Hydrocarbons

A hydrocarbon contains carbon and hydrogen only. Everything else in organic chemistry is a hydrocarbon with something bolted onto it, which is why this unit comes first among the families. Three families are studied: alkanes (CₙH₂ₙ₊₂, all single bonds, saturated), alkenes (CₙH₂ₙ, one C=C) and alkynes (CₙH₂ₙ₋₂, one C≡C). The plan is always the same — how each is made, how each reacts, and how you tell them apart in a test tube.

14.1 Alkanes: preparation from haloalkanes (reduction and Wurtz reaction), decarboxylation, catalytic hydrogenation of alkene and alkyne

Alkanes are also called paraffins (Latin parum affinis, "little affinity") because they are so unreactive. Four laboratory preparations are on the syllabus.

(a) Reduction of haloalkanes

Nascent hydrogen, generated in the flask by zinc and acid, replaces the halogen by hydrogen. The alkane keeps the same number of carbons as the halide.

CH₃CH₂Cl + H₂ --Zn/HCl--> CH₃CH₃ + HCl

CH₃CH₂I + HI --red P, 423 K--> CH₃CH₃ + I₂

Zinc and ethanoic acid, zinc–copper couple in ethanol, or lithium aluminium hydride LiAlH₄ in dry ether all do the same job. The order of ease is R—I > R—Br > R—Cl, because the C—I bond is the weakest.

(b) Wurtz reaction

Two molecules of haloalkane are coupled by metallic sodium in dry ether. The alkane produced has twice the number of carbons of the halide.

2CH₃CH₂Br + 2Na --dry ether--> CH₃CH₂CH₂CH₃ + 2NaBr

(c) Decarboxylation (soda-lime method)

The sodium salt of a carboxylic acid is heated with soda lime (NaOH + CaO, in the ratio 3:1). Carbon dioxide leaves as carbonate, so the alkane has one carbon fewer than the acid.

CH₃COONa + NaOH --CaO, Δ--> CH₄ + Na₂CO₃

CH₃CH₂COONa + NaOH --CaO, Δ--> CH₃CH₃ + Na₂CO₃

Calcium oxide is added to keep the very hygroscopic sodium hydroxide dry and to stop it attacking the glass.

(d) Catalytic hydrogenation (Sabatier–Senderens reduction)

An alkene or alkyne takes up hydrogen over a finely divided metal catalyst.

CH₂=CH₂ + H₂ --Ni, 523–573 K--> CH₃CH₃

CH≡CH + 2H₂ --Ni, 523–573 K--> CH₃CH₃

Nickel needs 250–300 °C; platinum or palladium work at room temperature. This is the reaction used industrially to turn liquid vegetable oils into vanaspati ghee.

Method	Reagent and condition	Carbon count	Note
Reduction of R—X	Zn/HCl, or Zn + CH₃COOH, or LiAlH₄	unchanged	ease R—I > R—Br > R—Cl
Wurtz reaction	2R—X + 2Na, dry ether	doubled	symmetrical alkanes only
Decarboxylation	RCOONa + NaOH/CaO, Δ	one fewer	good for CH₄ from CH₃COONa
Hydrogenation	H₂, Ni at 523–573 K (or Pt/Pd, 298 K)	unchanged	used for vanaspati ghee
Kolbe's electrolysis	conc. aq. RCOONa, electrolysis	doubled, minus 2	Section 14.9

14.2 Alkanes chemical properties: substitution (halogenation, nitration, sulphonation); oxidation of ethane

Alkanes have only strong, non-polar σ bonds, so they are attacked neither by electrophiles nor by nucleophiles. Their characteristic reaction is free-radical substitution, which needs heat or ultraviolet light to get started.

Halogenation

CH₄ + Cl₂ --hν / 573 K--> CH₃Cl + HCl

CH₃Cl + Cl₂ --hν--> CH₂Cl₂ + HCl → CHCl₃ → CCl₄

The reaction does not stop at the first product: with excess chlorine all four hydrogens are replaced in turn. Reactivity of the halogens falls sharply, F₂ > Cl₂ > Br₂ > I₂. Fluorination is explosive and iodination is reversible (HI reduces the product back), so iodination needs an oxidising agent such as HIO₃ or HNO₃ to destroy the HI as it forms.

