How to Use This Page
Attempt each NCERT question on your own before reading the solution below. Focus especially on reaction mechanisms and naming conventions — these are the most commonly tested areas in CBSE exams. Write out the reactions by hand for best retention.
Key Concepts (Quick Recap)
- Alkanes (CₙH₂ₙ₊₂): saturated hydrocarbons, sp³ hybridised, free radical substitution, combustion, pyrolysis/cracking
- Conformation: rotation about C–C single bond gives staggered (more stable, minimum torsional strain) and eclipsed (less stable, maximum torsional strain) conformers; represented by Newman projections and sawhorse structures
- Alkenes (CₙH₂ₙ): unsaturated, sp² hybridised, electrophilic addition reactions, Markovnikov's rule, anti-Markovnikov addition (peroxide effect/Kharasch effect), ozonolysis, polymerisation
- Alkynes (CₙH₂ₙ₋₂): triple bond (sp hybridised), addition reactions, terminal alkyne acidity (≈25 pKa), formation of acetylides with Na/NaNH₂
- Benzene (C₆H₆): aromatic, sp² hybridised, resonance stabilised, electrophilic substitution (SE) reactions
- Electrophilic substitution in benzene: nitration, sulfonation, halogenation, Friedel-Crafts alkylation, Friedel-Crafts acylation
- Directive influence: ortho/para directors (–OH, –NH₂, –OR, –CH₃, –Cl) activate the ring; meta directors (–NO₂, –CN, –COOH, –SO₃H) deactivate the ring
- Anti-Markovnikov addition: in presence of peroxides, HBr adds to propene to give 1-bromopropane (peroxide effect; does not work with HCl or HI)
Intext Questions
Page 361–363 (Alkanes, Conformation)
Q1. What are hydrocarbons? Give their classification.
Answer
Hydrocarbons are compounds of carbon and hydrogen only. They are classified as: 1. Open-chain (aliphatic): (a) Alkanes (CₙH₂ₙ₊₂) — single bonds, saturated; (b) Alkenes (CₙH₂ₙ) — double bond(s), unsaturated; (c) Alkynes (CₙH₂ₙ₋₂) — triple bond(s), unsaturated 2. Closed-chain (cyclic): (a) Alicyclic — cycloalkanes, cycloalkenes; (b) Aromatic — benzene and derivatives (contain benzene ring)
Q2. Explain the conformational isomerism in ethane.
Answer
In ethane (CH₃–CH₃), rotation about the C–C bond produces different conformations. The two extreme conformations are: 1. Staggered conformation: The H atoms on the front and back carbon are maximally apart (dihedral angle = 60°). This is more stable (lower energy) because the electron clouds of the C–H bonds are as far apart as possible, minimising torsional strain. 2. Eclipsed conformation: The H atoms on the front and back carbon are aligned (dihedral angle = 0°). This is less stable (higher energy) due to maximum torsional strain from overlapping electron clouds.
The energy difference between staggered and eclipsed is about 12.5 kJ mol⁻¹. At room temperature, ethane molecules rapidly interconvert between conformations, so conformers cannot be isolated.
Q3. Draw Newman projections for the staggered and eclipsed conformations of ethane.
Answer
Staggered: Front carbon has 3 H atoms at 12, 4, and 8 o'clock positions. Back carbon has 3 H atoms at 2, 6, and 10 o'clock positions — each back H sits between two front H atoms.
Eclipsed: Front carbon has 3 H atoms at 12, 4, and 8 o'clock. Back carbon has 3 H atoms at the same 12, 4, and 8 o'clock positions — each back H is directly behind a front H.
Q4. What is the difference between conformational isomerism and structural isomerism?
Answer
Conformational isomerism arises from rotation about C–C single bonds. Conformers interconvert rapidly at room temperature and cannot be isolated — they are not distinct compounds. Structural isomerism involves different connectivity of atoms. Structural isomers are distinct compounds with different physical and chemical properties. They cannot interconvert without breaking and reforming bonds.
