Hydrocarbons Class 11: Alkanes, Alkenes, Alkynes and Benzene with Reactions

What Hydrocarbons is about

Weightage: This chapter is Unit IX, Hydrocarbons, which the syllabus gives 10 of the 70 theory marks in the school annual exam.Source: CBSE Chemistry (043) syllabus 2026-27, Class XI course structure, Unit IX Hydrocarbons

Key concepts

Classification; alkanes: nomenclature, isomerism and preparation

NCERT section 9.1 Classification; 9.2.1 Nomenclature and Isomerism; 9.2.2 Preparation

Hydrocarbons are open-chain saturated (alkanes), open-chain unsaturated (alkenes, alkynes), alicyclic (cyclohexane) or aromatic (benzene, naphthalene). LPG (mainly butane), CNG and petrol are mixtures of hydrocarbons.

Alkanes: general formula CₙH₂ₙ₊₂; all carbons sp³, tetrahedral, C–C 154 pm, C–H 112 pm. Isomerism starts from butane (two chain isomers) and grows quickly (C₅H₁₂ has three: pentane, 2-methylbutane and 2,2-dimethylpropane). A carbon is primary (1°), secondary (2°), tertiary (3°) or quaternary (4°) depending on whether it is bonded to one, two, three or four other carbons.

Preparation:
1. Hydrogenation of alkenes or alkynes with H₂ over finely divided Pt, Pd or Ni: $C_2{H}_4+H_2\to C_2{H}_6$.
2. Reduction of alkyl halides (not fluorides) with zinc and dilute HCl: $C{H}_{3}Cl+H_2\to C{H}_4+HCl$ (overall).
3. Wurtz reaction: alkyl halide + sodium in dry ether gives an alkane with twice the carbons: $2C{H}_{3}Br+2Na\to C_2{H}_6+2NaBr$. It cannot make methane, and mixing two different halides gives a mixture of three alkanes.
4. Decarboxylation: heating the sodium salt of a carboxylic acid with soda lime (NaOH + CaO) removes CO₂ and gives an alkane with one carbon fewer: $C{H}_{3}COONa+NaOH\to C{H}_4+N{a}_{2}C{O}_3$.
5. Kolbe's electrolytic method: electrolysis of an aqueous solution of the sodium or potassium salt of a carboxylic acid gives an alkane with an even number of carbons at the anode: $2C{H}_{3}COONa+2H_{2}O\to C_2{H}_6+2C{O}_2+H_2+2NaOH$. Methane cannot be made this way.

Worked example

Name the alkane formed in each case and write the equation: (a) sodium propanoate is heated with soda lime (b) an aqueous solution of potassium butanoate is electrolysed (c) 2-bromopropane is treated with sodium in dry ether. (3 marks)

Try it first, then show the solutionSolution

1. (a) Decarboxylation removes one carbon: $C{H}_{3}C{H}_{2}COONa+NaOH\to C_2{H}_6+N{a}_{2}C{O}_3$; ethane. (1)
2. (b) Kolbe electrolysis joins two propyl groups at the anode: $2C{H}_{3}C{H}_{2}C{H}_{2}COOK+2H_{2}O\to C_6{H}_{14}+2C{O}_2+H_2+2KOH$; hexane. (1)
3. (c) Wurtz reaction joins two isopropyl groups: $2(C{H}_3{)}_{2}CHBr+2Na\to (C{H}_3{)}_{2}CH-CH(C{H}_3{)}_2+2NaBr$; 2,3-dimethylbutane. (1)

Answer: (a) ethane (b) hexane (c) 2,3-dimethylbutane

Properties of alkanes and free-radical halogenation

NCERT section 9.2.3 Properties

Physical properties: alkanes are non-polar, held by weak van der Waals forces. C₁–C₄ are gases, C₅–C₁₇ liquids and higher ones solids at 298 K. Boiling point rises with molar mass; among isomers, branching lowers the boiling point because the molecule becomes more spherical, with less surface area for attraction (pentane 309 K > 2-methylbutane 301 K > 2,2-dimethylpropane 282.5 K). Alkanes dissolve in non-polar solvents, not in water ('like dissolves like').

