ChemistryISC Class 11 · Chapter 10

Hydrocarbons ISC Class 11: Alkanes, Alkenes, Alkynes and Benzene

Updated 15 October 2026 · Fact-checked

Hydrocarbons is a chapter in Organic Chemistry for ISC Class 11. It covers alkanes (conformations, preparations such as Wurtz and Kolbe, free-radical halogenation), alkenes (preparation, Saytzeff's rule, Markownikoff additions and the peroxide effect, ozonolysis), alkynes (acidity, acetylides, additions) and benzene (aromaticity, electrophilic substitution, directive influence).

What Hydrocarbons is about

Hydrocarbons contain only carbon and hydrogen. They are saturated (alkanes, CₙH₂ₙ₊₂), unsaturated (alkenes CₙH₂ₙ with C=C; alkynes CₙH₂ₙ₋₂ with C≡C) or aromatic (benzene and its relatives).

Alkanes have only σ bonds, so rotation about C–C bonds gives conformations (staggered and eclipsed ethane). They are made by decarboxylation, Kolbe's electrolysis, the Wurtz and Corey-House reactions, from aldehydes by reduction, and from Grignard reagents. Being unreactive, they react mainly by free-radical substitution (halogenation in light), combustion and pyrolysis (cracking).

Alkenes have a σ + π double bond; the π electrons attract electrophiles, so alkenes undergo addition. Unsymmetrical additions follow Markownikoff's rule, except HBr with peroxides (anti-Markownikoff). Ozonolysis locates the double bond; Baeyer's reagent tests for unsaturation. Elimination reactions that make alkenes follow Saytzeff's rule.

Alkynes have a triple bond (1σ + 2π). Terminal alkynes are weakly acidic and form acetylides, which distinguishes them from alkenes.

Benzene is unusually stable: its six π electrons are delocalised (resonance, aromaticity), so it prefers electrophilic substitution (halogenation, nitration, sulphonation, Friedel-Crafts) to addition. A substituent already on the ring directs the next group to ortho/para or meta positions. Some polycyclic aromatic hydrocarbons are carcinogenic.

Weightage: Hydrocarbons belongs to unit 9 of the ISC Class XI Chemistry syllabus. CISCE weights units 8-9 together (Organic Chemistry) at 23 of the 70 marks in Paper I (Theory). Class XI is examined by the school, so your school sets the paper within this weightage.Source: CISCE ISC Chemistry syllabus, examination year 2028, Class XI unit weightage

