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State Board · Class 11 · Chemistry

Lesson 15: Hydrocarbons

7 concepts · 56 practice questions with answers · Free to read

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Classification, isomerism and preparation of alkanes

Link to Std 9–10 : In Std 9–10 you met carbon compounds; here we study the hydrocarbon families.

From daily life

LPG, petrol and candle wax are all hydrocarbons; they follow patterns such as CₙH₂ₙ₊₂.

Simple definition

Hydrocarbons contain only C and H. Alkanes CₙH₂ₙ₊₂ (single bonds), alkenes CₙH₂ₙ (C=C), alkynes CₙH₂ₙ₋₂ (C≡C), aromatic (benzene ring).

Pentane isomersThree isomers of C₅H₁₂ and boiling points309 Kn-pentanestraight chain301 Kisopentaneone branch283 Kneopentanecompact, two branchesMore branching ⇒ less surface ⇒ weaker van der Waals ⇒ lower b.p.

In the figure : The three isomers of C₅H₁₂.

Rule 1: Chain isomerism — Same molecular formula, different carbon skeleton. More branching lowers the boiling point (weaker van der Waals forces).

  • Easy : C₄H₁₀: 2 isomers.
  • Medium : C₅H₁₂: 3 (309, 301, 283 K).
  • Exam-type : C₆H₁₄: 5 isomers.

Rule 2: Preparation of alkanes — Hydrogenation of alkenes (H₂/Ni, Sabatier–Senderens); Wurtz: 2R–X + 2Na → R–R + 2NaX (dry ether); decarboxylation: RCOONa + NaOH (CaO) → R–H + Na₂CO₃; Kolbe electrolysis of RCOONa.

  • Easy : CH₃COONa + NaOH → CH₄ + Na₂CO₃ (CaO).
  • Medium : 2CH₃Br + 2Na → C₂H₆ + 2NaBr.
  • Exam-type : Kolbe: 2CH₃COONa + 2H₂O → C₂H₆ + 2CO₂ + 2NaOH + H₂.

Rule 3: Degree of unsaturation — DBE = (2C + 2 − H) ÷ 2: 0 = alkane, 1 = alkene/ring, 2 = alkyne, 4 = benzene.

  • Easy : C₆H₆: (14 − 6) ÷ 2 = 4.
  • Medium : C₄H₈: (10 − 8) ÷ 2 = 1.
  • Exam-type : C₂H₂: (6 − 2) ÷ 2 = 2.

Why does a Wurtz reaction with two different R–X give three products?

  • Step 1: CH₃Br + C₂H₅Br + Na.
  • Step 2: Possible couplings: CH₃–CH₃, CH₃–C₂H₅, C₂H₅–C₂H₅.
  • Step 3: A mixture results; use a single R–X for a pure alkane.

CET level (MCQ)

  • CET 1) Heating CH₃COONa with soda lime gives? 1) C₂H₆ 2) CH₄ 3) C₂H₄ 4) C₂H₂ — Answer: 2).
  • CET 2) Number of isomers of C₅H₁₂? 1) 2 2) 3 3) 4 4) 5 — Answer: 2).

Real-life use: LPG (C₃–C₄), petrol (C₅–C₁₀), diesel, wax.

Remember

CₙH₂ₙ₊₂, CₙH₂ₙ, CₙH₂ₙ₋₂; more branches ⇒ lower b.p.; Wurtz doubles the carbons.

Common mistake

Decarboxylation loses one carbon, Wurtz doubles them; do not mix them up.

CET trap

Kolbe gives CO₂ and H₂ along with the alkane.

Shortcut

Memory clue: "Wurtz joins (R–R), decarboxylation drops one".

