Chapter 9: Hydrocarbons Long Questions | chemistry Class 11 CBSE

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Chapter 9: Hydrocarbons Long Questions | chemistry Class 11 CBSE | ByteNotes.in

Long Answer

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Question 1

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Compare and contrast the three classes of hydrocarbons — alkanes, alkenes, and alkynes — with respect to their general formulas, hybridisation, bond parameters, and characteristic chemical reactions.

Alkanes (CnH2n+2) have sp3 hybridised carbons with tetrahedral geometry (bond angle 109.5°), C-C bond length 154 pm, and C-H bond length 112 pm. Alkenes (CnH2n) have sp2 hybridised carbons at the double bond (bond angle 120°) and C=C bond length 134 pm. Alkynes (CnH2n-2) have sp hybridised carbons at the triple bond (bond angle 180°) and C≡C bond length 120 pm.
Bond enthalpies increase from C-C (348 kJ mol-1) to C=C (681 kJ mol-1) to C≡C (823 kJ mol-1), meaning triple bonds are the strongest and shortest.
Alkanes are inert (no reactive pi bonds) and undergo free radical substitution reactions (halogenation), combustion, isomerisation, and pyrolysis. Alkenes are reactive due to the pi bond and predominantly undergo electrophilic addition reactions (with H2, X2, HX, H2SO4, H2O, KMnO4, O3).
Alkynes also undergo electrophilic addition to both pi bonds (two stages), plus they show acidic character due to sp hybridisation — terminal alkynes react with Na and NaNH2 to liberate H2.
Aromatic stability prevents benzene (an unsaturated cyclic hydrocarbon) from undergoing addition; instead it favours electrophilic substitution to retain the delocalised pi system.
In summary: reactivity order is alkyne ≈ alkene >> alkane for addition reactions, while acid character follows alkyne > alkene > alkane due to increasing s-character of the hybrid orbital in the C-H bond.

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Question 2

Long Form

Describe the free radical chain mechanism of chlorination of methane. Explain how ethane is formed as a byproduct in this reaction.

Halogenation of alkanes proceeds by a three-step free radical chain mechanism. In the initiation step, the Cl-Cl bond is homolytically cleaved by heat or UV light (hv) to generate two chlorine free radicals: Cl-Cl → 2Cl•. The Cl-Cl bond is cleaved preferentially as it is weaker than C-C or C-H bonds.
In the propagation step (a), a Cl• radical abstracts a hydrogen atom from methane to form HCl and a methyl free radical: CH4 + Cl• → •CH3 + HCl. In step (b), the methyl radical reacts with another Cl2 molecule: •CH3 + Cl2 → CH3Cl + Cl•. These two steps repeat in a chain fashion.
Further propagation steps explain poly-chlorination: CH3Cl + Cl• → •CH2Cl + HCl; •CH2Cl + Cl2 → CH2Cl2 + Cl•, leading to di-, tri-, and tetrachloromethane.
In the termination step, the chain is broken when two free radicals combine: Cl• + Cl• → Cl2; •CH3 + •CH3 → C2H6; •CH3 + Cl• → CH3Cl. The combination of two methyl radicals in the termination step explains the formation of ethane as a byproduct.
The rate of replacement of H atoms in halogenation follows the order 3° > 2° > 1°, reflecting the relative stability of the corresponding alkyl radicals.
The overall reaction CH4 + Cl2 → CH3Cl + HCl can be accompanied by further chlorination if excess Cl2 is present, ultimately forming CCl4 in a stepwise manner.

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Question 3

Long Form

Explain the structure of benzene using valence bond theory and resonance concept. How does the resonance structure explain the observed uniform C-C bond length in benzene?

Kekulé proposed that benzene has a cyclic structure with six carbon atoms, alternating single and double bonds, and one hydrogen attached to each carbon. He later suggested that the double bonds oscillate between two equivalent positions to explain why only one ortho disubstituted product is formed.
According to modern valence bond theory, benzene is a resonance hybrid of two Kekulé structures (A and B) and several other minor contributing structures. The hybrid is represented by a regular hexagon with a circle (or dotted circle) inside, symbolising the delocalised pi electron cloud.
All six carbon atoms in benzene are sp2 hybridised. Each carbon forms three sigma bonds (two C-C and one C-H) using sp2 orbitals in the hexagonal plane. Each carbon retains one unhybridised 2p orbital perpendicular to the ring plane.
There are two equally probable sets of lateral overlaps for the 2p orbitals: C1-C2, C3-C4, C5-C6 or C2-C3, C4-C5, C6-C1. Since both are equally probable, all six 2p orbitals overlap simultaneously, forming two continuous doughnut-shaped electron clouds above and below the ring — this represents complete delocalisation.
X-ray diffraction confirms that all six C-C bond lengths in benzene are equal at 139 pm, which is intermediate between C-C single bond (154 pm) and C=C double bond (133 pm). This uniformity is the direct result of resonance/delocalisation: no individual bond is purely single or double.
The delocalised pi electrons are attracted more strongly by all six nuclei than localised electrons would be by just two, giving benzene a lower energy (extra stability) than the hypothetical cyclohexatriene. This resonance stabilisation energy explains benzene's exceptional stability and its preference for substitution over addition reactions.

