NCERT Plus One (Class 11) Chemistry • Chapter 9

Hydrocarbons

Q: State Markovnikov's rule. What is the peroxide (Kharasch) effect? Why is the peroxide effect NOT observed in the addition of HCl and HI to unsymmetrical alkenes?

Markovnikov's Rule: When an unsymmetrical reagent adds to an unsymmetrical alkene, the negative part of the addendum attaches to the doubly bonded carbon atom having fewer hydrogen atoms (e.g., Propene + HBr → 2-bromopropane). Peroxide / Kharasch Effect: In the presence of benzoyl peroxide, HBr adds to propene anti-Markovnikov via a free radical chain mechanism to give 1-bromopropane . Why not HCl / HI? The H–Cl bond ( 430.5 kJ mol −1 ) is too strong to be cleaved homolytically by free radicals, whereas the H–I bond ( 296.8 kJ mol −1 ) is weak and iodine free radicals combine to form I 2 molecules instead of adding to the double bond.

Q: How will you convert: (a) Sodium ethanoate to methane, and (b) Ethyne to benzene?

(a) Soda-lime decarboxylation: Heat sodium ethanoate with soda lime (NaOH + CaO): CH 3 COONa + NaOH —[CaO, Δ]→ CH 4 + Na 2 CO 3 . (b) Cyclic polymerisation: Pass ethyne gas through a red-hot iron tube at 873 K : 3 HC≡CH —[Red-hot Fe, 873 K]→ C 6 H 6 (Benzene) .

Q: State the three necessary conditions for a compound to be aromatic according to Hückel's rule. Explain why cyclopentadiene is non-aromatic whereas cyclopentadienyl anion is aromatic.

Hückel's Conditions: (1) Planarity, (2) Complete delocalisation of π electrons in the ring, and (3) Presence of (4n + 2) π electrons (n = 0, 1, 2...). Cyclopentadiene (C 5 H 6 ): Contains an sp 3 hybridised –CH 2 – carbon which interrupts complete π-delocalisation (and has only 4π electrons) → Non-aromatic . Cyclopentadienyl anion (C 5 H 5 − ): All 5 carbons are sp 2 planar, fully conjugated, with 4π + 2π (lone pair) = 6π electrons (n = 1 in 4n+2) → Aromatic .

Q: Explain the Wurtz reaction with a chemical equation. Why is it NOT preferred for preparing alkanes with an odd number of carbon atoms?

Wurtz Reaction: Alkyl halides react with sodium metal in dry ether to form higher symmetrical alkanes: 2 C 2 H 5 Br + 2 Na —[dry ether]→ CH 3 CH 2 CH 2 CH 3 (n-Butane) + 2 NaBr. Odd-carbon limitation: Preparing an odd-carbon alkane (like propane) requires mixing two different alkyl halides (CH 3 Br + C 2 H 5 Br), which produces a mixture of three alkanes (ethane, propane, and butane) with close boiling points that are difficult to separate.

Q: Draw the cis and trans isomers of but-2-ene. Which isomer has a higher dipole moment and why? How can cis-but-2-ene be prepared from but-2-yne?

In cis-but-2-ene , both –CH 3 groups lie on the same side of C=C; in trans-but-2-ene , they lie on opposite sides. Dipole Moment: cis-but-2-ene is polar (μ = 0.33 D) because the two C–CH 3 bond dipoles reinforce, whereas in trans-but-2-ene (μ = 0 D) the opposite dipoles cancel out. Preparation from but-2-yne: Partial hydrogenation using H 2 over Lindlar's catalyst (Pd/C deactivated with S-compounds/quinoline) gives cis-but-2-ene.

Q: Arrange ethyne, ethene, and ethane in decreasing order of acidic strength. Give the reason and write one reaction proving the acidic nature of ethyne.

Order: HC≡CH > CH 2 =CH 2 > CH 3 –CH 3 . Reason: Carbon in ethyne is sp hybridised (50% s-character) compared to sp 2 (33.3%) in ethene and sp 3 (25%) in ethane. Greater s-character increases carbon electronegativity, pulling C–H electrons closer to C so H + is lost more easily. Reaction: HC≡CH + Na → HC≡C − Na + + ½ H 2 ↑ (Sodium ethynide + H 2 gas).

Alkanes & Conformations, Alkenes, Markownikoff vs Peroxide Effect, Ozonolysis, Alkynes, Benzene Resonance & Huckel's Aromaticity

📄 Official NCERT Textbook PDF

Topic 9.1 – 9.2.3: Classification, Alkane Isomerism, Preparation & Free Radical Halogenation

Every morning when a kitchen stove lights up with LPG (liquefied petroleum gas) or an auto-rickshaw runs cleanly on CNG (compressed natural gas), you are witnessing hydrocarbons in action—compounds built purely from Carbon and Hydrogen! Even plastics like polythene, petrol, kerosene, paint solvents, and life-saving drugs trace back to these molecular skeletons. Why did early chemists name alkanes paraffins (Latin: parum affinis, meaning little affinity)? Because single C–C and C–H sigma bonds sit calm and unreactive toward ordinary acids or bases—until UV sunlight or heat wakes them up!

