This MCQ module is based on: Classification Alkanes
TOPIC 9 OF 13
Classification Alkanes
🎓 Class 11
Chemistry
CBSE
Theory
Ch 9 – Hydrocarbons
⏱ ~14 min
🌐 Language: [gtranslate]
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Classification of Hydrocarbons and Alkanes
9.1 Hydrocarbons — An Introduction
The term 'hydrocarbon' is self-explanatory: it means a compound made up of only carbon and hydrogen. Hydrocarbons are the simplest class of organic compounds and constitute the parent skeleton of every functional class. Petrol, diesel, kerosene, LPG, CNG, polythene, polypropylene, asphalt — all are derived from naturally occurring hydrocarbons (mainly petroleum and natural gas).
The hydrocarbons used in our day-to-day life come from two main natural sources: petroleum (a complex liquid mixture trapped in sedimentary rocks) and natural gas (mostly methane). On heating crude petroleum in oil refineries, fractional distillation separates it into useful fractions — gasoline, kerosene, diesel, lubricating oil and bitumen.
9.2 Classification of Hydrocarbons
Hydrocarbons are of different types. Depending on the types of carbon-carbon bonds present, they are classified into three principal categories:
- Saturated hydrocarbons — only single C–C bonds (alkanes & cycloalkanes)
- Unsaturated hydrocarbons — at least one C=C double or C≡C triple bond (alkenes & alkynes)
- Aromatic hydrocarbons — special class containing benzene-like rings with delocalised π electrons
9.3 Alkanes
Alkanes are saturated open-chain hydrocarbons containing only carbon-carbon single bonds. Methane (CH₄) is the first member of the family; replacing one H of methane by –CH₃ gives ethane (C₂H₆), then propane, butane, and so on. The general molecular formula is CₙH₂ₙ₊₂ where n = 1, 2, 3, … Each carbon atom is sp³ hybridised with tetrahedral geometry (bond angle 109.5°).
9.3.1 Structural Isomerism in Alkanes
Alkanes with four or more carbons exhibit chain isomerism. Both n-butane (straight chain) and isobutane (branched) share C₄H₁₀ but differ in connectivity. C₅H₁₂ has 3 isomers (n-pentane, isopentane, neopentane); C₆H₁₄ has 5 isomers; C₇H₁₆ has 9.
9.3.2 IUPAC Nomenclature of Alkanes
The systematic naming of alkanes follows four IUPAC rules:
- Longest chain: Identify the longest continuous chain of carbons → this gives the parent name (meth-, eth-, prop-, but-, pent-, hex-, hept-, oct-, non-, dec-).
- Lowest locants: Number the chain from the end that gives the substituent(s) the lowest set of locants.
- Substituents alphabetically: Name substituents (ethyl before methyl) and prefix them to the parent name.
- Multiplying prefixes: Use di, tri, tetra for two, three, four identical substituents; these are NOT considered for alphabetisation.
| Structure | IUPAC name | Common name |
|---|---|---|
| CH₃–CH₂–CH₂–CH₂–CH₃ | Pentane | n-Pentane |
| (CH₃)₂CH–CH₂–CH₃ | 2-Methylbutane | Isopentane |
| (CH₃)₄C | 2,2-Dimethylpropane | Neopentane |
| CH₃–CH(C₂H₅)–CH₂–CH₂–CH₃ | 3-Methylhexane | — |
| CH₃–C(CH₃)₂–CH(CH₃)–CH₂–CH₃ | 2,2,3-Trimethylpentane | — |
Tip: When two locant sets are equally low at the first point of difference, choose the set with the alphabetically earlier substituent at the lower number. Always write multiple locants separated by commas (e.g. 2,2,3-) and locants–words separated by hyphens (e.g. 2-methyl-).
IUPAC Name Builder
Pick the parent chain length and one substituent position, then read off the IUPAC name in real time.