Key concepts

Classification, nomenclature and isomerism of alkanes: Hydrocarbons are acyclic (alkanes, alkenes, alkynes), alicyclic (cycloalkanes) or aromatic (arenes). Alkanes CₙH₂ₙ₊₂ are named from the longest chain with alkyl substituents numbered from the end giving lowest locants. From C₄ onwards they show chain isomerism: butane has 2 isomers, pentane (C₅H₁₂) 3, hexane (C₆H₁₄) 5. Physical properties: boiling points rise with molar mass (stronger dispersion forces) and fall with branching (smaller surface area); alkanes are non-polar, insoluble in water and less dense than water.; Worked example
Draw and name the isomers of C₅H₁₂ and arrange them by boiling point.
Try it first, then show the solutionSolution
Pentane CH₃(CH₂)₃CH₃; 2-methylbutane (CH₃)₂CHCH₂CH₃; 2,2-dimethylpropane C(CH₃)₄
More branching gives a more compact, nearly spherical molecule with less surface contact
Boiling point: pentane > 2-methylbutane > 2,2-dimethylpropane
Answer: Three isomers; boiling points 36 °C > 28 °C > 10 °C (approximately)
Conformations of ethane: Free rotation about the C–C σ bond produces conformations. In the staggered form the H atoms on the two carbons are as far apart as possible (dihedral angle 60°); in the eclipsed form they are directly behind each other (0°). Repulsion between eclipsed bonds (torsional strain) makes the eclipsed form about 12.5 kJ mol⁻¹ less stable. The barrier is small, so at room temperature rotation is rapid and the conformers cannot be separated. They are drawn as sawhorse projections (C–C bond seen obliquely) and Newman projections (looking along the C–C bond: front carbon as a point, back carbon as a circle).; Worked example
Which conformation of ethane is more stable, and why can't the two be separated?
Try it first, then show the solutionSolution
Staggered: minimum repulsion between C–H bond pairs on adjacent carbons
The energy difference (about 12.5 kJ mol⁻¹) is easily supplied by collisions at room temperature, so they interconvert rapidly
Answer: Staggered; low rotation barrier
Preparation of alkanes: Decarboxylation: sodium salt + soda lime: $C H_{3} C O O N a + N a O H \to C H_{4} + N a_{2} C O_{3}$ (one carbon fewer). Kolbe's electrolysis: aqueous sodium carboxylate gives the alkane at the anode: $2 C H_{3} C O O N a + 2 H_{2} O \to C_{2} H_{6} + 2 C O_{2} + 2 N a O H + H_{2}$ (even-numbered, symmetrical alkanes). Wurtz reaction: alkyl halide + sodium in dry ether: $2 C H_{3} B r + 2 N a \to C_{2} H_{6} + 2 N a B r$ (doubles the chain; poor for unsymmetrical alkanes). Corey-House synthesis: lithium dialkylcuprate + alkyl halide gives R–R′, good for unsymmetrical alkanes: $(C H_{3})_{2} C uL i + C_{2} H_{5} B r \to C_{3} H_{8} + C H_{3} C u + L i B r$ . From aldehydes: Clemmensen reduction (Zn-Hg/conc. HCl) of the C=O to CH₂: $C H_{3} C H O + 4 [H] \to C_{2} H_{6} + H_{2} O$ . From Grignard reagents: $C H_{3} M g B r + H_{2} O \to C H_{4} + M g (O H) B r$ .; Worked example
Why is the Wurtz reaction unsuitable for making propane, and what method would you use?
Try it first, then show the solutionSolution
CH₃Br + C₂H₅Br + Na gives a mixture of ethane, propane and butane, hard to separate
Corey-House: (CH₃)₂CuLi + C₂H₅Br gives propane cleanly
Answer: Mixtures form; use the Corey-House synthesis
Reactions of alkanes: halogenation, combustion, pyrolysis and others: Free-radical halogenation ( $C H_{4} + C l_{2} \to C H_{3} C l + H C l$ in light) proceeds by initiation (Cl₂ → 2Cl• in UV light), propagation (Cl• + CH₄ → •CH₃ + HCl; •CH₃ + Cl₂ → CH₃Cl + Cl•) and termination (two radicals combine: Cl• + Cl•, •CH₃ + Cl•, •CH₃ + •CH₃). Further substitution gives CH₂Cl₂, CHCl₃ and CCl₄. Reactivity: F₂ > Cl₂ > Br₂ > I₂; 3° H > 2° H > 1° H. Combustion: $C_{3} H_{8} + 5 O_{2} \to 3 C O_{2} + 4 H_{2} O$ , highly exothermic. Pyrolysis (cracking): heating without air breaks large molecules: $C_{6} H_{14} \to C_{4} H_{8} + C_{2} H_{6}$ . Aromatisation (cyclisation + dehydrogenation): n-hexane over Cr₂O₃/Al₂O₃ at about 773 K: $C_{6} H_{14} \to C_{6} H_{6} + 4 H_{2}$ . Isomerisation: n-alkanes with anhydrous AlCl₃/HCl give branched isomers (n-hexane → 2-methylpentane).; Worked example
Write the mechanism of chlorination of methane.
Try it first, then show the solutionSolution
Initiation: Cl–Cl → 2Cl• (homolysis by UV light)
Propagation: Cl• + CH₄ → •CH₃ + HCl; •CH₃ + Cl₂ → CH₃Cl + Cl• (chain repeats)
Termination: Cl• + Cl• → Cl₂; •CH₃ + Cl• → CH₃Cl; •CH₃ + •CH₃ → C₂H₆
Answer: Free-radical chain mechanism (ethane formed in termination is evidence for •CH₃)
Alkenes: structure, isomerism and preparation: In an alkene each double-bonded carbon is sp² (planar, 120°); the double bond is 1σ + 1π, shorter (134 pm) and stronger than C–C but more reactive because π electrons are exposed. Alkenes show structural and geometrical (cis-trans) isomerism. Preparation: dehydration of alcohols with conc. H₂SO₄ at 443 K, $C_{2} H_{5} O H \to C_{2} H_{4} + H_{2} O$ ; dehydrohalogenation with alcoholic KOH, $C_{2} H_{5} B r + K O H \to C_{2} H_{4} + K B r + H_{2} O$ ; Kolbe's electrolysis of sodium succinate, $N a O O C C H_{2} C H_{2} C O O N a + 2 H_{2} O \to C_{2} H_{4} + 2 C O_{2} + 2 N a O H + H_{2}$ ; partial hydrogenation of alkynes with Lindlar's catalyst (gives cis-alkenes), $C_{2} H_{2} + H_{2} \to C_{2} H_{4}$ . Saytzeff's rule: in elimination, the more substituted alkene (with fewer H on the double-bond carbons) is the major product.; Worked example
Predict the major product when 2-bromobutane is heated with alcoholic KOH.
Try it first, then show the solutionSolution
H can be removed from C1 (giving but-1-ene) or C3 (giving but-2-ene)
By Saytzeff's rule the more substituted alkene predominates
Answer: But-2-ene (major); but-1-ene (minor)
Addition reactions of alkenes: Hydrogen (Ni, Pt or Pd): alkene → alkane. Halogens: $C_{2} H_{4} + B r_{2} \to C_{2} H_{4} B r_{2}$ ; the red-brown colour of bromine water disappears (test for unsaturation). Hydrogen halides (HI > HBr > HCl): Markownikoff's rule: the negative part adds to the carbon carrying fewer hydrogens, because the more stable (more substituted) carbocation forms first: CH₃CH=CH₂ + HBr → CH₃CHBrCH₃. Peroxide (Kharasch) effect: with HBr in the presence of peroxide, a free-radical mechanism gives the anti-Markownikoff product CH₃CH₂CH₂Br, because Br• adds first to give the more stable secondary radical (not shown by HCl or HI). Sulphuric acid: $C_{2} H_{4} + H_{2} S O_{4} \to C_{2} H_{5} H S O_{4}$ , which hydrolyses to an alcohol; acid-catalysed hydration: $C_{2} H_{4} + H_{2} O \to C_{2} H_{5} O H$ . Ozonolysis: O₃ then Zn/H₂O cleaves C=C into two carbonyl compounds (propene → ethanal + methanal), locating the double bond. Baeyer's reagent (cold dilute alkaline KMnO₄): purple colour disappears, giving a diol: $3 C_{2} H_{4} + 2 K M n O_{4} + 4 H_{2} O \to 3 C_{2} H_{6} O_{2} + 2 M n O_{2} + 2 K O H$ . Polymerisation: n CH₂=CH₂ → (–CH₂–CH₂–)ₙ, polythene.; Worked example