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Question 4

Long Form

Discuss the electrophilic substitution reactions of benzene. Give the mechanism for nitration of benzene.

Benzene, despite being unsaturated, undergoes electrophilic substitution reactions because these reactions preserve its aromatic stability. The main electrophilic substitution reactions are: nitration, halogenation, sulphonation, Friedel-Crafts alkylation, and Friedel-Crafts acylation.
Nitration: Benzene is heated with a mixture of concentrated HNO3 and concentrated H2SO4 (nitrating mixture) at 323-333 K to give nitrobenzene. The role of H2SO4 is to generate the electrophile, nitronium ion (NO2+): HNO3 + H2SO4 → H2NO3+ + HSO4-; H2NO3+ → NO2+ + H2O.
Mechanism — Step 1 (Generation of electrophile): H2SO4 protonates HNO3 to form protonated nitric acid, which loses water to give the nitronium ion (NO2+), the active electrophile.
Step 2 (Formation of arenium ion): The nitronium ion attacks the pi electron cloud of benzene ring to form a sigma complex (arenium ion) in which one carbon becomes sp3 hybridised. The charge is delocalised over the ring by resonance.
Step 3 (Restoration of aromaticity): The arenium ion loses a proton (H+) to the HSO4- ion, restoring the aromatic pi system and regenerating H2SO4. The overall product is nitrobenzene.
Other SE reactions: Halogenation uses Lewis acids (AlCl3, FeCl3) to generate electrophilic X+ (halonium ion). Sulphonation uses fuming H2SO4 and generates SO3 as the electrophile. Friedel-Crafts reactions use AlCl3 to generate carbocations (alkylation) or acylium ions (acylation) as electrophiles.

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Question 5

Long Form

Analyse the conformations of ethane using Newman and Sawhorse projections. Discuss the relative stability of eclipsed and staggered conformations with reference to torsional strain.

Conformations are different spatial arrangements of atoms arising from rotation around a C-C single bond; they can interconvert without bond breaking. For ethane (C2H6), rotation around the C-C single bond generates an infinite number of conformations, but two extreme cases are the eclipsed and staggered forms.
In the Sawhorse projection, the molecule is viewed along the molecular axis; the C-C bond is drawn as a longer line tilted on paper, with the front carbon at the lower end and the rear carbon at the upper end, each bearing three lines representing H atoms at 120° to each other.
In the Newman projection, the C-C bond is viewed head-on. The front carbon is a dot with three H atoms drawn at 120°; the rear carbon is a circle with three H atoms on shorter lines at 120°. In the eclipsed conformation, rear H atoms are directly behind the front H atoms. In the staggered conformation, they are offset by 60°.
In the staggered conformation, the H atoms on adjacent carbons are maximally separated (dihedral angle = 60°), minimising electron cloud repulsion. This results in minimum torsional strain, minimum potential energy, and maximum stability.
In the eclipsed conformation, the H atoms on adjacent carbons are as close as possible (dihedral angle = 0°), causing maximum electron cloud repulsion between C-H bond electrons. This results in maximum torsional strain and maximum potential energy — it is the least stable conformation.
The energy difference between the two extreme forms is approximately 12.5 kJ mol-1. At room temperature, molecules have sufficient thermal energy to overcome this small barrier; hence rotation around C-C is nearly free. Staggered conformers are preferred and different conformers of ethane cannot be isolated separately.

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Question 6

Long Form

Discuss the addition reactions of alkenes with HBr under different conditions. Explain why Markovnikov's rule holds under normal conditions but is reversed in the presence of peroxides.

