Classification, nomenclature and the C–X bond
Link to Std 11 : In Std 11 you named alkanes and learnt electronegativity and polar bonds; halogens are more electronegative than carbon, so C–X is polar.
Many refrigerants, dry-cleaning liquids and anaesthetics are halogen compounds, and old insect sprays such as DDT were too.
Haloalkanes have a halogen on an sp³ carbon (R–X); haloarenes have it directly on an aromatic ring (Ar–X). By the number of halogen atoms they are mono-, di- or poly-halogen compounds. By the carbon bearing X, haloalkanes are primary (1°), secondary (2°) or tertiary (3°). The C–X bond is polar: Cδ⁺–Xδ⁻.
In the figure : Where the halogen sits decides reactivity: sp³ C–X (alkyl, allyl, benzyl) is reactive; sp² C–X (vinyl, aryl) is much less reactive.
Rule 1: Types by carbon bearing X — Alkyl halides: 1° (CH₃CH₂Br), 2° ((CH₃)₂CHBr), 3° ((CH₃)₃CBr). Allylic halides have X on an sp³ carbon next to C=C; benzylic halides on an sp³ carbon next to an aromatic ring; vinylic and aryl halides have X on an sp² carbon.
- Easy : CH₂=CH–CH₂Cl is allyl chloride; C₆H₅CH₂Cl is benzyl chloride.
- Medium : CH₂=CHCl (vinyl chloride) and C₆H₅Cl (chlorobenzene) are unreactive types.
- Exam-type : Classify (CH₃)₃CBr: three alkyl groups on the carbon with Br, so it is a 3° (tertiary) haloalkane.
Rule 2: IUPAC and common names — Common name: alkyl + halide (isopropyl bromide). IUPAC: halo + alkane, with the longest chain and the lowest locants (2-bromopropane). For haloarenes, the ortho/meta/para or numbers are used: 1-chloro-4-methylbenzene (p-chlorotoluene). Dihalides: gem (same carbon, 1,1-) and vic (adjacent carbons, 1,2-).
- Easy : CH₃CH₂CH₂Cl is 1-chloropropane (n-propyl chloride).
- Medium : CH₃CHBrCH₃ is 2-bromopropane (isopropyl bromide).
- Exam-type : CH₃CHCl₂ is a gem-dihalide (1,1-dichloroethane) and ClCH₂CH₂Cl is a vic-dihalide (1,2-dichloroethane).
Rule 3: Nature of the C–X bond — Down the group the C–X bond becomes longer and weaker (C–F strongest, C–I weakest), so bond enthalpy falls: C–F > C–Cl > C–Br > C–I. The dipole moment is the product of charge and distance, and it does not follow bond length alone: CH₃Cl has the largest dipole moment of the methyl halides, 1.86 D (Cl), while CH₃F is 1.847 D.
- Easy : C–I is the longest and weakest of the four bonds.
- Medium : Bond enthalpy values (kJ mol⁻¹): CH₃–F 452, CH₃–Cl 351, CH₃–Br 293, CH₃–I 234.
- Exam-type : Why is iodide a better leaving group than chloride? The C–I bond is longer and weaker, and I⁻ is a larger, more stable anion.
C–X bond: longest and weakest for iodine; reactivity of R–X in substitution goes R–I > R–Br > R–Cl > R–F.
Treating chlorobenzene as an alkyl-type halide; the C–Cl is on an sp² carbon and is much less reactive.
Memory clue: "Bigger halogen, weaker bond, better leaving group" — I⁻ leaves most easily.
Spot the difference
| Point | Haloalkane | Haloarene |
|---|---|---|
| C bonded to X | sp³ | sp² (aromatic) |
| Example | CH₃CH₂Br | C₆H₅Cl |
| Nucleophilic substitution | Easy | Very difficult |
Practice questions
- 1Classify (CH₃)₃CBr as primary, secondary or tertiary.
- 2Write the IUPAC name of CH₃CHBrCH₃.
- 3Which has the weakest bond to carbon: C–F or C–I?
- 4What is the difference between a gem-dihalide and a vic-dihalide? Give one example each.
- 5Draw the structure of benzyl chloride and say what type of halide it is.
- 6Why is the C–X bond polar?
- 7Arrange CH₃F, CH₃Cl, CH₃Br, CH₃I in order of increasing C–X bond length.
