ChemistryCBSE Class 11 · Hydrocarbons

Alkenes and Alkynes: Addition Reactions, Markovnikov Rule and Ozonolysis Class 11

Updated 17 October 2026 · Fact-checked

Alkenes and alkynes contain π bonds and give addition reactions. Unsymmetrical reagents HX add to an unsymmetrical alkene by Markovnikov's rule (hydrogen goes to the carbon with more hydrogens, X to the carbon that gives the more stable carbocation); HBr in the presence of peroxide adds in the opposite sense by a free-radical path. Ozonolysis followed by reductive work-up (Zn, water) cuts the C=C bond and gives aldehydes or ketones, which identifies the position of the double bond. Terminal alkynes are weakly acidic and give a white precipitate with ammoniacal silver nitrate.

Notice that Br lands on the middle carbon by Markovnikov's rule, but on the end carbon when peroxide is present with HBr.

Illustration drawn for QuestionBuddy.

Understand Alkenes and Alkynes: Addition Reactions, Markovnikov Rule and Ozonolysis

Electrophilic addition to alkenes. The π electrons attack an electrophile, forming a carbocation, and then a nucleophile adds. For $H X$ : $C = C + H^{+} \to$ carbocation $\to$ product. The more stable carbocation (tertiary > secondary > primary) is formed, which gives Markovnikov's rule: in the addition of $H X$ to an unsymmetrical alkene, the hydrogen attaches to the carbon already carrying more hydrogens. The reaction with $B r_{2}$ (in $C C l_{4}$ or water) is an addition that decolourises the reddish-brown colour, and cold dilute alkaline $K M n O_{4}$ (Baeyer's reagent) gives a diol and loses its purple colour: both are tests for unsaturation.

Peroxide (Kharasch) effect. Only for HBr, with a peroxide present, bromine radical adds first to the less substituted carbon so that the more stable radical forms, giving the anti-Markovnikov product. HCl and HI do not show this effect.

Other alkene reactions. Hydrogenation with Ni, Pd or Pt gives the alkane. Hydration with dilute $H_{2} S O_{4}$ follows Markovnikov's rule. Hot acidic $K M n O_{4}$ cleaves the double bond to give carboxylic acids or ketones.

Ozonolysis. Alkene + $O_{3}$ gives an ozonide, which on treatment with Zn and water gives carbonyl compounds: each $= C H_{2}$ gives methanal, each $= C H R$ an aldehyde, and each $= C R_{2}$ a ketone. Working backwards from the products locates the double bond.

Geometrical isomerism. A $C = C$ bond cannot rotate. If each doubly bonded carbon has two different groups the alkene exists as cis (like groups on the same side) and trans forms (different dipole moments and melting points).

Alkynes. Hydrogenation with Lindlar's catalyst (partially poisoned Pd) gives the cis alkene; sodium in liquid ammonia gives the trans alkene; complete hydrogenation gives the alkane. Hydration with $H g S O_{4}$ /dilute $H_{2} S O_{4}$ follows Markovnikov's rule and the enol rearranges: ethyne gives ethanal and every other alkyne a ketone. HX adds twice. Terminal alkynes ( $R - C \equiv C - H$ ) lose $H^{+}$ to a strong base such as $N a N H_{2}$ because the sp carbon is more electronegative, and give a white precipitate with ammoniacal silver nitrate (and a red one with ammoniacal cuprous chloride).

Alkenes and Alkynes: Addition Reactions, Markovnikov Rule and Ozonolysis: formulas and key facts

General formulas: $C_{n} H_{2 n} (alkene), C_{n} H_{2 n - 2} (alkyne)$; Ring count or π bonds shift the formula by 2H each.
Combustion of an alkene: $C_{n} H_{2 n} + \frac{3 n}{2} O_{2} \to n C O_{2} + n H_{2} O$; For C₄H₈ that is 6 mol O₂ per mole.
Hydration of ethyne: $H C \equiv C H + H_{2} O H g S O_{4}, H_{2} S O_{4} C H_{3} C H O$; Enol to aldehyde.
Ozonolysis rule: $= C H_{2} \to H C H O, = C H R \to R C H O, = C R_{2} \to R_{2} C = O$; Reductive work-up (Zn, water).

How to solve Alkenes and Alkynes: Addition Reactions, Markovnikov Rule and Ozonolysis questions

Identify which carbon of the double bond is more substituted, decide whether the conditions are ionic or radical, and then place the groups.

