Educational Galaxy

Chapter 6: Haloalkanes and Haloarenes — EG's own summary, in simple words

Book: Chemistry — Part II (CBSE, Class 12)  |  Chapter 6  |  Medium: English

Replacing a hydrogen of a hydrocarbon with a halogen creates a versatile family: haloalkanes react by substitution and elimination, haloarenes resist substitution but undergo electrophilic attack on the ring. This chapter teaches their names and preparations, then the two great mechanisms — SN1 and SN2 — and the reactions that make these compounds both useful and environmentally dangerous.

Diagram (labels on the image): Haloalkane, Haloarene, C–X bond (labels as drawn). Labels added by Educational Galaxy on an originally generated illustration.

EG schematic (drawn and labelled by Educational Galaxy): SN1 vs SN2: SN1 goes in two steps through a flat carbocation (racemisation, favoured by tertiary substrates and polar protic solvents); SN2 goes in one back-side step with inversion of configuration (favoured by primary substrates).

6.1 Classification and nomenclature

Haloalkanes (alkyl halides) have halogen on an sp³ carbon; haloarenes (aryl halides) have halogen on an sp² ring carbon. By the number of halogens: mono-, di-, tri- and polyhalo compounds. By the carbon carrying the halogen: primary (1°), secondary (2°) and tertiary (3°) — counted by how many other carbons attach to that carbon.

Common names survive for simple compounds (chloroform CHCl₃, carbon tetrachloride CCl₄, methyl chloride), but IUPAC names are systematic: choose the longest chain containing the halogen, number to give the halogen the lowest locant, and name halogens as prefixes (fluoro, chloro, bromo, iodo) alphabetically: 2-bromopropane, 1-chloro-2-methylpropane, chlorobenzene.

The C–X bond is polar (halogen more electronegative than carbon), so the carbon is slightly positive and open to nucleophilic attack. Bond strength falls down the group (C–F > C–Cl > C–Br > C–I), so reactivity rises in the reverse order: iodides react fastest, fluorides slowest. Haloalkanes are denser than water and insoluble in it, but soluble in organic solvents.

6.2 Nature of the C–X bond and preparation

Haloalkanes are made: (1) from alcohols — with HCl/ZnCl₂ (Lucas reagent), PCl₃, PCl₅, or SOCl₂ (thionyl chloride is best because the by-products SO₂ and HCl are gases, leaving pure alkyl chloride); (2) from hydrocarbons — free-radical halogenation of alkanes (mixture of products, poor selectivity); (3) by halogen exchange — Finkelstein reaction (R–Cl + NaI in acetone → R–I + NaCl; driven by NaCl precipitating) and Swarts reaction (for fluorides, using AgF or Hg₂F₂).

Haloarenes are made: (1) electrophilic substitution of arenes — Cl₂/Br₂ with FeCl₃ or AlCl₃ (FeBr₃) gives chlorobenzene/bromobenzene; (2) from diazonium salts — Sandmeyer reaction (CuCl/CuBr gives chlorobenzene/bromobenzene; CuCN gives benzonitrile) and Gattermann reaction (Cu/HCl or Cu/HBr); iodobenzene comes from diazonium salt + KI; fluorobenzene from diazonium tetrafluoroborate on heating (Balz–Schiemann reaction).

Direct halogenation of an alkane gives a mixture because all hydrogens compete; the method is used industrially (chloromethane) but not for a single pure product in the lab.

6.3 Nucleophilic substitution — SN1 and SN2

The key reaction of haloalkanes: a nucleophile (OH⁻, CN⁻, RO⁻, NH₃, I⁻) replaces the halogen. Two mechanisms compete.

• SN2 (substitution, nucleophilic, bimolecular): one step. The nucleophile attacks from the back side while the halogen leaves, passing through a five-coordinate transition state. Rate = k[R–X][Nu⁻]. It inverts the configuration (Walden inversion) like an umbrella flipping in the wind. Favoured by primary substrates (least hindrance), strong nucleophiles, and polar aprotic solvents (DMSO, acetone) that leave the nucleophile free.
• SN1 (unimolecular): two steps. First the C–X bond breaks slowly to a planar carbocation (rate-determining step); then the nucleophile attacks from either face. Rate = k[R–X] only. The product is racemised (equal d and l forms) because attack from both faces is equally likely — with a little inversion product because the leaving group briefly shields one face. Favoured by tertiary substrates (stable carbocation), weak nucleophiles, and polar protic solvents (water, alcohol) that stabilise ions.

Reactivity order: for SN2, CH₃X > 1° > 2° > 3° (steric hindrance). For SN1, 3° > 2° > 1° > CH₃X (carbocation stability). Allylic and benzylic halides are unusually reactive in both, because their carbocations are resonance-stabilised.

Haloarenes do NOT undergo nucleophilic substitution easily: the C–X bond has partial double-bond character (lone pairs of halogen conjugate with the ring), the ring carbon is sp² and hard to attack from behind, and a phenyl cation is very unstable. Only with strong electron-withdrawing groups ortho/para to the halogen (like NO₂) and vigorous conditions does substitution occur.

6.4 Elimination, reactions with metals, and electrophilic substitution

Elimination (dehydrohalogenation): heating a haloalkane with alcoholic KOH removes HX and forms an alkene. Saytzeff’s rule: the hydrogen is removed from the carbon with fewer hydrogens, giving the more substituted (more stable) alkene. Example: 2-bromobutane gives mainly but-2-ene, not but-1-ene. Substitution (aqueous KOH → alcohol) and elimination (alcoholic KOH → alkene) compete; the reagent decides.