Write the mechanism of addition of HBr to propene (no peroxide).
Try it first, then show the solutionSolution
H⁺ adds to the CH₂ end, giving the secondary carbocation CH₃C⁺HCH₃ (more stable than the primary CH₃CH₂CH₂⁺)
Br⁻ attacks the carbocation
Product: CH₃CHBrCH₃
Answer: 2-bromopropane (Markownikoff product)
Alkynes: structure, preparation, acidity and reactions: Each triple-bonded carbon is sp (linear, 180°); the triple bond is 1σ + 2π and short (120 pm). Preparation: from calcium carbide, $C a C_{2} + 2 H_{2} O \to C_{2} H_{2} + C a (O H)_{2}$ ; from natural gas at about 1500 °C, $2 C H_{4} \to C_{2} H_{2} + 3 H_{2}$ ; by double dehydrohalogenation of vicinal dihalides, $C H_{2} B r C H_{2} B r + 2 K O H \to C_{2} H_{2} + 2 K B r + 2 H_{2} O$ (alcoholic KOH, then NaNH₂); by Kolbe's electrolysis of potassium maleate. Acidic character: the sp carbon (50% s character) holds electrons closely, so the ≡C–H hydrogen is weakly acidic: $2 C_{2} H_{2} + 2 N a \to 2 C_{2} H N a + H_{2}$ ; terminal alkynes give a white precipitate with ammoniacal AgNO₃ and a red one with ammoniacal Cu₂Cl₂ (acetylides). Additions: H₂ (to alkane, or to alkene with Lindlar's catalyst), halogens, HX (Markownikoff, twice), water with HgSO₄/H₂SO₄, $C_{2} H_{2} + H_{2} O \to C H_{3} C H O$ . Polymerisation: through a red-hot iron tube, $3 C_{2} H_{2} \to C_{6} H_{6}$ . Oxidation with KMnO₄ cleaves the triple bond.; Worked example
Ethyne is made from natural gas: $2 C H_{4} \to C_{2} H_{2} + 3 H_{2}$ (about 1500 °C). What mass of methane gives 5.2 g of ethyne, and what volume of hydrogen at STP forms with it?
Try it first, then show the solutionSolution
n(C₂H₂) = 5.2/26 = 0.20 mol
n(CH₄) = 2 × 0.20 = 0.40 mol = 0.40 × 16 = 6.4 g
n(H₂) = 3 × 0.20 = 0.60 mol = 0.60 × 22.4 = 13.44 dm³
Answer: 6.4 g of methane; 13.44 dm³ of hydrogen
Distinguishing alkanes, alkenes and alkynes: Bromine water (red-brown): decolourised by alkenes and alkynes (addition), not by alkanes. Baeyer's reagent (purple): decolourised by alkenes and alkynes, not by alkanes. Ammoniacal silver nitrate: white precipitate of silver acetylide only with terminal alkynes (ethyne, propyne), not with alkenes or internal alkynes. Ammoniacal copper(I) chloride: red precipitate with terminal alkynes. Alkanes give none of these at room temperature.; Worked example
How would you distinguish ethane, ethene and ethyne?
Try it first, then show the solutionSolution
Bromine water: ethane no change; ethene and ethyne decolourise it
Ammoniacal AgNO₃: only ethyne gives a white precipitate
Answer: Bromine water separates ethane; ammoniacal AgNO₃ separates ethyne from ethene
Benzene: structure, resonance, aromaticity, naming and preparation: Kekulé proposed a six-membered ring with alternating single and double bonds, oscillating between two forms. In fact all C–C bonds are equal (139 pm, between C–C 154 and C=C 134 pm): benzene is a resonance hybrid of the two Kekulé structures, with six π electrons delocalised above and below the planar ring of sp² carbons. Aromaticity (Hückel's rule): a planar, cyclic, fully conjugated system with (4n + 2) π electrons is aromatic and extra stable (benzene: n = 1). Hence benzene resists addition and prefers substitution. Naming: monosubstituted benzenes (chlorobenzene, nitrobenzene, methylbenzene = toluene); disubstituted with locants 1,2- (ortho), 1,3- (meta), 1,4- (para). Preparation: from sodium benzoate with soda lime, $C_{6} H_{5} C O O N a + N a O H \to C_{6} H_{6} + N a_{2} C O_{3}$ ; from phenol with zinc dust, $C_{6} H_{5} O H + Z n \to C_{6} H_{6} + Z n O$ ; and by trimerising ethyne.; Worked example
Use Hückel's rule to show that benzene is aromatic, and give evidence that its C–C bonds are equal.
Try it first, then show the solutionSolution
Planar, cyclic, every carbon sp² with a p orbital; 6 π electrons = 4n + 2 with n = 1
X-ray data show all six C–C bonds are 139 pm
Only one 1,2-disubstituted isomer exists, which would not be so with fixed alternate bonds
Answer: Aromatic (6 π electrons); equal bond lengths
Electrophilic substitution, directive influence and other reactions of benzene: Mechanism: (1) generation of an electrophile E⁺; (2) E⁺ attacks the π cloud to form a σ-complex (arenium ion), losing aromaticity; (3) loss of H⁺ restores the aromatic ring. Nitration (conc. HNO₃ + conc. H₂SO₄): $H N O_{3} + 2 H_{2} S O_{4} \to N O_{2}^{+} + H_{3} O^{+} + 2 H S O_{4}^{-}$ , then $C_{6} H_{6} + H N O_{3} \to C_{6} H_{5} N O_{2} + H_{2} O$ . Halogenation (Lewis acid FeCl₃): $C_{6} H_{6} + C l_{2} \to C_{6} H_{5} C l + H C l$ . Sulphonation (fuming H₂SO₄, E = SO₃): $C_{6} H_{6} + H_{2} S O_{4} \to C_{6} H_{5} S O_{3} H + H_{2} O$ . Friedel-Crafts alkylation: $C_{6} H_{6} + C H_{3} C l \to C_{6} H_{5} C H_{3} + H C l$ (anhydrous AlCl₃); acylation: $C_{6} H_{6} + C H_{3} C O C l \to C_{6} H_{5} C O C H_{3} + H C l$ . Directive influence: ortho/para directors (activating: –OH, –NH₂, –OCH₃, –CH₃; deactivating but o/p: halogens) donate electron density to o/p positions; meta directors (deactivating: –NO₂, –CHO, –COOH, –CN, –SO₃H) withdraw it most from o/p, leaving meta relatively richer. Addition (under forcing conditions): $C_{6} H_{6} + 3 H_{2} \to C_{6} H_{12}$ ; $C_{6} H_{6} + 3 C l_{2} \to C_{6} H_{6} C l_{6}$ (UV light). Oxidation: side chains are oxidised by KMnO₄ to –COOH (toluene → benzoic acid). Combustion: $2 C_{6} H_{6} + 15 O_{2} \to 12 C O_{2} + 6 H_{2} O$ (sooty flame). Pyrolysis of hydrocarbons gives aromatic compounds. Toxicity: benzene is toxic, and polycyclic aromatic hydrocarbons such as benzo[a]pyrene (in soot and tobacco smoke) are carcinogenic.; Worked example
Predict the main products of nitrating (i) toluene (ii) nitrobenzene.
Try it first, then show the solutionSolution
(i) –CH₃ is an activating ortho/para director: 2-nitrotoluene and 4-nitrotoluene
(ii) –NO₂ is a deactivating meta director: 1,3-dinitrobenzene (needs harsher conditions)
Answer: o- and p-nitrotoluene; m-dinitrobenzene

