Exercises
Q 9.1: How does ethane form during chlorination of methane?
Chlorination of methane proceeds by a free radical chain mechanism:
Initiation: Cl–Cl → 2Cl• (by UV light or heat)
Propagation
Cl• + CH₄ → CH₃• + HCl
CH₃• + Cl₂ → CH₃Cl + Cl•
Termination (coupling of radicals)
CH₃• + CH₃• → CH₃–CH₃ (ethane)
When two methyl radicals encounter each other (combination step) instead of reacting with Cl₂, ethane is formed. This is a minor product of the termination step.
Q 9.2: IUPAC names of compounds
(a) CH₃CH=C(CH₃)₂: Parent chain = 4C with one C=C. Numbering from end closest to double bond: 2-methylbut-2-ene = 2-methylbut-2-ene
(b) CH₂=CH–C≡C–CH₃: Longest chain with both double and triple bond = 5C. C=C at C-1, C≡C at C-3: pent-1-en-3-yne
(c) Cyclohexane ring with –CH₂–CH₂–CH=CH₂ substituent: (3-butenyl)cyclohexane or cyclohexylbut-3-ene
(d) –CH₂–CH₂–CH=CH₂: As a chain: but-3-en-1-yl group or parent compound butene
(e) Cyclopentane ring: If substituted, name accordingly.
(f) CH₃(CH₂)₄CH(CH₂CH(CH₃)₂)–(CH₂)₃CH₃: Longest chain analysis: 2-propylheptane or based on longest chain = 4-propylundecane (count all carbons in longest chain)
(g) CH₃–CH=CH–CH₂–CH=CH–CH(C₂H₅)–CH₂–CH=CH₂: Longest chain with maximum double bonds: identify longest C chain containing all three double bonds. 3-ethyldeca-1,4,7-triene or similar — identify 10C chain with ethyl substituent and three double bonds.
Q 9.3: Structural formulas and IUPAC names of isomers
(a) C₄H₈ (one double bond):
But-1-ene: CH₂=CHCH₂CH₃
But-2-ene: CH₃CH=CHCH₃ (cis and trans)
2-Methylpropene (isobutylene): (CH₃)₂C=CH₂
Cyclobutane (C₄H₈ without double bond, cyclic): cyclo-C₄H₈
Methylcyclopropane: CH(CH₃)–cyclopropyl ring
(b) C₅H₈ (one triple bond):
Pent-1-yne: HC≡C–CH₂CH₂CH₃
Pent-2-yne: CH₃–C≡C–CH₂CH₃
3-Methylbut-1-yne: HC≡C–CH(CH₃)₂
Q 9.4: IUPAC names of ozonolysis products
(i) Pent-2-ene (CH₃CH=CHCH₂CH₃):
CH₃CH=CHCH₂CH₃ → CH₃CHO (ethanal) + CH₃CH₂CHO (propanal)
(ii) 3,4-Dimethylhept-3-ene:
CH₃CH₂C(CH₃)=C(CH₃)CH₂CH₂CH₃ → CH₃CH₂COCH₃ (butan-2-one) + CH₃COCH₂CH₂CH₃ (pentan-2-one)
(iii) 2-Ethylbut-1-ene (CH₂=C(C₂H₅)CH₂CH₃):
→ HCHO (methanal) + C₂H₅COCH₂CH₃ (pentan-3-one)
(iv) 1-Phenylbut-1-ene (C₆H₅CH=CHCH₂CH₃):
→ C₆H₅CHO (benzaldehyde) + CH₃CH₂CHO (propanal)
Q 9.5: Alkene 'A' from ozonolysis gives ethanal and pentan-3-one
Ethanal: CH₃CHO (from CH₃CH=); Pentan-3-one: CH₃CH₂COCH₂CH₃ (from =C(C₂H₅)CH₂CH₃)
Alkene A: CH₃CH=C(C₂H₅)CH₂CH₃
IUPAC name: 3-ethylpent-2-ene
6: Alkene 'A' with three C–C σ bonds, eight C–H σ bonds, one C=C π bond; ozonolysis gives 2 mol aldehyde of M = 44