Computed IUPAC name:
2-methylpentane
Lowest locants applied: numbered from end nearer the substituent.
9.3.3 Preparation of Alkanes
(a) From unsaturated hydrocarbons — Sabatier–Senderens reaction
Catalytic hydrogenation of alkenes/alkynes in the presence of finely divided Ni (or Pt, Pd) at 523–573 K converts them into alkanes:
CH₂=CH₂ + H₂ →Ni/Δ CH₃–CH₃HC≡CH + 2 H₂ →Ni/Δ CH₃–CH₃
(b) From alkyl halides — Wurtz reaction
Alkyl halides (except fluorides) react with sodium metal in dry ether to form symmetrical alkanes containing twice the number of carbons of the alkyl halide:
2 CH₃Br + 2 Na →dry ether CH₃–CH₃ + 2 NaBrLimitation: Wurtz reaction with two different alkyl halides gives a mixture of three alkanes and is therefore not preparatively useful for unsymmetrical alkanes. It also fails to give alkanes containing an odd number of carbon atoms when only one halide is used.
(c) From carboxylic acids — Decarboxylation
Sodium salts of carboxylic acids on heating with soda lime (NaOH + CaO, 3:1) lose CO₂ and give alkane with one carbon less:
CH₃COONa + NaOH →CaO/Δ CH₄ + Na₂CO₃(d) Kolbe's electrolytic method
Electrolysis of an aqueous solution of sodium or potassium salt of carboxylic acid produces alkane at the anode:
2 CH₃COONa + 2 H₂O →electrolysis CH₃–CH₃ + 2 CO₂ + H₂ + 2 NaOH9.3.4 Physical Properties
The first four members (C₁–C₄) are gases, C₅–C₁₇ are liquids and higher members (C₁₈ onwards) are waxy solids at room temperature. They are colourless, odourless and almost insoluble in water (non-polar) but soluble in benzene, ether, CCl₄. Boiling points rise steadily with chain length (about 20–30 K per CH₂), and for chain isomers the b.p. decreases with branching because branched molecules pack less efficiently and have less surface for van-der-Waals contact.
9.3.5 Chemical Reactions
Alkanes are generally inert towards acids, bases and oxidising agents at room temperature, but they undergo:
(i) Substitution — Halogenation
In presence of UV light or at 520–670 K, alkanes react with chlorine or bromine to give haloalkanes (the order of reactivity is F > Cl > Br > I; iodination is reversible and very slow).
CH₄ + Cl₂ →hν CH₃Cl + HCl → CH₂Cl₂ → CHCl₃ → CCl₄Free-radical chain mechanism for chlorination of methane:
Step 1 — Initiation: UV light cleaves Cl–Cl homolytically:
Cl–Cl →hν Cl• + Cl•Step 2 — Propagation (chain-carrying steps):
Cl• + CH₄ → •CH₃ + HCl•CH₃ + Cl₂ → CH₃Cl + Cl•
The Cl• regenerated continues the chain, so a single photon can chlorinate thousands of methane molecules.
Step 3 — Termination (radical–radical combination):
Cl• + Cl• → Cl₂•CH₃ + Cl• → CH₃Cl
•CH₃ + •CH₃ → CH₃–CH₃
(ii) Combustion
Alkanes burn completely in excess of air/oxygen with a non-luminous blue flame to give CO₂ and H₂O along with a large amount of heat:
CH₄ + 2 O₂ → CO₂ + 2 H₂O ΔH = –890 kJ/molC₄H₁₀ + 13/2 O₂ → 4 CO₂ + 5 H₂O ΔH = –2875.84 kJ/mol
This is the basis of LPG, CNG and petrol as fuels. With limited supply of air, incomplete combustion gives carbon monoxide (CO) and carbon (soot) — extremely dangerous in poorly ventilated heaters.
