Hydrocarbons are compounds containing only carbon and hydrogen. They are the most important sources of energy (LPG, CNG, petrol, diesel, kerosene) and are used in manufacture of polymers, dyes, drugs and solvents.
LPG = Liquefied Petroleum Gas · CNG = Compressed Natural Gas · LNG = Liquefied Natural Gas
🎨
AI Image: Classification of Hydrocarbons – Mind Map
📄 AI Generate
A complete hierarchical tree / mind map showing the classification of hydrocarbons into saturated, unsaturated, and aromatic, with sub-types and examples.
🤖 AI Image Prompt
A clean professional chemistry mind map on a pure white background showing the classification of hydrocarbons. Root node: "Hydrocarbons" in a dark navy rectangle. Three main branches: (1) "Saturated — Alkanes (CnH2n+2)" in teal, sub-branches: "Open chain (alkanes)" e.g. CH4, C2H6 and "Cyclic (cycloalkanes)" e.g. Cyclopropane. (2) "Unsaturated" in purple, sub-branches: "Alkenes (CnH2n)" e.g. C2H4, "Alkynes (CnH2n-2)" e.g. C2H2, "Dienes" e.g. Buta-1,3-diene. (3) "Aromatic" in dark green, sub-branches: "Benzenoid" e.g. Benzene, Toluene, Naphthalene and "Non-benzenoid" e.g. Tropone. All connecting arrows in amber/gold. Clean sans-serif fonts. White background only. No decorative patterns.
Fig. 9.0 — Classification tree of hydrocarbons
Type
General Formula
Bonds Present
Examples
Alkanes (Saturated)
CnH2n+2
C–C and C–H single bonds only
CH4, C2H6, C3H8
Cycloalkanes
CnH2n
C–C single bonds in a ring
Cyclopropane, Cyclohexane
Alkenes (Unsaturated)
CnH2n
At least one C=C double bond
C2H4, C3H6
Alkynes (Unsaturated)
CnH2n–2
At least one C≡C triple bond
C2H2, C3H4
Aromatic
Variable
Delocalised π electrons; (4n+2)π rule
Benzene C6H6, Toluene, Naphthalene
9.2 Alkanes
Definition
Alkanes are saturated open-chain hydrocarbons with only C–C and C–H single bonds. General formula: CnH2n+2. Earlier called paraffins (Latin: parum = little, affinis = affinity) because they are largely inert.
First member: Methane (CH4) — found in coal mines and marshy places.
9.2.1 Nomenclature and Isomerism
Methane, ethane, and propane each have only one possible structure. From butane onwards, multiple structural isomers exist.
3D ball-and-stick model of methane showing carbon at centre and four hydrogen atoms at corners of a regular tetrahedron. All H–C–H bond angles = 109.5°.
Fig. 9.1 — Structure of methane (tetrahedral, sp³ hybridised C)
9.2.2 Preparation of Alkanes
#
Method
Reaction / Key Details
Equation No.
1
From Unsaturated Hydrocarbons (Hydrogenation)
Alkenes/alkynes + H2 → Alkanes in presence of finely divided Pt, Pd or Ni catalyst. Pt/Pd work at room temp; Ni needs higher temp and pressure.
2R–X + 2Na → R–R + 2NaX (dry ether). Gives alkanes with even number of C atoms (same alkyl halide). If two different halides are used, mixture of 3 products forms.
9.7 – 9.8
3(i)
Decarboxylation
R–COONa + NaOH → R–H + Na2CO3 (soda lime, CaO/NaOH, heat). Alkane has one C less than the acid.
9.9
3(ii)
Kolbe's Electrolytic Method
Electrolysis of aq. solution of sodium/potassium salt of carboxylic acid. At anode: 2R–COO⁻ → R–R + 2CO₂ + 2e⁻. Gives alkane with even number of C atoms. Methane CANNOT be prepared by this method.
Wurtz reaction: Use same alkyl halides → one product. Mixed halides → mixture of three alkanes (not useful for preparation).
Kolbe's method: Methane CANNOT be prepared (acetate gives ethane at anode — C1 has no pair to form).
Decarboxylation: To get propane (C3), use sodium butanoate (C4 acid).
9.2.3 Properties of Alkanes
Physical Properties
Almost non-polar (very small electronegativity difference between C and H); possess weak van der Waals forces.
State at 298 K: C1–C4 = gases; C5–C17 = liquids; C18+ = solids.
Boiling point: increases steadily with molecular mass (more surface area → stronger van der Waals forces). Branched chain isomers have lower bp than straight chain (spherical shape → smaller contact area).
Colourless and odourless; insoluble in water; like dissolves like (polar in polar, non-polar in non-polar).