Page 373–375 (Alkenes, Markovnikov's Rule)
Q5. Why do alkenes undergo addition reactions?
Answer
Alkenes contain a carbon-carbon double bond (C=C), which consists of a strong σ-bond and a weak π-bond. The π-electron cloud is above and below the plane of the molecule, making it easily accessible to electrophiles. The π-bond can be broken relatively easily (bond dissociation energy ≈ 270 kJ mol⁻¹) to form two new σ-bonds. Since this provides a thermodynamically favourable pathway, alkenes undergo addition reactions rather than substitution.
Q6. Explain Markovnikov's rule with an example.
Answer
Markovnikov's rule: When an unsymmetrical reagent adds to an unsymmetrical alkene, the hydrogen of the reagent attaches to the carbon that already has more hydrogen atoms.
Example: Addition of HBr to propene (CH₃CH=CH₂)
- H⁺ adds to C-1 (which has more H atoms) → forms secondary carbocation CH₃–C⁺H–CH₃ (more stable)
- Br⁻ adds to C-2 → product is 2-bromopropane (CH₃CHBrCH₃)
The alternative (H⁺ adding to C-2) would give a less stable primary carbocation, so Markovnikov addition is preferred.
Q7. What is the peroxide effect (anti-Markovnikov addition)?
Answer
In the presence of organic peroxides (ROOR), HBr adds to an unsymmetrical alkene in an anti-Markovnikov fashion. This is called the peroxide effect or Kharasch effect.
Example: HBr + propene (with peroxides) → 1-bromopropane (CH₃CH₂CH₂Br)
Mechanism: The peroxide undergoes homolytic cleavage to generate free radicals (RO•). The radical adds to the double bond, forming the more stable radical intermediate (2° or 3°), and then abstracts H from HBr. Since the radical addition step is the reverse of electrophilic addition, the regiochemistry is reversed.
Important: The peroxide effect works only with HBr, not with HCl or HI. HCl has a strong H–Cl bond (radical formation is endothermic), and HI adds so rapidly that the radical pathway cannot compete.
Page 385–386 (Alkynes, Acidity)
Q8. Why are terminal alkynes acidic?
Answer
Terminal alkynes (RC≡CH) are acidic because the sp-hybridised carbon has 50% s-character, which holds the electron pair closer to the nucleus. This makes the C–H bond more polar and the hydrogen more easily lost as H⁺. The pKa of terminal alkynes is about 25 (much more acidic than alkenes, pKa ≈ 44, or alkanes, pKa ≈ 50). Terminal alkynes react with strong bases like NaNH₂ to form acetylide ions: RC≡CH + NaNH₂ → RC≡C⁻Na⁺ + NH₃
Q9. Explain the acidic character of alkynes with a comparison to alkenes and alkanes.
Answer
Acidity order: HC≡CH (pKa ≈ 25) > H₂C=CH₂ (pKa ≈ 44) > CH₃–CH₃ (pKa ≈ 50) The reason is the s-character of the orbital holding the H atom: sp (50% s) > sp² (33% s) > sp³ (25% s). Higher s-character means the electron pair is held closer to the nucleus, stabilising the conjugate base (carbanion) and making the proton more acidic. Only terminal alkynes react with NaNH₂ or Na/NH₃ to form acetylides; alkenes and alkanes do not.
Q10. Describe the addition reactions of alkynes.
Answer
Alkynes undergo addition reactions similar to alkenes, but in two stages (since there are two π-bonds): 1. Addition of H₂ (hydrogenation): With Lindlar's catalyst (Pd/BaSO₄, quinoline) → cis-alkene; With Na/NH₃ → trans-alkene; With excess H₂/Ni → alkane 2. Addition of halogens: Br₂ → trans-dibromoalkene → tetrabromoalkane (with excess Br₂) 3. Addition of HX: HBr → vinyl bromide → gem-dibromoalkane (with excess HBr) 4. Addition of water (in presence of HgSO₄/H₂SO₄): → enol → keto form (Markovnikov addition)
Page 399–402 (Benzene, Aromaticity, Electrophilic Substitution)
Q11. What is aromaticity? What are the rules for aromaticity?