1Draw the alkene and count the hydrogens on each doubly bonded carbon.
2For HX under ordinary conditions put H on the carbon with more hydrogens (Markovnikov) and X on the other.
3If the reagent is HBr and a peroxide is present, reverse the orientation.
4For ozonolysis join the carbonyl carbons of the products with a double bond to rebuild the alkene.
5For alkynes decide the reagent: Lindlar's catalyst for cis, Na/liquid NH₃ for trans, HgSO₄ for the carbonyl compound.
6Check the carbon count and the hydrogen count of the final structure.

Alkenes and Alkynes: Addition Reactions, Markovnikov Rule and Ozonolysis 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.

Ripening and polythene bags

Suppose a factory converts 28 kg of ethene into polythene by addition polymerisation (C₂H₄ = 28 g mol⁻¹).

How many moles of ethene are used and what mass of polymer forms if the conversion is complete?

Try it first, then show the solutionSolution

Moles of ethene = 28000/28 = 1000 mol.
In addition polymerisation no atom is lost, so the polymer mass = 28 kg.

Answer: 1000 mol; 28 kg of polymer.

Memory hook: Addition keeps every atom.

Welding gas from carbide

Suppose a welder passes 2.27 L of ethyne (measured at STP, 22.7 L mol⁻¹) through a trap of bromine water.

Find the mass of bromine (Br = 80) consumed for complete addition.

Try it first, then show the solutionSolution

Moles of ethyne = 2.27/22.7 = 0.10 mol.
Each ethyne adds two Br₂ to give CHBr₂–CHBr₂: moles of Br₂ = 0.20 mol, mass = 0.20 × 160 = 32 g.

Answer: 32 g of bromine.

Memory hook: A triple bond takes two moles of reagent.

Tricks to remember Alkenes and Alkynes: Addition Reactions, Markovnikov Rule and Ozonolysis

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

The rich get richer
In Markovnikov addition H goes to the carbon that already has more H.
Use it when: Orientation of HX.
Peroxide reverses only HBr
HBr alone shows the peroxide effect, not HCl or HI.
Use it when: Exam statements.
Rebuild from ozonolysis
Join the two carbonyl carbons with a double bond and remove the oxygens.
Use it when: Structure identification.
Lindlar is cis
Poisoned Pd gives cis; sodium in ammonia gives trans.
Use it when: Alkyne reduction.

Alkenes and Alkynes: Addition Reactions, Markovnikov Rule and Ozonolysis: solved examples

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

Example 1: Markovnikov and anti-Markovnikov addition · 2 marks · Very short answerBoard level

Write the major product and its IUPAC name when 2-methylbut-2-ene reacts with HBr (a) in the absence and (b) in the presence of a peroxide.

Try it first, then show the solutionSolution

(a) (CH₃)₂C=CH–CH₃: H adds to the CH carbon and Br to the more substituted carbon (tertiary carbocation): (CH₃)₂CBr–CH₂CH₃, 2-bromo-2-methylbutane. (1)
(b) With peroxide the orientation is reversed: (CH₃)₂CH–CHBr–CH₃, 2-bromo-3-methylbutane. (1)

Answer: (a) 2-bromo-2-methylbutane; (b) 2-bromo-3-methylbutane.

Example 2: Alkene from its ozonolysis products · 3 marks · Short answerHarder board level

An alkene C₅H₁₀ gives ethanal and propanal on ozonolysis (Zn, water). (a) Identify the alkene. (b) Does it show geometrical isomerism? Draw the two forms in words.

Try it first, then show the solutionSolution

Join the carbonyl carbons of CH₃CHO and CH₃CH₂CHO with a double bond: CH₃CH=CHCH₂CH₃, pent-2-ene (5 C and 10 H as required). (1)
Each doubly bonded carbon carries two different groups (H and CH₃; H and C₂H₅), so geometrical isomers exist. (1)
Cis form: CH₃ and C₂H₅ on the same side of the double bond; trans form: on opposite sides. (1)

Answer: Pent-2-ene; it exists as cis and trans isomers.

Example 3: From calcium carbide to propanone · 4 marks · Long answerHarder board level

Show how propanone can be prepared from calcium carbide in four steps, writing the equation or reagent for each.