(iii) Controlled oxidation
2 CH₄ + O₂ →Cu/523 K/100 atm 2 CH₃OHCH₄ + O₂ →Mo₂O₃ HCHO + H₂O
(iv) Isomerisation, Aromatisation and Pyrolysis
n-Hexane on heating with anhydrous AlCl₃/HCl rearranges to its branched isomers (isomerisation). On passing over Pt/V₂O₅ supported on Al₂O₃ at 773 K, n-hexane cyclises and dehydrogenates to give benzene (aromatisation or reforming). Higher alkanes on strong heating in the absence of air decompose into smaller alkanes, alkenes and H₂ (pyrolysis or cracking) — the basis of refinery cracking.
CH₃–(CH₂)₄–CH₃ →Pt/Al₂O₃, 773 K C₆H₆ + 4 H₂9.3.6 Conformations
Rotation about a C–C single bond is possible because the σ bond is cylindrically symmetric. Different spatial arrangements arising from such rotation are called conformations (or conformers / rotamers). For ethane, two extreme conformations are noteworthy:
- Staggered — the H atoms on adjacent carbons are as far apart as possible (dihedral angle 60°). Lowest energy, most stable.
- Eclipsed — the H atoms on adjacent carbons directly overlap (dihedral angle 0°). Higher energy by ~12.5 kJ/mol due to torsional strain.
Conformations are best represented by Newman projections — a view down the C–C axis. The front carbon is shown as a dot at the centre with three bonds at 120°; the back carbon is the surrounding circle with three bonds.
Conformation Energy Explorer
Drag the slider to rotate the back carbon of ethane and watch the potential energy curve change.
60°
Staggered — Energy: 0 kJ/mol
Activity 9.1 — Identify the most stable conformer
Setup: A teacher models n-butane and twists the central C2–C3 bond. Four conformations appear at dihedral angles 0°, 60°, 120°, 180°.
Predict: Order the four conformations by stability. Which is the lowest energy and why? Which is the highest and why?
For n-butane, four key conformations arise on rotation about the C2–C3 bond:
- Anti (180°) — two methyl groups on opposite sides; most stable.
- Gauche (60°/300°) — methyl groups 60° apart, slight steric strain (3.8 kJ/mol higher than anti).
- Eclipsed (120°/240°) — methyl eclipses H, ~16 kJ/mol higher.
- Fully eclipsed (0°) — methyl eclipses methyl; least stable (~19 kJ/mol higher than anti).
Reason: Strain order = torsional + steric. Anti minimises both; fully eclipsed maximises both.
Worked Example 1: IUPAC Naming
Write the IUPAC name of the alkane: (CH₃)₃C–CH₂–CH(CH₃)–CH₂–CH₃
Step 1: Find the longest chain. Starting from the terminal CH₃ on the right and going through the CH(CH₃) and CH₂ gives a 6-carbon chain — but expanding from the (CH₃)₃C group: C–C(CH₃)₂–CH₂–CH(CH₃)–CH₂–CH₃ is also 6 carbons. The longest = 6 C → hexane.
Step 2: Number to give lowest locants. Numbering from the (CH₃)₃C end: C1 = CH₃ (one of the three), C2 = C(CH₃)₂ (with two methyl substituents at C2), C3 = CH₂, C4 = CH(CH₃), C5 = CH₂, C6 = CH₃. Substituent locants: 2,2,4. From the other end they would be 3,5,5,5 — much higher.
Step 3: Substituents: three methyl groups at 2, 2, 4.
Name: 2,2,4-Trimethylhexane.
Step 2: Number to give lowest locants. Numbering from the (CH₃)₃C end: C1 = CH₃ (one of the three), C2 = C(CH₃)₂ (with two methyl substituents at C2), C3 = CH₂, C4 = CH(CH₃), C5 = CH₂, C6 = CH₃. Substituent locants: 2,2,4. From the other end they would be 3,5,5,5 — much higher.
Step 3: Substituents: three methyl groups at 2, 2, 4.