Answer
Aromaticity is a special stability associated with planar, cyclic, conjugated systems. A compound is aromatic if it satisfies Hückel's rule: (4n + 2) π-electrons in a conjugated, planar, cyclic system, where n = 0, 1, 2, 3...
Requirements for aromaticity: 1. Cyclic (ring) structure 2. Planar geometry (all atoms in one plane) 3. Fully conjugated (p-orbital on every ring atom) 4. (4n + 2) π-electrons (Hückel's rule)
Examples: Benzene (6 π-electrons, n=1) is aromatic; cyclobutadiene (4 π-electrons, n=0 gives 2 but actual count is 4) is anti-aromatic; cyclooctatetraene (8 π-electrons) is non-aromatic (tub-shaped, not planar).
Q12. Describe the electrophilic substitution reactions of benzene.
Answer
Benzene undergoes electrophilic substitution (SE) reactions where an electrophile replaces a hydrogen atom on the ring, preserving the aromatic system:
1. Nitration: C₆H₆ + HNO₃ →⁽ᴴ²ᔠᴼ⁴⁾ C₆H₅NO₂ + H₂O (electrophile: NO₂⁺) 2. Sulfonation: C₆H₆ + H₂SO₄ → C₆H₅SO₃H + H₂O (electrophile: SO₃) 3. Halogenation: C₆H₆ + Cl₂ →⁽ᶠᵉᶜˡ³⁾ C₆H₅Cl + HCl (electrophile: Cl⁺) 4. Friedel-Crafts alkylation: C₆H₆ + RCl →⁽ᴬˡᶜˡ³⁾ C₆H₅R + HCl (electrophile: R⁺) 5. Friedel-Crafts acylation: C₆H₆ + RCOCl →⁽ᴬˡᶜˡ³⁾ C₆H₅COR + HCl (electrophile: RCO⁺)
Q13. What is the directive influence of substituents in benzene?
Answer
A substituent on the benzene ring directs the incoming electrophile to specific positions:
Ortho/para directors (activate the ring, positions 2,4,6):
- Strongly activating: –OH, –NH₂, –NHR, –NR₂, –OR
- Moderately activating: –CH₃, –C₂H₅ (alkyl groups)
- Weakly activating: –C₆H₅ (phenyl)
- Weakly deactivating: –Cl, –Br, –I (deactivate but still ortho/para directing due to +R effect)
Meta directors (deactivate the ring, position 3):
- Strongly deactivating: –NO₂, –CN, –SO₃H, –CHO, –COOH, –COOR
- Moderately deactivating: –COR (acyl groups)
Chapter-End Exercise
Q1. Write the IUPAC names of: (a) CH₃CH₂CH₂CH₂CH₃ (b) (CH₃)₂CHCH₂CH₃ (c) (CH₃)₂CHCH(CH₃)₂ (d) CH₃CH₂C(CH₃)₂CH₂CH₃
Answer
(a) Pentane (b) 2-Methylbutane (c) 2,3-Dimethylbutane (or 2,3-dimethylbutane) (d) 3,3-Dimethylpentane
Q2. Write equations for: (a) Combustion of propane (b) Partial oxidation of methane (c) Pyrolysis of methane
Answer
(a) C₃H₈ + 5O₂ → 3CO₂ + 4H₂O (b) 2CH₄ + O₂ →⁽ᶜᵃᵗ/ᶜᵘ⁾ 2CO + 4H₂ (incomplete combustion) or CH₄ + O₂ →⁽ᴺᵃ⁾ HCHO + H₂O (controlled oxidation) (c) CH₄ →⁽¹²⁷³ ᴷ⁾ C + 2H₂ (thermal decomposition/pyrolysis)
Q3. Explain the mechanism of free radical substitution in methane with chlorine.
