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

These hydrocarbons class 11 notes compress NCERT Chapter 9 into one revision-ready page: classification of hydrocarbons, then alkanes, alkenes, alkynes and benzene — with every preparation method, key reaction, mechanism and isomerism rule you need.

Read them top to bottom once to rebuild the chapter mental map; then use the table of contents to jump straight to any weak spot, from Markovnikov’s rule to benzene’s electrophilic substitution mechanism.

The page follows the NCERT Class 11 Chemistry Part II textbook (pages 296–327), so the theory matches your board syllabus exactly. You can verify each reaction against the official NCERT textbook PDF before you trust it in an answer.

How Hydrocarbons Are Classified

A hydrocarbon is a compound of carbon and hydrogen only (NCERT, p. 296). These compounds matter because the fuels you use daily — LPG, CNG, LNG, petrol, diesel and kerosene — are hydrocarbon mixtures, and because hydrocarbons supply the starting materials for polymers (polythene, polypropene, polystyrene), dyes and drugs.

• Saturated hydrocarbons — contain only C–C and C–H single bonds. Open-chain members are alkanes; closed-chain (ring) members are cycloalkanes.
• Unsaturated hydrocarbons — contain C=C double bonds, C≡C triple bonds, or both: alkenes and alkynes.
• Aromatic hydrocarbons — a special class of cyclic compounds (benzene-type) with delocalised electrons.

You can model open and closed chains with toothpicks for bonds and plasticine balls for atoms, keeping carbon tetravalent and hydrogen monovalent; spring models work for multiple bonds (NCERT, p. 296). For the hybridisation and VSEPR logic behind these shapes, revise the organic chemistry fundamentals notes or browse the full Class 11 chemistry hub.

Alkanes: Naming, Isomerism and Preparation

Alkanes are saturated open-chain hydrocarbons with general formula \( \mathrm{C}_n\mathrm{H}_{2n+2} \). They were once called paraffins — from Latin parum (little) and affinis (affinity) — because they show little affinity for acids, bases and other reagents (NCERT, p. 297).

• Geometry: methane is tetrahedral, H–C–H angle \( 109.5^\circ \); C–C bond length 154 pm, C–H 112 pm; all bonds are σ bonds formed by head-on overlap of sp³ orbitals of carbon with 1s orbitals of hydrogen.
• Inertness: alkanes resist acids, bases, oxidising and reducing agents under normal conditions — reactivity appears only with light, heat or a catalyst.

Chain isomerism

Alkanes with four or more carbons can join the same atoms in different chains. These are chain isomers — structural isomers that differ in the carbon skeleton (NCERT, p. 297).

Primary, secondary, tertiary and quaternary carbons

Removing one hydrogen from an alkane gives an alkyl group, general formula \( \mathrm{C}_n\mathrm{H}_{2n+1} \): –CH₃ methyl, –C₂H₅ ethyl, –C₃H₇ propyl.

IUPAC naming in four steps

1. Select the longest continuous chain containing the principal functional group or multiple bond.
2. Number it so that substituents get the lowest possible locant sum.
3. Write substituents in alphabetical order (di-, tri- prefixes are ignored alphabetically; complex groups like isopropyl are counted as one word).
4. To write a structure from a name: draw the parent chain, number it, attach substituents at the correct carbons, then satisfy every carbon valence with hydrogen.

A classic trap: “2-ethylpentane” is wrong because the longest chain is six carbons — the correct name is 3-methylhexane (NCERT, p. 300).

Preparation of alkanes — five routes

• Hydrogenation of alkenes and alkynes over Pt, Pd or Ni: \( \mathrm{CH_2{=}CH_2 + H_2 \xrightarrow{Pt/Pd/Ni} CH_3{-}CH_3} \) (NCERT, eq. 9.1–9.3).
• Reduction of alkyl halides with zinc and dilute HCl: \( \mathrm{CH_3Cl + H_2 \xrightarrow{Zn,\ H^+} CH_4 + HCl} \).
• Wurtz reaction — alkyl halide with sodium in dry ether gives a higher alkane: \( \mathrm{2CH_3Br + 2Na \xrightarrow{dry\ ether} CH_3{-}CH_3 + 2NaBr} \). Only alkanes with an even number of carbons can be made this way.
• Decarboxylation — sodium salt of a carboxylic acid heated with soda lime (NaOH + CaO) loses CO₂, giving an alkane with one carbon less: \( \mathrm{CH_3COONa + NaOH \xrightarrow[\Delta]{CaO} CH_4 + Na_2CO_3} \).
• Kolbe’s electrolysis — electrolysis of a sodium or potassium carboxylate gives an even-carbon alkane at the anode: \( \mathrm{2CH_3COONa + 2H_2O \rightarrow CH_3{-}CH_3 + 2CO_2 + H_2 + 2NaOH} \). Methane cannot be prepared by this method.