Chemistry in real life: Hydrocarbons

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.

Cracking for petrol

Suppose a refinery cracks heavy hexane-rich fractions: $C_{6} H_{14} \to C_{4} H_{8} + C_{2} H_{6}$ .

Why is cracking useful, and what mass of butene comes from 86 kg of hexane?

Try it first, then show the solutionSolution

It turns less useful large molecules into smaller fuels and alkenes for plastics
86 kg hexane = 1000 mol (M = 86); butene formed = 1000 mol × 56 g = 56 kg

Answer: 56 kg of butene (with 30 kg of ethane), assuming this single route.

Memory hook: Cracking: big chains in, small alkanes and alkenes out.

Ripening fruit with ethene

Imagine a fruit warehouse that releases a little ethene gas to ripen bananas evenly.

How could a worker show that a gas sample contains an alkene rather than only alkanes?

Try it first, then show the solutionSolution

Bubble the gas through bromine water
Ethene adds Br₂ ( $C_{2} H_{4} + B r_{2} \to C_{2} H_{4} B r_{2}$ ) and decolourises it; alkanes do not

Answer: Red-brown bromine water turns colourless.

Memory hook: Unsaturation decolourises bromine.

Carbide lamps

Suppose an old miner's lamp drips water onto calcium carbide (M = 64) to make ethyne, and the flame uses 14 dm³ of ethyne (measured at STP) every hour.

What mass of carbide is used up in one hour, and why does ethyne burn with a bright flame?

Try it first, then show the solutionSolution

n(C₂H₂) = 14/22.4 = 0.625 mol; CaC₂ + 2H₂O → C₂H₂ + Ca(OH)₂ is 1 : 1, so 0.625 mol CaC₂ = 0.625 × 64 = 40 g
Ethyne has a high carbon content; the incandescent carbon particles glow brightly

Answer: 40 g of carbide per hour; a bright, sooty flame.

Memory hook: 1 mol carbide → 1 mol ethyne.

Topics in this chapter

Hydrocarbons formula sheet

General formulas: $alkane C_{n} H_{2 n + 2}; alkene C_{n} H_{2 n}; alkyne C_{n} H_{2 n - 2}$; Cycloalkanes are also CₙH₂ₙ.
Wurtz reaction: $2 C H_{3} B r + 2 N a \to C_{2} H_{6} + 2 N a B r$; Dry ether; doubles the chain.
Decarboxylation: $C H_{3} C O O N a + N a O H \to C H_{4} + N a_{2} C O_{3}$; Soda lime; one carbon fewer.
Kolbe's electrolysis: $2 C H_{3} C O O N a + 2 H_{2} O \to C_{2} H_{6} + 2 C O_{2} + 2 N a O H + H_{2}$; Alkane at the anode.
Ethyne from carbide: $C a C_{2} + 2 H_{2} O \to C_{2} H_{2} + C a (O H)_{2}$; Laboratory and industrial method.
Nitration of benzene: $C_{6} H_{6} + H N O_{3} \to C_{6} H_{5} N O_{2} + H_{2} O$; Electrophile NO₂⁺; conc. H₂SO₄.
Friedel-Crafts: $C_{6} H_{6} + C H_{3} C l \to C_{6} H_{5} C H_{3} + H C l$; Anhydrous AlCl₃.
Hückel's rule: $π electrons = 4 n + 2$; Planar, cyclic, conjugated.

Hydrocarbons: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

Newman and sawhorse projections of ethane

Showing staggered and eclipsed conformations.

