Hydrocarbons

A hydrocarbon is a compound of carbon and hydrogen only. Our fuels are hydrocarbons: LPG, CNG, petrol, diesel and kerosene come from petroleum and natural gas, and coal gives coal gas and coal tar.

NCERT classifies hydrocarbons by the carbon–carbon bonds present. Saturated hydrocarbons have only C–C single bonds: open-chain alkanes (C_nH_2n+2) and ring-shaped cycloalkanes (C_nH_2n). Unsaturated hydrocarbons contain at least one C=C double bond (alkenes, C_nH_2n) or C≡C triple bond (alkynes, C_nH_2n−2). Aromatic hydrocarbons (arenes) contain a benzene ring or a similar delocalised ring; benzene and its homologues are C_nH_2n−6. A second division is by chain shape: acyclic or aliphatic (open chain) versus cyclic, and cyclic compounds are either alicyclic (a ring but not aromatic, like cyclohexane) or aromatic.

Notice that cyclohexane and hex-1-ene share the formula C₆H₁₂: a ring costs the same two hydrogens as a double bond. The one-mark question 'to which class does cyclohexane belong' is set because of this trap; it is a saturated, alicyclic hydrocarbon.

Alkanes are saturated open-chain hydrocarbons, C_nH_2n+2. Every carbon is sp³ hybridised and tetrahedral with bond angles near 109.5°; the C–H bond in methane is 112 pm and the C–C bond in ethane 154 pm. These strong, non-polar σ bonds make alkanes unreactive under ordinary conditions, hence the old name paraffins (little affinity).

Names follow the IUPAC rules of Chapter 8: longest continuous chain, lowest locants, substituents as prefixes in alphabetical order. Carbons are classified by the number of other carbons attached: primary (1°) one, secondary (2°) two, tertiary (3°) three, quaternary (4°) four; a hydrogen takes the label of its carbon, so a 3° carbon carries one 3° hydrogen and a 4° carbon carries none.

Chain isomerism starts at C₄: butane has two isomers (butane and 2-methylpropane), C₅H₁₂ three (pentane, 2-methylbutane, 2,2-dimethylpropane), C₆H₁₄ five (hexane, 2-methylpentane, 3-methylpentane, 2,2-dimethylbutane, 2,3-dimethylbutane). To list them without missing any, draw the straight chain, then shorten it by one carbon and place a methyl at each distinct position, then shorten again and place two methyls.

NCERT gives three routes. Write the equation, the reagent and the condition for each; a marking scheme rewards all three.

1. From unsaturated hydrocarbons. Dihydrogen adds to an alkene or alkyne over finely divided Pt, Pd or Ni: CH₂=CH₂ + H₂ → CH₃–CH₃; HC≡CH + 2H₂ → CH₃–CH₃.

2. From alkyl halides. (a) Reduction with zinc and dilute HCl: CH₃Cl + H₂ → CH₄ + HCl (Zn, H⁺). (b) Wurtz reaction: an alkyl halide with sodium in dry ether gives an alkane with twice the carbons, 2CH₃Br + 2Na → CH₃–CH₃ + 2NaBr. It suits symmetrical, even-carbon alkanes only. An odd-carbon alkane would need two different halides, and then all three coupling products form (R–R, R–R′, R′–R′) with close boiling points that are hard to separate (exercise 9.25).

3. From carboxylic acids. (a) Decarboxylation: the sodium salt heated with soda lime (NaOH + CaO) gives an alkane with one carbon less: CH₃COONa + NaOH → CH₄ + Na₂CO₃ (CaO, heat). (b) Kolbe's electrolytic method: electrolysis of an aqueous solution of the sodium or potassium salt gives an even-carbon alkane at the anode: 2CH₃COO⁻Na⁺ + 2H₂O → CH₃–CH₃ + 2CO₂ + H₂ + 2NaOH. The mechanism is free-radical: acetate ions lose an electron at the anode to give CH₃COO• radicals, which lose CO₂ to give CH₃• radicals, which dimerise to ethane; at the cathode 2H₂O + 2e⁻ → H₂ + 2OH⁻. Methane cannot be made by either Wurtz or Kolbe, since both join two alkyl groups.

Physical. Alkanes are colourless and odourless; at 298 K, C₁–C₄ are gases, C₅–C₁₇ liquids and C₁₈ onwards solids. Being almost non-polar they dissolve in non-polar solvents, not in water. Boiling point rises with molar mass (about 20–30 K per CH₂) as van der Waals forces grow, and falls with branching because a branched isomer is more compact, with a smaller surface area and weaker forces: pentane 309 K, 2-methylbutane 301 K, 2,2-dimethylpropane 282.5 K.