Nitration

In the vapour phase at 673–773 K an alkane reacts with nitric acid vapour. The reaction is also free-radical, and it gives a mixture, because C—C bonds break as well.

CH₄ + HNO₃ --673–773 K--> CH₃NO₂ + H₂O

CH₃CH₃ + HNO₃ --673 K--> CH₃CH₂NO₂ + H₂O

Sulphonation

An alkane heated with oleum (fuming sulphuric acid, H₂SO₄ + SO₃) gives a sulphonic acid. Only alkanes from hexane upwards, and especially those with a tertiary hydrogen, react at a useful rate:

CH₃(CH₂)₄CH₃ + HO—SO₃H --oleum, Δ--> CH₃(CH₂)₄CH₂—SO₃H + H₂O

Methane, ethane and propane do not sulphonate — a favourite one-mark point.

Oxidation of ethane

Condition	Equation	Product
Excess O₂, ignite	2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O	complete combustion, ΔH negative
O₂, Cu tube, 473 K, 100 atm	2C₂H₆ + O₂ → 2CH₃CH₂OH	ethanol
O₂, Mo₂O₃ catalyst	C₂H₆ + O₂ → CH₃CHO + H₂O	ethanal
O₂, manganese(II) ethanoate, Δ	2C₂H₆ + 3O₂ → 2CH₃COOH + 2H₂O	ethanoic acid

Combustion is the reason alkanes matter economically: LPG (a mixture of propane and butane) is the commonest cooking fuel in Nepali kitchens. The complete combustion of butane is

2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

With a limited air supply the products are carbon monoxide and soot instead, which is why a badly ventilated kitchen is dangerous.

14.3 Alkenes: preparation by dehydration of alcohol, dehydrohalogenation, catalytic hydrogenation of alkyne

Alkenes, or olefins, are made by elimination — pulling two groups off adjacent carbons so that a π bond can form.

(a) Dehydration of alcohols

An alcohol loses water over concentrated sulphuric acid at 443 K, or over alumina at 623 K.

CH₃CH₂OH --conc. H₂SO₄, 443 K--> CH₂=CH₂ + H₂O

The ease of dehydration follows carbocation stability: 3° > 2° > 1°. A tertiary alcohol dehydrates with 20 % H₂SO₄ at 358 K, a primary one needs 95 % acid at 443 K.

(b) Dehydrohalogenation

A haloalkane loses HX when warmed with alcoholic potassium hydroxide.

CH₃CH₂Br + KOH (alcoholic) --Δ--> CH₂=CH₂ + KBr + H₂O

When more than one alkene is possible, Saytzeff's rule decides:

CH₃CH₂CHBrCH₃ + KOH (alc.) → CH₃CH=CHCH₃ (but-2-ene, 81 %) + CH₂=CHCH₂CH₃ (but-1-ene, 19 %)

The ease of elimination is again 3° > 2° > 1°, and R—I > R—Br > R—Cl.

(c) Catalytic hydrogenation of an alkyne

Adding only one molecule of hydrogen stops at the alkene, provided the catalyst is poisoned so that it cannot go further.

CH≡CH + H₂ --Lindlar's catalyst--> CH₂=CH₂

Lindlar's catalyst is palladium deposited on calcium carbonate and partly deactivated with quinoline or lead ethanoate. Both hydrogens arrive on the same face of the triple bond, so an internal alkyne gives the cis alkene. Sodium in liquid ammonia at 195 K does the opposite and gives the trans alkene.

14.4 Alkenes chemical properties: addition with HX (Markovnikov's rule and peroxide effect), H₂O, O₃, H₂SO₄

The π bond of an alkene is a loose, exposed cloud of electrons above and below the C=C axis. It is therefore a nucleophile, and the characteristic reaction of an alkene is electrophilic addition.

