing the curly arrows.

Show worked answer →

Apply Markownikoff's rule via the more stable carbocation, then give the mechanism.

(a) The major product is 2-bromobutane (1). Markownikoff addition forms the more stable secondary carbocation rather than the less stable primary one, and the $\text{Br}^-$ then bonds to that carbon (1).

(b) Electrophilic addition (1). The $\text{C=C}$ pi electrons attack the $\delta+$ H of $\text{H-Br}$ (curly arrow from the double bond to H, and from the $\text{H-Br}$ bond to Br), forming a secondary carbocation and $\text{Br}^-$ ; the bromide ion's lone pair then attacks the carbocation (curly arrow from $\text{Br}^-$ to the positive carbon) to give 2-bromobutane (1).

Related dot points

Sources & how we know this

Pearson Edexcel A-Level Chemistry (9CH0) specification — Pearson Edexcel (2015)