- 8Explain why alkyl iodides are more reactive than alkyl chlorides in substitution.
Show answers (try them yourself first)
- 1Tertiary (3°).
- 22-Bromopropane.
- 3C–I.
- 4Gem: both halogens on the same carbon (CH₃CHCl₂); vic: on adjacent carbons (ClCH₂CH₂Cl).
- 5C₆H₅–CH₂–Cl; a benzylic halide (X on an sp³ carbon next to the ring).
- 6Halogen is more electronegative than carbon, so electrons shift towards X, giving Cδ⁺–Xδ⁻.
- 7CH₃F < CH₃Cl < CH₃Br < CH₃I.
- 8The C–I bond is longer and weaker, and iodide is a better (more stable) leaving group.
Preparation of haloalkanes and haloarenes
Link to Std 11 : In Std 11 you learnt substitution of alkane hydrogens by halogens and addition of HX to alkenes; here those reactions are used as preparations.
Chemists choose a preparation by its by-products: a method that gives only gases such as SO₂ and HCl leaves a clean product.
Haloalkanes are made from alcohols (best), alkenes and alkanes. Haloarenes are made by electrophilic substitution in benzene or from diazonium salts (Sandmeyer reaction).
In the figure : Alcohols are the usual source. Thionyl chloride is the best reagent because the by-products SO₂ and HCl are gases and escape, leaving pure R–Cl.
Rule 1: From alcohols — ROH + SOCl₂ → RCl + SO₂↑ + HCl↑ (pure product). ROH + HX with ZnCl₂ (Lucas reagent) works best for 3° > 2° > 1°. ROH + PCl₅ → RCl + POCl₃ + HCl; 3ROH + PBr₃ → 3RBr + H₃PO₃; red P + I₂ forms PI₃ in situ for iodides. H₂SO₄ is not used with KI because it oxidises HI to I₂.
- Easy : CH₃CH₂OH + SOCl₂ → CH₃CH₂Cl + SO₂ + HCl.
- Medium : 10 g of 1-butanol (74.12 g mol⁻¹) can give at most 10 ÷ 74.12 × 92.57 = 12.5 g of 1-chlorobutane (92.57 g mol⁻¹).
- Exam-type : At 80 % yield the actual mass is 10.0 g.
Rule 2: From alkenes and alkanes; halogen exchange — Alkene + HX gives the Markovnikov product (HBr with peroxide gives anti-Markovnikov); alkene + X₂ gives the vic-dihalide. Alkane + X₂ in light gives a mixture of products (free radical), so it is a poor method. Finkelstein: R–Cl or R–Br + NaI in dry acetone → R–I (NaCl/NaBr precipitates). Swarts: R–Cl + AgF (or Hg₂F₂, CoF₂, SbF₃) → R–F.
- Easy : Propene + HBr → 2-bromopropane (Markovnikov).
- Medium : Propene + HBr with peroxide → 1-bromopropane (anti-Markovnikov).
- Exam-type : CH₃CH₂Br + NaI (dry acetone) → CH₃CH₂I + NaBr↓: the precipitate drives the reaction.
Rule 3: Haloarenes — Electrophilic substitution: benzene + Cl₂/Br₂ with FeCl₃ or FeBr₃ gives ArCl/ArBr (iodination needs an oxidising agent; fluorination is made through diazonium fluoroborate, the Balz–Schiemann reaction). Sandmeyer: ArN₂⁺Cl⁻ + CuCl/HCl (or CuBr/HBr) → ArCl (or ArBr); with Cu powder it is the Gattermann reaction. With KI the diazonium group is replaced by iodine directly.
- Easy : Benzene + Cl₂ (FeCl₃) → chlorobenzene + HCl.
- Medium : C₆H₅N₂⁺Cl⁻ + CuBr/HBr → C₆H₅Br + N₂.
- Exam-type : Toluene + Cl₂ in sunlight gives benzyl chloride (side chain), but with FeCl₃ in the cold it gives o- and p-chlorotoluene (ring).
Alcohol + SOCl₂ is the cleanest method: both by-products are gases.
Using H₂SO₄ with KI to make alkyl iodides; HI is oxidised to I₂ by concentrated H₂SO₄.
Memory clue: "SOCl₂ takes SO₂ and HCl away" — gases escape, the product stays pure.
