ChemistryISC Class 11 · Hydrocarbons

Alkanes and Alkenes: Preparation, Halogenation and Addition Reactions

Updated 17 October 2026 · Fact-checked

Alkanes (CₙH₂ₙ₊₂) are saturated and undergo free radical substitution; alkenes (CₙH₂ₙ) are unsaturated and undergo electrophilic addition. Addition of HX follows Markovnikov's rule, which is reversed for HBr in the presence of peroxide. Ozonolysis breaks the double bond to give carbonyl compounds and fixes its position.

Notice that Br lands on the middle carbon with HBr alone, but on the end carbon when peroxide is present.

Illustration drawn for QuestionBuddy.

Understand Alkanes and Alkenes: Preparation and Reactions

Preparation of alkanes. (1) Hydrogenation of alkenes or alkynes with H₂/Ni (or Pt, Pd). (2) Decarboxylation: sodium salt of a carboxylic acid heated with soda lime, CH₃COONa + NaOH →(CaO, heat) CH₄ + Na₂CO₃ (the alkane has one carbon fewer). (3) Wurtz reaction: 2R–X + 2Na →(dry ether) R–R + 2NaX (for example 2CH₃CH₂Br gives butane). (4) Kolbe electrolysis of sodium or potassium salts of acids.

Reactions of alkanes. Combustion: CₙH₂ₙ₊₂ + (3n + 1)/2 O₂ → nCO₂ + (n + 1)H₂O. Halogenation (substitution) in light or heat by a free radical chain: initiation Cl₂ → 2Cl·; propagation CH₄ + Cl· → CH₃· + HCl, CH₃· + Cl₂ → CH₃Cl + Cl·; termination 2Cl· → Cl₂, CH₃· + Cl· → CH₃Cl, 2CH₃· → C₂H₆. Reactivity of halogens: F₂ > Cl₂ > Br₂ > I₂; ease of H replacement: 3° > 2° > 1°. Pyrolysis (cracking) breaks large alkanes into smaller alkanes and alkenes. In ethane, the staggered conformation is more stable than the eclipsed by about 12.5 kJ mol⁻¹.

Preparation of alkenes. Dehydration of alcohols (conc. H₂SO₄, 443 K), dehydrohalogenation (alcoholic KOH; Saytzeff rule: the more substituted alkene is the major product), and partial hydrogenation of alkynes.

Addition reactions of alkenes. With H₂/Ni → alkane; with Br₂ in CCl₄ or Br₂ water → vicinal dibromide (reddish-brown colour disappears: the test for unsaturation); with HX → alkyl halide, following Markovnikov's rule (the negative part adds to the carbon with fewer hydrogens, via the more stable carbocation); with HBr in the presence of peroxide → anti-Markovnikov product (free radical mechanism, for HBr only); with water/H⁺ → alcohol. Cold dilute alkaline KMnO₄ (Baeyer's reagent) gives a glycol and the purple colour fades. Ozonolysis (O₃, then Zn/H₂O) cleaves C=C: CH₃CH=CHCH₃ gives 2CH₃CHO; (CH₃)₂C=CH₂ gives CH₃COCH₃ and HCHO.

Polymerisation. n CH₂=CH₂ → (–CH₂–CH₂–)ₙ (polythene).

Alkanes and Alkenes: Preparation and Reactions: formulas and key facts

Combustion of an alkane: $C_{n} H_{2 n + 2} + (3 n + 1) /2 O_{2} \to n C O_{2} + (n + 1) H_{2} O$; Oxygen needed: (3n + 1)/2 volumes per volume of gas.
Wurtz and decarboxylation: $2 R X + 2 N a \to R - R + 2 N a X ∣ R C O O N a + N a O H \to R H + N a_{2} C O_{3}$; Dry ether; soda lime (CaO) and heat.
Markovnikov addition: $C H_{3} C H = C H_{2} + H B r \to C H_{3} C H B r C H_{3}$; With peroxide: CH₃CH₂CH₂Br.
Ozonolysis: $R_{2} C = C R'_{2} \to R_{2} C = O + O = C R'_{2}$; Position of the double bond.

Alkanes and Alkenes: Preparation and Reactions in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Which bromide is formed?

Suppose a chemist adds HBr to propene, first in the dark in the absence of peroxide and then in the presence of benzoyl peroxide.

Name the product in each case and give the reasoning.