1. Classification of Hydrocarbons (9.1)

Saturated Hydrocarbons: Contain only C–C and C–H single bonds. Open-chain compounds are alkanes (C_nH_2n+2); closed-ring saturated compounds are cycloalkanes (C_nH_2n).
Unsaturated Hydrocarbons: Contain carbon–carbon multiple bonds—double bonds (alkenes), triple bonds (alkynes), or both.
Aromatic Hydrocarbons (Arenes): Special cyclic conjugated systems (benzenoid rings like benzene, naphthalene; or non-benzenoid aromatic rings).

2. Alkane Structure, Isomerism & Boiling Point Trend (9.2.1 & 9.2.3)

Geometry: Every carbon is sp³ hybridised and tetrahedral (H–C–H angle = 109.5°; C–C bond = 154 pm; C–H bond = 112 pm).
Carbon Types: Primary (1°, attached to 0 or 1 C), Secondary (2°, attached to 2 C), Tertiary (3°, attached to 3 C), Quaternary (4°, attached to 4 C).
Chain Isomerism: Butane (C₄H₁₀) has 2 isomers; Pentane (C₅H₁₂) has 3; Hexane (C₆H₁₄) has 5 (NCERT Problem 9.1); Decane (C₁₀H₂₂) has 75!
Physical Properties: Non-polar molecules with weak van der Waals forces (C₁–C₄ gases, C₅–C₁₇ liquids, ≥C₁₈ solids at 298 K). Boiling point increases with molecular mass, but decreases with branching because spherical shapes reduce contact surface area:
Isomeric Pentanes b.p. order: n-Pentane (309.1 K) > 2-Methylbutane (300.9 K) > 2,2-Dimethylpropane (282.5 K)

3. Key Preparation Methods of Alkanes (9.2.2)

Method	Equation & Reagents	Key Exam Note
1. Hydrogenation	CH₂=CH₂ + H₂ → CH₃–CH₃ (Pt/Pd at room temp; Ni at higher T, P)	Also converts alkynes + 2H₂ → alkanes.
2. Reduction of R–X	R–Cl + H₂ → R–H + HCl (Zn + dil. HCl)	Works for Cl, Br, I halides (except fluorides).
3. Wurtz Reaction	2 R–Br + 2 Na —[dry ether]→ R–R + 2 NaBr	Best for even-carbon symmetrical alkanes. Mixing 2 different halides gives 3 alkanes (hard to separate!).
4. Soda-Lime Decarboxylation	RCOO⁻Na⁺ + NaOH —[CaO, Δ]→ R–H + Na₂CO₃	Step-down! Product alkane has 1 carbon less (e.g., Sodium butanoate → Propane).
5. Kolbe's Electrolysis	2 CH₃COO⁻Na⁺ + 2 H₂O → C₂H₆ + 2 CO₂ (anode) + H₂ + 2 NaOH (cathode)	Gives even-carbon alkane at anode; Methane cannot be prepared!

4. Chemical Reactions of Alkanes (9.2.3)

Free Radical Halogenation (hν or 573–773 K): CH₄ + Cl₂ → CH₃Cl → CH₂Cl₂ → CHCl₃ → CCl₄.
- Three steps: (i) Initiation (Cl–Cl homolysis → 2 Cl•), (ii) Propagation (CH₄ + Cl• → CH₃• + HCl; CH₃• + Cl₂ → CH₃Cl + Cl•), (iii) Termination (Cl•+Cl•, CH₃•+Cl•, and CH₃• + CH₃• → C₂H₆, explaining ethane byproduct!).
- Reactivity: Halogens F₂ > Cl₂ > Br₂ > I₂; Hydrogen replacement rate 3° > 2° > 1°. Iodination is reversible and needs HIO₃ or HNO₃ to oxidise HI back to I₂.
Combustion: C_nH_2n+2 + ((3n+1)/2) O₂ → n CO₂ + (n+1) H₂O (exothermic; incomplete combustion yields carbon black for ink).
Controlled Oxidation: CH₄ + O₂ —[Cu/523K/100 atm]→ CH₃OH; CH₄ + O₂ —[Mo₂O₃, Δ]→ HCHO + H₂O; 2 C₂H₆ + 3 O₂ —[(CH₃COO)₂Mn, Δ]→ 2 CH₃COOH + 2 H₂O; Alkanes with 3° H oxidise with KMnO₄ to 3° alcohols!
Isomerisation: n-Hexane —[anhyd. AlCl₃/HCl, Δ]→ 2-Methylpentane + 3-Methylpentane.
Aromatisation (Reforming): n-Hexane —[Cr₂O₃ / V₂O₅ / Mo₂O₃ on Al₂O₃, 773 K, 10–20 atm]→ Benzene + 4 H₂ (n-Heptane gives Toluene).
Steam Reaction & Pyrolysis: CH₄ + H₂O —[Ni, 1273 K]→ CO + 3 H₂ (industrial H₂); Dodecane (C₁₂H₂₆) —[Pt/Pd/Ni, 973 K]→ C₇H₁₆ + C₅H₁₀ (cracking).

Two giant exam traps in Alkanes: (1) Neither Wurtz reaction nor Kolbe's electrolysis can prepare methane (CH₄)—both build C–C bonds of at least 2 carbons! (2) In Soda-lime decarboxylation, remember you LOSE 1 carbon (sodium propanoate gives ethane, not propane).