Spot the difference
| Point | Halogenation of alkane | From alcohol with SOCl₂ |
|---|---|---|
| Product | Mixture | One pure product |
| By-products | HX | SO₂ and HCl gases |
| Preferred? | No | Yes |
Practice questions
- 1Name the best reagent to convert an alcohol into an alkyl chloride.
- 2Write the equation for the Finkelstein reaction of bromoethane.
- 3What is the product of benzene with Cl₂ and FeCl₃?
- 4Why is SOCl₂ preferred to PCl₅ for converting alcohols to chlorides?
- 5How does propene give 1-bromopropane instead of 2-bromopropane?
- 6Why is sulphuric acid not used with KI to make iodoalkanes?
- 7How much 1-chlorobutane can theoretically be made from 10 g of 1-butanol, and what mass at 80 % yield? (74.12 and 92.57 g mol⁻¹)
- 8How would you prepare p-bromotoluene from p-toluidine?
Show answers (try them yourself first)
- 1Thionyl chloride, SOCl₂.
- 2CH₃CH₂Br + NaI → CH₃CH₂I + NaBr (dry acetone).
- 3Chlorobenzene.
- 4The by-products SO₂ and HCl escape as gases, so the alkyl chloride is obtained pure.
- 5By adding HBr in the presence of a peroxide (free-radical addition, anti-Markovnikov).
- 6H₂SO₄ is an oxidising agent: it oxidises HI to I₂, so no alkyl iodide forms.
- 7Theoretical = 12.5 g; at 80 % = 10.0 g.
- 8Diazotise p-toluidine with NaNO₂/HCl at 273–278 K to the diazonium chloride and treat with CuBr/HBr (Sandmeyer).
Physical properties of haloalkanes and haloarenes
Link to Std 11 : In Std 11 you saw how intermolecular forces control boiling point and solubility; the same ideas apply here.
Chloroform feels cool and evaporates fast on the hand, but it sinks in water: a dense liquid that does not mix with water.
Haloalkanes are colourless liquids (the lower ones are gases) with higher boiling points than alkanes of similar mass, because of dipole–dipole and London forces. They are sparingly soluble in water and soluble in organic solvents. Bromo- and iodo-compounds are denser than water.
In the figure : The boiling point rises from fluoride to iodide as molecular size and mass increase. In isomers, branching makes the molecule more compact, reduces the contact area, and lowers the boiling point.
Rule 1: Boiling point — For the same alkyl group: RI > RBr > RCl > RF. For isomeric alkyl halides: 1° > 2° > 3° (branching reduces surface area). Para-isomers of dihaloarenes have higher melting points than ortho or meta because they pack better in the crystal.
- Easy : CH₃CH₂CH₂CH₂Br boils higher than (CH₃)₃CBr.
- Medium : The order for n-butyl halides: bromide > chloride, since a bigger halogen gives more London force.
- Exam-type : p-dichlorobenzene melts at a higher temperature than o- or m-dichlorobenzene because the symmetrical molecule fits better in the crystal lattice.
Rule 2: Solubility and density — Water molecules hold each other by strong hydrogen bonds, and an alkyl halide cannot form hydrogen bonds of equal strength, so little energy is regained on dissolving. Density increases with the number of halogens and with halogen mass.
- Easy : CHCl₃ and CCl₄ sink below water.
- Medium : Ethyl iodide (density about 1.9 g mL⁻¹) is heavier than ethyl chloride (0.9 g mL⁻¹).
- Exam-type : Why is chloroethane only slightly soluble in water? Breaking water's hydrogen bonds costs more energy than the weak attraction between water and R–Cl gives back.
Rule 3: Comparison with alkanes and alcohols — Haloalkanes boil higher than alkanes of similar mass but lower than alcohols (no hydrogen bonding). Alcohols dissolve in water; haloalkanes do not.
- Easy : CH₃CH₂Cl (about 285 K) boils higher than C₂H₆ (about 184 K).
- Medium : CH₃CH₂Cl boils lower than CH₃CH₂OH (about 351 K) because of H bonding in the alcohol.
- Exam-type : Arrange C₂H₅OH, C₂H₅Br, C₂H₆ in increasing boiling point: C₂H₆ < C₂H₅Br < C₂H₅OH.