Try it first, then show the solutionSolution

Without peroxide: electrophilic addition via the more stable secondary carbocation CH₃CH⁺CH₃, giving 2-bromopropane (Markovnikov).
With peroxide: Br· adds first to the terminal carbon to give the more stable secondary radical CH₃CH·CH₂Br, giving 1-bromopropane.
The peroxide effect holds for HBr only.

Answer: 2-Bromopropane without peroxide; 1-bromopropane with peroxide.

Memory hook: The more stable intermediate (cation or radical) decides.

Tricks to remember Alkanes and Alkenes: Preparation and Reactions

Short memory aids. Say each one aloud once, then use it on a question the same day.

Salt, soda lime, one carbon less
Sodium ethanoate gives methane, not ethane.
Use it when: Counting carbon atoms in decarboxylation.
Rich get richer
In Markovnikov addition, H goes to the carbon that already has more hydrogens.
Use it when: Predicting products of HX addition.

Alkanes and Alkenes: Preparation and Reactions: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Markovnikov addition · 1 mark · MCQBoard level

The addition of HBr to propene (no peroxide) gives mainly:
(a) 1-bromopropane
(b) 2-bromopropane
(c) 1,2-dibromopropane
(d) propane

Try it first, then show the solutionSolution

H adds to the carbon with more hydrogens, so Br ends up on C2.

Answer: (b) 2-bromopropane

Example 2: Preparation of alkanes · 3 marks · Short answerBoard level

Write equations to prepare (i) methane from sodium ethanoate, (ii) butane from bromoethane, and (iii) ethane from ethene.

Try it first, then show the solutionSolution

(i) CH₃COONa + NaOH →(CaO, heat) CH₄ + Na₂CO₃. (1 mark)
(ii) 2CH₃CH₂Br + 2Na →(dry ether) CH₃CH₂CH₂CH₃ + 2NaBr. (1 mark)
(iii) CH₂=CH₂ + H₂ →(Ni, 573 K) CH₃CH₃. (1 mark)

Answer: (i) methane by decarboxylation; (ii) butane by the Wurtz reaction; (iii) ethane by hydrogenation.

Example 3: Mechanism of chlorination · 4 marks · Short answerHarder board level

Write the mechanism of the chlorination of methane in sunlight, naming each stage. How many monochloro derivatives can propane give, and what are they?

Try it first, then show the solutionSolution

Initiation: Cl–Cl →(hν) 2Cl·. (1 mark)
Propagation: CH₄ + Cl· → CH₃· + HCl; CH₃· + Cl₂ → CH₃Cl + Cl·. (1 mark)
Termination: Cl· + Cl· → Cl₂; CH₃· + Cl· → CH₃Cl; CH₃· + CH₃· → C₂H₆. (1 mark)
Propane gives two monochloro isomers: 1-chloropropane and 2-chloropropane. (1 mark)

Answer: Free radical chain with initiation, propagation and termination; propane gives two monochloro derivatives.

Example 4: Addition and cleavage of alkenes · 4 marks · Short answerHarder board level

Give the products of (i) propene + HBr with peroxide, (ii) but-2-ene + ozone followed by Zn/H₂O, (iii) ethene + cold dilute alkaline KMnO₄, and explain the product in (i).

Try it first, then show the solutionSolution

(i) CH₃CH₂CH₂Br (1-bromopropane): in the free radical mechanism Br· adds to the carbon that gives the more stable secondary radical. (1.5 marks)
(ii) 2CH₃CHO (two moles of ethanal). (1 mark)
(iii) Ethane-1,2-diol (glycol), HOCH₂CH₂OH; the purple colour fades. (1.5 marks)

Answer: (i) 1-bromopropane; (ii) ethanal; (iii) ethane-1,2-diol.

Example 5: Identifying two alkenes · 5 marks · Long answerHOTS / competency

Two isomers A and B of molecular formula C₄H₈ both decolourise bromine water. On ozonolysis, A gives only ethanal, while B gives methanal and propanone. Identify A and B and give the products of their hydrogenation and of the addition of HBr to B.

Try it first, then show the solutionSolution

A gives two molecules of ethanal, so the double bond is in the middle of a straight chain: A = but-2-ene, CH₃CH=CHCH₃. (1.5 marks)
B gives HCHO and (CH₃)₂C=O, so B = 2-methylpropene, (CH₃)₂C=CH₂. (1.5 marks)
Hydrogenation: A gives butane; B gives 2-methylpropane. (1 mark)
HBr adds to B by Markovnikov's rule through the tertiary carbocation (CH₃)₃C⁺, giving 2-bromo-2-methylpropane. (1 mark)

Answer: A = but-2-ene; B = 2-methylpropene; products: butane, 2-methylpropane and 2-bromo-2-methylpropane.