Addition of hydrogen halides

CH₂=CH₂ + HBr → CH₃CH₂Br

Reactivity falls in the order HI > HBr > HCl, which is the order of decreasing acid strength — the easier the H—X bond gives up H⁺, the faster the addition.

With an unsymmetrical alkene there are two possible products, and only one of them forms in quantity.

CH₃CH=CH₂ + HBr → CH₃CHBrCH₃ (2-bromopropane, major)

The rule is not a rule at all but a consequence of carbocation stability: the proton adds so as to give the more stable carbocation, and a secondary cation beats a primary one because two +I alkyl groups spread its charge (Section 13.7).

Peroxide effect (Kharasch effect)

In the presence of benzoyl peroxide or simply of air-formed peroxides, the addition of HBr goes the other way round:

CH₃CH=CH₂ + HBr --(C₆H₅CO)₂O₂--> CH₃CH₂CH₂Br (1-bromopropane)

The peroxide provides radicals, so the mechanism becomes free-radical instead of ionic. Br• adds first, and it adds to the terminal carbon because that gives the more stable secondary free radical. The result looks like anti-Markovnikov addition.

Addition of water (hydration)

Water adds across the double bond in the presence of an acid catalyst, and the addition follows Markovnikov's rule.

CH₂=CH₂ + H₂O --H₃PO₄, 573 K, 60 atm--> CH₃CH₂OH

CH₃CH=CH₂ + H₂O --dil. H₂SO₄--> CH₃CH(OH)CH₃ (propan-2-ol)

Addition of sulphuric acid

Cold concentrated sulphuric acid adds to give an alkyl hydrogen sulphate, which on boiling with water hydrolyses to the alcohol. The two steps together are the industrial indirect hydration of alkenes.

CH₂=CH₂ + H₂SO₄ (cold, conc.) → CH₃CH₂—OSO₃H (ethyl hydrogen sulphate)

CH₃CH₂—OSO₃H + H₂O --Δ--> CH₃CH₂OH + H₂SO₄

With propene, Markovnikov's rule puts the —OSO₃H on C2, so the alcohol obtained is propan-2-ol, not propan-1-ol.

Ozonolysis

Ozone adds across the double bond to give an unstable ozonide, which is then split by zinc dust and water. The zinc is essential — it destroys the hydrogen peroxide that would otherwise oxidise the products further.

CH₃CH=CHCH₃ --(i) O₃ (ii) Zn/H₂O--> 2CH₃CHO

CH₂=CH₂ --(i) O₃ (ii) Zn/H₂O--> 2HCHO

CH₃CH=CH₂ --(i) O₃ (ii) Zn/H₂O--> CH₃CHO + HCHO

Because each doubly bonded carbon becomes a carbonyl carbon, ozonolysis is the classical method for locating a double bond: identify the two carbonyl compounds and join them back together at the oxygen atoms.

14.5 Alkynes: preparation from carbon and hydrogen, 1,2-dibromoethane, chloroform/iodoform

(a) From the elements

An electric arc is struck between two carbon electrodes in an atmosphere of hydrogen. This is the only common organic compound made straight from its elements.

2C + H₂ --electric arc, 3000 °C--> C₂H₂

The mixture must be chilled at once, because ethyne is thermodynamically unstable with respect to its elements (ΔH°f is positive, +227 kJ mol⁻¹) and would decompose again if left hot.

(b) From 1,2-dibromoethane

Two molecules of HBr are eliminated by alcoholic potassium hydroxide.

CH₂Br—CH₂Br + 2KOH (alc.) --Δ--> CH≡CH + 2KBr + 2H₂O

The first elimination is easy; the second is difficult, because a vinyl halide is unreactive, so sodamide NaNH₂ at 423 K is often used for the second step:

CH₂Br—CH₂Br + KOH (alc.) → CH₂=CHBr + KBr + H₂O

CH₂=CHBr + NaNH₂ --Δ--> CH≡CH + NaBr + NH₃

(c) From chloroform or iodoform

Heating trihalomethane with silver powder removes all six halogen atoms:

2CHCl₃ + 6Ag --Δ--> CH≡CH + 6AgCl

2CHI₃ + 6Ag --Δ--> CH≡CH + 6AgI

(d) From calcium carbide — the industrial and laboratory route

Most ethyne actually used, including the gas in a gas-welding cylinder, comes from calcium carbide and water:

CaO + 3C --2300 K, electric furnace--> CaC₂ + CO

CaC₂ + 2H₂O → C₂H₂↑ + Ca(OH)₂

14.6 Alkynes chemical properties: addition with H₂, HX, H₂O; acidic nature (sodium, ammoniacal AgNO₃, ammoniacal Cu₂Cl₂)

An alkyne has two π bonds, so it adds two molecules of most reagents, one after the other.

