Hydrocarbons • Topic 2 of 3

Alkenes & Alkynes

Alkenes (general formula CnH2n) contain a carbon–carbon double bond, and alkynes (CnH2n-2) contain a triple bond. Both are unsaturated and far more reactive than alkanes because the π bonds are exposed, electron-rich regions that attract electrophiles. The double bond is one σ + one π (sp2 carbon, planar, 120°); the triple bond is one σ + two π (sp carbon, linear, 180°).

Nomenclature and isomerism

The parent chain must include the multiple bond, which gets the lowest locant: ethene (C2H4), propene, but-1-ene, but-2-ene; ethyne (C2H2, acetylene), propyne. Alkenes show geometrical (cis–trans) isomerism because rotation about C=C is restricted. When the two higher-priority groups lie on the same side it is cis; on opposite sides, trans. For example, but-2-ene exists as cis and trans forms. Geometrical isomerism is impossible if either doubly bonded carbon carries two identical groups.

Preparation

  • Alkenes: dehydrohalogenation of alkyl halides (alc. KOH), dehydration of alcohols (conc. H2SO4, Δ), or controlled hydrogenation of alkynes (Lindlar → cis-alkene; Na/liq. NH3 → trans-alkene).
  • Alkynes: from calcium carbide, CaC2 + 2H2O → HC≡CH + Ca(OH)2; or double dehydrohalogenation of vicinal dihalides with alc. KOH/NaNH2.

Addition reactions

The π bond breaks and two groups add across it.

  • Hydrogenation: C=C + H2 ⟶[Ni] C–C.
  • Halogenation: CH2=CH2 + Br2 → CH2Br–CH2Br; the decolourisation of orange bromine water is a test for unsaturation.
  • Addition of HX — Markovnikov’s rule: the H adds to the doubly bonded carbon already bearing more hydrogens, so the halogen goes to the more substituted carbon (via the more stable carbocation). Thus propene + HBr → 2-bromopropane.
  • Anti-Markovnikov (peroxide / Kharasch effect): with HBr in the presence of peroxides the addition reverses (free-radical), giving 1-bromopropane. This is observed only with HBr.
  • Hydration: H2O adds (dil. H2SO4) following Markovnikov to give an alcohol; hydroboration–oxidation (BH3 then H2O2/OH-) gives anti-Markovnikov alcohol.
  • Ozonolysis: O3 then Zn/H2O cleaves C=C to two carbonyl compounds, locating the double bond.

Tests and special properties

Baeyer’s test: cold dilute alkaline KMnO4 (purple) is decolourised by alkenes/alkynes, forming a diol — a test for unsaturation. Acidic character of terminal alkynes: the H on an sp carbon (≡C–H) is weakly acidic, so terminal alkynes give precipitates with ammoniacal AgNO3 (white) and Cu2Cl2 (red); internal alkynes do not. Polymerisation: ethene polymerises to polythene; this is the basis of many plastics.

Comparison of alkanes, alkenes and alkynes
FeatureAlkanesAlkenesAlkynes
General formulaCnH2n+2CnH2nCnH2n-2
Key bondC–C singleC=C doubleC≡C triple
Hybridisationsp3sp2sp
Bond angle109.5°120°180°
Saturationsaturatedunsaturatedunsaturated
Baeyer's testno reactiondecolourisesdecolourises
Typical reactionsubstitutionadditionaddition
Solved Examples
1
Worked Example
Predict the major product of the reaction of propene (CH3CH=CH2) with HBr in the absence of peroxide.
Solution
  1. Protonation of the double bond can give either a 1° or a 2° carbocation.
  2. Markovnikov’s rule: H adds to the carbon with more H atoms, forming the more stable (2°) carbocation CH3–C+H–CH3.
  3. Br- then attacks that carbon.

Answer: The major product is 2-bromopropane, CH3CHBrCH3.

2
Worked Example
What is the major product when propene reacts with HBr in the presence of benzoyl peroxide? Name the effect.
Solution
  1. Peroxides switch the mechanism to a free-radical chain (only with HBr).
  2. The Br radical adds first to the terminal CH2, giving the more stable secondary carbon radical.
  3. The H then completes the addition, so Br ends up on the terminal carbon.

