Hydrocarbons

🎯 Key Points

• Alkanes CₙH₂ₙ₊₂ (saturated, single bonds only); Alkenes CₙH₂ₙ (one C=C); Alkynes CₙH₂ₙ₋₂ (one C≡C)
• Markovnikov's rule: H adds to the carbon with more H already; Anti-Markovnikov (only HBr + peroxide): opposite, via free radical mechanism
• Saytzeff rule: dehydrohalogenation/dehydration favours the MORE substituted (more stable) alkene as major product
• Benzene needs 4n+2 π electrons (Hückel's rule) for aromaticity; reacts via electrophilic aromatic substitution (EAS), not addition (preserves the stable aromatic ring)
• -OH, -NH₂, -CH₃, -OCH₃ are ortho/para directors (activating); -NO₂, -COOH, -CHO, -SO₃H are meta directors (deactivating)
• Lindlar's catalyst + H₂ on alkyne → cis-alkene; Na/liquid NH₃ on alkyne → trans-alkene

Alkanes (CₙH₂ₙ₊₂)

Saturated hydrocarbons with only single bonds. Unreactive under normal conditions; react via free-radical halogenation.

• Methane (CH₄), Ethane (C₂H₆), Propane (C₃H₈)
• Main reaction: CH₄ + Cl₂ → CH₃Cl + HCl (UV light; free-radical mechanism)
• Physical state: C1-C4 = gas; C5-C17 = liquid; C18+ = solid

Alkenes (CₙH₂ₙ)

Unsaturated with one double bond. Undergo electrophilic addition reactions.

• Markovnikov's rule: In HX addition, H attaches to the carbon with more H (halide goes to carbon with fewer H)
• Reactions: hydrogenation, halogenation, hydrohalogenation, hydration
• Ozonolysis: Cleavage of double bond to give aldehydes or ketones

Alkynes (CₙH₂ₙ₋₂)

Unsaturated with one triple bond. More reactive than alkenes.

• Ethyne (acetylene): HC≡CH; burns with a hot sooty flame; used in welding
• Terminal alkynes have weakly acidic H (reacts with Na, AgNO₃/NH₃ to give white precipitate)
• Lindlar's catalyst gives cis-alkene; Na/liq. NH₃ gives trans-alkene from alkynes

Aromatic Hydrocarbons

Benzene (C₆H₆): Six-membered ring with delocalised pi electrons; Huckel's rule: 4n+2 pi electrons for aromaticity.

• Electrophilic aromatic substitution (EAS): halogenation, nitration, sulphonation, Friedel-Crafts
• Activating groups (ortho/para directors): -OH, -NH₂, -CH₃, -OCH₃
• Deactivating groups (meta directors): -NO₂, -COOH, -CHO, -SO₃H

Benzene as a resonance hybrid: the two equivalent Kekulé structures (with alternating double bonds) are not separate real molecules — the six π electrons are fully delocalised, shown by the circle inside the hexagon. This delocalisation gives benzene its extra stability (resonance energy) and its characteristic aromatic behaviour. Image: Edgar181 / Rodolopezdato, Public Domain, via Wikimedia Commons.

Preparation of Alkanes

• Wurtz reaction: 2RX + 2Na (dry ether) → R-R + 2NaX; gives alkanes with an even number of carbons
• Decarboxylation: Sodium salt of carboxylic acid + soda lime, heated → alkane + Na₂CO₃
• Kolbe's electrolysis: Electrolysis of sodium/potassium salt of carboxylic acid gives alkane at the anode along with CO₂ and H₂ at the cathode
• Catalytic hydrogenation: Alkenes/alkynes + H₂ (Ni/Pt/Pd catalyst) → alkanes

Preparation of Alkenes and Alkynes

• Dehydrohalogenation: Alkyl halide + alcoholic KOH → alkene (follows Saytzeff rule: more substituted alkene is the major product)
• Dehydration of alcohols: Alcohol + conc. H₂SO₄, heat → alkene + water
• From calcium carbide: CaC₂ + 2H₂O → Ca(OH)₂ + C₂H₂ (lab preparation of ethyne)
• Vicinal dihalide + alcoholic KOH (excess): gives alkynes by double elimination

Markovnikov vs Anti-Markovnikov Addition

• Markovnikov's rule: In ionic addition of HX to an unsymmetrical alkene, H adds to the carbon already bearing more hydrogens; the mechanism proceeds via the more stable carbocation
• Anti-Markovnikov (Kharasch/peroxide effect): With HBr in the presence of peroxides, addition proceeds by a free-radical mechanism, so Br adds to the carbon with more hydrogens (only applies to HBr, not HCl or HI)