Alkenes undergo addition of HBr by two pathways: the Markovnikov (ionic) pathway under normal conditions, and the anti-Markovnikov (free radical) pathway in the presence of peroxides.
Under normal conditions (ionic mechanism), HBr adds according to Markovnikov's rule. HBr ionises and H+ (electrophile) attacks the double bond. For propene, H+ adds to the terminal CH2 (more H atoms) generating a secondary carbocation (CH3-CH+-CH3), which is more stable than the primary carbocation. Br- then attacks the secondary carbocation to give 2-bromopropane as the major product.
In the presence of peroxides (Kharash effect, free radical mechanism): Peroxides are homolytically cleaved by heat or UV light to give phenyl radicals (C6H5•). These abstract H from HBr to generate bromine radicals (Br•). The bromine radical adds to the less substituted carbon of propene (terminal CH2), since this forms the more stable secondary free radical (CH3-•CH-CH2Br vs primary CH3-CH=CH2Br).
This gives 1-bromopropane as the major product — the opposite of Markovnikov's rule. The reaction is driven by the formation of the more stable secondary radical intermediate.
The peroxide effect is observed only with HBr, not HCl or HI. H-Cl bond (430.5 kJ mol-1) is too strong to be homolytically cleaved by radicals, while H-I bond (296.8 kJ mol-1) is weak but iodine radicals are stable and tend to recombine rather than add to the double bond.
In summary, the nature of the intermediate (carbocation in ionic; carbon radical in free radical) determines the regioselectivity; more stable intermediates govern the preferred reaction pathway in both mechanisms.

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Question 7

Long Form

Explain the directive influence of substituents in monosubstituted benzene, clearly distinguishing between ortho/para and meta directing groups with electronic reasoning.

When monosubstituted benzene undergoes a second electrophilic substitution, the nature of the existing substituent controls the position of the incoming group. Groups are classified as ortho/para directors or meta directors based on their electronic effects.
Ortho/para directing groups: Substituents like -OH, -NH2, -OCH3, -CH3 increase electron density at ortho and para positions through resonance (+M effect). For example, in phenol, the lone pair on oxygen delocalises into the ring, and the resulting resonance structures show increased electron density (negative charge) at the o- and p-positions. Hence, the electrophile attacks these positions preferentially.
Halogens are ortho/para directors but deactivating: their lone pairs donate by resonance (+M) to o/p positions, but their -I effect (withdrawing electrons by induction) reduces overall electron density. Net result: halogenated benzene still undergoes o/p substitution, but more slowly than benzene.
Meta directing groups: Electron-withdrawing groups like -NO2, -CN, -CHO, -COOH, -SO3H reduce electron density at all positions, but the reduction is greater at o- and p-positions than at the meta position due to the resonance structures of nitrobenzene showing positive charge at o/p.
As a result, the meta position is relatively electron-rich compared to o/p, so the electrophile attacks the meta position. Meta directors are also ring deactivating groups because they reduce the overall pi electron density.
The practical implication is that activating substituents (-OH, -NH2, -CH3) facilitate further electrophilic substitution giving predominantly o/p products, while deactivating substituents (-NO2, -CN, -COOH) hinder further substitution and predominantly give meta products.

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Question 8

Long Form

Discuss the preparation of alkynes from (i) calcium carbide and (ii) vicinal dihalides. Also explain the acidic character of terminal alkynes with equations.

Preparation from calcium carbide (industrial method): Calcium carbonate is first heated to give calcium oxide (CaO + CO2). CaO is then heated with coke at very high temperature to give calcium carbide: CaO + 3C → CaC2 + CO. Treatment of calcium carbide with water gives ethyne: CaC2 + 2H2O → Ca(OH)2 + C2H2. This is the main industrial route to acetylene.
Preparation from vicinal dihalides: Vicinal dihalides (halides on adjacent carbons) treated with alcoholic KOH undergo dehydrohalogenation (elimination) to first give a vinyl halide, which on further treatment with sodamide (NaNH2) gives the alkyne. For example: CH2Br-CH2Br + alc. KOH → CH2=CHBr + KBr + H2O; CH2=CHBr + NaNH2 → HC≡CH + NaBr + NH3.
Acidic character of terminal alkynes: Terminal alkynes (those with H bonded to the triply bonded carbon) are acidic because the sp hybridised carbon (50% s-character) has high electronegativity and strongly attracts the C-H bond electrons.
Evidence: Ethyne reacts with sodium metal: HC≡CH + Na → HC≡C-Na+ + 1/2 H2 (monosodium ethynide); HC≡C-Na + Na → Na-C≡C-Na + 1/2 H2 (disodium ethynide). Propyne reacts with sodamide: CH3-C≡C-H + NaNH2 → CH3-C≡C-Na+ + NH3.
These reactions are not observed with ethane or ethene, confirming that only alkynes with terminal triple bond H atoms are acidic. Acidic character order: HC≡CH > H2C=CH2 > H3C-CH3.
Internal alkynes (e.g., but-2-yne) do not show this acidic character as neither triply bonded carbon bears a hydrogen atom directly attached to it.

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Medium

Question 9

Long Form

Describe the preparation of alkenes from alkyl halides and from alcohols. Explain why the rate of dehydrohalogenation follows the order tertiary > secondary > primary for alkyl groups.

