Hydrocarbons Class 11 Notes

• From alkenes/alkynes: hydrogenation with a metal catalyst (Ni, Pd, Pt).
• From alkyl halides: Wurtz reaction 2RX + 2Na → R–R + 2NaX.
• From carboxylic acids: Kolbe's electrolysis of sodium salts.
• From Grignard reagents: RMgX with water gives RH.
• Physical trend: boiling points rise with carbon number; branched isomers boil lower.

• Combustion: CH₄ + 2O₂ → CO₂ + 2H₂O, releasing heat — alkanes are fuel.
• Halogenation (free-radical substitution): heat or UV light drives CH₄ + Cl₂ → CH₃Cl + HCl, proceeding through initiation, propagation and termination.
• Chlorination can over-halogenate; reactivity order F₂ > Cl₂ > Br₂ > I₂.
• Controlled oxidation gives alcohols, aldehydes or acids depending on conditions.

• Cracking of hydrocarbons, and elimination of water from alcohols (dehydration with H₂SO₄).
• Dehydrohalogenation of alkyl halides with alcoholic KOH (Saytzeff rule: the more substituted alkene forms).
• Geometric isomerism: cis/trans around the double bond because rotation is blocked by π bonding.

• Catalytic hydrogenation adds H₂ across the double bond.
• Halogenation: bromine water is decolourised — a simple test for unsaturation.
• Hydration with acid gives alcohols; hydrohalogenation gives alkyl halides.
• Ozonolysis cleaves the double bond to carbonyl compounds, revealing the alkene's structure.
• Polymerisation: many alkenes join to give polyethene, polypropene, PVC.

• Preparation: dehydrohalogenation of vicinal dihalides with alcoholic KOH.
• Acidity: terminal H (R–C≡C–H) is the most acidic of hydrocarbon hydrogens.
• Addition reactions mirror alkenes but can add two molar equivalents — e.g. 2H₂, 2Br₂.
• Calcium carbide + water gives acetylene in the laboratory.

• Electrophilic aromatic substitution (EAS): nitration, halogenation, sulphonation, Friedel–Crafts alkylation and acylation.
• A substituent on benzene directs and activates: –OH, –CH₃, –NH₂ are ortho/para directors; –NO₂, –CN, –CHO are meta directors.
• Benzene does not normally undergo addition reactions (unlike alkenes), showing its resonance stability.
• Toxicity: benzene is carcinogenic; toluene is used as a safer substitute in solvents.

Example: Predict the addition product of propene (CH₃–CH=CH₂) with HBr.

Solution: By Markovnikov's rule, H goes to the terminal carbon (more hydrogens), and Br joins the middle carbon, giving 2-bromopropane. In the presence of peroxides, the product is 1-bromopropane.

Example: Why does benzene prefer substitution over addition?

Solution: Addition would break the stable 6π aromatic system. Substitution keeps the aromatic ring intact, so benzene undergoes electrophilic aromatic substitution far more readily than addition.

Exercises (NCERT Q9.1–Q9.25)

