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NCERT Exercises and Solutions: Hydrocarbons

🎓 Class 11 Chemistry CBSE Theory Ch 9 – Hydrocarbons ⏱ ~8 min
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NCERT Exercises and Solutions: Hydrocarbons

Chapter Summary — Hydrocarbons at a Glance

Classification

  • Saturated — alkanes (CₙH₂ₙ₊₂), only single C–C bonds, sp³ C, tetrahedral, 109.5°.
  • Unsaturated — alkenes (CₙH₂ₙ, sp², 120°, 1σ+1π), alkynes (CₙH₂ₙ₋₂, sp, 180°, 1σ+2π).
  • Aromatic — planar cyclic conjugated systems with (4n+2) π electrons (Hückel's rule); benzene = parent.

Alkanes — Key reactions

  • Preparation: Sabatier-Senderens (alkene + H₂/Ni); Wurtz (2 RX + 2 Na, dry ether); decarboxylation (RCOONa + NaOH/CaO/Δ); Kolbe electrolysis.
  • Halogenation: free-radical chain (initiation–propagation–termination); reactivity F > Cl > Br > I.
  • Combustion → CO₂ + H₂O; pyrolysis (cracking); aromatisation (n-hexane → benzene over Pt/Al₂O₃, 773 K).
  • Conformations of ethane: staggered (stable) vs eclipsed (12.5 kJ/mol higher).

Alkenes — Key reactions

  • Preparation: from alkynes (partial H₂); β-elimination of alkyl halides (KOH/alc, Zaitsev); dehydration of alcohols (H⁺/Δ); dehalogenation of vicinal dihalides (Zn).
  • Geometrical isomerism: cis/trans; E/Z (CIP priority).
  • Electrophilic addition: H₂/Ni; X₂ (bromine water test); HX (Markovnikov via carbocation; anti-Markovnikov via radical with peroxide — only HBr).
  • Hydration (H₂O/H⁺); cold dilute KMnO₄ (Baeyer's test) → vicinal diol; ozonolysis → 2 carbonyl compounds.
  • Polymerisation → polythene, polypropene.

Alkynes — Key reactions

  • Preparation: CaC₂ + H₂O → HC≡CH; vicinal dihalides + alc KOH/NaNH₂.
  • Acidic terminal H (pKa 25); white precipitate with AgNO₃/NH₃, red with Cu₂Cl₂/NH₃.
  • Addition: H₂ (full or partial via Lindlar/Na-NH₃ for cis/trans alkene); Br₂; HX (Markovnikov, gem-dihalide); H₂O/HgSO₄ → carbonyl.
  • Cyclic trimerisation: 3 HC≡CH → C₆H₆.

Aromatic Hydrocarbons

  • Benzene: Kekulé structure ⇌ resonance hybrid; 6 equivalent C–C bonds (139 pm); resonance energy ~150 kJ/mol; Hückel (4n+2) rule.
  • Electrophilic substitution mechanism: arenium ion (σ-complex) → loss of H⁺ → aromatic product.
  • Reactions: nitration (HNO₃/H₂SO₄, NO₂⁺); halogenation (X₂/Fe-X or Al-X); sulphonation (oleum); Friedel-Crafts alkylation/acylation (RX or RCOX/AlCl₃).
  • Directive effects: o/p directors (–OH, –OR, –NH₂, –NHR, –NHCOR, alkyl, halogens); m-directors (–NO₂, –COOH, –CHO, –COR, –SO₃H, –CN).
  • PAHs: benzo[a]pyrene and dibenzanthracene from incomplete combustion → carcinogenic.

Key Terms

Hydrocarbon, Alkane, Alkene, Alkyne — saturated/unsaturated open-chain hydrocarbons with general formulas CₙH₂ₙ₊₂, CₙH₂ₙ, CₙH₂ₙ₋₂.

Conformation — different spatial arrangements arising from C–C bond rotation; staggered & eclipsed for ethane.

Geometrical (cis–trans) isomerism — restricted rotation about C=C; same atoms, different spatial arrangement.

Markovnikov's Rule — H of HX adds to C of C=C bearing more H; X to C with fewer H. Mechanism: most stable carbocation.

Anti-Markovnikov / Peroxide effect (Kharasch) — radical mechanism in presence of peroxide; reverses Markovnikov; only with HBr.

