NCERT Solutions for Class 11 Chemistry Chapter 9: Hydrocarbons

These Class 11 Chemistry Chapter 9 solutions cover Hydrocarbons from the NCERT textbook (session 2026–27). The chapter studies the chemistry of compounds made of carbon and hydrogen only — alkanes, alkenes, alkynes and aromatic hydrocarbons — including their IUPAC naming, isomerism, methods of preparation, physical and chemical properties, reaction mechanisms (free-radical halogenation, electrophilic addition, electrophilic aromatic substitution) and the special stability of benzene. Every numbered NCERT exercise (9.1–9.25) is reproduced and fully solved below.

Class: 11 Subject: Chemistry Unit: 9 Chapter: Hydrocarbons Exercises: 9.1 – 9.25 Session: 2026–27
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Class 11 Chemistry Chapter 9 – Overview

Hydrocarbons are compounds of carbon and hydrogen only and are the chief sources of energy (LPG, CNG, petrol, diesel). They are classified as saturated (alkanes, CnH2n+2, only C–C single bonds), unsaturated (alkenes CnH2n with a C=C double bond, alkynes CnH2n−2 with a C≡C triple bond) and aromatic (benzene and its derivatives). Alkanes are largely inert and undergo free-radical substitution, combustion and pyrolysis; alkenes and alkynes are electron-rich and undergo electrophilic addition (governed by Markovnikov’s rule, reversed by the peroxide effect); benzene, a resonance hybrid satisfying Hückel’s (4n+2)π rule, is exceptionally stable and undergoes electrophilic substitution. The chapter also covers conformations of ethane, cis–trans (geometrical) isomerism, ozonolysis and the directive (ortho/para vs meta) influence of substituents.

Key Concepts & Definitions

Homologous series: a family of compounds with the same general formula in which successive members differ by –CH2–, e.g. alkanes CnH2n+2.

Chain / position / functional isomerism: structural isomers differing in the carbon skeleton, in the position of the multiple bond, or in the functional group respectively.

Conformations: the infinite spatial arrangements obtained by rotation about a C–C single bond; ethane has a staggered (most stable, least torsional strain) and an eclipsed (least stable) form, differing by about 12.5 kJ mol−1.

Geometrical (cis–trans) isomerism: arises from restricted rotation about C=C; cis has identical groups on the same side, trans on opposite sides.

Markovnikov’s rule: in the addition of an unsymmetrical reagent (e.g. HBr) to an unsymmetrical alkene, the negative part adds to the carbon bearing fewer hydrogen atoms (via the more stable carbocation).

Peroxide (Kharasch) effect: in the presence of peroxide, HBr adds anti-Markovnikov to alkenes by a free-radical mechanism.

Aromaticity: a planar, cyclic, fully conjugated ring with delocalised (4n+2)π electrons (Hückel’s rule).

Ozonolysis: reaction of an alkene with O3 followed by reductive work-up (Zn/H2O) to cleave the C=C bond into two carbonyl compounds; used to locate the double bond.

Important Formulae & Reactions

General formulae: Alkanes CnH2n+2  |  Alkenes CnH2n  |  Alkynes CnH2n−2  |  Alkyl group CnH2n+1

Combustion: CnH2n+2 + (3n+1)/2 O2 → n CO2 + (n+1) H2O

Halogenation (free radical): CH4 + Cl2 ⟶(hν) CH3Cl + HCl (initiation → propagation → termination)

Hydrogenation: alkene/alkyne + H2 ⟶(Pt/Pd/Ni) alkane

Markovnikov addition: CH3–CH=CH2 + HBr → CH3–CHBr–CH3 (2-bromopropane)

Reactivity orders: halogenation F2 > Cl2 > Br2 > I2; HX addition HI > HBr > HCl; acidity of terminal H: HC≡CH > C6H6 > C6H14.

NCERT Exercise Solutions (9.1 – 9.25)

Questions reproduced verbatim from the NCERT textbook; answers are original, exam-ready and cross-checked with the NCERT answer key.

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

ANSWER Chlorination of methane proceeds by a free-radical chain mechanism. In the termination step, two methyl free radicals (formed during propagation) can combine with each other: 2 •CH3 → CH3–CH3 (ethane) Thus ethane appears as a by-product because of this side reaction in which two methyl radicals couple together.