Alkane Physical Properties and Chemical Reactions

Alkanes are nearly non-polar (C and H have almost equal electronegativity), so only weak van der Waals forces act between molecules (NCERT, p. 302).

Chemical reactions of alkanes

• Halogenation (substitution): \( \mathrm{CH_4 + Cl_2 \xrightarrow{h\nu} CH_3Cl + HCl} \), then successive chlorination gives CH₂Cl₂, CHCl₃ and CCl₄. Conditions: 573–773 K or diffused sunlight/UV. Reactivity order \( \mathrm{F_2 \gt Cl_2 \gt Br_2 \gt I_2} \); hydrogen reactivity \( 3^\circ \gt 2^\circ \gt 1^\circ \). Iodination is reversible, so it needs an oxidising agent such as HIO₃ or HNO₃ to remove HI (NCERT, p. 303).
• Free-radical mechanism: initiation (Cl–Cl homolysis under light), propagation (Cl• pulls H from CH₄ forming •CH₃, then •CH₃ attacks Cl₂ giving CH₃Cl and a new Cl•), termination (Cl• + Cl•, •CH₃ + •CH₃, •CH₃ + Cl•). The •CH₃ + •CH₃ step explains why ethane appears as a by-product during chlorination of methane.
• Combustion: \( \mathrm{C}_n\mathrm{H}_{2n+2} + \left(\frac{3n+1}{2}\right)\mathrm{O}_2 \rightarrow n\mathrm{CO}_2 + (n+1)\mathrm{H}_2\mathrm{O} \). Enthalpy values: \( \Delta_cH^\circ \) for CH₄ is –890 kJ mol⁻¹ and for C₄H₁₀ is –2875.84 kJ mol⁻¹. Incomplete combustion gives carbon black (used in inks and black pigments).
• Controlled oxidation: with a regulated supply of O₂ and a catalyst, methane gives methanol or methanal, ethane gives ethanoic acid, and an alkane with a tertiary H gives a tertiary alcohol with KMnO₄.
• Isomerisation: n-hexane heated with anhydrous AlCl₃ and HCl gives 2-methylpentane and 3-methylpentane.
• Aromatisation (reforming): n-alkanes with six or more carbons, heated to 773 K at 10–20 atm over Cr₂O₃, V₂O₅ or Mo₂O₃, cyclise and dehydrogenate to benzene and toluene.
• Reaction with steam: \( \mathrm{CH_4 + H_2O \xrightarrow[\Delta]{Ni} CO + 3H_2} \) at 1273 K — the industrial route to dihydrogen.
• Pyrolysis (cracking): higher alkanes decompose on heating; dodecane at 973 K over Pt/Pd/Ni gives heptane and pentene (NCERT, p. 305).

Conformations of Ethane: Staggered vs Eclipsed

Rotation about a C–C σ bond changes the spatial arrangement of atoms; each arrangement is a conformation (also called conformer or rotamer). Rotation is not completely free — a small barrier of 1–20 kJ mol⁻¹ arises from torsional strain, the weak repulsion between adjacent bond electron clouds (NCERT, p. 306).

• Eclipsed: hydrogens on the two carbons are as close together as possible — maximum torsional strain.
• Staggered: hydrogens are as far apart as possible — minimum strain, so this is the preferred conformation.
• Skew: any intermediate arrangement.

Two projection styles are used. The Sawhorse projection draws the C–C bond as a tilted straight line with three bonds at 120° on each carbon. The Newman projection looks down the C–C axis: the front carbon is a point and the rear carbon is a circle, each carrying three 120° bonds (NCERT, p. 306, Figs 9.2 and 9.3).

