Hydrocarbons
Welcome to HSLC Guru. This page presents complete English-medium notes for Class 11 Chemistry Chapter 13 — Hydrocarbons, prepared strictly according to the ASSEB (Assam State School Education Board) syllabus. Hydrocarbons form the backbone of organic chemistry; petrol, diesel, LPG, polythene and even the natural rubber on your shoe sole are all hydrocarbons or their direct products. The chapter introduces the four major classes of hydrocarbons, their methods of preparation, the chemistry of their characteristic reactions, and the modern theoretical pictures (such as Newman projections and the Hückel rule) used to describe them.
Each section below moves from definition to preparation to reactions, with particular attention to mechanism (free-radical halogenation, electrophilic addition, electrophilic substitution) and to selectivity rules (Markovnikov, peroxide effect, ortho/para vs meta directors). Solved questions, MCQs, fill-in-the-blanks, true/false statements and a glossary table are added at the end so that you can revise quickly before the half-yearly and final examinations.
Summary
Classification. Hydrocarbons are compounds of carbon and hydrogen only. They are broadly divided into open-chain (acyclic) and closed-chain (cyclic) types. The open-chain group contains the saturated alkanes (general formula CₙH₂ₙ₊₂, e.g. CH₄, C₂H₆), the unsaturated alkenes carrying one C=C double bond (CₙH₂ₙ, e.g. C₂H₄), and the alkynes carrying one C≡C triple bond (CₙH₂ₙ₋₂, e.g. C₂H₂). The closed-chain group is further split into alicyclic hydrocarbons (cycloalkanes, cycloalkenes) and aromatic hydrocarbons such as benzene that show special stability arising from delocalised π electrons.
Alkanes. Alkanes are prepared by the Wurtz reaction (R–X + 2Na + X–R → R–R), Kolbe’s electrolysis of sodium salts of carboxylic acids, decarboxylation of sodium carboxylates with soda-lime, catalytic hydrogenation of alkenes/alkynes and reduction of alkyl halides. Being saturated, they undergo substitution rather than addition. Free-radical halogenation in the presence of UV light proceeds through initiation, propagation and termination steps to give haloalkanes. At very high temperatures (>773 K) larger alkanes break into smaller ones — a process called pyrolysis or cracking, used industrially to manufacture petrol from heavy oil fractions. Two carbon atoms in ethane can rotate about the C–C single bond to give different conformations; the eclipsed form (torsional angle 0°) is least stable and the staggered form (60°) is most stable, with skew/gauche conformations in between. Newman projections are the easiest way to visualise these conformers.
Alkenes and Alkynes. Alkenes are commonly prepared by acid-catalysed dehydration of alcohols and by dehydrohalogenation of alkyl halides with alcoholic KOH (β-elimination). Their main reactions are electrophilic addition: addition of H₂, X₂ and HX, the last following Markovnikov’s rule (H goes to the carbon already bearing more H atoms) under ionic conditions but reversed (anti-Markovnikov / Kharasch peroxide effect) when peroxides supply free radicals. Alkenes also undergo addition polymerisation (ethene → polythene) and oxidative cleavage by ozone (ozonolysis) to identify the position of the double bond. Alkynes are made by dehalogenation of vicinal/geminal dihalides and from CaC₂ (calcium carbide) plus water. Terminal alkynes have an acidic hydrogen (pKa ≈ 25) that reacts with sodium or ammoniacal AgNO₃/Cu₂Cl₂ to give acetylides — a useful test. Alkynes too undergo electrophilic addition, but more slowly than alkenes.
Aromatic hydrocarbons. Benzene (C₆H₆), discovered by Faraday and structurally explained by Kekulé through alternating single and double bonds, is the parent of the aromatic family. The modern picture treats benzene as a planar regular hexagon with six delocalised π electrons satisfying the Hückel rule for aromaticity. Benzene is prepared by decarboxylation of sodium benzoate, by trimerisation of acetylene at red-hot iron tube, and from phenol (Zn dust). Its characteristic reactions are electrophilic substitutions — nitration (HNO₃/H₂SO₄), halogenation (X₂/AlX₃), sulphonation (oleum) and Friedel–Crafts alkylation/acylation (RCl or RCOCl with anhydrous AlCl₃). Substituents on the ring control the position of further substitution: –OH, –NH₂, –OR, –CH₃, –X are ortho/para directors, while –NO₂, –CN, –COOH, –SO₃H are meta directors. Many polynuclear aromatic hydrocarbons (benzo[a]pyrene from coal-tar, tobacco smoke, charred meat) are carcinogenic and pose serious health hazards.
