Hydrocarbons | ISC Class 11 Chemistry Notes

This note covers hydrocarbon classification, nomenclature, isomerism, bonding, ethane conformations, physical properties, preparation and reactions of alkanes, alkenes and alkynes, benzene structure, aromaticity, substitution mechanisms, directing effects, distinguishing tests and uses.

How are hydrocarbons classified and named?

Definition: Hydrocarbons are compounds containing carbon and hydrogen only. Carbon is tetravalent, meaning that it forms four covalent bonds; hydrogen is monovalent and forms one.

Saturated hydrocarbons contain only single bonds between carbon atoms, with no carbon-carbon double or triple bonds. Open-chain saturated hydrocarbons are alkanes; saturated carbon rings are cycloalkanes. Unsaturated hydrocarbons contain carbon-carbon double or triple bonds. Aromatic hydrocarbons, also called arenes, form a special class of cyclic compounds.

In formulae, C represents carbon and H hydrogen; a subscript gives the number of atoms. Here n is the number of carbon atoms. A homologous series is a family whose successive members differ by a CH₂ unit and have similar chemical properties.

Other element symbols used below are O (oxygen), N (nitrogen), Na (sodium), K (potassium), Ca (calcium), Mg (magnesium), Zn (zinc), Cu (copper), Li (lithium), S (sulphur), Cl (chlorine), Br (bromine) and I (iodine). Superscript + and − signs indicate ionic charges.

How does systematic naming work?

IUPAC means International Union of Pure and Applied Chemistry. Choose the longest suitable carbon chain, including the multiple bond when present. Number an alkene or alkyne chain from the end nearer that bond. The endings are -ane, -ene and -yne.

An alkyl group is derived from an alkane by removal of one hydrogen atom: methyl is CH₃ and ethyl is C₂H₅. A substituent is an atom or group replacing hydrogen on the parent structure. Numbers in names identify its attachment position.

Structural isomers have the same molecular formula but different connectivity. Butane, CH₃CH₂CH₂CH₃, and 2-methylpropane, CH₃CH(CH₃)CH₃, show chain isomerism. But-1-ene and but-2-ene show position isomerism because the double bond occupies different positions on the same carbon skeleton.

Worked example 1. Name CH₃C(CH₃)₂CH₂C(CH₃)₂CH₃. The longest chain contains five carbon atoms. Two methyl groups occur at carbon 2 and two at carbon 4. The name is 2,2,4,4-tetramethylpentane; the prefix tetra- indicates four methyl substituents.

How do bonding and rotation determine hydrocarbon shapes?

A sigma bond, written σ, forms by head-on orbital overlap along the line joining two nuclei. A pi bond, written π, forms by sideways overlap. An orbital is a region described by an electron wavefunction; hybridisation combines atomic orbitals into suitable bonding orbitals.

sp³ hybridisation combines one s orbital and three p orbitals; sp² combines one s and two p orbitals; sp combines one s and one p orbital. Here s and p name atomic orbital types. Methane has tetrahedral geometry, with hydrogen-carbon-hydrogen angles of 109.5°.

Ethene has sp² carbon atoms and a carbon-carbon double bond containing one σ and one π bond. Ethyne has sp carbon atoms and a triple bond containing one σ and two π bonds. Ethyne is linear, with a hydrogen-carbon-carbon angle of 180°.

The drawing shows sideways p-orbital overlap, a π-electron cloud above and below the carbon-carbon bond, and a separate labelled drawing of bond angles and lengths.

See Fig. 9.5 in your NCERT textbook

Why is staggered ethane more stable?

Conformations are spatial arrangements interconverted by rotation around a carbon-carbon single bond. Ethane has infinitely many such arrangements. In the eclipsed form, hydrogen atoms on adjacent carbons are as close together as possible; in the staggered form they are as far apart as possible.

Torsional strain is the repulsive interaction between adjacent bond electron clouds. Staggered ethane has minimum torsional strain and is more stable. Energy per amount of substance is expressed here in kJ mol⁻¹, kilojoules per mole. The energy difference is of the order of 12.5 kJ mol⁻¹.

Rotation is not completely free, but the barrier is small enough for rotation to be almost free for practical purposes at ordinary temperatures. Bond lengths and bond angles remain the same in the different conformations. Intermediate arrangements are called skew conformations.

Sawhorse drawings show an inclined carbon-carbon bond and three hydrogen bonds at each end. Newman drawings view the bond end-on, with the front carbon at a point and the rear carbon represented by a circle. Both eclipsed and staggered arrangements are shown.

See Figs. 9.2 and 9.3 in your NCERT textbook

When is geometrical isomerism possible?

Geometrical isomerism results from restricted rotation around a double bond when each double-bonded carbon bears two different groups. In cis-but-2-ene the methyl groups lie on the same side; in trans-but-2-ene they lie on opposite sides. Propene does not meet this condition.

How do physical properties vary across the families?

Intermolecular forces are attractions between molecules. Alkanes are almost non-polar, meaning that their charge distribution has little separation into positive and negative regions. Their weak van der Waals attractions increase with molecular size and surface area, raising boiling points along the series.

K after a temperature denotes kelvin. At 298 K, the alkanes containing one to four carbon atoms are gases, those containing five to seventeen are liquids, and those containing eighteen or more are solids. Kelvin is the unit of thermodynamic temperature. Alkanes are colourless and odourless.

Boiling point is the temperature at which a liquid boils at the stated pressure. Melting point is the solid-to-liquid transition temperature. For a pure substance at the same pressure, freezing and melting occur at the same equilibrium temperature. Density is mass per unit volume.

Branching makes an alkane more compact, reducing intermolecular contact and lowering its boiling point relative to a less-branched isomer. The three pentane isomers illustrate this. Their melting points do not follow the same ordering; do not transfer the boiling-point trend mechanically to melting.