Name
Formula
Mol. mass (u)
bp (K)
mp (K)
Methane
CH4
16
111.0
90.5
Ethane
C2H6
30
184.4
101.0
Propane
C3H8
44
230.9
85.3
Butane
C4H10
58
272.4
134.6
2-Methylpropane
C4H10
58
261.0
114.7
Pentane
C5H12
72
309.1
143.3
2-Methylbutane
C5H12
72
300.9
113.1
2,2-Dimethylpropane
C5H12
72
282.5
256.4
Hexane
C6H14
86
341.9
178.5
Decane
C10H22
142
447.1
243.3
Chemical Properties
Reaction
Conditions
Products / Key Points
1. Halogenation (Substitution)
hν (UV/diffuse sunlight) or 573–773 K
H replaced by halogen. Rate: F2 > Cl2 > Br2 > I2. Rate of H replacement: 3° > 2° > 1°. Proceeds by free radical chain mechanism (initiation → propagation → termination).
2. Combustion
Air/O2, heat
Complete: CnH2n+2 + O2 → CO2 + H2O + large heat. CH4: ΔcH = –890 kJ/mol; C4H10: ΔcH = –2875.84 kJ/mol. Incomplete: gives C (carbon black) used in inks, pigments.
During termination, two methyl free radicals can combine to form ethane (reaction b above). This is confirmed by the free radical chain mechanism.
9.2.4 Conformations of Ethane
Key Definitions
Conformations (Conformers / Rotamers): Different spatial arrangements of atoms in a molecule that can be interconverted by rotation around a C–C single bond. Alkanes can have infinite conformations, but rotation is NOT completely free — hindered by torsional strain (1–20 kJ/mol).
Torsional strain: Weak repulsive interaction between electron clouds of adjacent C–H bonds that resist rotation.
Dihedral angle (torsional angle): The angle of rotation about the C–C bond.
Eclipsed Conformation
H atoms on adjacent carbons are as close together as possible (0° dihedral angle). Maximum electron cloud repulsion → maximum torsional strain → least stable.
Staggered Conformation
H atoms are as far apart as possible (60° dihedral angle). Minimum repulsion → minimum torsional strain → most stable. Preferred conformation.
Any intermediate conformation = Skew conformation. Energy difference between eclipsed and staggered = 12.5 kJ/mol (small enough that free rotation occurs at room temperature).
Representation Methods
Sawhorse Projection
View along C–C axis. C–C drawn as a long diagonal line. Front carbon at lower end, rear carbon at upper end. Each C has 3 lines for H at 120°.
Newman Projection
View at C–C bond head-on. Front carbon = central point; rear carbon = circle. Each has 3 H-bonds at 120°. Eclipsed: H overlaps; Staggered: H at 60° to each other.
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Fig. 9.2 & 9.3 – Sawhorse and Newman Projections of Ethane Conformations
📄 NCERT Page 305
Fig 9.2: Sawhorse projections showing (i) Eclipsed conformation — H atoms aligned on adjacent carbons; (ii) Staggered conformation — H atoms maximally separated.
Fig 9.3: Newman projections showing same two conformations viewed head-on along C–C bond. Front carbon = point, rear carbon = circle. Eclipsed: bonds overlap; Staggered: bonds offset by 60°.
Fig. 9.2 & 9.3 — Sawhorse and Newman projections of eclipsed and staggered ethane conformations
💡
Stability Rule
Staggered > Skew > Eclipsed (in order of decreasing stability). Staggered = minimum torsional strain = preferred. Ethane conformers CANNOT be isolated (energy barrier too small at room temp).
9.3 Alkenes
Definition
Alkenes are unsaturated hydrocarbons containing at least one C=C double bond. General formula: CnH2n. Also called olefins (oil-forming) because ethylene forms an oily liquid with Cl2. First stable member: Ethene (C2H4).
9.3.1 Structure of the Double Bond
C=C consists of: one σ bond (sp² head-on overlap, enthalpy ≈ 397 kJ/mol) + one π bond (lateral overlap of 2p orbitals, enthalpy ≈ 284 kJ/mol)
C=C bond length = 134 pm (shorter than C–C single bond 154 pm)
Total C=C bond enthalpy = 681 kJ/mol > C–C single bond (348 kJ/mol)
π bond is weaker → alkenes are electron-rich sources (easily attacked by electrophiles)
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Fig. 9.4 & 9.5 – Orbital Diagrams of Ethene (σ bonds, π bond, π cloud, bond angles)
📄 NCERT Pages 306–307
Fig 9.4: Orbital picture showing only σ bonds in ethene — sp²–sp² C–C sigma bond in the centre and sp²–s C–H sigma bonds.