Addition of hydrogen

CH≡CH + H₂ --Ni, 473 K--> CH₂=CH₂ --H₂, Ni--> CH₃CH₃

Addition of hydrogen halides

Markovnikov's rule applies at each stage, so both halogens end up on the same carbon:

CH≡CH + HBr → CH₂=CHBr (bromoethene) --HBr--> CH₃CHBr₂ (1,1-dibromoethane)

CH≡CH + HCl --Hg²⁺, 470 K--> CH₂=CHCl (vinyl chloride, the monomer of PVC)

Addition of water

Water adds only in the presence of mercury(II) sulphate in dilute sulphuric acid. The first product is an enol, which tautomerises at once to the carbonyl compound (Section 13.4).

CH≡CH + H₂O --40 % H₂SO₄, 1 % HgSO₄, 333 K--> [CH₂=CHOH] → CH₃CHO

CH₃C≡CH + H₂O --HgSO₄/H₂SO₄--> CH₃COCH₃ (propanone, by Markovnikov addition)

Acidic nature of terminal alkynes

The hydrogen on a triply bonded carbon is weakly acidic. The reason is hybridisation: an sp carbon has 50 % s character, so its electrons are held much closer to the nucleus and it behaves as a more electronegative atom, which makes the C—H bond more polar and the resulting carbanion more stable.

Acidity: HC≡CH > CH₂=CH₂ > CH₃—CH₃ (pKa ≈ 25, 44, 50; s character 50 %, 33 %, 25 %)

Three reactions demonstrate it, and all three need a terminal ≡C—H:

1. With sodium metal at 473 K, hydrogen is displaced:

2CH≡CH + 2Na → 2CH≡C—Na (monosodium acetylide) + H₂↑

CH≡C—Na + Na --473 K--> Na—C≡C—Na (disodium acetylide) + ½H₂

2. With ammoniacal silver nitrate (Tollens' reagent) — a white precipitate of silver acetylide:

CH≡CH + 2[Ag(NH₃)₂]OH → Ag—C≡C—Ag↓ + 4NH₃ + 2H₂O

3. With ammoniacal cuprous chloride — a red precipitate of copper acetylide:

CH≡CH + Cu₂Cl₂ + 2NH₄OH → Cu—C≡C—Cu↓ + 2NH₄Cl + 2H₂O

14.7 Test of unsaturation (ethene and ethyne): bromine water test and Baeyer's test

Both tests work because the π bond is attacked and destroyed, taking the colour of the reagent with it.

Bromine water test

Bromine water is reddish-brown. An unsaturated hydrocarbon decolourises it without any hydrogen bromide being given off:

CH₂=CH₂ + Br₂ → CH₂Br—CH₂Br (1,2-dibromoethane, colourless)

CH≡CH + 2Br₂ → CHBr₂—CHBr₂ (1,1,2,2-tetrabromoethane, colourless)

Note that ethyne needs two molecules of bromine and therefore decolourises twice as much bromine water per mole.