9.2 Write IUPAC names of the following compounds: (a) CH3CH=C(CH3)2 (b) CH2=CH–C≡C–CH3 (c) a buta-1,3-diene type chain (as shown in the text) (d) CH2=CH–CH2–CH2– attached to a benzene ring (as shown) (e) the substituted phenol shown in the text (f) CH3(CH2)4CH(CH2CH(CH3)2)(CH2)3CH3 (g) CH3–CH=CH–CH2–CH=CH–CH(C2H5)–CH2–CH=CH2

ANSWER (a) 2-Methylbut-2-ene (b) Pent-1-en-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

9.3 For the following compounds, write structural formulas and IUPAC names for all possible isomers having the number of double or triple bonds as indicated: (a) C4H8 (one double bond) (b) C5H8 (one triple bond)

ANSWER (a) C4H8 (one C=C): (i) CH2=CH–CH2–CH3 — But-1-ene (ii) CH3–CH=CH–CH3 — But-2-ene (cis & trans) (iii) CH2=C(CH3)–CH3 — 2-Methylprop-1-ene (isobutylene) (b) C5H8 (one C≡C): (i) HC≡C–CH2–CH2–CH3 — Pent-1-yne (ii) CH3–C≡C–CH2–CH3 — Pent-2-yne (iii) (CH3)2CH–C≡CH — 3-Methylbut-1-yne

9.4 Write IUPAC names of the products obtained by the 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

ANSWER Ozonolysis cleaves C=C, replacing it with two C=O groups. (i) Pent-2-ene (CH3CH=CHCH2CH3) → Ethanal (CH3CHO) and Propanal (CH3CH2CHO). (ii) 3,4-Dimethylhept-3-ene → Butan-2-one and Pentan-2-one. (iii) 2-Ethylbut-1-ene (CH2=C(C2H5)2) → Methanal (HCHO) and Pentan-3-one. (iv) 1-Phenylbut-1-ene (C6H5CH=CHCH2CH3) → Benzaldehyde (C6H5CHO) and Propanal (CH3CH2CHO).

9.5 An alkene ‘A’ on ozonolysis gives a mixture of ethanal and pentan-3-one. Write structure and IUPAC name of ‘A’.

ANSWER Ethanal = CH3CHO (from CH3CH=) and pentan-3-one = (C2H5)2C=O (from =C(C2H5)2). Joining the two carbonyl carbons by a double bond gives ‘A’: CH3–CH=C(C2H5)–C2H5 IUPAC name: 3-Ethylpent-2-ene.

9.6 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’.

ANSWER Three C–C σ bonds + one C–C π bond means four carbon atoms with one C=C; with eight C–H bonds the molecular formula is C4H8. An aldehyde of molar mass 44 u is ethanal, CH3CHO. Two moles of CH3CHO on rejoining give CH3–CH=CH–CH3. IUPAC name: But-2-ene.

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

ANSWER Propanal = CH3CH2CHO (fragment CH3CH2CH=) and pentan-3-one = (C2H5)2C=O (fragment =C(C2H5)2). Joining the carbonyl carbons by C=C: CH3–CH2–CH=C(C2H5)–C2H5  (4-Ethylhex-3-ene).

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

ANSWER (i) 2 C4H10 + 13 O2 → 8 CO2 + 10 H2O   (i.e. C4H10 + 13/2 O2 → 4 CO2 + 5 H2O) (ii) C5H10 + 15/2 O2 → 5 CO2 + 5 H2O (iii) C6H10 + 17/2 O2 → 6 CO2 + 5 H2O (iv) C7H8 + 9 O2 → 7 CO2 + 4 H2O

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

ANSWER Hex-2-ene is CH3–CH=CH–CH2–CH2–CH3. cis-hex-2-ene: the –CH3 and the –C3H7 groups lie on the same side of the C=C. trans-hex-2-ene: the –CH3 and the –C3H7 groups lie on opposite sides of the C=C. The cis isomer has the higher boiling point. Being more polar (the bond dipoles do not cancel), it has stronger intermolecular dipole–dipole interactions, so more heat energy is needed to separate the molecules.

9.10 Why is benzene extraordinarily stable though it contains three double bonds?

ANSWER Benzene is extra stable because of resonance. Its six π electrons are not fixed in three localised double bonds but are completely delocalised over the planar ring, forming two continuous π clouds above and below the plane. This delocalisation lowers the energy (resonance energy ≈ 150 kJ mol−1), making benzene a stable resonance hybrid rather than a reactive triene.