Draw

Newman: a dot (front carbon) with three bonds in a Y; a circle (back carbon) with three bonds from its edge.
Staggered: back bonds midway between front bonds; eclipsed: back bonds directly behind (draw slightly offset).
Sawhorse: a diagonal C–C bond with three H on each carbon, staggered and eclipsed versions.

Label

front carbon
back carbon
H
dihedral angle 60° (staggered)
0° (eclipsed)
more stable: staggered

Watch out: Mark the energy difference (about 12.5 kJ mol⁻¹) beside the two forms.

Orbital picture of ethene and ethyne

Explaining σ and π bonds in C=C and C≡C.

Draw

Ethene: two sp² carbons with σ bonds in a plane (120°), and one pair of p orbitals above and below overlapping sideways.
Ethyne: two sp carbons in a line (180°), and two pairs of p orbitals overlapping at right angles.
Shade the π clouds.

Label

sp² C
sp C
σ bond
π bond
120°
180°

Watch out: Two π clouds in ethyne merge into a cylinder of electron density.

Resonance and the delocalised ring of benzene

Representing benzene's structure.

Draw

The two Kekulé structures with a double-headed resonance arrow between them.
A hexagon with an inscribed circle representing the hybrid.
An orbital sketch: six p orbitals above and below the ring forming two doughnut-shaped π clouds.

Label

Kekulé I
Kekulé II
resonance hybrid
delocalised π electrons
C–C 139 pm

Watch out: Use ↔ for resonance, never ⇌.

How the school exam asks Hydrocarbons

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Alkane chemistry	Isomers and boiling points; conformations; preparations; halogenation mechanism; cracking and aromatisation.	Write balanced equations with conditions; show radicals with dots.
Alkene chemistry	Preparations, Saytzeff products, Markownikoff and peroxide products with mechanisms, ozonolysis products.	Find the more stable carbocation or radical; split C=C into two C=O for ozonolysis.
Alkyne chemistry	Preparations, acidity and acetylides, hydration, polymerisation; stoichiometry with CaC₂.	Relate acidity to s character; terminal vs internal alkynes.
Distinguishing tests	Distinguish alkanes, alkenes and alkynes; terminal from internal alkynes.	Bromine water/Baeyer's reagent, then ammoniacal AgNO₃.
Benzene	Resonance and aromaticity; preparations; electrophilic substitution mechanism; directive influence.	Three-step mechanism; classify substituents as o/p or m directors.

Tricks to remember Hydrocarbons

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

Rich get richer
Markownikoff: H adds to the carbon that already has more H.
Use it when: Addition of HX.
Peroxide flips only bromide
Anti-Markownikoff with peroxide works for HBr only, not HCl or HI.
Use it when: Peroxide effect.
Saytzeff likes crowds
Elimination gives the more substituted alkene.
Use it when: Predicting elimination products.
Ozone cuts and caps
Ozonolysis cuts C=C and puts =O on each end.
Use it when: Locating double bonds.
Donors o/p, withdrawers meta
–OH, –NH₂, –CH₃, –X direct ortho/para; –NO₂, –CHO, –COOH, –CN direct meta.
Use it when: Directive influence.

Hydrocarbons: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Decolourising bromine water · 1 mark · MCQExam level

Which compound does NOT decolourise bromine water?
(a) ethene
(b) ethyne
(c) propene
(d) propane

Try it first, then show the solutionSolution

Only unsaturated hydrocarbons add bromine; propane is saturated.

Answer: (d)

Example 2: Acidity of ethyne · 1 mark · Assertion-reasonExam level

Assertion: Ethyne reacts with sodium to liberate hydrogen, but ethene does not.
Reason: The sp-hybridised carbon in ethyne has greater s character and holds the C–H bonding electrons more tightly, making the hydrogen acidic.
(a) Both Assertion and Reason are true, and Reason is the correct explanation for Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation for Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Try it first, then show the solutionSolution

50% s character (sp) vs 33% (sp²): the conjugate base of ethyne is more stable, so ethyne is more acidic.

Answer: (a)

Example 3: Kolbe's electrolysis · 2 marks · Short answerHarder exam level

Write the equation for Kolbe's electrolysis of aqueous sodium propanoate and name the product formed at the anode. [2]

Try it first, then show the solutionSolution

$2 C_{2} H_{5} C O O N a + 2 H_{2} O \to C_{4} H_{10} + 2 C O_{2} + 2 N a O H + H_{2}$
Two ethyl groups join at the anode: butane (with CO₂); H₂ and NaOH form at the cathode side

Answer: Butane

Example 4: Peroxide effect · 3 marks · Short answerHarder exam level

Give the products of adding HBr to propene (i) without peroxide (ii) with benzoyl peroxide, and explain the difference. [3]

Try it first, then show the solutionSolution

(i) Ionic mechanism: H⁺ adds first to give the 2° carbocation; product 2-bromopropane (Markownikoff)
(ii) Free-radical mechanism: Br• (from peroxide + HBr) adds to CH₂ first, giving the more stable 2° radical CH₃C•HCH₂Br
This radical takes H from HBr: 1-bromopropane (anti-Markownikoff)

Answer: (i) CH₃CHBrCH₃ (ii) CH₃CH₂CH₂Br

Example 5: Ozonolysis to find structure · 3 marks · Short answerHarder exam level

An alkene C₅H₁₀ on ozonolysis gives propanone and ethanal. Identify the alkene. [3]