Baeyer's test

Baeyer's reagent is cold dilute alkaline potassium permanganate, which is purple. An alkene turns it colourless and throws down a brown precipitate of manganese dioxide:

CH₂=CH₂ + H₂O + [O] --cold dil. alk. KMnO₄--> CH₂OH—CH₂OH (ethane-1,2-diol)

CH≡CH + 4[O] --cold dil. alk. KMnO₄--> HOOC—COOH (oxalic acid)

Test	Reagent	Alkane	Alkene	Alkyne (terminal)
Bromine water	Br₂(aq), reddish-brown	no change	decolourised	decolourised (2 mol Br₂)
Baeyer's test	cold dil. alkaline KMnO₄, purple	no change	decolourised, brown MnO₂	decolourised, brown MnO₂
Ammoniacal AgNO₃	[Ag(NH₃)₂]OH	no change	no change	white ppt Ag₂C₂
Ammoniacal Cu₂Cl₂	CuCl in NH₄OH	no change	no change	red ppt Cu₂C₂

14.8 Comparative physical properties of alkane, alkene and alkyne

Property	Alkane	Alkene	Alkyne
General formula	CₙH₂ₙ₊₂	CₙH₂ₙ	CₙH₂ₙ₋₂
Hybridisation of C	sp³	sp²	sp
Bond angle	109.5°	120°	180°
C—C bond length	154 pm	134 pm	120 pm
C—C bond enthalpy	348 kJ mol⁻¹	614 kJ mol⁻¹	839 kJ mol⁻¹
s character	25 %	33.3 %	50 %
Physical state (C₁–C₄)	gas	gas	gas
Boiling point of the C₂ member	−88.6 °C	−103.7 °C	−84 °C (sublimes)
Density	less than water	less than water	less than water
Solubility	insoluble in water, soluble in benzene, ether, CCl₄	same	slightly more soluble (weakly polar C—H)
Dipole moment	≈ 0	≈ 0 (trans), small (cis)	≈ 0
Chemical character	inert; free-radical substitution	electrophilic addition	electrophilic addition + acidic H
Flame	blue, non-sooty	slightly luminous	bright, smoky (high C content)

Within any one family the boiling point rises steadily with the number of carbons, because the van der Waals attraction grows with molecular size, and falls with branching, because a branched molecule is more nearly spherical and its molecules touch over a smaller area.

14.9 Kolbe's electrolysis method

For sodium ethanoate the overall change is

2CH₃COONa + 2H₂O --electrolysis--> CH₃CH₃↑ + 2CO₂↑ + H₂↑ + 2NaOH

The mechanism, which NEB asks for in full, is:

At the anode (oxidation).

2CH₃COO⁻ → 2CH₃COO• + 2e⁻

2CH₃COO• → 2CH₃• + 2CO₂↑

CH₃• + CH₃• → CH₃CH₃

At the cathode (reduction).

2Na⁺ + 2e⁻ → 2Na

2Na + 2H₂O → 2NaOH + H₂↑

So ethane and carbon dioxide come off the anode and hydrogen off the cathode, and the solution slowly turns alkaline.

2CH₃CH₂COONa + 2H₂O --electrolysis--> CH₃CH₂CH₂CH₃ + 2CO₂ + H₂ + 2NaOH

Electrolysing a mixture of two different salts gives three alkanes, exactly as the Wurtz reaction does. A related use: the potassium salt of a dicarboxylic acid ester gives long-chain diesters, which is how some polymer feedstocks are made.

Chapter summary

• Alkanes are made by reducing haloalkanes (same carbon count), by the Wurtz reaction with Na in dry ether (double the carbons), by heating RCOONa with soda lime (one carbon fewer) and by hydrogenating alkenes or alkynes over Ni.
• Alkanes undergo free-radical substitution — halogenation (hν), vapour-phase nitration (673–773 K) and sulphonation with oleum — and are oxidised stepwise to alcohol, aldehyde, acid and finally CO₂ + H₂O.
• Alkenes are made by elimination: dehydration of alcohols (conc. H₂SO₄, 443 K), dehydrohalogenation with alcoholic KOH, and partial hydrogenation of an alkyne over Lindlar's catalyst. Saytzeff's rule gives the more substituted alkene.
• Alkenes undergo electrophilic addition. HX follows Markovnikov's rule because the more stable carbocation forms; HBr with peroxide goes the other way. Hydration gives alcohols; H₂SO₄ gives alkyl hydrogen sulphates; ozonolysis followed by Zn/H₂O cleaves the molecule at the double bond.
• Alkynes come from C + H₂ in an electric arc, from 1,2-dibromoethane with alcoholic KOH, from CHCl₃ or CHI₃ with silver, and industrially from CaC₂ + H₂O. They add H₂, HX and H₂O (giving CH₃CHO from ethyne, ketones from all others).
• A terminal alkyne has an acidic hydrogen because sp carbon has 50 % s character: it gives H₂ with Na, a white Ag₂C₂ with ammoniacal AgNO₃ and a red Cu₂C₂ with ammoniacal Cu₂Cl₂.
• Bromine water and Baeyer's reagent are both decolourised by alkenes and alkynes but not alkanes; only ammoniacal AgNO₃ or Cu₂Cl₂ separates ethyne from ethene.
• Kolbe's electrolysis of concentrated aqueous RCOONa gives an alkane of 2(n − 1) carbons at the anode along with CO₂, plus H₂ and NaOH at the cathode.