Fig 9.5: (a) π-bond formation by sideways overlap of p orbitals; (b) π-cloud above and below the plane; (c) Bond angles (116.6° and 121.7°) and bond lengths (C=C = 134 pm, C–H = 110 pm).
Fig. 9.4 & 9.5 — Orbital structure of ethene showing σ bonds only (9.4) and π-bond, π-cloud, and bond parameters (9.5)
9.3.2 Nomenclature
Suffix: –ene replaces –ane of alkane
Longest chain containing the C=C is the parent chain
Numbering from end nearer to the C=C double bond
Position of double bond indicated by lower numbered C of the pair
Structure
IUPAC Name
CH3–CH=CH2
Propene
CH3–CH2–CH=CH2
But-1-ene
CH3–CH=CH–CH3
But-2-ene
CH2=CH–CH=CH2
Buta-1,3-diene
CH2=C(CH3)–CH3
2-Methylprop-1-ene
CH2=CH–CH(CH3)–CH3
3-Methylbut-1-ene
9.3.3 Isomerism in Alkenes
Structural Isomerism (C₄H₈ — 3 isomers)
Structure
Name
Type of Isomerism
CH2=CH–CH2–CH3
But-1-ene
Position isomers (I vs II)
CH3–CH=CH–CH3
But-2-ene
Position isomers (I vs II)
CH2=C(CH3)2
2-Methylprop-1-ene
Chain isomers (I or II vs III)
Geometrical (cis-trans) Isomerism
Definition
Occurs when rotation about C=C is restricted. If each doubly bonded carbon has two different groups, two spatial arrangements are possible:
cis: identical groups on the SAME side of the double bond
trans: identical groups on OPPOSITE sides
Condition for cis-trans isomerism: Both doubly bonded carbons must each carry two DIFFERENT substituents. If either carbon carries two identical groups → no cis-trans isomerism.
cis-But-2-ene
μ = 0.33 D · bp 277 K More polar · Higher mp (solids)
⇌
trans-But-2-ene
μ ≈ 0 D · bp 274 K Non-polar · Lower bp · Higher mp
cis vs trans — Key Differences
Polarity: cis is more polar (bond dipoles don't cancel); trans ≈ non-polar (dipoles cancel)
Melting point of solids: trans > cis (more symmetric → better packing in crystal)
Dipole moment: cis-but-2-ene = 0.33 D; trans-but-2-ene ≈ 0 D
9.3.4 Preparation of Alkenes
#
Method
Reaction & Key Details
1
From Alkynes (Partial Hydrogenation)
Lindlar's catalyst (palladised charcoal + poison like quinoline/sulphur) → cis-alkene. Na in liquid NH3 → trans-alkene.
2
From Alkyl Halides (Dehydrohalogenation)
R–X + alc. KOH → alkene + HX. β-elimination (H removed from β-carbon). Rate for halogens: I > Br > Cl. Rate for alkyl: tertiary > secondary > primary.
3
From Vicinal Dihalides (Dehalogenation)
X–CH2–CH2–X + Zn → CH2=CH2 + ZnX2
4
Acidic Dehydration of Alcohols
R–OH + conc. H2SO4 → alkene + H2O (β-elimination). –OH removes H from β-carbon.
9.3.5 Properties of Alkenes
First three members = gases; next 14 = liquids; higher = solids. Colourless, odourless (except ethene which has faint sweet smell). Insoluble in water; soluble in non-polar solvents.
Chemical Properties — Addition Reactions
Reaction
Reagent / Conditions
Product
Key Notes
1. Addition of H2 (Hydrogenation)
H2, Ni/Pt/Pd
Alkane
Catalytic hydrogenation
2. Addition of Halogens
Br2 or Cl2 in CCl4
Vicinal dihalide
Reddish-orange Br2 decolourised = test for unsaturation. I2 does NOT add under normal conditions.
Markovnikov's Rule (1869): In the addition of an unsymmetrical reagent (H–X) to an unsymmetrical alkene, the negative part of the addendum (X⁻) attaches to the carbon atom that has the lesser number of hydrogen atoms.
Mechanistic explanation: H⁺ attacks the C=C to form two possible carbocations. The more stable carbocation (2° > 1°) forms faster and predominates → product determined by stability of intermediate carbocation.
CH3–CH=CH2 + HBr →
H⁺ adds to C3 (more H): forms CH3–CH⁺–CH3 (2° carbocation — more stable)
Br⁻ attacks C2: → CH3–CHBr–CH3 (2-Bromopropane — MAJOR product ✓)
H⁺ adds to C2: forms CH3–CH2–CH2⁺ (1° carbocation — less stable)
Br⁻ attacks C3: → CH3–CH2–CH2Br (1-Bromopropane — MINOR product)
In the presence of peroxide, HBr adds to unsymmetrical alkenes in anti-Markovnikov fashion (Br goes to the carbon with MORE H). Observed ONLY with HBr (not HCl or HI). Free radical mechanism (not ionic).