Answer
Step 1 — Initiation: Cl₂ →ˣʰᵛ 2Cl• (homolytic cleavage of Cl₂ by UV light or heat)
Step 2 — Propagation:
- Cl• + CH₄ → CH₃• + HCl (chlorine radical abstracts H from methane)
- CH₃• + Cl₂ → CH₃Cl + Cl• (methyl radical attacks Cl₂, forming methyl chloride and regenerating Cl•)
Step 3 — Termination (radicals combine):
- Cl• + Cl• → Cl₂
- CH₃• + CH₃• → C₂H₆ (ethane)
- CH₃• + Cl• → CH₃Cl
The chain reaction can produce CH₂Cl₂, CHCl₃, and CCl₄ with excess Cl₂.
Q4. Explain Markovnikov's rule and anti-Markovnikov addition with examples.
Answer
Markovnikov addition: When HBr adds to propene in the absence of peroxides: CH₃–CH=CH₂ + HBr → CH₃–CHBr–CH₃ (2-bromopropane) The H adds to C-1 (more H atoms), Br adds to C-2, giving the more stable 2° carbocation intermediate.
Anti-Markovnikov addition (peroxide effect): When HBr adds to propene in the presence of peroxides: CH₃–CH=CH₂ + HBr →⁽ᴿᴼᴼᴿ⁾ CH₃–CH₂–CH₂Br (1-bromopropane) The radical mechanism reverses the regiochemistry: the RO• radical adds to the less substituted carbon first, and then H is abstracted from HBr. This effect is observed only with HBr.
Q5. Explain the ozonolysis of alkenes with an example.
Answer
Ozonolysis cleaves the C=C double bond using ozone (O₃) to form carbonyl compounds. The reaction proceeds through an ozonide intermediate: 1. Alkene + O₃ → molozonide → ozonide (1,2,4-trioxolane) 2. Ozonide + Zn/H₂O (reductive workup) → two carbonyl compounds
Example: Ozonolysis of 2-butene: CH₃–CH=CH–CH₃ →⁽ᴼ³⁾ CH₃CHO + CH₃CHO (two molecules of ethanal)
Ozonolysis of ethene: CH₂=CH₂ →⁽ᴼ³⁾ HCHO + HCHO (two molecules of formaldehyde)
Ozonolysis helps identify the position of the double bond in an unknown alkene.
Q6. What is polymerisation? Give examples.
Answer
Polymerisation is the process of combining many small molecules (monomers) into a large molecule (polymer). In addition polymerisation, alkene monomers add to each other:
- Ethene → polythene: nCH₂=CH₂ → (–CH₂–CH₂–)ₙ
- Propene → polypropylene: nCH₃CH=CH₂ → (–CH(CH₃)–CH₂–)ₙ
- Vinyl chloride → PVC: nCH₂=CHCl → (–CH₂–CHCl–)ₙ
- Tetrafluoroethene → Teflon: nCF₂=CF₂ → (–CF₂–CF₂–)ₙ
Q7. Describe the preparation of alkanes from: (a) Alkenes (b) Alkyl halides (c) Carboxylic acids
Answer
(a) From alkenes — catalytic hydrogenation: RCH=CHR' + H₂ →⁽ᴺⁱ/ᴾᵈ⁾ RCH₂–CH₂R' (alkane) (b) From alkyl halides — reduction with LiAlH₄: R–X + LiAlH₄ → R–H (alkane); or with Zn/HCl: R–X + Zn + HCl → R–H + ZnCl₂ (c) From carboxylic acids — decarboxylation: RCOONa + NaOH →⁽ᶜᴀ/ᴺᵃ ᴼᴴ, ᴰʳʸ⁾ RH + Na₂CO₃ (soda lime decarboxylation)
Q8. Write the reactions for the preparation of alkenes from: (a) Alkyl halides (b) Alcohols
Answer
(a) From alkyl halides — elimination (dehydrohalogenation): RCH₂CH₂X + KOH (alcoholic) → RCH=CH₂ + KX + H₂O (β-elimination, follows Saytzeff rule: more substituted alkene is major product) (b) From alcohols — dehydration: RCH₂CH₂OH →⁽ᴴ²ˢᴼ⁴/⁴⁷³ ᴷ⁾ RCH=CH₂ + H₂O (E1 mechanism, Saytzeff rule applies)
Q9. Explain the acidity of terminal alkynes. How are acetylides formed?
