Hydrocarbons
On mobile, swipe inside the PDF to read all pages and pinch to zoom.
1. Hydrocarbons: Introduction
An organic compound made only of carbon and hydrogen atoms.
Hydrocarbons form the simplest family of organic compounds and provide the structural basis for many more complex organic substances.
Saturated
Contain only C–C single bonds. Main family: alkanes.
Unsaturated
Contain C=C or C≡C bonds. Main families: alkenes and alkynes.
Diagram 1: Main hydrocarbon classes
2. Alkane, Alkene and Alkyne Comparison
| Family | Characteristic bond | General formula | First common member | Main reaction type |
|---|---|---|---|---|
| Alkane | C–C | CₙH₂ₙ₊₂ | CH₄ | Substitution |
| Alkene | C=C | CₙH₂ₙ | C₂H₄ | Addition |
| Alkyne | C≡C | CₙH₂ₙ₋₂ | C₂H₂ | Addition |
The π component of a multiple bond is more exposed and generally more reactive than a C–C σ bond, so alkenes and alkynes readily undergo addition reactions.
Diagram 2: Bond composition in the three hydrocarbon families
3. Alkanes: Saturated Hydrocarbons
Acyclic saturated hydrocarbons containing only C–C and C–H single bonds, with general formula CₙH₂ₙ₊₂.
Examples:
- CH₄ — methane
- C₂H₆ — ethane
- C₃H₈ — propane
- C₄H₁₀ — butane
Alkanes are comparatively less reactive than alkenes and alkynes because they contain strong σ bonds and no π bond.
4. Preparation of Alkanes
4.1 Reduction of Haloalkanes
Example:
C₂H₅Cl + 2[H] → C₂H₆ + HCl4.2 Wurtz Reaction
Example:
2CH₃Cl + 2Na → C₂H₆ + 2NaClWurtz reaction is most useful for preparing symmetrical alkanes with an even number of carbon atoms from identical haloalkanes.
4.3 Decarboxylation
Sodium salts of carboxylic acids heated with soda lime give an alkane with one fewer carbon atom:
Example:
CH₃COONa + NaOH → CH₄ + Na₂CO₃4.4 Catalytic Hydrogenation of Alkenes
CH₂=CH₂ + H₂ → CH₃–CH₃ (Ni/Pt/Pd)4.5 Catalytic Hydrogenation of Alkynes
HC≡CH + 2H₂ → CH₃–CH₃Diagram 3: Syllabus methods for preparing alkanes
5. Chemical Properties of Alkanes
5.1 Halogenation
Under light or heat, alkanes undergo free-radical substitution.
CH₄ + Cl₂ → CH₃Cl + HCl (hν)Further substitution can produce CH₂Cl₂, CHCl₃ and CCl₄.
At Grade 11 level, remember that halogenation is a substitution reaction promoted by light/heat; the detailed radical-chain mechanism belongs naturally with reaction-mechanism study.
5.2 Nitration
RH + HNO₃ → RNO₂ + H₂O (high temperature)5.3 Sulphonation
RH + H₂SO₄ → RSO₃H + H₂O5.4 Oxidation of Ethane
Complete combustion:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂OCombustion is highly exothermic and is one reason hydrocarbons are important fuels.
6. Alkenes: Unsaturated Hydrocarbons
Acyclic hydrocarbons containing at least one C=C double bond; for one double bond the general formula is CₙH₂ₙ.
The double bond contains one σ bond and one π bond. The π bond is responsible for the characteristic addition reactions of alkenes.
Diagram 4: π-bond reactivity in alkenes
7. Preparation of Alkenes
7.1 Dehydration of Alcohol
CH₃CH₂OH → CH₂=CH₂ + H₂O (conc. H₂SO₄, heat)Removal of H₂O from an alcohol to form an alkene.
7.2 Dehydrohalogenation of Haloalkane
CH₃CH₂Br + KOH(alc.) → CH₂=CH₂ + KBr + H₂OElimination of HX from a haloalkane, commonly using alcoholic KOH.