Why not HCl? H–Cl bond (430.5 kJ/mol) too strong for free radical cleavage. Why not HI? I• free radicals combine to form I2 instead of adding to double bond (H–I bond = 296.8 kJ/mol, too weak).
HBr (363.7 kJ/mol) = just right.
If ozonolysis gives two specific aldehydes/ketones, we can work backwards to find the original alkene structure. This is the primary use of ozonolysis.
9.4 Alkynes
Definition
Alkynes are unsaturated hydrocarbons with at least one C≡C triple bond. General formula: CnH2n–2. First stable member: Ethyne (C2H2) — commonly called acetylene. Used for oxyacetylene welding.
9.4.2 Structure of the Triple Bond (Ethyne)
Each C in ethyne: sp hybridised (2 sp orbitals + 2 unhybridised p orbitals)
C≡C: one σ (sp–sp head-on overlap) + two π bonds (lateral overlap of 2px and 2py orbitals)
C–H bonds: sp–1s overlap
H–C–C bond angle: 180° (linear molecule)
C≡C bond length: 120 pm (shorter than C=C 134 pm and C–C 154 pm)
Electron cloud is cylindrically symmetrical about the internuclear axis
🖼️
Fig. 9.6 – Orbital Picture of Ethyne (σ and π overlaps)
📄 NCERT Page 315
(a) σ-bond picture: H–C–C–H linear arrangement, showing sp hybrid orbitals (1s on H, sp on each C) forming two C–H and one C–C σ bonds. All atoms in a straight line.
(b) π-bond picture: Two sets of p orbitals (2px and 2py) on each C overlap laterally to form two π bonds. The 2px orbitals overlap in the plane of the paper (π bond), and 2py orbitals overlap perpendicular to the paper (second π bond).
Fig. 9.6 — Orbital picture of ethyne: (a) sigma overlaps forming the C–C and C–H bonds; (b) two pi bonds formed by lateral overlap of p orbitals
9.4.1 Nomenclature and Isomerism
Suffix: –yne replaces –ane. In common system: derivatives of acetylene (e.g., methylacetylene).
First 3 = gases; next 8 = liquids; higher = solids. All colourless; ethyne has characteristic odour. Weakly polar. Lighter than water; immiscible with water; soluble in organic solvents.
A. Acidic Character
KEY CONCEPT
Terminal alkynes (H attached to triply bonded C) are acidic — they react with strong bases like Na metal and NaNH2 to liberate H2.
Reason: sp hybridised C (50% s character) is most electronegative → attracts C–H shared electrons more strongly → H is released as H⁺ more easily than in alkenes (sp²) or alkanes (sp³).
Linear: → Polyacetylene (conducting polymer used in batteries) Cyclic: 3 C2H2 → Benzene (red hot iron tube, 873 K)
Gem vs Vicinal Dihalide
Gem dihalide: two halogens on the same carbon (from HX + alkyne)
Vicinal dihalide: two halogens on adjacent carbons (from X2 + alkene)
Cyclic Polymerisation — Benzene from Ethyne
3 HC≡CH —(red hot Fe tube, 873 K)→ C6H6 (Benzene)
This is the best route from aliphatic to aromatic compounds.
🎨
AI Image: Cyclic Polymerisation of Ethyne to Benzene
📄 AI Generate (NCERT p.318)
Reaction diagram showing 3 molecules of ethyne (HC≡CH) combining in a red hot iron tube at 873 K to form benzene ring.
🤖 AI Image Prompt
A clean chemistry reaction diagram on pure white background. Shows 3 structural formulas of ethyne (H–C≡C–H) on the left side with a large arrow pointing right, labelled "Red hot iron tube, 873 K" above the arrow. On the right side, show the benzene ring structure (hexagon with a circle inside, in amber/gold colour). Below the benzene, write "Benzene (C₆H₆)". The three ethyne molecules should be arranged diagonally or in a triangular arrangement to suggest they come together. All molecular structures drawn clearly with proper bond lines. Professional chemistry diagram style. White background only. No text other than the labels mentioned.
Cyclic polymerisation of ethyne → benzene (aromatisation route from aliphatic to aromatic)
9.5 Aromatic Hydrocarbons (Arenes)
Aromatic hydrocarbons are also called arenes (from Greek aroma = pleasant smell). Most contain benzene ring. Benzenoids contain benzene ring; non-benzenoids don't (e.g. Tropone).
