Classification of Hydrocarbons
Hydrocarbons are compounds containing only carbon and hydrogen.
| Type | Bond Type | Hybridisation | Examples |
|---|---|---|---|
| Alkanes | C-C single | sp^3 | CH4, C2H6 |
| Alkenes | C=C double | sp^2 | C2H4, C3H6 |
| Alkynes | C(triple)C triple | sp | C2H2, C3H4 |
| Arenes | Aromatic | sp^2 | C6H6, C6H5CH3 |
Alkanes (C_n H_(2n+2))
Sources
Natural gas and petroleum. Petroleum refining separates alkanes by fractional distillation.
Preparation Methods
- From unsaturated hydrocarbons (hydrogenation):
C2H4 + H2 -> C2H6(Ni catalyst, 200-300 C). - From alkyl halides (Wurtz reaction):
2CH3Br + 2Na -> CH3-CH3 + 2NaBr. - From carboxylic acids (Decarboxylation):
CH3COONa + NaOH -> CH4 + Na2CO3(soda lime). - Kolbe's electrolytic method: Electrolysis of sodium acetate gives ethane.
Physical Properties
- Non-polar, insoluble in water.
- Soluble in organic solvents.
- Boiling point increases with molecular mass.
- C1-C4: gases, C5-C17: liquids, C18+: solids.
Chemical Properties
Substitution reactions (free radical halogenation):
CH4 + Cl2 -> CH3Cl + HCl (UV light or heat).
Mechanism: Initiation, propagation, termination steps.
Reactivity: F2 > Cl2 > Br2 > I2.
Combustion: CH4 + 2O2 -> CO2 + 2H2O (complete).
2CH4 + 3O2 -> 2CO + 4H2O (incomplete).
Pyrolysis (Cracking): C4H10 -> C2H4 + C2H6 (higher temperature).
Aromatisation: CH3(CH2)4CH3 -> C6H6 + 4H2 (600 C, Cr2O3/Al2O3).
Alkenes (C_n H_(2n))
Preparation Methods
- Dehydration of alcohols:
C2H5OH -> C2H4 + H2O(conc. H2SO4, 170 C). - Dehydrohalogenation of alkyl halides:
CH3-CH2Br + KOH(alc) -> C2H4 + KBr + H2O. - From alkynes (partial reduction): Lindlar's catalyst (Pd/BaSO4 + quinoline) gives cis-alkene. Na/NH3 gives trans-alkene.
Physical Properties
- C2-C4: gases, C5-C18: liquids.
- Insoluble in water, soluble in organic solvents.
- More polar than alkanes.
Chemical Properties
Addition reactions (electrophilic addition):
Hydrogenation: C2H4 + H2 -> C2H6.
Halogenation: C2H4 + Br2 -> CH2Br-CH2Br (bromine water decolourised).
Hydrohalogenation: C2H4 + HBr -> C2H5Br.
Hydration: C2H4 + H2O -> C2H5OH (dilute H2SO4, 300 C, high pressure).
Markovnikov's Rule
In addition of HX to an unsymmetrical alkene, the hydrogen adds to the carbon with more hydrogens (the less substituted carbon).
CH3-CH=CH2 + HBr -> CH3-CHBr-CH3 (2-bromopropane, major product).
Peroxide effect (Anti-Markovnikov): In presence of peroxides, HBr adds opposite to Markovnikov's rule.
CH3-CH=CH2 + HBr -> CH3-CH2-CH2Br (1-bromopropane).
Alkynes (C_n H_(2n-2))
Preparation Methods
- From calcium carbide:
CaC2 + 2H2O -> C2H2 + Ca(OH)2. - Dehydrohalogenation of vicinal dihalides:
CH2Br-CH2Br + 2KOH(alc) -> C2H2 + 2KBr + 2H2O.
Physical Properties
- C2: gas, higher alkynes: liquids.
- Insoluble in water, soluble in organic solvents.
- More polar than alkenes.