Bigger halogen, higher boiling point; more branching, lower boiling point.
Thinking haloalkanes dissolve in water because the C–X bond is polar; they cannot hydrogen-bond with water.
Memory clue: "Big and straight boils late" — large halogen and straight chain raise the boiling point.
Spot the difference
| Point | Haloalkane | Alcohol of similar size |
|---|---|---|
| Hydrogen bonding | No | Yes |
| Boiling point | Lower | Higher |
| Solubility in water | Low | High (small ones) |
Practice questions
- 1Which has the higher boiling point: CH₃CH₂Cl or CH₃CH₂I?
- 2Name two haloalkanes denser than water.
- 3Are haloalkanes soluble in water?
- 4Arrange 1-bromobutane, 2-bromobutane and 2-bromo-2-methylpropane in increasing boiling point.
- 5Why is p-dichlorobenzene a solid with a higher melting point than the ortho isomer?
- 6Why do haloalkanes have higher boiling points than alkanes of similar mass?
- 7Arrange C₂H₆, C₂H₅Br and C₂H₅OH in increasing order of boiling point, with reasons.
- 8Explain why haloalkanes are insoluble in water although their C–X bonds are polar.
Show answers (try them yourself first)
- 1CH₃CH₂I.
- 2Chloroform (CHCl₃) and carbon tetrachloride (CCl₄) (also bromo- and iodoalkanes).
- 3No, they are only slightly soluble.
- 42-bromo-2-methylpropane < 2-bromobutane < 1-bromobutane (more branching, lower boiling point).
- 5The symmetrical para isomer packs better in the crystal.
- 6They have stronger dipole–dipole and London forces because of the polar C–X bond and larger mass.
- 7C₂H₆ < C₂H₅Br < C₂H₅OH: alkane has weak London forces only, the bromide has dipole–dipole forces, the alcohol has hydrogen bonds.
- 8They cannot form hydrogen bonds with water, so the energy needed to break water's hydrogen bonds is not repaid.
Nucleophilic substitution: SN2 and SN1
Link to Std 11 : In Std 11 you learnt about electrophiles, nucleophiles and carbocations; haloalkanes show how a nucleophile replaces a leaving group.
A crowded door lets one person enter easily but blocks a group: a bulky tertiary carbon is hard for a nucleophile to approach from behind, but it can still lose its halide to form a stable carbocation.
In nucleophilic substitution a nucleophile (Nu⁻) replaces the halide (the leaving group). SN2: one step, bimolecular, rate = k[RX][Nu⁻], inversion of configuration. SN1: two steps through a carbocation, unimolecular, rate = k[RX], racemisation.
Nucleophilic substitution: SN2 and SN1
In the app you can move this figure with your finger.
Open in the appIn the figure : In SN2 the nucleophile attacks from the side opposite the leaving group, turning the carbon like an umbrella in the wind (inversion). In SN1 the leaving group departs first, leaving a flat carbocation that the nucleophile can attack from either face, giving equal amounts of both mirror images.
Rule 1: SN2 reactions — Favoured by 1° (and methyl) halides because the carbon is less crowded; a high concentration of a strong nucleophile and a polar aprotic solvent help. Relative reactivity: CH₃X > 1° > 2° > 3°. The reaction goes through a single transition state in which Nu and X are both partly bonded.
- Easy : CH₃Br + OH⁻ → CH₃OH + Br⁻ is SN2 (rate = k[CH₃Br][OH⁻]).
- Medium : If both [CH₃Br] and [OH⁻] are doubled, the rate rises 2 × 2 = 4 times.
- Exam-type : (R)-2-Bromobutane + OH⁻ via SN2 gives (S)-butan-2-ol: complete inversion of the configuration.
Rule 2: SN1 reactions — Favoured by 3° halides (stable carbocation), a polar protic solvent (water, alcohol) and a good leaving group. The slow step is ionisation; the nucleophile does not appear in the rate law. Relative reactivity: 3° > 2° > 1° > CH₃X; allyl and benzyl halides are also very reactive.
- Easy : (CH₃)₃CBr + H₂O → (CH₃)₃COH + HBr is SN1 (rate = k[(CH₃)₃CBr]).
- Medium : If [RX] is tripled and [Nu] is doubled, the SN1 rate becomes 3 times (no dependence on [Nu]).