Chemical properties (generally unreactive, hence 'paraffins'):
1. Halogenation (substitution) in diffused sunlight or at high temperature: $C{H}_4+C{l}_2\to C{H}_{3}Cl+HCl$, continuing to CH₂Cl₂, CHCl₃ and CCl₄. Reactivity of halogens: F₂ > Cl₂ > Br₂ > I₂. Rate of replacement of H: 3° > 2° > 1°.
Free-radical mechanism:

• Initiation: Cl–Cl breaks homolytically in light: Cl₂ → 2Cl•.
• Propagation: Cl• + CH₄ → •CH₃ + HCl; •CH₃ + Cl₂ → CH₃Cl + Cl•.
• Termination: radicals combine: Cl• + Cl• → Cl₂; •CH₃ + •CH₃ → C₂H₆; •CH₃ + Cl• → CH₃Cl. (This explains the small amount of ethane formed.)

2. Combustion: complete combustion gives CO₂ and water with much heat: $C_3{H}_8+5O_2\to 3C{O}_2+4H_{2}O$. Incomplete combustion gives carbon black or CO.
3. Controlled oxidation: with suitable catalysts, methane gives methanol (Cu, 523 K, 100 atm) or methanal (Mo₂O₃); alkanes with a tertiary H are oxidised by KMnO₄ to alcohols.
4. Isomerisation: n-alkanes with anhydrous AlCl₃ and HCl give branched isomers.
5. Aromatisation: n-hexane over oxides of V, Mo or Cr at 773 K and 10-20 atm gives benzene; heptane gives toluene.
6. Reaction with steam: methane + steam over Ni at 1273 K gives CO and H₂ (used to make hydrogen).
7. Pyrolysis (cracking): heating higher alkanes to 773 K in the absence of air breaks them into smaller alkanes and alkenes, a free-radical process used to make oil gas and petrol gas.

Worked example

Chlorination of propane in light gives both 1-chloropropane and 2-chloropropane. (a) Write the propagation steps that lead to 2-chloropropane. (b) Propane has six primary and only two secondary hydrogens, yet a large share of the product is 2-chloropropane. Explain. (c) Name one alkane formed in a termination step. (3 marks)

Try it first, then show the solutionSolution

1. (a) $C{l}^{\bullet }+C{H}_{3}C{H}_{2}C{H}_3\to C{H}_3\dot{C}HC{H}_3+HCl$; then $C{H}_3\dot{C}HC{H}_3+C{l}_2\to C{H}_{3}CHClC{H}_3+C{l}^{\bullet }$. (1)
2. (b) The secondary radical is more stable than the primary one (more electron release and hyperconjugation from the two CH₃ groups), so a 2° C–H is broken faster; the higher rate per hydrogen (3° > 2° > 1°) makes up for the smaller number of 2° hydrogens. (1)
3. (c) Two isopropyl radicals combining give 2,3-dimethylbutane (two propyl radicals give hexane). (1)

Answer: (a) Cl• takes a 2° H, then the radical takes Cl from Cl₂ (b) 2° hydrogens react faster because the 2° radical is more stable (c) for example 2,3-dimethylbutane

Conformations of ethane

NCERT section 9.2.4 Conformations

Rotation about a C–C σ bond gives different spatial arrangements called conformations (conformers or rotamers). For ethane there are infinitely many; two are important:

• Staggered: the hydrogens on the two carbons are as far apart as possible (dihedral angle 60°). Minimum repulsion, most stable.
• Eclipsed: the hydrogens on the front carbon are directly in line with those on the back carbon (dihedral angle 0°). Maximum repulsion (torsional strain), least stable.
• Any other arrangement is a skew conformation.

The energy difference is only about 12.5 kJ mol⁻¹, so at room temperature molecules interconvert rapidly; the conformers cannot be isolated.

Two ways to draw them:
1. Sawhorse projection: the C–C bond is drawn as a long diagonal line; the front carbon is at the lower left and the back carbon at the upper right, each with three H atoms.
2. Newman projection: you look straight down the C–C bond. The front carbon is a point with three bonds meeting at the centre; the back carbon is a circle with three bonds coming from its edge.

Worked example

Draw (in words) the Newman projections of the staggered and eclipsed conformations of ethane. Which is more stable and why? Why can they not be separated? (3 marks)

Try it first, then show the solutionSolution

1. Staggered: front C–H bonds as a 'Y' meeting at the centre; back C–H bonds from the circle pointing between the front ones (60° apart). Eclipsed: back C–H bonds directly behind the front ones (drawn slightly offset). (1)
2. Staggered is more stable: the electron clouds of the C–H bonds are farthest apart, so torsional strain is minimum. (1)
3. The energy barrier is only about 12.5 kJ mol⁻¹, so free rotation interconverts them rapidly at room temperature. (1)

Answer: Staggered is more stable (less torsional strain); the small barrier lets them interconvert, so they cannot be isolated.