Chemical Properties
Addition reactions:
C2H2 + H2 -> C2H4 -> C2H6(stepwise).C2H2 + Br2 -> C2H2Br2 -> C2H2Br4.C2H2 + HCl -> CH2=CHCl(vinyl chloride, monomer for PVC).
Acidic nature: Terminal alkynes are weakly acidic.
2C2H2 + 2Na -> 2C2HNa + H2.
C2H2 + AgNO3 + NH3 -> C2Ag2 + NH4NO3 (white precipitate, test for terminal alkyne).
Benzene: Structure and Aromaticity
Kekule Structure
Cyclic, alternating single and double bonds. Planar hexagon.
Resonance Structure
Two Kekule structures contribute equally. All C-C bonds are equivalent (bond length = 139 pm, intermediate between single and double).
Aromaticity Criteria (Huckel's Rule)
- Planar cyclic structure.
- Complete delocalisation of pi electrons.
(4n + 2)pielectrons (Huckel's rule). For benzene, n = 1 (6 pi electrons).
Aromatic Electrophilic Substitution
Benzene undergoes substitution (not addition) due to aromatic stability.
Mechanism:
- Generation of electrophile (E+).
- Formation of sigma-complex (arenium ion).
- Loss of H+ to restore aromaticity.
Examples: Nitration (C6H6 + HNO3 -> C6H5NO2), Halogenation, Sulphonation, Friedel-Crafts alkylation/acylation.
Worked Examples
Example 1: Write the product of CH3-CH=CH2 + HBr (with and without peroxide).
Solution: Without peroxide: CH3-CHBr-CH3 (Markovnikov). With peroxide: CH3-CH2-CH2Br (Anti-Markovnikov).
Example 2: Test for distinguishing terminal alkyne from internal alkyne. Solution: Terminal alkyne (C2H2, CH3-C(triple)CH) gives white precipitate with Tollen's reagent (AgNO3 + NH3). Internal alkyne (CH3-C(triple)C-CH3) does not react.
Common Mistakes
- Wurtz reaction gives only symmetric products:
2RX + 2Na -> R-R + 2NaX. Mixed Wurtz gives mixture. - Peroxide effect only with HBr: HCl and HI do not show anti-Markovnikov addition with peroxides.
- Benzene stability: Benzene does not undergo addition reactions readily due to aromatic stabilisation.
- Saytzeff rule: In elimination, the more substituted alkene (more stable) is the major product.
ISC Exam Focus
- Theory (70%): Preparation methods, reaction mechanisms, Markovnikov's rule, aromaticity, electrophilic substitution.
- Application (30%): Predicting products, distinguishing tests, writing reaction mechanisms.
- ISC frequently asks: "Write the mechanism of ..." and "Predict the product of ...".
- Distinction between alkane, alkene, alkyne, and benzene is essential.
Self-Test Questions
Q1: Write the preparation of methane from sodium acetate.
Answer: CH3COONa + NaOH -> CH4 + Na2CO3 (soda lime decarboxylation).
Q2: State Markovnikov's rule with an example.
Answer: In addition of HX to unsymmetrical alkene, H adds to carbon with more H atoms. Example: CH3-CH=CH2 + HBr -> CH3-CHBr-CH3.
Q3: How will you distinguish between ethene and ethyne? Answer: Ethyne reacts with Cu2Cl2/NH3 to give red precipitate (Cu2C2). Ethene does not.
Q4: Write the resonance structures of benzene. Answer: Two Kekule structures with alternating single and double bonds, and a circle representing delocalised pi electrons.
Q5: State Huckel's rule for aromaticity. How many pi electrons does benzene have?
Answer: (4n+2)pi electrons in planar cyclic conjugated system. Benzene has 6 pi electrons (n=1).
Q6: Write the Friedel-Crafts alkylation of benzene.
Answer: C6H6 + CH3Cl -> C6H5CH3 + HCl (anhydrous AlCl3 catalyst).