Name: 2,2,4-Trimethylhexane.
Worked Example 2: Halogenation Mechanism
Write the three steps of the free-radical mechanism for the monochlorination of ethane (C₂H₆ + Cl₂ → C₂H₅Cl + HCl).
Initiation (homolytic cleavage of Cl₂ by hν or heat):
Cl–Cl →hν Cl• + Cl•
Propagation (the chain-carrying steps):
Cl• + CH₃–CH₃ → •CH₂–CH₃ + HCl
•CH₂–CH₃ + Cl–Cl → CH₃–CH₂Cl + Cl•
The Cl• regenerated re-enters the cycle so one initiation event yields many product molecules.
Termination (radicals combine, ending the chain):
Cl• + Cl• → Cl₂
•CH₂–CH₃ + Cl• → CH₃CH₂Cl
•CH₂–CH₃ + •CH₂–CH₃ → CH₃–CH₂–CH₂–CH₃ (n-butane!)
Cl–Cl →hν Cl• + Cl•
Propagation (the chain-carrying steps):
Cl• + CH₃–CH₃ → •CH₂–CH₃ + HCl
•CH₂–CH₃ + Cl–Cl → CH₃–CH₂Cl + Cl•
The Cl• regenerated re-enters the cycle so one initiation event yields many product molecules.
Termination (radicals combine, ending the chain):
Cl• + Cl• → Cl₂
•CH₂–CH₃ + Cl• → CH₃CH₂Cl
•CH₂–CH₃ + •CH₂–CH₃ → CH₃–CH₂–CH₂–CH₃ (n-butane!)
Worked Example 3: Wurtz Reaction
Predict the product(s) of the Wurtz reaction of (a) bromoethane alone, (b) bromomethane + bromoethane.
(a) Two molecules of bromoethane couple:
2 CH₃CH₂Br + 2 Na → CH₃CH₂–CH₂CH₃ + 2 NaBr → n-butane.
(b) A mixture of bromomethane and bromoethane gives THREE alkanes (statistical mixture):
CH₃Br + CH₃Br → CH₃–CH₃ (ethane)
CH₃Br + C₂H₅Br → CH₃–C₂H₅ (propane)
C₂H₅Br + C₂H₅Br → C₂H₅–C₂H₅ (n-butane)
This is why Wurtz reaction is unsuitable for synthesising unsymmetrical alkanes.
2 CH₃CH₂Br + 2 Na → CH₃CH₂–CH₂CH₃ + 2 NaBr → n-butane.
(b) A mixture of bromomethane and bromoethane gives THREE alkanes (statistical mixture):
CH₃Br + CH₃Br → CH₃–CH₃ (ethane)
CH₃Br + C₂H₅Br → CH₃–C₂H₅ (propane)
C₂H₅Br + C₂H₅Br → C₂H₅–C₂H₅ (n-butane)
This is why Wurtz reaction is unsuitable for synthesising unsymmetrical alkanes.
Competency-Based Questions
Q1. Which of the following is NOT a saturated hydrocarbon? L1 Remember
Answer: (c) Ethene (CH₂=CH₂) contains a C=C double bond, hence is unsaturated. Methane, cyclohexane and n-pentane have only C–C single bonds.
Q2. Explain why neopentane (2,2-dimethylpropane) has a lower boiling point than n-pentane although both have the formula C₅H₁₂. L4 Analyse
Answer: Neopentane has a near-spherical, highly branched structure that minimises the molecular surface area available for van der Waals (London dispersion) interactions. n-Pentane is an extended zig-zag chain with much larger surface contact between molecules → stronger intermolecular attractions → higher boiling point (36 °C vs 9.5 °C).
Q3. Write the IUPAC name of (CH₃)₂CH–CH₂–CH(C₂H₅)–CH₃ L3 Apply
Answer: Longest chain through the C₂H₅: 6 C → hexane. Numbering from the right: C1 = CH₃ (of ethyl became part of chain), C2 = CH (with methyl), C3 = CH₂, C4 = CH (with methyl), C5,C6 = CH₃. Methyls at C2 and C4 → 2,4-dimethylhexane.