Answer
Molar mass of aldehyde = 44 = C₂H₄O
⇒ ethanal (CH₃CHO)
Ozonolysis gives 2 moles of ethanal
⇒ symmetrical alkene: CH₃CH=CHCH₃
Check: but-2-ene has 3 C–C σ bonds (C1–C2, C2=C3, C3–C4 counting: C1–C2 σ, C2–C3 σ+π, C3–C4 σ = 3 C–C σ bonds), 8 C–H bonds ✓
IUPAC name: but-2-ene (CH₃CH=CHCH₃)
Q 9.7: Alkene from propanal and pentan-3-one ozonolysis
Propanal: CH₃CH₂CHO → came from CH₃CH₂CH=
Pentan-3-one: CH₃CH₂COCH₂CH₃ → came from =C(C₂H₅)CH₂CH₃
Alkene: CH₃CH₂CH=C(C₂H₅)CH₂CH₃ (3-ethylhex-3-ene)
Q 9.8: Combustion reactions
(i) Butane: 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
(ii) Pentene (pent-1-ene): 2C₅H₁₀ + 15O₂ → 10CO₂ + 10H₂O
(iii) Hexyne (hex-1-yne): 2C₆H₁₀ + 17O₂ → 12CO₂ + 10H₂O
(iv) Toluene: C₇H₈ + 9O₂ → 7CO₂ + 4H₂O (or 2C₇H₈ + 18O₂ → 14CO₂ + 8H₂O)
9: Cis and trans structures of hex-2-ene; which has higher b.p.?
Answer
Cis-hex-2-ene: Both CH₃ and CH₂CH₂CH₃ groups on same side of C=C double bond.
Trans-hex-2-ene: Groups on opposite sides.
Structure: H₃C–CH=CH–CH₂CH₂CH₃ with cis = both substituents on same side.
Cis isomer has higher boiling point. Reason: Cis isomers are more polar (dipole moments add) and have stronger intermolecular dipole-dipole interactions, requiring more energy to separate. Trans isomers are less polar (dipoles cancel) with lower b.p.
10: Why is benzene extraordinarily stable despite three double bonds?
Answer
Benzene's stability results from aromatic delocalisation. The six π electrons are completely delocalised over the entire ring (all six carbons), forming a π electron cloud above and below the ring. This resonance/delocalisation energy (approximately 152 kJ mol⁻¹ extra stability, calculated from theoretical cyclohexatriene vs actual benzene) makes benzene much more stable than expected for a triene. Therefore benzene undergoes electrophilic substitution (preserving aromaticity) rather than addition (which would disrupt the π system).
Q 9.11: Necessary conditions for a system to be aromatic
Hückel's rule: A system is aromatic if it is:
(i) Cyclic
(ii) Planar
(iii) Fully conjugated (continuous p orbitals around ring)
(iv) Contains (4n + 2) π electrons, where n = 0, 1, 2, 3, ... (i.e., 2, 6, 10, 14... π electrons)
Q 9.12: Why the following systems are not aromatic?
(i) Cyclopentadienyl cation (C₅H₅⁺): Has 4 π electrons (4n where n=1) → antiaromatic, not aromatic.
(ii) Cyclohexadiene (non-conjugated ring): Not fully conjugated (sp³ carbons break conjugation) → not aromatic.
(iii) Cyclopentadienyl anion or a system with 4 π electrons: Does not satisfy Hückel's rule if 4n electrons.