Topic 9.2.4: Conformations of Ethane: Sawhorse, Newman Projections & Torsional Strain

Picture holding a spinning umbrella with two 3-bladed propellers mounted on one central axle. Because a C–C sigma (σ) bond has cylindrically symmetrical electron density around the internuclear axis, the front –CH₃ group of ethane can spin relative to the rear –CH₃ group! When the front blades line up directly behind the rear blades (Eclipsed), they crowd each other uncomfortably. Twist the wheel by 60° so every blade sits comfortably in the gaps between the back blades (Staggered), and the tension relaxes!

Infinite spatial arrangements of atoms obtained by rotation around a C–C single bond are called conformations (or conformers / rotamers). Bond lengths and bond angles remain unchanged across all conformations.

1. Two Ways to Draw Ethane (C₂H₆)

Sawhorse Projection: Viewed slightly tilted along the C–C axis; the front carbon is drawn at the lower left and rear carbon at the upper right, each carrying three C–H lines at 120° to each other.
Newman Projection: Viewed head-on straight down the C–C axis! The front carbon is a central dot with three 120° bonds; the rear carbon is a circle with three shorter 120° bonds radiating from its circumference.

2. Eclipsed vs Staggered vs Skew Conformations

Conformation	Dihedral Angle (θ)	C–H Cloud Repulsion (Torsional Strain)	Energy & Stability
Staggered	60°, 180°, 300°	Minimum (C–H bonds farthest apart)	Lowest energy (0 kJ/mol) → Most Stable!
Eclipsed	0°, 120°, 240°	Maximum (C–H electron clouds align closely)	Highest energy (+12.5 kJ/mol) → Least Stable
Skew	Any intermediate angle	Intermediate torsional strain	Intermediate energy & stability

Torsional Energy Barrier in Ethane = 12.5 kJ mol⁻¹
Thermal molecular collisions at room temperature easily supply 12.5 kJ mol⁻¹ → rotation is virtually free and individual conformers cannot be isolated!

Remember: Conformational isomers of ethane are NOT separable in bottles at room temperature because the energy gap is tiny (only 12.5 kJ mol⁻¹)! Also, never say bond angles change during rotation—C–C–H angles remain tetrahedral (109.5°); only the dihedral (torsional) angle θ changes!

Topic 9.3.1 – 9.3.4: Alkenes: Double Bond Structure, Cis–Trans Isomerism & Preparations

Take two wooden blocks and hammer just ONE nail straight through their centres—the blocks spin freely like ethane's single bond. Now hammer TWO nails side-by-side through both blocks—can you twist one block without ripping out the second nail? Impossible! That second nail is the sideways pi (π) bond in an alkene's C=C double bond. Because a π bond locks rotation, cis-but-2-ene and trans-but-2-ene become two totally distinct molecules with different polarity, boiling points, and melting points!

Alkenes (C_nH_2n) are unsaturated hydrocarbons containing at least one C=C double bond—formerly called olefins (oil-forming, because ethene + Cl₂ forms an oily liquid).

1. Structure of the Double Bond (9.3.1)

C=C = 1 strong σ bond (397 kJ mol⁻¹, sp²–sp²) + 1 weak π bond (284 kJ mol⁻¹, sideways 2p–2p overlap).
Total C=C enthalpy = 681 kJ mol⁻¹ (stronger than C–C 348 kJ mol⁻¹), and shorter bond length = 134 pm (vs 154 pm for C–C).
The loosely held, exposed π-electron cloud attracts electrophiles (electron-seeking reagents).

2. Structural & Cis–Trans Geometrical Isomerism (9.3.3)

Structural Isomers of C₄H₈: But-1-ene and But-2-ene (position isomers); 2-Methylprop-1-ene (chain isomer).
Geometrical (Cis–Trans) Isomerism: Arises from restricted rotation about C=C when each doubly bonded carbon holds two different groups (e.g., XYC=CXY, XYC=CXZ, or XYC=CZW — NCERT Problem 9.11).
Cis vs Trans Properties (But-2-ene):
- cis-But-2-ene: Both –CH₃ groups on the same side; bond dipoles reinforce → μ = 0.33 D (Polar) → higher boiling point (277 K).
- trans-But-2-ene: Identical –CH₃ groups on opposite sides; C–CH₃ dipoles cancel → μ = 0 D (Non-polar)! Better symmetrical crystal packing → higher melting point (167 K vs 134 K)!

3. Four Standard Preparations of Alkenes (9.3.4)

Stereoselective Partial Reduction of Alkynes:
- Alkyne + H₂ with Lindlar's catalyst (palladised charcoal partially poisoned with S-compounds or quinoline) → cis-alkene.
- Alkyne + Na in liquid NH₃ → trans-alkene!
Dehydrohalogenation of Alkyl Halides (β-elimination): Heating R–X with alcoholic KOH eliminates HX from α and β carbons. Rate: I > Br > Cl and 3° > 2° > 1° alkyl group.
Dehalogenation of Vicinal Dihalides: CH₂Br–CH₂Br + Zn → CH₂=CH₂ + ZnBr₂.
Acidic Dehydration of Alcohols (β-elimination): CH₃CH₂OH —[conc. H₂SO₄, Δ]→ CH₂=CH₂ + H₂O.

Remember the crystal-clear distinction: Lindlar's catalyst gives cis-alkenes, whereas Na / liquid NH₃ gives trans-alkenes! Also check Problem 9.11: (CH₃)₂C=CH–C₂H₅ and CH₂=CBr₂ CANNOT show cis–trans isomerism because one of the double-bonded carbons carries two identical groups!