Saytzeff (Zaitsev) rule — major elimination product is the more substituted alkene.

Aromaticity (Hückel's rule) — planar, fully conjugated cyclic system with (4n+2) π electrons (n = 0,1,2,…).

Resonance energy — extra stability of a delocalised π system over a hypothetical localised one; ~150 kJ/mol for benzene.

Electrophilic aromatic substitution — replacement of an H on an aromatic ring by an E⁺ via arenium ion intermediate.

Activator/Deactivator; o/p- vs m-director — describes how an existing substituent affects ring reactivity and the position of the next E⁺.

NCERT Exercises — Step-by-Step Solutions

Exercise 9.1: How do you account for the formation of ethane during chlorination of methane?

During chlorination of methane, the propagation step generates methyl radicals (•CH₃). In the termination step, two •CH₃ radicals can combine: •CH₃ + •CH₃ → CH₃–CH₃ (ethane). Hence small amounts of ethane appear as a side-product whenever radical concentration is high.

Exercise 9.2: Write IUPAC names of the following compounds: (a) (CH₃)₃CCH₂C(CH₃)₃ (b) (CH₃)₂C(C₂H₅)₂ (c) CH₃C(CH₃)₂CH₂CH₂CH(CH₃)₂

(a) (CH₃)₃CCH₂C(CH₃)₃ — longest chain: C1=C(CH₃)₃ → CH₂ → C(CH₃)₃ = 5 carbons (pentane); methyls at C2 (×2) and C4 (×2). Name: 2,2,4,4-tetramethylpentane.
(b) (CH₃)₂C(C₂H₅)₂ — central C attached to two CH₃ and two C₂H₅; longest chain through one ethyl, central C, the other ethyl = 5 C (pentane); two methyls both at C3. Name: 3,3-dimethylpentane.
(c) CH₃C(CH₃)₂CH₂CH₂CH(CH₃)₂ — longest chain: 7 C (heptane). Numbering from end nearer (CH₃)₂C: C1, C2 (CH₃)₂, C3, C4, C5, C6 (CH₃), C7. Substituents: methyls at 2,2 and 5. Name: 2,2,5-trimethylhexane. Note: actually 6 C chain → hexane (recount: (CH₃)₃C-CH₂-CH₂-CH(CH₃)-CH₃ = 6 carbons → hexane with three methyls at 2,2,5).

Exercise 9.3: For the following compounds, write structural formulas and IUPAC names of all possible isomers: (a) C₄H₁₀ (b) C₅H₁₀ (alkene) (c) C₅H₈ (alkyne)

(a) C₄H₁₀ — 2 isomers:
1. CH₃CH₂CH₂CH₃ — n-butane
2. (CH₃)₃CH — 2-methylpropane (isobutane)

(b) C₅H₁₀ alkenes — 5 structural + geometrical isomers:
1. CH₂=CHCH₂CH₂CH₃ — pent-1-ene
2. CH₃CH=CHCH₂CH₃ — pent-2-ene (cis & trans)
3. (CH₃)₂C=CHCH₃ — 2-methylbut-2-ene
4. CH₂=C(CH₃)CH₂CH₃ — 2-methylbut-1-ene
5. (CH₃)₂CHCH=CH₂ — 3-methylbut-1-ene

(c) C₅H₈ alkynes:
1. CH≡CCH₂CH₂CH₃ — pent-1-yne
2. CH₃C≡CCH₂CH₃ — pent-2-yne
3. CH≡CCH(CH₃)CH₃ — 3-methylbut-1-yne

Exercise 9.4: Give the IUPAC names of the following compounds: (a) (CH₃)₂C=CHCH₃ (b) (CH₃)₂CHCH₂C≡CH (c) CH₃CH=C(CH₃)CH₂CH(CH₃)CH₃

(a) (CH₃)₂C=CHCH₃ = (CH₃)₂C=CHCH₃ → 4-C chain with C=C between C2 and C3, methyl at C2: 2-methylbut-2-ene.
(b) (CH₃)₂CHCH₂C≡CH → 5-C chain with C≡C at C1; methyl at C4: 4-methylpent-1-yne.
(c) CH₃CH=C(CH₃)CH₂CH(CH₃)CH₃ → 7-C chain (CH₃-CH=C(CH₃)-CH₂-CH(CH₃)-CH₃ counted = 6 C + methyl branch). Longest chain: 6 C; C=C at C2; methyl at C3 and C5 (numbering from C=C end): 3,5-dimethylhex-2-ene.

