S5 Chemistry – Unit 3: Alkenes and Alkynes

3.1

Structure of Alkenes and Alkynes

Alkenes Hydrocarbons containing at least one C=C double bond. General formula: C_nH_2n (for one double bond). They are unsaturated — they can undergo addition reactions.

Alkynes Hydrocarbons containing at least one C≡C triple bond. General formula: C_nH_2n−2 (for one triple bond). More unsaturated than alkenes.

Bonding in Alkenes — The C=C Double Bond

The C=C double bond consists of:

One σ (sigma) bond — formed by direct head-on overlap of sp² hybrid orbitals. This bond lies along the internuclear axis and can rotate freely.
One π (pi) bond — formed by sideways overlap of unhybridised p orbitals above and below the plane of the molecule. This bond prevents free rotation around the C=C axis, giving rise to geometric (cis/trans) isomerism.

Each carbon in C=C is sp² hybridised — trigonal planar, bond angles approximately 120°.

Bonding in Alkynes — The C≡C Triple Bond

The C≡C triple bond consists of:

One σ bond (sp hybrid orbital overlap, head-on).
Two π bonds (sideways overlap of two pairs of p orbitals, in perpendicular planes).

Each carbon in C≡C is sp hybridised — linear geometry, bond angle 180°.

Feature	Alkane (C–C)	Alkene (C=C)	Alkyne (C≡C)
Bond type	Single (σ)	Double (σ + π)	Triple (σ + 2π)
Hybridisation	sp³	sp²	sp
Bond angle	109.5°	~120°	180°
Geometry	Tetrahedral	Trigonal planar	Linear
C–C bond length	154 pm	134 pm	120 pm
C–C bond energy	347 kJ/mol	612 kJ/mol	837 kJ/mol
Rotation about bond?	Yes (free)	No (restricted)	No
General formula	C_nH_2n+2	C_nH_2n	C_nH_2n−2

3.2

Nomenclature

IUPAC Rules for Alkenes

Find the longest chain containing the C=C bond. Replace –ane with –ene.
Number the chain from the end nearest the double bond to give the double bond the lowest locant.
Position of double bond: cite the lower-numbered carbon of the C=C (e.g. but-1-ene, but-2-ene).
Substituents: named and numbered as with alkanes, with alphabetical ordering.
If two double bonds: –diene (e.g. buta-1,3-diene).

IUPAC Rules for Alkynes

Same as alkenes but replace –ane with –yne. The triple bond gets the lowest locant. If a molecule has both a double and triple bond, number to give the double bond the lower locant; suffix is –en–yne (e.g. pent-1-en-4-yne).

Name	Formula	Structure	Class
Ethene	C₂H₄	CH₂=CH₂	Alkene
Propene	C₃H₆	CH₂=CHCH₃	Alkene
But-1-ene	C₄H₈	CH₂=CHCH₂CH₃	Alkene
But-2-ene	C₄H₈	CH₃CH=CHCH₃	Alkene
2-methylpropene	C₄H₈	CH₂=C(CH₃)₂	Alkene
Ethyne (acetylene)	C₂H₂	HC≡CH	Alkyne
Propyne	C₃H₄	HC≡CCH₃	Alkyne
But-1-yne	C₄H₆	HC≡CCH₂CH₃	Alkyne
But-2-yne	C₄H₆	CH₃C≡CCH₃	Alkyne

Example 1

Name the Alkene: CH₃CH=CHCH₂CH₃

Longest chain containing C=C: 5 carbons → parent: pent-

Number from left: C=C is between C2 and C3 (locant = 2). From right: locant = 3. Choose lower → 2.

No substituents.

✓

Name: pent-2-ene.

Example 2

Name: CH₃C(CH₃)=CHCH₃

Longest chain through C=C: C1–C2=C3–C4 → 4 carbons → but-

Number from right to give C=C lowest locant: C=C is between C2 and C3; from left = locant 2, from right = locant 2. Either gives 2.

Methyl branch at C2 → 2-methyl.

✓

Name: 2-methylbut-2-ene.

3.3

Isomerism in Alkenes and Alkynes

A. Structural (Position) Isomerism

Structural isomers of alkenes can differ in the position of the double bond along the chain (position isomers) or in the carbon skeleton (chain isomers).

Example: C₄H₈ has four structural isomers: but-1-ene, but-2-ene, 2-methylpropene, and cyclobutane (cyclic isomer).