Alkenes: structure, isomerism and preparation

NCERT section 9.3 Alkenes (9.3.1 to 9.3.4)

Alkenes (CₙH₂ₙ, olefins) contain a C=C made of one strong σ bond and one weaker π bond (C=C 134 pm; bond enthalpy 681 kJ mol⁻¹ vs 348 for C–C). The π electrons are loosely held and exposed, so alkenes attract electrophiles.

Isomerism:

• Structural: chain and position isomers (C₄H₈: but-1-ene, but-2-ene, 2-methylpropene).
• Geometrical (cis-trans): because rotation about C=C is restricted, an alkene with two different groups on each double-bonded carbon (like CH₃CH=CHCH₃) has two forms: cis (similar groups on the same side) and trans (on opposite sides). Cis isomers of such compounds are often polar; trans are often non-polar (trans-but-2-ene has μ = 0) and have higher melting points because they pack better.

Preparation:
1. From alkynes by partial reduction: H₂ with Lindlar's catalyst (Pd on charcoal deactivated with sulphur compounds or quinoline) gives the cis alkene; sodium in liquid ammonia gives the trans alkene.
2. From alkyl halides (dehydrohalogenation): heating with alcoholic KOH removes H and X from adjacent carbons (β-elimination): CH₃CH₂Br + KOH(alc) → CH₂=CH₂ + KBr + H₂O.
3. From vicinal dihalides (halogens on adjacent carbons) with zinc: CH₂Br–CH₂Br + Zn → CH₂=CH₂ + ZnBr₂ (dehalogenation).
4. From alcohols (acidic dehydration): heating with concentrated H₂SO₄ at about 443 K: $C_2{H}_{5}OH\to C_2{H}_4+H_{2}O$.

Worked example

Which of the following show geometrical isomerism: (a) propene (b) but-2-ene (c) 2-methylbut-2-ene (d) pent-2-ene? Draw (in words) the cis and trans forms of one, and say which is more polar. (3 marks)

Try it first, then show the solutionSolution

1. Each double-bonded carbon must carry two different groups. Propene (CH₂= end has two H) and 2-methylbut-2-ene (one carbon has two CH₃) fail. (1)
2. But-2-ene and pent-2-ene show geometrical isomerism. (1)
3. cis-But-2-ene: both CH₃ on the same side of C=C; trans: on opposite sides. The cis form is more polar (dipoles add); in trans they cancel (μ = 0). (1)

Answer: (b) and (d); cis-but-2-ene is polar, trans is non-polar

Reactions of alkenes: addition, Markovnikov's rule, peroxide effect, oxidation and ozonolysis

NCERT section 9.3.5 Properties (of alkenes)

1. Hydrogen: with Ni, Pt or Pd, alkenes give alkanes.
2. Halogens: Br₂ in CCl₄ adds across C=C to give a vicinal dibromide; the reddish-orange colour of bromine disappears, a test for unsaturation.
3. Hydrogen halides (reactivity HI > HBr > HCl): with symmetrical alkenes there is one product. With unsymmetrical alkenes, Markovnikov's rule applies: the negative part of the reagent adds to the carbon carrying fewer hydrogens. CH₃CH=CH₂ + HBr → CH₃CHBrCH₃ (2-bromopropane, major).
Mechanism: H⁺ (electrophile) adds first to give the more stable carbocation (2° CH₃C⁺HCH₃ rather than 1° CH₃CH₂CH₂⁺), then Br⁻ attacks it.
4. Peroxide (Kharasch) effect: in the presence of an organic peroxide, HBr only adds against Markovnikov's rule: CH₃CH=CH₂ + HBr → CH₃CH₂CH₂Br (1-bromopropane). This goes by a free-radical mechanism, in which Br• adds to the terminal carbon to give the more stable 2° radical. HCl and HI do not show it (the H–Cl bond is too strong to break homolytically; the I• radical combines to I₂ instead of adding).
5. Sulphuric acid: cold concentrated H₂SO₄ adds by Markovnikov's rule to give alkyl hydrogen sulphates.
6. Water: in the presence of a few drops of concentrated H₂SO₄, alkenes form alcohols (Markovnikov).
7. Oxidation: cold, dilute aqueous KMnO₄ (Baeyer's reagent) adds two –OH groups to give a diol (vicinal glycol); its purple colour disappears, another test for unsaturation. Acidic KMnO₄ or K₂Cr₂O₇ breaks the double bond to give ketones and/or acids.
8. Ozonolysis: O₃ adds across C=C to give an ozonide, which with Zn and water splits into carbonyl compounds. Each C=C becomes two C=O; joining the carbonyl carbons back together with a double bond reveals the original alkene. Propene gives ethanal and methanal.
9. Polymerisation: many ethene molecules join at high temperature and pressure with a catalyst to give polythene.