The Hückel rule states that a planar, fully-conjugated, monocyclic ring is aromatic if it contains:
$4n + 2 \pi \text{ electrons}$
where n = 0, 1, 2, … . For benzene, n = 1 and the count is six π electrons.
1-Mark Questions
Q1. Write the general formula of an alkane.
Answer: CₙH₂ₙ₊₂.
Q2. Name the simplest alkyne.
Answer: Ethyne (acetylene), HC≡CH.
Q3. What is the hybridisation of carbon in benzene?
Answer: sp² hybridisation.
Q4. Which conformation of ethane is most stable?
Answer: The staggered conformation.
Q5. State Markovnikov’s rule in one sentence.
Answer: When an unsymmetrical reagent HX adds to an unsymmetrical alkene, the hydrogen attaches to the doubly-bonded carbon that already carries more hydrogen atoms.
Q6. Which catalyst is used in Friedel–Crafts alkylation?
Answer: Anhydrous aluminium chloride (AlCl₃).
Q7. Name a carcinogenic polynuclear aromatic hydrocarbon found in tobacco smoke.
Answer: Benzo[a]pyrene.
Q8. Write the IUPAC name of CH₃–CH=CH–CH₃.
Answer: But-2-ene.
Q9. What is the product when ethyne is passed through a red-hot iron tube?
Answer: Benzene (three molecules of acetylene trimerise).
Q10. Why is the C–H bond in terminal alkynes acidic?
Answer: Because the carbon is sp-hybridised, has 50% s-character, holds the bonding electrons more tightly and stabilises the conjugate acetylide anion.
2 to 3-Mark Questions
Q1. Explain the Wurtz reaction with an example.
Answer: The Wurtz reaction couples two molecules of an alkyl halide in dry ether with metallic sodium to give a higher symmetrical alkane:
2 CH₃–Br + 2 Na → CH₃–CH₃ + 2 NaBr
It is useful for preparing alkanes containing an even number of carbons but is unsuitable when two different alkyl halides are used because a mixture of three products is obtained.
Q2. What is decarboxylation? Give an example.
Answer: Decarboxylation is the removal of a CO₂ molecule from a carboxylate. When sodium acetate is heated with soda-lime (NaOH + CaO):
CH₃COONa + NaOH (CaO, Δ) → CH₄ + Na₂CO₃
It is a standard laboratory method for methane and lower alkanes.
Q3. Distinguish between eclipsed and staggered conformations of ethane.
Answer: In the eclipsed conformation the front and back C–H bonds overlap exactly when looked at along the C–C axis (dihedral angle 0°), giving maximum torsional strain (≈12.5 kJ mol⁻¹). In the staggered conformation the C–H bonds are rotated 60° apart, the bonds are as far from each other as possible, and the molecule is at an energy minimum. Therefore at room temperature ethane spends most of its time in the staggered form.
Q4. What is the peroxide effect (Kharasch effect)?
Answer: When HBr (only HBr, not HCl or HI) is added to an unsymmetrical alkene in the presence of organic peroxides, the addition follows the anti-Markovnikov direction — Br attaches to the carbon bearing more hydrogen atoms. The reaction proceeds through a free-radical mechanism, where the more stable secondary radical determines the orientation, opposite to that of the ionic, Markovnikov addition.
Q5. Define ozonolysis and write its use.
Answer: Ozonolysis is the reaction of an alkene with ozone (O₃) to form an unstable ozonide that, on hydrolysis with Zn/H₂O, breaks down to give two carbonyl fragments. It is used to determine the position of the double bond in an unknown alkene; for example, but-2-ene gives two molecules of acetaldehyde.
Q6. State and explain Hückel’s rule of aromaticity.
Answer: A cyclic, planar, fully-conjugated molecule is aromatic if it contains (4n + 2) delocalised π electrons, where n = 0, 1, 2, … . Benzene with six π electrons (n = 1) satisfies the rule and shows aromatic stability. Cyclobutadiene (4 π) and cyclooctatetraene (8 π) do not, so they are non-aromatic / antiaromatic.
5 to 7-Mark Questions
Q1. Describe the free-radical mechanism of chlorination of methane.
Answer: Chlorination of methane in UV light is a chain reaction with three steps.
(i) Initiation. The Cl–Cl bond is homolytically cleaved by light: Cl₂ → 2 Cl·
(ii) Propagation. A chlorine radical abstracts a hydrogen from methane forming a methyl radical, which in turn attacks another Cl₂ molecule:
Cl· + CH₄ → ·CH₃ + HCl
·CH₃ + Cl₂ → CH₃Cl + Cl·
The Cl· regenerated keeps the chain going.
