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Hydrocarbons

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Chapter Notes

Overview
This chapter covers the classification, nomenclature, preparation, properties, and reactions of hydrocarbons—alkanes, alkenes, alkynes, and aromatic hydrocarbons. It is essential in the syllabus because hydrocarbons form the backbone of organic chemistry and are fundamental to fuels, polymers, and industrial chemistry. The one big idea a student must walk away with is that the presence of sigma (σ) and pi (π) bonds determines the reactivity of hydrocarbons, with alkenes and alkynes undergoing electrophilic addition due to their π bonds, while alkanes undergo free radical substitution.
Key Concepts (24)
Hydrocarbons
Compounds of carbon and hydrogen only.
e.g. Methane (CH₄) is the simplest hydrocarbon.
Saturated hydrocarbons
Hydrocarbons containing only carbon-carbon single bonds.
e.g. Alkanes like ethane (C₂H₆) are saturated hydrocarbons.
Unsaturated hydrocarbons
Hydrocarbons containing carbon-carbon multiple bonds (double or triple bonds).
e.g. Ethene (C₂H₄) has a double bond and is unsaturated.
Aromatic hydrocarbons
A special type of cyclic compounds that follow Hückel's rule (4n+2 π electrons) and exhibit aromaticity.
e.g. Benzene (C₆H₆) is the simplest aromatic hydrocarbon.
Alkanes
Saturated open chain hydrocarbons containing carbon-carbon single bonds, with general formula CₙH₂ₙ₊₂.
e.g. Methane (CH₄), ethane (C₂H₆), propane (C₃H₈).
Alkenes
Unsaturated hydrocarbons containing at least one carbon-carbon double bond, with general formula CₙH₂ₙ.
e.g. Ethene (C₂H₄), propene (C₃H₆).
Alkynes
Unsaturated hydrocarbons containing at least one carbon-carbon triple bond, with general formula CₙH₂ₙ₋₂.
e.g. Ethyne (C₂H₂), propyne (C₃H₄).
Structural isomerism
Isomers that have the same molecular formula but different structural arrangements of atoms.
e.g. Butane (C₄H₁₀) has two chain isomers: n-butane and isobutane.
Chain isomers
Structural isomers that differ in the arrangement of the carbon chain (straight vs branched).
e.g. Pentane (C₅H₁₂) has three chain isomers: n-pentane, isopentane, and neopentane.
Geometrical isomerism (cis-trans isomerism)
A type of stereoisomerism where atoms or groups are arranged differently around a double bond due to restricted rotation.
e.g. But-2-ene exists as cis-but-2-ene (same side) and trans-but-2-ene (opposite sides).
Conformations (conformers/rotamers)
Different spatial arrangements of atoms that can be converted into one another by rotation around a C-C single bond.
e.g. Ethane has staggered and eclipsed conformations.
Markovnikov rule
When an unsymmetrical reagent adds to an unsymmetrical alkene, the negative part of the addendum gets attached to that carbon atom which possesses lesser number of hydrogen atoms.
e.g. Addition of HBr to propene gives 2-bromopropane as the major product.
Peroxide effect (Kharash effect)
In the presence of peroxides, addition of HBr to unsymmetrical alkenes takes place contrary to Markovnikov rule, giving anti-Markovnikov product.
e.g. Addition of HBr to propene in presence of peroxide gives 1-bromopropane.
Electrophilic addition reaction
A reaction in which an electrophile attacks the π bond of an alkene or alkyne, leading to addition of the reagent.
e.g. Addition of Br₂ to ethene gives 1,2-dibromoethane.
Decarboxylation
Elimination of carbon dioxide from a carboxylic acid, typically by heating its sodium salt with soda lime.
e.g. Sodium ethanoate on heating with soda lime gives methane.
Wurtz reaction
Alkyl halides on treatment with sodium metal in dry ethereal solution give higher alkanes containing even number of carbon atoms.
e.g. Bromomethane with sodium in dry ether gives ethane.
Dehydrohalogenation
Elimination of a hydrogen halide from an alkyl halide on heating with alcoholic potash to form an alkene.
e.g. Chloroethane on heating with alcoholic KOH gives ethene.
Dehalogenation
Elimination of halogen atoms from vicinal dihalides using zinc metal to form an alkene.
e.g. 1,2-Dibromoethane with zinc gives ethene.
Ozonolysis
Cleavage of an alkene by ozone followed by treatment with Zn-H₂O to give smaller carbonyl compounds, used to locate the position of the double bond.
e.g. Ozonolysis of but-2-ene gives two molecules of ethanal.
Aromatization (reforming)