Topic 9.3.5 (Part 1): Electrophilic Addition in Alkenes: Markovnikov Rule vs Peroxide (Kharasch) Effect

When hydrogen bromide (HBr) meets an unsymmetrical alkene like propene (CH₃–CH=CH₂), the rich carbon vs poor carbon rule decides who wins the hydrogen! Under normal dark conditions, Markovnikov's rule works like "the rich get richer": the CH₂ carbon that already holds more H atoms grabs H⁺ so the stable 2° carbocation forms, sending Br⁻ to the middle carbon (2-bromopropane). Add a dash of benzoyl peroxide, however, and a free-radical plot twist flips the outcome completely—Br• attacks first, reversing the product to 1-bromopropane!

Unsaturation Test (Br₂/CCl₄): Reddish-orange bromine in CCl₄ is decolourised by alkenes forming colourless vicinal dibromides (CH₂Br–CH₂Br). Note: I₂ does not add under normal conditions.
Reactivity of Hydrogen Halides: HI > HBr > HCl.

1. Markovnikov Rule (1869) — Ionic / Electrophilic Mechanism

When an unsymmetrical reagent adds to an unsymmetrical alkene, the negative part of the addendum attaches to the doubly bonded carbon atom carrying fewer hydrogen atoms.

Propene + HBr (No Peroxide):
• Step 1: Electrophile H⁺ adds to CH₂ to give the more stable 2° carbocation CH₃–C⁺H–CH₃ (instead of unstable 1° CH₃CH₂CH₂⁺).
• Step 2: Br⁻ attacks quickly → 2-Bromopropane (major product).
• Also obeys Markovnikov: Cold conc. H₂SO₄ (forms alkyl hydrogen sulphate) & acid-catalysed hydration H₂O / H⁺ (forms 2-methylpropan-2-ol from 2-methylpropene).

2. Anti-Markovnikov / Peroxide / Kharasch Effect (1933)

In the presence of an organic peroxide such as benzoyl peroxide, (C₆H₅CO)₂O₂, addition of HBr to unsymmetrical alkenes occurs opposite to Markovnikov's rule via a free radical chain mechanism:

CH₃–CH=CH₂ + HBr —[(C₆H₅CO)₂O₂]→ CH₃–CH₂–CH₂Br (1-Bromopropane)

Peroxide homolysis generates C₆H₅• radicals, which abstract H from HBr to release Br• radical.
Br• attacks the terminal CH₂ carbon to form the more stable 2° carbon free radical (CH₃–C•H–CH₂Br), which abstracts H from another HBr to give 1-bromopropane.

NCERT Problem 9.12 — Hex-1-ene + HBr:
• Absence of peroxide → 2-Bromohexane (Markovnikov).
• Presence of peroxide → 1-Bromohexane (Anti-Markovnikov).

BOARD EXAM FAVORITE: Why does the Peroxide / Kharasch effect work ONLY with HBr and NOT with HCl or HI?
• HCl: The H–Cl bond is too strong (430.5 kJ mol⁻¹ vs 363.7 kJ mol⁻¹ for H–Br) to be cleaved homolytically by phenyl/alkyl free radicals.
• HI: The H–I bond is weak (296.8 kJ mol⁻¹), so iodine radicals rapidly combine with each other to form I₂ molecules instead of adding to C=C!

Topic 9.3.5 (Part 2): Ozonolysis, Baeyer's Unsaturation Test & Polymerisation of Alkenes

Suppose an unknown alkene arrives in a lab bottle—you know it has one C=C double bond, but WHERE along the carbon chain is it hiding? Chemists use ozone (O₃) followed by zinc dust and water (Zn/H₂O) as molecular scissors! Ozonolysis snips right across the C=C bond and caps both sliced ends with =O oxygen atoms. Just examine the two resulting aldehyde or ketone fragments, lay their =O noses nose-to-nose, delete the oxygens, and snap the double bond back together—instant molecular detective work!

1. Oxidation Tests of Alkenes

Baeyer's Test (Cold, dilute, aqueous KMnO₄): Discharges purple colour of KMnO₄ (with brown MnO₂) while adding two –OH groups across C=C to produce colourless vicinal glycols (e.g., Ethene → Ethane-1,2-diol; Propene → Propane-1,2-diol). Standard lab test for unsaturation!
Strong Oxidation (Acidic KMnO₄ / H⁺ or K₂Cr₂O₇ / H⁺): Cleaves C=C completely into ketones and/or carboxylic acids (e.g., But-2-ene → 2 CH₃COOH; 2-Methylpropene → Propanone + CO₂ + H₂O).

2. Reductive Ozonolysis (O₃, then Zn + H₂O)

Alkene + O₃ → Cyclic Ozonide —[Zn + H₂O]→ Aldehydes and/or Ketones + H₂O₂ (Zn prevents over-oxidation of aldehydes!)

Propene (CH₃–CH=CH₂): → Ethanal (CH₃CHO) + Methanal (HCHO).
2-Methylpropene ((CH₃)₂C=CH₂): → Propanone ((CH₃)₂C=O) + Methanal (HCHO).

NCERT Exercises 9.5 & 9.6 (Reverse Ozonolysis):
• Ex 9.5: Alkene 'A' gives Ethanal (CH₃CHO) + Pentan-3-one ((C₂H₅)₂C=O). Remove both O's and join the C=O carbons: (C₂H₅)₂C=CH–CH₃ → 3-Ethylpent-2-ene!
• Ex 9.6: Alkene with 3 C–C σ, 8 C–H σ, 1 C=C π bond gives 2 moles of aldehyde (mass 44 u = CH₃CHO). Joining two CH₃CH= fragments gives But-2-ene (CH₃–CH=CH–CH₃)!