Try it first, then show the solutionSolution

Join the two carbonyl carbons with a double bond, removing the oxygens
(CH₃)₂C=O and O=CHCH₃ ⇒ (CH₃)₂C=CHCH₃
Check formula: C₅H₁₀ ✓

Answer: 2-methylbut-2-ene

Example 6: Preparing and nitrating benzene · 5 marks · Long answerHarder exam level

(a) Write equations for preparing benzene from sodium benzoate and from phenol. [2]
(b) Write the mechanism of nitration of benzene. [2]
(c) 7.8 g of benzene gave 9.84 g of nitrobenzene. Find the percentage yield. [1]

Try it first, then show the solutionSolution

(a) $C_{6} H_{5} C O O N a + N a O H \to C_{6} H_{6} + N a_{2} C O_{3}$ (soda lime, heat); $C_{6} H_{5} O H + Z n \to C_{6} H_{6} + Z n O$
(b) Electrophile: $H N O_{3} + 2 H_{2} S O_{4} \to N O_{2}^{+} + H_{3} O^{+} + 2 H S O_{4}^{-}$ ; NO₂⁺ attacks the ring to give a σ-complex
HSO₄⁻ removes H⁺ from the σ-complex, restoring aromaticity: C₆H₅NO₂
(c) 7.8 g = 0.10 mol; theoretical 0.10 × 123 = 12.3 g; yield = 9.84/12.3 × 100 = 80%

Answer: (c) 80%

Example 7: Saytzeff versus Markownikoff · 3 marks · Short answerHOTS / competency

Starting from 2-bromo-2-methylbutane, give the major product of (i) heating with alcoholic KOH (ii) adding HBr to that product. [3]

Try it first, then show the solutionSolution

(i) Saytzeff: H is lost from C3 to give the more substituted alkene, 2-methylbut-2-ene (CH₃)₂C=CHCH₃
(ii) Markownikoff: H adds to C3 (the CH), giving the 3° carbocation at C2
Br⁻ adds at C2: 2-bromo-2-methylbutane is regenerated

Answer: (i) 2-methylbut-2-ene (ii) 2-bromo-2-methylbutane

Example 8: Directing effects in sequence · 3 marks · Short answerHOTS / competency

Starting from benzene, give the order of steps to make (i) m-bromonitrobenzene (ii) p-bromonitrobenzene, and explain. [3]

Try it first, then show the solutionSolution

(i) Nitrate first: –NO₂ is a meta director; then brominate (Br₂/FeBr₃) to get the meta product
(ii) Brominate first: –Br is an ortho/para director; then nitrate and separate the para isomer
The group already on the ring decides where the next one goes

Answer: (i) Nitration then bromination (ii) bromination then nitration

Case-study practice (Section D style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

Making ethyne in the lab

Suppose a student adds water dropwise to 12.8 g of calcium carbide (M = 64) and collects the gas.

(i) · 1 mark
Write the equation.
Answer
Answer: $C a C_{2} + 2 H_{2} O \to C_{2} H_{2} + C a (O H)_{2}$
Carbide ion takes two protons from water.
(ii) · 1 mark
Find the volume of ethyne at STP.
Answer
Answer: 4.48 dm³
0.20 mol × 22.4.
(iii) · 1 mark
How would you show that the gas is a terminal alkyne?
Answer
Answer: White precipitate with ammoniacal silver nitrate
Silver acetylide forms only with ≡C–H.
(iv) · 2 marks
What is formed when the gas is passed through a red-hot iron tube, and what mass of it forms (assume 100%)?
Answer
Answer: Benzene; 5.2 g
$3 C_{2} H_{2} \to C_{6} H_{6}$ : 0.20 mol ethyne gives 0.0667 mol benzene × 78 g mol⁻¹ = 5.2 g.

From propene to products

Suppose propene is supplied to a lab for a series of addition reactions.

(i) · 1 mark
Name the product with HCl.
Answer
Answer: 2-chloropropane
Markownikoff: Cl adds to the middle carbon.
(ii) · 1 mark
Name the product with HBr in the presence of peroxide.
Answer
Answer: 1-bromopropane
Free-radical (anti-Markownikoff) addition.
(iii) · 1 mark
What are the ozonolysis products?
Answer
Answer: Ethanal and methanal
CH₃CH= becomes CH₃CHO; =CH₂ becomes HCHO.
(iv) · 2 marks
What does cold, dilute alkaline KMnO₄ give, and what is observed?
Answer
Answer: Propane-1,2-diol; the purple colour disappears and a brown precipitate (MnO₂) forms
Baeyer's test for unsaturation: two –OH groups add across the double bond.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Write the Wurtz reaction for preparing butane and state its limitation. (2 marks)

Step	Marks
$2 C_{2} H_{5} B r + 2 N a \to C_{4} H_{10} + 2 N a B r$ (dry ether)	1
Limitation: with two different halides a mixture of three alkanes forms; methane cannot be made	1

Write the steps of the free-radical monochlorination of ethane, and name the termination product that shows ethyl radicals were present. (3 marks)

Step	Marks
Initiation: Cl₂ → 2Cl• in UV light	1
Propagation: Cl• + C₂H₆ → •C₂H₅ + HCl; •C₂H₅ + Cl₂ → C₂H₅Cl + Cl•	1
Termination: two radicals combine (Cl₂, C₂H₅Cl); •C₂H₅ + •C₂H₅ → C₄H₁₀, and the butane found in the product is evidence for ethyl radicals	1

Explain the directive influence of the –OH group in phenol and of the –NO₂ group in nitrobenzene towards electrophilic substitution. (5 marks)