7.3 Controlled Hydrogenation of Alkynes
HC≡CH + H₂ → H₂C=CH₂Partial hydrogenation stops at the alkene stage under suitable catalytic conditions.
8. Markovnikov’s Rule
In the addition of HX to an unsymmetrical alkene, H generally adds to the double-bond carbon already bearing more H atoms, while X adds to the more substituted carbon.
Example: Propene + HBr
CH₃–CH=CH₂ + HBr → CH₃–CHBr–CH₃2-bromopropane is the major Markovnikov product under ordinary ionic conditions.
For simple HX addition, “H goes to the carbon that already has more H” is a useful Grade 11 shortcut.
Diagram 5: Markovnikov orientation in HBr addition
9. Peroxide Effect / Anti-Markovnikov Addition
In the presence of organic peroxide, HBr can add to an unsymmetrical alkene in the opposite orientation.
CH₃–CH=CH₂ + HBr → CH₃–CH₂–CH₂Br (peroxide)The radical-promoted anti-Markovnikov addition of HBr to an unsymmetrical alkene in the presence of peroxide.
The classical peroxide effect applies to HBr; it is not ordinarily used as a general anti-Markovnikov rule for HCl or HI.
| Propene + HBr | Major product |
|---|---|
| Without peroxide | 2-bromopropane |
| With peroxide | 1-bromopropane |
10. Important Addition Reactions of Alkenes
10.1 Hydrogenation
CH₂=CH₂ + H₂ → CH₃–CH₃ (Ni/Pt/Pd)10.2 Addition of HX
CH₂=CH₂ + HBr → CH₃CH₂Br10.3 Hydration
CH₂=CH₂ + H₂O → CH₃CH₂OH (acid catalyst)10.4 Addition of H₂SO₄
CH₂=CH₂ + H₂SO₄ → CH₃CH₂OSO₃HHydrolysis of the alkyl hydrogen sulfate gives an alcohol.
10.5 Halogen Addition
CH₂=CH₂ + Br₂ → BrCH₂–CH₂BrDecolorization of bromine solution is used as a test for unsaturation.
11. Ozonolysis of Alkenes
Reaction of an alkene with ozone followed by work-up, cleaving the C=C bond to form carbonyl compounds.
Example: Ethene
CH₂=CH₂ → 2HCHO (O₃, then reductive work-up)Use
Ozonolysis products can help identify the positions of double bonds in an unknown alkene.
Diagram 6: Structural information from ozonolysis
12. Alkynes
Acyclic hydrocarbons containing a carbon–carbon triple bond; for one triple bond the general formula is CₙH₂ₙ₋₂.
The C≡C bond contains one σ and two π bonds. Ethyne (acetylene), HC≡CH, is the first member.
The carbon atoms of a simple C≡C unit are sp-hybridized and the bond angle is approximately 180°.
13. Preparation of Alkynes
13.1 From Carbon and Hydrogen
At very high temperature, carbon and hydrogen can form ethyne:
2C + H₂ → C₂H₂13.2 From 1,2-Dibromoethane
Successive dehydrohalogenation removes two molecules of HBr:
BrCH₂–CH₂Br → HC≡CH + 2HBrStrong basic conditions are required for complete double elimination.
13.3 From Chloroform / Iodoform
Under suitable dehalogenating conditions, haloforms can be converted to ethyne; this is treated as a syllabus preparation route.
For haloform preparation, learn the exact reagent/equation used in your prescribed classroom text and handwritten source notes; the current syllabus requires the preparation route but not a detailed mechanism.
14. Chemical Properties of Alkynes
14.1 Hydrogenation
HC≡CH + H₂ → CH₂=CH₂ CH₂=CH₂ + H₂ → CH₃CH₃14.2 Addition of HX
HC≡CH + HBr → CH₂=CHBrFurther HBr addition can give a geminal dihalide.