- Exam-type : An optically active 2° halide giving a racemic product points to an SN1 path through a planar carbocation.
Rule 3: Stereochemistry and chirality — A chiral carbon has four different groups; the two non-superimposable mirror images are enantiomers (rotate plane-polarised light equally and oppositely, d or l). A 1:1 mixture is a racemic mixture (±), optically inactive. Retention: the spatial arrangement is unchanged; inversion: it is reversed; racemisation: both forms result.
- Easy : CH₃CHBrCH₂CH₃ has a chiral carbon (H, CH₃, C₂H₅, Br).
- Medium : (CH₃)₂CHBr has no chiral carbon: it has two identical CH₃ groups.
- Exam-type : A racemic mixture is optically inactive because the equal and opposite rotations cancel (external compensation).
SN2: 1° fastest, inversion, strong Nu. SN1: 3° fastest, racemisation, polar protic solvent.
Using the carbocation (SN1) argument for a primary halide; primary carbocations are too unstable, so 1° halides go by SN2.
Memory clue: "SN1 = 1 step slow, racemic; SN2 = 2 species, inversion" — SN2 rate depends on two species and flips the configuration.
Spot the difference
| Point | SN1 | SN2 |
|---|---|---|
| Rate law | k[RX] | k[RX][Nu⁻] |
| Steps | Two (carbocation) | One (transition state) |
| Stereochemistry | Racemisation | Inversion |
| Reactivity | 3° > 2° > 1° | 1° > 2° > 3° |
Practice questions
- 1What is the rate law of an SN2 reaction?
- 2Name the intermediate in an SN1 reaction.
- 3Define a chiral carbon.
- 4Which of CH₃CH₂Br and (CH₃)₃CBr reacts faster by SN2 and why?
- 5If both concentrations are doubled, what happens to the rate of an SN2 reaction?
- 6Why is a racemic mixture optically inactive?
- 7Predict the product and stereochemistry when (R)-2-bromobutane reacts with OH⁻ by SN2.
- 8Why do tertiary halides prefer SN1 whereas primary halides prefer SN2?
Show answers (try them yourself first)
- 1Rate = k[RX][Nu⁻].
- 2A planar carbocation.
- 3A carbon attached to four different atoms or groups.
- 4CH₃CH₂Br, because the primary carbon is less crowded and the nucleophile can attack from behind.
- 5It becomes 2 × 2 = 4 times.
- 6It contains equal amounts of the two enantiomers; their opposite rotations cancel.
- 7(S)-Butan-2-ol: the configuration is inverted by backside attack.
- 8A tertiary carbocation is stabilised by three alkyl groups, and the tertiary carbon is too crowded for backside attack; a primary carbocation is unstable and the primary carbon is open to attack.
Elimination, reaction with metals and behaviour of haloarenes
Link to Std 11 : In Std 11 you studied elimination of water from alcohols and the Saytzeff idea of more substituted alkenes; the same pattern appears with halides.
Cooking oil that sits in a hot pan sometimes smokes and breaks apart; likewise haloalkanes heated with a strong base lose HX and form alkenes.
Elimination (dehydrohalogenation): haloalkane + alcoholic KOH, heat → alkene + KX + H₂O. Saytzeff's rule: the more substituted (more stable) alkene is the major product. Haloalkanes react with Mg in dry ether to give Grignard reagents and with Na to give alkanes (Wurtz). Haloarenes resist nucleophilic substitution but undergo electrophilic substitution.
In the figure : Chlorobenzene is unreactive because of resonance, the sp² carbon, the repulsion between the ring electrons and the nucleophile, and the unstable phenyl cation. Strong electron-withdrawing groups (NO₂) at the ortho or para position activate the ring to nucleophilic attack.
Rule 1: Elimination — Alcoholic KOH removes H from a β-carbon and X from the α-carbon. Aqueous KOH gives substitution (alcohol), alcoholic KOH gives elimination (alkene). Major product by Saytzeff: the alkene with more alkyl groups on the double bond.
- Easy : CH₃CH₂Br + alcoholic KOH → CH₂=CH₂ + KBr + H₂O.
- Medium : 2-Bromobutane + alc. KOH → but-2-ene (major, more substituted) + but-1-ene (minor).