Where students lose marks in Alkanes and Alkenes: Preparation and Reactions

The slips examiners see most often, and what to write instead.

Mistake: Wrong carbon count in decarboxylation
Saying sodium ethanoate gives ethane.
Do this: The alkane has one carbon fewer than the acid salt: methane.
Usually costs: 2 marks.
Mistake: Reversing Markovnikov addition
Putting Br on the carbon with more hydrogens.
Do this: The negative part goes to the carbon with fewer hydrogens.
Usually costs: 1 to 2 marks.
Mistake: Missing conditions
Writing reactions without catalysts or solvent.
Do this: Include Ni/573 K, dry ether, soda lime, alcoholic KOH, as needed.
Usually costs: 1 mark.

Exam tips

Write the conditions above each arrow.
Name the mechanism: free radical substitution or electrophilic addition.
For Markovnikov questions, show the carbocation intermediate.
For ozonolysis, work backwards by joining the carbonyl carbons with a double bond.

Practice questions on Alkanes and Alkenes: Preparation and Reactions

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The most stable conformation of ethane is:
(a) eclipsed
(b) staggered
(c) skew
(d) all are equally stable
Hint
The hydrogen atoms are farthest apart in this form.
Answer
(b) staggered
Q2 · 2 marks · Short answerBoard level
How can ethane and ethene be distinguished by a simple chemical test?
Hint
Think about unsaturation.
Answer
Ethene decolourises bromine water (or the purple colour of Baeyer's reagent) while ethane does not.
Q3 · 3 marks · Short answerHarder board level
2-Butanol is heated with concentrated H₂SO₄. Name the major and the minor products and state the rule used.
Hint
Saytzeff rule.
Answer
The major product is but-2-ene (the more substituted alkene), and the minor product is but-1-ene, by the Saytzeff rule.
Q4 · 3 marks · Short answerHarder board level
Write the equation for the complete combustion of propane and find the volumes of oxygen needed and carbon dioxide formed for 2 L of propane (measured at the same T and p).
Hint
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.
Answer
Per 1 L of C₃H₈: 5 L O₂ and 3 L CO₂. For 2 L: 10 L of O₂ and 6 L of CO₂.
Q5 · 3 marks · Short answerHarder board level
In the chlorination of 2-methylpropane, the relative rates of replacing one tertiary and one primary hydrogen are 5 : 1. Find the percentage of the tertiary chloride among the monochloro products.
Hint
There are 9 primary H and 1 tertiary H.
Answer
Relative amounts: tertiary 1 × 5 = 5, primary 9 × 1 = 9. Tertiary chloride = 5/14 = 35.7%, primary chloride = 64.3%.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a student studies the reactions of ethene.
(i) Give the product with Br₂ in CCl₄ and the observation. (1 mark)
(ii) Give the product with cold dilute alkaline KMnO₄. (1 mark)
(iii) Give the product of hydration with H₂O/H⁺. (1 mark)
(iv) Write the polymerisation reaction. (1 mark)
Hint
Ethene reacts at the C=C bond.
Answer
(i) 1,2-dibromoethane; the reddish-brown colour disappears. (ii) Ethane-1,2-diol. (iii) Ethanol. (iv) n CH₂=CH₂ → (–CH₂–CH₂–)ₙ, polythene.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Product of Wurtz reaction of CH₃Br?Show answer
Ethane.
2.What is the test for unsaturation?Show answer
Bromine water decolourised (or Baeyer's reagent).
3.Which acid shows the peroxide effect?Show answer
HBr only.
4.Ozonolysis of ethene gives?Show answer
Two molecules of methanal.

Alkanes and Alkenes: Preparation and Reactions: frequently asked questions

Why do alkanes undergo substitution while alkenes undergo addition?

Alkanes have only strong, nonpolar σ bonds and no electrons available for attack by electrophiles, so a radical substitution of H occurs. Alkenes have a weaker, exposed π bond, which is easily attacked.

Why is a tertiary carbocation more stable than a primary one?

Three alkyl groups release electrons (+I) and give nine hyperconjugative structures, spreading the positive charge.

Why does the peroxide effect not occur with HCl or HI?

The H–Cl bond is too strong to be broken homolytically in the chain, and the I· radical is not reactive enough to add to the double bond.