What changes with unsaturation and aromatic rings?

The first three members of the alkene series are gases, the next fourteen are liquids and higher members are solids. Ethene has a faint sweet smell. Alkenes are insoluble in water but fairly soluble in non-polar solvents. Straight-chain alkenes have higher boiling points than their branched-chain isomers.

Dipole moment measures charge separation in a molecule. Cis-but-2-ene is more polar than trans-but-2-ene, whose dipole moment is almost zero. The opposite arrangement of its methyl groups allows their bond dipole contributions to cancel.

The first three members of the alkyne series are gases, the next eight liquids and higher members solids. Alkynes are weakly polar, lighter than water and immiscible with it; immiscible liquids do not mix into one uniform liquid phase. Their melting points, boiling points and densities increase with molar mass, the mass per mole. Aromatic hydrocarbons are usually colourless liquids or solids and are immiscible with water.

How are alkanes prepared?

Petroleum and natural gas are major sources of alkanes. Laboratory preparations differ in whether they retain, shorten or join carbon chains. A catalyst changes reaction rate without being consumed overall. In equations, → means “forms”; conditions written alongside an equation apply to that reaction.

Which reactions retain the carbon skeleton?

Hydrogenation adds hydrogen across a multiple bond. Ethene forms ethane: CH₂=CH₂ + H₂ → CH₃CH₃, using platinum, palladium or nickel. Platinum and palladium work at room temperature; nickel requires relatively higher temperature and pressure.

An alkyl halide has a halogen atom attached to an alkyl group. Except for fluorides, alkyl halides can be reduced with zinc and dilute hydrochloric acid. Reduction here replaces the halogen by hydrogen, as chloromethane gives methane.

For the following general equations, R and R′ represent alkyl groups, which may differ; X represents a halogen. A Grignard reagent, RMgX, is an organomagnesium halide. Water converts it to an alkane: RMgX + H₂O → RH + Mg(OH)X.

Alcohols, compounds containing a hydroxyl group, OH, can be reduced to alkanes by heating with excess hydrogen iodide and red phosphorus. Aldehydes, compounds containing the CHO group, undergo Clemmensen reduction with zinc amalgam and concentrated hydrochloric acid, converting RCHO to RCH₃.

Which methods shorten or join chains?

Decarboxylation removes the carboxyl carbon from a carboxylic acid salt. A carboxylic acid contains the COOH group. Heating sodium ethanoate with soda lime, a mixture of sodium hydroxide and calcium oxide, gives methane: CH₃COONa + NaOH → CH₄ + Na₂CO₃. The alkane contains one carbon fewer than the acid.

Wurtz reaction couples alkyl halides using sodium in dry ether: 2CH₃Br + 2Na → C₂H₆ + 2NaBr. Here Na is sodium and Br is bromine. Using different alkyl halides produces a mixture, so this method is unsuitable for obtaining a pure unsymmetrical coupling product.

Corey-House synthesis uses lithium dialkylcuprate, R₂CuLi, to couple an alkyl group with a suitable alkyl halide: R₂CuLi + R′X → R-R′ + RCu + LiX. The reagent contains copper and lithium; this route permits preparation of unsymmetrical alkanes.

Kolbe electrolysis electrolyses aqueous sodium or potassium carboxylates. At the anode, the electrode where oxidation occurs, carboxylate ions lose electrons and carbon dioxide; the resulting alkyl radicals couple. The cathode, where reduction occurs, evolves hydrogen from water.

For sodium ethanoate: 2CH₃COONa + 2H₂O → C₂H₆ + 2CO₂ + H₂ + 2NaOH. This gives ethane, whereas soda-lime decarboxylation of the same salt gives methane. A free radical is a species with an unpaired electron.

Worked example 2. To prepare propane by soda-lime decarboxylation, choose sodium butanoate, containing four carbons: CH₃CH₂CH₂COONa + NaOH → CH₃CH₂CH₃ + Na₂CO₃. Removal of the carboxyl carbon leaves the required three-carbon alkane.

How do alkanes undergo substitution, oxidation and cracking?

Alkanes are generally inert towards acids, bases, oxidising agents and reducing agents. They nevertheless react under suitable conditions. A substitution reaction replaces an atom or group; in methane chlorination, chlorine replaces hydrogen in diffused sunlight or ultraviolet light.

How does the radical chain mechanism work?

Homolysis splits a covalent bond so that each fragment receives one bonding electron. A dot, •, denotes an unpaired electron. Halogenation is supposed to proceed through initiation, propagation and termination.

1. Initiation: Light splits chlorine, Cl₂ → 2Cl•, forming chlorine radicals.
2. First propagation step: Cl• + CH₄ → HCl + CH₃•. Hydrogen chloride and a methyl radical form.
3. Second propagation step: CH₃• + Cl₂ → CH₃Cl + Cl•. Chloromethane forms and a chlorine radical is regenerated.
4. Termination: Radicals combine, for example CH₃• + CH₃• → C₂H₆. This explains ethane formation as a by-product.

Other termination reactions are Cl• + Cl• → Cl₂ and CH₃• + Cl• → CH₃Cl. Further substitution can produce dichloromethane, CH₂Cl₂, trichloromethane, CHCl₃, and tetrachloromethane, CCl₄. Formation of chloromethane is therefore not necessarily the endpoint.

How does oxygen supply affect the products?

Complete combustion forms carbon dioxide and water with release of much heat: CH₄ + 2O₂ → CO₂ + 2H₂O. With insufficient oxygen, incomplete combustion can form carbon black. Controlled oxidation uses regulated oxygen, suitable catalysts and specified conditions to obtain oxygen-containing organic compounds.

An atmosphere is a pressure unit. Methanol is an alcohol, methanal an aldehyde, and ethanoic acid a carboxylic acid. Products depend on conditions, so “oxidation of methane” alone does not specify a unique outcome.