(i) Halogenation (substitution). In diffused sunlight or at 573–773 K chlorine replaces hydrogens one at a time, CH₄ → CH₃Cl → CH₂Cl₂ → CHCl₃ → CCl₄, releasing HCl at each step. Reactivity: F₂ > Cl₂ > Br₂ > I₂; fluorination is explosive, iodination is reversible and needs an oxidising agent (HIO₃ or HNO₃) to remove HI. The mechanism is a free-radical chain. Initiation: homolysis in light, Cl₂ → 2Cl•. Propagation: Cl• + CH₄ → CH₃• + HCl; CH₃• + Cl₂ → CH₃Cl + Cl•, and the regenerated Cl• continues the chain. Termination: Cl• + Cl• → Cl₂; CH₃• + Cl• → CH₃Cl; CH₃• + CH₃• → C₂H₆, which is why a little ethane is always found (exercise 9.1). In the dark no radicals form, so there is no reaction.

(ii) Combustion. CH₄ + 2O₂ → CO₂ + 2H₂O, Δ_cH^⊖ = −890 kJ mol⁻¹; with insufficient air, incomplete combustion gives carbon black, CH₄ + O₂ → C(s) + 2H₂O. (iii) Controlled oxidation. 2CH₄ + O₂ → 2CH₃OH (Cu, 523 K, 100 atm); CH₄ + O₂ → HCHO + H₂O (Mo₂O₃); 2CH₃CH₃ + 3O₂ → 2CH₃COOH + 2H₂O ((CH₃COO)₂Mn); a tertiary C–H is oxidised by KMnO₄ to a tertiary alcohol, (CH₃)₃CH → (CH₃)₃COH. (iv) Isomerisation. n-Hexane with anhydrous AlCl₃ and HCl gives 2-methylpentane and 3-methylpentane. (v) Aromatisation. n-Hexane over V₂O₅, Cr₂O₃, Mo₂O₃ or Al₂O₃ at 773 K and 10–20 atm gives benzene (n-heptane gives toluene). (vi) Reaction with steam. CH₄ + H₂O → CO + 3H₂ (Ni, 1273 K), the industrial source of dihydrogen. (vii) Pyrolysis (cracking). Higher alkanes heated strongly break by C–C cleavage into smaller alkanes and alkenes; NCERT's example is dodecane at 973 K over Pt, Pd or Ni giving heptane and pentene, or hexane and hexene.

The C–C σ bond in ethane has cylindrical symmetry, so one CH₃ group can rotate about it relative to the other. Rotation is not completely free: as the C–H bonds on the two carbons pass one another their electron clouds repel weakly, an effect called torsional strain. The spatial arrangements produced by rotation are conformations (conformers, rotamers). Because the barrier is tiny they interconvert millions of times a second and cannot be isolated at room temperature.

Ethane has two extreme conformations. In the eclipsed form the hydrogens on the two carbons are as close as possible (dihedral angle 0°); in the staggered form they are as far apart as possible (dihedral angle 60°); anything between is a skew conformation. Staggered ethane is the most stable because torsional strain is least; the eclipsed form is least stable. The difference is only about 12.5 kJ mol⁻¹.

Two projections are used. In the sawhorse projection the C–C bond is a longer tilted line, front carbon lower left, back carbon upper right, with three bonds drawn from each end. In the Newman projection you look along the C–C axis: the front carbon is a point with its three bonds at 120°, the back carbon is a circle with its bonds drawn from the rim. In the eclipsed projection the back bonds sit directly behind the front ones (drawn slightly offset); in the staggered projection they bisect the angles between the front bonds.

Alkenes, C_nH_2n, contain one C=C double bond; the old name olefins (oil-forming) comes from ethene giving oily 1,2-dichloroethane with chlorine. The doubly bonded carbons are sp² hybridised: three σ bonds in a plane at about 120°, and the unhybridised p orbitals overlap sideways to give a π bond above and below that plane. The C=C bond (134 pm, 681 kJ mol⁻¹) is shorter and stronger than C–C (154 pm, 348 kJ mol⁻¹) but not twice as strong, because the π bond formed by sideways overlap is weaker than the σ bond. That exposed π cloud is why alkenes add electrophiles, and the sideways overlap stops rotation about C=C, which is the origin of geometrical isomerism.