Spot the difference

MethodReagentCarbons
WurtzR–X + Nadouble
DecarboxylationRCOONa + NaOH/CaOone less
Kolbeelectrolysis2 × (n − 1)

Practice questions

Easy
  1. 1Write the general formula of alkanes.
  2. 2Draw the isomers of C₄H₁₀.
  3. 3What is the Wurtz reaction?
Medium
  1. 4Why is neopentane's b.p. lowest?
  2. 5How is CH₄ made from CH₃COONa?
  3. 6DBE of C₆H₆?
Challenging
  1. 7Write the products of the Kolbe reaction.
  2. 8Why do two different R–X not suit Wurtz?
Show answers (try them yourself first)
  1. 1CₙH₂ₙ₊₂.
  2. 2n-butane, isobutane.
  3. 32R–X + 2Na → R–R + 2NaX.
  4. 4Compact shape ⇒ less surface.
  5. 5Heat with soda lime.
  6. 64.
  7. 7C₂H₆ + CO₂ + NaOH + H₂.
  8. 8A mixture of three products.

Properties of alkanes: conformations, combustion and halogenation

From daily life

LPG burning on a stove gives heat, CO₂ and water; that is alkane combustion.

Simple definition

Alkanes are fairly unreactive (paraffins = 'little affinity'). Main reactions: combustion, free-radical halogenation, pyrolysis. Conformations arise from rotation about the C–C axis.

Ethane conformationsEthane conformations: Newman and energystaggeredlowest energyeclipsed+ 12 kJ/moldihedral angle 0° → 360°barrier ≈ 12 kJ/molRotation about C–C is easy at room temperature

In the figure : Conformations of ethane.

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Properties of alkanes: conformations, combustion and halogenation

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HalogenationCH₄ + Cl₂: free-radical chain reactionInitiation: heat/lightCl–Cl → 2Cl•Propagation: chain continuesCl• + CH₄ → HCl + •CH₃•CH₃ + Cl₂ → CH₃Cl + Cl•Termination: radicals combineCl• + Cl• → Cl₂ ; •CH₃ + Cl• → CH₃Cl•CH₃ + •CH₃ → C₂H₆Overall: CH₄ → CH₃Cl → CH₂Cl₂ → CHCl₃ → CCl₄Reactivity F₂ > Cl₂ > Br₂ > I₂; selectivity Br₂ > Cl₂

In the figure : Free-radical chain reaction.

Rule 1: Conformations — Ethane: staggered is most stable; eclipsed is about 12 kJ/mol higher. Butane: anti is the most stable.

  • Easy : Dihedral 60° ⇒ staggered.
  • Medium : Dihedral 0° ⇒ eclipsed.
  • Exam-type : Small barrier ⇒ free rotation at room temperature.

Rule 2: Combustion — CₙH₂ₙ₊₂ + (3n + 1)/2 O₂ → n CO₂ + (n + 1) H₂O; limited air gives CO or soot.

  • Easy : CH₄ + 2O₂ → CO₂ + 2H₂O; 16 g CH₄ ⇒ 44 g CO₂.
  • Medium : C₃H₈ + 5O₂ → 3CO₂ + 4H₂O; 22 g C₃H₈ ⇒ 66 g CO₂.
  • Exam-type : 1 L CH₄ needs 2 L O₂.

Rule 3: Halogenation — Initiation → propagation → termination. Reactivity F₂ > Cl₂ > Br₂ > I₂; ease of H abstraction 3° > 2° > 1°.

  • Easy : CH₄ + Cl₂ → CH₃Cl + HCl (light).
  • Medium : Propagation: Cl• + CH₄ → HCl + •CH₃.
  • Exam-type : 2-methylpropane + Br₂ ⇒ mainly the tertiary bromide.

Calculation: CO₂ from burning 16 g CH₄

  • Step 1: n(CH₄) = 16 ÷ 16 = 1 mol.
  • Step 2: CH₄ : CO₂ = 1 : 1 ⇒ 1 mol CO₂.
  • Step 3: Mass = 1 × 44 = 44 g.

CET level (MCQ)

  • CET 1) Lowest-energy conformation of ethane? 1) eclipsed 2) staggered 3) skew 4) half-eclipsed — Answer: 2).
  • CET 2) O₂ needed for complete combustion of 1 L C₃H₈? 1) 3 L 2) 4 L 3) 5 L 4) 7 L — Answer: 3).

Real-life use: Fuels; making chloroform/CCl₄; solvents.