How can the carbon skeleton change?

Isomerisation converts straight-chain alkanes to branched isomers on heating with anhydrous aluminium chloride and hydrogen chloride. Anhydrous means free from water. Aromatisation combines cyclisation, or ring formation, with dehydrogenation, or hydrogen removal.

Alkanes with six or more carbons undergo aromatisation at 773 K and 10 to 20 atmospheres over suitable metal oxides on alumina. Hexane forms benzene. Pyrolysis, or cracking, decomposes higher alkanes into smaller hydrocarbons by heat; it is believed to be a free-radical reaction.

How are alkenes prepared by elimination and reduction?

Elimination removes atoms or groups from adjacent carbons to create a multiple bond. It must be distinguished from addition, which joins atoms or groups across a multiple bond. The reagent and reaction conditions determine the useful direction of a conversion.

Dehydration removes water. Dehydrohalogenation removes a hydrogen halide such as hydrogen bromide. A vicinal dihalide bears halogens on adjacent carbon atoms. Its reaction with zinc removes both halogens, so it is dehalogenation, not dehydrohalogenation.

Which alkene becomes the major product?

In elimination, the beta carbon is adjacent to the carbon bearing the leaving group. Saytzeff’s rule predicts that the more substituted alkene is generally the major product when alternative eliminations are possible. Heating 2-bromobutane with alcoholic potassium hydroxide gives mainly but-2-ene rather than but-1-ene.

Lindlar’s catalyst is a partially deactivated palladium catalyst used for partial hydrogenation. Alkynes giving geometrical isomers form cis-alkenes by this route. Sodium in liquid ammonia gives trans-alkenes. Ethyne and propyne cannot produce a cis-trans pair because the resulting terminal double bond has two hydrogens on one carbon.

Kolbe electrolysis can also prepare ethene from an aqueous succinate salt. The succinate ion, ⁻OOCCH₂CH₂COO⁻, is the doubly charged ion of butanedioic acid. Loss of two carbon dioxide molecules at the anode creates ethene: ⁻OOCCH₂CH₂COO⁻ → CH₂=CH₂ + 2CO₂ + 2e⁻. Here e⁻ denotes an electron.

How do alkenes undergo addition reactions?

An electrophile accepts an electron pair. Alkene π electrons are relatively loosely held, so electrophiles can attack the double bond. A carbocation is an organic ion with a positively charged carbon; a nucleophile donates an electron pair.

Hydrogenation gives an alkane. Bromine addition gives a vicinal dibromide: CH₂=CH₂ + Br₂ → CH₂BrCH₂Br. The reddish-orange colour of bromine solution in carbon tetrachloride disappears. Halogen addition involves a cyclic halonium ion, an intermediate in which a halogen bridges the two carbons.

What does Markownikoff’s rule predict?

Markownikoff’s rule, also spelt Markovnikov’s rule, states that the negative part of an adding molecule attaches to the double-bonded carbon bearing fewer hydrogen atoms. With propene and hydrogen bromide, HBr, the principal product is 2-bromopropane.

1. The π bond reacts with a proton, H⁺, supplied by hydrogen bromide.
2. Proton attachment to the terminal carbon produces CH₃CH⁺CH₃, a secondary carbocation, whose charged carbon is attached to two other carbons.
3. This secondary carbocation predominates because it is more stable and forms faster than the alternative primary carbocation, whose charged carbon has one carbon neighbour.
4. The bromide ion, Br⁻, donates an electron pair to the carbocation, giving CH₃CHBrCH₃.

Why does peroxide reverse HBr addition?

The peroxide effect is anti-Markownikoff addition of HBr through a radical chain. A peroxide contains an oxygen-oxygen single bond whose cleavage initiates radical formation. This effect occurs with HBr, not with hydrogen chloride or hydrogen iodide.

A bromine radical adds to propene to produce CH₃CH•CH₂Br, the more stable secondary radical. It removes hydrogen from HBr, giving CH₃CH₂CH₂Br, 1-bromopropane, and regenerating Br•. The carbon radical determines the preferred orientation; no carbocation is required in this pathway.

Cold concentrated sulphuric acid adds according to Markownikoff’s rule to form an alkyl hydrogen sulphate, containing the OSO₃H group. Water in the presence of a few drops of concentrated sulphuric acid gives an alcohol by hydration, meaning addition of water.

Worked example 3. Hex-1-ene, CH₂=CHCH₂CH₂CH₂CH₃, gives mainly 2-bromohexane with HBr without peroxide. With peroxide it gives mainly 1-bromohexane. The carbon chain is unchanged; the difference is the position at which bromine becomes attached.

How do oxidation, ozonolysis and polymerisation change alkenes?

Baeyer’s reagent is cold, dilute alkaline potassium permanganate solution, KMnO₄. Alkenes decolourise it while forming vicinal glycols. A glycol is a dihydric alcohol; a vicinal glycol has hydroxyl groups on adjacent carbons. Ethene gives ethane-1,2-diol, HOCH₂CH₂OH.

Stronger oxidation can cleave the carbon-carbon double bond. Hot alkaline permanganate produces carboxylate salts from double-bonded carbons bearing one hydrogen; acidification gives the corresponding carboxylic acids. A terminal =CH₂ carbon is oxidised further to carbonate, which releases carbon dioxide on acidification. But-2-ene gives ethanoate, which gives ethanoic acid after acidification. Carbon atoms without attached hydrogen can give ketones.

How does ozonolysis locate a double bond?

Ozonolysis adds ozone, O₃, to form an ozonide, an oxygen-containing intermediate. Treatment with zinc and water cleaves it to carbonyl compounds. A carbonyl group is C=O; aldehydes have a hydrogen attached to its carbon, whereas ketones have two carbon groups attached.

To predict products, split the original double bond and replace it at each carbon by a carbonyl bond. Retain the groups already attached to each carbon. This makes ozonolysis useful for identifying the original double-bond position.