Nomenclature. Replace -ane by -ene and give the double bond the lowest locant: but-1-ene, but-2-ene, 2-methylpropene, buta-1,3-diene. CH₂=CH– is vinyl, CH₂=CH–CH₂– is allyl.

Structural isomerism. C₄H₈ has three structural isomers: but-1-ene and but-2-ene (position isomers) and 2-methylpropene (chain isomer).

Geometrical (cis–trans) isomerism. When each doubly bonded carbon carries two different groups, restricted rotation gives two isomers: cis (like groups on the same side) and trans (opposite sides). But-2-ene shows it; but-1-ene and 2-methylpropene do not, since one carbon carries two identical groups. The cis form is more polar (cis-but-2-ene 0.33 D, trans almost zero; cis-1,2-dichloroethene 1.90 D, trans zero), so the cis isomer has the higher boiling point, while the more symmetrical trans isomer packs better and has the higher melting point.

1. From alkynes. Partial hydrogenation with a calculated amount of H₂ over palladised charcoal partially deactivated with sulphur compounds or quinoline (Lindlar's catalyst) gives the cis-alkene; reduction with sodium in liquid ammonia gives the trans-alkene. But-2-yne gives cis- or trans-but-2-ene accordingly.

2. From alkyl halides (dehydrohalogenation). Heating with alcoholic KOH removes hydrogen from the β-carbon and halogen from the α-carbon, a β-elimination: CH₃CH₂Br + KOH(alc) → CH₂=CH₂ + KBr + H₂O. Rate order for the halogen is I > Br > Cl and for the alkyl group tertiary > secondary > primary. When two alkenes can form, the more substituted one is the major product (Saytzeff rule): 2-bromobutane gives mainly but-2-ene.

3. From vicinal dihalides (dehalogenation). Halogens on adjacent carbons are removed by zinc in methanol or ethanol: CH₂Br–CH₂Br + Zn → CH₂=CH₂ + ZnBr₂.

4. From alcohols (acidic dehydration). Concentrated H₂SO₄ at 443 K: CH₃CH₂OH → CH₂=CH₂ + H₂O, also a β-elimination.

If asked to define β-elimination, write: 'a reaction in which two atoms or groups are removed from adjacent carbon atoms, the α- and β-carbons, to form a multiple bond'.

Physical. The first three members are gases, the next fourteen liquids, the higher ones solids; ethene is a colourless gas with a faint sweet smell. Alkenes are insoluble in water, soluble in non-polar solvents, with boiling points rising with molar mass.

Chemical. The loosely held π electrons make alkenes give electrophilic addition reactions.

(i) H₂ (Pt, Pd or Ni) gives the alkane. (ii) Halogens: Br₂ in CCl₄ gives the vicinal dibromide, CH₂=CH₂ + Br₂ → BrCH₂–CH₂Br, through a cyclic halonium ion; the reddish-brown colour disappears, the standard test for unsaturation. Fluorine reacts too violently and iodine does not add under normal conditions.

(iii) Hydrogen halides (HI > HBr > HCl). A symmetrical alkene gives one product: CH₂=CH₂ + HBr → CH₃CH₂Br. For an unsymmetrical alkene Markovnikov's rule applies: the negative part of the addendum attaches to the carbon with fewer hydrogens, so propene with HBr gives 2-bromopropane. Mechanism: H⁺ adds first to the CH₂ carbon, giving the secondary carbocation CH₃–CH⁺–CH₃, more stable (hyperconjugation, +I effect) than the primary CH₃CH₂CH₂⁺; then Br⁻ attacks it. Peroxide (Kharasch) effect: with benzoyl peroxide, HBr adds the other way and propene gives 1-bromopropane, by a free-radical chain. The peroxide homolyses to C₆H₅COO•, which loses CO₂ to give C₆H₅•; C₆H₅• + HBr → C₆H₆ + Br•; Br• adds to the terminal CH₂ to give the more stable secondary radical CH₃–CH•–CH₂Br, which abstracts H from HBr to give CH₃CH₂CH₂Br and a new Br•. Only HBr shows the effect: H–Cl (431 kJ mol⁻¹) is too strong for the phenyl radical to break, and though H–I breaks easily the iodine radical is too unreactive to add and simply recombines to I₂.

(iv) H₂SO₄: cold concentrated acid adds by Markovnikov's rule, CH₃CH=CH₂ → CH₃CH(OSO₃H)CH₃. (v) Water with a few drops of concentrated H₂SO₄: Markovnikov hydration, propene → propan-2-ol.