Remember

Staggered < eclipsed energy; combustion n CO₂ + (n + 1) H₂O; radical: initiation-propagation-termination.

Common mistake

Alkane halogenation is radical, not ionic.

CET trap

Iodination does not go easily (HI reduces, reversing it).

Shortcut

Memory clue: "I-P-T" (initiation, propagation, termination).

Spot the difference

ConformationDihedralEnergy
Staggered60°lowest
Eclipsed0°+ 12 kJ/mol

Practice questions

Easy
  1. 1What are conformations?
  2. 2Why is the staggered form more stable?
  3. 3Write the combustion equation of CH₄.
Medium
  1. 4CO₂ from 16 g CH₄?
  2. 5Name the three steps of a radical reaction.
  3. 6Why is iodination difficult?
Challenging
  1. 7CO₂ from 22 g C₃H₈?
  2. 8Why is a 3° H abstracted easily?
Show answers (try them yourself first)
  1. 1Shapes from rotation about C–C.
  2. 2Less repulsion.
  3. 3CH₄ + 2O₂ → CO₂ + 2H₂O.
  4. 444 g.
  5. 5Initiation, propagation, termination.
  6. 6HI reduces the product.
  7. 766 g.
  8. 8The 3° radical is more stable.

Alkenes: preparation and Markovnikov's rule

From daily life

Ethene ripens fruit and is the raw material of plastic bags.

Simple definition

Alkenes: C=C (sp², 120°, 1σ + 1π, 134 pm), readily undergo addition. Markovnikov: in HX addition to an unsymmetrical alkene, H goes to the carbon with more H.

MarkovnikovCarbocation stability and Markovnikov's ruleCH₃⁺methyl1°primary2°secondary3°tertiarystability increases →CH₃–CH=CH₂ + HBr → CH₃–CHBr–CH₃H goes to the carbon with more H; Br to the one with fewer H(stable 2° cation ⇒ major product 2-bromopropane)

In the figure : Carbocation stability and Markovnikov.

Rule 1: Preparation — Dehydration of alcohols: conc. H₂SO₄, 443 K; dehydrohalogenation: alcoholic KOH; partial reduction of alkynes: H₂/Lindlar (cis).

  • Easy : C₂H₅OH → C₂H₄ + H₂O (443 K).
  • Medium : C₂H₅Br + KOH(alc) → C₂H₄ + KBr + H₂O.
  • Exam-type : Zaitsev: the more substituted alkene is major.

Rule 2: Carbocations and Markovnikov — Stability 3° > 2° > 1° > CH₃⁺ (hyperconjugation and +I). The more stable cation forms first.

  • Easy : CH₃CH=CH₂ + HBr → CH₃CHBrCH₃.
  • Medium : (CH₃)₂C=CH₂ + HCl → (CH₃)₃CCl.
  • Exam-type : CH₃CH=CH₂ + H₂O/H⁺ → CH₃CH(OH)CH₃.

Rule 3: Peroxide effect — Only HBr with a peroxide (R–O–O–R) ⇒ radical path, anti-Markovnikov product.

  • Easy : CH₃CH=CH₂ + HBr (peroxide) → CH₃CH₂CH₂Br.
  • Medium : HCl and HI show no such effect.
  • Exam-type : Reason: Br• adds first.

Derivation: major product of propene + HBr

  • Step 1: H⁺ adds to give a 1° or a 2° cation.
  • Step 2: The 2° cation (CH₃–C⁺H–CH₃) is more stable.
  • Step 3: Br⁻ attacks it ⇒ 2-bromopropane.

CET level (MCQ)

  • CET 1) Major product of propene + HBr? 1) 1-bromopropane 2) 2-bromopropane 3) propane 4) allyl bromide — Answer: 2).
  • CET 2) Propene + HBr with peroxide? 1) 2-bromopropane 2) 1-bromopropane 3) 1,2-dibromide 4) none — Answer: 2).

Real-life use: Making ethanol, plastics, ripening fruit.

Remember

Markovnikov: H to the carbon with more H; peroxide effect only with HBr; Zaitsev: more substituted.