9.11 What are the necessary conditions for any system to be aromatic?

ANSWER A system is aromatic if it is: (i) cyclic and planar; (ii) has a fully conjugated (continuous) system of π electrons / p-orbitals around the ring; and (iii) contains (4n+2)π electrons (Hückel’s rule), where n is an integer (0, 1, 2, …).

9.12 Explain why the following systems are not aromatic? (the three ring systems shown in the text)

ANSWER These systems are not aromatic because they fail Hückel’s rule — there is a lack of delocalisation of (4n+2)π electrons in the cyclic system. Either the ring is not fully conjugated (an sp3 carbon breaks the conjugation) or the number of π electrons does not equal 4n+2, so a continuous delocalised π cloud cannot form and the system is non-aromatic.

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

ANSWER (i) p-Nitrobromobenzene: first brominate benzene (Br2/FeBr3) to bromobenzene; Br is an ortho/para director, so nitration (conc. HNO3+conc. H2SO4) gives mainly p-nitrobromobenzene. (ii) m-Nitrochlorobenzene: first nitrate benzene to nitrobenzene; –NO2 is a meta director, so chlorination (Cl2/FeCl3) gives m-nitrochlorobenzene. (Doing the steps in this order places the groups meta.) (iii) p-Nitrotoluene: first methylate benzene by Friedel–Crafts (CH3Cl/anhyd. AlCl3) to toluene; –CH3 is ortho/para directing, so nitration gives mainly p-nitrotoluene. (iv) Acetophenone: Friedel–Crafts acylation of benzene with acetyl chloride (CH3COCl) in the presence of anhyd. AlCl3 gives acetophenone (C6H5COCH3).

9.14 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 of these.

ANSWER The molecule is 2,2,4-trimethylpentane. Counting all carbons: Primary (1°) carbons: 5 such carbons (the five CH3 groups) → 15 H attached to 1° carbons. Secondary (2°) carbons: 2 such CH2 carbons → 4 H attached to 2° carbons. Tertiary (3°) carbon: 1 such CH carbon → 1 H attached to a 3° carbon. (The remaining quaternary carbon bears no hydrogen.)

9.15 What effect does branching of an alkane chain has on its boiling point?

ANSWER More branching lowers the boiling point. Branching makes the molecule more spherical, reducing the surface area of contact between molecules; this weakens the van der Waals (London) forces, which are then overcome at a lower temperature. Hence, among isomers, the most branched one has the lowest 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 and give mechanism.

ANSWER Without peroxide (ionic, Markovnikov): HBr gives H+, which adds to the terminal carbon of CH3CH=CH2 to form the more stable secondary carbocation CH3–+CH–CH3; Br− then attacks it → 2-bromopropane (Br on the C with fewer H, Markovnikov). With benzoyl peroxide (free-radical, anti-Markovnikov / peroxide effect): the peroxide generates Br• radicals. Br• adds to the terminal CH2 to give the more stable secondary radical CH3–•CH–CH2Br; this abstracts H from HBr → 1-bromopropane, regenerating Br• to continue the chain. So the change in mechanism (ionic vs free-radical) reverses the regiochemistry of addition.

9.17 Write down the products of ozonolysis of 1,2-dimethylbenzene (o-xylene). How does the result support Kekulé structure for benzene?

ANSWER Because the position of the double bonds in benzene is not fixed, o-xylene exists as two resonating Kekulé structures, and ozonolysis cleaves both. The products obtained are a mixture of methylglyoxal (2-oxopropanal, CH3CO–CHO), glyoxal (OHC–CHO) and butane-2,3-dione (diacetyl, CH3CO–COCH3). No single Kekulé structure can give all three products at once; obtaining all of them shows the double bonds are not localised but oscillate — i.e. benzene is a resonance hybrid of the two Kekulé structures.

9.18 Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Also give reason for this behaviour.

ANSWER Order of decreasing acidity: HC≡CH > C6H6 > C6H14 (ethyne > benzene > n-hexane). Reason: acidity of a C–H bond increases with the s-character of the carbon’s hybrid orbital, because greater s-character holds the bonding electrons closer to carbon, stabilising the resulting anion. The C of ethyne is sp (50% s), of benzene sp2 (33% s) and of n-hexane sp3 (25% s). Hence ethyne is the most acidic and n-hexane the least.