(iii) Termination. Two radicals combine and the chain stops:
Cl· + Cl· → Cl₂; ·CH₃ + Cl· → CH₃Cl; ·CH₃ + ·CH₃ → C₂H₆.
Excess chlorine and longer exposure converts CH₃Cl successively to CH₂Cl₂, CHCl₃ and CCl₄.
Q2. Discuss the methods of preparation and the four important reactions of ethene.
Answer: Preparation. (a) Dehydration of ethanol with conc. H₂SO₄ at 443 K: C₂H₅OH → C₂H₄ + H₂O. (b) Dehydrohalogenation of ethyl bromide with alc. KOH: C₂H₅Br + KOH (alc.) → C₂H₄ + KBr + H₂O. (c) Dehalogenation of 1,2-dibromoethane with Zn dust.
Reactions. (i) Hydrogenation: CH₂=CH₂ + H₂ (Ni) → C₂H₆. (ii) Addition of halogens: CH₂=CH₂ + Br₂ → CH₂Br–CH₂Br (decolorises bromine water — test for unsaturation). (iii) Addition of HX: CH₂=CH₂ + HBr → CH₃–CH₂Br. (iv) Polymerisation: n CH₂=CH₂ → (–CH₂–CH₂–)ₙ, the polymer polythene used in plastic bags. (v) Ozonolysis gives 2 HCHO; combustion gives CO₂ + H₂O with a luminous flame.
Q3. Why is benzene exceptionally stable? Discuss its structure and aromatic character.
Answer: Benzene C₆H₆ has a planar regular hexagonal ring of six sp²-hybridised carbon atoms. Each carbon contributes one electron to a p-orbital perpendicular to the ring; these six p-orbitals overlap sideways forming two continuous π-clouds (one above, one below) that delocalise the six electrons over all six carbons. Because all C–C bonds are equivalent (139 pm — between single 154 and double 134 pm) and the resonance energy is large (≈150 kJ mol⁻¹), benzene resists addition reactions and prefers electrophilic substitution.
Aromaticity demands: (i) the molecule must be cyclic, (ii) every ring atom must be sp²-hybridised so the ring is planar, (iii) the π-system must be fully conjugated, and (iv) the count of π electrons must equal (4n + 2). Benzene satisfies every condition with six π electrons and is therefore strongly aromatic.
Q4. Explain electrophilic substitution in benzene with the mechanism of nitration.
Answer: A mixture of conc. HNO₃ and conc. H₂SO₄ at about 330 K nitrates benzene to nitrobenzene.
Step 1 — Generation of electrophile. HNO₃ + 2 H₂SO₄ → NO₂⁺ + H₃O⁺ + 2 HSO₄⁻. The nitronium ion NO₂⁺ is the active electrophile.
Step 2 — Formation of σ-complex (arenium ion). The π-cloud of benzene attacks NO₂⁺, giving a non-aromatic carbocation in which the positive charge is delocalised over the ortho and para positions of the ring.
Step 3 — Loss of proton. HSO₄⁻ removes the proton from the carbon bearing –NO₂, restoring the aromatic ring and giving nitrobenzene C₆H₅NO₂. The overall reaction is C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O.
Q5. Explain directing effects of substituents on the benzene ring with examples.
Answer: A group already present on benzene controls the position taken by the next incoming group through its electronic effect (resonance + inductive).
(i) Ortho/para directing groups push electron density into the ring through resonance and so stabilise the σ-complex formed by attack at o- and p-positions. Examples: –OH, –NH₂, –OR, –NR₂, –CH₃ (alkyl, activating) and the halogens –F, –Cl, –Br, –I (deactivating but still o/p-directing because of lone-pair donation).
(ii) Meta directing groups withdraw electron density through resonance/induction; the σ-complex from o- or p-attack would place positive charge directly on the carbon bearing the group, which is destabilised. Substitution therefore occurs at the meta position. Examples: –NO₂, –CN, –COOH, –SO₃H, –CHO, –COR. All meta directors are deactivators.
For instance, nitration of toluene gives mainly o- and p-nitrotoluene, while nitration of nitrobenzene gives m-dinitrobenzene as the major product.
Multiple Choice Questions (MCQs)
Q1. The general formula of an alkyne is —
(a) CₙH₂ₙ₊₂ (b) CₙH₂ₙ (c) CₙH₂ₙ₋₂ (d) CₙH₂ₙ₋₆
Answer: (c) CₙH₂ₙ₋₂.
Q2. Wurtz reaction needs —
(a) Zn / HCl (b) Na / dry ether (c) Pd / H₂ (d) AlCl₃
Answer: (b) Na / dry ether.
Q3. The most stable conformation of n-butane is —
(a) eclipsed (b) gauche (c) anti (d) skew
Answer: (c) anti.