Step	Marks
–OH donates a lone pair into the ring by resonance (+R)	1
Resonance structures place negative charge at the ortho and para carbons, so electrophiles attack there; the ring is activated	1
–NO₂ withdraws electrons by resonance (−R) and induction (−I)	1
Resonance structures place positive charge at ortho and para, so these positions are most deactivated	1
The meta position is least deactivated, so substitution occurs at meta, slowly	1

Where students lose marks in Hydrocarbons

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

Mistake: Applying the peroxide effect to HCl or HI
Students generalise from HBr.
Wrong working
Propene + HCl + peroxide → 1-chloropropane.
Right working
Only HBr shows the peroxide effect; HCl gives 2-chloropropane.
Do this: Remember: peroxide effect = HBr only.
Usually costs: 1 mark.
Mistake: Wrong elimination product
Students remove H from the nearest CH₃.
Wrong working
2-bromobutane + alc. KOH → but-1-ene (major).
Right working
Saytzeff: but-2-ene is major.
Do this: Pick the alkene with more alkyl groups on the double bond.
Usually costs: 1 mark.
Mistake: Drawing benzene with fixed double bonds only
Kekulé structures are taken as real.
Wrong working
Benzene has three short and three long C–C bonds.
Right working
All six bonds are equal (139 pm); benzene is a resonance hybrid.
Do this: Show both Kekulé forms with ↔ or the circle-in-hexagon.
Usually costs: 1 mark.
Mistake: Testing internal alkynes with ammoniacal AgNO₃
Students think all alkynes give acetylides.
Wrong working
But-2-yne gives a white precipitate.
Right working
Only terminal alkynes (≡C–H) react.
Do this: Check for a hydrogen on the triple-bond carbon.
Usually costs: 1 mark.
Mistake: Unbalanced combustion and Kolbe equations
Students balance carbon and forget oxygen or hydrogen.
Wrong working
C₆H₆ + 7O₂ → 6CO₂ + 3H₂O.
Right working
$2 C_{6} H_{6} + 15 O_{2} \to 12 C O_{2} + 6 H_{2} O$ .
Do this: Balance C, then H, then O last; use whole numbers.
Usually costs: 1 mark.

Practice questions on Hydrocarbons

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 · MCQExam level
Which conformation of ethane is the most stable?
(a) eclipsed
(b) staggered
(c) skew
(d) all equal
Hint
Least repulsion.
Answer
(b)
Q2 · 1 mark · MCQExam level
The reagent used to convert sodium ethanoate into methane is:
(a) Na in dry ether
(b) soda lime
(c) Zn/HCl
(d) alcoholic KOH
Hint
Decarboxylation.
Answer
(b)
Q3 · 1 mark · Very short answerExam level
Name the product of hydration of ethyne in the presence of HgSO₄ and dilute H₂SO₄.
Hint
Enol rearranges.
Answer
Ethanal (acetaldehyde)
Q4 · 1 mark · Very short answerExam level
How many π electrons does benzene have?
Hint
4n + 2.
Answer
6
Q5 · 2 marks · Short answerExam level
Write the equation for the preparation of ethyne from calcium carbide and give one use of ethyne.
Hint
Add water.
Answer
$C a C_{2} + 2 H_{2} O \to C_{2} H_{2} + C a (O H)_{2}$ ; used in oxy-acetylene welding torches.
Q6 · 2 marks · Short answerHarder exam level
Write equations for (i) Friedel-Crafts acylation of benzene with ethanoyl chloride (ii) sulphonation of benzene.
Hint
AlCl₃; fuming H₂SO₄.
Answer
(i) $C_{6} H_{6} + C H_{3} C O C l \to C_{6} H_{5} C O C H_{3} + H C l$ (ii) $C_{6} H_{6} + H_{2} S O_{4} \to C_{6} H_{5} S O_{3} H + H_{2} O$
Q7 · 2 marks · Short answerHarder exam level
Give the product of 2-bromopropane with alcoholic KOH and of propene with water in acid.
Hint
Elimination; Markownikoff hydration.
Answer
Propene; propan-2-ol
Q8 · 2 marks · Short answerHarder exam level
An alkene gives only propanone on ozonolysis. Identify it.
Hint
Join two propanone carbons.
Answer
(CH₃)₂C=C(CH₃)₂, 2,3-dimethylbut-2-ene
Q9 · 3 marks · Short answerHarder exam level
How will you distinguish (i) but-1-yne from but-2-yne (ii) cyclohexane from cyclohexene?
Hint
Acetylide test; bromine water.
Answer
(i) Ammoniacal AgNO₃: but-1-yne (terminal, ≡C–H) gives a white silver acetylide precipitate; but-2-yne (internal) does not. (ii) Bromine water or Baeyer's reagent: cyclohexene decolourises it by addition; cyclohexane, being saturated, does not.
Q10 · 2 marks · Short answerHarder exam level
What mass of bromine adds to 1.40 g of ethene, and what is the product?
Hint
1 : 1 addition.
Answer
1.40 g = 0.050 mol; Br₂ = 0.050 × 160 = 8.0 g; 1,2-dibromoethane
Q11 · 3 marks · Short answerHarder exam level
Explain why benzene undergoes substitution rather than addition, although it is unsaturated.
Hint
Resonance energy.
Answer
Its six π electrons are delocalised over the ring (aromatic, 4n + 2), giving extra stability. Addition would destroy the aromatic system, while substitution restores it in the last step, so substitution is favoured.
Q12 · 3 marks · Short answerHarder exam level
Classify these substituents as ortho/para or meta directors: –CH₃, –NO₂, –Cl, –CHO, –NH₂, –COOH.
Hint
Electron donors vs withdrawers (halogens are special).
Answer
Ortho/para: –CH₃, –Cl, –NH₂. Meta: –NO₂, –CHO, –COOH.
Q13 · 5 marks · Long answerHarder exam level
(a) Write the equation for the decarboxylation of sodium propanoate with soda lime. [1]
(b) Explain Markownikoff's rule using the addition of HCl to 2-methylpropene, with the mechanism. [2]
(c) Write the equation for the aromatisation of n-hexane and state the conditions. [2]
Hint
One carbon fewer; carbocation stability; cyclisation and dehydrogenation.
Answer
(a) $C_{2} H_{5} C O O N a + N a O H \to C_{2} H_{6} + N a_{2} C O_{3}$ (ethane). (b) H⁺ adds to the =CH₂ end, giving the tertiary carbocation (CH₃)₃C⁺, more stable than the primary alternative; Cl⁻ then adds to give 2-chloro-2-methylpropane, (CH₃)₃CCl. (c) $C_{6} H_{14} \to C_{6} H_{6} + 4 H_{2}$ , by heating with Cr₂O₃ (or V₂O₅/Mo₂O₃) supported on alumina at about 773 K and 10–20 atm.
Q14 · 3 marks · Short answerHOTS / competency
Ethene and ethyne both have π bonds, but ethyne is less reactive than ethene towards electrophilic addition of Br₂. Suggest a reason.
Hint
sp carbon and the intermediate.
Answer
The sp carbons of ethyne hold their π electrons more tightly (greater s character, more electronegative carbon), so they are less available to electrophiles; also the vinyl-type cation formed from an alkyne is less stable than the alkyl carbocation from an alkene.
Q15 · 3 marks · Short answerHOTS / competency
A hydrocarbon A (C₆H₁₀) adds 2 mol of H₂, gives a white precipitate with ammoniacal AgNO₃, and has an unbranched chain. Identify A and write its reaction with excess HBr.
Hint
Terminal alkyne.
Answer
A is hex-1-yne, HC≡C(CH₂)₃CH₃ (CₙH₂ₙ₋₂, terminal). With excess HBr (Markownikoff twice): CH₃CBr₂(CH₂)₃CH₃, 2,2-dibromohexane.