Answer
Terminal alkynes (RC≡CH) have acidic hydrogen because the sp-hybridised carbon (50% s-character) strongly holds the electron pair, polarising the C–H bond. The resulting carbanion (RC≡C⁻) is stabilised by the high s-character.
Acetylide formation: RC≡CH + NaNH₂ → RC≡C⁻Na⁺ + NH₃ (sodium acetylide) 2RC≡CH + 2Na → 2RC≡C⁻Na⁺ + H₂ (with sodium metal)
Acetylides are used to lengthen the carbon chain: RC≡C⁻Na⁺ + R'X → RC≡CR' + NaX
Q10. Explain Friedel-Crafts alkylation and acylation with mechanisms.
Answer
Friedel-Crafts Alkylation: C₆H₆ + RCl →⁽ᴬˡᶜˡ³⁾ C₆H₅R + HCl Mechanism: AlCl₃ polarises R–Cl → R⁺ (carbocation) is the electrophile. R⁺ attacks benzene ring → arenium ion → loss of H⁺ → alkylbenzene. Limitation: Polyalkylation can occur (first alkyl group activates the ring for further substitution).
Friedel-Crafts Acylation: C₆H₆ + RCOCl →⁽ᴬˡᶜˡ³⁾ C₆H₅COR + HCl Mechanism: AlCl₃ polarises RCOCl → RCO⁺ (acylium ion) is the electrophile. RCO⁺ attacks benzene → arenium ion → loss of H⁺ → acylbenzene (ketone). Advantage over alkylation: The acyl group is a meta director that deactivates the ring, so polyacylation does not occur. The product can be reduced (Clemmensen or Wolff-Kishner) to give the alkylbenzene without rearrangement.
Q11. Explain Saytzeff's rule with an example.
Answer
Saytzeff's rule (Zaitsev's rule): In elimination reactions, the more substituted alkene (the one with more alkyl groups on the double-bond carbons) is the major product, because it is more stable.
Example: Dehydrohalogenation of 2-bromobutane with alcoholic KOH: CH₃–CHBr–CH₂–CH₃ + KOH (alc.) → CH₃–CH=CH–CH₃ (but-2-ene, 80%, more substituted) + CH₂=CH–CH₂–CH₃ (but-1-ene, 20%, less substituted)
But-2-ene has two alkyl groups on the double bond carbons (more stable), while but-1-ene has only one.
Q12. Describe the ozonolysis of: (a) Ethene (b) Propene (c) 2-Methylbut-2-ene
Answer
(a) CH₂=CH₂ →⁽ᴼ³/ᴼⁿ/ᴴ²ᴼ⁾ HCHO + HCHO (formaldehyde) (b) CH₃–CH=CH₂ →⁽ᴼ³/ᴼⁿ/ᴴ²ᴼ⁾ CH₃CHO + HCHO (ethanal + formaldehyde) (c) (CH₃)₂C=CHCH₃ →⁽ᴼ³/ᴼⁿ/ᴴ²ᴼ⁾ (CH₃)₂C=O + CH₃CHO (acetone + ethanal)
Q13. Explain the Kharasch effect with an example.