(vi) Oxidation. Cold, dilute, alkaline KMnO₄ (Baeyer's reagent) gives a vicinal diol, CH₂=CH₂ → HOCH₂–CH₂OH (ethane-1,2-diol), and its pink colour is discharged, a second test for unsaturation. Hot acidic KMnO₄ or K₂Cr₂O₇ cleaves the double bond: propene gives ethanoic acid + CO₂ + H₂O; 2-methylpropene gives propanone + CO₂ + H₂O.

(vii) Ozonolysis. Ozone gives an ozonide, which Zn and water cleave to aldehydes and ketones: CH₂=CH₂ → 2HCHO; CH₃CH=CH₂ → CH₃CHO + HCHO. Each doubly bonded carbon becomes a C=O carbon, so the products locate the double bond; zinc destroys the H₂O₂ formed so the aldehydes are not oxidised.

(viii) Polymerisation. At high temperature and pressure with a catalyst, n CH₂=CH₂ → –(CH₂–CH₂)_n– (polythene); propene gives polypropene; used in bags, pipes, insulation and milk crates.

Alkynes, C_nH_2n−2, contain a C≡C triple bond; the first member is ethyne (acetylene), burnt in oxy-acetylene welding torches. Names replace -ane with -yne: but-1-yne and but-2-yne are position isomers; C₅H₈ has three isomers, pent-1-yne, pent-2-yne and 3-methylbut-1-yne. A terminal alkyne has H–C≡C– at the chain end; but-2-yne is non-terminal, and only terminal alkynes have an acidic hydrogen.

Structure. Each triply bonded carbon is sp hybridised: two sp orbitals at 180°, one bonding to H (C–H 106 pm) and one to the other carbon, so the molecule is linear. The two remaining p orbitals on each carbon, perpendicular to each other and to the axis, overlap sideways to give two π bonds; the triple bond is one σ plus two π, and the four π electrons form a cylindrical cloud around the C–C axis. C≡C is 120 pm with bond enthalpy 823 kJ mol⁻¹, shorter and stronger than C=C (134 pm, 681) and C–C (154 pm, 348).

Preparation. (1) From calcium carbide, the industrial route: CaCO₃ → CaO + CO₂ (heat); CaO + 3C → CaC₂ + CO; CaC₂ + 2H₂O → Ca(OH)₂ + C₂H₂. (2) From vicinal dihalides by double dehydrohalogenation: alcoholic KOH removes the first HX, and the harder second one needs sodamide. CH₂Br–CH₂Br + KOH(alc) → CH₂=CHBr (bromoethene) + KBr + H₂O; CH₂=CHBr + NaNH₂ → HC≡CH + NaBr + NH₃. Since the vicinal dihalide comes from adding Br₂ to an alkene, this is also the ethene-to-ethyne route used in conversion questions.

Physical. The first three alkynes are gases, the next eight liquids, the rest solids; all colourless, and pure ethyne is odourless. They are weakly polar, lighter than water, insoluble in water but soluble in organic solvents; melting point, boiling point and density rise with molar mass.

A. Acidic character. An sp carbon has 50% s-character, so it is more electronegative than sp² (33.3%) or sp³ (25%) carbon and holds the C–H electrons closer; the hydrogen of a terminal alkyne is weakly acidic. Ethyne reacts with sodium: HC≡CH + Na → HC≡C⁻Na⁺ (monosodium ethynide) + ½H₂, and then to disodium ethynide. Acidity order: HC≡CH > CH₂=CH₂ > CH₃–CH₃, following s-character. Alkynes are still far weaker acids than water, so they do not react with NaOH or Na₂CO₃, only with very strong bases (NaNH₂) or sodium metal; but-2-yne, with no hydrogen on an sp carbon, shows no acidity. School papers also use the classical tests: a terminal alkyne gives a white precipitate with ammoniacal AgNO₃ (silver acetylide) and a red precipitate with ammoniacal Cu₂Cl₂; but-2-yne gives neither.