Q4. The chlorination of n-butane gives a mixture of 1-chlorobutane and 2-chlorobutane. Predict which is the major product and justify. L5 Evaluate
Answer: 2-chlorobutane is the major product. n-Butane has 6 primary H (–CH₃ end) and 4 secondary H (–CH₂– middle). Although primary H's are more numerous, the relative reactivity of secondary H is ~3.8× that of primary at 25 °C with Cl₂. Statistical product distribution (rate × number of H): 1-chloro = 6 × 1 = 6; 2-chloro = 4 × 3.8 ≈ 15.2. Ratio ≈ 1 : 2.5 in favour of 2-chlorobutane.
Q5. HOT (Create): Design a complete synthetic route from acetic acid to n-butane using only standard alkane preparations. L6 Create
Sample route:
Step 1: Decarboxylation: CH₃COONa + NaOH/CaO → CH₄ (methane).
Step 2: Halogenation: CH₄ + Cl₂/hν → CH₃Cl.
Step 3: Conversion to ethyl halide via doubling first: 2 CH₃Cl + 2 Na (Wurtz) → CH₃–CH₃ (ethane). Halogenate: C₂H₆ + Cl₂/hν → C₂H₅Cl.
Step 4: Wurtz again: 2 C₂H₅Cl + 2 Na → CH₃CH₂CH₂CH₃ (n-butane).
Alternative single step: 2 C₂H₅Br + 2 Na → n-butane (if ethyl halide is already available). The route demonstrates how decarboxylation, halogenation and Wurtz coupling complement one another.
Step 1: Decarboxylation: CH₃COONa + NaOH/CaO → CH₄ (methane).
Step 2: Halogenation: CH₄ + Cl₂/hν → CH₃Cl.
Step 3: Conversion to ethyl halide via doubling first: 2 CH₃Cl + 2 Na (Wurtz) → CH₃–CH₃ (ethane). Halogenate: C₂H₆ + Cl₂/hν → C₂H₅Cl.
Step 4: Wurtz again: 2 C₂H₅Cl + 2 Na → CH₃CH₂CH₂CH₃ (n-butane).
Alternative single step: 2 C₂H₅Br + 2 Na → n-butane (if ethyl halide is already available). The route demonstrates how decarboxylation, halogenation and Wurtz coupling complement one another.
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: Staggered conformation of ethane is more stable than eclipsed.
R: In the staggered form the C–H bond pairs of adjacent carbons are 60° apart, minimising electron-pair repulsion.
Answer: (A). Both true; R correctly explains A. Staggered conformation has 12.5 kJ/mol lower torsional strain than eclipsed.
A: Wurtz reaction is preferred for synthesising symmetrical alkanes only.
R: When two different alkyl halides are used, a statistical mixture of three alkanes is obtained, which is difficult to separate.
Answer: (A). Both true; R explains A.
A: Iodination of methane with I₂/hν is a useful synthesis of CH₃I.
R: Iodination is reversible because HI formed is a strong reducing agent that converts CH₃I back to CH₄.
Answer: (D). Assertion is FALSE — iodination is NOT a useful synthesis. Reason is TRUE — that's exactly why it isn't useful! HI must be removed (e.g. by HIO₃ oxidant) for the reaction to go forward.
Frequently Asked Questions — Classification of Hydrocarbons and Alkanes
How are hydrocarbons classified?
