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Hydrocarbons: Class 12 Chemistry

Written by Dr. Ayesha Nadeem (Chemistry)

Alkanes, alkenes, alkynes and benzene: general formulas, nomenclature, isomerism, preparation, physical properties, characteristic reactions, aromaticity, electrophilic substitution and tests for unsaturation.

Key points

Hydrocarbons are compounds of carbon and hydrogen only. Alkanes (CₙH₂ₙ₊₂) are saturated and have only single bonds. Alkenes (CₙH₂ₙ) contain a C=C double bond and alkynes (CₙH₂ₙ₋₂) contain a C≡C triple bond. Both are unsaturated. Aromatic hydrocarbons such as benzene contain a stable ring with delocalised electrons.

In IUPAC naming, find the longest carbon chain that holds the functional group, number it so the double or triple bond gets the lowest number, then name and number the side groups in alphabetical order. Examples: CH₃CH₂CH=CH₂ is but-1-ene and CH₃C≡CCH₃ is but-2-yne. Branched alkanes are named as alkyl derivatives, as in 2-methylpropane.

Types of isomerism:

  • Chain isomerism: different carbon skeletons, such as butane and 2-methylpropane (both C₄H₁₀).
  • Position isomerism: the double bond is at a different place, such as but-1-ene and but-2-ene.
  • Geometric (cis-trans) isomerism: in but-2-ene each carbon of the double bond carries two different groups, and rotation about C=C is not possible. The cis form has similar groups on the same side and the trans form has them on opposite sides.

In alkanes carbon is sp³ hybridised with bond angles of 109.5°. In alkenes carbon is sp² with a planar shape and angle 120°, and the double bond is one sigma and one pi bond. In alkynes carbon is sp with a linear shape, 180°, and the triple bond is one sigma and two pi bonds. The C=C bond length is about 134 pm, which is shorter than C-C (154 pm).

Preparation of alkanes, alkenes and alkynes:

  • Alkanes: hydrogenation of alkenes with a nickel catalyst, reduction of alkyl halides, or Wurtz reaction, 2R-X + 2Na → R-R + 2NaX in dry ether.
  • Alkenes: dehydration of alcohols with hot concentrated H₂SO₄ (C₂H₅OH → C₂H₄ + H₂O at about 170 °C), or dehydrohalogenation of alkyl halides with alcoholic KOH.
  • Alkynes: ethyne from calcium carbide, CaC₂ + 2H₂O → C₂H₂ + Ca(OH)₂, or by dehydrohalogenation of vicinal dihalides.

Alkanes are fairly unreactive because C-C and C-H bonds are strong and non-polar. Their main reactions are combustion, CH₄ + 2O₂ → CO₂ + 2H₂O, and free radical substitution with halogens in sunlight. Substitution goes through initiation (Cl₂ splits into two Cl· radicals), propagation (Cl· + CH₄ → HCl + CH₃·, then CH₃· + Cl₂ → CH₃Cl + Cl·) and termination (two radicals combine).

Alkenes and alkynes are reactive because the pi bond is weaker and electron rich. They undergo electrophilic addition. Examples: addition of H₂ (Ni, giving an alkane), Br₂ (giving a dibromoalkane), HBr and water (with acid catalyst). Ethyne adds two moles of reagent, first forming an alkene derivative and then a saturated product.

Markovnikov's rule: when HX adds to an unsymmetrical alkene, the hydrogen goes to the carbon that already has more hydrogen atoms. Reason: the more substituted carbocation is more stable. So propene plus HBr gives 2-bromopropane as the major product.

Tests for unsaturation:

  • Bromine water (orange or brown) is decolourised by alkenes and alkynes.
  • Cold dilute alkaline KMnO₄ (Baeyer's reagent, purple) is decolourised and a brown precipitate of MnO₂ may form. Alkenes give a diol.
  • Terminal alkynes such as ethyne form a precipitate with ammoniacal silver nitrate (white) or ammoniacal cuprous chloride (red).

Benzene C₆H₆ is a planar ring of six sp² carbons, all C-C bonds equal (139 pm), with delocalised pi electrons above and below the ring. It is stable and follows Hückel's rule (4n + 2 pi electrons, here 6). It prefers substitution to addition. Kekulé's structure shows alternating double bonds, but the real molecule is a resonance hybrid.

Electrophilic substitution of benzene keeps the ring intact. Nitration uses conc. HNO₃ and conc. H₂SO₄ at about 50 °C to give nitrobenzene. Halogenation uses Cl₂ or Br₂ with FeCl₃ or FeBr₃. Friedel-Crafts alkylation with CH₃Cl and anhydrous AlCl₃ gives toluene. Addition happens only under severe conditions, such as hydrogenation to cyclohexane.

A group already on the ring directs the next substituent. The CH₃, OH and halogen groups direct to ortho and para positions. The NO₂, CHO and COOH groups direct to the meta position. Most groups that direct ortho and para activate the ring, while the others deactivate it (halogens deactivate slightly but still direct ortho and para).

Study tip: Learn each reaction with its reagent, condition and product, as a short chain: reagent, then condition, then product. Practise naming five or six alkenes and alkynes every day. For mechanism questions write each step on its own line and show the movement of the species clearly.

Definitions and flashcards

Say the meaning first, then open the card to check.

Saturated hydrocarbon

A hydrocarbon with only single bonds between carbon atoms.

Unsaturated hydrocarbon

A hydrocarbon with at least one double or triple carbon-carbon bond.

Isomers

Compounds with the same molecular formula but different structures.