14.3 Hydration of Ethyne
HC≡CH + H₂O → CH₂=CHOH → CH₃CHOThe initially formed enol rearranges to ethanal (acetaldehyde).
Hydration of an alkyne may first form an unstable enol that tautomerizes to a carbonyl compound.
15. Acidic Nature of Terminal Alkynes
Hydrogen directly attached to an sp-hybridized carbon in a terminal alkyne is more acidic than ordinary alkene/alkane hydrogen.
15.1 Reaction with Sodium
2HC≡CH + 2Na → 2HC≡CNa + H₂15.2 Ammoniacal Silver Nitrate Test
Terminal alkynes form silver acetylide-type precipitates under suitable ammoniacal AgNO₃ conditions.
15.3 Ammoniacal Cuprous Chloride Test
Terminal alkynes form copper(I) acetylide-type precipitates under ammoniacal CuCl/Cu₂Cl₂ test conditions.
Internal alkynes do not contain the acidic terminal ≡C–H group and do not give these terminal-alkyne precipitation tests.
Diagram 7: Structural basis of terminal-alkyne acidity
16. Tests of Unsaturation
16.1 Bromine Water / Bromine Solution
Alkenes and alkynes decolorize bromine because Br₂ adds across the multiple bond.
CH₂=CH₂ + Br₂ → BrCH₂–CH₂BrBromine color disappears in the presence of a reactive C=C or C≡C bond under test conditions.
16.2 Baeyer’s Test
Cold dilute alkaline KMnO₄ is decolorized by unsaturated hydrocarbons while the multiple bond is oxidized.
The purple permanganate color disappears; brown MnO₂ may be observed depending on the conditions.
| Hydrocarbon | Bromine test | Baeyer test |
|---|---|---|
| Alkane | No rapid decolorization in dark/ordinary test conditions | Negative under ordinary conditions |
| Alkene | Positive | Positive |
| Alkyne | Positive | Positive |
Diagram 8: Bromine decolorization test
17. Kolbe Electrolysis and Hydrocarbon Preparation
Electrolysis of an aqueous solution of a carboxylate salt in which decarboxylation occurs at the anode and carbon radicals can couple to form hydrocarbons.
Anodic Idea
RCOO⁻ → R· + CO₂ + e⁻ 2R· → R–RThus suitable carboxylate salts can produce alkanes. The curriculum also asks for Kolbe electrolysis approaches connected with preparation of alkenes and alkynes using appropriate dicarboxylate systems.
Learn the exact balanced equations and salt choices used in your course text for alkane, alkene and alkyne Kolbe preparations. The central concept is anodic decarboxylation followed by formation of the carbon skeleton.
Diagram 9: Conceptual Kolbe electrolysis pathway
18. Comparative Physical Properties of Alkanes, Alkenes and Alkynes
| Property | Alkanes | Alkenes | Alkynes |
|---|---|---|---|
| Polarity | Mostly non-polar | Mostly non-polar | Mostly non-polar |
| Water solubility | Very low | Very low | Very low |
| Organic-solvent solubility | Generally good in non-polar solvents | Generally good | Generally good |
| Boiling point trend | Generally rises with molecular size | Generally rises with molecular size | Generally rises with molecular size |
| Density | Usually lower than water for common members | Usually lower than water for common members | Usually lower than water for common members |
| Characteristic chemical reactivity | Substitution/combustion | Addition | Addition; terminal-alkyne acidity |
As molecular mass and surface area increase, London dispersion forces generally become stronger, so boiling points tend to rise within each homologous series.
19. Worked Examples
C₄H₁₀ matches CₙH₂ₙ₊₂ → alkane.
C₅H₁₀ may match the open-chain monoalkene formula CₙH₂ₙ. Structural information is still needed because cyclic isomers can share this formula.
Sodium propanoate gives an alkane with one less carbon:
CH₃CH₂COONa + NaOH → C₂H₆ + Na₂CO₃Two CH₃Br molecules couple:
2CH₃Br + 2Na → C₂H₆ + 2NaBrPropene + HCl under ordinary conditions gives mainly 2-chloropropane.