Conversion of n-alkanes having six or more carbon atoms into benzene and its homologues by heating at 773 K under pressure in presence of catalysts like V₂O₅, Mo₂O₃, or Cr₂O₃ supported over alumina.
e.g. n-Hexane on heating with Cr₂O₃/Al₂O₃ at 773 K gives benzene.
Pyrolysis (cracking)
Decomposition of higher alkanes into lower alkanes, alkenes, etc., by heating to high temperature.
e.g. Dodecane on heating to 973 K in presence of Pt/Pd/Ni gives heptane and pentene.
Lindlar's catalyst
Partially deactivated palladised charcoal (Pd/C) poisoned with sulphur compounds or quinoline, used for partial hydrogenation of alkynes to cis-alkenes.
e.g. Ethyne on reduction with Lindlar's catalyst gives cis-ethene.
Baeyer's reagent
Cold, dilute, aqueous solution of potassium permanganate (KMnO₄), used as a test for unsaturation as it decolorises on reaction with alkenes or alkynes.
e.g. Ethene decolorises Baeyer's reagent to form ethane-1,2-diol.
Torsional strain
Repulsive interaction between electron clouds of bonds on adjacent carbon atoms, which affects the stability of conformations.
e.g. In ethane, the eclipsed conformation has higher torsional strain than the staggered conformation.
Formulas & Equations (15)
CₙH₂ₙ₊₂
Reads as: C n H 2n plus 2
Variables: n = number of carbon atoms
Use when: To find the molecular formula of any alkane (saturated hydrocarbon).
⚠ Students often write CₙH₂ₙ for alkanes, which is the formula for alkenes.
CₙH₂ₙ
Reads as: C n H 2n
Variables: n = number of carbon atoms (n ≥ 2 for stable alkenes)
Use when: To find the molecular formula of any alkene with one double bond.
⚠ Applying this formula to cycloalkanes (which also have CₙH₂ₙ) without specifying unsaturation.
CₙH₂ₙ₋₂
Reads as: C n H 2n minus 2
Variables: n = number of carbon atoms
Use when: To find the molecular formula of any alkyne with one triple bond.
⚠ Confusing this with the formula for dienes (which also have CₙH₂ₙ₋₂).
CH₄ + Cl₂ → CH₃Cl + HCl (hv)
Reads as: Methane plus chlorine gives chloromethane plus hydrogen chloride, in the presence of light
Variables: hv = light or heat energy
Use when: For free radical substitution reactions of alkanes with halogens.
⚠ Forgetting to write 'hv' or 'heat' as the reaction condition.
CH₂=CH₂ + H₂ → CH₃-CH₃ (Pt/Pd/Ni)
Reads as: Ethene plus hydrogen gives ethane, in the presence of platinum, palladium, or nickel catalyst
Variables: Catalyst = Pt, Pd, or Ni
Use when: For hydrogenation of alkenes to alkanes.
⚠ Writing the product as CH₂=CH₂-H₂ instead of CH₃-CH₃.
CH₃CH₂Br + 2Na + BrCH₂CH₃ → CH₃CH₂CH₂CH₃ + 2NaBr (dry ether)
Reads as: Ethyl bromide plus sodium plus ethyl bromide gives n-butane plus sodium bromide, in dry ether
Variables: Dry ether = anhydrous diethyl ether solvent
Use when: For Wurtz reaction to prepare higher alkanes with even number of carbon atoms.
⚠ Using wet ether or forgetting 'dry' condition, leading to side reactions.
CH₃COONa + NaOH → CH₄ + Na₂CO₃ (CaO, heat)
Reads as: Sodium acetate plus sodium hydroxide gives methane plus sodium carbonate, in presence of calcium oxide and heat
Variables: CaO = soda lime component; heat = high temperature
Use when: For decarboxylation of sodium salts of carboxylic acids to alkanes.
⚠ Writing the product as CH₃COONa + NaOH → CH₃OH + Na₂CO₃ (wrong).
CH₃-CH=CH₂ + HBr → CH₃-CHBr-CH₃ (without peroxide)
Reads as: Propene plus hydrogen bromide gives 2-bromopropane, in the absence of peroxide
Variables: HBr = hydrogen bromide
Use when: For Markovnikov addition of HBr to unsymmetrical alkenes.
⚠ Writing the product as 1-bromopropane when no peroxide is mentioned.
CH₃-CH=CH₂ + HBr → CH₃-CH₂-CH₂Br (with peroxide)
Reads as: Propene plus hydrogen bromide gives 1-bromopropane, in the presence of peroxide
Variables: (C₆H₅CO)₂O₂ = benzoyl peroxide (initiator)
Use when: For anti-Markovnikov addition (peroxide effect) of HBr to unsymmetrical alkenes.
⚠ Applying peroxide effect to HCl or HI addition (it works only with HBr).
CH≡CH + H₂ → CH₂=CH₂ (Lindlar's catalyst)
Reads as: Ethyne plus hydrogen gives ethene, using Lindlar's catalyst
Variables: Lindlar's catalyst = Pd/C poisoned with sulphur or quinoline
Use when: For partial hydrogenation of alkynes to cis-alkenes.
⚠ Writing the product as ethane (over-reduction) instead of ethene.
CH≡CH + 2H₂ → CH₃-CH₃ (Ni, heat)
Reads as: Ethyne plus two molecules of hydrogen gives ethane, in presence of nickel and heat
Variables: Ni = nickel catalyst; heat = high temperature