3. Addition Polymerisation

Under high temperature, high pressure, and catalyst, thousands of alkene monomers link into long-chain polymers: n(CH₂=CH₂) → Polythene –(CH₂–CH₂)–_n; n(CH₃–CH=CH₂) → Polypropene (milk crates, buckets, toys).

Watch the difference between mild Baeyer's KMnO₄ (cold dilute → stops at vicinal diols / glycols) and acidic hot KMnO₄/H⁺ (cleaves right through C=C into ketones / carboxylic acids). Also in ozonolysis, =CH₂ ends ALWAYS give Methanal (HCHO), =CHR gives aldehydes (RCHO), and =CR₂ gives ketones (R₂C=O)!

Topic 9.4: Alkynes: sp Hybridisation, Acidic Character, Preparation from CaC₂ & Reactions

Think of an oxyacetylene welding torch slicing smoothly through thick steel girders—its fuel is ethyne (acetylene, HC≡CH)! Not only does ethyne pack enormous energy into its 120 pm short triple bond, it also hides a surprise chemical personality: unlike ethane and ethene, terminal alkynes bubble hydrogen gas when dropped onto sodium metal! Why? Because an sp carbon holds 50% s-character, gripping C–H electrons so tightly toward itself that the terminal hydrogen pops off easily as an acidic proton (H⁺)!

Alkynes (C_nH_2n−2) contain a carbon–carbon triple bond (1 strong σ + 2 perpendicular π bonds). In ethyne, each carbon is sp hybridised (linear, 180°), bond length = 120 pm (shorter than C=C 134 pm & C–C 154 pm), and bond enthalpy = 823 kJ mol⁻¹.

1. Preparation of Alkynes (9.4.3)

Industrial from Calcium Carbide (CaC₂): CaCO₃ —[Δ]→ CaO + CO₂; CaO + 3 C → CaC₂ + CO; CaC₂ + 2 H₂O → Ca(OH)₂ + C₂H₂ (Ethyne).
From Vicinal Dihalides (2-Step Elimination): Step 1: CH₂Br–CH₂Br + alc. KOH → CH₂=CH–Br (vinyl bromide).
Step 2: Requires stronger base sodamide (NaNH₂) → HC≡CH + NaBr + NH₃.

2. Acidic Character of Terminal Alkynes (9.4.4 A)

Because sp orbitals have 50% s-character (vs 33.3% in sp² ethene and 25% in sp³ ethane), sp carbons are most electronegative and pull the C–H bond pair strongly, making terminal ≡C–H protons acidic:

HC≡CH + Na → HC≡C⁻Na⁺ + ½ H₂ ↑ | CH₃–C≡C–H + NaNH₂ → CH₃–C≡C⁻Na⁺ + NH₃
Acidic Strength Order: HC≡CH > CH₂=CH₂ > CH₃–CH₃ & HC≡CH > CH₃–C≡CH >> CH₃–C≡C–CH₃

3. Addition & Polymerisation Reactions (9.4.4 B)

Hydrogen, Halogens & HX (2 moles add via vinylic cation): Ethyne + 2 HBr → CH₃–CHBr₂ (1,1-dibromoethane, gem-dihalide); Propyne + 2 HBr → 2,2-dibromopropane (Markovnikov). Decolourises Br₂/CCl₄ to 1,1,2,2-tetrabromoalkane.
Addition of Water (Hydration, HgSO₄ + dil. H₂SO₄ at 333 K):
- Ethyne + H₂O → [CH₂=CH–OH] —[tautomerises]→ CH₃CHO (Ethanal — an aldehyde!)
- Propyne + H₂O → [CH₃–C(OH)=CH₂] —[tautomerises]→ CH₃COCH₃ (Propanone — a ketone!)
Linear Polymerisation: Gives polyacetylene –(CH=CH–CH=CH)_n–, an electrically conducting polymer used in thin-film battery electrodes!
Cyclic Polymerisation (Trimerisation): Passing ethyne through a red-hot iron tube at 873 K converts 3 C₂H₂ → Benzene (C₆H₆)—the famous bridge from aliphatic to aromatic chemistry!

Watch two classic lab/exam distinctions in Alkynes: (1) But-1-yne reacts with Na / NaNH₂ (has 1 acidic terminal ≡C–H), whereas But-2-yne (CH₃–C≡C–CH₃) has NO acidic ≡C–H and does NOT react! (2) Hydration (HgSO₄/H₂SO₄, 333 K) gives an aldehyde ONLY from ethyne; EVERY higher alkyne gives a ketone!

Topic 9.5.1 – 9.5.3: Aromatic Hydrocarbons: Structure of Benzene, Resonance & Hückel (4n+2)π Rule

In 1865, German chemist August Kekulé fell asleep beside his fireplace while wrestling with benzene's baffling formula, C₆H₆. He dreamed of dancing chains of carbon atoms—until suddenly one snake bit its own tail and whirled around like a ring! That dream cracked benzene's cyclic hexagon. Yet X-ray diffraction later uncovered an even deeper marvel: benzene does not have three short double bonds and three long single bonds—all six C–C bonds are an identical 139 pm, wrapped in two glowing π-electron doughnuts above and below the ring!