• 9.1) How do you account for the formation of ethane during the chlorination of methane?
• 9.2) Write the IUPAC names of the following compounds: (a) CH₃CH=C(CH₃)₂. (b) CH₂=CH–C≡C–CH₃. (c) CH₂=CH–CH=CH₂. (d) C₆H₅–CH₂–CH₂–CH=CH₂. (e) o-CH₃–C₆H₄–OH, with the methyl and hydroxyl groups adjacent. (f) CH₃(CH₂)₄CH(CH₂)₃CH₃, with a –CH₂–CH(CH₃)₂ group attached to the central CH. (g) CH₃–CH=CH–CH₂–CH=CH–CH(C₂H₅)–CH₂–CH=CH₂.
• 9.3) Write structural formulas and IUPAC names for all possible isomers having the indicated number of double or triple bonds: (a) C₄H₈, one double bond. (b) C₅H₈, one triple bond.
• 9.4) Write the IUPAC names of the products obtained by ozonolysis of: (i) pent-2-ene; (ii) 3,4-dimethylhept-3-ene; (iii) 2-ethylbut-1-ene; and (iv) 1-phenylbut-1-ene.
• 9.5) An alkene A on ozonolysis gives a mixture of ethanal and pentan-3-one. Write the structure and IUPAC name of A.
• 9.6) An alkene A contains three C–C σ bonds, eight C–H σ bonds and one C–C π bond. Ozonolysis of A gives two moles of an aldehyde of molar mass 44 u. Write the IUPAC name of A.
• 9.7) Propanal and pentan-3-one are the ozonolysis products of an alkene. What is the structural formula of the alkene?
• 9.8) Write balanced chemical equations for complete combustion of: (i) butane; (ii) pentene; (iii) hexyne; and (iv) toluene.
• 9.9) Draw the cis and trans structures of hex-2-ene, using condensed formulae and stating the relative positions of the two larger groups. Which isomer has the higher boiling point and why?
• 9.10) Why is benzene extraordinarily stable although it contains three carbon–carbon double bonds?
• 9.11) What are the necessary conditions for a system to be aromatic?
• 9.12) Explain why the following systems are not aromatic: (i) the displayed system with six π-electrons and an sp³-hybridised carbon; (ii) the displayed system with four π-electrons and an sp³-hybridised carbon; and (iii) cyclooctatetraene.
• 9.13) How will you convert benzene into: (i) p-nitrobromobenzene; (ii) m-nitrochlorobenzene; (iii) p-nitrotoluene; and (iv) acetophenone?
• 9.14) In CH₃–CH₂–C(CH₃)₂–CH₂–CH(CH₃)₂, identify the primary, secondary and tertiary carbon atoms and give the number of hydrogens bonded to each of them.
• 9.15) What effect does branching of an alkane chain have on its boiling point?
• 9.16) Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide the same reaction yields 1-bromopropane. Explain the difference and give the mechanisms.
• 9.17) Write down the products of ozonolysis of 1,2-dimethylbenzene (o-xylene). How does the result support the Kekulé structure for benzene?
• 9.18) Arrange benzene, n-hexane and ethyne in decreasing order of acidic character and explain the order.
• 9.19) Why does benzene undergo electrophilic substitution easily but nucleophilic substitution with difficulty?
• 9.20) How would you convert: (i) ethyne into benzene; (ii) ethene into benzene; and (iii) hexane into benzene?
• 9.21) Write the structures of all alkenes which on hydrogenation give 2-methylbutane.
• 9.22) Arrange each set in decreasing order of relative reactivity with an electrophile E⁺: (a) chlorobenzene, p-nitrochlorobenzene and 2,4-dinitrochlorobenzene; (b) toluene, p-nitrotoluene and p-dinitrobenzene.
• 9.23) Out of benzene, m-dinitrobenzene and toluene, which undergoes nitration most easily and why?
• 9.24) Suggest a Lewis acid other than anhydrous aluminium chloride that can be used during ethylation of benzene.
• 9.25) Why is the Wurtz reaction not preferred for preparing alkanes containing an odd number of carbon atoms? Illustrate the answer with one example.

Answers at a glance

1. 1
   Q9.1 → Ethane is formed in a termination step when two methyl free radicals combine: CH₃• + CH₃• → CH₃CH₃.
2. 2
   Q9.2 → (a) 2-methylbut-2-ene; (b) pent-1-en-3-yne (pent-1-ene-3-yne); (c) buta-1,3-diene; (d) 4-phenylbut-1-ene; (e) 2-methylphenol; (f) 5-(2-methylpropyl)decane; (g) 4-ethyldeca-1,5,8-triene.
3. 3
   Q9.3 → (a) CH₂=CH–CH₂–CH₃, but-1-ene; CH₃–CH=CH–CH₃, but-2-ene; CH₂=C(CH₃)₂, 2-methylprop-1-ene. (b) HC≡C–CH₂–CH₂–CH₃, pent-1-yne; CH₃–C≡C–CH₂–CH₃, pent-2-yne; HC≡C–CH(CH₃)–CH₃, 3-methylbut-1-yne.
4. 4
   Q9.4 → (i) Ethanal and propanal; (ii) butan-2-one and pentan-2-one; (iii) methanal and pentan-3-one; (iv) benzaldehyde and propanal.
5. 5
   Q9.5 → A is CH₃–CH=C(CH₂CH₃)₂, and its IUPAC name is 3-ethylpent-2-ene.
6. 6
   Q9.6 → A is but-2-ene.
7. 7
   Q9.7 → The alkene is CH₃CH₂–CH=C(CH₂CH₃)₂, named 4-ethylhex-3-ene.
8. 8
   Q9.8 → The balanced equations are 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O; C₅H₁₀ + 15/2 O₂ → 5CO₂ + 5H₂O; 2C₆H₁₀ + 17O₂ → 12CO₂ + 10H₂O; and C₇H₈ + 9O₂ → 7CO₂ + 4H₂O.
9. 9
   Q9.9 → Cis-hex-2-ene has the higher boiling point because its polar bond dipoles add to give a larger molecular dipole moment and therefore stronger intermolecular attraction.
10. 10
    Q9.10 → Benzene is unusually stable because its six π electrons are delocalised over a planar, fully conjugated ring, giving resonance stabilisation and six equivalent C–C bonds.
11. 11
    Q9.11 → An aromatic system is cyclic, planar, fully conjugated through overlapping p orbitals at every ring atom, and contains 4n+2 delocalised π electrons.
12. 12
    Q9.12 → All three are non-aromatic: (i) has incomplete conjugation caused by an sp³ carbon despite six π-electrons; (ii) has an sp³ carbon and only four π-electrons; and (iii) is non-planar with eight π-electrons.
13. 13
    Q9.13 → Use bromination then nitration for p-nitrobromobenzene; nitration then chlorination for m-nitrochlorobenzene; methylation then nitration for p-nitrotoluene; and Friedel–Crafts acetylation for acetophenone.
14. 14
    Q9.14 → There are five primary carbons with 15 H atoms, two secondary carbons with 4 H atoms, one tertiary carbon with 1 H atom, and one quaternary carbon with no H.
15. 15
    Q9.15 → Greater branching lowers the boiling point because the compact molecules have less surface contact and weaker London dispersion forces.
16. 16
    Q9.16 → The ionic mechanism follows Markovnikov's rule through a secondary carbocation and gives 2-bromopropane; the peroxide-initiated radical mechanism gives the more stable secondary radical orientation and ultimately 1-bromopropane, the anti-Markovnikov product.
17. 17
    Q9.17 → The products are glyoxal, methylglyoxal and butane-2,3-dione (diacetyl). Their formation supports benzene as a resonance hybrid of equivalent Kekulé structures.
18. 18
    Q9.18 → Decreasing acidic character is ethyne > benzene > n-hexane, because the order of carbon hybridisation is sp > sp² > sp³ and acidity increases with s-character.
19. 19
    Q9.19 → Benzene's electron-rich π cloud readily attracts electrophiles, while nucleophiles are repelled and substitution by a nucleophile would require loss of aromaticity.
20. 20
    Q9.20 → Ethyne and ethene each cyclotrimerise in a red-hot iron tube, while hexane is aromatised over Cr₂O₃/Al₂O₃ at high temperature and pressure.
21. 21
    Q9.21 → The three alkenes are 2-methylbut-1-ene, 2-methylbut-2-ene and 3-methylbut-1-ene, with the structures shown above.
22. 22
    Q9.22 → (a) Chlorobenzene > p-nitrochlorobenzene > 2,4-dinitrochlorobenzene. (b) Toluene > p-nitrotoluene > p-dinitrobenzene.
23. 23
    Q9.23 → Toluene nitrates most easily. The order of reactivity is toluene > benzene > m-dinitrobenzene because CH₃ activates the ring whereas NO₂ groups deactivate it.
24. 24
    Q9.24 → Anhydrous ferric chloride, FeCl₃, can be used; boron trifluoride, BF₃, is another acceptable Lewis acid.
25. 25
    Q9.25 → Odd-carbon alkanes require two different alkyl halides in Wurtz coupling, so self-coupling gives a mixture. Ethyl bromide plus propyl bromide gives pentane together with butane and hexane.