Common mistake

Applying the peroxide effect to HCl or HI.

CET trap

Markovnikov does not matter for a symmetrical alkene such as ethene.

Shortcut

Memory clue: "the rich get richer" (H goes to the C with more H).

Spot the difference

ItemMarkovnikovPeroxide
Mechanismionicradical
Position of Br2°1°
Applies toall HXHBr only

Practice questions

Easy
  1. 1State Markovnikov's rule.
  2. 2Order carbocations by stability.
  3. 3How is ethene obtained from ethanol?
Medium
  1. 4What is the peroxide effect?
  2. 5Write the product of propene + HBr.
  3. 6State Zaitsev's rule.
Challenging
  1. 7How many acyclic isomers of C₄H₈?
  2. 8Product of (CH₃)₂C=CH₂ + HCl?
Show answers (try them yourself first)
  1. 1H to the carbon with more H.
  2. 23° > 2° > 1° > CH₃⁺.
  3. 3Conc. H₂SO₄, 443 K.
  4. 4Anti-Markovnikov addition of HBr.
  5. 52-bromopropane.
  6. 6More substituted alkene is major.
  7. 74 (including cis/trans).
  8. 8(CH₃)₃CCl.

Reactions of alkenes and polymerisation

From daily life

A polythene bag is a chain of thousands of ethene molecules.

Simple definition

Alkenes show addition (H₂, X₂, HX, H₂O), oxidation (KMnO₄, O₃) and polymerisation. Decolourising Br₂ is the test for unsaturation.

Ethene mapReaction map of ethene (C₂H₄)CH₂=CH₂etheneH₂/Ni → C₂H₆Br₂/CCl₄ → CH₂Br–CH₂BrHBr → C₂H₅BrH₂O/H⁺ → C₂H₅OHcold KMnO₄ → diolO₃ → HCHO (ozonolysis)n CH₂=CH₂ → (–CH₂–CH₂–)ₙ polythene (high P/catalyst)Brown Br₂ decolourises = test for unsaturation

In the figure : Reactions of ethene.

Rule 1: Addition reactions — H₂/Ni ⇒ alkane; X₂ (CCl₄) ⇒ vicinal dihalide (anti addition); HX ⇒ Markovnikov; H₂O/H⁺ ⇒ alcohol.

  • Easy : CH₂=CH₂ + Br₂ → CH₂Br–CH₂Br.
  • Medium : 2.8 g of ethene needs 16 g Br₂ (0.1 mol).
  • Exam-type : CH₂=CH₂ + H₂O (H⁺) → C₂H₅OH.

Rule 2: Oxidation — Cold dilute alkaline KMnO₄ (Baeyer) ⇒ 1,2-diol, purple fades; hot KMnO₄ ⇒ acids/ketones; O₃ ⇒ ozonide ⇒ Zn/H₂O ⇒ aldehydes/ketones.

  • Easy : Propene + O₃ ⇒ CH₃CHO + HCHO.
  • Medium : Hot KMnO₄: CH₃CH=CH₂ ⇒ CH₃COOH + CO₂.
  • Exam-type : Ethene + cold KMnO₄ ⇒ ethylene glycol.

Rule 3: Polymerisation — n CH₂=CH₂ → (–CH₂–CH₂–)ₙ polythene; propene ⇒ polypropylene.

  • Easy : High pressure + peroxide ⇒ LDPE.
  • Medium : Ziegler–Natta catalyst ⇒ HDPE.
  • Exam-type : These are addition polymers.

Calculation: 2.8 g ethene + Br₂ (CCl₄)

  • Step 1: n(C₂H₄) = 2.8 ÷ 28 = 0.1 mol.
  • Step 2: 1 : 1 ⇒ Br₂ = 0.1 mol = 0.1 × 159.8 = 16 g.
  • Step 3: Product C₂H₄Br₂ = 18.8 g.