B. Geometric (Cis/Trans) Isomerism

Geometric Isomerism Arises in alkenes when rotation about the C=C bond is restricted (by the π bond) and each carbon of the double bond carries two different groups. The two isomers differ in the spatial arrangement of groups around the double bond.

cis- isomer

The two identical (or similar priority) groups are on the same side of the double bond.

Example: cis-but-2-ene — both CH₃ groups on the same side. Higher boiling point than trans isomer (molecular dipoles reinforce).

trans- isomer

The two identical (or similar priority) groups are on opposite sides of the double bond.

Example: trans-but-2-ene — the two CH₃ groups on opposite sides. More stable (less steric strain).

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Condition for Geometric Isomerism Each carbon of the C=C must have two different substituents. If either carbon has two identical groups (e.g. CH₂=), geometric isomers do not exist. For example: but-1-ene (CH₂=CHCH₂CH₃) has no geometric isomers because C1 carries two H atoms.

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E/Z Notation (IUPAC) Modern IUPAC uses E (from German entgegen, opposite) and Z (from zusammen, together) based on Cahn-Ingold-Prelog priority rules. Higher priority groups on the same side → Z; on opposite sides → E. For simple cases, Z = cis and E = trans, but not always.

Example 3

Identify Geometric Isomers of But-2-ene

Does but-2-ene (CH₃CH=CHCH₃) show geometric isomerism? If yes, draw and name both isomers.

Check condition: C2 carries H and CH₃ → two different groups ✔. C3 carries H and CH₃ → two different groups ✔. Geometric isomers exist.

cis-but-2-ene: both CH₃ on same side of C=C. B.P. = 3.7°C.

trans-but-2-ene: CH₃ groups on opposite sides. B.P. = 0.9°C.

✓

cis isomer has higher B.P. because its net molecular dipole is non-zero (dipoles don't cancel), giving slightly stronger intermolecular forces.

3.4

Preparation of Alkenes and Alkynes

Preparation of Alkenes

Method 1: Dehydration of Alcohols

An alcohol is heated with concentrated H₂SO₄ or Al₂O₃ at high temperature. An H and OH are eliminated from adjacent carbons (elimination reaction).

CH3CH2OH --conc. H2SO4, 170 degC--> CH2=CH2 + H2O (ethene) CH3CH(OH)CH3 --conc. H2SO4, 170 degC--> CH3CH=CH2 + H2O (propene)

At lower temperatures (~140°C with H₂SO₄), the ether is formed preferentially. At higher temperatures (~170°C), the alkene is formed.

Method 2: Dehydrohalogenation of Halogenoalkanes

A halogenoalkane is heated with an alcoholic KOH solution (elimination conditions). HX is removed.

CH3CH2Br + KOH(alc) --heat--> CH2=CH2 + KBr + H2O (ethene)

Aqueous KOH gives substitution (alcohol); alcoholic KOH gives elimination (alkene). Zaitsev's rule: the more substituted (more stable) alkene is the major product.

Method 3: Cracking of Alkanes

Thermal or catalytic cracking of large alkanes gives a mixture of alkenes and shorter alkanes (see Unit 2).

C4H10 --high T or catalyst--> C2H4 + C2H6

Preparation of Alkynes

Method 1: Double Dehydrohalogenation

A vicinal (on adjacent C) or geminal (on same C) dihalide reacts with excess alcoholic KOH:

CH3CHBrCHBrCH3 + 2KOH(alc) --> CH3C≡CCH3 + 2KBr + 2H2O (2,3-dibromobutane) (but-2-yne)

Method 2: Preparation of Ethyne (Acetylene)

Industrially, ethyne is made by the reaction of calcium carbide with water:

CaC2 + 2H2O --> Ca(OH)2 + HC≡CH (calcium carbide) (ethyne / acetylene)

Calcium carbide is itself made by heating CaO with coke (carbon) at 2000°C: CaO + 3C → CaC₂ + CO.

Example 4

Preparation of Propene from Propanol

Write the equation and state the conditions for preparing propene from propan-1-ol.

Reaction type: dehydration (elimination of water).

Reagent: concentrated H₂SO₄ (or Al₂O₃). Temperature: ~170°C.

CH₃CH₂CH₂OH →^{conc. H₂SO₄, 170°C} CH₃CH=CH₂ + H₂O

✓

Product: propene ✔

3.5

Physical Properties

States, Boiling Points, and Solubility

Alkenes and alkynes have similar physical properties to alkanes of the same carbon number: non-polar (or very slightly polar for alkynes due to the sp C–H bond), low boiling points, insoluble in water, soluble in non-polar solvents.