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.Difference in stability between staggered and eclipsed ethane?Show answer
About 12.5 kJ mol⁻¹.
2.Reaction that doubles an alkyl chain using sodium?Show answer
Wurtz reaction.
3.Best method for unsymmetrical alkanes?Show answer
Corey-House synthesis.
4.Saytzeff's rule?Show answer
The more substituted alkene is the major elimination product.
5.Peroxide effect?Show answer
Anti-Markownikoff addition of HBr in the presence of peroxide (free radical).
6.Test for terminal alkynes?Show answer
White precipitate with ammoniacal AgNO₃.
7.Electrophile in nitration?Show answer
NO₂⁺ (nitronium ion).
8.Two meta-directing groups?Show answer
–NO₂ and –COOH (also –CHO, –CN).

Best order to study it

Learn classification, naming, isomers and physical properties of alkanes; draw ethane conformations.
Learn alkane preparations and reactions, including the halogenation mechanism.
Learn alkene structure, preparation, Saytzeff's rule and all addition reactions with mechanisms.
Learn alkyne structure, preparation, acidity and reactions; then the distinguishing tests.
Learn benzene's structure, aromaticity, naming and preparations.
Finish with electrophilic substitution mechanisms, Friedel-Crafts, directive influence and toxicity.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you draw Newman and sawhorse projections of ethane?
Can you write balanced equations for Wurtz, Kolbe and decarboxylation?
Can you write the chlorination mechanism of methane?
Can you predict Saytzeff and Markownikoff products?
Can you explain the peroxide effect with a mechanism?
Can you deduce an alkene from its ozonolysis products?
Can you distinguish alkanes, alkenes and terminal alkynes?
Can you predict the position of substitution in monosubstituted benzenes?

Hydrocarbons: frequently asked questions

Which reactions need mechanisms in ISC Class 11 Hydrocarbons?

The ISC 2028 syllabus lists the free-radical halogenation of alkanes, addition of hydrogen halides to alkenes (Markownikoff's rule and the peroxide effect) and electrophilic substitution in benzene (halogenation, nitration, sulphonation) with mechanisms.

Why does the peroxide effect work for HBr but not HCl or HI?

For HCl the H–Cl bond is too strong for the chain step that regenerates Cl•, and for HI the I• radical is too unreactive to add to the alkene, so only HBr has both propagation steps energetically favourable.

Why are halogens deactivating but ortho/para directing?

Their strong −I effect withdraws electron density and slows the reaction, but their lone pairs donate by resonance to the ortho and para positions, making those the least deactivated.

Why is benzene considered harmful?

Benzene is toxic and long-term exposure is linked to blood disorders, and polycyclic aromatic hydrocarbons such as benzo[a]pyrene, found in soot and smoke, are carcinogenic.

What is the difference between Kolbe's method for alkanes, alkenes and alkynes?

The same electrolysis is used with different salts: sodium ethanoate gives ethane, sodium succinate gives ethene, and potassium maleate or fumarate gives ethyne.

Why do branched alkanes boil lower than straight-chain isomers?

Branching makes the molecule more compact, reducing the surface area in contact with neighbours, so dispersion forces are weaker.

Based on: CISCE ISC 2028 Chemistry syllabus (Classes XI and XII); CISCE Regulations for the ISC Examination Year 2028 (Class XI internal examination)