- Explain sp² and sp hybridisation and count σ and π bonds and hybrid orbitals the DİM way.
- Write equations for making alkenes (Zaitsev’s rule) and for their additions (Markovnikov’s rule, KMnO₄, Br₂, polymerisation).
- Know how divinyl, isoprene, chloroprene and acetylene are made and how they react (Kucherov reaction, acetylides, trimerisation).
- Solve formula and mixture problems from bromine and hydrogen additions and from hydrogen-count links between series.
Put an unripe pear in a bag with an apple and it ripens faster: the apple gives off ethylene gas. Plastic bags are made from ethylene too, and a welder’s flame is burning acetylene. These substances have double or triple bonds between carbon atoms — they are unsaturated hydrocarbons. The June 2026 DİM exams had four tasks on this topic: recognising ethylene with KMnO₄, an alkene by Zaitsev’s rule, the formula CₙH₂ₙ₋₂ and an alkene–diene hydrogen-count task.
C=C and C≡C: structure, σ and π bonds
Alkenes (CₙH₂ₙ, n ≥ 2) have one C=C bond. A double bond is one σ and one π bond. The carbons of the double bond are sp² hybridised: three hybrid orbitals lie in one plane at 120°, and the unhybridised p orbitals overlap side by side to form the π bond. A π bond is weaker than a σ bond — that is why unsaturated compounds are reactive; and since there is no rotation about C=C, cis–trans isomerism is possible. In alkynes (CₙH₂ₙ₋₂, n ≥ 2) C≡C = σ + 2π, the carbons are sp and this part of the molecule is linear (180°). Dienes (CₙH₂ₙ₋₂, n ≥ 3) have two C=C bonds.
| Hybridisation | Hybrid orbitals (1 C) | Angle | Bond, length | Example |
|---|---|---|---|---|
| sp³ | 4 | 109°28′ | C–C, 0.154 nm | ethane |
| sp² | 3 | 120° | C=C (σ + π), 0.134 nm | ethene (ethylene) |
| sp | 2 | 180° | C≡C (σ + 2π), 0.120 nm | ethyne (acetylene) |
- N(hybrid)total number of hybrid orbitals of the carbon atoms
- N(sp³), N(sp²), N(sp)numbers of carbons in each hybrid state
This is how DİM counts. σ bonds: every single bond plus one per multiple bond; π: C=C — 1, C≡C — 2.
Heating 3-bromopentane with alcoholic alkali gives alkene X. Choose the true statements about X.
1. Its name is pent-2-ene
2. It has no cis–trans isomers
3. Its molecule has 14 σ bonds in total
4. Its carbon atoms have 18 hybrid orbitals in total
5. It is the fifth member of its homologous series
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1 — true: pent-2-ene. 2 — false: the carbons carry H/CH₃ and H/C₂H₅, so cis and trans forms exist.
3 — true: 4 C–C (one from the double bond) + 10 C–H = 14.
4 — true: 2 sp² · 3 + 3 sp³ · 4 = 6 + 12 = 18.
5 — false: the series starts with ethene (C₂ — first), so C₅ is the fourth member.
Answer: 1, 3, 4.
Making alkenes: Zaitsev’s rule
Industry makes alkenes by cracking oil products and by dehydrogenating alkanes: C₂H₆ → C₂H₄ + H₂ (Ni, t°). In the laboratory elimination reactions are used: dehydration of alcohols — C₂H₅OH → CH₂=CH₂ + H₂O (conc. H₂SO₄, t > 140 °C); dehydrohalogenation of haloalkanes with an alkali dissolved in alcohol — C₂H₅Br + KOH → CH₂=CH₂ + KBr + H₂O; dehalogenation of dihaloalkanes with zinc — CH₂Br–CH₂Br + Zn → CH₂=CH₂ + ZnBr₂. Over Pd, alkynes are hydrogenated only as far as alkenes: C₂H₂ + H₂ → C₂H₄.
When HX or H₂O is eliminated, the hydrogen leaves from the neighbouring carbon that has fewer hydrogens, so the more substituted alkene is the main product. 2-Bromobutane CH₃–CHBr–CH₂–CH₃ → mainly but-2-ene CH₃–CH=CH–CH₃ (a little but-1-ene).
Which is the main product when 3-methylpentan-3-ol (C₂H₅)₂C(OH)–CH₃ is heated with conc. H₂SO₄?
A) 3-methylpent-1-ene B) 3-methylpent-2-ene C) 2-ethylbut-1-ene D) 3-methylpentane E) 2-methylpent-2-ene
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Main chain 5 C, C=C at 2, methyl at 3 → 3-methylpent-2-ene. Loss of H from CH₃ would give the minor product 2-ethylbut-1-ene; D is an alkane, E has another skeleton and A cannot form from this alcohol.