Worked example 4. Pent-2-ene is CH₃CH=CHCH₂CH₃. Ozonolysis followed by zinc and water gives ethanal, CH₃CHO, and propanal, CH₃CH₂CHO. The two-carbon and three-carbon fragments together account for the five carbons in the starting alkene.

What happens during polymerisation?

A monomer is a small molecule that joins with others to form a polymer, a large molecule containing repeating units. At high temperature and pressure with a catalyst, ethene forms polythene. The double bond opens to allow chain formation without removing the carbon skeleton.

Propene similarly forms polypropene, used for milk crates, buckets and moulded articles. Hydrocarbons therefore serve as starting materials for products as well as fuels. Polymerisation joins molecules; ozonolysis divides a molecule at its unsaturation site.

How are alkynes prepared, and why are terminal alkynes acidic?

Ethyne, also called acetylene, is the first stable alkyne. Industrial preparation uses calcium carbide, CaC₂: CaC₂ + 2H₂O → C₂H₂ + Ca(OH)₂. Calcium hydroxide is the other product. Calcium carbide is obtained by heating quicklime, calcium oxide, with coke: CaO + 3C → CaC₂ + CO.

From methane, rapid high-temperature cracking followed by quenching gives ethyne: 2CH₄ → C₂H₂ + 3H₂. Quenching means rapid cooling, used here to limit further decomposition. Natural gas supplies methane for this preparation.

How does double elimination create a triple bond?

Vicinal dihalides first lose a hydrogen halide with alcoholic potassium hydroxide, forming an alkenyl halide, whose halogen is attached to a double-bonded carbon. A stronger base, sodamide, NaNH₂, removes a second hydrogen halide to form the alkyne.

Electrolysis of suitable unsaturated dicarboxylate salts also gives ethyne. The maleate or fumarate ion has the formula ⁻OOCCH=CHCOO⁻. At the anode: ⁻OOCCH=CHCOO⁻ → HC≡CH + 2CO₂ + 2e⁻. Removal of the two carboxyl groups increases the bond order between the remaining carbons.

Which hydrogen atoms are replaceable?

A terminal alkyne has hydrogen directly attached to a triply bonded carbon, as in propyne, CH₃C≡CH. An internal alkyne, such as but-2-yne, CH₃C≡CCH₃, has carbon groups at both ends of the triple bond.

The sp orbital has 50% s-character, meaning half its character comes from the s orbital. It attracts the shared electron pair more strongly than sp² or sp³ carbon. Consequently, the terminal hydrogen is more readily released as a proton than hydrogen in ethene or ethane.

Sodium forms an acetylide, a salt derived by replacing an acidic alkyne hydrogen: 2HC≡CH + 2Na → 2HC≡CNa + H₂. Sodamide gives ammonia, NH₃: CH₃C≡CH + NaNH₂ → CH₃C≡CNa + NH₃. Not every hydrogen in an alkyne has this acidity.

How do alkyne reactions and distinguishing tests work?

Alkynes can add two molecules of hydrogen, halogen or hydrogen halide across a triple bond. Addition in unsymmetrical alkynes follows Markownikoff’s rule. The majority of alkyne reactions are examples of electrophilic addition; avoid treating every reaction as the same mechanism.

A geminal dihalide has two halogens on the same carbon, unlike a vicinal dihalide. Under suitable conditions ethyne also forms polyethyne by linear polymerisation. Under special conditions this polymer conducts electricity.

What happens during oxidation?

Ethyne burns in sufficient oxygen: 2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O. Its mixture with oxygen gives the oxyacetylene flame used in welding. Oxidation by permanganate decolourises the reagent; prolonged oxidation of ethyne gives oxalate in alkaline solution, and oxalic acid after acidification.

Vigorous oxidation of internal alkynes cleaves the triple bond to carboxylate salts in alkaline solution. Acidification gives carboxylic acids. Ozonolysis followed by hydrolysis likewise cleaves an internal alkyne to acids. But-2-yne gives two molecules of ethanoic acid.

Which tests distinguish the families?

Bromine solution and Baeyer’s reagent test for unsaturation: both alkenes and alkynes decolourise them. These observations alone cannot distinguish the two families. Alkanes do not give these addition tests under ordinary test conditions; light must be excluded from the bromine comparison.

Terminal alkynes form precipitates with ammoniacal silver nitrate or ammoniacal cuprous chloride, meaning solutions containing ammonia. Ethyne gives white silver acetylide or red copper(I) acetylide. Alkenes and alkanes do not. Internal alkynes lack the terminal hydrogen and do not give this acetylide test.

Note: Interpret a positive acetylide test as evidence for a terminal alkyne. A negative result does not exclude an internal alkyne. Keep the reagent and conditions alongside the observation.

Why is benzene aromatic, and how is it prepared?

Benzene, C₆H₆, is a planar six-carbon ring. Each carbon is sp² hybridised and has an unhybridised p orbital perpendicular to the ring. The two Kekulé structures alternate single and double bonds in different positions, but neither alone represents the actual molecule.

Resonance describes a molecule using several contributing electron arrangements with the same atomic positions. Benzene is a resonance hybrid: its six π electrons are delocalised, spread around the ring rather than confined between individual pairs of carbons. All six carbon-carbon bonds are equivalent.

The carbon-carbon distance is 139 pm, where pm means picometre, one trillionth of a metre. This is intermediate between the single-bond and double-bond values discussed for comparison. Delocalisation increases stability and helps explain benzene’s reluctance to undergo addition under normal conditions.

One drawing shows p orbitals arranged around the six-carbon ring; the other represents the electron cloud as rings above and below the plane of the carbon framework.

Reference: NCERT Class 11 Figure 9.7(c) and (d)

What are the conditions for aromaticity?