B. Addition reactions, one π bond at a time. H₂: HC≡CH → CH₂=CH₂ → CH₃CH₃ (Pt/Pd/Ni). Halogens: HC≡CH + Br₂ → CHBr=CHBr (1,2-dibromoethene) → CHBr₂–CHBr₂ (1,1,2,2-tetrabromoethane); bromine water is decolourised. Hydrogen halides: HC≡CH + HBr → CH₂=CHBr → CH₃CHBr₂ (1,1-dibromoethane); the second addition follows Markovnikov's rule, so both bromines go to the same carbon. Water: with 40% dilute H₂SO₄ and 1% HgSO₄ at 333 K ethyne gives vinyl alcohol, CH₂=CH–OH, an unstable enol that tautomerises to ethanal, CH₃CHO. Propyne gives propanone by Markovnikov addition; every alkyne except ethyne gives a ketone.

C. Polymerisation. Linear: ethyne gives polyethyne (polyacetylene), –(CH=CH–CH=CH)_n–, a conjugated polymer that conducts electricity when doped and is used as a thin-film electrode in batteries. Cyclic: through a red-hot iron tube at 873 K, 3HC≡CH → C₆H₆, the bridge from aliphatic to aromatic chemistry; propyne under the same conditions gives 1,3,5-trimethylbenzene (mesitylene).

Aromatic hydrocarbons, or arenes, were named for their pleasant smell. Most contain a benzene ring and are called benzenoids; those with a delocalised ring other than benzene are non-benzenoids. Substituted benzenes are named as derivatives: methylbenzene (toluene), ethylbenzene. A monosubstituted benzene has only one isomer because all six positions are equivalent; a disubstituted benzene has three, 1,2 (ortho), 1,3 (meta) and 1,4 (para), as in the three xylenes. C₆H₅– is phenyl and C₆H₅CH₂– is benzyl.

Kekulé structure (1865): a planar six-membered ring with alternate single and double bonds and one hydrogen on each carbon. It could not explain why only one 1,2-disubstituted isomer exists (the structure predicts two, with the substituents across a C–C or a C=C) or why benzene refuses the addition reactions three C=C bonds should give. Kekulé suggested the two structures oscillate rapidly.

Resonance: the two Kekulé structures are canonical forms of one resonance hybrid. All six C–C bonds are identical, 139 pm, between a single (154 pm) and a double (134 pm) bond, as X-ray diffraction confirms. Resonance stabilisation (the resonance energy) makes benzene unusually stable and unwilling to lose that stability by addition.

Orbital picture: each carbon is sp² hybridised and forms three σ bonds (two to carbons, one to hydrogen) in one plane at 120°, giving a regular planar hexagon. The six unhybridised p orbitals stand perpendicular to the ring and overlap sideways with both neighbours, so the six π electrons are delocalised in doughnut-shaped clouds above and below the plane; there is no localised C=C bond.

Aromaticity (Hückel's rule): a compound is aromatic if it is (i) planar, (ii) has complete delocalisation of π electrons in the ring, and (iii) contains (4n+2) π electrons, n = 0, 1, 2, .... Benzene has 6 (n = 1), naphthalene 10, anthracene 14. The cyclopentadienyl anion and the tropylium (cycloheptatrienyl) cation have 6 π electrons and are aromatic; cyclooctatetraene has 8 and is non-planar, so it is not; a ring containing an sp³ carbon breaks delocalisation and cannot be aromatic. Exercise 9.12 asks exactly this: test each structure against all three conditions in order.

Preparation. (i) Cyclic polymerisation of ethyne in a red-hot iron tube at 873 K. (ii) Decarboxylation of sodium benzoate with soda lime: C₆H₅COONa + NaOH → C₆H₆ + Na₂CO₃ (CaO, heat). (iii) Reduction of phenol vapour over heated zinc dust: C₆H₅OH + Zn → C₆H₆ + ZnO.

Physical properties. Arenes are non-polar, colourless liquids or solids with a characteristic aroma, immiscible with water, miscible with organic solvents, and burn with a sooty flame because of their high carbon content.

Electrophilic substitution. The π cloud makes benzene electron-rich, so electrophiles attack it; but instead of adding (which would destroy aromaticity) benzene replaces a hydrogen and keeps the ring. Nitration: C₆H₆ + conc. HNO₃ → C₆H₅NO₂ + H₂O (conc. H₂SO₄, 323–333 K). Halogenation: C₆H₆ + Cl₂ → C₆H₅Cl + HCl (anhydrous FeCl₃ or AlCl₃, no sunlight). Sulphonation: C₆H₆ + fuming H₂SO₄ (containing SO₃) → C₆H₅SO₃H + H₂O. Friedel–Crafts alkylation: C₆H₆ + CH₃Cl → C₆H₅CH₃ + HCl (anhydrous AlCl₃). Friedel–Crafts acylation: C₆H₆ + CH₃COCl → C₆H₅COCH₃ (acetophenone) + HCl (anhydrous AlCl₃; acetic anhydride also works). Excess reagent gives polysubstitution: benzene with excess conc. HNO₃/H₂SO₄ gives 1,3,5-trinitrobenzene.