Use when: For complete hydrogenation of alkynes to alkanes.
⚠ Forgetting to balance the equation with 2H₂.
CₙH₂ₙ₊₂ + (3n+1)/2 O₂ → nCO₂ + (n+1)H₂O + heat
Reads as: Alkane plus oxygen gives carbon dioxide plus water plus heat
Variables: n = number of carbon atoms in the alkane
Use when: For writing the general combustion equation of alkanes.
⚠ Writing (n+1)CO₂ instead of nCO₂.
CH₄ + O₂ → C + 2H₂O (incomplete combustion)
Reads as: Methane plus oxygen gives carbon black plus water
Variables: Insufficient oxygen supply
Use when: For incomplete combustion of alkanes to produce carbon black.
⚠ Writing CO instead of C for carbon black formation.
CH₃-CH=CH-CH₃ + O₃ → Ozonide → 2CH₃CHO (Zn-H₂O)
Reads as: But-2-ene plus ozone gives ozonide, then on treatment with zinc and water gives two molecules of ethanal
Variables: O₃ = ozone; Zn-H₂O = reductive workup
Use when: For ozonolysis to determine the position of the double bond.
⚠ Writing the product as CH₃COOH instead of CH₃CHO (oxidation occurs only with oxidative workup).
n(CH₂=CH₂) → -(CH₂-CH₂)-ₙ (high temp/pressure, catalyst)
Reads as: n molecules of ethene give polythene, under high temperature and pressure with a catalyst
Variables: n = large number (degree of polymerization); catalyst = typically peroxide or Ziegler-Natta
Use when: For polymerization of ethene to polythene.
⚠ Writing the product as CH₂=CH₂-CH₂=CH₂ instead of showing the repeating unit.
How It Works
Free radical halogenation of alkanes (e.g., chlorination of methane)
  1. Step 1 — Initiation: Homolysis of Cl₂ molecule in presence of light (hv) gives two chlorine free radicals (Cl•).
  2. Step 2 — Propagation: Chlorine free radical attacks methane, abstracting a hydrogen atom to form HCl and methyl free radical (•CH₃).
  3. Step 3 — Propagation: Methyl free radical attacks another Cl₂ molecule, forming CH₃Cl and regenerating a chlorine free radical.
  4. Step 4 — Termination: Free radicals combine to form stable molecules (e.g., Cl• + Cl• → Cl₂, •CH₃ + •CH₃ → C₂H₆, •CH₃ + Cl• → CH₃Cl).
💡 The termination step can produce ethane as a byproduct, which students often forget to mention.
Electrophilic addition of HBr to unsymmetrical alkenes (Markovnikov rule)
  1. Step 1 — The π bond of the alkene acts as a nucleophile and attacks the electrophile (H⁺ from HBr), forming a carbocation.
  2. Step 2 — The more stable carbocation (secondary > primary, tertiary > secondary) is formed preferentially.
  3. Step 3 — The nucleophile (Br⁻) attacks the carbocation to form the final product, with Br attaching to the more substituted carbon.
💡 The stability of carbocations determines the product: tertiary > secondary > primary > methyl.
Peroxide effect (anti-Markovnikov addition of HBr)
  1. Step 1 — Peroxide (e.g., benzoyl peroxide) undergoes homolysis to form free radicals (C₆H₅•).
  2. Step 2 — The free radical abstracts a hydrogen atom from HBr, generating a bromine free radical (Br•).
  3. Step 3 — Br• adds to the double bond of the alkene, forming a more stable free radical (secondary > primary).
  4. Step 4 — This free radical abstracts a hydrogen from another HBr molecule, giving the anti-Markovnikov product and regenerating Br•.
💡 This mechanism works only with HBr, not with HCl or HI, due to bond strength differences.
Ozonolysis of alkenes
  1. Step 1 — Ozone (O₃) adds to the double bond to form a cyclic ozonide (1,2,3-trioxolane).
  2. Step 2 — The ozonide is cleaved by reductive workup (Zn-H₂O) to give carbonyl compounds (aldehydes or ketones).
💡 Reductive workup (Zn-H₂O) gives aldehydes/ketones; oxidative workup (H₂O₂) gives carboxylic acids/ketones.
Dehydrohalogenation of alkyl halides to alkenes
  1. Step 1 — Alkyl halide is heated with alcoholic KOH (potassium hydroxide in ethanol).
  2. Step 2 — The base (OH⁻) abstracts a β-hydrogen atom (from the carbon adjacent to the halogen-bearing carbon).
  3. Step 3 — A double bond forms between the α and β carbons, and the halide ion leaves as HX.
💡 The reaction follows Saytzeff's rule: the more substituted alkene (more alkyl groups on double bond) is the major product.
Acidic dehydration of alcohols to alkenes
  1. Step 1 — Alcohol is heated with concentrated H₂SO₄ (or other dehydrating agents like H₃PO₄).
  2. Step 2 — The -OH group is protonated, making it a better leaving group (H₂O).
  3. Step 3 — A carbocation is formed after loss of water.
  4. Step 4 — A β-hydrogen is eliminated to form the alkene.