Aromaticity requires a planar ring, complete π-electron delocalisation around it and the appropriate electron count. Hückel’s rule specifies 4n + 2 π electrons, where n is a non-negative integer in this rule, not the carbon count used in general hydrocarbon formulae.

Benzene has six π electrons, so n = 1. Aromaticity is an electronic property; pleasant odour is not a test. Benzenoid aromatic compounds contain benzene rings; non-benzenoid aromatic compounds do not.

Which preparations give benzene?

Heating sodium benzoate with soda lime removes its carboxyl carbon: C₆H₅COONa + NaOH → C₆H₆ + Na₂CO₃. Passing phenol vapour over heated zinc dust removes its oxygen: C₆H₅OH + Zn → C₆H₆ + ZnO. Phenol has a hydroxyl group directly attached to a benzene ring.

Cyclic polymerisation gives 3C₂H₂ → C₆H₆. Hexane aromatisation provides another route, C₆H₁₄ → C₆H₆ + 4H₂, under the reforming conditions already described. Benzene is commercially isolated from coal tar.

How does benzene undergo substitution and other reactions?

Benzene characteristically undergoes electrophilic substitution, replacing ring hydrogen while restoring the aromatic system. A Lewis acid accepts an electron pair; anhydrous aluminium chloride and iron(III) halides help generate electrophiles in several aromatic reactions.

An acyl group is RCO; acetyl is CH₃CO. Acetyl chloride is CH₃COCl. Alkylation introduces an alkyl group; acylation introduces an acyl group. These are different transformations despite their shared catalyst.

What are the steps of electrophilic substitution?

Electrophilic substitution reactions are supposed to proceed through electrophile generation, carbocation formation and proton removal. In nitration, sulphuric acid protonates nitric acid; loss of water produces the nitronium ion, NO₂⁺, the attacking electrophile.

The ring donates π electrons to the electrophile, forming a sigma complex, also called an arenium ion. One carbon becomes sp³ hybridised, interrupting aromatic delocalisation. Loss of a proton from this carbon restores aromaticity and gives the substituted benzene.

In halogenation the Lewis acid polarises the halogen molecule. In sulphonation, sulphur trioxide, SO₃, acts as an electrophile. Both reactions follow ring attack and proton loss, with the appropriate electrophile and acid-base steps.

When do addition, oxidation and pyrolysis occur?

Under vigorous conditions with nickel, hydrogenation gives cyclohexane: C₆H₆ + 3H₂ → C₆H₁₂. Ultraviolet light promotes chlorine addition: C₆H₆ + 3Cl₂ → C₆H₆Cl₆, hexachlorocyclohexane. Photochemical bromine addition can similarly give hexabromocyclohexane, C₆H₆Br₆, under suitable conditions.

Complete combustion gives carbon dioxide and water with a sooty flame. Catalytic oxidation with air over vanadium(V) oxide gives maleic anhydride, a cyclic derivative of a dicarboxylic acid. Ozonolysis followed by reductive work-up gives glyoxal, OHCCHO, a compound with two aldehyde groups.

At high temperature, benzene can form biphenyl, two benzene rings joined by a single bond: 2C₆H₆ → C₆H₅C₆H₅ + H₂. These reactions require their own conditions; benzene must not be treated as an ordinary alkene.

How do substituents direct reactions, and why do uses and toxicity matter?

In a monosubstituted benzene, one ring hydrogen has been replaced. Relative to that substituent, ortho means positions 1,2; meta means 1,3; and para means 1,4. Further substitution does not usually produce the three isomers in equal amounts.

How are position and reactivity controlled?

Directive influence is the effect of an existing substituent on where an incoming group attaches. An activating group increases ring reactivity towards electrophilic substitution; a deactivating group decreases it. Direction and activation describe different aspects of the reaction.

The hydroxyl group donates an electron pair through resonance, increasing electron density at ortho and para positions. Its inductive effect, the displacement of electron density through sigma bonds, withdraws electron density and slightly reduces it there, but resonance donation predominates.

Hydroxyl, amino, NH₂, methoxy, OCH₃, and methyl groups are ortho/para directors. Halogens are also ortho/para directors but are moderately deactivating: inductive withdrawal reduces overall ring electron density, while resonance makes ortho and para positions relatively richer than meta.

The nitro group, NO₂, withdraws electrons and directs electrophiles mainly to meta. Carboxyl and cyano, CN, groups are other meta directors. The reason is relative electron density and intermediate stability, not a physical obstruction placed in front of the electrophile.

What changes in nucleophilic aromatic substitution?

In nucleophilic aromatic substitution, an electron-pair donor replaces a leaving group on an aromatic carbon. For an aryl halide, a nitro group ortho or para to the halogen facilitates the addition-elimination pathway by stabilising the negatively charged intermediate through resonance.

The nucleophile adds at the carbon bearing halogen, forming an anionic intermediate; halide elimination then restores aromaticity. A meta nitro group does not provide the same resonance stabilisation. This directing effect concerns nucleophilic replacement and must not be confused with nitro’s meta direction in electrophilic substitution.

Where are hydrocarbons useful?

Alkanes provide fuels and solvents. Alkenes provide monomers for plastics; ethyne provides a welding fuel and a starting material for organic synthesis. Aromatic hydrocarbons supply starting materials for dyes, drugs and other organic products. Uses follow from both physical properties and chemical convertibility.

Carcinogenic means capable of causing cancer; toxic means harmful to living systems. Benzene and polynuclear hydrocarbons containing more than two fused benzene rings are toxic and are said to possess carcinogenic properties. Fused rings share adjacent carbon atoms.

Such polynuclear hydrocarbons form during incomplete combustion of materials such as tobacco, coal and petroleum. After entering the body they undergo biochemical reactions that damage DNA, deoxyribonucleic acid, the hereditary material, and cause cancer. Useful industrial applications do not remove these toxic properties.