Why staggered wins: in the staggered form the C–H electron clouds are farthest apart, so repulsion is least. The energy gap between the extreme forms is only 12.5 kJ mol⁻¹. At ordinary temperature, molecular collisions supply this energy, so rotation is effectively free and conformers of ethane cannot be isolated.

Alkenes: The Double Bond, Naming and Isomerism

Alkenes are unsaturated hydrocarbons with at least one C=C bond, general formula \( \mathrm{C}_n\mathrm{H}_{2n} \). They are called olefins (oil-forming) because the first member, ethene, reacted with chlorine to give an oily liquid (NCERT, p. 307).

• Double-bond structure: one σ bond (sp² head-on overlap, about 397 kJ mol⁻¹) plus one π bond (lateral 2p overlap, about 284 kJ mol⁻¹); total bond enthalpy 681 kJ mol⁻¹, bond length 134 pm. The loosely held π electrons make alkenes a target for electrophiles (electron-seeking reagents).
• Naming: choose the longest chain containing the double bond, number from the end nearer the double bond, and replace the suffix -ane with -ene. Examples: propene, but-1-ene, but-2-ene, buta-1,3-diene, 2-methylprop-1-ene, 3-methylbut-1-ene.

Isomerism in alkenes

For \( \mathrm{C_4H_8} \) there are three structural isomers: but-1-ene and but-2-ene are position isomers (the double bond moves), while 2-methylprop-1-ene is a chain isomer of but-1-ene (the skeleton changes).

Geometrical (cis-trans) isomerism arises because rotation about C=C is restricted. The condition: each doubly bonded carbon must carry two different atoms or groups. If identical groups lie on the same side of the double bond it is cis; on opposite sides it is trans (NCERT, p. 309).

In but-2-ene, cis is more polar (\( \mu = 0.33\) D) than trans (\( \mu = 0 \)) because the C–CH₃ bond dipoles cancel in the trans form; in solids, the trans isomer usually has the higher melting point.

Preparation of alkenes — four routes

• Partial hydrogenation of alkynes: Lindlar’s catalyst (Pd/C poisoned with sulphur compounds or quinoline) gives cis alkenes; sodium in liquid ammonia gives trans alkenes.
• Dehydrohalogenation: alkyl halide + alcoholic KOH eliminates HX. This is a β-elimination because the hydrogen comes from the β-carbon. Rate order: \( \mathrm{I \gt Br \gt Cl} \) and \( 3^\circ \gt 2^\circ \gt 1^\circ \).
• Dehalogenation: a vicinal dihalide (halogens on adjacent carbons) with zinc metal gives the alkene.
• Acidic dehydration: alcohol heated with concentrated H₂SO₄ loses water — also a β-elimination.

Alkene Addition Reactions: Markovnikov’s Rule and the Peroxide Effect

• Addition of H₂: alkene \( \xrightarrow{Pt/Pd/Ni} \) alkane.
• Addition of halogens: Br₂ or Cl₂ gives a vicinal dihalide. The reddish-orange bromine-in-CCl₄ solution is decolourised — this is the standard test for unsaturation.
• Addition of HX: reactivity \( \mathrm{HI \gt HBr \gt HCl} \). For an unsymmetrical alkene the product follows Markovnikov’s rule: the negative part of the addendum adds to the carbon with the fewer hydrogen atoms (NCERT, p. 312).

Memory device for Markovnikov’s rule: “The rich get richer” — the hydrogen atom of H–X joins the carbon that already has more hydrogens, and the halogen lands on the carbon with fewer.

Mechanism of HX addition: the π bond acts as a base and accepts H⁺, forming a carbocation. From propene, two carbocations are possible — the more stable secondary carbocation forms faster than the primary one. The Br⁻ then attacks this carbocation, giving 2-bromopropane as the major product.

• Peroxide (Kharash) effect: in the presence of benzoyl peroxide, HBr adds against Markovnikov’s rule. It is a free-radical chain reaction: Br• adds first to give the more stable secondary radical, so the final major product is 1-bromopropane. Only HBr shows this effect.
• Addition of H₂SO₄: cold concentrated H₂SO₄ gives an alkyl hydrogen sulphate (Markovnikov addition).
• Hydration: alkene + water in the presence of H⁺ gives an alcohol (Markovnikov).
• Oxidation: cold dilute KMnO₄ (Baeyer’s reagent) gives a vicinal glycol and is decolourised — another unsaturation test. Acidic KMnO₄ cleaves the double bond to ketones and/or acids.
• Ozonolysis: O₃ followed by Zn/H₂O cleaves the C=C. Propene gives ethanal + methanal; 2-methylpropene gives propan-2-one + methanal. The products reveal the position of the double bond.
• Polymerisation: ethene → polythene, propene → polypropene. The small molecules are monomers; the large product is a polymer.