Benzene
C₆H₆
Toluene
Methylbenzene
Naphthalene
C₁₀H₈ · Two fused rings
9.5.2 Structure of Benzene
Historical Facts
Benzene isolated by Michael Faraday (1825)
Molecular formula C6H6 — indicates high degree of unsaturation
Benzene forms a triozonide → 3 double bonds present
Benzene gives only one monosubstituted product → all 6 C and H atoms are identical
Kekulé (1865): proposed cyclic structure with alternating single and double bonds
Problem with Kekulé: predicts 2 ortho-disubstituted isomers, but only ONE is observed → solved by "oscillating double bonds"
Modern explanation: resonance (delocalisation)
Electronic Structure of Benzene
All 6 C atoms are sp² hybridised
Each C forms σ bonds with 2 adjacent C atoms and 1 H atom → hexagonal planar framework
Each C has one unhybridised 2p orbital perpendicular to the ring plane
Equal probability of p orbitals of each C overlapping with neighbours → 6 π electrons completely delocalised over all 6 C atoms
Represented as: two doughnut-shaped electron clouds above and below the ring
Kekulé A
Structure A
⇌
Kekulé B
Structure B
→
Resonance Hybrid (C)
Actual structure
C–C = 139 pm (all equal)
🖼️
Fig. 9.7 (a), (b), (c), (d) – Orbital Overlap in Benzene & Delocalised π Cloud
📄 NCERT Pages 320–321
Fig 9.7(a): p orbitals overlapping as in Kekulé structure A (C1–C2, C3–C4, C5–C6 π bonds).
Fig 9.7(b): p orbitals overlapping as in Kekulé structure B (C2–C3, C4–C5, C6–C1 π bonds).
Fig 9.7(c): Actual equal overlap — each C's p orbital overlaps equally with both neighbours; dashed lines showing equal probability.
Fig 9.7(d): The resulting delocalised π electron cloud — two doughnut-shaped lobes, one above and one below the plane of the hexagonal ring.
Fig. 9.7 — Orbital overlap and delocalised π electron cloud in benzene: (a) Kekulé A, (b) Kekulé B, (c) equal overlap (actual), (d) doughnut-shaped π clouds above and below the ring
KEY CONCEPT
Why benzene prefers substitution over addition:
Benzene's delocalised π electrons are attracted more strongly by the 6 carbon nuclei than localised electrons between 2 carbons. This extra stability (resonance energy) means benzene resists addition reactions (which would destroy delocalisation) and instead undergoes electrophilic substitution (which retains the aromatic ring).
X-Ray data: all C–C = 139 pm (intermediate between single 154 and double 134). No pure double bonds → explains resistance to addition.
9.5.3 Aromaticity
KEY CONCEPT
Hückel's Rule of Aromaticity
A compound is aromatic if it satisfies ALL THREE conditions:
Planarity: the molecule is flat (planar)
Complete delocalisation of π electrons in the ring
(4n + 2) π electrons in the ring where n = 0, 1, 2, 3, … (Hückel Rule)
Compound
n value
π electrons
Aromatic?
Benzene
n=1
6π
✅ Yes
Cyclopentadienyl anion (C5H5⁻)
n=1
6π
✅ Yes
Cycloheptatrienyl cation (C7H7⁺)
n=1
6π
✅ Yes
Naphthalene
n=2
10π
✅ Yes
Anthracene / Phenanthrene
n=3
14π
✅ Yes
Cyclobutadiene
—
4π (4n, n=1)
❌ Anti-aromatic
Cyclohexadiene (non-planar ring)
—
Not fully delocalised
❌ Not aromatic
9.5.4 Preparation of Benzene
Method
Reaction
Commercial
Isolated from coal tar (fractional distillation)
Cyclic polymerisation of ethyne
3 HC≡CH →(Fe, 873 K)→ C6H6
Decarboxylation of benzoic acid
C6H5COONa + NaOH →(CaO/Δ)→ C6H6 + Na2CO3
Reduction of phenol
C6H5OH + Zn →(Δ)→ C6H6 + ZnO
9.5.5 Electrophilic Substitution Reactions of Benzene
Definition
Arenes undergo electrophilic substitution (SE) reactions in which an electrophile (E⁺) replaces a hydrogen atom on the ring, retaining aromaticity. Three steps: (a) Generation of E⁺, (b) Formation of σ-complex (arenium ion), (c) Removal of H⁺ to restore aromaticity.
Reaction
Reagents & Conditions
Electrophile Generated
Product
Eq. No.
