By the end of this chapter you'll be able to…

  • 1Prepare alkanes, alkenes, and alkynes by standard routes (Wurtz, dehydrohalogenation, etc.)
  • 2Apply Markovnikov's and Saytzeff's rules and the peroxide effect
  • 3Explain the free-radical halogenation mechanism of alkanes
  • 4Describe electrophilic substitution of benzene and directive effects
  • 5Distinguish alkanes, alkenes, alkynes, and aromatics by chemical tests
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Why this chapter matters
Hydrocarbons are the simplest organic compounds and the basis of fuels, polymers, and synthetic chemistry. Mastering the preparation, properties, and reaction mechanisms of alkanes, alkenes, alkynes, and aromatics -- including Markovnikov's rule and electrophilic substitution -- is essential for all of organic chemistry.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

Hydrocarbons

'Hydrocarbons are the simplest organic compounds — composed only of carbon and hydrogen — yet they are the foundation of fuels and organic chemistry.' — Organic Chemistry

1. Chapter Overview

HYDROCARBONS are compounds containing ONLY CARBON and HYDROGEN. They are classified as ALKANES (single bonds), ALKENES (double bonds), ALKYNES (triple bonds), and AROMATIC hydrocarbons (benzene and derivatives). This chapter covers the PREPARATION, PHYSICAL PROPERTIES, CHEMICAL REACTIONS, and MECHANISMS for each class, with special emphasis on the UNIQUE stability of aromatic compounds.


2. Alkanes (CₙH₂ₙ₊₂)

General

  • sp³ hybridised carbon, tetrahedral geometry
  • Saturated hydrocarbons (all single bonds)
  • General formula: CₙH₂ₙ₊₂

Preparation

  1. From unsaturated hydrocarbons: Catalytic hydrogenation (Ni, Pt, or Pd catalyst)
    • C₂H₄ + H₂ → C₂H₆ (nickel catalyst)
  2. From alkyl halides: Wurtz reaction (2RX + 2Na → R—R + 2NaX)
    • 2CH₃Br + 2Na → CH₃—CH₃ + 2NaBr
    • Limitation: For different alkyl halides, MIXTURE of products
  3. From carboxylic acids: Decarboxylation (soda lime, heat)
    • CH₃COONa + NaOH → CH₄ + Na₂CO₃
  4. From Grignard reagents: R—Mg—X + H₂O → R—H + Mg(OH)X

Physical Properties

  • Non-polar, INSOLUBLE in water, soluble in organic solvents
  • Boiling point INCREASES with chain length
  • Branched alkanes have LOWER boiling points than straight chain
  • C₁-C₄: Gases; C₅-C₁₇: Liquids; C₁₈+: Solids

Chemical Reactions

  1. Substitution (Free Radical Halogenation):

    • CH₄ + Cl₂ → CH₃Cl + HCl (UV light or heat, 300°C)
    • Mechanism: Chain reaction — Initiation (Cl₂ → 2Cl•), Propagation, Termination
    • Reactivity: F₂ > Cl₂ > Br₂ > I₂
    • Selectivity: Tertiary H > Secondary H > Primary H (for Br₂)
  2. Combustion:

    • CₙH₂ₙ₊₂ + (3n+1)/2 O₂ → nCO₂ + (n+1)H₂O + HEAT
    • Incomplete combustion: gives CO + C (soot)
  3. Pyrolysis (Cracking):

    • Higher alkanes heated → mixture of smaller hydrocarbons

3. Alkenes (CₙH₂ₙ)

General

  • One or more C=C DOUBLE bonds (sp² hybridised)
  • Unsaturated hydrocarbons
  • General formula: CₙH₂ₙ

Preparation

  1. Dehydrohalogenation of alkyl halides:

    • R—CH₂—CH₂—X + alc. KOH → R—CH=CH₂ + KX + H₂O
    • Follows SAYTZEFF rule: MORE substituted alkene is the MAJOR product
  2. Dehydration of alcohols:

    • R—CH₂—CH₂—OH + conc. H₂SO₄ → R—CH=CH₂ + H₂O (170°C)
  3. From alkynes: Partial hydrogenation (Lindlar's catalyst → cis, Na/liq NH₃ → trans)

Chemical Reactions (Electrophilic Addition)