How are complete combustion equations balanced?

Complete combustion converts hydrocarbon carbon into carbon dioxide and hydrogen into water when sufficient oxygen is available. Balance carbon first, hydrogen second and oxygen last. The resulting coefficients give mole ratios; they do not give equal masses of reactants and products.

Derivation: How is the general alkane combustion equation balanced?

For an alkane, $\mathrm{C}_{n}\mathrm{H}_{2n+2}$, let $a$, $b$ and $c$ be the coefficients of carbon dioxide, water and oxygen respectively, with the alkane coefficient fixed at one.

1. Balance carbon: each carbon dioxide molecule contains one carbon atom, so $a = n$.
2. Balance hydrogen: each water molecule contains two hydrogen atoms, so $2b = 2n+2$, giving $b = n+1$.
3. Balance oxygen: the products contain $2n+(n+1)$ oxygen atoms. Hence $2c = 3n+1$, giving $c = (3n+1)/2$.

Balanced alkane equation: $$\mathrm{C}_{n}\mathrm{H}_{2n+2}+\frac{3n+1}{2}\mathrm{O}_{2}\rightarrow n\mathrm{CO}_{2}+(n+1)\mathrm{H}_{2}\mathrm{O}$$ Multiply all coefficients by two if whole numbers are needed.

Derivation: How does the balance extend to any hydrocarbon?

Write the hydrocarbon as $\mathrm{C}_{x}\mathrm{H}_{y}$, where $x$ and $y$ count its carbon and hydrogen atoms. Keep its coefficient at one and use the same product and oxygen coefficient symbols.

1. Carbon balance gives $a = x$, since all $x$ carbon atoms become carbon dioxide.
2. Hydrogen balance gives $b = y/2$, since each water molecule contains two hydrogen atoms.
3. The products contain $2x+y/2$ oxygen atoms. Dividing by two atoms per oxygen molecule gives $c = x+y/4$.

Balanced hydrocarbon equation: $$\mathrm{C}_{x}\mathrm{H}_{y}+\left(x+\frac{y}{4}\right)\mathrm{O}_{2}\rightarrow x\mathrm{CO}_{2}+\frac{y}{2}\mathrm{H}_{2}\mathrm{O}$$ This balance applies to complete combustion, including alkenes, alkynes and arenes. It does not describe incomplete combustion to carbon black.

Worked example 5. Balance the complete combustion of methane, $\mathrm{C}\mathrm{H}_{4}$, and interpret the smallest whole-number coefficients in moles.

Formula: $a = x$, $b = y/2$ and $c = x+y/4$, where $a$, $b$ and $c$ are the carbon dioxide, water and oxygen coefficients for one hydrocarbon molecule.

Substitute: $x = 1$ and $y = 4$. Carbon balance gives $a = 1$; hydrogen balance gives $b = 4/2 = 2$; oxygen balance gives $c = 1+4/4 = 2$. All coefficients are already whole numbers.

$$\mathrm{C}\mathrm{H}_{4}+2\mathrm{O}_{2}\rightarrow \mathrm{CO}_{2}+2\mathrm{H}_{2}\mathrm{O}$$

Answer: 1 mol of methane reacts with 2 mol of oxygen to form 1 mol of carbon dioxide and 2 mol of water. Check: each side contains 1 carbon, 4 hydrogen and 4 oxygen atoms per balanced equation.

Worked example 6. Balance the complete combustion of butane, $\mathrm{C}_{4}\mathrm{H}_{10}$, and interpret the smallest whole-number coefficients in moles.

Formula: $a = x$, $b = y/2$ and $c = x+y/4$, where $a$, $b$ and $c$ are the carbon dioxide, water and oxygen coefficients for one hydrocarbon molecule.

Substitute: $x = 4$ and $y = 10$. Carbon balance gives $a = 4$; hydrogen balance gives $b = 10/2 = 5$; oxygen balance gives $c = 4+10/4 = 13/2$. Multiply every coefficient by two to remove the fractional oxygen coefficient.

$$2\mathrm{C}_{4}\mathrm{H}_{10}+13\mathrm{O}_{2}\rightarrow 8\mathrm{CO}_{2}+10\mathrm{H}_{2}\mathrm{O}$$

Answer: 2 mol of butane reacts with 13 mol of oxygen to form 8 mol of carbon dioxide and 10 mol of water. Check: each side contains 8 carbon, 20 hydrogen and 26 oxygen atoms per balanced equation.

Worked example 7. Balance the complete combustion of pentene, $\mathrm{C}_{5}\mathrm{H}_{10}$, and interpret the smallest whole-number coefficients in moles.

Formula: $a = x$, $b = y/2$ and $c = x+y/4$, where $a$, $b$ and $c$ are the carbon dioxide, water and oxygen coefficients for one hydrocarbon molecule.

Substitute: $x = 5$ and $y = 10$. Carbon balance gives $a = 5$; hydrogen balance gives $b = 10/2 = 5$; oxygen balance gives $c = 5+10/4 = 15/2$. Multiply every coefficient by two to remove the fractional oxygen coefficient.

$$2\mathrm{C}_{5}\mathrm{H}_{10}+15\mathrm{O}_{2}\rightarrow 10\mathrm{CO}_{2}+10\mathrm{H}_{2}\mathrm{O}$$

Answer: 2 mol of pentene reacts with 15 mol of oxygen to form 10 mol of carbon dioxide and 10 mol of water. Check: each side contains 10 carbon, 20 hydrogen and 30 oxygen atoms per balanced equation.

Worked example 8. Balance the complete combustion of hexyne, $\mathrm{C}_{6}\mathrm{H}_{10}$, and interpret the smallest whole-number coefficients in moles.

Formula: $a = x$, $b = y/2$ and $c = x+y/4$, where $a$, $b$ and $c$ are the carbon dioxide, water and oxygen coefficients for one hydrocarbon molecule.