Answer: B.
Reactions of alkenes: Markovnikov’s rule
Addition is typical of alkenes: the π bond breaks and two new σ bonds form. Hydrogenation: CH₂=CH–CH₃ + H₂ → CH₃–CH₂–CH₃ (Ni, t°). Halogenation: CH₂=CH₂ + Br₂ → CH₂Br–CH₂Br — yellow-orange bromine water is decolourised. Hydrohalogenation: CH₂=CH₂ + HCl → CH₃–CH₂Cl (chloroethane). Hydration (H⁺, t°, p): CH₂=CH₂ + H₂O → C₂H₅OH — industrial ethanol is made this way.
When HX or H₂O adds to an unsymmetrical alkene, the hydrogen goes to the carbon of the double bond that already has more hydrogens, and X (OH) to the other one: CH₂=CH–CH₃ + HBr → CH₃–CHBr–CH₃ (2-bromopropane); CH₂=CH–CH₃ + H₂O → CH₃–CH(OH)–CH₃ (propan-2-ol).
Oxidation. Aqueous KMnO₄ oxidises alkenes to diols — the purple colour disappears and brown MnO₂ settles (the Wagner reaction): 3CH₂=CH₂ + 2KMnO₄ + 4H₂O → 3HO–CH₂–CH₂–OH + 2MnO₂↓ + 2KOH. This and bromine water are the tests for unsaturation; alkanes give neither. Ethylene burns with a luminous flame: C₂H₄ + 3O₂ → 2CO₂ + 2H₂O. Polymerisation: nCH₂=CH₂ → (–CH₂–CH₂–)ₙ (polyethylene); propylene gives polypropylene (–CH₂–CH(CH₃)–)ₙ.
- n(alkene)amount of alkene that adds bromine, mol
- 160molar mass of Br₂, g/mol
One C=C adds one Br₂ (a diene or an alkyne adds two). Since an alkene has M = 14n, the formula follows from m / n.
1) 2.8 g of an alkene adds 8 g of bromine. Find its formula.
2) 4.48 L (STP) of a propane–propene mixture was passed through bromine water and 16 g of bromine was taken up. Find the volume fraction (%) of propene.
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2) Only propene takes up bromine: n(C₃H₆) = 16 / 160 = 0.1 mol. n(mixture) = 4.48 / 22.4 = 0.2 mol.
φ(C₃H₆) = 0.1 / 0.2 · 100% = 50%.
Which reaction gives 2-bromopropane as the main product?
A) CH₂=CH–CH₃ + HBr → B) CH₂=CH–CH₃ + Br₂ → C) CH₃–CH₂–CH₂OH + HBr → D) CH≡C–CH₃ + 2HBr → E) CH₂=CH₂ + CH₃Br →
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B gives 1,2-dibromopropane; C replaces OH by Br — 1-bromopropane; D adds both HBr to the same carbon (Markovnikov) — 2,2-dibromopropane; E does not happen.
Answer: A.
Dienes and rubber
Dienes have two C=C bonds. They can be next to each other (cumulated: CH₂=C=CH₂), separated by one single bond (conjugated: CH₂=CH–CH=CH₂) or further apart (isolated: CH₂=CH–CH₂–CH=CH₂). The important ones are conjugated: divinyl (buta-1,3-diene), isoprene (2-methylbuta-1,3-diene) CH₂=C(CH₃)–CH=CH₂ and chloroprene (2-chlorobuta-1,3-diene) CH₂=CCl–CH=CH₂. They give 1,2- and 1,4-addition: CH₂=CH–CH=CH₂ + Br₂ → CH₂Br–CH=CH–CH₂Br (1,4-dibromobut-2-ene); excess bromine gives CH₂Br–CHBr–CHBr–CH₂Br.
Divinyl is made from ethanol by S. V. Lebedev’s process (2C₂H₅OH → CH₂=CH–CH=CH₂ + 2H₂O + H₂; Al₂O₃, ZnO, t°) and by dehydrogenating butane (C₄H₁₀ → C₄H₆ + 2H₂); isoprene comes from 2-methylbutane, and chloroprene from adding HCl to vinylacetylene (CH≡C–CH=CH₂ + HCl → CH₂=CCl–CH=CH₂). Dienes polymerise to rubbers: nCH₂=CH–CH=CH₂ → (–CH₂–CH=CH–CH₂–)ₙ. Natural rubber is a polymer of isoprene; heating rubber with sulfur (vulcanisation) gives elastic rubber goods. Polymers have their own lesson.