Propene + HBr in peroxide gives mainly 1-bromopropane.
Ethene decolorizes bromine solution because Br₂ adds across C=C.
Ethene gives methanal after reductive ozonolysis:
CH₂=CH₂ → 2HCHOCH₃–C≡CH contains ≡C–H, so it can show terminal-alkyne acidity tests; CH₃–C≡C–CH₃ cannot.
C≡C contains 1σ + 2π; C=C contains 1σ + 1π.
20. High-Yield Reaction Summary
| Starting material | Reaction / reagent | Main product idea |
|---|---|---|
| Haloalkane | Reduction | Alkane |
| Haloalkane | Na / dry ether | Higher alkane by Wurtz coupling |
| Sodium carboxylate | Soda lime, heat | Alkane with one fewer C |
| Alkene | H₂ / catalyst | Alkane |
| Alcohol | Dehydration | Alkene |
| Haloalkane | Alcoholic KOH | Alkene |
| Alkene | HX | Haloalkane |
| Alkene | H₂O / H⁺ | Alcohol |
| Alkene | O₃, work-up | Carbonyl fragments |
| Alkyne | H₂ | Alkene then alkane |
| Alkyne | H₂O | Enol → carbonyl compound |
| Terminal alkyne | Na / ammoniacal Ag⁺ / Cu⁺ tests | Acetylide formation |
21. Common Exam Mistakes
- Calling every CₙH₂ₙ compound an alkene; cycloalkanes can have the same molecular formula.
- Calling every CₙH₂ₙ₋₂ compound an alkyne; other unsaturation patterns can also give this formula.
- Confusing saturated with “contains lots of hydrogen”; saturated specifically means no C=C/C≡C in the standard hydrocarbon context.
- Using Wurtz reaction without considering that mixed haloalkanes can give product mixtures.
- Forgetting decarboxylation gives an alkane with one fewer carbon atom than the carboxylate.
- Writing alkane halogenation as an addition reaction; it is substitution.
- Calling alkene hydrogenation substitution; it is addition.
- Applying Markovnikov orientation to a symmetrical alkene where both orientations are equivalent.
- Applying peroxide effect to HCl or HI as if it were a general rule; classical peroxide effect is associated with HBr.
- Forgetting that HBr + propene gives different major products with and without peroxide.
- Writing bromine-water decolorization for an ordinary alkane under normal unsaturation-test conditions.
- Confusing bromine-water test with Baeyer’s test.
- Forgetting Baeyer’s reagent is cold dilute alkaline KMnO₄ in the usual qualitative test.
- Thinking ozonolysis adds ozone without cleavage; its analytical value comes from double-bond cleavage after work-up.
- Calling an alkyne double bond; it contains a triple bond.
- Writing C≡C as 3σ bonds; it is 1σ + 2π.
- Assuming all alkynes have acidic hydrogen; only terminal alkynes contain ≡C–H.
- Using terminal-alkyne silver/copper tests for internal alkynes.
- Forgetting that alkyne hydration initially forms an enol that can tautomerize.
- Confusing Kolbe electrolysis with soda-lime decarboxylation; both can lose CO₂ but use different processes.
22. Important Exam Questions
Very Short / Short Questions
- Define hydrocarbon.
- Differentiate saturated and unsaturated hydrocarbons.
- Write the general formulae of alkane, alkene and alkyne series.
- Define alkane and give two examples.
- Explain reduction of haloalkane to alkane.
- State Wurtz reaction with an equation.
- Explain soda-lime decarboxylation.
- Explain catalytic hydrogenation of alkene/alkyne.
- State halogenation, nitration and sulphonation of alkanes.
- Write the combustion equation of ethane.
- Define alkene and explain its π-bond reactivity.
- Explain dehydration of alcohol to alkene.
- Explain dehydrohalogenation.
- State Markovnikov’s rule with an example.
- Explain peroxide effect with propene + HBr.