Exercise 9.5: Although alkenes and alkynes both contain π bonds, alkenes generally undergo electrophilic addition while alkynes show both electrophilic and nucleophilic addition. Why?

The C atoms of an alkyne are sp-hybridised (50% s character), holding the bonding electrons closer to nucleus → more electronegative C. The π electrons of the triple bond are therefore less available for electrophiles than those of an alkene's C=C (sp², 33% s). Consequently alkynes react somewhat slower with electrophiles. At the same time the increased electronegativity of the sp C makes the C δ⁺ in an alkyne polar enough to allow nucleophiles (e.g., RO⁻, CN⁻) to attack the C, especially of activated alkynes.

Exercise 9.6: Draw cis and trans isomers of the following compounds and write their IUPAC names: (a) CHCl=CHCl (b) C₂H₅C(CH₃)=C(CH₃)C₂H₅

(a) CHCl=CHCl (1,2-dichloroethene):
cis-1,2-dichloroethene (Z): both Cl on same side.
trans-1,2-dichloroethene (E): Cl on opposite sides.
IUPAC: (Z)- and (E)-1,2-dichloroethene.
(b) C₂H₅C(CH₃)=C(CH₃)C₂H₅ (3,4-dimethylhex-3-ene):
cis: both ethyl on same side, methyls on the other.
trans: ethyl groups on opposite sides.
IUPAC: (Z)- and (E)-3,4-dimethylhex-3-ene.

Exercise 9.7: Why is benzene extra ordinarily stable though it contains three double bonds?

The six π electrons of benzene are completely delocalised over all six C atoms in two doughnut-shaped clouds above and below the ring plane. This delocalisation gives a resonance energy ~150 kJ/mol — i.e., benzene is 150 kJ/mol more stable than a hypothetical cyclohexa-1,3,5-triene with localised double bonds. Equivalently, the heat of hydrogenation of benzene (–208 kJ/mol) is much less negative than 3× that of cyclohexene (–360 kJ/mol). The Hückel (4n+2) condition is satisfied with n=1 (6 π electrons), confirming benzene's aromaticity.

Exercise 9.8: What are the necessary conditions for any system to be aromatic?

A system is aromatic if it satisfies ALL three Hückel conditions:
1. Cyclic — closed ring of atoms.
2. Planar — all atoms of the ring in one plane (so p orbitals can overlap).
3. Fully conjugated — every ring atom has a p-orbital perpendicular to the plane, contributing to one continuous π cloud.
4. (4n+2) π electrons — number of delocalised π electrons must be 2, 6, 10, 14, … (n = 0, 1, 2, 3,…).
Examples: benzene (6 π, n=1), naphthalene (10 π, n=2), cyclopentadienyl anion (6 π), tropylium cation (6 π).

Exercise 9.9: Explain why following systems are not aromatic? (i) cyclooctatetraene (ii) cyclohexa-1,3,5-triene (in tub form)

(i) Cyclooctatetraene (C₈H₈): contains 8 π electrons. 8 ≠ (4n+2) — fits 4n with n=2 → would be anti-aromatic if planar. To avoid the anti-aromatic destabilisation, the molecule adopts a non-planar tub (boat) shape, which prevents continuous p-orbital overlap → behaves as a non-aromatic polyene.
(ii) "Cyclohexa-1,3,5-triene" if drawn with localised double bonds and non-planar geometry — would lack the continuous overlap that makes benzene aromatic. Real benzene IS aromatic; the non-aromatic localised triene form is purely hypothetical and is exactly what we compare benzene to when calculating resonance energy.

Exercise 9.10: How will you convert benzene into (i) p-nitrobromobenzene (ii) m-nitrochlorobenzene (iii) p-nitrotoluene (iv) acetophenone?