Electrophile

A species that is short of electrons and attacks electron-rich sites.

Free radical

An atom or group with an unpaired electron.

Markovnikov's rule

In addition of HX to an unsymmetrical alkene, H goes to the carbon with more hydrogens.

Aromatic compound

A cyclic, planar compound with delocalised pi electrons and 4n + 2 of them.

Resonance

Representing a molecule by two or more structures differing only in the position of electrons.

Formula sheet

NameFormulaWhat it means
AlkanesCₙH₂ₙ₊₂General formula of saturated chain hydrocarbons
AlkenesCₙH₂ₙGeneral formula with one C=C bond
AlkynesCₙH₂ₙ₋₂General formula with one C≡C bond
Hückel's ruleNumber of pi electrons = 4n + 2Condition for aromaticity (n = 0, 1, 2 ...)

Short questions with model answers

These are the kind of questions asked for 2, 3 and 5 marks. Try each one before opening the answer.

Why do alkenes show addition reactions while alkanes show substitution? (2 marks)

Alkenes have a pi bond that is weaker and rich in electrons, so electrophiles attack it and add across the double bond. Alkanes have only strong sigma bonds, so they react only with a free radical that replaces a hydrogen atom.

State Markovnikov's rule with an example. (2 marks)

When a molecule HX adds to an unsymmetrical alkene, the hydrogen attaches to the doubly bonded carbon with more hydrogen atoms. Example: CH₃CH=CH₂ + HBr gives CH₃CHBrCH₃ (2-bromopropane) as the main product.

How can you distinguish ethene from ethane in the laboratory? (2 marks)

Shake each gas with bromine water or with dilute alkaline KMnO₄. Ethene decolourises the orange bromine water (or the purple permanganate) because it adds across the double bond. Ethane does not change the colour.

Describe the free radical chlorination of methane in three steps. (3 marks)

Initiation: in sunlight Cl₂ splits into two chlorine radicals, Cl₂ → 2Cl·. Propagation: Cl· + CH₄ → HCl + CH₃·, then CH₃· + Cl₂ → CH₃Cl + Cl·. This repeats as a chain. Termination: radicals combine to end the chain, for example Cl· + Cl· → Cl₂ or CH₃· + Cl· → CH₃Cl.

Explain the structure of benzene. (5 marks)

Benzene is a planar hexagon of six sp² carbons. Each carbon forms sigma bonds with two carbons and one hydrogen, and the left over p orbitals overlap sideways to give a cloud of six delocalised pi electrons above and below the ring. All C-C bonds are equal at 139 pm, between a single and a double bond. This is shown by resonance between two Kekulé structures, and it explains the high stability and the preference for substitution.

Write the nitration of benzene with conditions. (3 marks)

Benzene is warmed at about 50 °C with a mixture of concentrated nitric acid and concentrated sulfuric acid. The nitronium ion NO₂⁺ is the electrophile. C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O. The product is nitrobenzene, a pale yellow oily liquid.

Numericals with worked steps

Calculate the volume of oxygen at STP needed to burn 11 g of propane completely, and the mass of CO₂ formed. (C = 12, H = 1, O = 16; 1 mol of gas = 22.4 dm³)
  1. C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.
  2. Molar mass of C₃H₈ = 36 + 8 = 44 g/mol.
  3. Moles of propane = 11 / 44 = 0.25 mol.
  4. Moles of O₂ = 5 × 0.25 = 1.25 mol; volume = 1.25 × 22.4 = 28 dm³.
  5. Moles of CO₂ = 3 × 0.25 = 0.75 mol; mass = 0.75 × 44 = 33 g.

Answer: 28 dm³ of O₂ and 33 g of CO₂

A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. Its molar mass is 42 g/mol. Find its molecular formula.
  1. Moles of C = 85.7 / 12 = 7.14; moles of H = 14.3 / 1 = 14.3.
  2. Divide by the smaller: C = 1, H = 2.
  3. Empirical formula = CH₂, mass 14.
  4. n = 42 / 14 = 3.
  5. Molecular formula = C₃H₆ (propene).

Answer: C₃H₆

Find the mass of bromine that reacts with 2.8 g of ethene. (C = 12, H = 1, Br = 80)
  1. C₂H₄ + Br₂ → C₂H₄Br₂.
  2. Molar mass of C₂H₄ = 28 g/mol, so moles = 2.8 / 28 = 0.1 mol.
  3. The ratio is 1 : 1, so moles of Br₂ = 0.1 mol.
  4. Molar mass of Br₂ = 160 g/mol; mass = 0.1 × 160 = 16 g.

Answer: 16 g of bromine

Calculate the percentage of carbon by mass in ethyne (C₂H₂).
  1. Molar mass of C₂H₂ = 24 + 2 = 26 g/mol.
  2. Mass of carbon = 24 g.
  3. Percentage = 24 / 26 × 100 = 92.3%.

Answer: 92.3% carbon

Common mistakes

  • Writing the general formula of alkynes as CₙH₂ₙ₊₂ or of alkanes as CₙH₂ₙ.
  • Applying Markovnikov's rule to the wrong carbon, or to symmetrical alkenes such as ethene where it does not matter.
  • Showing benzene undergoing addition under normal conditions. It gives substitution.
  • Forgetting the catalyst or condition in a reaction, such as Ni for hydrogenation, FeBr₃ for bromination of benzene or alcoholic KOH for elimination.
  • Numbering the chain from the wrong end, so the double bond does not get the lowest number.

Practise this chapter

Practice questions and timed test Printable worksheet

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