CET level (MCQ)

  • CET 1) Baeyer's reagent? 1) cold dilute alkaline KMnO₄ 2) hot KMnO₄ 3) K₂Cr₂O₇ 4) Br₂ — Answer: 1).
  • CET 2) Ozonolysis of propene gives? 1) 2 CH₃CHO 2) CH₃CHO + HCHO 3) 2 HCHO 4) CH₃COCH₃ — Answer: 2).

Real-life use: Polythene, polypropylene, ethylene glycol (antifreeze).

Remember

Br₂/CCl₄: colour fades; Baeyer: purple fades; O₃: aldehydes/ketones; addition polymers.

Common mistake

Mixing up the products of cold and hot KMnO₄.

CET trap

Without Zn/H₂O the H₂O₂ oxidises aldehydes to acids.

Shortcut

Memory clue: "cold = diol, hot = cleaves".

Spot the difference

ReagentProduct
H₂/Nialkane
Br₂/CCl₄dibromide
cold KMnO₄diol
O₃ ; Zn/H₂Oaldehyde/ketone

Practice questions

Easy
  1. 1What is the test for unsaturation?
  2. 2What is Baeyer's reagent?
  3. 3Equation of ethene + Br₂.
Medium
  1. 4Br₂ needed for 2.8 g ethene?
  2. 5Ozonolysis products of propene.
  3. 6How is polythene formed?
Challenging
  1. 7Products of hot KMnO₄ + propene?
  2. 8What is anti addition?
Show answers (try them yourself first)
  1. 1Decolourisation of Br₂ or KMnO₄.
  2. 2Cold dilute alkaline KMnO₄.
  3. 3CH₂=CH₂ + Br₂ → CH₂Br–CH₂Br.
  4. 416 g.
  5. 5CH₃CHO + HCHO.
  6. 6By addition polymerisation of ethene.
  7. 7CH₃COOH + CO₂.
  8. 8Two groups add from opposite faces.

Alkynes: structure, acidity and reactions

From daily life

A welding torch burns acetylene (ethyne) for a very hot flame (≈ 3200 °C).

Simple definition

Alkynes: C≡C (sp, linear 180°, 1σ + 2π, 120 pm). The terminal alkyne H is acidic (sp has 50 % s-character).

EthyneEthyne (acetylene): sp hybridisationCC120 pm106 pm106 pmH–C≡C–H: 180°, 1σ + 2πAcidic H (sp, 50 % s)HC≡CH + Na → HC≡CNaTest: AgNO₃/NH₃ white; Cu₂Cl₂ red(terminal alkynes only)Hydration: HgSO₄/H₂SO₄→ CH₃CHOTrimerisation 873 K (red-hot Fe)3 C₂H₂ → C₆H₆Acidity: C₂H₂ > C₂H₄ > C₂H₆

In the figure : Structure and reactions of ethyne.

Rule 1: Preparation — CaC₂ + 2H₂O → C₂H₂ + Ca(OH)₂; vicinal dihalide + alc. KOH/NaNH₂ ⇒ alkyne.

  • Easy : 64 g CaC₂ (1 mol) ⇒ 26 g C₂H₂.
  • Medium : CH₂Br–CH₂Br + 2KOH(alc) → C₂H₂ + 2KBr + 2H₂O.
  • Exam-type : Fruit ripened with calcium carbide is unsafe.

Rule 2: Acidity and tests — Terminal alkynes react with Na or NaNH₂ and give a white ppt with ammoniacal AgNO₃ and a red ppt with Cu₂Cl₂/NH₃. Acidity C₂H₂ > C₂H₄ > C₂H₆.

  • Easy : HC≡CH + 2Na → NaC≡CNa + H₂.
  • Medium : CH₃C≡CH is acidic; CH₃C≡CCH₃ is not.
  • Exam-type : Acetylide precipitate = terminal alkyne test.

Rule 3: Addition, hydration, trimerisation — H₂ (Lindlar ⇒ cis-alkene; Na/NH₃ ⇒ trans); HX twice; H₂O/HgSO₄ ⇒ aldehyde/ketone; 3 C₂H₂ → C₆H₆ (873 K).