Alkynes: Triple-Bond Chemistry and Acidic Character

Alkynes contain at least one C≡C triple bond, general formula \( \mathrm{C}_n\mathrm{H}_{2n-2} \). The first member, ethyne (acetylene), burns in oxygen to give the oxyacetylene flame used for arc welding (NCERT, p. 315).

• Naming: suffix -yne, numbered from the first triply bonded carbon. But-1-yne and but-2-yne are position isomers; 3-methylbut-1-yne is a chain isomer. Common names treat alkynes as acetylene derivatives (methylacetylene, ethylacetylene).
• Structure: each carbon is sp-hybridised and linear, H–C–C angle 180°. The triple bond is one σ + two π bonds, bond enthalpy 823 kJ mol⁻¹, bond length 120 pm, with a cylindrically symmetrical electron cloud.
• Preparation: industrial route from calcium carbide — \( \mathrm{CaCO_3 \rightarrow CaO + CO_2} \), \( \mathrm{CaO + 3C \rightarrow CaC_2 + CO} \), then \( \mathrm{CaC_2 + 2H_2O \rightarrow Ca(OH)_2 + C_2H_2} \). Laboratory route: vicinal dihalide + alcoholic KOH gives a haloalkene, then sodamide (NaNH₂) gives the alkyne.

Acidic character of terminal alkynes — the s-character logic

Hydrogen attached to a triply bonded carbon is weakly acidic. Why? The sp-hybridised carbon has 50% s character, the highest of any hybrid, so its orbital is the most electronegative. It pulls the shared C–H electron pair strongly towards carbon, letting hydrogen leave as H⁺ (NCERT, p. 317).

Memory device — “more s character, more acidic”: \( \mathrm{CH\equiv CH \gt CH_2{=}CH_2 \gt CH_3{-}CH_3} \) for acidity, and terminal alkynes \( \gt \) internal alkynes. Ethyne reacts with sodium to give monosodium ethynide, and with excess sodium gives disodium ethynide; propyne reacts with NaNH₂ to give sodium propynide.

Alkenes and alkanes do not react this way, so this test distinguishes a terminal alkyne from the other families.

Addition reactions — two molecules add

• H₂: alkyne → alkene → alkane (two molecules of H₂).
• Br₂: gives a dihalide, then a tetrabromide; the Br₂/CCl₄ solution is decolourised.
• HX: first molecule gives a vinyl halide, the second gives a gem dihalide (both halogens on the same carbon), following Markovnikov’s rule.
• Water (Hg²⁺/H⁺, 333 K): adds as H–OH to form an enol, which isomerises to a carbonyl compound — ethyne gives ethanal, propyne gives propanone.
• Polymerisation: linear polymerisation gives polyacetylene, which conducts electricity and can be used as battery electrodes; passing ethyne through a red-hot iron tube at 873 K gives benzene by cyclic trimerisation.

Benzene: Structure, Resonance and Aromaticity

Aromatic hydrocarbons are called arenes; those containing a benzene ring are benzenoids, the rest are non-benzenoids. Benzene (\( \mathrm{C_6H_6} \)) was isolated by Faraday in 1825. Despite a high degree of unsaturation it is unusually stable, and it forms only one monosubstituted product — proof that all six hydrogens are equivalent (NCERT, p. 320).

• Kekulé structure (1865): a six-membered ring with alternate single and double bonds. Its failure: it predicts two different 1,2-dibromobenzenes, but only one exists. Kekulé added “oscillating double bonds” to fix this, but that still did not explain benzene’s stability.
• Resonance: benzene is a hybrid of the two Kekulé structures. The hybrid is drawn as a hexagon with an inscribed circle, representing six π electrons delocalised over all six carbons.
• Orbital picture: all six carbons are sp². Six C–C σ bonds and six C–H σ bonds lie in the plane; each carbon leaves one unhybridised p orbital, and these overlap laterally to form a delocalised π cloud shaped like two doughnuts, one above and one below the ring (NCERT, p. 321, Fig. 9.7).
• X-ray evidence: benzene is planar and all six C–C bond lengths are equal at 139 pm — between a C–C single bond (154 pm) and a C=C double bond (133 pm). There is no localised double bond.