  1. Addition of H₂ (Hydrogenation): C₂H₄ + H₂ → C₂H₆ (Ni catalyst)

  2. Addition of Halogens (X₂):

    • C₂H₄ + Br₂ → CH₂Br—CH₂Br (decolourises bromine water — TEST for unsaturation)
  3. Addition of Hydrogen Halides (HX):

    • Markovnikov's Rule: H adds to the C with MORE H atoms already; X adds to the other
    • CH₃—CH=CH₂ + HBr → CH₃—CHBr—CH₃ (major, 2-bromopropane)
    • Peroxide effect (Kharasch effect): With HBr + peroxide → ANTI-Markovnikov product
  4. Addition of H₂O (Acid-catalysed):

    • Follows Markovnikov's rule
    • CH₃—CH=CH₂ + H₂O → CH₃—CHOH—CH₃
  5. Oxidation:

    • Cold KMnO₄ (Baeyer's reagent): Diol formation (cis-dihydroxylation)
    • Hot KMnO₄/acid: Cleavage at C=C → ketones/carboxylic acids

Polymerisation

  • nCH₂=CH₂ → (—CH₂—CH₂—)ₙ (Polythene, under high pressure/temperature)

4. Alkynes (CₙH₂ₙ₋₂)

General

  • One or more C≡C TRIPLE bonds (sp hybridised)
  • General formula: CₙH₂ₙ₋₂

Preparation

  1. Dehydrohalogenation of vicinal dihalides: alc. KOH, then NaNH₂
  2. From calcium carbide: CaC₂ + 2H₂O → C₂H₂ + Ca(OH)₂

Chemical Reactions (Electrophilic Addition)

  1. Addition of H₂: Stepwise → alkene then alkane
  2. Addition of X₂: Decolourises Br₂ water faster than alkenes
  3. Addition of HX: Follows Markovnikov's rule
    • HC≡CH + HCl → CH₂=CHCl (vinyl chloride, used for PVC)
  4. Addition of H₂O (Hydration): HgSO₄/H₂SO₄ catalyst → enol → KETONE (Keto-enol tautomerism)
    • HC≡CH + H₂O → CH₃CHO (for ethyne: acetaldehyde, exception — terminal alkyne gives aldehyde)

Acidic Nature of Terminal Alkynes

  • HC≡CH + NaNH₂ → HC≡C⁻Na⁺ + NH₃ (terminal H is ACIDIC)
  • Reaction with AgNO₃/ammoniacal: White precipitate (identification test)

5. Aromatic Hydrocarbons

Benzene Structure

  • C₆H₆, planar hexagonal, sp² hybridised
  • 6 p-orbitals overlap to form a DELOCALISED π-system (aromatic sextet)
  • Huckel's Rule: Aromatic if planar, cyclic, conjugated, with (4n+2)π electrons

Preparation

  1. From acetylene: 3C₂H₂ → C₆H₆ (cyclisation, red hot tube)
  2. From phenol: Reduction (Zn dust, heat)

Chemical Reactions (Electrophilic Substitution — SEAr)

  1. Nitration: C₆H₆ + HNO₃/H₂SO₄ → C₆H₅NO₂ + H₂O
  2. Halogenation: C₆H₆ + Br₂/FeBr₃ → C₆H₅Br + HBr
  3. Sulphonation: C₆H₆ + H₂SO₄ (fuming) → C₆H₅SO₃H + H₂O
  4. Friedel-Crafts Alkylation: C₆H₆ + RCl/AlCl₃ → C₆H₅R + HCl
  5. Friedel-Crafts Acylation: C₆H₆ + RCOCl/AlCl₃ → C₆H₅COR + HCl

Mechanism of SEAr

  • Step 1: Formation of ELECTROPHILE (e.g., NO₂⁺ from HNO₃ + H₂SO₄)
  • Step 2: Electrophile attacks π-system → σ-complex (arenium ion)
  • Step 3: Loss of H⁺ → regeneration of aromaticity

Directive Effects

SubstituentOrtho/Para directing (Activating)Meta directing (Deactivating)
Strong—OH, —NH₂, —OCH₃—NO₂, —CN
Moderate—CH₃, —C₂H₅—CHO, —COOH
Weak—F, —Cl, —Br, —I—SO₃H, —COR

6. Common Mistakes

  1. Wurtz reaction with mixed alkyl halides gives ALL three possible products: Not useful for preparing specific alkanes
  2. Markovnikov's rule is about H+ addition: The proton goes to the more substituted carbon (more H already)
  3. Benzene does NOT undergo addition reactions easily: It prefers substitution to preserve aromaticity
  4. Alkynes are more ACIDIC than alkenes and alkanes: Due to greater s-character in sp hybridisation
  5. Keto-enol tautomerism is EQUILIBRIUM, not resonance: It involves bond rearrangement, not electron delocalisation