(i) p-nitrobromobenzene — Brominate first (–Br is o/p director), then nitrate.
  C₆H₆ + Br₂/Fe → C₆H₅Br → +HNO₃/H₂SO₄ → p-O₂N-C₆H₄-Br (major) + ortho (minor).
(ii) m-nitrochlorobenzene — Nitrate first (–NO₂ is meta director), then chlorinate.
  C₆H₆ + HNO₃/H₂SO₄ → C₆H₅NO₂ → +Cl₂/Fe → m-Cl-C₆H₄-NO₂.
(iii) p-nitrotoluene — Methylate first (–CH₃ is o/p director), then nitrate.
  C₆H₆ + CH₃Cl/AlCl₃ → C₆H₅CH₃ → +HNO₃/H₂SO₄ → p-CH₃-C₆H₄-NO₂.
(iv) Acetophenone (C₆H₅COCH₃) — Friedel-Crafts acylation:
  C₆H₆ + CH₃COCl/anhyd. AlCl₃ → C₆H₅COCH₃ + HCl.

Exercise 9.11: In the alkane H₃C–CH₂–C(CH₃)₂–CH₂–CH(CH₃)₂, identify 1°, 2°, 3° carbon atoms and give the number of H atoms bonded to each one.

Number the carbons: C1=CH₃, C2=CH₂, C3=C(CH₃)₂ (with two more methyls C3', C3''), C4=CH₂, C5=CH(CH₃) (with C5' methyl), C6=CH₃.
Primary (1°) C — bonded to one other C, hence with 3 H each: C1, C3', C3'', C5', C6 (five 1° carbons; total 5 × 3 = 15 H).
Secondary (2°) C — bonded to 2 other C, with 2 H each: C2, C4 (two 2° carbons; total 4 H).
Tertiary (3°) C — bonded to 3 other C, with 1 H each: C5 (one 3° carbon; 1 H).
Quaternary (4°) C — bonded to 4 other C, with 0 H: C3 (one 4° carbon; 0 H).
Total H = 15 + 4 + 1 + 0 = 20. Check formula: C₉H₂₀ — consistent.

Exercise 9.12: What effect does branching of an alkane chain have on its boiling point?

Boiling point decreases with branching for isomeric alkanes. Reason: branched molecules are more spherical/compact, with smaller surface area; van der Waals (London dispersion) attractions between molecules are weaker → less energy needed to separate molecules. Example for C₅H₁₂: n-pentane 36 °C > 2-methylbutane 28 °C > neopentane 9.5 °C.

Exercise 9.13: Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give mechanism.

Without peroxide (ionic, Markovnikov): HBr ionises; H⁺ adds to the C of C=C with more H (terminal CH₂), generating the more stable secondary carbocation CH₃–CH⁺–CH₃. Br⁻ then attacks → 2-bromopropane.
With peroxide (free-radical, anti-Markovnikov / Kharasch):
Initiation: (PhCO₂)₂ → 2 PhCO₂• → 2 Ph• + 2 CO₂; Ph• + HBr → PhH + Br•.
Propagation: Br• + CH₃-CH=CH₂ → CH₃-•CH-CH₂Br (more stable 2° radical, Br on terminal C); then CH₃-•CH-CH₂Br + HBr → CH₃-CH₂-CH₂Br + Br• → 1-bromopropane.
Termination: combinations of radicals.
The two mechanisms place Br on opposite carbons, giving opposite products. The peroxide effect is observed only with HBr (energetics of propagation are right only for HBr).

Exercise 9.14: Write IUPAC names of the products obtained by ozonolysis of the following compounds: (i) Pent-2-ene (ii) 3,4-Dimethylhept-3-ene (iii) 2-Ethylbut-1-ene (iv) 1-Phenylbut-1-ene

Ozonolysis cleaves C=C and replaces it with two C=O.
(i) Pent-2-ene CH₃-CH=CH-CH₂CH₃: → 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)C₂H₅: → HCHO (methanal) + CH₃CH₂COCH₂CH₃ (pentan-3-one).
(iv) 1-Phenylbut-1-ene C₆H₅-CH=CH-CH₂CH₃: → C₆H₅CHO (benzaldehyde) + CH₃CH₂CHO (propanal).

Exercise 9.15: An alkene 'A' on ozonolysis gives a mixture of ethanal and pentan-3-one. Write structure and IUPAC name of 'A'.

Reverse the ozonolysis: combine CH₃CHO (= CH₃-CH=) and pentan-3-one (CH₃CH₂-CO-CH₂CH₃ → CH₃CH₂-C(C₂H₅)=) into one alkene at the C=O carbons.
Structure: CH₃-CH=C(C₂H₅)-CH₂CH₃ → IUPAC name: 3-Ethylpent-2-ene.