  • Easy : C₂H₂ + H₂O (HgSO₄/H₂SO₄) → CH₃CHO.
  • Medium : Propyne + H₂O ⇒ CH₃COCH₃.
  • Exam-type : 3C₂H₂ → C₆H₆ (red-hot Fe tube).

Why is the alkyne H more acidic?

  • Step 1: sp carbon has 50 % s-character, sp² 33 %, sp³ 25 %.
  • Step 2: s electrons lie closer to the nucleus ⇒ carbon is more electronegative.
  • Step 3: C–H electrons shift to carbon ⇒ H⁺ leaves easily, acetylide ion is stable.

CET level (MCQ)

  • CET 1) CaC₂ + H₂O gives? 1) C₂H₄ 2) C₂H₂ 3) CH₄ 4) C₂H₆ — Answer: 2).
  • CET 2) Which gives a white ppt with ammoniacal AgNO₃? 1) CH₃C≡CCH₃ 2) CH₃C≡CH 3) C₂H₄ 4) C₂H₆ — Answer: 2).

Real-life use: Oxy-acetylene welding, PVC feedstock, acetaldehyde.

Remember

sp, 180°, 120 pm; terminal H acidic; HgSO₄ ⇒ CH₃CHO; 873 K ⇒ benzene.

Common mistake

Saying the internal alkyne also has an acidic H.

CET trap

Lindlar gives cis, Na/liq. NH₃ gives trans; both give alkenes.

Shortcut

Memory clue: "terminal = test" (Ag/Cu precipitate).

Spot the difference

ItemAlkaneAlkeneAlkyne
Hybridisationsp³sp²sp
C–C (pm)154134120
Shapetetrahedralplanar 120°linear 180°

Practice questions

Easy
  1. 1How is ethyne made in the lab?
  2. 2Why are alkynes acidic?
  3. 3State the test for a terminal alkyne.
Medium
  1. 4Product of C₂H₂ + H₂O/HgSO₄?
  2. 5Product of 3C₂H₂?
  3. 6C₂H₂ from 64 g CaC₂?
Challenging
  1. 7Difference between Lindlar and Na/NH₃.
  2. 8State the C≡C bond length and angle.
Show answers (try them yourself first)
  1. 1CaC₂ + 2H₂O.
  2. 2sp has 50 % s-character.
  3. 3White ppt with ammoniacal AgNO₃.
  4. 4CH₃CHO.
  5. 5Benzene.
  6. 626 g.
  7. 7cis versus trans alkene.
  8. 8120 pm, 180°.

Benzene: structure, resonance and aromaticity

From daily life

Naphthalene balls and many scented substances are aromatic; benzene itself is toxic.

Simple definition

Benzene C₆H₆: a planar hexagon, all C–C = 139 pm, 6 π electrons delocalised. Aromatic: planar, cyclic, fully conjugated, (4n + 2) π electrons (Hückel).

BenzeneBenzene: resonance and aromaticity↔=structure Istructure IIhybridAll C–C = 139 pm(between 154 and 134)Resonance energy ≈ 152 kJ/molhence very stableHückel rule: (4n + 2) π electrons, planar, cyclic, fully conjugatedBenzene: 6 π electrons (n = 1) ⇒ aromatic

In the figure : Resonance and aromaticity.

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Benzene: structure, resonance and aromaticity

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Rule 1: Resonance energy — Heat of hydrogenation: cyclohexene −120 kJ/mol ⇒ −360 for three double bonds; benzene actually −208; the difference is 152 kJ/mol.

  • Easy : 360 − 208 = 152 kJ/mol.
  • Medium : So benzene prefers substitution to addition.
  • Exam-type : Benzene does not decolourise Br₂/CCl₄.

Rule 2: Hückel's rule — Planar + cyclic + conjugated + (4n + 2) π electrons (n = 0, 1, 2…). 4n ⇒ anti-aromatic.

  • Easy : Benzene: 6 ⇒ n = 1 ⇒ aromatic.
  • Medium : Naphthalene: 10 π electrons ⇒ aromatic.
  • Exam-type : Cyclobutadiene: 4 ⇒ anti-aromatic.