7. CBSE Exam Focus

  1. Wurtz reaction, decarboxylation (3-mark)
  2. Markovnikov's and Saytzeff's rules — numericals/products (3-mark)
  3. Electrophilic substitution of benzene — mechanism (5-mark)
  4. Directive effects of substituents on benzene (3/5-mark)
  5. Distinction tests (alkane/alkene/alkyne/benzene)
  6. Free radical halogenation mechanism (3-mark)

8. Key Concepts

  • Alkanes: SP³, sigma bonds, substitution reactions
  • Alkenes: SP², pi bonds, addition reactions
  • Alkynes: SP, pi bonds, acidic terminal H
  • Aromatic: SP², (4n+2)π, substitution reactions
  • Stability: Benzene > Alkane > Alkene > Alkyne (in terms of heat of hydrogenation per π-bond)

9. Self-Test (5+ Q&A)

Q1: Predict product of CH₃—CH=CH₂ + H₂O (H⁺ catalyst). A: Markovnikov addition. H⁺ adds to CH₂ (more H) → carbocation CH₃—C⁺H—CH₃. OH⁻ adds to carbocation. Product: PROPAN-2-OL.

Q2: Why does benzene undergo electrophilic substitution rather than addition? A: Addition would BREAK the aromatic π-system (lose 152 kJ/mol stabilisation energy). Substitution PRESERVES aromaticity.

Q3: Identify A and B: CH₃COONa + NaOH → A. A + Cl₂/UV → B. A: A = CH₄ (decarboxylation). B = CH₃Cl (free radical chlorination).

Q4: What is the peroxide effect? Give an example. A: In presence of organic peroxides, HBr adds to alkene AGAINST Markovnikov's rule (ANTI-Markovnikov). CH₃—CH=CH₂ + HBr (peroxide) → CH₃—CH₂—CH₂Br.

Q5: How would you distinguish between ethane, ethene, and ethyne? A: Baeyer's test (cold KMnO₄): Ethane → no reaction (purple persists). Ethene → KMnO₄ decolourised (diol). Ethyne → KMnO₄ decolourised faster. Further: Ethyne gives white ppt with ammoniacal AgNO₃ (terminal alkyne test).


10. Conclusion

Hydrocarbons are the SIMPLEST organic compounds but form the BASIS of organic chemistry. Alkanes are relatively INERT (paraffins = little affinity). Alkenes are REACTIVE due to the π-bond — they undergo addition reactions. Alkynes have TWO π-bonds and show ACIDIC character at terminal positions. Aromatic compounds have UNIQUE stability (aromaticity) and undergo substitution, not addition. Understanding hydrocarbons is ESSENTIAL for fuels (petroleum), polymers, and synthetic organic chemistry — the foundation for Class 12 organic chemistry.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

General formulae
Alkanes CnH(2n+2); alkenes CnH(2n); alkynes CnH(2n-2)
Degree of unsaturation increases from alkane to alkyne.
Markovnikov's rule
In HX addition, H goes to the carbon with more H atoms; X to the other
Peroxide (Kharasch) effect reverses this for HBr only -- anti-Markovnikov.
Huckel's rule (aromaticity)
Aromatic if planar, cyclic, conjugated with (4n+2) pi-electrons
Benzene has 6 pi-electrons (n=1).
⚠️

Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT
Using the Wurtz reaction for a specific unsymmetrical alkane
Wurtz with two different alkyl halides gives a mixture of all three coupling products, so it is unsuitable for a single specific alkane.
WATCH OUT
Misapplying Markovnikov's rule
The hydrogen of HX adds to the double-bond carbon that already has more hydrogens, giving the more stable carbocation.
WATCH OUT
Expecting benzene to undergo addition like alkenes
Benzene prefers electrophilic substitution because addition would destroy its aromatic stabilisation.
WATCH OUT
Calling keto-enol tautomerism resonance
Tautomerism is a real equilibrium with bond and atom movement, not the electron delocalisation of resonance.