Exercise 9.16: An alkene 'A' contains three C–C, eight C–H σ bonds and one C–C π bond. 'A' on ozonolysis gives two moles of an aldehyde of molar mass 44 u. Write IUPAC name of 'A'.

Total bonds: 3 (C–C σ) + 1 (C–C π) + 8 (C–H) = 12 → molecular formula needs 4 C, 8 H → C₄H₈ alkene.
Aldehyde of M = 44 = CH₃CHO (acetaldehyde).
Two CH₃CHO from ozonolysis means the alkene has structure CH₃-CH=CH-CH₃ with CH₃ on each side of C=C.
Structure: but-2-ene (CH₃CH=CHCH₃); IUPAC name: but-2-ene.

Exercise 9.17: Propanal and pentan-3-one are the ozonolysis products of an alkene. What is the structural formula of the alkene?

Propanal (CH₃CH₂CHO) contributes CH₃CH₂-CH= ; pentan-3-one (C₂H₅COC₂H₅) contributes =C(C₂H₅)C₂H₅.
Combine: CH₃CH₂-CH=C(C₂H₅)C₂H₅. IUPAC: 3-Ethylpent-2-ene. (Alternate: 4-ethylhex-3-ene if ethyl is counted as part of the chain; correct IUPAC = 3-ethylpent-2-ene.)

Exercise 9.18: Write chemical equations for combustion reaction of the following hydrocarbons: (i) Butane (ii) Pentene (iii) Hexyne (iv) Toluene

General CₙHₘ + (n + m/4) O₂ → n CO₂ + (m/2) H₂O.
(i) Butane (C₄H₁₀): 2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O.
(ii) Pentene (C₅H₁₀): 2 C₅H₁₀ + 15 O₂ → 10 CO₂ + 10 H₂O.
(iii) Hexyne (C₆H₁₀): 2 C₆H₁₀ + 17 O₂ → 12 CO₂ + 10 H₂O.
(iv) Toluene (C₆H₅CH₃ = C₇H₈): C₇H₈ + 9 O₂ → 7 CO₂ + 4 H₂O.

Exercise 9.19: Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p. and why?

Hex-2-ene = CH₃-CH=CH-CH₂-CH₂-CH₃.
cis (Z)-: CH₃ and C₃H₇ on same side.
trans (E)-: CH₃ and C₃H₇ on opposite sides.
The cis isomer has a small dipole moment (μ ≠ 0) due to vector addition of two C–C dipoles → stronger dipole–dipole attractions in the bulk liquid. Hence cis-hex-2-ene has higher b.p. (~68.9 °C vs 67.9 °C for trans). However, trans is more thermodynamically stable (less steric strain) and packs better in the solid → higher melting point.

Exercise 9.20: Why is benzene extra ordinarily stable though it contains three double bonds?

See Exercise 9.7. Benzene's six π electrons are delocalised in a continuous ring, giving a resonance energy of ~150 kJ/mol — far greater than expected for any normal alkene. This delocalisation explains: (i) all six C–C bonds are equivalent (139 pm); (ii) heat of hydrogenation is only 208 kJ/mol (vs 360 kJ/mol for hypothetical cyclohexatriene); (iii) benzene undergoes substitution rather than addition.

Exercise 9.21: What are the necessary conditions for any system to be aromatic?

See Exercise 9.8: the system must be cyclic, planar, fully conjugated (all ring atoms sp² with a perpendicular p-orbital), and contain (4n+2) π electrons (Hückel's rule, n = 0, 1, 2, …).

Exercise 9.22: Explain why the following systems are not aromatic? (i) (cyclopentadiene) (ii) cyclohepta-1,3,5-triene (iii) cyclooctatetraene

(i) Cyclopentadiene (C₅H₆): has one sp³ CH₂ — breaks conjugation. The deprotonated cyclopentadienyl anion (C₅H₅⁻) IS aromatic (6 π).
(ii) Cyclohepta-1,3,5-triene: has one sp³ CH₂ — breaks conjugation. The cycloheptatrienyl cation (tropylium, C₇H₇⁺) IS aromatic (6 π).
(iii) Cyclooctatetraene (C₈H₈): has 8 π electrons → does not satisfy (4n+2). To avoid anti-aromatic destabilisation, it adopts a non-planar tub shape, which also prevents continuous π overlap → behaves as a normal non-aromatic polyene.