Arenes are pleasant-smelling aromatic hydrocarbons—either benzenoids (benzene, toluene, naphthalene, biphenyl) or non-benzenoid aromatic systems. All six hydrogens of benzene are equivalent (only 1 monosubstituted isomer; disubstituted forms give 1,2-ortho, 1,3-meta, and 1,4-para isomers).

1. Structure, Kekulé Evidence & Resonance Stability of Benzene (9.5.2)

Why Kekulé modified his structure: Ordinary alternating C=C/C–C rings predicted two different 1,2-dibromobenzenes, yet benzene forms ONLY ONE! So Kekulé proposed rapidly oscillating double bonds. (o-Xylene triozonolysis giving glyoxal, methylglyoxal & dimethylglyoxal also proves alternating double-bond character.)
Modern Resonance & Orbital Picture: All 6 carbons are sp² hybridised (planar hexagon, 120°). Six perpendicular 2p orbitals overlap continuously to form two delocalised π-electron cloud rings (doughnuts) above and below the hexagonal plane.
Why Benzene Resists Addition: X-ray diffraction shows ALL six C–C bonds have equal length 139 pm (intermediate between C–C 154 pm and C=C 133 pm). Resonance delocalisation makes benzene far more stable than hypothetical cyclohexatriene, so it prefers substitution (preserving the aromatic ring) over addition!

2. Hückel Rule of Aromaticity (9.5.3)

Three Necessary Conditions for Aromaticity:
1. Planarity (flat ring)
2. Complete delocalisation of π electrons throughout the ring (every ring atom has a p-orbital; no sp³ break!)
3. Presence of (4n + 2) π electrons in the ring, where n = 0, 1, 2, 3... (i.e. 2, 6, 10, 14 π e⁻).

Species	Planar & Conjugated?	π Electron Count	Verdict
Benzene (C₆H₆)	Yes (6 sp² C)	6π (4×1 + 2; n = 1)	Aromatic
Cyclopentadienyl Anion (C₅H₅⁻)	Yes (lone pair in p)	6π (4 + 2 lone pair; n = 1)	Aromatic
Cycloheptatrienyl / Tropylium Cation (C₇H₇⁺)	Yes (vacant p on C⁺)	6π (3 double bonds; n = 1)	Aromatic
Naphthalene (C₁₀H₈)	Yes (fused rings)	10π (4×2 + 2; n = 2)	Aromatic
Cyclopentadiene (C₅H₆)	✘ Has sp³ –CH₂– break!	4π	Non-Aromatic
Cyclooctatetraene (C₈H₈)	✘ Bends into non-planar tub!	8π (4n, n not integer)	Non-Aromatic

To check Hückel's rule on exams (like NCERT Ex 9.12), NEVER just count π electrons blindly! First check if an sp³ carbon (–CH₂–) breaks the conjugated loop (as in cyclopentadiene or cyclohepta-1,3,5-triene). Also remember: Cyclobutadiene (4π) is planar & anti-aromatic, while Cyclooctatetraene (8π) twists into a non-planar tub to stay non-aromatic!

Topic 9.5.4 – 9.6: Electrophilic Substitution in Benzene, Directive Influence & Carcinogenicity

Picture benzene's rich circular π-cloud as an inviting golden magnet for positively charged attackers called electrophiles (E⁺). An electrophile dives into the ring, briefly breaking aromaticity—but benzene desperately wants its 6π aromatic crown back, so it kicks off an H⁺ proton instead of welcoming a second nucleophile! And if a substituent is already sitting at position 1, it acts like a traffic cop: –OH or –CH₃ waves incoming electrophiles to the ortho and para lanes, while –NO₂ pushes them into the meta lane!

1. Preparation of Benzene (9.5.4)

Isolated from coal tar, or prepared in the lab by: (i) Cyclic polymerisation of ethyne (873 K, Fe), (ii) Decarboxylation of sodium benzoate (C₆H₅COONa + NaOH —[CaO, Δ]→ C₆H₆ + Na₂CO₃), and (iii) Reduction of phenol with heated zinc dust (C₆H₅OH + Zn —[Δ]→ C₆H₆ + ZnO).

2. Five Electrophilic Substitution (S_E) Reactions of Benzene (9.5.5)

Reaction	Reagents & Conditions	Electrophile (E⁺)	Product
Nitration	conc. HNO₃ + conc. H₂SO₄ (323–333 K)	NO₂⁺ (Nitronium ion)	Nitrobenzene
Halogenation	Cl₂ / anhyd. AlCl₃ or FeCl₃ (excess Cl₂ → C₆Cl₆)	Cl⁺	Chlorobenzene
Sulphonation	Fuming H₂SO₄ (Oleum)	SO₃	Benzenesulphonic acid
Friedel-Crafts Alkylation	R–Cl / anhyd. AlCl₃	R⁺ (rearranges!)	Alkylbenzene
Friedel-Crafts Acylation	CH₃COCl or (CH₃CO)₂O / anhyd. AlCl₃	CH₃C⁺=O (Acylium)	Acetophenone

3-Step S_E Mechanism: (a) Generation of electrophile E⁺ (in nitration, H₂SO₄ acts as acid and HNO₃ acts as a base to form NO₂⁺!), (b) Attack of E⁺ forming resonance-stabilised σ-complex (arenium ion) with one sp³ carbon, (c) Loss of H⁺ (via [AlCl₄]⁻ or [HSO₄]⁻) restores the aromatic ring!
Special Addition & Combustion: Benzene + 3 H₂ —[Ni, high T/P]→ Cyclohexane; Benzene + 3 Cl₂ —[UV light, 500 K]→ BHC / Gammaxane (C₆H₆Cl₆); burns with a sooty flame (C₆H₆ + 15/2 O₂ → 6 CO₂ + 3 H₂O).