1. Nitration
Conc. HNO3 + Conc. H2SO4 (nitrating mixture), 323–333 K
NO2⁺ (Nitronium ion)
Nitrobenzene (C6H5NO2)
9.72
2. Halogenation
Cl2 or Br2, anhydrous FeCl3/FeBr3/AlCl3 (Lewis acid)
Cl⁺ or Br⁺
Chlorobenzene / Bromobenzene
9.73
3. Sulphonation
Fuming H2SO4 (oleum), heat
SO3 (electrophilic)
Benzene sulphonic acid (C6H5SO3H)
9.74
4. Friedel-Crafts Alkylation
Alkyl halide (R–X) + anhydrous AlCl3
R⁺ (carbocation)
Alkylbenzene (e.g. Toluene with CH3Cl)
9.75–9.76
5. Friedel-Crafts Acylation
Acyl halide (R–COX) or acid anhydride + AlCl3
RCO⁺ (Acylium ion)
Acylbenzene (e.g. Acetophenone with CH3COCl)
9.77–9.78
Mechanism of Electrophilic Substitution (SE) — Detailed
(a) Generation of Electrophile E⁺:
For nitration: HNO3 + H2SO4 → H2NO3⁺ + HSO4⁻ → H2O + ⁺NO2 (Nitronium ion)
For halogenation: Cl–Cl + AlCl3 → Cl⁺ + [AlCl4]⁻
For F-C alkylation: CH3Cl + AlCl3 → CH3⁺ + [AlCl4]⁻
(b) Formation of σ-complex (Arenium ion):
Benzene + E⁺ → Arenium ion (σ-complex) — one C becomes sp³, aromaticity LOST; ion stabilised by resonance over remaining π system.
(c) Removal of H⁺ (Restoration of Aromaticity):
[AlCl4]⁻ or [HSO4]⁻ removes H⁺ from sp³ carbon → aromaticity restored → substituted benzene + HCl/H2SO4 + AlCl3 (Lewis acid regenerated)
AI Image: Mechanism of Electrophilic Substitution (Nitration of Benzene)
📄 AI Generate (NCERT p.323)
Step-by-step mechanism showing: (1) generation of NO₂⁺, (2) attack on benzene ring to form arenium ion (σ-complex) with sp³ carbon shown, (3) loss of H⁺ to restore benzene ring with NO₂ group.
🤖 AI Image Prompt
A clean three-step chemistry mechanism diagram on a pure white background showing electrophilic nitration of benzene. Step 1 (top): Shows HNO3 + H2SO4 producing nitronium ion (NO2+) + HSO4-. Step 2 (middle): Shows benzene ring (hexagon with circle) being attacked by NO2+ to form the sigma complex/arenium ion — show the benzene ring with one sp3 carbon (indicated by a small H on top of that carbon), NO2 attached, and a positive charge distributed over the ring shown by resonance brackets. Step 3 (bottom): Shows the arenium ion losing H+ to HSO4- to form nitrobenzene (benzene ring with NO2 group) + H2SO4 + restored aromaticity. Each step has a curved arrow showing electron movement. All drawn in black ink on white background. Professional organic chemistry mechanism style. No extra text or decorative elements.
Mechanism of electrophilic nitration of benzene: generation of NO₂⁺ → σ-complex (arenium ion) → loss of H⁺ → nitrobenzene
9.5.6 Directive Influence of Substituents in Monosubstituted Benzene
Definition
When monosubstituted benzene undergoes further substitution, the incoming group is directed to specific positions (ortho, meta or para) based on the nature of the substituent already present — not on the nature of the incoming group.
Ortho & Para Directing Groups (Activating)
Mechanism: These groups donate electrons into the ring via +R or +I effect → electron density increases at o- and p-positions → electrophile attacks o- and p-positions.
Halogens are o- and p-directing but moderately deactivating (strong –I but +R effect increases density at o/p)
These groups make benzene ring MORE reactive toward electrophilic substitution.
Meta Directing Groups (Deactivating)
Mechanism: These groups withdraw electrons from ring via –R or –I effect → electron density decreases more at o- and p-positions than at m-position → electrophile attacks the relatively electron-rich meta position.
These groups make benzene ring LESS reactive toward electrophilic substitution.
📝 Examples of Directive Effects
Phenol (–OH group): Resonance structures show high electron density at C-2, C-4, C-6 (ortho and para positions) → attack at o- and p-positions → o-nitrophenol + p-nitrophenol (major products)
Nitrobenzene (–NO2 group): –NO2 withdraws electrons strongly (–I and –R) → electron density decreases at o- and p-positions more than m → electrophile attacks m-position → m-dinitrobenzene (major)
Toluene (–CH3 group): +I effect of alkyl group + hyperconjugation → increased electron density at o- and p-positions → o-nitrotoluene + p-nitrotoluene (major)
🎨
AI Image: Resonance Structures of Phenol – Showing o- and p-Directing Nature
📄 AI Generate (NCERT p.324)
Five resonance structures of phenol connected by double-headed arrows, showing how electron density from the lone pair on oxygen delocalises into the ortho and para positions of the benzene ring.