9.19 Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty?

ANSWER Benzene has a delocalised cloud of six π electrons, making it a rich source of electrons. It is therefore readily attacked by electron-deficient species (electrophiles), so electrophilic substitution occurs easily. Conversely, an electron-rich ring repels incoming electron-rich species (nucleophiles), so nucleophilic substitution takes place only with great difficulty.

9.20 How would you convert the following compounds into benzene? (i) Ethyne    (ii) Ethene    (iii) Hexane

ANSWER (i) Ethyne → benzene: cyclic polymerisation — passing ethyne through a red-hot iron tube at 873 K trimerises three molecules to benzene. 3 HC≡CH ⟶(red-hot Fe, 873 K) C6H6. (ii) Ethene → benzene: first add Br2 to ethene to get 1,2-dibromoethane; dehydrohalogenate with alc. KOH to ethyne (then HC≡CH on partial dehydrohalogenation), and finally trimerise the ethyne (red-hot Fe, 873 K) to benzene. (i.e. CH2=CH2 → CH2Br–CH2Br → HC≡CH → C6H6.) (iii) Hexane → benzene: aromatisation (reforming) — heat n-hexane to 773 K at 10–20 atm over V2O5 (or Mo2O3/Cr2O3) on alumina; it is dehydrogenated and cyclised to benzene. C6H14 → C6H6 + 4 H2.

9.21 Write structures of all the alkenes which on hydrogenation give 2-methylbutane.

ANSWER 2-Methylbutane is (CH3)2CH–CH2–CH3. Placing a C=C between the various adjacent carbons of this skeleton gives: (i) CH2=C(CH3)–CH2–CH3 — 2-Methylbut-1-ene (ii) CH3–C(CH3)=CH–CH3 — 2-Methylbut-2-ene (iii) (CH3)2CH–CH=CH2 — 3-Methylbut-1-ene

9.22 Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile, E+: (a) Chlorobenzene, 2,4-dinitrochlorobenzene, p-nitrochlorobenzene (b) Toluene, p-H3C–C6H4–NO2, p-O2N–C6H4–NO2

ANSWER Electron-withdrawing groups (–NO2) deactivate the ring; more of them means lower reactivity toward an electrophile. (a) Chlorobenzene > p-nitrochlorobenzene > 2,4-dinitrochlorobenzene. (b) Toluene > p-nitrotoluene (p-CH3–C6H4–NO2) > p-dinitrobenzene (p-O2N–C6H4–NO2).

9.23 Out of benzene, m-dinitrobenzene and toluene which will undergo nitration most easily and why?

ANSWER Toluene undergoes nitration most easily. The –CH3 group is electron-releasing (+I, hyperconjugation), so it activates the ring and makes it electron-rich, favouring attack by the nitronium electrophile. Benzene has no activating group, while m-dinitrobenzene has two strongly deactivating –NO2 groups, making it the least reactive. Order: toluene > benzene > m-dinitrobenzene.

9.24 Suggest the name of a Lewis acid other than anhydrous aluminium chloride which can be used during ethylation of benzene.

ANSWER Anhydrous ferric chloride (FeCl3) can be used as the Lewis acid catalyst for the Friedel–Crafts ethylation of benzene. (Anhydrous FeBr3 or BF3 would also serve.)

9.25 Why is Wurtz reaction not preferred for the preparation of alkanes containing odd number of carbon atoms? Illustrate your answer by taking one example.

ANSWER In the Wurtz reaction two alkyl halides couple with sodium, so a symmetrical alkane with an even number of carbons forms readily. To make an alkane with an odd number of carbons, two different alkyl halides must be used; this produces a mixture of three alkanes (the two symmetrical products plus the desired unsymmetrical one), which is hard to separate — so the yield of the required alkane is poor. Example: to prepare heptane (7 C), 1-bromopropane and 1-bromobutane are taken. Besides heptane, the side products hexane (propyl–propyl) and octane (butyl–butyl) are also formed. Hence the Wurtz reaction is not preferred for odd-carbon alkanes.

Extra Practice Questions

Short Answer Type Questions

Q1. State Markovnikov’s rule.