Mechanism (three steps; the five-mark derivation). Step 1, generation of the electrophile: Cl₂ + AlCl₃ → Cl⁺ + AlCl₄⁻; CH₃Cl + AlCl₃ → CH₃⁺ + AlCl₄⁻; CH₃COCl + AlCl₃ → CH₃CO⁺ (acylium ion) + AlCl₄⁻; in nitration, HNO₃ + 2H₂SO₄ → NO₂⁺ (nitronium ion) + H₃O⁺ + 2HSO₄⁻, where H₂SO₄ acts as the acid and HNO₃ as the base. Step 2, formation of the carbocation: E⁺ attacks the π cloud and bonds to one ring carbon, which becomes sp³. This arenium ion (σ-complex) is resonance-stabilised over three structures with the positive charge at the ortho and para carbons, but it is not aromatic, so the step is slow and rate-determining. Step 3, loss of a proton: AlCl₄⁻ or HSO₄⁻ removes H⁺ from the sp³ carbon, aromaticity is restored, and the substituted benzene forms quickly.

Addition reactions need drastic conditions: C₆H₆ + 3H₂ → cyclohexane (Ni, high temperature); C₆H₆ + 3Cl₂ → C₆H₆Cl₆ in ultraviolet light, benzene hexachloride (BHC, gammaxene), an insecticide. Combustion: C₆H₆ + 15/2 O₂ → 6CO₂ + 3H₂O; in general C_nH_2n−6 + (3n−3)/2 O₂ → nCO₂ + (n−3)H₂O.

A group already on the ring decides where the next electrophile goes (ortho and para, or meta) and how fast the ring reacts compared with benzene (activating or deactivating). NCERT explains both through the electron density at the ring positions.

Ortho- and para-directing groups: –OH, –NH₂, –NHR, –NHCOCH₃, –OCH₃, –CH₃, –C₂H₅ and the halogens. In phenol the oxygen lone pair is delocalised into the ring (+R effect); the resonance structures put a negative charge at the ortho and para carbons and never at meta. Electron density is highest there, so the electrophile attacks ortho/para, and because the ring is richer than benzene the group is activating. –CH₃ does the same through hyperconjugation and its +I effect. Halogens are the exception: their −I effect outweighs their +R effect, so chlorobenzene reacts more slowly than benzene (deactivating), but the +R effect still places the electrons at ortho and para, so substitution still goes ortho/para.

Meta-directing groups: –NO₂, –CN, –CHO, –COR, –COOH, –COOR, –SO₃H. In nitrobenzene the nitro group withdraws electrons inductively and by resonance (−R effect); the resonance structures place a positive charge at the ortho and para carbons, so those positions are poorest in electrons and the electrophile attacks meta. The whole ring is depleted, so the group is deactivating. Hence toluene is nitrated more easily than benzene and m-dinitrobenzene far less easily (exercise 9.23), and reactivity towards E⁺ falls in the order chlorobenzene > p-nitrochlorobenzene > 2,4-dinitrochlorobenzene (exercise 9.22).

These rules decide the order of steps in every conversion. For p-nitrobromobenzene, brominate first (Br directs para); for m-nitrochlorobenzene, nitrate first (NO₂ directs meta), then chlorinate.

Benzene and the polynuclear (fused-ring) aromatic hydrocarbons are toxic and carcinogenic, that is, cancer-causing. Polynuclear hydrocarbons such as 1,2-benzanthracene, 3-methylcholanthrene, 1,2-benzpyrene and 1,2,5,6-dibenzanthracene contain several fused benzene rings. They form on incomplete combustion of organic material, carbon-containing fuels and tobacco; coal, petroleum and cigarette smoke are the usual sources.

NCERT's explanation is short and is what a two-mark answer needs: when these hydrocarbons enter the body they undergo various biochemical reactions, and the products damage DNA, causing cancer. Benzene itself is linked to leukaemia, which is why toluene has replaced it as a laboratory solvent.

The exam angle is small but reliable: name two carcinogenic hydrocarbons, state where they come from, and say why they are harmful. Do not mix this up with the ozone layer or the greenhouse effect, which belong to environmental chemistry.