Q 9.13: Converting benzene into specific derivatives
(i) p-Nitrobromobenzene:
Step 1: Benzene + Br₂/FeBr₃ → bromobenzene (electrophilic substitution)
Step 2: Bromobenzene + conc. HNO₃/H₂SO₄ (mild conditions, 50–60°C) → para-nitrobromobenzene (Br is ortho/para director; mainly p-product)
(ii) m-Nitrochlorobenzene:
Step 1: Benzene + Cl₂/AlCl₃ → chlorobenzene
Step 2: Chlorobenzene + HNO₃/H₂SO₄ (but Cl is o/p director → gives mostly o/p). For meta, nitrate first:
Better: Benzene + HNO₃/H₂SO₄ → nitrobenzene; Nitrobenzene + Cl₂/AlCl₃ → meta-nitrochlorobenzene (–NO₂ is meta director)
(iii) p-Nitrotoluene:
Step 1: Benzene + CH₃Cl/AlCl₃ → toluene
Step 2: Toluene + HNO₃/H₂SO₄ (dilute, below 30°C) → para-nitrotoluene (CH₃ is o/p director; at low temp para product predominates)
(iv) Acetophenone (C₆H₅COCH₃):
Benzene + CH₃COCl/AlCl₃ → acetophenone (Friedel-Crafts acylation)
Q 9.14: Primary, secondary, tertiary carbon atoms in H₃C–CH₂–C(CH₃)₂–CH₂–CH(CH₃)₂
Label each carbon:
C1: CH₃ — 1° (bonded to one C); C2: CH₂ — 2° (bonded to two C); C3: C(CH₃)₂ — 4° (bonded to four C); C3's methyls: 1°; C4: CH₂ — 2°; C5: CH — 3° (bonded to three C); C5's methyl groups: 1°
Primary C: C1(CH₃), two CH₃ on C3, two CH₃ on C5 — each bonded to 3H
Secondary C: C2 and C4 — bonded to 2H each
Tertiary C: C5 — bonded to 1H
Quaternary C: C3 — bonded to 0H
Q 9.15: Effect of branching on boiling point of alkanes
Branching decreases boiling point. Reason: Branched alkanes are more compact (globular) and have less surface area for contact between molecules. This reduces van der Waals (London dispersion) forces, requiring less energy to separate molecules → lower boiling point.
Example: n-pentane (b.p. 36°C) > isopentane (b.p. 28°C) > neopentane (b.p. 10°C)
Q 9.16: Markovnikov vs. anti-Markovnikov addition of HBr to propene
Normal (Markovnikov) addition: HBr adds to propene; H⁺ goes to carbon with more H atoms (C-1), Br⁻ to C-2 (more substituted). Product: 2-bromopropane (CH₃CHBrCH₃).
Mechanism: Step 1: H⁺ adds to C-1 → secondary carbocation at C-2 (more stable); Step 2: Br⁻ attacks C-2 → 2-bromopropane.
With benzoyl peroxide (anti-Markovnikov/peroxide effect): Benzoyl peroxide generates free radicals. Chain mechanism:
Initiation: (PhCOO)₂ → 2PhCOO• → 2Ph• + 2CO₂; Ph• + HBr → PhH + Br•
Propagation: Br• + CH₂=CHCH₃ → Br–CH₂–•CHCH₃ (secondary radical, more stable → Br goes to C-1, radical on C-2); •CHCH₃–CH₂Br + HBr → CH₃CHBrCH₂Br? No — Br• adds to C-1 to give more stable secondary radical (C-2 radical); radical at C-2 + HBr → CH₃CHBrCH₂Br... Actually: Br• adds to C-1: Br–CH₂–•CH–CH₃ (2° radical); This radical takes H from HBr: BrCH₂CH₂CH₃ (1-bromopropane) + Br•
Product: 1-bromopropane (anti-Markovnikov) because free radical mechanism favours more stable secondary radical at C-2 when Br adds to C-1.
Q 9.17: Ozonolysis of 1,2-dimethylbenzene (o-xylene) and Kekulé structure support
o-Xylene: 1,2-dimethylbenzene with two CH₃ on adjacent ring carbons.
If benzene has three C=C double bonds (Kekulé), ozonolysis of the double bond between C-1 and C-2 (which bears the two CH₃ groups) would give:
CH₃–CO–CHO + CHO–CO–CH₃ (glyoxal derivative)
Actual products of exhaustive ozonolysis of o-xylene include methylglyoxal (CH₃–CO–CHO) × 2 and glyoxal (OHC–CHO).
This is consistent with Kekulé's alternating double and single bond structure — both arrangements give the same product mixture, supporting the concept of two equivalent Kekulé structures (resonance).