Q4. Markovnikov’s rule applies to —
(a) CH₂=CH₂ + HBr (b) CH₃–CH=CH₂ + HBr (c) C₂H₂ + 2H₂ (d) C₆H₆ + Cl₂
Answer: (b) CH₃–CH=CH₂ + HBr (an unsymmetrical alkene).
Q5. Which of the following is the strongest acid?
(a) ethane (b) ethene (c) ethyne (d) methane
Answer: (c) ethyne (sp-C–H is most acidic).
Q6. The number of π electrons in benzene is —
(a) 2 (b) 4 (c) 6 (d) 8
Answer: (c) 6.
Q7. Friedel–Crafts acylation introduces which group into benzene?
(a) –NO₂ (b) –SO₃H (c) –COR (d) –Cl
Answer: (c) –COR.
Q8. Which substituent is a meta director?
(a) –NH₂ (b) –OH (c) –CH₃ (d) –NO₂
Answer: (d) –NO₂.
Q9. Ozonolysis of but-2-ene gives —
(a) 2 HCHO (b) HCHO + CH₃CHO (c) 2 CH₃CHO (d) CH₃COCH₃
Answer: (c) 2 CH₃CHO.
Q10. A well-known carcinogenic PAH is —
(a) naphthalene (b) benzo[a]pyrene (c) toluene (d) cumene
Answer: (b) benzo[a]pyrene.
Fill in the Blanks
Q1. Alkanes are also called __________ hydrocarbons.
Answer: saturated (paraffin).
Q2. Electrolysis of sodium acetate solution gives __________ at the anode.
Answer: ethane.
Q3. The C–C bond length in benzene is __________ pm.
Answer: 139.
Q4. Anti-Markovnikov addition of HBr requires the presence of __________.
Answer: organic peroxides.
Q5. Pyrolysis of higher alkanes is also called __________.
Answer: cracking.
True or False
Q1. Methane gives an addition reaction with chlorine.
Answer: False — alkanes give substitution, not addition.
Q2. Ethene decolorises bromine water.
Answer: True.
Q3. Benzene is a planar molecule with delocalised π electrons.
Answer: True.
Q4. –NO₂ is an ortho/para directing group.
Answer: False — it is a meta director.
Q5. Terminal alkynes form silver acetylides with ammoniacal AgNO₃.
Answer: True.
Glossary
| Term | Meaning |
|---|---|
| Hydrocarbon | Compound made up of carbon and hydrogen only. |
| Alkane | Saturated open-chain hydrocarbon, CₙH₂ₙ₊₂. |
| Alkene | Open-chain hydrocarbon with one C=C double bond, CₙH₂ₙ. |
| Alkyne | Open-chain hydrocarbon with one C≡C triple bond, CₙH₂ₙ₋₂. |
| Wurtz reaction | Coupling of two alkyl halides with Na/dry ether to give a higher alkane. |
| Kolbe’s electrolysis | Electrolysis of a sodium carboxylate giving an alkane at the anode. |
| Decarboxylation | Loss of CO₂ from a carboxylate, e.g. with soda-lime. |
| Pyrolysis (cracking) | Thermal breakdown of higher alkanes into smaller alkanes/alkenes. |
| Conformation | Different spatial arrangement obtained by rotation about a single bond. |
| Newman projection | Diagram showing the front and back carbons of a C–C bond as a dot inside a circle. |
| Markovnikov’s rule | In ionic addition of HX to an unsymmetrical alkene, H goes to the carbon with more H. |
| Peroxide effect | Anti-Markovnikov addition of HBr to alkenes in presence of peroxides (free-radical). |
| Polymerisation | Joining of many small monomers into a long-chain polymer molecule. |
| Ozonolysis | Cleavage of an alkene by O₃ followed by Zn/H₂O to give two carbonyl compounds. |
| Acetylide | Salt formed by replacement of the acidic H of a terminal alkyne by a metal. |
| Aromaticity | Special stability of cyclic, planar, conjugated systems with (4n+2) π electrons. |
| Hückel rule | (4n + 2) π-electron rule that decides whether a ring is aromatic. |
| Electrophilic substitution | Replacement of an –H of an aromatic ring by an electrophile (e.g., NO₂⁺). |
| Friedel–Crafts reaction | Alkylation/acylation of benzene using RX or RCOCl with anhyd. AlCl₃. |
| o/p-director | Substituent that directs the next group to ortho and para positions (e.g., –OH, –CH₃). |
| m-director | Substituent that directs the next group to the meta position (e.g., –NO₂, –COOH). |
| PAH | Polynuclear aromatic hydrocarbon; many are carcinogenic (e.g., benzo[a]pyrene). |