💡 The reaction follows Saytzeff's rule, and carbocation rearrangements are possible if a more stable carbocation can form.
Partial reduction of alkynes to alkenes (Lindlar's catalyst vs Na/NH₃)
  1. Step 1 — For cis-alkene: Alkyne is hydrogenated using Lindlar's catalyst (Pd/C poisoned with quinoline or sulphur).
  2. Step 2 — For trans-alkene: Alkyne is reduced using sodium in liquid ammonia (Na/NH₃).
💡 Lindlar's catalyst gives cis-alkene; Na/NH₃ gives trans-alkene. Students often mix up the stereochemistry.
Wurtz reaction
  1. Step 1 — Two molecules of alkyl halide (R-X) are treated with sodium metal in dry ether.
  2. Step 2 — Sodium metal donates electrons, forming alkyl radicals and NaX.
  3. Step 3 — Two alkyl radicals couple to form a higher alkane (R-R).
💡 Only works for symmetrical alkanes (even number of carbon atoms). With two different alkyl halides, a mixture of three products is obtained.
Kolbe's electrolytic method
  1. Step 1 — An aqueous solution of sodium or potassium salt of a carboxylic acid is electrolyzed.
  2. Step 2 — At the anode: Carboxylate ions lose electrons to form carboxyl radicals, which decarboxylate to give alkyl radicals.
  3. Step 3 — Two alkyl radicals couple to form an alkane with even number of carbon atoms.
  4. Step 4 — At the cathode: Water is reduced to H₂ gas and OH⁻ ions.
💡 Methane cannot be prepared by this method because methyl radicals would need to couple, but the reaction is inefficient for small radicals.
Exam Traps — Don't Make These Mistakes
⚠ Confusing Markovnikov and anti-Markovnikov products
✗ In the addition of HBr to propene, students write 1-bromopropane as the major product without specifying the condition.
→ Without peroxide, the major product is 2-bromopropane (Markovnikov). Only in the presence of peroxide does 1-bromopropane form (anti-Markovnikov).
✓ Specify the condition: 'In the absence of peroxide, addition of HBr to propene gives 2-bromopropane as the major product (Markovnikov rule). In the presence of peroxide, 1-bromopropane is the major product (peroxide effect).'
⚠ Forgetting the catalyst in hydrogenation
✗ CH₂=CH₂ + H₂ → CH₃-CH₃ (without mentioning catalyst)
→ Hydrogenation of alkenes requires a catalyst (Pt, Pd, or Ni) at room temperature or higher. Without catalyst, the reaction does not occur under normal conditions.
✓ CH₂=CH₂ + H₂ → CH₃-CH₃ (in the presence of Pt/Pd/Ni catalyst)
⚠ Writing incorrect products for ozonolysis
✗ Ozonolysis of but-2-ene gives CH₃COOH (ethanoic acid) directly.
→ Ozonolysis with reductive workup (Zn-H₂O) gives aldehydes or ketones, not carboxylic acids. Ethanoic acid would form only with oxidative workup (H₂O₂).
✓ Ozonolysis of but-2-ene with reductive workup (Zn-H₂O) gives two molecules of ethanal (CH₃CHO).
⚠ Misapplying the peroxide effect to HCl and HI
✗ Addition of HCl to propene in presence of peroxide gives 1-chloropropane.
→ The peroxide effect (anti-Markovnikov addition) works only with HBr, not with HCl or HI. HCl has a stronger bond that is not cleaved by free radicals, and HI radicals recombine to I₂ instead of adding.
✓ The peroxide effect is observed only with HBr. Addition of HCl or HI to alkenes follows Markovnikov rule regardless of the presence of peroxide.
⚠ Confusing cis-trans isomerism conditions
✗ All alkenes show geometrical isomerism.
→ Geometrical isomerism requires that each carbon of the double bond has two different substituents. If one carbon has two identical groups (e.g., CH₂=CBr₂), cis-trans isomers are not possible.
✓ Geometrical isomerism is shown by alkenes of the type abC=Cab, abC=Cac, or abC=Ccd, but not when one carbon has two identical groups (e.g., aaC=Cab).
⚠ Writing the wrong product for Wurtz reaction with two different alkyl halides
✗ CH₃Br + C₂H₅Br + 2Na → CH₃-C₂H₅ (propane) as the only product.
→ When two different alkyl halides are used, three products are possible: R-R, R-R', and R'-R'. For example, CH₃Br + C₂H₅Br gives ethane (CH₃-CH₃), propane (CH₃-CH₂-CH₃), and butane (C₂H₅-C₂H₅).
✓ CH₃Br + C₂H₅Br + 2Na → CH₃-CH₃ + CH₃-CH₂-CH₃ + CH₃-CH₂-CH₂-CH₃ (mixture of three alkanes)
Compare & Contrast
General formula
Alkanes
Alkenes
Alkanes: CₙH₂ₙ₊₂; Alkenes: CₙH₂ₙ
Type of bond
Alkanes
Alkenes
Alkanes contain only C-C single bonds (σ bonds); Alkenes contain at least one C=C double bond (one σ and one π bond).
Reactivity
Alkanes
Alkenes
