Hydrocarbons

What this chapter is about

Hydrocarbons are organic compounds containing only carbon and hydrogen atoms. This chapter introduces the major classes of hydrocarbons: alkanes, alkenes, alkynes, and aromatic compounds. You learn how to name them using IUPAC rules, write their structures, understand their bonding, and predict their physical and chemical behaviour.

At the Class 11 level, you have already met the basic ideas of organic chemistry—tetravalence of carbon, catenation, hybridisation, and structural isomerism. Hydrocarbons form the foundation on which all other organic compounds are built; alcohols, aldehydes, carboxylic acids and many polymers are derived from hydrocarbon skeletons. Understanding hydrocarbons now prepares you to study functional groups and reaction mechanisms in later chapters.

After working through this chapter, you should be able to draw and name hydrocarbons up to about ten carbon atoms, classify them as saturated or unsaturated, predict the products of their characteristic reactions, and appreciate why aromatic compounds behave differently from ordinary unsaturated compounds.

Key ideas

• Saturated vs unsaturated: Alkanes have only C–C single bonds (saturated). Alkenes have at least one C═C double bond; alkynes have at least one C≡C triple bond (both unsaturated). Unsaturated compounds undergo addition reactions.
• Hybridisation and shape: In alkanes every carbon is sp³-hybridised with bond angles near 109.5°. In alkenes the doubly-bonded carbons are sp²-hybridised (bond angle about 120°). In alkynes the triply-bonded carbons are sp-hybridised (bond angle 180°).
• Isomerism: Chain isomers differ in branching; positional isomers differ in the location of a functional group or multiple bond; geometrical (cis-trans) isomerism arises when rotation about a double bond is restricted.
• General formulas: Alkanes CₙH₂ₙ₊₂; cycloalkanes CₙH₂ₙ; alkenes CₙH₂ₙ; alkynes CₙH₂ₙ₋₂.
• Characteristic reactions: Alkanes mainly undergo free-radical substitution (e.g., halogenation in sunlight). Alkenes and alkynes undergo electrophilic addition (e.g., addition of H₂, HBr, H₂O). Aromatic compounds prefer electrophilic substitution over addition to preserve their stable ring.
• Markovnikov's rule: When an unsymmetrical reagent HX adds to an unsymmetrical alkene, the hydrogen attaches to the carbon already holding more hydrogen atoms.
• Aromaticity: Benzene and related compounds are aromatic because they have a planar, cyclic ring of sp²-hybridised atoms with (4n + 2) π electrons (where n is a non-negative integer). This delocalization gives extra stability.

Formulas and facts to remember

1. Alkane general formula: CₙH₂ₙ₊₂. Each successive member differs by CH₂ (homologous series).
2. Degree of unsaturation (DBE): For CₓHᵧ, DBE = (2x + 2 − y) / 2. One double bond or one ring equals one DBE; one triple bond equals two.
3. Markovnikov's rule: In the addition of HX to an unsymmetrical alkene, H goes to the carbon with more H atoms; the halide goes to the carbon with fewer H atoms.
4. Anti-Markovnikov (peroxide effect): With HBr in the presence of peroxides, the opposite orientation occurs via a free-radical mechanism. HCl and HI do not show this effect.
5. Benzene resonance energy: Benzene is about 150 kJ mol⁻¹ more stable than a hypothetical "cyclohexatriene" with localised double bonds. This extra stability explains its preference for substitution over addition.
6. Hückel's rule for aromaticity: A planar, cyclic, fully conjugated system is aromatic if it has (4n + 2) π electrons (n = 0, 1, 2, …).
7. Combustion of hydrocarbons: Complete combustion gives CO₂ and H₂O. Example: CH₄ + 2 O₂ → CO₂ + 2 H₂O.
8. Ozonolysis: Alkenes react with ozone followed by zinc/water to cleave the double bond and yield aldehydes or ketones; useful for locating the double bond position.

Worked examples

Example 1 – IUPAC naming of a branched alkene

Problem: Name the compound with the structure CH₃–CH(CH₃)–CH═CH–CH₃.

Solution:

1. Identify the longest carbon chain containing the double bond. The chain has five carbons, so the parent name is pentene.
2. Number from the end that gives the double bond the lowest locant. Numbering from the right gives the double bond between C-2 and C-3.
3. Locate substituents. A methyl group is on C-4.
4. Write the full name: 4-methylpent-2-ene.

Example 2 – Predicting the major product using Markovnikov's rule

Problem: Propene (CH₂═CH–CH₃) reacts with HBr in the absence of peroxides. Predict the major product.

Solution:

1. The double bond is between C-1 and C-2.
2. According to Markovnikov's rule, H attaches to C-1 (which has two H atoms) and Br attaches to C-2 (which has one H atom).
3. Product: CH₃–CHBr–CH₃ (2-bromopropane).

Example 3 – Degree of unsaturation

Problem: A hydrocarbon has the molecular formula C₆H₁₀. Calculate its degree of unsaturation and suggest a possible structure.

Solution:

1. DBE = (2 × 6 + 2 − 10) / 2 = (14 − 10) / 2 = 2.
2. Two DBE may mean two double bonds, one triple bond, one double bond plus one ring, or two rings.
3. Possible structures: hex-1-yne (CH≡C–CH₂–CH₂–CH₂–CH₃) or cyclohexene (a six-membered ring with one double bond).

Common mistakes

• Forgetting that geometric isomerism requires two different groups on each doubly-bonded carbon → check both carbons before labelling cis or trans.
• Numbering the parent chain without giving the principal functional group the lowest locant → always give double or triple bonds lower numbers than substituents.
• Applying Markovnikov's rule to symmetrical alkenes → it only affects orientation in unsymmetrical alkenes.
• Assuming alkenes and alkynes burn incompletely because they are unsaturated → with sufficient oxygen, they undergo complete combustion like alkanes.
• Treating benzene like a typical alkene and expecting easy addition → benzene prefers electrophilic substitution to preserve aromaticity.

Quick revision

• Alkanes: CₙH₂ₙ₊₂, sp³, single bonds, substitution reactions.
• Alkenes: CₙH₂ₙ, sp², double bonds, electrophilic addition, follow Markovnikov's rule.
• Alkynes: CₙH₂ₙ₋₂, sp, triple bonds, can add two moles of reagent.
• Aromatic compounds: planar ring with (4n + 2) π electrons, prefer substitution over addition.
• DBE = (2C + 2 − H) / 2 tells you the sum of rings and multiple bonds.