C₂–C₄: gases; C₅–C₁₇: liquids; C₁₈₊: solids.

Compound	Formula	B.P. (°C)	State
Ethene	C₂H₄	−104	Gas
Propene	C₃H₆	−47	Gas
But-1-ene	C₄H₈	−6	Gas
Pent-1-ene	C₅H₁₀	+30	Liquid
Ethyne	C₂H₂	−84	Gas
Propyne	C₃H₄	−23	Gas
But-1-yne	C₄H₆	+8	Gas/Liquid

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Acidic Character of Terminal Alkynes The sp C–H bond in terminal alkynes (HC≡C–R) is more polar than C–H bonds in alkanes or alkenes, making the H slightly acidic. Terminal alkynes react with strong bases (e.g. NaNH₂) or with silver/copper(I) ions to form precipitates: HC≡CR + AgNO₃(NH₃) → AgC≡CR (white ppt). This is a chemical test for terminal alkynes.

3.6

Chemical Reactions

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Key Difference from Alkanes The π electrons of C=C and C≡C make alkenes and alkynes electron-rich. They react readily with electrophiles (electron-seeking species) by electrophilic addition. They do NOT undergo free-radical substitution under normal conditions.

Reaction 1: Addition of Hydrogen (Hydrogenation)

Alkene: CH2=CH2 + H2 --Ni/Pt/Pd, heat--> CH3CH3 Alkyne: HC≡CH + H2 --Ni, heat--> CH2=CH2 (1 mole H2) HC≡CH + 2H2 --Ni, heat--> CH3CH3 (2 moles H2)

Alkynes undergo partial (1 mole H₂) or complete (2 moles H₂) hydrogenation. Using a Lindlar catalyst (Pd/CaCO₃/quinoline), only the cis alkene is formed.

Reaction 2: Addition of Halogens (Halogenation)

CH2=CH2 + Br2 --> CH2Br-CH2Br (1,2-dibromoethane) HC≡CH + Br2 --> CHBr=CHBr (1,2-dibromoethene) [1 mol Br2] HC≡CH + 2Br2 --> CHBr2-CHBr2 (1,1,2,2-tetrabromoethane) [2 mol Br2]

Test for unsaturation: alkenes and alkynes decolourise bromine water (orange/brown → colourless). Alkanes do not react with bromine water in the dark.

Reaction 3: Addition of Hydrogen Halides (HX)

CH2=CH2 + HBr --> CH3CH2Br (bromoethane) CH3CH=CH2 + HBr --> CH3CHBrCH3 (major, Markovnikov) or CH3CH2CH2Br (minor, anti-Markovnikov)

Markovnikov's Rule When HX adds across an unsymmetrical double bond, the hydrogen adds to the carbon already carrying more hydrogens (the less substituted carbon), and the halide adds to the more substituted carbon. Equivalently: the major product is the more substituted halogenoalkane.

Modern explanation: the more substituted carbocation intermediate is more stable (tertiary > secondary > primary). The reaction proceeds through the more stable intermediate.

Example 5

Apply Markovnikov's Rule to Propene + HBr

Predict the major product when HBr adds to propene (CH₃CH=CH₂).

Identify the two possible products:
(A) CH₃CHBrCH₃ — Br on C2 (secondary carbon)
(B) CH₃CH₂CH₂Br — Br on C3 (primary carbon)

Apply Markovnikov: H adds to CH₂ (2 H already); Br adds to CH (1 H already). → Major product: (A) 2-bromopropane.

Via carbocations: secondary carbocation (CH₃•⁺CHCH₃) is more stable than primary (CH₃CH₂•⁺CH₂), so (A) forms faster.

✓

Major product: 2-bromopropane ✔

Reaction 4: Addition of Water (Hydration)

CH2=CH2 + H2O --conc. H3PO4, 300 degC, 60 atm--> CH3CH2OH (ethanol -- industrial route)

In the laboratory, dilute H₂SO₄ or H₃PO₄ can catalyse hydration. Markovnikov's rule applies for unsymmetrical alkenes.

Reaction 5: Oxidation — KMnO₄ Test

Cold dilute KMnO₄ (Baeyer's reagent, purple) oxidises the C=C to give a diol, and the solution decolourises to colourless (MnO₄⁻ → MnO₂ or Mn²⁺). Alkanes do not decolourise cold KMnO₄.

CH2=CH2 + [O] + H2O --cold KMnO4--> CH2(OH)-CH2(OH) (ethane-1,2-diol)

Hot concentrated KMnO₄ cleaves the double bond, giving carboxylic acids and/or ketones — useful for structure determination.