Worked example

Predict the products of adding HBr to 2-methylpropene (a) without peroxide and (b) in the presence of benzoyl peroxide. Explain the first with the mechanism in words. (3 marks)

Try it first, then show the solutionSolution

1. (a) Markovnikov: H⁺ adds to the CH₂ end, giving the tertiary carbocation (CH₃)₃C⁺; Br⁻ attacks it: (CH₃)₃CBr, 2-bromo-2-methylpropane. (1½)
2. The tertiary carbocation forms because it is much more stable (+I effect and hyperconjugation of three methyl groups) than the primary alternative. (½)
3. (b) Peroxide effect: Br adds to the CH₂ end via the more stable tertiary radical, giving (CH₃)₂CHCH₂Br, 1-bromo-2-methylpropane. (1)

Answer: (a) 2-bromo-2-methylpropane (b) 1-bromo-2-methylpropane

Alkynes: structure, preparation and properties

NCERT section 9.4 Alkynes (9.4.1 to 9.4.4)

Alkynes (CₙH₂ₙ₋₂) have a C≡C made of one σ and two π bonds; the carbons are sp, the molecule is linear around the triple bond (H–C≡C–H 180°), C≡C is 120 pm and its bond enthalpy is 823 kJ mol⁻¹. The two π clouds merge into a cylinder of electron density around the axis.

Isomerism: chain and position (C₅H₈: pent-1-yne, pent-2-yne, 3-methylbut-1-yne).

Preparation:
1. From calcium carbide: $Ca{C}_2+2H_{2}O\to Ca(OH{)}_2+C_2{H}_2$. Calcium carbide is made by heating quicklime with coke.
2. From vicinal dihalides: two successive dehydrohalogenations, first with alcoholic KOH and then with sodamide (NaNH₂).

Properties:

• Acidic character: H attached to a triple-bonded carbon (sp, 50% s character, more electronegative) is weakly acidic. Ethyne reacts with sodium or sodamide to give sodium acetylide and H₂; but-2-yne, with no terminal H, does not. Acidity order: HC≡CH > CH₂=CH₂ > CH₃–CH₃.
• Addition reactions (they add two molecules of reagent, as there are two π bonds): hydrogen (to alkene, then alkane); halogens (Br₂ colour disappears; tetrabromide forms); hydrogen halides by Markovnikov's rule (gem-dihalides, both X on the same carbon); and water in the presence of HgSO₄ and dilute H₂SO₄ at 333 K: ethyne gives ethanal ($C_2{H}_2+H_{2}O\to C{H}_{3}CHO$); propyne gives propanone.
• Polymerisation: linear polymers such as polyacetylene (conducts electricity); cyclic: ethyne passed through a red-hot iron tube at 873 K trimerises to benzene: $3C_2{H}_2\to C_6{H}_6$.

Worked example

6.4 g of commercial calcium carbide gives 2.0 L of ethyne at STP (22.7 L mol⁻¹) with excess water. (a) Write the equation. (b) Calculate the percentage purity of the carbide. (c) Why is the hydrogen of ethyne acidic? (Ca = 40, C = 12) (3 marks)

Try it first, then show the solutionSolution

1. (a) $Ca{C}_2+2H_{2}O\to Ca(OH{)}_2+C_2{H}_2$. (½)
2. (b) Ethyne = 2.0 ÷ 22.7 = 0.0881 mol, so pure CaC₂ = 0.0881 × 64 = 5.64 g; purity = 5.64 ÷ 6.4 × 100 = 88%. (1½)
3. (c) The carbon is sp hybridised (50% s character), so it holds the C–H electrons closely and the H leaves as H⁺ more easily than from sp² or sp³ carbons. (1)

Answer: (b) about 88% pure (c) high s character of the sp carbon

Benzene: structure, aromaticity and preparation

NCERT section 9.5 Aromatic Hydrocarbon (9.5.1 to 9.5.4)

Benzene, C₆H₆, was isolated by Faraday (1825). Kekulé proposed a six-membered ring with alternate single and double bonds. But benzene gives only one ortho-disubstituted product, and it does not undergo addition readily, so Kekulé suggested the double bonds keep shifting.