MCQ Questions

1 Mark Questions

1. 1
   What is Markovnikov's rule? (Ans: in the addition of HX to an alkene, the hydrogen attaches to the carbon with the greater number of hydrogen atoms)
2. 2
   Which is the major product of the hydration of ethene? (Ans: ethanol)
3. 3
   Name the catalyst used in the industrial hydrogenation of alkenes. (Ans: nickel (Raney Ni); also Pt/Pd)
4. 4
   Define aromatic hydrocarbons. (Ans: hydrocarbons containing a planar ring of delocalised π electrons, e.g. benzene, following Hückel's rule (4n+2) π electrons)

2 Mark Questions

1. 1
   State one preparation and one important property of alkanes. (Ans: Wurtz/Hydrogenation of alkenes; free-radical halogenation to haloalkane)
2. 2
   Why are alkenes more reactive than alkanes? (Ans: the weaker π bond is readily attacked by electrophiles)
3. 3
   What is ozonolysis? (Ans: oxidative cleavage of an alkene with ozone followed by reductive workup, giving carbonyl compounds that reveal the position of the double bond)

3 Mark Questions

1. 1
   Explain the mechanism of free-radical halogenation of methane giving the initiation, propagation and termination steps. (Ans: Cl₂ → 2Cl•; CH₄ + Cl• → •CH₃ + HCl; •CH₃ + Cl₂ → CH₃Cl + Cl•; termination radicals combine)
2. 2
   Describe the addition of HBr to propene with a clear statement of Markovnikov's rule and the stability of the carbocation intermediate. (Ans: 2-bromopropane major)
3. 3
   Write the Friedel–Crafts alkylation and acylation of benzene with equations for the formation of the electrophile. (Ans: C₆H₆ + CH₃Cl → C₆H₅CH₃ + HCl (AlCl₃); with CH₃COCl gives acetophenone)

5 Mark Questions

1. 1
   Discuss the electrophilic substitution reactions of benzene — nitration, sulphonation, halogenation and Friedel–Crafts — giving the mechanism of nitration (formation of N⁺O₂ electrophile) and the character of the aromatic ring.
2. 2
   Compare the chemical properties of alkanes, alkenes and alkynes with reference to addition, combustion and halogenation, and explain why benzene undergoes substitution rather than addition despite the presence of π electrons (aromatic stabilisation).