Reaction 6: Combustion

C2H4 + 3O2 --> 2CO2 + 2H2O (complete) 2C2H2 + 5O2 --> 4CO2 + 2H2O (complete combustion of ethyne) Note: alkenes and alkynes burn with a smokier, more luminous flame than alkanes due to higher carbon content (lower H:C ratio).

Reaction 7: Polymerisation (Alkenes only)

Under high pressure and temperature with a catalyst, alkene molecules join together to form addition polymers:

n CH2=CH2 --high P, T, Ziegler-Natta catalyst--> [-CH2-CH2-]n (ethene) (poly(ethene) / polyethylene)

Other important polymers: poly(propene) from propene; PVC from chloroethene; PTFE from tetrafluoroethene; polystyrene from styrene (phenylethene).

Example 6

Complete Combustion of Propyne

Write the balanced equation for the complete combustion of propyne (C₃H₄).

General formula for alkynes: C_nH_2n−2. For n=3: C₃H₄.

Products: 3 CO₂ and 2 H₂O. O on right: 3×2 + 2×1 = 8 O atoms → 4 O₂.

✓

C₃H₄ + 4 O₂ → 3 CO₂ + 2 H₂O

3.7

Uses of Alkenes and Alkynes

Compound	Use	Notes
Ethene (CH₂=CH₂)	Manufacture of polyethylene (LDPE, HDPE)	Most widely produced organic chemical
Ethene	Production of ethanol (hydration), ethylene oxide, ethanal	Key petrochemical feedstock
Ethene	Fruit ripening agent (plant hormone)	Accelerates ripening in stored fruit
Propene	Polypropylene manufacture	Used in packaging, textiles, automotive parts
Propene	Manufacture of acetone (propanone) and glycerol	Industrial chemical synthesis
Buta-1,3-diene	Synthetic rubber (polybutadiene, styrene-butadiene rubber SBR)	Tyres, footwear
Chloroethene (vinyl chloride)	PVC (poly(vinyl chloride)) manufacture	Pipes, flooring, insulation
Ethyne (HC≡CH)	Oxy-acetylene welding and cutting (flame ~3100°C)	Extremely high flame temperature
Ethyne	Synthesis of vinyl chloride (then PVC), vinyl acetate, acrylonitrile	Important industrial precursor
Ethyne	Formerly used as anaesthetic (now replaced)	Historical use only

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Exercises

Give the IUPAC name for each compound:
(a) CH₂=CHCH₂CH₃ (b) CH₃CH=CHCH₂CH₃ (c) HC≡CCH₂CH₃

(a) but-1-ene (C=C at C1; 4-carbon chain)
(b) pent-2-ene (C=C at C2; 5-carbon chain; number from left gives C2)
(c) but-1-yne (triple bond at C1; 4-carbon chain)
Does pent-2-ene show geometric isomerism? Explain your answer and, if yes, name and describe both isomers.

Yes. In CH₃CH=CHCH₂CH₃ (pent-2-ene): C2 carries H and CH₃ (two different groups ✔); C3 carries H and CH₂CH₃ (two different groups ✔). Rotation about C=C is restricted by the π bond.

cis-pent-2-ene: CH₃ and CH₂CH₃ on the same side.
trans-pent-2-ene: CH₃ and CH₂CH₃ on opposite sides.
Write the equation for the dehydration of ethanol to produce ethene. State the reagent and conditions.

CH₃CH₂OH →^{conc. H₂SO₄, 170°C} CH₂=CH₂ + H₂O
Reagent: concentrated H₂SO₄ (or Al₂O₃). Temperature: ~170°C.
Predict the major product when HI adds to 2-methylpropene (CH₂=C(CH₃)₂). Justify using Markovnikov's rule.

Major product: 2-iodo-2-methylpropane, (CH₃)₃CI.
Markovnikov's rule: H adds to C1 (=CH₂, has 2 H already); I adds to C2 (=C(CH₃)₂, more substituted).
The tertiary carbocation (CH₃)₃C⁺ is more stable than the primary carbocation CH₂I–C(CH₃)₂, so the reaction proceeds through the more stable intermediate.
A student treats an unknown compound with bromine water. The orange colour is discharged immediately. What class of compound is likely present? What further test would distinguish an alkene from an alkyne?

Immediate decolourisation of bromine water indicates an unsaturated compound (alkene or alkyne) — the compound adds across the π bond(s).