ANSWERWhen an unsymmetrical reagent adds to an unsymmetrical alkene, the negative part of the reagent attaches to the doubly bonded carbon bearing the smaller number of hydrogen atoms (because it proceeds via the more stable carbocation).

Q2. Why are alkanes called paraffins?

ANSWERFrom Latin parum (little) and affinis (affinity), because alkanes are chemically very inert — they do not react with acids, bases or common reagents under normal conditions.

Q3. Why is the staggered conformation of ethane more stable than the eclipsed?

ANSWERIn the staggered form the C–H bonds of the two carbons are as far apart as possible, giving minimum electron-cloud repulsion (minimum torsional strain) and hence minimum energy, so it is more stable.

Q4. What is Lindlar’s catalyst and what is it used for?

ANSWERPalladised charcoal partially deactivated (poisoned) with sulphur compounds or quinoline. It is used for the partial hydrogenation of alkynes to cis-alkenes.

Q5. Give the test for unsaturation using bromine.

ANSWERThe reddish-orange colour of bromine in CCl4 is discharged (decolourised) when added to an alkene or alkyne, because the halogen adds across the multiple bond — a positive test for unsaturation.

Long Answer Type Questions

Q1. Describe the free-radical mechanism of the chlorination of methane.

ANSWERIt is a chain reaction in three stages. Initiation: Cl2 absorbs UV light and undergoes homolysis to give two chlorine radicals, Cl2 ⟶(hν) 2 Cl•. Propagation: Cl• abstracts H from CH4 to give CH3• + HCl; CH3• then reacts with Cl2 to give CH3Cl + Cl•, regenerating the radical so the chain continues; further substitution gives CH2Cl2, CHCl3 and CCl4. Termination: radicals combine (Cl•+Cl•→Cl2, CH3•+CH3•→C2H6, CH3•+Cl•→CH3Cl), ending the chain. The coupling of two methyl radicals explains the ethane by-product.

Q2. Explain why cis-but-2-ene is polar while trans-but-2-ene is non-polar.

ANSWERIn cis-but-2-ene both methyl groups lie on the same side of the C=C, so the two C–CH3 bond dipoles point in similar directions and add up, giving a net dipole moment (about 0.33 D). In trans-but-2-ene the methyl groups lie on opposite sides; the equal bond dipoles point in opposite directions and cancel, so the net dipole moment is almost zero and the molecule is non-polar. This is why the cis isomer is more polar and boils higher.

Q3. Explain the directive influence of substituents in electrophilic substitution of benzene, with examples.

ANSWERA group already on the ring decides where the next electrophile attacks and whether the ring is activated. Ortho/para directors (electron-releasing: –CH3, –OH, –NH2, –OCH3, halogens) increase electron density at the ortho and para positions, so substitution occurs there; most of them also activate the ring (halogens are deactivating ortho/para directors). Meta directors (electron-withdrawing: –NO2, –COOH, –CHO, –SO3H, –CN) reduce electron density at ortho/para, leaving the meta position relatively richer, so substitution occurs at meta, and they deactivate the ring. For example, nitration of toluene gives mainly p- (and o-) nitrotoluene, while nitration of nitrobenzene gives m-dinitrobenzene.

MCQs & Assertion–Reason

1. The general formula of alkynes is:

(a) CnH2n+2    (b) CnH2n    (c) CnH2n−2    (d) CnH2n+1

2. Addition of HBr to propene in the presence of peroxide gives mainly:

(a) 2-bromopropane    (b) 1-bromopropane    (c) 1,2-dibromopropane    (d) propan-2-ol

3. The most stable conformation of ethane is:

(a) eclipsed    (b) skew    (c) staggered    (d) gauche only

4. According to Hückel’s rule, a system is aromatic if it has:

(a) 4n π electrons    (b) (4n+2) π electrons    (c) any number of π electrons    (d) only σ electrons

5. Ozonolysis of pent-2-ene gives:

(a) two molecules of ethanal    (b) ethanal and propanal    (c) methanal and butanal    (d) propanone and ethanal

6. Lindlar’s catalyst is used to convert an alkyne into:

(a) an alkane    (b) a trans-alkene    (c) a cis-alkene    (d) an aldehyde

7. Which is the strongest acid (most acidic C–H)?

(a) n-hexane    (b) benzene    (c) ethene    (d) ethyne

8. The correct order of rate of halogenation of alkanes is:

(a) I2 > Br2 > Cl2 > F2    (b) F2 > Cl2 > Br2 > I2    (c) Cl2 > F2 > Br2 > I2    (d) Br2 > Cl2 > I2 > F2