The modern picture:

• Resonance: benzene is a hybrid of the two Kekulé structures. All six C–C bonds are equal (139 pm), between C–C (154 pm) and C=C (133 pm).
• Orbital picture: each carbon is sp², bonded to two carbons and one hydrogen in a flat hexagon; the six unhybridised p orbitals overlap sideways all round the ring, forming two doughnut-shaped π clouds above and below the plane with six delocalised electrons. This delocalisation makes benzene very stable (it is reluctant to add).

Aromaticity (Hückel's rule): a compound is aromatic if it is (i) planar, (ii) cyclic with complete delocalisation of π electrons, and (iii) has (4n + 2) π electrons (n = 0, 1, 2 …): 2, 6, 10 …. Benzene (6), naphthalene (10) and the cyclopentadienyl anion (6) are aromatic; cyclooctatetraene (8 π electrons, non-planar) is not.

Preparation:
1. Cyclic polymerisation of ethyne in a red-hot iron tube at 873 K.
2. Decarboxylation of sodium benzoate with soda lime: $C_6{H}_{5}COONa+NaOH\to C_6{H}_6+N{a}_{2}C{O}_3$.
3. Reduction of phenol by heating with zinc dust: C₆H₅OH + Zn → C₆H₆ + ZnO.

Worked example

Apply Hückel's rule to decide whether these are aromatic: (a) benzene (b) cyclopentadienyl anion, C₅H₅⁻ (c) cyclobutadiene (C₄H₄, planar). (3 marks)

Try it first, then show the solutionSolution

1. (a) Planar, cyclic, fully conjugated, 6 π electrons = 4(1) + 2: aromatic. (1)
2. (b) Planar ring; four π electrons from two C=C plus the lone pair on the CH⁻ carbon = 6 π electrons, delocalised: aromatic. (1)
3. (c) 4 π electrons; 4n + 2 = 4 has no whole-number solution: not aromatic. (1)

Answer: (a) aromatic (b) aromatic (c) not aromatic

Properties of benzene: electrophilic substitution, directive influence and carcinogenicity

NCERT section 9.5.5 Properties; 9.5.6 Directive Influence of a Functional Group in Monosubstituted Benzene; 9.6 Carcinogenicity and Toxicity

Benzene's delocalised π electrons attract electrophiles, but addition would destroy the aromatic sextet, so benzene prefers electrophilic substitution:
1. Nitration: with a mixture of concentrated HNO₃ and H₂SO₄ (nitrating mixture) at about 323-333 K: $C_6{H}_6+HN{O}_3\to C_6{H}_{5}N{O}_2+H_{2}O$.
2. Halogenation: with Cl₂ and anhydrous FeCl₃ (Lewis acid) in the dark: chlorobenzene.
3. Sulphonation: with fuming H₂SO₄ (oleum): benzenesulphonic acid.
4. Friedel-Crafts alkylation: with an alkyl halide and anhydrous AlCl₃: $C_6{H}_6+C{H}_{3}Cl\to C_6{H}_{5}C{H}_3+HCl$ (toluene).
5. Friedel-Crafts acylation: with an acyl chloride or acid anhydride and anhydrous AlCl₃: acetophenone, C₆H₅COCH₃.

Mechanism (three steps):

• (a) Generation of the electrophile: e.g. HNO₃ + 2H₂SO₄ → NO₂⁺ (nitronium ion) + H₃O⁺ + 2HSO₄⁻; Cl₂ + FeCl₃ → Cl⁺ + FeCl₄⁻.
• (b) Formation of the carbocation (arenium ion, σ-complex): the electrophile attacks the ring; one carbon becomes sp³ and aromaticity is lost; the positive charge is spread by resonance.
• (c) Removal of a proton: a base (HSO₄⁻ or AlCl₄⁻) removes H⁺ from the sp³ carbon, restoring aromaticity.

Addition reactions happen only under drastic conditions: with H₂ and Ni at high temperature and pressure, benzene gives cyclohexane; with Cl₂ in UV light, benzene hexachloride (C₆H₆Cl₆, gammaxane). Combustion gives a sooty flame (high C : H ratio).

Directive influence of substituents:

• Ortho/para directing (and activating): –OH, –NH₂, –NHR, –OCH₃, –CH₃: they push electron density into the ring (+R or +I), especially at ortho and para positions.
• Meta directing (and deactivating): –NO₂, –CN, –CHO, –COOH, –SO₃H: they withdraw electron density (−R, −I) mainly from ortho and para, so meta is the least deactivated.
• Halogens are an exception: deactivating (strong −I) but ortho/para directing (+R).