To distinguish: add ammoniacal silver nitrate solution (Tollens’ reagent without the aldehyde context — use AgNO₃/NH₃). A terminal alkyne gives a white/cream precipitate (silver acetylide, AgC≡CR). An alkene gives no precipitate.
Alternatively: if the compound reacts with only 1 mole of Br₂ it is an alkene; if 2 moles of Br₂ it is an alkyne.
Write the balanced equation for the preparation of ethyne from calcium carbide. State one industrial use of ethyne.

CaC₂ + 2H₂O → Ca(OH)₂ + HC≡CH

Industrial use: oxy-acetylene welding and cutting (flame temperature ~3100°C). Also: production of vinyl chloride (monomer for PVC).

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Unit Test

ℹ️

Instructions Total: 50 marks | Time: 50 minutes | Attempt all questions | Show all working.

Section A — Short Answer

30 marks

Q1 [4 marks]

Give the IUPAC name for each compound:
(a) CH₂=C(CH₃)CH₂CH₃ (b) CH₃CH=C(CH₃)CH₂CH₃ (c) HC≡CCH(CH₃)₂ (d) CH₃C≡CCH₂CH₃

(a) 2-methylbut-1-ene (longest chain = 4 C through double bond; methyl on C2; double bond at C1)
(b) 3-methylpent-2-ene (longest chain = 5 C; double bond between C2–C3; methyl on C3)
(c) 3-methylbut-1-yne (longest chain = 4 C; triple bond at C1; methyl on C3)
(d) pent-2-yne (longest chain = 5 C; triple bond between C2–C3)

Q2 [4 marks]

Explain, with reference to orbital theory, why:

(a) free rotation is possible around the C–C single bond in ethane but not around the C=C bond in ethene. [2]

(b) the bond angle in ethene (~120°) differs from that in ethane (~109.5°). [2]

(a) In ethane, the C–C σ bond is cylindrically symmetrical — rotation does not break or weaken it. In ethene, the C=C contains both a σ bond AND a π bond. The π bond is formed by sideways p-orbital overlap above and below the molecular plane; rotation would break this overlap and destroy the π bond. Therefore rotation is restricted.

(b) In ethane, each carbon is sp³ hybridised (4 electron pairs arranged tetrahedrally → 109.5°). In ethene, each carbon is sp² hybridised (3 electron groups arranged in a plane → ~120°).

Q3 [4 marks]

(a) State the condition required for geometric isomerism to exist in an alkene. [1]

(b) Draw and name the two geometric isomers of but-2-ene. [2]

(a) Each carbon of the C=C must bear two different substituents (and rotation about C=C is restricted).

(b) cis-but-2-ene: both CH₃ groups on the same side of the C=C.
trans-but-2-ene: CH₃ groups on opposite sides of the C=C.

(c) In cis-but-2-ene, the molecular dipoles from the two C–CH₃ bonds reinforce each other (net dipole ≠ 0), giving slightly stronger intermolecular attractions and a higher boiling point. In the trans isomer, the dipoles cancel (molecule is symmetric, net dipole ≈ 0).

Q4 [5 marks]

Write equations, with reagents and conditions, for three different methods of preparing alkenes from other organic compounds.

1. Dehydration of alcohol:
CH₃CH₂OH →^{conc. H₂SO₄, 170°C} CH₂=CH₂ + H₂O

2. Dehydrohalogenation:
CH₃CH₂Br + KOH(alc) →^heat CH₂=CH₂ + KBr + H₂O

3. Cracking of alkanes:
C₄H₁₀ →^{high T, cat.} C₂H₄ + C₂H₆

Q5 [5 marks]

Predict the major and minor products when HBr reacts with but-1-ene (CH₂=CHCH₂CH₃). Write equations for both and explain the product distribution using carbocation stability.

Major product: 2-bromobutane (CH₃CHBrCH₂CH₃)
CH₂=CHCH₂CH₃ + HBr → CH₃CHBrCH₂CH₃

Minor product: 1-bromobutane (CH₂BrCH₂CH₂CH₃)
CH₂=CHCH₂CH₃ + HBr → CH₂BrCH₂CH₂CH₃

Explanation: The intermediate for the major product is a secondary carbocation (CH₃•⁺CHCH₂CH₃), which is more stable than the primary carbocation (CH₂•⁺CH₂CH₂CH₃) for the minor product. Stability: 3° > 2° > 1° carbocation. The reaction proceeds preferentially through the more stable intermediate.