Hydrocarbons are compounds containing only carbon and hydrogen and are classified by their bonding and structure in NCERT Class 11 Chemistry Chapter 9: (1) saturated — only single bonds (alkanes, cycloalkanes); (2) unsaturated — with double bonds (alkenes), triple bonds (alkynes), or both; (3) aromatic — containing benzene rings or similar; (4) acyclic (open-chain) vs cyclic (closed-ring). The general formulas: alkanes CₙH₂ₙ₊₂, alkenes CₙH₂ₙ, alkynes CₙH₂ₙ₋₂, cycloalkanes CₙH₂ₙ. Hydrocarbons are the parent class of all organic compounds and are major fuels (petrol, diesel, natural gas).
What is the IUPAC nomenclature of alkanes?
IUPAC nomenclature of alkanes follows these steps in NCERT Class 11 Chemistry Chapter 9: (1) find the longest continuous carbon chain — this is the parent; (2) name parent: meth (1), eth (2), prop (3), but (4), pent (5), hex (6), hept (7), oct (8), non (9), dec (10) + suffix '-ane'; (3) number from the end nearest the first substituent; (4) name substituents (methyl, ethyl, etc.) with locants and prefix in alphabetical order. Example: CH₃-CH(CH₃)-CH₂-CH₂-CH₃ is 2-methylpentane. Use commas between numbers and hyphens between numbers and words.
What are the preparation methods of alkanes?
NCERT Class 11 Chemistry Chapter 9 describes major preparation methods for alkanes: (1) hydrogenation of alkenes/alkynes using Ni, Pt or Pd catalyst (Sabatier-Senderens reaction) — alkene + H₂ → alkane; (2) reduction of alkyl halides using Zn/HCl, LiAlH₄ or hydrolysis of Grignard reagent — R-X → R-H; (3) Wurtz reaction — 2R-X + 2Na → R-R + 2NaX (gives symmetrical alkane); (4) Kolbe electrolysis — electrolysis of sodium salt of carboxylic acid → R-R + CO₂ + H₂; (5) Frankland reaction with R₂Zn. Each method is examined for yield, side reactions and applicability.
What is the Wurtz reaction?
The Wurtz reaction is a synthesis of symmetrical alkanes by reaction of an alkyl halide with sodium metal in dry ether: 2R-X + 2Na → R-R + 2NaX. NCERT Class 11 Chemistry Chapter 9 examples: 2CH₃Br + 2Na → CH₃-CH₃ + 2NaBr (ethane from methyl bromide). The reaction works best with primary alkyl halides. Limitations: it gives only symmetrical alkanes when one alkyl halide is used; with two different alkyl halides, three products are obtained (R-R, R'-R', R-R') making separation difficult; it does not work for tertiary halides which give alkenes by elimination.
What is the free radical mechanism of halogenation of alkanes?
Halogenation of alkanes occurs by a free-radical chain mechanism in NCERT Class 11 Chemistry Chapter 9 with three steps: (1) initiation — Cl₂ → 2Cl• under UV light or heat (homolytic cleavage); (2) propagation — Cl• + CH₄ → •CH₃ + HCl, then •CH₃ + Cl₂ → CH₃Cl + Cl•; (3) termination — Cl• + Cl• → Cl₂, or •CH₃ + Cl• → CH₃Cl, or •CH₃ + •CH₃ → C₂H₆. The product is a mixture (CH₃Cl, CH₂Cl₂, CHCl₃, CCl₄). Order of reactivity: F₂ > Cl₂ > Br₂ > I₂. Selectivity: tertiary > secondary > primary H atoms.
What are conformations of ethane?
Conformations are the different spatial arrangements of atoms in a molecule arising from rotation around single bonds. NCERT Class 11 Chemistry Chapter 9 covers ethane (CH₃-CH₃) which has two extreme conformations: (1) staggered — H atoms on the two carbons are as far apart as possible (60° apart, viewed along C-C axis) — energy minimum; (2) eclipsed — H atoms directly overlap each other — energy maximum, ~12 kJ/mol higher than staggered. Intermediate conformations between these are called skew or gauche. Newman and sawhorse projections are used to depict these 3D arrangements. Rotation is essentially free at room temperature.
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