Mechanism of Electrophilic Aromatic Substitution

• Step 1: Electrophile (E⁺) generated (e.g., NO₂⁺ in nitration, using conc. HNO₃ + conc. H₂SO₄)
• Step 2: Electrophile attacks the electron-rich benzene ring forming a resonance-stabilised carbocation intermediate (arenium ion/sigma complex)
• Step 3: Loss of H⁺ restores aromaticity, giving the substituted product
• Friedel-Crafts alkylation/acylation uses AlCl₃ as a Lewis acid catalyst to generate the electrophile (R⁺ or RCO⁺)

Conformations of Ethane

• Conformations are the different spatial arrangements of atoms that arise from rotation about a single C-C sigma bond; they are interconvertible and cannot be isolated
• Staggered conformation: the C-H bonds of the two carbons are as far apart as possible (dihedral angle 60°); it has minimum torsional strain and is the most stable
• Eclipsed conformation: the C-H bonds directly face each other (dihedral angle 0°); maximum torsional strain and least stable
• The energy difference (torsional/rotational barrier) between the two forms in ethane is only about 12.5 kJ/mol, so rotation is essentially free at room temperature
• These are conveniently represented by Newman projections (looking down the C-C axis) and sawhorse diagrams

Isomerism in Hydrocarbons

• Chain (skeletal) isomerism: same molecular formula but different carbon-skeleton branching, e.g., n-butane and isobutane (2-methylpropane) both C₄H₁₀
• Position isomerism: same skeleton but the multiple bond or substituent is at a different position, e.g., but-1-ene and but-2-ene
• Functional and geometric isomerism: alkenes with two different groups on each doubly-bonded carbon show cis-trans (geometrical) isomerism due to restricted rotation about the C=C bond
• The number of possible chain isomers rises sharply with carbon count (pentane has 3, hexane has 5, heptane has 9)

Acidic Character of Terminal Alkynes

• The hydrogen on a triply-bonded (sp) carbon is weakly acidic because the sp carbon has high s-character (50%), holding the bonding electrons close to the nucleus and stabilising the resulting carbanion (acetylide ion)
• Acidity order: HC≡CH > H₂C=CH₂ > H₃C-CH₃ (sp > sp² > sp³ s-character)
• Terminal alkynes react with sodium metal to release H₂, and with ammoniacal AgNO₃ to give a white precipitate of silver acetylide, or with ammoniacal Cu₂Cl₂ to give a red precipitate of copper acetylide (a test to distinguish terminal from internal alkynes)
• Alkenes and alkanes do NOT show this acidic behaviour, so the reaction is diagnostic of a terminal ≡C-H

Directive Influence and Carcinogenicity of Aromatic Hydrocarbons

• Ortho/para directors (-OH, -NH₂, -OCH₃, -CH₃, halogens) donate electron density by resonance/hyperconjugation, stabilising the arenium ion when the electrophile attacks at the ortho/para positions
• Meta directors (-NO₂, -CN, -CHO, -COOH, -SO₃H) withdraw electron density, destabilising ortho/para attack most and so directing the incoming group to the meta position
• Halogens are exceptional: they are deactivating (electron-withdrawing by induction) yet still ortho/para directing (electron-donating by resonance)
• Carcinogenicity: polynuclear (fused-ring) aromatic hydrocarbons such as benzo[a]pyrene and 1,2-benzanthracene, formed in incomplete combustion of tobacco, coal and petroleum, are potent carcinogens; benzene itself is toxic and carcinogenic (causes leukaemia)

🚀 JEE Advanced Edge

Why Friedel-Crafts fails on nitrobenzene: Strongly deactivated rings (with a meta-director like -NO₂) are too electron-poor to attack the electrophile, AND the -NO₂ group can complex with/destroy the AlCl₃ catalyst — so Friedel-Crafts reactions simply don't proceed on nitrobenzene or other strongly deactivated arenes.

Ozonolysis for structure determination: Ozonolysis followed by hydrolysis cleaves a C=C bond into two carbonyl fragments — working BACKWARD from the carbonyl products lets you deduce the exact structure and position of the original double bond, a common JEE structure-elucidation question type.

Aromaticity edge cases: A ring is aromatic only if it is cyclic, planar, fully conjugated, AND has 4n+2 π electrons. Cyclopentadienyl anion (6 π electrons in a 5-membered ring) IS aromatic; cyclobutadiene (4 π electrons, 4n not 4n+2) is anti-aromatic and highly unstable — these exceptions to "just count ring size" are favourite JEE Advanced traps.

Worked problem: An alkene on ozonolysis gives only one product, CH₃COCH₃ (acetone), with no other carbonyl fragment. Deduce the alkene. Approach: If both fragments are identical, the original alkene must have been symmetric: (CH₃)₂C=C(CH₃)₂ (2,3-dimethyl-2-butene) — ozonolysis cleaves the central C=C, and since both halves are -C(CH₃)₂ groups, both give the same acetone product.