Practice questions

Group A — Multiple choice (1 mark each)

1. The Wurtz reaction of bromoethane gives [1]aethanebpropanecbutanedethene
2. Heating sodium propanoate with soda lime gives [1]apropanebethanecmethanedbutane
3. Propene reacts with HBr in the presence of benzoyl peroxide to give mainly [1]a2-bromopropaneb1-bromopropanec1,2-dibromopropanedpropan-2-ol
4. The catalyst used to stop the hydrogenation of an alkyne at the alkene stage is [1]aRaney nickelbLindlar's catalystcanhydrous AlCl₃dHgSO₄
5. Ethyne can be distinguished from ethene by [1]abromine waterbBaeyer's reagentcammoniacal AgNO₃dburning it
6. Hydration of propyne with HgSO₄/H₂SO₄ gives [1]apropanalbpropanonecpropan-1-oldpropanoic acid
7. Kolbe's electrolysis of a concentrated solution of sodium ethanoate gives, at the anode, [1]amethane and CO₂bethane and CO₂cethene and H₂dethyne and O₂
8. Which one does not undergo sulphonation with oleum? [1]ahexanebheptanec2-methylpropanedmethane
9. The C—C bond length is shortest in [1]aethanebethenecethynedall are equal

Answers to Group A

1. (c) — Wurtz coupling doubles the carbon count, 2 × C₂ = C₄.

2. (b) — decarboxylation removes one carbon from the three-carbon acid.

3. (b) — the peroxide effect makes the addition anti-Markovnikov.

4. (b) — Pd on CaCO₃ poisoned with quinoline.

5. (c) — only the terminal alkyne has an acidic H and gives a white Ag₂C₂ precipitate.

6. (b) — Markovnikov addition puts the OH on C2, and the enol tautomerises to the ketone.

7. (b) — 2CH₃COO⁻ → 2CH₃• + 2CO₂ + 2e⁻, and the methyl radicals couple.

8. (d) — the first three members, methane, ethane and propane, do not sulphonate.

9. (c) — 120 pm, against 134 pm for C=C and 154 pm for C—C.

Group B — Short answer (5 marks each)

1. How is ethane prepared from (i) bromoethane by the Wurtz reaction, (ii) sodium propanoate, (iii) ethene and (iv) ethyne? Give equations with conditions. [5]
2. Describe the free-radical mechanism of the chlorination of methane, naming the three stages. Why is a trace of ethane always found in the product? [5]
3. State Markovnikov's rule and explain it on the basis of carbocation stability, using the addition of HBr to propene. [5]
4. What is the peroxide effect? Why is it shown by HBr but not by HCl or HI? [5]
5. What is ozonolysis? An alkene C₄H₈ gives only ethanal on ozonolysis. Identify it and write the equations. [5]
6. How would you distinguish between ethane, ethene and ethyne in the laboratory? Give the reagent, the observation and one equation in each case. [5]
7. Explain the acidic character of ethyne. Write the reactions with sodium, ammoniacal silver nitrate and ammoniacal cuprous chloride, giving the colour of each precipitate. [5]
8. What is Kolbe's electrolytic method? Write the anode and cathode reactions for a concentrated solution of sodium ethanoate, and state two limitations of the method. [5]

Answers to Group B

1. (i) 2CH₃Br + 2Na --dry ether--> C₂H₆ + 2NaBr (note bromomethane, not bromoethane, is needed for ethane). (ii) CH₃CH₂COONa + NaOH --CaO, Δ--> C₂H₆ + Na₂CO₃. (iii) C₂H₄ + H₂ --Ni, 523–573 K--> C₂H₆. (iv) C₂H₂ + 2H₂ --Ni--> C₂H₆.