Answer: 1-Bromopropane (anti-Markovnikov); this is the peroxide or Kharasch effect.

3
Worked Example
An alkene on ozonolysis (O3, then Zn/H2O) gives only CH3CHO. Identify the alkene.
Solution
  1. Ozonolysis cleaves C=C and puts =O on each carbon that was doubly bonded.
  2. Both fragments are the same (CH3CHO), so the two ends of the double bond were identical: CH3CH=.
  3. Joining the two fragments back at the double bond gives CH3CH=CHCH3.

Answer: The alkene is but-2-ene, CH3CH=CHCH3.

4
Worked Example
Why does acetylene (HC≡CH) form a white precipitate with ammoniacal silver nitrate, while ethene does not?
Solution
  1. The ≡C–H bond uses an sp carbon, which is highly electronegative and holds the bonding electrons tightly.
  2. This makes the terminal alkyne hydrogen weakly acidic, so it can be replaced by a metal ion.
  3. HC≡CH + 2[Ag(NH3)2]+ → AgC≡CAg↓ (silver acetylide). Ethene has only sp2 C–H bonds, which are not acidic.

Answer: The acidic sp C–H of acetylene gives silver acetylide; ethene’s sp2 C–H is not acidic, so no precipitate.

5
Worked Example
Does but-1-ene (CH2=CHCH2CH3) show geometrical isomerism? Explain.
Solution
  1. Geometrical isomerism needs each doubly bonded carbon to carry two different groups.
  2. The terminal carbon (=CH2) bears two identical hydrogen atoms.
  3. Because one carbon has two like groups, cis and trans forms are identical.

Answer: No — but-1-ene cannot show cis–trans isomerism (the =CH2 carbon has two identical H atoms).

6
Worked Example
Predict the product of hydroboration-oxidation of propene (BH3 then H2O2/OH-) and contrast it with acid-catalysed hydration.
Solution
  1. Hydroboration adds boron to the less substituted (terminal) carbon; after oxidation the OH ends up there (anti-Markovnikov).
  2. So propene gives propan-1-ol, CH3CH2CH2OH.
  3. Acid-catalysed hydration follows Markovnikov via the 2° carbocation, giving propan-2-ol, CH3CHOHCH3.

Answer: Hydroboration-oxidation → propan-1-ol; acid hydration → propan-2-ol.

Key Points

  • Alkenes (CnH2n, sp2, 120°) have one π bond; alkynes (CnH2n-2, sp, 180°) have two π bonds — both undergo addition.
  • Geometrical (cis–trans) isomerism arises from restricted rotation about C=C, but only if each double-bond carbon carries two different groups.
  • Markovnikov: H adds to the carbon with more H (more stable carbocation); with HBr + peroxide the addition is anti-Markovnikov (free-radical, Kharasch effect).
  • Ozonolysis (O3/Zn-H2O) cleaves C=C to carbonyls and locates the double bond; Baeyer's test (cold dil. KMnO4) confirms unsaturation.
  • Terminal alkynes are weakly acidic (sp C–H): they give precipitates with ammoniacal AgNO3 (white) and Cu2Cl2 (red); ethene polymerises to polythene.
Quick Check — 4 MCQs
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Q1.Markovnikov's rule predicts that propene + HBr (no peroxide) gives mainly:
Explanation: H adds to the carbon with more H atoms, forming the stabler 2° cation, so Br lands on C-2: 2-bromopropane.
Q2.The peroxide (anti-Markovnikov) addition of HBr to propene gives:
Explanation: With peroxides the free-radical mechanism puts Br on the terminal carbon, giving 1-bromopropane.
Q3.Which reagent decolourises when shaken with an alkene (Baeyer's test)?
Explanation: Cold dilute alkaline KMnO4 (purple) is decolourised by unsaturation, forming a diol.
Q4.A terminal alkyne can be distinguished from an internal alkyne because only the terminal one:
Explanation: Only terminal alkynes have an acidic ≡C–H, so they form silver acetylide with ammoniacal AgNO3.
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