Answer
The Kharasch effect (peroxide effect) is the anti-Markovnikov addition of HBr to unsymmetrical alkenes in the presence of peroxides.
Without peroxide (Markovnikov): CH₃CH=CH₂ + HBr → CH₃CHBrCH₃ (2-bromopropane) With peroxide (anti-Markovnikov): CH₃CH=CH₂ + HBr →⁽ᴿᴼᴼᴿ⁾ CH₃CH₂CH₂Br (1-bromopropane)
Mechanism (radical addition): 1. ROOR → 2RO• (initiation) 2. RO• + HBr → ROH + Br• 3. Br• + CH₃CH=CH₂ → CH₃CH•CH₂Br (more stable 2° radical) 4. CH₃CH•CH₂Br + HBr → CH₃CH₂CH₂Br + Br• (chain propagation)
Q14. What is Wurtz reaction? Give its limitations.
Answer
Wurtz reaction: 2RX + 2Na → R–R + 2NaX (dry ether) Used to prepare alkanes with an even number of carbon atoms by coupling alkyl halides with sodium metal.
Limitations: 1. Only symmetrical alkanes (even number of C atoms) are best prepared. If two different alkyl halides are used, a mixture of three products (R–R, R'–R', R–R') is formed, difficult to separate. 2. Cannot prepare methane (lowest alkane). 3. Not suitable for tertiary alkyl halides (elimination competes). 4. The reaction involves free radical intermediates, so side reactions (elimination) can occur.
Q15. Explain the structure and stability of benzene.
Answer
Benzene (C₆H₆) has a planar hexagonal structure with all C–C bond lengths equal (139 pm, between typical C–C single bond 154 pm and C=C double bond 134 pm). Each carbon is sp² hybridised with one unhybridised p-orbital. The six p-orbitals overlap to form a delocalised π-electron cloud above and below the ring plane (π-sextet). The resonance energy (delocalisation energy) of benzene is about 150 kJ mol⁻¹, making it about 150 kJ mol⁻¹ more stable than the hypothetical cyclohexatriene with three localised double bonds. This extra stability explains why benzene prefers substitution over addition — addition would destroy the aromatic stabilisation.
Q16. Write the products of the following reactions: (a) Benzene + Cl₂ (FeCl₃) (b) Benzene + HNO₃ (H₂SO₄) (c) Benzene + CH₃Cl (AlCl₃) (d) Benzene + CH₃COCl (AlCl₃)
Answer
(a) C₆H₆ + Cl₂ →⁽ᶠᵉᶜˡ³⁾ C₆H₅Cl + HCl (chlorobenzene, halogenation) (b) C₆H₆ + HNO₃ →⁽ᴴ²ˢᴼ⁴⁾ C₆H₅NO₂ + H₂O (nitrobenzene, nitration) (c) C₆H₆ + CH₃Cl →⁽ᴬˡᶜˡ³⁾ C₆H₅CH₃ + HCl (toluene, Friedel-Crafts alkylation) (d) C₆H₆ + CH₃COCl →⁽ᴬˡᶜˡ³⁾ C₆H₅COCH₃ + HCl (acetophenone, Friedel-Crafts acylation)
Q17. Explain the directive influence of –NO₂ and –CH₃ groups in electrophilic substitution of benzene.
Answer
–NO₂ group (meta director, deactivating): The –NO₂ group is a strong electron-withdrawing group (–I and –R effects). It withdraws electron density from the ortho and para positions more than the meta position, making the ring less reactive. The ortho and para positions bear partial positive charges, so the electrophile attacks at the meta position. Nitrobenzene is less reactive than benzene.
–CH₃ group (ortho/para director, activating): The –CH₃ group is electron-donating (+I effect and hyperconjugation). It increases electron density at the ortho and para positions, making these positions more nucleophilic. The electrophile preferentially attacks at ortho and para positions. Toluene is more reactive than benzene.