Carcinogenicity and toxicity: benzene and many polynuclear hydrocarbons (more than two fused benzene rings), formed by incomplete combustion of tobacco, coal and petroleum, are toxic and carcinogenic (cancer-causing); for example 1,2-benzanthracene and 3,4-benzopyrene damage DNA after metabolic activation.

Worked example

Write the steps in the mechanism of nitration of benzene, naming the electrophile and the intermediate. (3 marks)

Try it first, then show the solutionSolution

1. Generation of the electrophile: HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻; the electrophile is the nitronium ion, NO₂⁺. (1)
2. NO₂⁺ attacks the π cloud to form the arenium ion (σ-complex), a resonance-stabilised carbocation in which one ring carbon is sp³ and aromaticity is lost. (1)
3. HSO₄⁻ removes the H⁺ from that sp³ carbon, restoring the aromatic ring and giving nitrobenzene. (1)

Answer: Electrophile NO₂⁺; intermediate arenium ion (σ-complex); loss of H⁺ restores aromaticity.

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.

Topics in this chapter

• Alkenes and Alkynes: Addition Reactions, Markovnikov Rule and Ozonolysis
• Benzene: Aromaticity, Electrophilic Substitution and Directing Effects

Hydrocarbons formula sheet

General formulas

$\text{alkane }{C}_n{H}_{2n+2},\text{alkene }{C}_n{H}_{2n},\text{alkyne }{C}_n{H}_{2n-2}$

Cycloalkanes also have CₙH₂ₙ.

Wurtz reaction

$2RX+2Na\stackrel{\text{dry ether}}{\to }R-R+2NaX$

Doubles the number of carbons.

Decarboxylation

$RCOONa+NaOH\stackrel{CaO,\Delta }{\to }R-H+N{a}_{2}C{O}_3$

One carbon fewer.

Combustion of a hydrocarbon

$C_x{H}_y+\left(x+\frac{y}{4}\right)O_2\to xC{O}_2+\frac{y}{2}{H}_{2}O$

Use for volume and mass questions on burning.

Hückel's rule

$\pi \text{electrons}=4n+2,n=0,1,2,\dots$

Also planar, cyclic and fully conjugated.

Hydration of ethyne

$C_2{H}_2+H_{2}O\stackrel{H{g}^{2+},H_{2}S{O}_4}{\to }C{H}_{3}CHO$

Propyne gives propanone (Markovnikov).

Nitronium ion formation

$HN{O}_3+2H_{2}S{O}_4\to N{O}_2^{+}+H_3{O}^{+}+2HS{O}_4^{-}$

Electrophile in nitration.

Hydrocarbons: diagrams to practise

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

How the school exam asks Hydrocarbons

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

Tricks to remember Hydrocarbons

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

• MaRkovnikov: the Rich get Richer

  H goes to the carbon already rich in hydrogen; the halide goes to the carbon with fewer H.

  Use it when: Predicting HX addition products.

• Peroxide only Bothers Bromine

  The anti-Markovnikov peroxide effect works only with HBr.

  Use it when: MCQs and reasoning.

• Wurtz doubles, decarboxylation drops, Kolbe evens

  Wurtz gives 2n carbons; soda lime gives n − 1; Kolbe electrolysis gives an even number.

  Use it when: Choosing a preparation method.

• OH, NH₂, alkyl, halogen: Ortho-Para

  Groups with lone pairs or alkyls direct o/p; groups with C=O, N=O, C≡N direct meta.

  Use it when: Directive influence questions.

• 4n + 2: 2, 6, 10, 14

  Aromatic π counts are 2, 6, 10, 14; 4 and 8 are anti-aromatic or non-aromatic.

  Use it when: Hückel's rule.

Hydrocarbons: exam-level solved questions

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

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 2 marks.

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.

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.

  ✓Do this: Only HBr adds anti-Markovnikov in presence of peroxide.

  Usually costs: 1 mark.

• ✗Mistake: Wrong product in ozonolysis

  Students break single bonds too.

  ✓Do this: Only C=C bonds break; each carbon of the double bond becomes a C=O.

  Usually costs: 1-2 marks.

• ✗Mistake: Drawing trans-but-2-ene as polar

  Each C–CH₃ bond is slightly polar.