Misconception autopsy: why benzene does not decolourise bromine water

Students write: benzene has three double bonds, so it should add Br₂ like an alkene and decolourise bromine water. The correction: benzene’s six π electrons are delocalised over the whole ring, not parked between two carbons. X-ray data shows every C–C bond is identical (139 pm), so there is no localised double bond for bromine to add across.

Under normal conditions benzene does not decolourise bromine water; instead it undergoes electrophilic substitution, which preserves the delocalised cloud. Only under vigorous conditions (UV light, Cl₂) does it add — giving benzene hexachloride. This extra stability is what aromaticity means.

Aromaticity (Hückel rule): a ring is aromatic if it is (i) planar, (ii) has complete π-electron delocalisation, and (iii) contains \( (4n+2)\pi \) electrons, where \( n = 0, 1, 2, \dots \) (NCERT, p. 322). Benzene has six π electrons (\( n = 1 \)).

Preparation of benzene

• Cyclic polymerisation of ethyne at 873 K (red-hot iron tube).
• Decarboxylation of sodium benzoate with soda lime: \( \mathrm{C_6H_5COONa + NaOH \xrightarrow[\Delta]{CaO} C_6H_6 + Na_2CO_3} \).
• Reduction of phenol by passing its vapours over heated zinc dust.
• Commercial source: coal tar.

Electrophilic Substitution and Directive Influence in Benzene

Arenes characteristically undergo electrophilic substitution reactions (EAS). The five you must know:

• Nitration: conc. HNO₃ + conc. H₂SO₄ (nitrating mixture) introduces –NO₂; the electrophile is the nitronium ion \( \mathrm{NO_2^+} \).
• Halogenation: Cl₂ or Br₂ with a Lewis acid (anhydrous FeCl₃, FeBr₃ or AlCl₃) gives a haloarene.
• Sulphonation: fuming H₂SO₄ (oleum) introduces –SO₃H (benzenesulphonic acid).
• Friedel–Crafts alkylation: alkyl halide + anhydrous AlCl₃ gives an alkylbenzene. Benzene with 1-chloropropane gives isopropylbenzene, not n-propylbenzene, because the propyl carbocation rearranges to the more stable secondary form.
• Friedel–Crafts acylation: acyl halide or acid anhydride + AlCl₃ gives an acylbenzene. With excess chlorine and AlCl₃, benzene can be perchlorinated to hexachlorobenzene.

Mechanism — three steps that earn the marks

1. Generation of the electrophile: AlCl₃ (a Lewis acid) helps form Cl⁺, R⁺ or the acylium ion RCO⁺; for nitration, H₂SO₄ protonates HNO₃ to produce NO₂⁺.
2. Formation of the arenium ion (σ-complex): the electrophile attacks the ring; one carbon becomes sp³, the ring loses aromaticity, and the positive charge is stabilised by resonance.
3. Removal of a proton: [AlCl₄]⁻ (or [HSO₄]⁻ for nitration) removes H⁺ from the sp³ carbon, restoring aromaticity.

Addition and combustion under forcing conditions

• Hydrogenation over Ni at high temperature/pressure gives cyclohexane.
• Three molecules of Cl₂ under UV light add to give benzene hexachloride, \( \mathrm{C_6H_6Cl_6} \), called gammaxane or BHC.
• Benzene burns with a sooty flame: \( \mathrm{C_6H_6 + \frac{15}{2}O_2 \rightarrow 6CO_2 + 3H_2O} \).

Directive influence — where the next group goes

In phenol, resonance forms place negative charge on the ortho and para carbons, guiding the electrophile there. In nitrobenzene, resonance forms place positive charge at o/p positions — so those positions are electron-poor and attack shifts to meta (NCERT, p. 325).

Carcinogenicity: benzene and polynuclear fused-ring hydrocarbons (1,2-benzanthracene, 1,2-benzpyrene, 3-methylcholanthrene, dibenzanthracenes) form during incomplete combustion of tobacco, coal and petroleum. They enter the body, damage DNA and cause cancer (NCERT, p. 326).