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1MEDIUM· Addition
Predict the product of CH3-CH=CH2 + H2O with an acid catalyst.
Show solution
By Markovnikov's rule, H+ adds to the terminal CH2 (more hydrogens), forming the more stable secondary carbocation CH3-C+H-CH3. Water then adds to give propan-2-ol (CH3-CHOH-CH3).
Q2HARD· Aromaticity
Why does benzene undergo electrophilic substitution rather than addition?
Show solution
Addition would break the delocalised aromatic pi-system and lose about 152 kJ/mol of resonance (aromatic) stabilisation energy. Substitution replaces a hydrogen while preserving the aromatic sextet, so it is strongly favoured.
Q3MEDIUM· Reactions
Identify A and B: CH3COONa + NaOH(soda lime) -> A; A + Cl2/UV -> B.
Show solution
A = CH4 (decarboxylation of sodium acetate). B = CH3Cl (free-radical chlorination of methane).
Q4MEDIUM· Peroxide Effect
What is the peroxide effect? Give an example.
Show solution
In the presence of organic peroxides, HBr adds to an alkene against Markovnikov's rule (anti-Markovnikov), via a free-radical mechanism. Example: CH3-CH=CH2 + HBr (peroxide) -> CH3-CH2-CH2Br (1-bromopropane).
Q5HARD· Distinction
How would you distinguish ethane, ethene, and ethyne?
Show solution
Baeyer's test (cold dilute KMnO4): ethane gives no reaction (purple persists); ethene and ethyne both decolourise it. To separate the latter two, ethyne (a terminal alkyne) gives a white precipitate with ammoniacal AgNO3, while ethene does not.

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

  • Alkanes (sp3) are saturated, fairly inert, and undergo free-radical substitution and combustion.
  • Alkenes (sp2) undergo electrophilic addition; Markovnikov's rule governs HX addition.
  • Peroxide effect gives anti-Markovnikov addition with HBr only.
  • Alkynes (sp) add electrophiles and have acidic terminal hydrogens (react with NaNH2, ammoniacal AgNO3).
  • Benzene is aromatic (Huckel's 4n+2 rule) and undergoes electrophilic substitution: nitration, halogenation, sulphonation, Friedel-Crafts.
  • Directing groups: -OH, -NH2, -CH3 are ortho/para directors; -NO2, -COOH are meta directors.
  • Baeyer's test and ammoniacal AgNO3 distinguish alkenes/alkynes from alkanes.

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 7-9 marks across the chapter

Question typeMarks eachTypical countWhat it tests
Mechanisms (SEAr / free radical)3-51Electrophilic substitution and halogenation mechanisms
Addition reactions / rules31Markovnikov, Saytzeff, peroxide effect
Preparation / distinction tests2-31Wurtz, decarboxylation, identification tests
Prep strategy
  • Learn each preparation route with conditions
  • Master Markovnikov, anti-Markovnikov, and Saytzeff with examples
  • Practise the SEAr and free-radical mechanisms step by step
  • Memorise ortho/para vs meta directing groups

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Fuels and petroleum

Alkanes are the main components of petrol, diesel, LPG, and natural gas that power transport and homes.

Polymers

Alkenes polymerise into plastics like polythene and PVC (from vinyl chloride made from ethyne).

Aromatic chemicals

Benzene and its derivatives are feedstocks for dyes, drugs, detergents, and explosives.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

  1. Always state reaction conditions (catalyst, temperature, reagent)
  2. Show the carbocation to justify Markovnikov products
  3. Write SEAr mechanism in three labelled steps
  4. Use Baeyer's and silver tests for distinction questions

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

  • Predict major and minor products using carbocation rearrangements in addition reactions.
  • Rationalise directive effects quantitatively using sigma and resonance contributions.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 11 Chemistry examHigh
JEE Main and Advanced (Hydrocarbons)Very High
NEET ChemistryVery High

Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

The terminal C-H of an alkyne is attached to an sp-hybridised carbon, which has 50% s-character. The greater s-character holds the bonding electrons closer to the carbon nucleus, stabilising the negative charge of the resulting carbanion. So terminal alkynes can lose this proton to strong bases (giving acetylide ions), whereas the sp2 (alkene) and sp3 (alkane) carbons hold their hydrogens far less acidically.

Electron-donating groups (like -OH, -NH2, -CH3) increase electron density at the ortho and para positions through resonance/inductive effects, so the next electrophile attacks there (ortho/para directors, usually activating). Electron-withdrawing groups (like -NO2, -COOH) pull electron density away, leaving the meta position relatively richer, so they are meta directors and deactivating.
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