Substitute: $x = 6$ and $y = 10$. Carbon balance gives $a = 6$; hydrogen balance gives $b = 10/2 = 5$; oxygen balance gives $c = 6+10/4 = 17/2$. Multiply every coefficient by two to remove the fractional oxygen coefficient.

$$2\mathrm{C}_{6}\mathrm{H}_{10}+17\mathrm{O}_{2}\rightarrow 12\mathrm{CO}_{2}+10\mathrm{H}_{2}\mathrm{O}$$

Answer: 2 mol of hexyne reacts with 17 mol of oxygen to form 12 mol of carbon dioxide and 10 mol of water. Check: each side contains 12 carbon, 20 hydrogen and 34 oxygen atoms per balanced equation.

Worked example 9. Balance the complete combustion of toluene, $\mathrm{C}_{7}\mathrm{H}_{8}$, and interpret the smallest whole-number coefficients in moles.

Formula: $a = x$, $b = y/2$ and $c = x+y/4$, where $a$, $b$ and $c$ are the carbon dioxide, water and oxygen coefficients for one hydrocarbon molecule.

Substitute: $x = 7$ and $y = 8$. Carbon balance gives $a = 7$; hydrogen balance gives $b = 8/2 = 4$; oxygen balance gives $c = 7+8/4 = 9$. All coefficients are already whole numbers.

$$\mathrm{C}_{7}\mathrm{H}_{8}+9\mathrm{O}_{2}\rightarrow 7\mathrm{CO}_{2}+4\mathrm{H}_{2}\mathrm{O}$$

Answer: 1 mol of toluene reacts with 9 mol of oxygen to form 7 mol of carbon dioxide and 4 mol of water. Check: each side contains 7 carbon, 8 hydrogen and 18 oxygen atoms per balanced equation.

Worked example 10. Balance the complete combustion of benzene, $\mathrm{C}_{6}\mathrm{H}_{6}$, and interpret the smallest whole-number coefficients in moles.

Formula: $a = x$, $b = y/2$ and $c = x+y/4$, where $a$, $b$ and $c$ are the carbon dioxide, water and oxygen coefficients for one hydrocarbon molecule.

Substitute: $x = 6$ and $y = 6$. Carbon balance gives $a = 6$; hydrogen balance gives $b = 6/2 = 3$; oxygen balance gives $c = 6+6/4 = 15/2$. Multiply every coefficient by two to remove the fractional oxygen coefficient.

$$2\mathrm{C}_{6}\mathrm{H}_{6}+15\mathrm{O}_{2}\rightarrow 12\mathrm{CO}_{2}+6\mathrm{H}_{2}\mathrm{O}$$

Answer: 2 mol of benzene reacts with 15 mol of oxygen to form 12 mol of carbon dioxide and 6 mol of water. Check: each side contains 12 carbon, 12 hydrogen and 30 oxygen atoms per balanced equation.

Glossary

• Hydrocarbon — A compound containing carbon and hydrogen only, classified by its carbon bonding and overall molecular structure.
• Homologous series — A family of related compounds whose successive members differ by a CH₂ unit and have similar chemical properties.
• Structural isomerism — The occurrence of compounds with the same molecular formula but different arrangements of atom-to-atom connectivity.
• Conformation — A spatial arrangement of atoms interconverted with another by rotation around a carbon-carbon single bond.
• Torsional strain — Repulsive interaction between adjacent bond electron clouds that affects the relative stability of molecular conformations.
• Electrophile — A species that accepts an electron pair when forming a new covalent bond during a reaction.
• Nucleophile — A species that donates an electron pair to form a new covalent bond with another species.
• Free radical — A chemical species containing an unpaired electron, often formed through homolytic cleavage of a covalent bond.
• Decarboxylation — Removal of the carboxyl carbon, giving an alkane with one fewer carbon in soda-lime preparation.
• Ozonolysis — Reaction with ozone followed by work-up that cleaves unsaturation and helps identify multiple-bond positions.
• Acetylide — A salt formed by replacement of an acidic hydrogen directly attached to a triply bonded carbon.
• Aromaticity — The character associated with a planar, cyclic, completely delocalised π-electron system satisfying the appropriate electron-count rule.
• Directive influence — The effect of an existing substituent on the position predominantly occupied by an incoming group.

Common errors and misconceptions

• Misconception: Every compound with formula CₙH₂ₙ is an alkene. Correct: This formula also fits a saturated monocyclic hydrocarbon; inspect the bonding and structure before classifying it.
• Misconception: Ethane rotates without an energy barrier. Correct: Rotation is not completely free, although the small barrier makes it almost free for practical purposes.
• Misconception: All alkenes show cis-trans isomerism. Correct: Each double-bonded carbon must have two different attached groups; propene fails this condition.
• Misconception: Peroxide reverses every hydrogen-halide addition. Correct: The peroxide effect occurs with HBr, not HCl or HI; it proceeds through radicals.
• Misconception: Sodamide and sodium release the same gas from a terminal alkyne. Correct: Sodium releases hydrogen; sodamide forms ammonia while producing the acetylide.
• Misconception: Bromine decolourisation distinguishes an alkene from an alkyne. Correct: Both give this unsaturation test; acetylide formation specifically identifies terminal alkynes.
• Misconception: Every ortho/para director activates benzene. Correct: Halogens direct ortho/para but moderately deactivate the ring towards further electrophilic substitution.

Exam-style questions with model answers

Q1. Compare the relative stability and torsional strain of staggered and eclipsed ethane. [2 marks]

1. Staggered ethane is more stable because its carbon-hydrogen bond electron clouds are farther apart.
2. Staggered ethane has minimum torsional strain, whereas eclipsed ethane has maximum torsional strain and higher energy.