- Exam-type : 2-Bromo-2-methylbutane gives 2-methylbut-2-ene (major) rather than 2-methylbut-1-ene, by Saytzeff's rule.
Rule 2: Reactions with metals — Grignard: R–X + Mg (dry ether) → RMgX; it must be free of water, or it turns into alkane (RMgX + H₂O → RH + Mg(OH)X). Wurtz: 2R–X + 2Na (dry ether) → R–R + 2NaX. Wurtz–Fittig: Ar–X + R–X + 2Na → Ar–R; Fittig: 2Ar–X + 2Na → Ar–Ar.
- Easy : CH₃CH₂Br + Mg (dry ether) → CH₃CH₂MgBr.
- Medium : 2CH₃Br + 2Na → CH₃–CH₃ + 2NaBr (ethane).
- Exam-type : Why must Grignard reagent be made in dry ether? Water destroys it: CH₃MgBr + H₂O → CH₄ + Mg(OH)Br.
Rule 3: Haloarene reactions — Nucleophilic substitution needs very drastic conditions (chlorobenzene + NaOH at 623 K, 300 atm gives phenol). With NO₂ at o- or p- it is easier (2,4-dinitrochlorobenzene reacts with warm aqueous NaOH; 2,4,6-trinitrochlorobenzene reacts with water). Electrophilic substitution: halogens are deactivating but o,p-directing, so nitration of chlorobenzene gives mainly o- and p-nitrochlorobenzene.
- Easy : Chlorobenzene + conc. HNO₃/H₂SO₄ → o- and p-nitrochlorobenzene (para major).
- Medium : p-Nitrochlorobenzene reacts with OH⁻ more easily than chlorobenzene because NO₂ withdraws electrons and stabilises the intermediate.
- Exam-type : Why are halogens o,p-directing yet deactivating? The lone pair donated by resonance directs to o,p; the –I effect withdraws electrons and slows the ring reactions.
Aqueous KOH: substitution. Alcoholic KOH: elimination. Haloarenes: electrophilic substitution is easy, nucleophilic is difficult.
Using aqueous KOH when an alkene is wanted from a haloalkane.
Memory clue: "Alcoholic gives alkene" — alcoholic KOH removes HX; aqueous KOH makes the alcohol.
Spot the difference
| Point | Haloalkane | Haloarene |
|---|---|---|
| Nucleophilic substitution | Easy (SN1/SN2) | Very hard |
| Electrophilic substitution | Not applicable | Easy, o/p directing |
| Reason | sp³ carbon | Resonance and sp² carbon |
Practice questions
- 1What reagent converts 2-bromopropane to propene?
- 2Write the equation for the Wurtz reaction of bromomethane.
- 3Is the C–Cl bond in chlorobenzene shorter or longer than in chloroethane?
- 4Give the major product of 2-bromobutane with alcoholic KOH. Name the rule.
- 5Why must Grignard reagents be prepared in dry ether?
- 6Which of chlorobenzene and p-nitrochlorobenzene reacts faster with aqueous NaOH, and why?
- 7List four reasons why chlorobenzene is less reactive to nucleophilic substitution than chloroethane.
- 8Explain why halogens are o,p-directing but deactivating.
Show answers (try them yourself first)
- 1Alcoholic KOH (heat).
- 22CH₃Br + 2Na → CH₃CH₃ + 2NaBr (dry ether).
- 3Shorter, because of the partial double-bond character from resonance.
- 4But-2-ene is the major product (Saytzeff's rule: the more substituted alkene).
- 5Water decomposes the Grignard reagent to an alkane (RMgX + H₂O → RH + Mg(OH)X).
- 6p-Nitrochlorobenzene, because NO₂ at the para position withdraws electrons and stabilises the intermediate carbanion.
- 7(i) resonance gives the C–Cl partial double-bond character; (ii) C is sp² and more electronegative; (iii) the ring π electrons repel the nucleophile; (iv) the phenyl cation is unstable, so no SN1.
- 8The lone pair on the halogen donates by resonance to the o and p positions (directing); the strong –I effect withdraws electrons from the ring overall, reducing reactivity (deactivating).
Polyhalogen compounds and their effects
Link to Std 11 : In Std 11 you learnt about environmental chemistry and the ozone layer; several halogen compounds are directly linked to ozone damage.