Q 9.18: Acidic behaviour order: benzene, n-hexane, ethyne
Order of decreasing acidity: Ethyne > benzene > n-hexane
Reason: Acidity depends on the stability of the conjugate base (carbanion) and the s-character of the C–H bond.
Ethyne (HC≡CH): sp hybrid, 50% s-character → electrons held closer to nucleus → more electronegative C → H is more acidic (pKa ≈ 25)
Benzene: sp² hybrid, 33% s-character (pKa ≈ 43)
n-Hexane: sp³ hybrid, 25% s-character (pKa ≈ 50) → least acidic
19: Why benzene undergoes electrophilic substitution easily but nucleophilic substitution with difficulty?
Answer
Benzene has a ring of π electrons above and below the plane — an electron-rich system. This attracts electrophiles (positive or partially positive species) but repels nucleophiles (electron-rich species).
After electrophilic attack, benzene restores aromaticity by losing H⁺ (substitution, not addition), preserving the stable aromatic π system.
For nucleophilic substitution, the electron-rich π cloud repels nucleophiles; high-energy conditions are required.
Q 9.20: Convert compounds to benzene
(i) Ethyne → Benzene: 3HC≡CH → C₆H₆ (benzene); trimerisation over activated charcoal at 873 K
(ii) Ethene → Benzene: Ethene → ethyne (dehydrogenation); then trimerisation; or direct: 3C₂H₄ → C₆H₆ + 3H₂ (high temp, catalyst)
(iii) Hexane → Benzene: n-C₆H₁₄ → C₆H₆ + 4H₂ (catalytic reforming/dehydrogenation over Pt/Al₂O₃ at 773 K, aromatisation)
Q 9.21: Alkenes giving 2-methylbutane on hydrogenation
2-Methylbutane: (CH₃)₂CHCH₂CH₃ (2-methylbutane = isopentane)
All alkenes that on adding H₂ give (CH₃)₂CHCH₂CH₃:
(i) (CH₃)₂C=CHCH₃ (2-methylbut-2-ene)
(ii) (CH₃)₂CHCH=CH₂ (3-methylbut-1-ene)
(iii) CH₃C(=CH₂)CH₂CH₃ (2-methylbut-1-ene)
(iv) (CH₃)₂C=CH₂ would give 2-methylpropane... No — that's 4C. Correct structures must be 5C with correct branching.
Q 9.22: Decreasing reactivity with electrophile E⁺
(a) Chlorobenzene, 2,4-dinitrochlorobenzene, p-nitrochlorobenzene:
–NO₂ groups withdraw electrons (meta directors, deactivators). More NO₂ = more deactivated.
Order (decreasing reactivity): Chlorobenzene > p-nitrochlorobenzene > 2,4-dinitrochlorobenzene
(b) Toluene, p-CH₃C₆H₄–NO₂, p-O₂NC₆H₄–NO₂:
–CH₃ is activating (+I); –NO₂ is deactivating.
Order: Toluene > p-nitrotoluene > p-dinitrobenzene
23: Which undergoes nitration most easily: benzene, m-dinitrobenzene, toluene?
Answer
Toluene undergoes nitration most easily. The –CH₃ group is electron-donating (activating group, +I and hyperconjugation effect), increasing the electron density of the ring and making it more reactive towards the electrophile NO₂⁺. Benzene is next. m-Dinitrobenzene is least reactive (two deactivating –NO₂ groups).
Q 9.24: Lewis acid other than AlCl₃ for ethylation of benzene
Lewis acids used as Friedel-Crafts catalysts: FeCl₃, BF₃, ZnCl₂, SnCl₄. These accept electron pairs from the halogen of the alkyl or acyl halide to generate the carbocation electrophile.
25: Why Wurtz reaction not preferred for alkanes with odd carbon numbers?
Answer
Wurtz reaction: 2RX + 2Na
⇒ R–R + 2NaX. It joins two identical alkyl halides. For an odd-carbon alkane (e.g., propane, C₃H₈), we need to join CH₃X and C₂H₅X (two different halides). The reaction gives a mixture of three products:
CH₃X + CH₃X → C₂H₆ (ethane)
C₂H₅X + C₂H₅X → C₄H₁₀ (butane)
CH₃X + C₂H₅X → C₃H₈ (propane)
Separation of propane from this mixture is difficult and yield is poor. Therefore Wurtz reaction is not preferred for odd-carbon alkanes.