Q18. Why does benzene not undergo addition reactions under normal conditions?
Answer
Benzene has a special stability due to the delocalised π-electron cloud (resonance stabilisation of ≈150 kJ mol⁻¹). Addition reactions would destroy this aromatic stabilisation by breaking the conjugated π-system. Instead, benzene undergoes electrophilic substitution, which preserves the aromatic ring. Addition products are less stable than the aromatic starting material. However, under forcing conditions (e.g., H₂/Ni at high pressure and temperature, or Cl₂/hν), benzene can undergo addition.
Q19. How would you convert: (a) Ethene to ethane (b) Ethyne to ethene (c) Propene to 1-propanol (d) Benzene to nitrobenzene
Answer
(a) CH₂=CH₂ + H₂ →⁽ᴺⁱ⁾ CH₃–CH₃ (catalytic hydrogenation) (b) HC≡CH + H₂ →⁽ᴸⁱⁿᵈˡᵃʳ'ˢ ᶜᵃᵗ/ᴮᵃˢᴼ⁴/qᵘⁱⁿᵒˡⁱⁿᵉ⁾ CH₂=CH₂ (cis-ethene, partial hydrogenation) (c) CH₃CH=CH₂ + H₂O →⁽ᴴ³ᴾᴼ⁴⁾ CH₃CH(OH)CH₃ (2-propanol via Markovnikov addition) — For 1-propanol, use anti-Markovnikov: CH₃CH=CH₂ + HBr →⁽ᴾᵉʳᴼˣ⁾ CH₃CH₂CH₂Br → CH₃CH₂CH₂OH (1-propanol) (d) C₆H₆ + HNO₃ →⁽ᴴ²ˢᴼ⁴⁾ C₆H₅NO₂ + H₂O (nitration)
Q20. Explain why the following reactions will not work: (a) Friedel-Crafts alkylation of nitrobenzene (b) Friedel-Crafts alkylation of benzene with vinyl chloride
Answer
(a) Nitrobenzene cannot undergo Friedel-Crafts alkylation because the –NO₂ group is a strong deactivating group. It withdraws electron density from the ring, making it less nucleophilic. The ring becomes so deactivated that it cannot be attacked by even a strong electrophile. Additionally, AlCl₃ (the Lewis acid catalyst) complexes with the –NO₂ group itself.
(b) Vinyl chloride (CH₂=CHCl) cannot undergo Friedel-Crafts alkylation because the vinyl carbocation (CH₂=CH⁺) is extremely unstable. It is a primary carbocation with no hyperconjugative stabilisation from sp-hybridised carbons. AlCl₃ cannot generate a stable electrophile from vinyl chloride.
Q21. Write a note on the carcinogenicity of hydrocarbons.
Answer
Polynuclear aromatic hydrocarbons (PAHs) with four or more fused benzene rings are carcinogenic (cancer-causing). Examples include benzo[a]pyrene (found in tobacco smoke, charred meat, and automobile exhaust), benzanthracene, and dibenzanthracene. These compounds are formed during incomplete combustion of organic matter. They are metabolically activated in the body to form epoxides that bind to DNA, causing mutations that can lead to cancer. Benzene itself is also carcinogenic, linked to leukaemia with long-term exposure.
Q22. What are petrochemicals? Name some important hydrocarbon petrochemicals.
Answer
Petrochemicals are chemicals derived from petroleum (crude oil) and natural gas. Important hydrocarbon petrochemicals include:
- Methane → formaldehyde, methanol, chloromethane
- Ethene → polythene, ethylene glycol, ethanol, PVC
- Propene → polypropylene, isopropanol, acetone, acrylonitrile
- Butadiene → synthetic rubber (SBR, nitrile rubber)
- Benzene → styrene, nylon, phenol, aniline, detergents
- Toluene → TNT, benzaldehyde, polyurethane
- Xylene → PET, Dacron, terephthalic acid
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