Chloroform was once used to put patients to sleep, but it is converted in air and light into a poisonous gas, so its use as an anaesthetic stopped.
Polyhalogen compounds have more than one halogen atom. They are useful as solvents, refrigerants, anaesthetics and pesticides, but many are toxic and do not break down easily in the environment.
In the figure : Each compound has a main use and a main hazard. Note that chloroform needs special storage, and chlorofluorocarbons (freons) and CCl₄ damage the ozone layer.
Rule 1: Methane derivatives — CH₂Cl₂: solvent, paint remover, propellant; harmful to the nervous system. CHCl₃: solvent and earlier anaesthetic; in air and sunlight it oxidises to poisonous phosgene (COCl₂), so it is stored in dark brown, completely filled bottles; 1 % ethanol is added to convert phosgene to harmless diethyl carbonate. CHI₃: antiseptic through the release of iodine. CCl₄: cleaning agent, fire extinguishers; harms the liver and depletes ozone.
- Easy : CHCl₃ + ½O₂ (light) → COCl₂ + HCl, so store it in dark bottles.
- Medium : CCl₄ vapour was used in fire extinguishers (pyrene) but is now avoided because it can form phosgene on heating and harms ozone.
- Exam-type : Why is ethanol added to chloroform? It converts poisonous phosgene into harmless ethyl carbonate.
Rule 2: Freons and DDT — Freons (CCl₂F₂, CCl₃F) are chlorofluorocarbons used in refrigerators and air conditioners. They rise to the stratosphere and release Cl atoms, which destroy ozone. DDT (p,p′-dichlorodiphenyltrichloroethane) is a powerful insecticide against mosquitoes and crop pests, but it is not biodegradable and accumulates in fatty tissues up the food chain; many countries banned its agricultural use.
- Easy : Freon-12 is CCl₂F₂ and was used as a refrigerant.
- Medium : Cl atoms from CFCs react with ozone: Cl + O₃ → ClO + O₂, and the cycle repeats.
- Exam-type : Why was DDT banned? It is stable, accumulates in fatty tissues (bioaccumulation) and harms birds and other animals.
Rule 3: Safe use — Halogen compounds should be used where needed and in limited quantity; non-ozone-depleting substitutes (HFCs, hydrocarbons) are replacing CFCs, and less persistent pesticides are used in place of DDT.
- Easy : Modern refrigerators use HFC or hydrocarbon refrigerants instead of CFCs.
- Medium : Care is taken in the disposal of solvents such as CCl₄ and CH₂Cl₂ because they harm the environment.
- Exam-type : Write the formula of the product of phosgene formation from chloroform: COCl₂ (carbonyl chloride).
Chloroform: dark bottles. CFCs: ozone damage. DDT: persistent and bioaccumulating.
Saying that CFCs are harmful because they are poisonous; the main problem is ozone destruction by chlorine atoms.
Memory clue: "Chloroform turns to phosgene, so keep it dark and full".
Spot the difference
| Compound | Use | Hazard |
|---|---|---|
| CHCl₃ | Solvent, old anaesthetic | Forms phosgene in light |
| CCl₄ | Cleaning, extinguisher | Liver damage, ozone |
| CFCs | Refrigerants | Ozone depletion |
| DDT | Insecticide | Persistent, bioaccumulates |
Practice questions
- 1Write the formula of chloroform and of freon-12.
- 2Why is chloroform kept in dark brown bottles?
- 3Name an insecticide that is a polyhalogen compound.
- 4What is phosgene and how is it formed?
- 5How do CFCs damage the ozone layer?
- 6Why is DDT considered harmful?
- 7Why is a small quantity of ethanol added to chloroform?
- 8Name two substitutes used in place of CFCs.
Show answers (try them yourself first)
- 1CHCl₃ and CCl₂F₂.
- 2Light and air convert chloroform to poisonous phosgene.
- 3DDT.
- 4COCl₂, a poisonous gas formed when chloroform is oxidised by air in light.
- 5UV light in the stratosphere breaks C–Cl bonds, releasing Cl atoms that catalytically convert O₃ to O₂.
- 6It does not break down, accumulates in fat and moves up the food chain, harming animals.
- 7Ethanol converts any phosgene that forms into harmless ethyl carbonate.
- 8Hydrofluorocarbons (HFCs) and hydrocarbons such as propane or isobutane.