Key Definitions at a Glance

Scan this table the night before the exam — every entry is a one-mark question waiting to happen.

Formulas and Reaction Orders to Memorise

Bond angles: \( 109.5^\circ \) (alkane), about \( 120^\circ \) (alkene), \( 180^\circ \) (alkyne). Hückel condition: \( (4n+2)\pi \) electrons. Staggered–eclipsed gap in ethane: 12.5 kJ mol⁻¹.

• Halogenation reactivity: \( \mathrm{F_2 \gt Cl_2 \gt Br_2 \gt I_2} \); hydrogen replacement: \( 3^\circ \gt 2^\circ \gt 1^\circ \).
• HX addition to alkenes: \( \mathrm{HI \gt HBr \gt HCl} \); alkyl halide elimination: \( \mathrm{I \gt Br \gt Cl} \), \( 3^\circ \gt 2^\circ \gt 1^\circ \).
• C–H acidity: \( \mathrm{sp \gt sp^2 \gt sp^3} \); terminal alkyne \( \gt \) internal alkyne.
• Combustion of any hydrocarbon: \( \mathrm{C}_x\mathrm{H}_y + \left(x + \frac{y}{4}\right)\mathrm{O}_2 \rightarrow x\mathrm{CO}_2 + \frac{y}{2}\mathrm{H}_2\mathrm{O} \).

Worked Examples: Practise These

Example 1 — IUPAC naming and σ/π bond count

Method: find the longest chain containing the double bond, number from the nearer end, then count σ bonds as C–C (n−1) plus C–H (2n) for a formula \( \mathrm{C}_n\mathrm{H}_{2n} \).

Example 2 — Markovnikov vs peroxide addition of HBr to but-1-ene

Method: without peroxide, follow carbocation stability (Markovnikov); with benzoyl peroxide, follow the free-radical pathway (anti-Markovnikov).

Example 3 — identifying an alkene from ozonolysis products

Method: ozonolysis cleaves the double bond and puts oxygen at each broken end. To work backwards, join the carbonyl carbons through a double bond.

Common Mistakes in Hydrocarbons Questions

One-line memory device: more s character → more acidic (sp \( \gt \) sp² \( \gt \) sp³ hydrogens).

Exam Notes: What Earns the Mark

• Naming questions: always show the longest-chain selection and the numbering logic. The reasoning step earns the mark, not just the final name.
• Two-product reactions: state the rule (Markovnikov) or the condition (peroxide) before writing the product, and label it “major”.
• Ordering questions (boiling points, acidity, electrophile reactivity): attach a one-line reason — van der Waals forces and surface area for b.p., s character for acidity, electron density for E⁺ attack.
• Mechanism questions: write initiation → propagation → termination for halogenation and the peroxide effect; for benzene EAS write electrophile generation → arenium ion → proton loss.
• Distinguishing questions: name the test and the expected observation — Br₂/CCl₄ or Baeyer’s reagent decolourisation for unsaturation; Na or NaNH₂ reaction for a terminal alkyne.
• Ozonolysis in reverse: given the carbonyl products, join the carbonyl carbons through a double bond to find the alkene.
• Boiling point logic you can reuse: branched alkanes have lower b.p. because compact, spherical molecules have a smaller contact area and therefore weaker van der Waals forces.

How the chapter’s exercises split: direct-concept questions (9.1, 9.10, 9.11, 9.15, 9.18, 9.24) test definitions and reasons; naming and structure questions (9.2, 9.3, 9.4, 9.14, 9.21) test IUPAC steps; mechanism and reasoning questions (9.16, 9.17, 9.19, 9.22, 9.23) test rule application; calculation-style logic questions (9.5, 9.6, 9.7, 9.8, 9.9, 9.20, 9.25) test reverse reasoning, combustion balancing and conversions.

Revision Summary: Hydrocarbons Class 11 Notes on One Screen

If you remember only one table from this chapter, remember this one — it is the whole chapter in four rows.