Rule 3: Preparation and properties — 3C₂H₂ → C₆H₆ (873 K); phenol + Zn; C₆H₅COONa + soda lime. Benzene: colourless liquid, b.p. 353 K, burns with a sooty flame.

  • Easy : C₆H₅COONa + NaOH → C₆H₆ + Na₂CO₃.
  • Medium : C₆H₅OH + Zn → C₆H₆ + ZnO.
  • Exam-type : 2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O.

Calculation: resonance energy

  • Step 1: Expected (3 C=C) = 3 × (−120) = −360 kJ/mol.
  • Step 2: Observed = −208 kJ/mol.
  • Step 3: Difference = 360 − 208 = 152 kJ/mol (extra stability).

CET level (MCQ)

  • CET 1) C–C bond length in benzene? 1) 154 pm 2) 139 pm 3) 134 pm 4) 120 pm — Answer: 2).
  • CET 2) Which is aromatic? 1) cyclobutadiene 2) benzene 3) cyclohexane 4) cyclooctatetraene — Answer: 2).

Real-life use: Source of styrene, phenol, drugs and dyes.

Remember

139 pm; 6 π electrons; (4n + 2); 152 kJ/mol; substitution > addition.

Common mistake

Writing that Kekulé's two structures are different compounds; they are resonance forms.

CET trap

Resonance does not mean the molecule flips between forms; it is one hybrid.

Shortcut

Memory clue: "6, 10, 14 = aromatic" (4n + 2).

Spot the difference

Compoundπ e⁻Type
Benzene6aromatic
Naphthalene10aromatic
Cyclobutadiene4anti-aromatic

Practice questions

Easy
  1. 1State Hückel's rule.
  2. 2Why is benzene's C–C 139 pm?
  3. 3What is the resonance energy?
Medium
  1. 4Why does benzene not decolourise Br₂/CCl₄?
  2. 5How is benzene made from phenol?
  3. 6Why is naphthalene aromatic?
Challenging
  1. 7Equation for combustion of 2C₆H₆.
  2. 8What is anti-aromatic?
Show answers (try them yourself first)
  1. 1(4n + 2) π electrons, planar, cyclic, conjugated.
  2. 2Between 154 and 134 due to resonance.
  3. 3≈ 152 kJ/mol.
  4. 4It keeps its aromatic stability.
  5. 5C₆H₅OH + Zn → C₆H₆ + ZnO.
  6. 610 π electrons (4n + 2, n = 2).
  7. 72C₆H₆ + 15O₂ → 12CO₂ + 6H₂O.
  8. 8A cyclic conjugated system with 4n π electrons, unstable.

Electrophilic substitution and directing groups

From daily life

Making drugs and dyes from benzene means swapping an H for groups like NO₂ or Cl.

Simple definition

EAS: an electrophile E⁺ attacks the ring ⇒ σ-complex ⇒ H⁺ leaves and aromaticity is restored. Directing groups: an existing group decides where the next one goes.

EASElectrophilic aromatic substitution (EAS)C₆H₆ + E⁺σ-complex(arenium ion)C₆H₅–E+ H⁺Slow step: attack by E⁺; aromaticity is restoredCl₂/FeCl₃→ C₆H₅Clconc. HNO₃/H₂SO₄→ C₆H₅NO₂conc. H₂SO₄→ C₆H₅SO₃HRCl/AlCl₃Friedel–Craftso/p directors (activating)–OH, –NH₂, –CH₃, –X*m directors (deactivating)–NO₂, –CN, –CHO, –COOH*halogens deactivate yet direct o/p

In the figure : EAS mechanism and directors.

Rule 1: Five main EAS reactions — Halogenation (X₂/FeX₃), nitration (conc. HNO₃ + conc. H₂SO₄ ⇒ NO₂⁺), sulphonation, Friedel–Crafts alkylation and acylation (AlCl₃).

  • Easy : C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O.
  • Medium : C₆H₆ + CH₃Cl (AlCl₃) → C₆H₅CH₃ + HCl.
  • Exam-type : C₆H₆ + CH₃COCl (AlCl₃) → C₆H₅COCH₃ + HCl.