  ✓Do this: In trans isomers the two dipoles point opposite and cancel.

  Usually costs: 1 mark.

• ✗Mistake: Using aqueous KOH for dehydrohalogenation

  Aqueous and alcoholic KOH are confused.

  Wrong working

  CH₃CH₂Br + KOH(aq) → CH₂=CH₂

  Right working

  Alcoholic KOH gives the alkene; aqueous KOH gives the alcohol (substitution).

  ✓Do this: Write 'alc.' above the arrow for elimination.

  Usually costs: 1 mark.

• ✗Mistake: Introducing groups in the wrong order in conversions

  Directing effects are ignored.

  Wrong working

  Benzene → chlorobenzene → nitration to get m-chloronitrobenzene

  Right working

  Nitrate first (NO₂ is meta directing), then chlorinate to get m-chloronitrobenzene.

  ✓Do this: Ask which group must be present first to steer the second.

  Usually costs: 2 marks.

• ✗Mistake: Saying benzene readily decolourises bromine water

  Benzene is drawn with double bonds.

  ✓Do this: Benzene does not add Br₂ under ordinary conditions; it undergoes substitution with Br₂/FeBr₃.

  Usually costs: 1 mark.

NCERT exercise guide: Hydrocarbons

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

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.

1. Q1 · 1 mark · MCQExam level

   Which of the following does not show geometrical isomerism?
   (a) but-2-ene (b) pent-2-ene (c) but-1-ene (d) 1,2-dichloroethene

   Hint

   Check the terminal carbon.

   Answer

   (c) but-1-ene: C1 carries two H atoms.

2. Q2 · 1 mark · MCQExam level

   The hybridisation of carbon atoms in benzene is:
   (a) sp (b) sp² (c) sp³ (d) sp³d

   Hint

   Flat hexagon, 120°.

   Answer

   (b) sp²

3. Q3 · 2 marks · Very short answerExam level

   Write two tests to distinguish ethene from ethane. (2 marks)

   Hint

   Colour changes.

   Answer

   Ethene decolourises bromine water (1) and cold dilute alkaline KMnO₄ (Baeyer's reagent) (1); ethane does neither.

4. Q4 · 2 marks · Very short answerHarder exam level

   Write the products of ozonolysis of hex-2-ene and of 2-methylpent-2-ene. (2 marks)

   Hint

   Cut the C=C.

   Answer

   Hex-2-ene (CH₃CH=CHCH₂CH₂CH₃): ethanal + butanal (1). 2-Methylpent-2-ene ((CH₃)₂C=CHCH₂CH₃): propanone + propanal (1).

5. Q5 · 2 marks · Very short answerHarder exam level

   Why is the C–H bond of ethyne more acidic than that of ethene? Give a reaction to show the acidity. (2 marks)

   Hint

   s character.

   Answer

   The sp carbon (50% s) is more electronegative than sp² (33%), so the C–H electrons are held closer and H⁺ is released more easily (1). HC≡CH + NaNH₂ → HC≡CNa + NH₃ (or with Na giving H₂) (1).

6. Q6 · 3 marks · Short answerHarder exam level

   Suppose 0.10 mol of a hydrocarbon C₅H₈ adds 32 g of bromine, and on ozonolysis gives two moles of methanal and one mole of propanedial, OHC–CH₂–CHO, per mole. (a) How many C=C bonds does it have? (b) Identify it. (Br = 80) (3 marks)

   Hint

   Br₂ = 160 g mol⁻¹; each C=C takes one Br₂.

   Answer

   (a) Br₂ = 32 ÷ 160 = 0.20 mol, two Br₂ per molecule, so two C=C bonds (1). (b) Joining each end carbon of OHC–CH₂–CHO to a CH₂ from methanal gives CH₂=CH–CH₂–CH=CH₂ (1), penta-1,4-diene (1).

7. Q7 · 2 marks · Very short answerHarder exam level

   Explain why benzene undergoes substitution rather than addition reactions, although it is unsaturated. (2 marks)

   Hint

   Aromatic stability.

   Answer

   Benzene's six π electrons are delocalised, giving extra resonance stability (1). Addition would destroy this aromatic sextet, while substitution keeps it, so substitution is favoured (1).

8. Q8 · 3 marks · Short answerHarder exam level

   Write the mechanism of addition of HBr to propene (no peroxide), and name the more stable intermediate. (3 marks)

   Hint

   Electrophile first.