Exercise 9.23: How will you convert benzene into (i) p-nitrobromobenzene (ii) m-nitrochlorobenzene (iii) p-nitrotoluene (iv) acetophenone?

See Exercise 9.10. Strategy: install activating groups (–CH₃, –Br) before the second reaction if you want o/p, install deactivating groups (–NO₂) first if you want m.

Exercise 9.24: In the following compounds, identify the more reactive towards electrophilic substitution: (i) toluene vs nitrobenzene (ii) phenol vs benzene (iii) chlorobenzene vs bromobenzene

(i) Toluene >> nitrobenzene (–CH₃ activates; –NO₂ strongly deactivates).
(ii) Phenol >>> benzene (–OH is a strong activator due to lone-pair donation by O; phenol is ~10⁶× more reactive).
(iii) Chlorobenzene ≈ bromobenzene (both are weak deactivators by –I; +M effect is weaker for Br as 4p–2p overlap is poor compared to 3p–2p of Cl). In practice chlorobenzene is slightly more reactive towards EAS than bromobenzene.

Exercise 9.25: Suggest the name of a Lewis acid other than anhydrous AlCl₃ which can be used during ethylation of benzene.

Other anhydrous Lewis acids that catalyse Friedel-Crafts alkylation: FeCl₃, BF₃, ZnCl₂, SbCl₅, HF, H₂SO₄, AlBr₃. These all generate the carbocation by abstracting halide from the alkyl halide.

Exercise 9.26: Why is Wurtz reaction not preferred for the preparation of alkanes containing odd number of carbon atoms? Illustrate.

Wurtz reaction (2 RX + 2 Na → R-R + 2 NaX) couples two identical alkyl groups, so it always gives an alkane with an EVEN number of C atoms. To prepare an alkane with an ODD number of C, one would need two DIFFERENT alkyl halides (e.g., CH₃Br + C₂H₅Br to get propane). But this gives a statistical mixture of three alkanes:
  CH₃Br + CH₃Br → ethane (C₂H₆)
  CH₃Br + C₂H₅Br → propane (C₃H₈)
  C₂H₅Br + C₂H₅Br → n-butane (C₄H₁₀)
The mixture is hard to separate by distillation (low yield of desired propane). Hence Wurtz is unsuitable for unsymmetrical / odd-carbon alkanes.
Activity 9.5 — Synthesis design challenge

Setup: You are asked to convert benzene to 1,3,5-trinitrobenzene (TNB).

Predict: What conditions and number of nitrations are required? Why is each subsequent nitration progressively harder?

Three nitrations are needed:

  1. C₆H₆ + HNO₃/H₂SO₄ (333 K) → C₆H₅NO₂ (nitrobenzene). –NO₂ is m-director.
  2. C₆H₅NO₂ + HNO₃/H₂SO₄ (warm) → m-dinitrobenzene. The first –NO₂ already deactivates the ring → harsher conditions.
  3. m-dinitrobenzene + HNO₃/H₂SO₄ (fuming, 95–100 °C, prolonged) → 1,3,5-trinitrobenzene. With two –NO₂ groups already present, the ring is extremely deactivated; both groups direct meta, so the third –NO₂ enters at the only mutually meta position (C5).

Each successive –NO₂ withdraws more electron density and slows the next substitution by ~10⁵ times. TNB is a useful but very expensive explosive precursor; TNT (trinitrotoluene) is far easier to make from toluene because –CH₃ activates and points o/p, partly counteracting the –NO₂ groups.

Reaction Identifier Quiz

Pick a reactant + reagent pair; the simulator names the reaction.

Reaction: Free-radical halogenation

Cl₂ under UV light splits homolytically; the chain mechanism (initiation–propagation–termination) chlorinates the alkane.

Competency-Based Questions

Q1. The major product when 1-butene reacts with HBr in the presence of peroxide is: L1 Remember

  • (a) 2-bromobutane (b) 1-bromobutane (c) 2-bromopropane (d) 1,2-dibromobutane
Answer: (b) 1-bromobutane — anti-Markovnikov addition via radical mechanism (Kharasch effect).