🤖 AI Image Prompt
A clean chemistry resonance diagram on a pure white background showing five resonance structures of phenol (C6H5OH). Structure I: normal phenol with lone pairs on oxygen and the benzene ring represented as a hexagon with alternating bonds. Structure II: curved arrow from O lone pair into ring, positive charge on O, negative charge at ortho carbon (C2). Structure III: negative charge moves to para carbon (C4), positive on O. Structure IV: different ortho position (C6). Structure V: another resonance form. All five structures connected by double-headed arrows. Below each structure, label the position of the partial negative charge (ortho-2, para-4, ortho-6). At the bottom, add the summary: "High electron density at ortho (2,6) and para (4) positions → o- and p-substitution". Black structural drawing on white background. Professional organic chemistry style.
Resonance structures of phenol: electron density builds up at o- and p-positions → ortho and para directing
Substituent already on ring
Directing Effect
Activating/Deactivating?
Effect on ring reactivity
–OH, –NH2, –OCH3, –CH3
o/p directing
Activating
Ring MORE reactive
–F, –Cl, –Br, –I (halogens)
o/p directing
Moderately deactivating
Ring less reactive
–NO2, –CN, –COOH, –CHO, –SO3H
m directing
Strongly deactivating
Ring much less reactive
9.6 Carcinogenicity and Toxicity
Important
Benzene and polynuclear hydrocarbons (containing more than two fused benzene rings) are toxic and possess carcinogenic (cancer-causing) properties.
They are formed during incomplete combustion of organic materials like tobacco, coal and petroleum. They enter the human body, undergo biochemical reactions, damage DNA and cause cancer.
Carcinogenic Hydrocarbon
Notes
1,2-Benzanthracene
Polynuclear aromatic hydrocarbon; fused ring system
3-Methylcholanthrene
Very potent carcinogen
1,2-Benzpyrene
Found in cigarette smoke
1,2,5,6-Dibenzanthracene
Strong carcinogen
9,10-Dimethyl-1,2-benzanthracene
Potent carcinogen
Use of polythene bags and polypropylene is also a concern — excessive use raises environmental issues.
⚡ Comprehensive Comparison: Alkanes vs Alkenes vs Alkynes
Property
Alkanes
Alkenes
Alkynes
General Formula
CnH2n+2
CnH2n
CnH2n–2
Hybridisation
sp³
sp² (at C=C)
sp (at C≡C)
Bond angle
109.5°
120°
180°
C–C bond length
154 pm (single)
134 pm (double)
120 pm (triple)
Molecular shape
Tetrahedral at each C
Planar around C=C
Linear around C≡C
Characteristic reactions
Substitution (free radical)
Addition (electrophilic)
Addition (electrophilic); Acidic character
Test for unsaturation
No reaction with Br2/KMnO4
Decolourises Br2/KMnO4
Decolourises Br2/KMnO4
Acidic character
None
None
Terminal H is acidic (sp C)
Isomerism
Chain, position
Chain, position, geometrical (cis-trans)
Chain, position
First stable member
Methane (CH4)
Ethene (C2H4)
Ethyne (C2H2)
✏️ Practice Questions
Q1
How do you account for the formation of ethane as a by-product during chlorination of methane?
During the termination step of free radical chain mechanism, two methyl free radicals (•CH₃) combine to form ethane: •CH₃ + •CH₃ → CH₃–CH₃. This is one of the chain-terminating steps which consumes the free radicals and produces ethane as a by-product.
Q2
An alkene A on ozonolysis gives a mixture of ethanal and pentan-3-one. Write the structure and IUPAC name of A.
Ethanal = CH₃CHO (from C1=C2 part), Pentan-3-one = CH₃CH₂COCH₂CH₃ (from C2=C part with ethyl groups). Working back: A = CH₃CH=C(CH₂CH₃)₂ → actually CH₃CH=CHCOCH₂CH₃... Let's think: ethanal gives –CH=, pentan-3-one gives =C(CH₂CH₃)– side. So A = CH₃–CH=C(CH₂CH₃)–CH₂CH₃ ... IUPAC: The alkene is 3-ethylpent-2-ene: CH₃–CH=C(C₂H₅)–CH₂CH₃. On ozonolysis: CH₃CHO + CH₃CH₂CO–CH₂CH₃ (pentan-3-one). ✓
Q3
An alkene A contains three C–C bonds, eight C–H σ bonds and one C–C π bond. On ozonolysis it gives two moles of an aldehyde of molar mass 44 u. Write the IUPAC name of A.