Preparation of alkenes from alkyl halides (dehydrohalogenation): Alkyl halides are heated with alcoholic KOH (KOH dissolved in ethanol). One molecule of HX is eliminated to form an alkene. This is a beta-elimination reaction because the H atom is removed from the beta carbon (adjacent to the carbon bearing the halogen).
Example: CH3-CH2-Cl + alc. KOH → CH2=CH2 + KCl + H2O. The rate order for halogens is I > Br > Cl, reflecting the ease of C-X bond breaking. For alkyl groups, the rate is: tertiary > secondary > primary.
Preparation from alcohols (acidic dehydration): Alcohols are heated with concentrated H2SO4. Water is eliminated (one H from the beta-carbon and the -OH group), giving an alkene. Example: CH3CH2OH + conc. H2SO4 → CH2=CH2 + H2O. This is also a beta-elimination reaction.
Rate order for dehydrohalogenation (tertiary > secondary > primary): In E2 (or E1) elimination mechanisms, the alkyl group can stabilise the developing positive charge or partial positive charge on the alpha-carbon in the transition state. Tertiary carbons with more alkyl groups provide better stabilisation (hyperconjugation and inductive effect), lowering the activation energy and increasing the reaction rate.
Additionally, tertiary and secondary systems can readily form more stable carbocation intermediates (in E1 mechanism) compared to primary alkyl halides, further favouring their elimination.
The regioselectivity of elimination from unsymmetrical alkyl halides follows Saytzeff's rule: the more substituted (more stable) alkene is the major product.

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Question 10

Long Form

Discuss aromaticity with reference to Hückel's rule. Evaluate whether benzene, naphthalene, cyclopentadienyl anion, and cyclooctatetraene are aromatic, giving electronic reasoning.

Aromaticity is the special stability possessed by certain cyclic, planar, conjugated systems. Hückel's rule states that a planar monocyclic system with complete delocalisation of pi electrons is aromatic if it contains (4n+2) pi electrons, where n = 0, 1, 2, 3, ...
Benzene (C6H6): Has 6 pi electrons. Putting 4n+2 = 6 gives n = 1 (an integer). It is planar, fully conjugated, and satisfies Hückel's rule. Benzene is aromatic with exceptional stability.
Naphthalene (C10H8): Has 10 pi electrons; 4n+2 = 10 gives n = 2. It is planar, completely delocalised, and satisfies Hückel's rule. Naphthalene is aromatic.
Cyclopentadienyl anion (C5H5-): Has 6 pi electrons (5 from the 5 sp2 carbons' p orbitals, one from the carbanion); 4n+2 = 6, n = 1. It is planar with a complete conjugated system. The anion is aromatic.
Cyclooctatetraene (C8H8): Has 8 pi electrons; 4(1)+2 = 6 and 4(2)+2 = 10, so 8 does not satisfy (4n+2) for any integer n. Moreover, it is tub-shaped (non-planar) and the pi system is not fully delocalised. Cyclooctatetraene is NOT aromatic.
In summary: aromaticity requires simultaneous satisfaction of planarity, complete delocalisation, and the (4n+2) electron count. Absence of any one criterion means the compound is non-aromatic or anti-aromatic.

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Chapter 9: Hydrocarbons | Long Questions

Compare and contrast the three classes of hydrocarbons — alkanes, alkenes, and alkynes — with respect to their general formulas, hybridisation, bond parameters, and characteristic chemical reactions.

Describe the free radical chain mechanism of chlorination of methane. Explain how ethane is formed as a byproduct in this reaction.

Explain the structure of benzene using valence bond theory and resonance concept. How does the resonance structure explain the observed uniform C-C bond length in benzene?

Discuss the electrophilic substitution reactions of benzene. Give the mechanism for nitration of benzene.

Analyse the conformations of ethane using Newman and Sawhorse projections. Discuss the relative stability of eclipsed and staggered conformations with reference to torsional strain.

Discuss the addition reactions of alkenes with HBr under different conditions. Explain why Markovnikov's rule holds under normal conditions but is reversed in the presence of peroxides.

Explain the directive influence of substituents in monosubstituted benzene, clearly distinguishing between ortho/para and meta directing groups with electronic reasoning.

Discuss the preparation of alkynes from (i) calcium carbide and (ii) vicinal dihalides. Also explain the acidic character of terminal alkynes with equations.

Describe the preparation of alkenes from alkyl halides and from alcohols. Explain why the rate of dehydrohalogenation follows the order tertiary > secondary > primary for alkyl groups.

Discuss aromaticity with reference to Hückel's rule. Evaluate whether benzene, naphthalene, cyclopentadienyl anion, and cyclooctatetraene are aromatic, giving electronic reasoning.

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