- Write alkene addition reactions with H₂, HX, H₂O, O₃ and H₂SO₄.
- Define ozonolysis and state its use.
- Define alkyne and give the formula of ethyne.
- State syllabus methods for preparation of ethyne/alkynes.
- Explain addition of H₂, HX and H₂O to alkynes.
- Why are terminal alkynes acidic?
- Explain action of terminal alkynes with sodium.
- State the ammoniacal AgNO₃ and Cu(I) tests of terminal alkynes.
- Explain bromine test for unsaturation.
- Explain Baeyer’s test.
- Compare the physical properties of alkanes, alkenes and alkynes.
- Explain the principle of Kolbe electrolysis.
Long / Descriptive Questions
- Discuss preparation and chemical properties of alkanes according to the syllabus.
- Discuss preparation and important addition reactions of alkenes.
- Explain Markovnikov’s rule and peroxide effect with suitable equations.
- Explain ozonolysis and its structural significance.
- Discuss preparation and chemical properties of alkynes.
- Explain the acidic nature of terminal alkynes and their characteristic tests.
- Describe bromine and Baeyer tests for unsaturation.
- Compare alkanes, alkenes and alkynes in bonding, formula, reactivity and physical properties.
- Explain Kolbe electrolysis as a hydrocarbon-preparation method.
Study Unit 14 in three major blocks: alkanes → alkenes → alkynes. For each family learn general formula, preparation, characteristic reactions and tests. Give special attention to Wurtz/decarboxylation, Markovnikov vs peroxide effect, ozonolysis, terminal-alkyne acidity, bromine/Baeyer tests and Kolbe electrolysis.
23. One-Minute Revision
- Unit 14: Hydrocarbons — 8 teaching hours.
- Hydrocarbons contain only C and H.
- Alkanes are saturated; alkenes and alkynes are unsaturated.
- Alkane formula: CₙH₂ₙ₊₂.
- Alkene formula: CₙH₂ₙ for one open-chain C=C.
- Alkyne formula: CₙH₂ₙ₋₂ for one open-chain C≡C.
- Alkanes mainly undergo substitution.
- Alkenes and alkynes mainly undergo addition.
- Wurtz: haloalkanes couple using Na in dry ether.
- Decarboxylation removes CO₂/carboxyl carbon and gives one fewer C in the alkane.
- Hydrogenation converts unsaturated hydrocarbons toward alkanes.
- Alcohol dehydration gives an alkene.
- Alcoholic KOH removes HX from haloalkane to give alkene.
- Markovnikov: H goes to the double-bond carbon with more H.
- Peroxide effect reverses HBr orientation in suitable unsymmetrical alkenes.
- Classical peroxide effect applies to HBr.
- Alkenes add H₂, HX, H₂O, Br₂ and H₂SO₄.
- Ozonolysis cleaves C=C to carbonyl compounds after work-up.
- C=C = 1σ + 1π.
- C≡C = 1σ + 2π.
- Ethyne is HC≡CH.
- Alkyne hydrogenation can proceed alkyne → alkene → alkane.
- Alkyne hydration can form an enol that tautomerizes.
- Terminal alkynes contain ≡C–H and show weak acidity.
- Terminal alkynes react with Na and characteristic ammoniacal Ag/Cu reagents.
- Alkenes and alkynes decolorize bromine solution.
- Baeyer’s test uses dilute alkaline KMnO₄.
- Hydrocarbons are generally poorly soluble in water.
- Boiling points generally rise with molecular size.
- Kolbe electrolysis uses anodic decarboxylation to build hydrocarbon products.
24. Diagram Practice
- Classification of hydrocarbons.
- C–C vs C=C vs C≡C bonding.
- Routes for preparation of alkanes.
- π-bond addition concept.
- Markovnikov addition to propene.
- Ozonolysis cleavage.
- Terminal vs internal alkyne.
- Bromine test for unsaturation.
- Kolbe electrolysis pathway.
Discussion
Share a helpful question, idea, or explanation with other students.