Aldehyde with molar mass 44 u = CH₃CHO (ethanal, MW = 44). Two moles of same aldehyde → A is symmetrical alkene. Both sides give CH₃CHO → A is CH₃–CH=CH–CH₃. IUPAC name: But-2-ene.
Q4
Addition of HBr to propene gives 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction gives 1-bromopropane. Explain.
Without peroxide: Electrophilic addition via Markovnikov rule. H⁺ attacks C1 (more H), forming more stable 2° carbocation at C2. Br⁻ attacks C2 → 2-bromopropane. With peroxide (Kharash effect): Free radical mechanism. Peroxide generates Br• radical which attacks C1 (less substituted) forming more stable 2° carbon radical at C2. H• then attacks C2 from HBr → 1-bromopropane (anti-Markovnikov).
Q5
Why is benzene extraordinarily stable though it contains three double bonds?
Benzene is a resonance hybrid of two Kekulé structures. All six π electrons are completely delocalised over all 6 carbon atoms forming a stable aromatic ring. The delocalised electrons are attracted more strongly by the 6 carbon nuclei than localised electrons between 2 carbons would be. This resonance stabilisation (resonance energy) makes benzene much more stable than the hypothetical cyclohexatriene. X-ray data confirms: all C–C bonds are equal at 139 pm — no pure double bonds exist.
Q6
What are the necessary conditions for any system to be aromatic?
(i) Planarity — the molecule must be flat. (ii) Complete delocalisation of π electrons in the ring — continuous p-orbital overlap. (iii) Presence of (4n + 2) π electrons where n = 0, 1, 2, 3,... (Hückel Rule). Examples: Benzene (n=1, 6π), Naphthalene (n=2, 10π), Cyclopentadienyl anion (n=1, 6π).
Q7
Draw the cis and trans structures of hex-2-ene. Which isomer will have higher boiling point and why?
cis-Hex-2-ene: CH₃ and CH₂CH₂CH₃ on same side of C=C. trans-Hex-2-ene: on opposite sides. cis isomer has higher boiling point because it is more polar (dipole moments of the C-C bonds don't fully cancel), resulting in stronger intermolecular dipole-dipole interactions compared to the non-polar trans isomer.
Q8
Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Give reason.
Ethyne > Benzene > n-Hexane. Reason: C–H acidity depends on electronegativity of the C atom (i.e., s-character). Ethyne C is sp hybridised (50% s) → most electronegative → H is most acidic. Benzene C is sp² (33% s) → intermediate. n-Hexane C is sp³ (25% s) → least electronegative → H is least acidic.
Q9
Why is the Wurtz reaction not preferred for preparation of alkanes with an odd number of carbon atoms?
Wurtz reaction: 2R–X + 2Na → R–R + 2NaX. Using the same alkyl halide always gives a product with an even number of carbons. Using two different alkyl halides (R–X + R'–X) gives a mixture of THREE products: R–R, R–R', and R'–R'. Separation of the desired product from this mixture is impractical. Therefore Wurtz reaction cannot be used for specific synthesis of odd-carbon alkanes.
Q10
Out of benzene, m-dinitrobenzene and toluene, which will undergo nitration most easily and why?
Toluene undergoes nitration most easily. –CH₃ group is an electron-donating (+I effect) ortho/para director — it increases electron density in the ring, activating it for electrophilic attack. Benzene has no substituent (reference). m-Dinitrobenzene has two –NO₂ groups (strongly deactivating, –I and –R effects) → ring electron density greatly decreased → most resistant to further nitration. Order: Toluene > Benzene > m-Dinitrobenzene.
Q11
Write all the alkenes which on hydrogenation give 2-methylbutane.
2-Methylbutane = (CH₃)₂CHCH₂CH₃. The alkenes that give this on hydrogenation must have the same carbon skeleton with one double bond at different positions:
(i) 2-Methylbut-1-ene: CH₂=C(CH₃)CH₂CH₃
(ii) 2-Methylbut-2-ene: CH₃C(CH₃)=CHCH₃
(iii) 3-Methylbut-1-ene: CH₂=CHCH(CH₃)CH₃
Q12
Methane cannot be prepared by Kolbe's electrolytic method. Why?
In Kolbe's method, at the anode: 2RCOO⁻ → 2RCOO• → 2R• + 2CO₂ → R–R. To prepare methane (CH₄), we would need •CH₃ radical to pair with another •CH₃ to give C₂H₆ (ethane), not CH₄. To get CH₄, we would need •H radical — but formate (HCOO⁻) would give CO₂ only at the anode. There is no carboxylic acid that can produce •CH₃ which can combine to give methane directly. So methane (C₁) cannot be made by this method which always gives even-numbered alkanes.