Try it first, then show the solutionSolution

CaC₂ + 2H₂O → Ca(OH)₂ + HC≡CH (ethyne). (1)
HC≡CH + NaNH₂ → HC≡C⁻Na⁺ (sodium ethynide) + NH₃, because the sp C–H bond is acidic. (1)
HC≡C⁻Na⁺ + CH₃Br → HC≡C–CH₃ (propyne) + NaBr. (1)
HC≡C–CH₃ + H₂O (HgSO₄, dilute H₂SO₄) → CH₃–CO–CH₃ via the enol; Markovnikov addition puts OH on the internal carbon. (1)

Answer: CaC₂ to ethyne to sodium ethynide to propyne to propanone.

Example 4: Combustion and identification of an alkene · 4 marks · Long answerHarder board level

5.6 g of a gaseous alkene (vapour density 28) is burnt completely. (a) Find its molecular formula. (b) Find the mass of CO₂ and the volume of O₂ at STP (22.7 L mol⁻¹) involved. (c) How many alkene isomers fit the formula, and which shows geometrical isomerism?

Try it first, then show the solutionSolution

(a) Molar mass = 2 × 28 = 56 g mol⁻¹; CₙH₂ₙ = 14n, so n = 4 and the formula is C₄H₈. (1)
(b) Moles = 5.6/56 = 0.10 mol; C₄H₈ + 6O₂ → 4CO₂ + 4H₂O gives 0.40 mol CO₂ = 17.6 g and 0.60 mol O₂ = 13.6 L. (1)
(c) Four alkenes: but-1-ene, cis-but-2-ene, trans-but-2-ene, 2-methylpropene (three structural isomers counting but-2-ene once). (1)
Only but-2-ene shows geometrical isomerism, because only there does each doubly bonded carbon carry two different groups. (1)

Answer: C₄H₈; 17.6 g CO₂ and 13.6 L O₂; but-2-ene is the geometrical isomer.

Example 5: Unsaturation in a ring compound · 4 marks · Long answerHOTS / competency

Compound Z is an unsubstituted monocyclic alkene (a ring with no side chain) of formula C₆H₁₀. It decolourises bromine water, takes up 1 mol of H₂ per mole to give C₆H₁₂, and on ozonolysis (Zn, water) gives a single compound with the formula C₆H₁₀O₂. (a) Identify Z and the ozonolysis product. (b) Why is only one product formed? (c) What product forms with hot acidic KMnO₄? (d) Find the volume of H₂ at STP (22.7 L mol⁻¹) taken up by 8.2 g of Z (C₆H₁₀ = 82 g mol⁻¹).

Try it first, then show the solutionSolution

(a) C₆H₁₀ has two degrees of unsaturation (CₙH₂ₙ₋₂); one H₂ is added, so one is a C=C and one is a ring. Z is unsubstituted, so all six carbons are in the ring, which makes it cyclohexene (a methyl-substituted ring such as 1-methylcyclopentene is ruled out because it has a side chain). Ozonolysis gives the unbranched dialdehyde hexanedial, OHC–(CH₂)₄–CHO (C₆H₁₀O₂). (1)
(b) The double bond is inside the ring, so cleavage opens the ring and both carbonyl groups stay in the same molecule. (1)
(c) Hot acidic KMnO₄ oxidises each –CHO end to –COOH: hexanedioic acid, HOOC–(CH₂)₄–COOH. (1)
(d) Moles = 8.2/82 = 0.10 mol, so H₂ = 0.10 mol = 2.27 L. (1)

Answer: (a) Cyclohexene, hexanedial; (b) ring opening; (c) hexanedioic acid; (d) 2.27 L.

Where students lose marks in Alkenes and Alkynes: Addition Reactions, Markovnikov Rule and Ozonolysis

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

Mistake: Writing the anti-Markovnikov product for HBr without a peroxide.
Applying the peroxide effect everywhere.
Do this: Check the conditions: peroxide and HBr only.
Usually costs: Product marks.
Mistake: Stating that every alkene with a double bond shows cis-trans isomerism.
Not checking the groups on each carbon.
Do this: Each doubly bonded carbon must carry two different groups.
Usually costs: Concept marks.
Mistake: Writing an aldehyde for the hydration of propyne or but-1-yne.
Applying the ethyne result to all alkynes.
Do this: Markovnikov addition gives a ketone for every alkyne except ethyne.
Usually costs: Product marks.
Mistake: Counting one mole of reagent for a triple bond.
Forgetting the second π bond.
Do this: Complete addition to an alkyne takes two moles of H₂, Br₂ or HX.
Usually costs: The numerical.