Q2. Give the initiation step, both propagation equations and one termination equation for methane chlorination in ultraviolet light. Explain how ethane forms. [3 marks]

1. Initiation occurs by homolysis: Cl₂ → 2Cl•. Ultraviolet light breaks the chlorine bond, giving each chlorine atom one electron from the shared pair.
2. Propagation is Cl• + CH₄ → HCl + CH₃•, followed by CH₃• + Cl₂ → CH₃Cl + Cl•. Regeneration of chlorine radicals maintains the chain.
3. Termination includes CH₃• + CH₃• → C₂H₆. Two methyl radicals combine, explaining ethane formation while removing radicals from the reacting mixture.

Q3. Propene, CH₃CH=CH₂, reacts with HBr. State the principal product without peroxide and explain its formation. Then state the principal product with peroxide and the mechanism type. [4 marks]

1. Without peroxide, the principal product is 2-bromopropane, CH₃CHBrCH₃, in accordance with Markownikoff’s rule for the unsymmetrical alkene.
2. Proton addition preferentially produces the more stable secondary carbocation, CH₃CH⁺CH₃, rather than the primary alternative.
3. Bromide attacks this positively charged carbon, completing electrophilic addition and forming the carbon-bromine bond.
4. With peroxide, 1-bromopropane, CH₃CH₂CH₂Br, is the principal product. This anti-Markownikoff addition proceeds by a free-radical chain mechanism.

Q4. Pent-2-ene has the structure CH₃CH=CHCH₂CH₃. Explain its ozonolysis with reductive zinc-water work-up, name both products and account for all five carbons. [3 marks]

1. Ozone adds at the carbon-carbon double bond to form an ozonide. Zinc-water work-up cleaves this intermediate into smaller carbonyl compounds.
2. The CH₃CH side gives ethanal, CH₃CHO, while the CHCH₂CH₃ side gives propanal, CH₃CH₂CHO. Each original double-bonded carbon becomes a carbonyl carbon.
3. Ethanal contains two carbons and propanal contains three. Their total is five, matching the complete carbon skeleton of the starting pent-2-ene.

Q5. Separate samples are known to be ethane, ethene and ethyne. Explain how bromine solution and ammoniacal silver nitrate distinguish them, including the test limitation for internal alkynes. [5 marks]

1. Test separate portions with bromine solution under ordinary conditions excluding light. Ethane does not decolourise it through the addition reaction used to detect unsaturation.
2. Ethene decolourises bromine because addition occurs across its double bond. This establishes unsaturation but does not by itself identify the hydrocarbon as an alkene.
3. Ethyne also decolourises bromine because bromine adds across its triple bond. Thus, the two positive samples need a further test.
4. Use ammoniacal silver nitrate on fresh portions. Ethyne forms white silver acetylide because it has hydrogen directly attached to triply bonded carbon; ethene does not.
5. Combine the observations to identify all three samples. The silver test identifies terminal alkynes; an internal alkyne lacks the required terminal hydrogen and gives a negative result.

Q6. Explain benzene’s aromatic stability and its preference for electrophilic substitution. Include the three stages of nitration and identify the electrophile. [5 marks]

1. Benzene is planar, with six delocalised π electrons around its ring. Delocalisation stabilises the molecule and makes all six carbon-carbon bonds equivalent.
2. Its six π electrons satisfy Hückel’s 4n + 2 rule with n equal to one. Addition would disrupt this stabilised aromatic system.
3. The first nitration stage generates the nitronium ion, NO₂⁺, from concentrated nitric acid in the presence of concentrated sulphuric acid.
4. The ring attacks this electrophile to form an arenium ion, or sigma complex. One carbon becomes sp³ hybridised, temporarily interrupting aromatic delocalisation.
5. Loss of a proton restores aromaticity and gives nitrobenzene. Substitution is favoured because the final product retains the aromatic ring system.

Key takeaways

• Classify hydrocarbons by both connectivity and bond type; a molecular formula alone may not identify the family uniquely.
• Staggered ethane minimises torsional strain, while restricted double-bond rotation permits geometrical isomerism when suitable groups are attached.
• Track carbon atoms through preparations: soda-lime decarboxylation shortens a chain, while Wurtz coupling joins alkyl groups.
• Reaction conditions determine products, especially for controlled oxidation, alkene elimination, hydrogenation and peroxide-assisted hydrogen bromide addition.
• Ozonolysis reveals double-bond position by converting the two original double-bonded carbons into separate carbonyl centres.
• Terminal alkyne acidity depends on hydrogen attached directly to sp carbon; internal alkynes do not give terminal-acetylide tests.
• Benzene’s delocalised π electrons stabilise its ring, and electrophilic substitution restores aromaticity after a temporary interruption.
• Directing position and ring activation are distinct: halogens direct ortho/para while moderately deactivating electrophilic substitution.

Test yourself

Why does but-2-ene show geometrical isomerism while propene does not?

Each double-bonded carbon in but-2-ene has two different groups. In propene, one double-bonded carbon has two identical hydrogen atoms.

What different alkanes do soda-lime treatment and Kolbe electrolysis give from sodium ethanoate?

Soda-lime decarboxylation gives methane by removing the carboxyl carbon. Kolbe electrolysis gives ethane by coupling two methyl radicals.

Why is a methyl-radical combination a termination step?

Two methyl radicals combine to form ethane without generating a new radical, removing reactive chain carriers.

Which gas forms when propyne reacts with sodamide?

Ammonia forms as sodamide removes the terminal hydrogen and produces sodium propynide; this is distinct from hydrogen evolution with sodium.

What is the directing behaviour of chlorine already attached to benzene?

Chlorine directs further electrophilic substitution mainly to ortho and para positions while moderately deactivating the ring overall.

Why must “benzene reacts with chlorine” include conditions?

A Lewis acid promotes substitution to chlorobenzene, whereas ultraviolet light promotes addition to form hexachlorocyclohexane.