Alkanes are relatively inert (paraffins) and undergo substitution reactions (e.g., halogenation); Alkenes are reactive due to π bonds and undergo electrophilic addition reactions.
Product formed in addition of HBr to unsymmetrical alkene
Markovnikov addition
Anti-Markovnikov addition (peroxide effect)
Markovnikov: Br attaches to the carbon with fewer hydrogen atoms (more substituted carbon); Anti-Markovnikov: Br attaches to the carbon with more hydrogen atoms (less substituted carbon).
Dipole moment
cis-But-2-ene
trans-But-2-ene
cis-But-2-ene has a dipole moment of 0.33 D (polar); trans-But-2-ene has a dipole moment of almost zero (non-polar) because the two C-CH₃ bond dipoles cancel each other.
Stereochemistry of alkene product
Lindlar's catalyst reduction of alkyne
Na/NH₃ reduction of alkyne
Lindlar's catalyst gives cis-alkene (syn addition); Na/NH₃ gives trans-alkene (anti addition).
Exam Focus
Marks pattern: 2 MCQs (1 mark each), 1 SA-I (2 marks), 1 SA-II (3 marks), 1 LA (5 marks) — total approximately 11-12 marks in board exam
Most asked: Nomenclature and isomerism of alkanes, alkenes, and alkynes · Markovnikov rule and peroxide effect (with mechanism) · Electrophilic addition reactions of alkenes · Ozonolysis to determine position of double bond · Wurtz reaction and its limitations · Free radical halogenation of alkanes (mechanism) · Geometrical isomerism (cis-trans) conditions · Preparation of alkanes, alkenes, and alkynes · Combustion and controlled oxidation of alkanes · Conformations of ethane (staggered vs eclipsed)
Don't skip: Write the mechanism of free radical chlorination of methane (5 marks); Explain Markovnikov rule with mechanism and give an example of peroxide effect (5 marks); How will you distinguish between alkanes, alkenes, and alkynes using chemical tests? (3 marks); What is ozonolysis? How can it be used to locate the position of a double bond? (3 marks); Write the IUPAC names and structures of all isomers of C₅H₁₀ (alkenes) (3 marks)
One-Line Revision (read aloud before the exam)
  1. Hydrocarbons are compounds of carbon and hydrogen only, classified as saturated (alkanes), unsaturated (alkenes, alkynes), and aromatic.
  2. Alkanes have the general formula CₙH₂ₙ₊₂ and undergo free radical substitution reactions like halogenation.
  3. Alkenes have the general formula CₙH₂ₙ and undergo electrophilic addition reactions due to the presence of a π bond.
  4. Alkynes have the general formula CₙH₂ₙ₋₂ and undergo addition reactions similar to alkenes but can add two molecules of reagent.
  5. Markovnikov rule: The negative part of the addendum attaches to the carbon with fewer hydrogen atoms in an unsymmetrical alkene.
  6. Peroxide effect (Kharash effect) causes anti-Markovnikov addition of HBr to alkenes in the presence of peroxides.
  7. Wurtz reaction: 2R-X + 2Na → R-R + 2NaX (dry ether), used to prepare higher alkanes with even number of carbon atoms.
  8. Lindlar's catalyst (Pd/C poisoned with quinoline) reduces alkynes to cis-alkenes, while Na/NH₃ gives trans-alkenes.
  9. Ozonolysis of alkenes with reductive workup (Zn-H₂O) gives aldehydes or ketones, used to locate the double bond position.
  10. Baeyer's reagent (cold dilute KMnO₄) decolorises when added to alkenes or alkynes, confirming unsaturation.
  11. Geometrical isomerism (cis-trans) occurs when each carbon of the double bond has two different substituents.
  12. Conformations of ethane: staggered (more stable, lower energy) and eclipsed (less stable, higher energy) with a 12.5 kJ mol⁻¹ energy barrier.
  13. Decarboxylation: RCOONa + NaOH → RH + Na₂CO₃ (CaO, heat), reduces the carbon chain by one carbon atom.
  14. Kolbe's electrolytic method: Electrolysis of sodium carboxylate gives alkanes with even number of carbon atoms at the anode.
  15. Combustion of alkanes: CₙH₂ₙ₊₂ + (3n+1)/2 O₂ → nCO₂ + (n+1)H₂O + heat, making them excellent fuels.
  16. Pyrolysis (cracking) breaks higher alkanes into lower alkanes and alkenes at high temperature.
  17. Aromatization (reforming) converts n-alkanes with 6 or more carbons into benzene and its homologues at 773 K with catalysts.
  18. Dehydrohalogenation: R-X + alcoholic KOH → alkene + KX + H₂O (elimination reaction).
  19. Acidic dehydration: R-OH + conc. H₂SO₄ → alkene + H₂O (follows Saytzeff's rule).
  20. Free radical halogenation of alkanes proceeds via initiation, propagation, and termination steps.