Q2. Why does ozonolysis of 2-methylbut-2-ene produce both an aldehyde and a ketone? L4 Analyse

Answer: The C=C of 2-methylbut-2-ene, (CH₃)₂C=CHCH₃, has different substituents on each carbon. Ozonolysis cleaves C=C and replaces it with two C=O. The C bearing two CH₃ becomes a ketone ((CH₃)₂C=O = acetone); the C bearing one H and one CH₃ becomes an aldehyde (CH₃CHO = acetaldehyde). Thus the unsymmetrical alkene yields two different carbonyls, while symmetrical 2,3-dimethylbut-2-ene would yield only acetone (×2).

Q3. Predict the product when toluene is treated with conc HNO₃ and conc H₂SO₄. L3 Apply

Answer: Toluene (C₆H₅CH₃) undergoes nitration at the o & p positions because –CH₃ is an o/p director and weak activator. Major products: 2-nitrotoluene (o) and 4-nitrotoluene (p); in practice the para isomer dominates due to less steric hindrance. Trinitration with excess fuming HNO₃ → 2,4,6-trinitrotoluene (TNT).

Q4. Compare the acidity of ethane, ethene, and ethyne. Provide quantitative reasoning using % s-character. L5 Evaluate

Acidity order: ethyne (sp, 50% s, pKa 25) > ethene (sp², 33% s, pKa 44) > ethane (sp³, 25% s, pKa ~50). Higher s-character in the C–H bonding orbital means the bonding electrons spend more time near C → C is effectively more electronegative → the conjugate base (carbanion) is more stable when the negative charge is on a high-s orbital. Hence sp C–H is the most acidic; ethyne reacts with Na metal to release H₂. Ethene and ethane do not lose H⁺ under standard conditions.

Q5. HOT (Create): Design a synthesis of n-hexane using only ethyl chloride and any inorganic reagents. L6 Create

Plan: n-hexane = CH₃(CH₂)₄CH₃ = 6 carbons. Three molecules of ethyl chloride contribute 6 carbons.
Step 1 (Wurtz to butane): 2 C₂H₅Cl + 2 Na (dry ether) → CH₃CH₂CH₂CH₃ + 2 NaCl (n-butane).
Step 2 (Halogenation, statistical): CH₃CH₂CH₂CH₃ + Cl₂/hν → mixture of 1-chlorobutane and 2-chlorobutane. Separate 1-chlorobutane.
Step 3 (Wurtz crossed): CH₃CH₂CH₂CH₂Cl + C₂H₅Cl + 2 Na → CH₃(CH₂)₄CH₃ (n-hexane) + by-products (n-butane and n-octane).
Yield is moderate but isolation of n-hexane by fractional distillation (b.p. 69 °C) is straightforward. Cleaner alternative: prepare 1-bromohexane separately and reduce, but the Wurtz approach uses only the requested ingredients.

Assertion–Reason Questions

Choose: (A) Both true, R explains A. (B) Both true, R doesn't explain A. (C) A true, R false. (D) A false, R true.

A: Branched alkanes have lower boiling points than straight-chain isomers.

R: Branched molecules are more compact and have smaller surface area, so van der Waals attractions between molecules are weaker.

Answer: (A). Both true; R correctly explains A. Example: n-pentane 36 °C > isopentane 28 °C > neopentane 9.5 °C.

A: Aniline (C₆H₅NH₂) gives Friedel-Crafts alkylation product easily.

R: The –NH₂ group is a strong activator that increases ring electron density.

Answer: (D). Assertion is FALSE — aniline does NOT undergo F-C alkylation cleanly because –NH₂ is a strong Lewis base that complexes with AlCl₃, deactivating the catalyst and the ring (now NH₂AlCl₃⁺ is meta-directing). Reason is TRUE in isolation: –NH₂ is indeed a strong activator. The protection (acetylation to acetanilide) is the standard workaround.

A: Polythene is obtained by catalytic polymerisation of ethene under high pressure.

R: π electrons of ethene undergo addition successively forming a long –(CH₂CH₂)–ₙ chain.

Answer: (A). Both true; R explains A. The Ziegler-Natta catalyst (TiCl₄/Al(C₂H₅)₃) and Phillips process (CrO₃) make this an industrial-scale reaction; polythene production exceeds 100 million tonnes per year worldwide.