Exam tips

Give the reason in terms of carbocation stability when you apply Markovnikov's rule.
Name the reagent and condition above every arrow in a conversion.
For ozonolysis write the ozonide step and then the Zn/H₂O step.
Show a test for unsaturation with the observation (decolourisation).
Write the hydrogen count of each product so the examiner can verify the formula.

Practice questions on Alkenes and Alkynes: Addition Reactions, Markovnikov Rule and Ozonolysis

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
Reductive ozonolysis (Zn, water) of 2-methylbut-2-ene gives
(a) methanal and butanone (b) ethanal and propanone (c) two moles of propanone (d) ethanal and propanal
Hint
Rebuild the alkene (CH₃)₂C=CHCH₃ and cut at the double bond.
Answer
(b): (CH₃)₂C=O (propanone) from the CR₂ end and CH₃CHO (ethanal) from the CHR end.
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The peroxide effect is shown by HBr but not by HCl or HI.
Reason (R): The H–Cl bond is too strong for the radical chain to be favourable, and iodine atoms are not reactive enough to add to the double bond.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Radical chain steps must be energetically favourable.
Answer
(a).
Q3 · 2 marks · Very short answerBoard level
Write the products of the addition of (a) 1 mol and (b) 2 mol of HCl to propyne, with names.
Hint
Apply Markovnikov's rule at each step.
Answer
(a) CH₃–CCl=CH₂, 2-chloropropene. (b) CH₃–CCl₂–CH₃, 2,2-dichloropropane (the second Cl goes to the carbon that already carries one Cl).
Q4 · 3 marks · Short answerHarder board level
Which of the following show geometrical isomerism: (a) hex-3-ene (b) 3-methylpent-2-ene (c) 2-methylbut-2-ene (d) 1,1-dichloroethene (e) 1,2-dichloroethene? Give the reason for each answer in a few words.
Hint
Each doubly bonded carbon needs two different groups.
Answer
(a), (b) and (e) show it: each doubly bonded carbon has two different groups. (c) has two methyl groups on one carbon and (d) has two chlorines on one carbon, so neither exists as cis and trans.
Q5 · 4 marks · Long answerHOTS / competency
(a) How can but-2-yne be converted into cis-but-2-ene and into trans-but-2-ene? (b) Convert ethyne into ethanal and into benzene. (c) Name the product when but-2-yne is hydrated with HgSO₄ and dilute H₂SO₄, and explain why no aldehyde forms.
Hint
Only ethyne gives an aldehyde on hydration.
Answer
(a) H₂ with Lindlar's catalyst gives the cis alkene; sodium in liquid NH₃ gives the trans alkene. (b) Ethyne with HgSO₄/dilute H₂SO₄ gives ethanal via vinyl alcohol; heating ethyne in a red-hot iron tube trimerises it to benzene. (c) Butan-2-one; both triple-bonded carbons are internal, so the enol CH₃–C(OH)=CH–CH₃ rearranges to a ketone, and only ethyne can give an aldehyde.

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.Markovnikov's rule?Show answer
The negative part of an unsymmetrical reagent goes to the carbon with fewer hydrogens.
2.Which halogen acid shows the peroxide effect?Show answer
HBr only.
3.Product of ozonolysis of an =CH₂ group?Show answer
Methanal.
4.Catalyst for cis reduction of alkynes?Show answer
Lindlar's catalyst.
5.Hydration of ethyne gives?Show answer
Ethanal.
6.Test for a terminal alkyne?Show answer
White precipitate with ammoniacal silver nitrate.

Alkenes and Alkynes: Addition Reactions, Markovnikov Rule and Ozonolysis: frequently asked questions

Why do alkenes give addition and not substitution reactions?

The π bond is weaker than a σ bond and its electrons are exposed, so it reacts with electrophiles and the double bond turns into single bonds.

Why does the more stable carbocation decide the product?

It forms faster, so the nucleophile adds at that carbon and one product dominates.

What is the purpose of Zn in ozonolysis?

It prevents the aldehyde from being oxidised further by hydrogen peroxide formed in the hydrolysis.

Why are terminal alkynes acidic?

The sp-hybridised carbon is more electronegative, so the C–H bond polarises and the proton can be removed.

How can ethene and ethyne be distinguished?

Ethyne gives a white precipitate with ammoniacal silver nitrate; ethene does not.