Important Extra Questions
Q A: Describe the mechanism of free radical halogenation of methane. Why is this a chain reaction?
Free radical chlorination of methane: CH₄ + Cl₂ → CH₃Cl + HCl (in presence of UV light)
Initiation: Cl₂ → 2Cl• (photolysis, hν)
Propagation
Cl• + CH₄ → •CH₃ + HCl (chain carrier regenerated in next step)
•CH₃ + Cl₂ → CH₃Cl + Cl• (Cl• regenerated → repeats)
This is a chain reaction because Cl• is continuously regenerated. One initiation step can lead to thousands of propagation cycles before termination.
Termination: Cl• + Cl• → Cl₂; •CH₃ + Cl• → CH₃Cl; •CH₃ + •CH₃ → C₂H₆
Q B: Explain Markovnikov's rule with mechanism for addition of HBr to propene.
Markovnikov's rule: When HX adds to an unsymmetrical alkene, H goes to carbon with more H atoms (less substituted), and X goes to the more substituted carbon.
Mechanism for propene + HBr
Step 1: H⁺ from HBr adds to C-1 of CH₂=CH–CH₃ → forms secondary carbocation at C-2: CH₃⁺CHCH₃ (more stable than primary at C-1)
Step 2: Br⁻ attacks the carbocation at C-2 → CH₃CHBrCH₃ (2-bromopropane)
Explanation: Secondary carbocation is more stable than primary (hyperconjugation, inductive effect), so H adds to give the more stable carbocation, which then reacts with Br⁻.
Q C: Why do alkynes show higher acidity than alkenes and alkanes?
Acidity of C–H bond depends on the hybridisation of carbon.
sp hybrid (alkyne): 50% s-character, electrons in s orbital are closer to nucleus, carbon is more electronegative, holds electrons more tightly, making the C–H bond more polar. The alkynyl carbanion (RC≡C⁻) is stabilised by the high s-character.
Terminal alkynes (RC≡CH) react with strong bases (NaNH₂) to form sodium acetylides: RC≡CH + NaNH₂ → RC≡C⁻Na⁺ + NH₃. Alkenes and alkanes do not react similarly because their carbanions are much less stable (lower s-character, pKa ~44 and ~50 respectively).
Q D: Explain electrophilic aromatic substitution with nitration of benzene as example.
Electrophilic aromatic substitution (EAS): Benzene acts as nucleophile (via π electrons) attacking an electrophile. Aromaticity is preserved by loss of H⁺ in the final step (substitution, not addition).
Nitration of benzene
Step 1: Generate electrophile: HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O (nitronium ion)
Step 2: Attack: Benzene + NO₂⁺ → arenium ion (sigma complex) — sp³ carbon formed, aromaticity temporarily lost
Step 3: Restore aromaticity: HSO₄⁻ removes H⁺ from sp³ carbon → nitrobenzene + H₂SO₄
Overall: C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O (at 55°C with conc. H₂SO₄)
Q E: Explain cis-trans isomerism in alkenes. Which properties differ between cis and trans isomers?
Restricted rotation around C=C double bond (due to the π bond) leads to cis-trans (geometrical) isomerism when each doubly bonded carbon has two different substituents.
Example: but-2-ene: cis (CH₃ groups on same side) and trans (CH₃ groups on opposite sides).
Properties that differ
Boiling point: Cis isomers are generally more polar → higher intermolecular forces → higher b.p. Trans isomers are less polar (symmetrical dipole cancellation) → lower b.p.
Melting point: Trans isomers often have higher m.p. due to better packing in crystal lattice (symmetrical molecules pack better).
Dipole moment: Cis isomers have higher dipole moment; trans have lower (often zero for symmetric trans).
Chemical reactivity: Generally similar, but can differ in specific reactions (e.g., maleic acid/cis can form anhydride by heating; fumaric acid/trans cannot).