MCQs — 10 Questions with Answers

Q1. Which of the following is the correct IUPAC name for the compound with structure CH₃–CH(CH₃)–CH₂–CH₃?

  • A. 2-Methylbutane ✓
  • B. 3-Methylbutane
  • C. n-Pentane
  • D. 2-Ethylpropane

Answer: A — The longest chain contains 4 carbons (butane); the methyl branch is at position 2 from the nearest end, giving 2-methylbutane.

Q2. The general molecular formula for alkanes is CₙH₂ₙ₊₂. What is the formula for an alkane with 8 carbon atoms?

  • A. C₈H₁₆
  • B. C₈H₁₈ ✓
  • C. C₈H₂₀
  • D. C₈H₁₄

Answer: B — Using CₙH₂ₙ₊₂ with n = 8: C₈H₂(₈)₊₂ = C₈H₁₈.

Q3. Which statement correctly describes the difference between chain isomers and positional isomers?

  • A. Chain isomers differ in the position of functional groups; positional isomers differ in carbon skeleton.
  • B. Chain isomers differ in carbon skeleton arrangement; positional isomers differ in position of the same functional group. ✓
  • C. Both types have identical molecular formulas but differ only in 3D orientation.
  • D. Chain isomers belong to different homologous series.

Answer: B — Chain isomers result from different ways to arrange carbon atoms (e.g., n-butane vs isobutane); positional isomers have the same skeleton but different functional group locations.

Q4. A carbon atom bonded to exactly two other carbon atoms in an alkane is classified as:

  • A. Primary carbon
  • B. Secondary carbon ✓
  • C. Tertiary carbon
  • D. Quaternary carbon

Answer: B — A secondary (2°) carbon is attached to exactly two carbon atoms; it is not a terminal carbon.

Q5. Consider the structural isomers of C₆H₁₄. How many chain isomers are possible?

  • A. 3
  • B. 4
  • C. 5 ✓
  • D. 6

Answer: C — The five chain isomers of C₆H₁₄ are: n-hexane, 2-methylpentane, 3-methylpentane, 2,2-dimethylbutane, and 2,3-dimethylbutane.