3. Directive Influence of Substituents (9.5.6)

Ortho & Para Directing (Activating): –OH, –NH₂, –NHR, –NHCOCH₃, –OCH₃, –CH₃, –C₂H₅ increase electron density at o & p positions (+R / hyperconjugation), accelerating S_E faster than benzene.
Ortho & Para Directing (Moderately Deactivating Exception!): Halogens (–Cl, –Br, –I) pull electrons via strong −I effect (making the ring slower than benzene), yet donate lone pairs via resonance (+R) so o/p stay richer than meta → o/p-directing!
Meta Directing (Deactivating): –NO₂, –CN, –CHO, –COR, –COOH, –COOR, –SO₃H strongly withdraw electrons from o & p (−R, −I), leaving the meta site comparatively richer → m-directing!

NCERT Ex 9.13 — Order Matters in Synthesis!
• Benzene → p-Nitrobromobenzene: Brominate FIRST (Br₂/FeBr₃ → Bromobenzene, o/p-directing), THEN nitrate!
• Benzene → m-Nitrochlorobenzene: Nitrate FIRST (HNO₃/H₂SO₄ → Nitrobenzene, m-directing), THEN chlorinate!

9.6 Carcinogenicity & Toxicity: Benzene and fused polynuclear hydrocarbons (from incomplete combustion of tobacco, coal, petroleum—e.g., 1,2-benzpyrene, 1,2-benzanthracene, 3-methylcholanthrene) damage DNA and are carcinogenic.

Watch THREE massive exam traps here: (1) Reacting benzene with 1-chloropropane + anhyd. AlCl₃ gives isopropylbenzene (cumene), NOT n-propylbenzene, due to 1°→2° carbocation rearrangement! (2) Cl₂ + anhyd. AlCl₃ (dark) substitutes → Chlorobenzene (or C₆Cl₆), whereas Cl₂ + UV light (500 K) ADDS → BHC / Gammaxane (C₆H₆Cl₆)! (3) Halogens are deactivating yet o/p-directing!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March (Repeated) • 2 Marks): Draw the Newman projections for the eclipsed and staggered conformations of ethane. Which conformation is more stable and why?

In Newman projection, the front carbon is drawn as a dot and the rear carbon as a circle, each with three C–H bonds at 120°:
• Eclipsed (θ = 0°): Rear C–H bonds lie directly behind front C–H bonds.
• Staggered (θ = 60°): Rear C–H bonds sit equidistantly between front C–H bonds.

Stability: The staggered conformation is more stable (by 12.5 kJ mol⁻¹) because the C–H bond electron clouds are at maximum separation, resulting in minimum electron–cloud repulsion (minimum torsional strain).

Q2 (Kerala HSE SAY / NCERT • 3 Marks): State Markovnikov's rule. What is the peroxide (Kharasch) effect? Why is the peroxide effect NOT observed in the addition of HCl and HI to unsymmetrical alkenes?

Markovnikov's Rule: When an unsymmetrical reagent adds to an unsymmetrical alkene, the negative part of the addendum attaches to the doubly bonded carbon atom having fewer hydrogen atoms (e.g., Propene + HBr → 2-bromopropane).

Peroxide / Kharasch Effect: In the presence of benzoyl peroxide, HBr adds to propene anti-Markovnikov via a free radical chain mechanism to give 1-bromopropane.

Why not HCl / HI? The H–Cl bond (430.5 kJ mol⁻¹) is too strong to be cleaved homolytically by free radicals, whereas the H–I bond (296.8 kJ mol⁻¹) is weak and iodine free radicals combine to form I₂ molecules instead of adding to the double bond.

Q3 (Kerala HSE March • 2 Marks): How will you convert: (a) Sodium ethanoate to methane, and (b) Ethyne to benzene?

(a) Soda-lime decarboxylation: Heat sodium ethanoate with soda lime (NaOH + CaO):
CH₃COONa + NaOH —[CaO, Δ]→ CH₄ + Na₂CO₃.

(b) Cyclic polymerisation: Pass ethyne gas through a red-hot iron tube at 873 K:
3 HC≡CH —[Red-hot Fe, 873 K]→ C₆H₆ (Benzene).

Q4 (Kerala HSE Model / NCERT • 3 Marks): State the three necessary conditions for a compound to be aromatic according to Hückel's rule. Explain why cyclopentadiene is non-aromatic whereas cyclopentadienyl anion is aromatic.

Hückel's Conditions: (1) Planarity, (2) Complete delocalisation of π electrons in the ring, and (3) Presence of (4n + 2) π electrons (n = 0, 1, 2...).

Cyclopentadiene (C₅H₆): Contains an sp³ hybridised –CH₂– carbon which interrupts complete π-delocalisation (and has only 4π electrons) → Non-aromatic.
Cyclopentadienyl anion (C₅H₅⁻): All 5 carbons are sp² planar, fully conjugated, with 4π + 2π (lone pair) = 6π electrons (n = 1 in 4n+2) → Aromatic.