Rule 2: Directing groups — o/p directors (activating): –OH, –NH₂, –OR, –CH₃, –R; halogens deactivate yet direct o/p. m directors (deactivating): –NO₂, –CN, –CHO, –COOH, –SO₃H.

  • Easy : Toluene + Br₂ ⇒ o- and p-bromotoluene.
  • Medium : Nitrobenzene + Br₂ ⇒ m-bromonitrobenzene.
  • Exam-type : Phenol + Br₂ water ⇒ 2,4,6-tribromophenol (white).

Rule 3: Friedel–Crafts limitations — Fails on deactivated rings (–NO₂); alkylation may rearrange, acylation does not.

  • Easy : Nitrobenzene is inert to Friedel–Crafts.
  • Medium : n-propyl chloride ⇒ mainly isopropylbenzene.
  • Exam-type : Acylation + Zn/Hg, HCl ⇒ n-alkylbenzene.

Reasoning: why is –OH an o/p director?

  • Step 1: The lone pair on O delocalises into the ring.
  • Step 2: Electron density rises at the ortho and para positions.
  • Step 3: E⁺ attack there gives a more stable σ-complex.

CET level (MCQ)

  • CET 1) The attacking species in nitration? 1) NO₂⁺ 2) NO₂⁻ 3) NO₃⁻ 4) NO⁺ — Answer: 1).
  • CET 2) A meta director? 1) –OH 2) –CH₃ 3) –NO₂ 4) –Cl — Answer: 3).

Real-life use: TNT, dyes, drugs, detergents (sulphonation).

Remember

E⁺ ⇒ σ-complex ⇒ H⁺ lost; –OH, –NH₂ o/p; –NO₂ m; acylation has no rearrangement.

Common mistake

Saying halogens are m directors; they are deactivating yet o/p.

CET trap

Friedel–Crafts fails on nitrobenzene.

Shortcut

Memory clue: "electron donors are o/p; withdrawers are m".

Spot the difference

GroupEffectPosition
–OH, –NH₂, –CH₃activatingo/p
–Cl, –Brdeactivatingo/p
–NO₂, –CN, –COOHdeactivatingm

Practice questions

Easy
  1. 1What is EAS?
  2. 2What is the nitrating mixture?
  3. 3Write Friedel–Crafts acylation.
Medium
  1. 4Products of toluene + Br₂?
  2. 5Why is –NO₂ a m director?
  3. 6Why no Friedel–Crafts on nitrobenzene?
Challenging
  1. 7Phenol + bromine water?
  2. 8What is rearrangement in alkylation?
Show answers (try them yourself first)
  1. 1Electrophilic substitution on benzene.
  2. 2Conc. HNO₃ + conc. H₂SO₄.
  3. 3C₆H₆ + CH₃COCl → C₆H₅COCH₃ + HCl (AlCl₃).
  4. 4o- and p-bromotoluene.
  5. 5It withdraws electrons, lowering density at o/p.
  6. 6The ring is strongly deactivated.
  7. 72,4,6-tribromophenol.
  8. 8The cation shifts to a more stable form.

Quick revision of the whole lesson

ConceptKey formula / fact
AlkaneCₙH₂ₙ₊₂ ; Wurtz ; Kolbe
Conformationstaggered < eclipsed (+12 kJ/mol)
AlkeneMarkovnikov ; peroxide HBr
TestBr₂/CCl₄, Baeyer
Alkynesp, 180°, acidic H
Benzene139 pm, 152 kJ/mol
Hückel4n + 2
EASE⁺ → σ-complex → H⁺ lost
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🎯 Most important points for the exam

  • Classification and alkane preparation.
  • Conformations, combustion, halogenation.
  • Alkene preparation and Markovnikov.
  • Alkene reactions and polymerisation.
  • Alkynes: acidity and reactions.
  • Benzene and aromaticity.
  • EAS and directing groups.
Study mantra

understand the idea → remember the formula → solve the example → solve problems yourself.