Q6. Assertion: Alkanes are unreactive under normal conditions. Reason: Alkanes contain only single σ bonds, which are very strong and require high energy to break.

  • A. Both assertion and reason are correct; reason explains assertion. ✓
  • B. Both assertion and reason are correct; reason does NOT explain assertion.
  • C. Assertion is correct; reason is incorrect.
  • D. Assertion is incorrect; reason is correct.

Answer: A — Alkanes are inert (paraffinic) because their C–C and C–H σ bonds are very strong, requiring high activation energy for reaction.

Q7. Which of the following is NOT a correct characteristic of alkanes?

  • A. They react readily with dilute acids under normal conditions. ✓
  • B. They are called paraffins due to low reactivity.
  • C. Their boiling point increases with increase in carbon chain length.
  • D. Branched isomers have lower boiling points than straight-chain isomers.

Answer: A — Alkanes do NOT react with dilute acids under normal conditions; this is why they are classified as inert or paraffinic.

Q8. The bond angle in methane (CH₄) is approximately 109.5°. This is best explained by:

  • A. sp hybridisation of carbon
  • B. sp² hybridisation of carbon
  • C. sp³ hybridisation of carbon with tetrahedral geometry ✓
  • D. p orbital overlap without hybridisation

Answer: C — The 109.5° angle in methane arises from sp³ hybridisation of carbon, resulting in a tetrahedral geometry as predicted by VSEPR theory.

Q9. Among the following, identify the pair that represents chain isomers: (i) CH₃CH₂CH₂CH₃ and CH₃CH(CH₃)CH₃ (ii) n-pentane and 2-pentene. Which is correct?

  • A. Only (i) represents chain isomers. ✓
  • B. Only (ii) represents chain isomers.
  • C. Both (i) and (ii) represent chain isomers.
  • D. Neither (i) nor (ii) represents chain isomers.

Answer: A — Pair (i) both have formula C₄H₁₀ with different carbon skeletons (chain isomers). Pair (ii) are n-pentane (C₅H₁₂) and 2-pentene (C₅H₁₀), different formulas—not isomers.

Q10. Given that the C–H bond length in alkanes is 112 pm and the C–C bond length is 154 pm, which inference can be drawn about the relative strength of these bonds?

  • A. C–H bonds are stronger because they are shorter.
  • B. C–C bonds are stronger because they are longer.
  • C. The shorter C–H bond suggests higher bond dissociation energy compared to the longer C–C bond. ✓
  • D. Bond lengths have no relation to bond strength; only bond order matters.

Answer: C — Shorter bond length generally correlates with higher bond dissociation energy; thus, the 112 pm C–H bond is stronger than the 154 pm C–C bond.

Flashcards

What is the general molecular formula for alkanes?

CₙH₂ₙ₊₂, where n is the number of carbon atoms.

Define chain isomers with one example.

Structural isomers that differ in the arrangement of the carbon chain skeleton; example: n-butane and 2-methylpropane both have formula C₄H₁₀.

What is the bond angle in a tetrahedral alkane molecule like methane?

109.5° (degrees), determined by VSEPR theory for sp³ hybridised carbon.

Why are alkanes called 'paraffins'?

From Latin 'parum affinis' (little affinity) because alkanes are unreactive with acids, bases, and most reagents under normal conditions.

Distinguish between primary and tertiary carbon atoms.

Primary carbon is bonded to one other carbon atom (or none, as in methane); tertiary carbon is bonded to three other carbon atoms.

What does LPG stand for and what is its source?

Liquified Petroleum Gas; it is obtained from petroleum and natural gas deposits and used as a domestic fuel with minimal pollution.

Name the three main classes of hydrocarbons based on bonding.

Saturated (single bonds only), unsaturated (double or triple bonds), and aromatic (cyclic with delocalised electrons).

What are the bond lengths of C–C and C–H in alkanes?

C–C bond: 154 pm; C–H bond: 112 pm.

How many structural isomers does C₅H₁₂ have?

Three chain isomers: n-pentane, 2-methylbutane, and 2,2-dimethylpropane.

What orbitals overlap to form C–C and C–H bonds in alkanes?

sp³ hybrid orbitals of carbon atoms overlap with each other (for C–C) or with 1s orbitals of hydrogen (for C–H) via head-on σ bonding.