Q5 (Kerala HSE March • 2 Marks): Explain the Wurtz reaction with a chemical equation. Why is it NOT preferred for preparing alkanes with an odd number of carbon atoms?

Wurtz Reaction: Alkyl halides react with sodium metal in dry ether to form higher symmetrical alkanes:
2 C₂H₅Br + 2 Na —[dry ether]→ CH₃CH₂CH₂CH₃ (n-Butane) + 2 NaBr.

Odd-carbon limitation: Preparing an odd-carbon alkane (like propane) requires mixing two different alkyl halides (CH₃Br + C₂H₅Br), which produces a mixture of three alkanes (ethane, propane, and butane) with close boiling points that are difficult to separate.

Q6 (Kerala HSE SAY • 2 Marks): Draw the cis and trans isomers of but-2-ene. Which isomer has a higher dipole moment and why? How can cis-but-2-ene be prepared from but-2-yne?

In cis-but-2-ene, both –CH₃ groups lie on the same side of C=C; in trans-but-2-ene, they lie on opposite sides.

Dipole Moment: cis-but-2-ene is polar (μ = 0.33 D) because the two C–CH₃ bond dipoles reinforce, whereas in trans-but-2-ene (μ = 0 D) the opposite dipoles cancel out.

Preparation from but-2-yne: Partial hydrogenation using H₂ over Lindlar's catalyst (Pd/C deactivated with S-compounds/quinoline) gives cis-but-2-ene.

Q7 (Kerala HSE March / NCERT 9.5 • 3 Marks): An alkene 'A' on ozonolysis (O₃, Zn/H₂O) gives a mixture of ethanal and pentan-3-one. Write the structural formula and IUPAC name of 'A'. Also give the ozonolysis products of propene.

Align the two carbonyl products with their C=O groups facing:
CH₃–CH=O + O=C(C₂H₅)₂.
Removing both oxygens and joining with a C=C double bond gives CH₃–CH=C(C₂H₅)–CH₂–CH₃.
IUPAC Name of 'A': 3-Ethylpent-2-ene.

Propene Ozonolysis: CH₃–CH=CH₂ —[O₃; Zn/H₂O]→ CH₃CHO (Ethanal) + HCHO (Methanal).

Q8 (Kerala HSE Improvement • 2 Marks): Arrange ethyne, ethene, and ethane in decreasing order of acidic strength. Give the reason and write one reaction proving the acidic nature of ethyne.

Order: HC≡CH > CH₂=CH₂ > CH₃–CH₃.

Reason: Carbon in ethyne is sp hybridised (50% s-character) compared to sp² (33.3%) in ethene and sp³ (25%) in ethane. Greater s-character increases carbon electronegativity, pulling C–H electrons closer to C so H⁺ is lost more easily.
Reaction: HC≡CH + Na → HC≡C⁻Na⁺ + ½ H₂ ↑ (Sodium ethynide + H₂ gas).

Q9 (Kerala HSE March / NCERT 9.13 • 4 Marks): Explain the three-step mechanism of nitration of benzene. How will you convert benzene into: (i) p-nitrobromobenzene, and (ii) m-nitrochlorobenzene?

Mechanism of Nitration (conc. HNO₃ + conc. H₂SO₄):
1. Generation of electrophile: H₂SO₄ protonates HNO₃ to yield nitronium ion NO₂⁺ + HSO₄⁻ + H₂O.
2. Formation of σ-complex (arenium ion): NO₂⁺ attacks benzene to form a resonance-stabilised carbocation intermediate.
3. Loss of proton: [HSO₄]⁻ removes H⁺ from the sp³ carbon, restoring the aromatic ring to yield nitrobenzene.

Conversions:
(i) p-Nitrobromobenzene: First brominate benzene (Br₂/FeBr₃ → bromobenzene; –Br is o/p-directing), then nitrate (conc. HNO₃ + H₂SO₄) to get p-nitrobromobenzene (major).
(ii) m-Nitrochlorobenzene: First nitrate benzene (conc. HNO₃ + H₂SO₄ → nitrobenzene; –NO₂ is m-directing), then chlorinate (Cl₂ / anhyd. AlCl₃).

Q10 (Kerala HSE Model • 2 Marks): What happens when propyne is warmed with water in the presence of mercuric sulphate (HgSO₄) and dilute H₂SO₄ at 333 K? Give the equation.

Propyne undergoes Markovnikov addition of water to form an unstable enol (propen-2-ol), which tautomerises to give propanone (acetone):
CH₃–C≡CH + H₂O —[Hg²⁺ / H⁺, 333 K]→ [CH₃–C(OH)=CH₂] ⇌ CH₃–CO–CH₃ (Propanone).

Q11 (Kerala HSE March • 1 Marks): What is Baeyer's reagent? State its important analytical application.

Baeyer's reagent is a cold, dilute, aqueous solution of potassium permanganate (KMnO₄). Its purple colour is decolourised by alkenes and alkynes (forming colourless vicinal glycols), so it is used as a test for unsaturation.

Q12 (Kerala HSE SAY • 1 Marks): Write the major organic product formed when benzene reacts with 1-chloropropane in the presence of anhydrous AlCl₃.

Isopropylbenzene (Cumene), because the initially formed primary 1-propyl carbocation rearranges via a 1,2-hydride shift into the more stable secondary isopropyl carbocation.