   Answer

   H⁺ from HBr attacks the π bond (1). It can give a 2° carbocation CH₃C⁺HCH₃ or a 1° carbocation CH₃CH₂CH₂⁺; the 2° is more stable (1). Br⁻ attacks it to give 2-bromopropane (1).

9. Q9 · 2 marks · Very short answerHarder exam level

   Arrange in increasing order of reactivity towards electrophilic substitution and justify: benzene, toluene, nitrobenzene, phenol. (2 marks)

   Hint

   Activating vs deactivating.

   Answer

   Nitrobenzene < benzene < toluene < phenol (1). –NO₂ withdraws electrons; –CH₃ donates weakly (+I, hyperconjugation); –OH donates strongly (+R) (1).

10. Q10 · 3 marks · Short answerHOTS / competency

    Suppose a student mixes sodium ethanoate and sodium propanoate and electrolyses the aqueous solution. Which alkanes form at the anode? Why is this a poor way to make butane? (3 marks)

    Hint

    Radicals pair in all combinations.

    Answer

    CH₃• and C₂H₅• radicals form at the anode (1). They combine in three ways: ethane (CH₃–CH₃), propane (CH₃–C₂H₅) and butane (C₂H₅–C₂H₅) (1). The mixture must be separated, so butane is better made from sodium propanoate alone (1).

11. Q11 · 5 marks · Long answerHOTS / competency

    (a) Draw the Newman projections of the staggered and eclipsed conformations of ethane and mark the dihedral angles. (b) Why does the eclipsed form have higher energy? (c) Is it correct to call them isomers? (d) How does the number of conformations of ethane differ from the number of geometrical isomers of but-2-ene, and why? (5 marks)

    Hint

    Free rotation vs restricted rotation.

    Answer

    (a) Staggered 60°, eclipsed 0° as described (2). (b) In eclipsed ethane, the C–H bond electron clouds of the two carbons are closest, giving maximum repulsion (torsional strain) (1). (c) They are conformational isomers (conformers) that interconvert rapidly; they cannot be isolated (1). (d) Ethane has infinitely many conformations because rotation about C–C is free; but-2-ene has only two geometrical isomers because the π bond prevents rotation about C=C (1).

12. Q12 · 1 mark · Assertion-reasonHOTS / competency

    Assertion (A): Cyclooctatetraene (C₈H₈) is not aromatic.
    Reason (R): It has 8 π electrons, which do not satisfy (4n + 2), and the molecule is non-planar.
    (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

    Hückel's rule.

    Answer

    (a)

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. 1.Name the product of Kolbe's electrolysis of sodium propanoate.Show answer

   Butane.

2. 2.Why does 2,2-dimethylpropane boil lower than pentane?Show answer

   Branching makes it more spherical with less surface area, so van der Waals forces are weaker.

3. 3.What is the energy difference between staggered and eclipsed ethane?Show answer

   About 12.5 kJ mol⁻¹.

4. 4.Name the catalyst that reduces an alkyne to a cis alkene.Show answer

   Lindlar's catalyst.

5. 5.State Markovnikov's rule.Show answer

   The negative part of the adding reagent goes to the double-bonded carbon with fewer hydrogen atoms.

6. 6.What does propene give on ozonolysis?Show answer

   Ethanal and methanal.

7. 7.Name the reagent that converts ethyne to ethanal.Show answer

   Water with HgSO₄ and dilute H₂SO₄ (at 333 K).

8. 8.What is the electrophile in Friedel-Crafts alkylation with CH₃Cl?Show answer

   CH₃⁺ (from CH₃Cl and AlCl₃).

9. 9.Give the three conditions for aromaticity.Show answer

   Planar, cyclic with complete π delocalisation, and (4n + 2) π electrons.

10. 10.Name one carcinogenic polynuclear hydrocarbon.Show answer

    3,4-Benzopyrene (or 1,2-benzanthracene).

Quick revision checklist

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

• ☐Can you write the products of Wurtz, decarboxylation and Kolbe reactions for given starting materials?
• ☐Can you explain the boiling points of pentane isomers?
• ☐Can you draw staggered and eclipsed ethane in Newman and sawhorse forms?
• ☐Can you predict HBr addition products with and without peroxide and explain both?
• ☐Can you identify an alkene from its ozonolysis products?
• ☐Can you explain why ethyne is acidic but ethane is not?
• ☐Can you test any ring for aromaticity using Hückel's rule?
• ☐Can you write the mechanism of nitration and classify common groups as o/p or m directors?

Hydrocarbons: frequently asked questions