9. –NO2 group attached to benzene is:

(a) an ortho/para-directing activator    (b) a meta-directing deactivator    (c) an ortho/para-directing deactivator    (d) a meta-directing activator

10. Branching in an alkane, compared with its straight-chain isomer, generally:

(a) raises the boiling point    (b) lowers the boiling point    (c) does not change it    (d) raises the melting point only

Answer key: 1-(c), 2-(b), 3-(c), 4-(b), 5-(b), 6-(c), 7-(d), 8-(b), 9-(b), 10-(b).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: Benzene undergoes electrophilic substitution rather than addition.

Reason: Benzene is a resonance hybrid with a delocalised π cloud that makes it electron-rich and stable.

A-R 2. Assertion: cis-but-2-ene has a higher boiling point than trans-but-2-ene.

Reason: The cis isomer is more polar, so it has stronger dipole–dipole intermolecular forces.

A-R 3. Assertion: Addition of HBr to propene in the presence of peroxide gives 1-bromopropane.

Reason: The peroxide effect changes the mechanism to free-radical, giving anti-Markovnikov addition.

A-R 4. Assertion: Ethyne is more acidic than ethane.

Reason: The sp carbon of ethyne has greater s-character, holding the C–H bonding electrons closer to carbon.

A-R 5. Assertion: The Wurtz reaction is preferred for preparing alkanes with an odd number of carbon atoms.

Reason: Coupling of two different alkyl halides gives a single pure product.

Answer key: 1-(A), 2-(A), 3-(A), 4-(A), 5-(D).

Common Mistakes to Avoid

Watch out for these

  • Confusing the general formulae — alkanes CnH2n+2, alkenes CnH2n, alkynes CnH2n−2.
  • Applying Markovnikov’s rule when a peroxide is present — with HBr + peroxide the addition is anti-Markovnikov (peroxide effect works only for HBr, not HCl/HI).
  • Forgetting that cis (more polar) usually boils higher in liquids, but trans usually melts higher in solids.
  • Treating benzene’s bonds as three fixed double bonds — they are delocalised; ozonolysis of o-xylene proves this.
  • Writing the wrong direction for substituents: –CH3, –OH, –NH2 are ortho/para; –NO2, –COOH are meta.
  • Not balancing combustion equations (use 3n+1/2 O2 for alkanes and recount O for alkenes/alkynes/aromatics).
  • Recombining ozonolysis fragments wrongly — always join the two carbonyl carbons by a C=C bond.

How to score full marks in this chapter

For naming, always pick the longest chain that contains the multiple bond and give it the lowest locant. For mechanism questions (halogenation, HBr addition, peroxide effect), clearly label initiation, propagation and termination or the carbocation/radical stability. In ozonolysis problems work backwards: split each carbonyl, drop the two O atoms, and join the carbons by C=C. Remember the standard reactivity orders (F2>Cl2>Br2>I2; HI>HBr>HCl; acidity sp>sp2>sp3) and quote Hückel’s (4n+2)π rule for any aromaticity question.

Frequently Asked Questions

What is Class 11 Chemistry Chapter 9 about?

Chapter 9, Hydrocarbons, studies compounds of carbon and hydrogen — alkanes, alkenes, alkynes and aromatic hydrocarbons — covering their nomenclature, isomerism, preparation, properties, reaction mechanisms (free-radical halogenation, electrophilic addition and aromatic substitution) and the special stability of benzene.

How many exercises are there in Class 11 Chemistry Chapter 9?

There are 25 numbered exercise questions (9.1 to 9.25). All of them are reproduced and solved step by step on this page, with answers cross-checked against the NCERT answer key.

What is the peroxide effect in Chapter 9 Hydrocarbons?

The peroxide (Kharasch) effect is the anti-Markovnikov addition of HBr to an unsymmetrical alkene in the presence of a peroxide. It proceeds by a free-radical mechanism, so HBr adds to propene to give 1-bromopropane instead of 2-bromopropane.

Are these Class 11 Chemistry Chapter 9 solutions free?

Yes. All ClearStudy NCERT Solutions for Class 11 Chemistry are free and follow the official NCERT textbook for session 2026–27.

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