Tests to distinguish alkanes, alkenes and alkynes

• Add Br₂ in CCl₄ (or cold dilute KMnO₄): decolourisation shows an alkene or alkyne; an alkane gives no change.
• Add sodium metal or NaNH₂: brisk effervescence (H₂ or NH₃) identifies a terminal alkyne — alkanes, alkenes and internal alkynes stay silent.
• Burn a drop: a sooty flame with a sharp smell points to benzene, not a simple alkane.

Preparations checklist

• Alkanes: hydrogenation of alkene/alkyne; Zn/HCl reduction of alkyl halide; Wurtz; decarboxylation; Kolbe’s electrolysis.
• Alkenes: Lindlar’s catalyst (cis) or Na/liq NH₃ (trans); dehydrohalogenation with alc. KOH; dehalogenation with Zn; acidic dehydration of alcohols.
• Alkynes: calcium carbide + water; vicinal dihalide + alcoholic KOH, then NaNH₂.
• Benzene: ethyne trimerisation at 873 K; sodium benzoate decarboxylation; phenol reduction over zinc dust; coal tar.

Named reactions and effects checklist

Markovnikov rule · peroxide (Kharash) effect · Wurtz reaction · Kolbe electrolysis · decarboxylation · Lindlar’s catalyst · Friedel–Crafts alkylation and acylation · ozonolysis · Baeyer’s test · Hückel rule. For each one, you should be able to write the reagent, the condition, the product and one line of mechanism.

What to memorise vs what to derive: memorise general formulas, bond parameters, reactivity orders and the H–X bond enthalpies that explain the peroxide effect. Derive everything else — Markovnikov products from carbocation stability, benzene behaviour from resonance, acid strength from s character.

Every reaction here is taken from the NCERT Class 11 Chemistry Part II textbook, Chapter 9 (Hydrocarbons is Chapter 9 in that book, starting on textbook p. 296); you can also browse all CBSE notes or other Class 11 revision pages on this site.

Frequently Asked Questions

Why does benzene undergo substitution rather than addition reactions even though it has double bonds?

Because its six π electrons are delocalised over the whole ring, not localised between any two carbons. X-ray diffraction shows all six C–C bonds are equal at 139 pm — between a single bond (154 pm) and a double bond (133 pm) — so there is no localised double bond for an electrophile to add across.

Addition would destroy the delocalised cloud; substitution keeps it intact, so benzene is much more stable than the hypothetical cyclohexatriene.

Why does HBr show the peroxide (Kharash) effect while HCl and HI do not?

The free-radical chain needs a halogen atom that can both be generated and add to the alkene. The H–Cl bond is too strong (430.5 kJ mol⁻¹) for the phenyl radical to cleave, so no Cl• forms.

The H–I bond is weak (296.8 kJ mol⁻¹) and does break, but the iodine radicals recombine to I₂ instead of adding to the double bond. H–Br (363.7 kJ mol⁻¹) is the middle case: it is cleaved, and Br• adds to the alkene to give the anti-Markovnikov product.

How do you decide whether a compound will show geometrical (cis-trans) isomerism?

Check every carbon of the double bond: each one must carry two different atoms or groups. If either doubly bonded carbon has two identical groups, cis-trans isomerism is impossible — for example, \( \mathrm{(CH_3)_2C{=}CH{-}C_2H_5} \) and \( \mathrm{CH_2{=}CBr_2} \) show no geometrical isomerism. The arrangement also needs restricted rotation, which the π bond provides.

Why is staggered ethane more stable than eclipsed ethane?

In the staggered form, the C–H bond electron clouds on the two carbons are as far apart as possible, so torsional strain (electron-cloud repulsion) is minimum. In the eclipsed form the clouds are closest together, giving maximum repulsion and higher energy.

The energy gap is only 12.5 kJ mol⁻¹, so at ordinary temperatures the molecule easily rotates and the two conformers cannot be separated.

How would you distinguish a terminal alkyne from an alkene and an alkane in the laboratory?

First use the unsaturation test: add Br₂ in CCl₄. An alkene and an alkyne decolourise it; an alkane does not. Then use the acidity test: add sodium metal or sodamide (NaNH₂). Only a terminal alkyne (e.g., propyne) reacts, giving effervescence of hydrogen or ammonia and forming a sodium alkynide; alkenes and alkanes do not react.

Reference: NCERT Class 11 Chemistry textbook, chapter Hydrocarbons.

Official source: download the NCERT textbook free from ncert.nic.in.