Substance — Number of hydrogen atoms in the molecule
Alkane — a
Diene — a
The alkane is the 5th member of its homologous series. Calculate 1) a; 2) the number of carbon atoms in the diene.
3) Find the difference in molar mass between any alkyne and any alkane that have the same number of hydrogen atoms.
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2) Diene CₙH₂ₙ₋₂: 2n − 2 = 12 → n = 7 (C₇H₁₂).
3) Alkyne CₘH₂ₘ₋₂, alkane CₖH₂ₖ₊₂: 2m − 2 = 2k + 2 → m = k + 2. The mass difference is (14m − 2) − (14k + 2) = 14 · 2 − 4 = 24 — always the same (e.g. C₇H₁₂ = 96 and C₅H₁₂ = 72).
Alkynes: acetylene and its reactions
In the laboratory acetylene is made by adding water to calcium carbide: CaC₂ + 2H₂O → C₂H₂↑ + Ca(OH)₂ (the carbide itself comes from CaO + 3C → CaC₂ + CO in an electric furnace); industry cracks methane at 1500 °C: 2CH₄ → C₂H₂ + 3H₂. It also forms from dihaloalkanes: CH₂Br–CH₂Br + 2KOH → C₂H₂ + 2KBr + 2H₂O (in alcohol). Alkynes add in two steps: C₂H₂ + 2Br₂ → CHBr₂–CHBr₂ (bromine water is decolourised), C₂H₂ + HCl → CH₂=CHCl (vinyl chloride, the monomer of PVC). Unsymmetrical alkynes follow Markovnikov’s rule: CH≡C–CH₃ + 2HBr → CH₃–CBr₂–CH₃.
- Kucherov reaction (HgSO₄, H₂SO₄): C₂H₂ + H₂O → CH₃CHO (acetaldehyde); other alkynes give ketones: CH≡C–CH₃ + H₂O → CH₃–CO–CH₃ (acetone).
- Acetylides: the ≡C–H hydrogen is weakly acidic. With ammoniacal Ag₂O a precipitate forms: C₂H₂ + Ag₂O → Ag₂C₂↓ + H₂O; with sodium: 2C₂H₂ + 2Na → 2HC≡CNa + H₂. Internal alkynes (but-2-yne) do not react.
- Trimerisation (activated carbon, 600 °C): 3C₂H₂ → C₆H₆ — benzene (see “Aromatic hydrocarbons: benzene and its homologues”).
- Combustion: 2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O; in oxygen the flame is hotter than 3000 °C — used to weld and cut metals.
What volume of acetylene (STP) can be made from 80 g of technical calcium carbide that contains 20% impurities? What mass of acetaldehyde can this acetylene give by the Kucherov reaction? Mr(CaC₂) = 64, Mr(CH₃CHO) = 44
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CaC₂ + 2H₂O → C₂H₂ + Ca(OH)₂: n(C₂H₂) = 1 mol, V = 22.4 L.
C₂H₂ + H₂O → CH₃CHO: n(CH₃CHO) = 1 mol, m = 44 g.
0.5 mol of a propene–propyne mixture needs 0.8 mol of H₂ for complete hydrogenation. Calculate 1) the mass of propyne in the mixture (g); 2) the mass of precipitate (g) formed when the same mixture is passed through excess ammoniacal Ag₂O. Mr(C₃H₄) = 40, Mr(AgC≡C–CH₃) = 147
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1) m(C₃H₄) = 0.3 · 40 = 12 g.
2) Only propyne (it has ≡C–H) gives the precipitate: 2CH≡C–CH₃ + Ag₂O → 2AgC≡C–CH₃↓ + H₂O, n(precipitate) = 0.3 mol.
m = 0.3 · 147 = 44.1 g.
Key points
- Alkenes CₙH₂ₙ (C=C = σ + π, sp², 120°); alkynes and dienes CₙH₂ₙ₋₂ (C≡C = σ + 2π, sp, 180°); hybrid orbitals: sp³ — 4, sp² — 3, sp — 2.
- Preparation: dehydration of alcohols, haloalkane + alkali in alcohol (Zaitsev: H leaves the carbon with fewer hydrogens), cracking, dehydrogenation.
- Additions: H₂, Br₂ (bromine water decolourised), HX and H₂O (Markovnikov), KMnO₄ (a diol, decolourisation), polymerisation.
- Dienes give 1,2- and 1,4-addition and polymerise to rubbers (natural rubber is polyisoprene; vulcanisation gives rubber goods).
- Acetylene: CaC₂ + 2H₂O, 2CH₄ (1500 °C); Kucherov — an aldehyde from acetylene, ketones from other alkynes; alkynes with ≡C–H give acetylide precipitates with Ag₂O; 3C₂H₂ → C₆H₆.
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