Frequently Asked Questions — NCERT Exercises and Solutions: Hydrocarbons

What are the most common Chapter 9 questions in CBSE board exams?
Common CBSE Class 11 Chemistry Chapter 9 board questions: (1) IUPAC nomenclature of given hydrocarbons; (2) preparation and properties of methane, ethene, ethyne — equations only or with mechanisms; (3) Markovnikov vs anti-Markovnikov addition with peroxide effect; (4) ozonolysis to identify alkene structure from products; (5) electrophilic aromatic substitution mechanism (nitration, halogenation, sulphonation, Friedel-Crafts); (6) directive effects of substituents in benzene derivatives; (7) conformations of ethane (staggered vs eclipsed). The MyAiSchool exercise set covers all patterns with step-by-step model solutions.
How do you predict products of Markovnikov and anti-Markovnikov addition?
To predict products in NCERT Class 11 Chemistry Chapter 9 exercises: (1) for Markovnikov addition (HX without peroxide, H₂SO₄, H₂O/H⁺), H adds to the carbon with more H, X adds to the carbon with fewer H — gives more substituted product via more stable carbocation; (2) for anti-Markovnikov addition (HBr with peroxide only — Kharasch effect), H adds to the carbon with fewer H, Br adds to the carbon with more H — gives less substituted product via more stable free radical. Example: CH₃-CH=CH₂ + HBr (Markovnikov) → CH₃-CHBr-CH₃; with peroxide → CH₃-CH₂-CH₂Br.
How do you determine alkene structure from ozonolysis products?
To determine alkene structure from ozonolysis in NCERT Class 11 Chemistry Chapter 9: (1) identify the carbonyl products (aldehydes from =CH-, ketones from =CR₂ groups); (2) reconnect the carbon-carbon bond of the carbonyl groups to reconstruct the original C=C; the carbonyl oxygens replace the alkene carbons that were originally bonded. Example: ozonolysis gives CH₃-CHO + CH₃-CO-CH₃, meaning the original alkene was CH₃-CH=C(CH₃)-CH₃ (2-methylbut-2-ene). For diol product clue, oxidative ozonolysis gives carboxylic acid or ketone — adjust the analysis accordingly.
How do you predict the orientation of disubstituted benzene products?
To predict orientation in NCERT Class 11 Chemistry Chapter 9 EAS exercises: (1) identify the existing substituent on the ring; (2) classify as ortho/para directing (electron-donating: -OH, -NH₂, -OCH₃, -CH₃, -X) or meta directing (electron-withdrawing: -NO₂, -CN, -COOH, -SO₃H); (3) the incoming electrophile attaches at the predicted positions. Example: toluene (methylbenzene) with HNO₃/H₂SO₄ gives ortho- and para-nitrotoluene (major), little meta; nitrobenzene with HNO₃/H₂SO₄ gives mainly m-dinitrobenzene. Always cite the activating/deactivating nature and explain via resonance for full marks.
How do you draw Newman projections for ethane conformations?
To draw Newman projections in NCERT Class 11 Chemistry Chapter 9 exercises: (1) view the molecule along the C-C bond axis; (2) draw the front carbon as a dot with three bonds at 120°; (3) draw the back carbon as a larger circle with three bonds at 120° behind; (4) in staggered conformation, back-carbon bonds are offset by 60° from front-carbon bonds; (5) in eclipsed, back-carbon bonds align directly behind front-carbon bonds. Show the energy diagram with three eclipsed maxima (each ~12 kJ/mol) and three staggered minima (each at 0) per 360° rotation. Free rotation is essentially possible at room temperature.
What 5-mark questions appear in Chapter 9 CBSE board exams?
Common 5-mark CBSE Class 11 Chemistry Chapter 9 board questions: (1) describe preparation and chemical properties of methane (4 reactions) and acetylene (4 reactions); (2) explain Markovnikov rule, anti-Markovnikov (peroxide effect) and ozonolysis with examples; (3) discuss aromaticity and Hückel's rule; describe the structure of benzene with resonance; (4) explain electrophilic substitution mechanism (nitration of benzene) with stepwise reactions and intermediates; (5) describe directing effects with examples of activating and deactivating groups. The MyAiSchool exercise set provides model answers with marking schemes.
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