Important Board Questions

Define chain isomers and write the structures and IUPAC names of all chain isomers of C₅H₁₂. [2 marks]

Chain isomers differ in the arrangement of the carbon skeleton. Draw three possible arrangements of 5 carbons and identify longest chains; assign position numbers to any methyl branches.

Explain why alkanes are called 'paraffins' and justify their unreactivity under normal conditions using concepts of bonding. What type of bond is present between carbon atoms in alkanes, and why does this contribute to their low reactivity? [5 marks]

Paraffins = low affinity (from Latin). Discuss σ bonds formed by sp³ hybridisation, high bond dissociation energy, and lack of polar character. Show that strong σ bonds require high activation energy to break.

For the molecular formula C₆H₁₄, identify and draw all possible chain isomers. For each isomer, classify each carbon atom as primary (1°), secondary (2°), tertiary (3°), or quaternary (4°). Explain why branched isomers have lower boiling points than the straight-chain isomer despite having the same molecular formula. [6 marks]

Five chain isomers exist for C₆H₁₄. For each structure, count carbons bonded to each carbon atom to classify. Lower boiling points in branched isomers result from decreased surface area and weaker intermolecular van der Waals forces compared to extended linear chains.

Assertion-Reason Questions for Hydrocarbons

Q1. Assertion (A):

Alkanes undergo combustion reactions readily despite being classified as inert hydrocarbons.

Reason (R):

The C-C and C-H bonds in alkanes are very strong single bonds that require high activation energy to break.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — Both statements are true, but the reason does not correctly explain the assertion. Alkanes are 'inert' to acids, bases, and nucleophiles under normal conditions (due to strong bonds and lack of reactive sites), but they readily undergo high-temperature combustion because combustion is a free radical chain reaction providing sufficient energy to break these bonds. The reason describes why alkanes are unreactive with common reagents, not why they combust.

Q2. Assertion (A):

In the homologous series of straight-chain alkanes, boiling point increases with increasing number of carbon atoms.

Reason (R):

Increasing chain length increases the surface area of contact between molecules, leading to stronger van der Waals intermolecular forces.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true, and the reason correctly explains the assertion. As carbon chain length increases, molecular surface area increases, allowing greater van der Waals interactions between molecules, which increases the energy required for vaporization and thus increases boiling point.

Q3. Assertion (A):

Methane has a tetrahedral structure with H-C-H bond angles of 109.5°.

Reason (R):

Methane has sp hybridization of the central carbon atom.

Answer: C — A is true but R is false — The assertion is true—methane does have a tetrahedral structure with 109.5° bond angles. However, the reason is false. Methane uses sp³ (not sp) hybridization, which produces four equivalent sp³ hybrid orbitals arranged tetrahedrally. sp hybridization produces linear geometry with 180° angles.

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Case-Based Questions (CBQ) for Hydrocarbons

Case Study:

India's increasing reliance on Compressed Natural Gas (CNG) for automobiles has reduced urban air pollution significantly. CNG is primarily methane (CH4) with smaller amounts of ethane (C2H6) and propane (C3H8). These hydrocarbons burn efficiently in specially designed engines. The combustion of one mole of methane releases approximately 890 kJ of energy. Petrol, derived from fractional distillation of crude oil, contains a complex mixture of hydrocarbons ranging from C6H14 to C12H26. While petrol provides more energy per unit volume than CNG, CNG combustion produces fewer harmful pollutants such as nitrogen oxides and particulate matter.

Q1. Which hydrocarbon in CNG has the highest boiling point at standard conditions?

  • A. Methane
  • B. Ethane
  • C. Propane ✓
  • D. Butane

Answer: C

Q2. Why does petrol provide more energy per unit volume than CNG despite both being hydrocarbon mixtures?

  • A. Petrol contains only single bonds while CNG contains double bonds
  • B. Petrol is liquid with higher density and contains larger hydrocarbon molecules with more C-H bonds per unit volume ✓
  • C. CNG is a pure substance while petrol is a mixture
  • D. Petrol undergoes complete combustion while CNG does not

Answer: B

Q3. If methane combustion releases 890 kJ per mole, what would be the expected approximate energy released by ethane (C2H6) combustion per mole?

  • A. 445 kJ
  • B. 890 kJ
  • C. 1700-1800 kJ ✓
  • D. Cannot be determined without additional data

Answer: C

Q4. The fraction distillation of crude oil produces petrol hydrocarbons from C6H14 to C12H26. Which general formula best represents this range?

  • A. CnH2n
  • B. CnH2n+2 ✓
  • C. CnH2n-2
  • D. CnHn+1

Answer: B

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Class 11 Chemistry Formula Sheet

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