5. Ozonolysis is the addition of ozone across a C=C to give an ozonide, followed by cleavage with Zn/H₂O to give carbonyl compounds. If the only product is ethanal, both halves of the molecule must be CH₃CH=, so the alkene is but-2-ene: CH₃CH=CHCH₃ + O₃ → ozonide, then ozonide + H₂O --Zn--> 2CH₃CHO. (The isomer but-1-ene would give HCHO + CH₃CH₂CHO, and 2-methylpropene would give HCHO + propanone.)

6. Bromine water: no change with ethane, decolourised by ethene and ethyne. Ammoniacal AgNO₃: white precipitate with ethyne only, C₂H₂ + 2[Ag(NH₃)₂]OH → Ag₂C₂↓ + 4NH₃ + 2H₂O. So no change to bromine water = ethane; decolourises bromine water but gives no precipitate = ethene; decolourises bromine water and gives a white precipitate = ethyne.

8. Definition as in Section 14.9. Anode: 2CH₃COO⁻ → 2CH₃COO• + 2e⁻; 2CH₃COO• → 2CH₃• + 2CO₂; 2CH₃• → C₂H₆. Cathode: 2Na⁺ + 2e⁻ → 2Na; 2Na + 2H₂O → 2NaOH + H₂. Limitations: (i) only alkanes with an even number of carbons can be made, so methane and propane are impossible; (ii) a mixture of two salts gives a mixture of three alkanes; (iii) the solution must be concentrated, or oxygen is discharged at the anode instead.

Group C — Long answer (8 marks each)

1. (a) Describe, with equations and conditions, four laboratory methods of preparing an alkane. [4](b) What happens when ethane is (i) chlorinated in sunlight, (ii) heated with nitric acid vapour at 673 K, (iii) oxidised with O₂ over a Cu tube at 473 K and 100 atm, (iv) burnt in excess oxygen? [4]
2. (a) State Markovnikov's rule and the peroxide effect, and explain both in terms of the stability of the intermediate formed. [4](b) Complete: (i) CH₃CH=CH₂ + H₂O/H⁺ → (ii) CH₂=CH₂ + conc. H₂SO₄ → , then + H₂O, Δ → (iii) 2-methylbut-2-ene + O₃, then Zn/H₂O → (iv) CH₃CH₂CHBrCH₃ + alc. KOH → [4]
3. (a) How is ethyne prepared from calcium carbide and from 1,2-dibromoethane? Give equations. [3](b) Explain, with equations, the acidic nature of ethyne and the two precipitation tests based on it. [3](c) 5.6 dm³ of ethyne at STP is burnt completely. Calculate the mass of carbon dioxide formed. [2]

Answers to Group C

2. (b) (i) propan-2-ol, CH₃CH(OH)CH₃ — Markovnikov. (ii) CH₂=CH₂ + H₂SO₄ → CH₃CH₂OSO₃H, and CH₃CH₂OSO₃H + H₂O --Δ--> C₂H₅OH + H₂SO₄. (iii) (CH₃)₂C=CHCH₃ gives propanone (CH₃)₂CO and ethanal CH₃CHO. (iv) but-2-ene CH₃CH=CHCH₃ as the major product (Saytzeff) with but-1-ene as the minor product.

3. (c) The balanced equation is 2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O. Moles of ethyne = 5.6 / 22.4 = 0.25 mol. From the equation, 2 mol of C₂H₂ gives 4 mol of CO₂, i.e. 1 mol gives 2 mol, so moles of CO₂ = 2 × 0.25 = 0.50 mol. Mass of CO₂ = 0.50 × 44 = 22 g.