Q6 [4 marks]

Write balanced equations for the following reactions of ethyne:
(a) Complete combustion (b) Addition of 2 moles of HCl (c) Addition of 1 mole of Br₂ (d) Reaction with water (Markovnikov product only)

(a) 2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O
(b) HC≡CH + 2HCl → CH₃CHCl₂ (1,1-dichloroethane)
(c) HC≡CH + Br₂ → CHBr=CHBr (1,2-dibromoethene)
(d) HC≡CH + H₂O →^{H₂SO₄/HgSO₄} CH₃CHO (ethanal, via vinyl alcohol tautomerism)

Q7 [4 marks]

Describe two chemical tests that can be used to distinguish an alkene from an alkane. For each test, state the reagent, observation with the alkene, and observation with the alkane.

Test 1 — Bromine water:
Reagent: bromine water (orange/brown)
Alkene: immediate decolourisation (orange → colourless) — Br₂ adds across C=C.
Alkane: no change in colour (no reaction in the dark).

Test 2 — Cold dilute KMnO₄ (Baeyer’s reagent):
Reagent: cold dilute KMnO₄ solution (purple)
Alkene: purple colour discharged (decolourises) — diol formed by oxidation of C=C.
Alkane: no colour change (alkanes not oxidised by cold KMnO₄).

Section B — Extended Response

20 marks

Q8 [10 marks]

(a) Compare the structure, bonding, and geometry of ethene and ethyne. Include hybridisation, bond angles, bond lengths, and the number and type of bonds. [6 marks]

(b) Starting from ethyne, describe with equations how you could prepare: (i) ethene; (ii) 1,1,2,2-tetrabromoethane. [4 marks]

(a)
Ethene (CH₂=CH₂): each C is sp² hybridised. The C=C consists of one σ bond (sp²–sp² overlap) and one π bond (p–p sideways overlap). Bond angle ~120°, trigonal planar. C=C bond length = 134 pm. Molecule is planar; restricted rotation about C=C.

Ethyne (HC≡CH): each C is sp hybridised. The C≡C consists of one σ bond and two π bonds (in perpendicular planes). Bond angle = 180°, linear. C≡C bond length = 120 pm (shorter than C=C). No rotation issue (already symmetric).

Comparison: as bond order increases (1 → 2 → 3): bond length decreases, bond energy increases, bond angle changes (109.5° → 120° → 180°), geometry changes (tetrahedral → trigonal planar → linear).

(b)(i) Ethene from ethyne:
HC≡CH + H₂ →^{Lindlar catalyst (Pd/CaCO₃)} CH₂=CH₂
(Lindlar catalyst ensures only 1 mole H₂ adds, giving cis-alkene; regular Ni would give ethane with excess H₂.)

(b)(ii) 1,1,2,2-tetrabromoethane from ethyne:
HC≡CH + 2Br₂ → CHBr₂–CHBr₂ (1,1,2,2-tetrabromoethane)

Q9 [10 marks]

(a) State Markovnikov's rule and explain it in terms of carbocation stability. Use the addition of HBr to 2-methylpropene as your example. [6 marks]

(b) Alkenes undergo addition polymerisation. Write the repeat unit of poly(propene) and name two industrial uses of this polymer. [4 marks]

(a)
Markovnikov's rule: When HX adds to an unsymmetrical alkene, the hydrogen goes to the carbon already bearing more hydrogen atoms (the less substituted carbon), and X goes to the more substituted carbon.

Mechanism for 2-methylpropene + HBr:
2-methylpropene: CH₂=C(CH₃)₂
Step 1 (H⁺ adds): Two possible carbocations:
• H⁺ to C2 → CH₃–•⁺CH–CH₃ = secondary carbocation
• H⁺ to C1 → ⁺C(CH₃)₃ = tertiary carbocation ← more stable
Step 2 (Br⁻ attacks carbocation): Br⁻ + (CH₃)₃C⁺ → (CH₃)₃CBr
Major product: 2-bromo-2-methylpropane

Stability order: 3° > 2° > 1° > methyl carbocation. Tertiary carbocations have three alkyl groups donating electron density, dispersing the positive charge and increasing stability.

(b) Poly(propene) repeat unit: [–CH₂–CH(CH₃)–]_n
From: n CH₂=CHCH₃ → [–CH₂–CH(CH₃)–]_n

Industrial uses: (any two) food packaging (yogurt pots, crisp packets); carpet fibres and ropes; automotive bumpers and trim; laboratory containers; medical syringes and equipment.

← Unit 2: Alkanes Unit 4: Halogenoalkanes →