With metals: magnesium in dry ether gives Grignard reagents (R–MgX), the workhorse of organic synthesis — they add to carbonyls to build bigger molecules. With sodium in dry ether, two alkyl halides couple (Wurtz reaction: 2R–X + 2Na → R–R + 2NaX), useful only for symmetric alkanes.

Haloarenes undergo electrophilic substitution on the ring: the halogen is ortho/para directing but deactivating (it donates by resonance but withdraws by induction). Chlorobenzene with Cl₂/FeCl₃ gives mainly p-dichlorobenzene; with concentrated HNO₃/H₂SO₄ it gives p-chloronitrobenzene; with CH₃Cl/AlCl₃ (Friedel–Crafts) it gives p-chlorotoluene. The Wurtz–Fittig reaction (aryl halide + alkyl halide + Na) gives alkylbenzene; the Fittig reaction (two aryl halides + Na) gives biphenyl.

Polyhalogen compounds and the environment: chloroform (CHCl₃) is oxidised by air to poisonous phosgene, so it is stored in dark bottles with a little ethanol; carbon tetrachloride depletes ozone and is toxic to the liver; DDT (dichlorodiphenyltrichloroethane) was a powerful insecticide but persists in food chains (biomagnification) and is now banned in most countries; freons (chlorofluorocarbons) from refrigerators destroy stratospheric ozone, so they are being replaced.

SN1 vs SN2

Feature	SN1	SN2
Steps	Two (via carbocation)	One (concerted)
Rate law	Rate = k[R–X]	Rate = k[R–X][Nu⁻]
Substrate order	3° > 2° > 1°	CH₃ > 1° > 2° > 3°
Stereochemistry	Racemisation	Inversion (Walden)
Favoured solvent	Polar protic (water, alcohol)	Polar aprotic (DMSO, acetone)

Key Terms / Important Terms

• Haloalkane / haloarene: Halogen on an sp³ carbon / on an sp² ring carbon.
• Nucleophile: An electron-rich species that attacks a positive carbon.
• SN1 / SN2: Unimolecular / bimolecular nucleophilic substitution.
• Walden inversion: Inversion of configuration in SN2 back-side attack.
• Racemisation: Formation of equal d and l enantiomers, cancelling optical activity.
• Saytzeff’s rule: Elimination gives the more substituted alkene.
• Grignard reagent: R–MgX in dry ether; adds to carbonyls to extend chains.
• Finkelstein reaction: Halogen exchange: R–Cl + NaI → R–I + NaCl in acetone.

Practice set — try it yourself

1. Classify: CH₃CH₂Br, (CH₃)₃CBr, C₆H₅Cl as 1°/2°/3° or aryl, and predict which reacts fastest by SN1 and by SN2.
2. Why does thionyl chloride (SOCl₂) give the purest alkyl chloride from an alcohol?
3. Explain Walden inversion with a diagram in words.
4. Why does chlorobenzene not react with NaOH under ordinary conditions, while chloroethane does?
5. Write the Saytzeff product of 2-bromobutane with alcoholic KOH.

Questions with Answers — full answers

1. Arrange in increasing SN2 reactivity: (CH₃)₃CBr, CH₃Br, CH₃CH₂Br, (CH₃)₂CHBr. Justify. Answer: CH₃Br > CH₃CH₂Br > (CH₃)₂CHBr > (CH₃)₃CBr. SN2 needs back-side attack; bulkier groups block the nucleophile, so reactivity falls with substitution.
2. Why does SN1 of an optically active halide give a racemic mixture, not a single enantiomer? Answer: The intermediate carbocation is planar, so the nucleophile attacks from either face with equal probability, giving equal amounts of d and l products whose rotations cancel.
3. How would you convert bromoethane to (a) ethanol, (b) ethene, (c) butane? Answer: (a) Aqueous KOH (substitution). (b) Alcoholic KOH, heat (elimination). (c) Sodium in dry ether — Wurtz reaction couples two ethyl groups.
4. What is the Finkelstein reaction? Why does it work in acetone? Answer: R–Cl + NaI → R–I + NaCl in dry acetone. NaCl is insoluble in acetone and precipitates, pulling the equilibrium toward the iodide.
5. Chlorobenzene is ortho/para directing but deactivating. Explain. Answer: The halogen donates electrons by resonance (+R), stabilising the ortho/para arenium ions, so substitution goes ortho/para. But it withdraws by induction (−I) more strongly overall, so the ring is less reactive than benzene.
6. Why is chloroform stored in dark bottles with a little ethanol? Answer: Light and air oxidise chloroform to phosgene (COCl₂), a poison. Ethanol converts any phosgene formed into harmless diethyl carbonate.
7. Give the Sandmeyer reaction for making bromobenzene from aniline. Answer: Aniline is diazotised with NaNO₂/HCl at 0–5 °C to benzenediazonium chloride, then treated with CuBr/HBr to give bromobenzene with loss of N₂.
8. 2-Chlorobutane with alcoholic KOH gives but-2-ene as the major product. Why not but-1-ene? Answer: Saytzeff’s rule: the hydrogen is removed from the more substituted carbon, giving the more substituted, more stable alkene (but-2-ene).

Extension Questions with Answers

1. A tertiary halide reacts with water to give a racemic alcohol with slight excess of the inverted product. Explain the ‘slight excess’. Answer: The carbocation and the leaving halide stay briefly as an ion pair; the halide shields the front face, so back-side attack is slightly favoured — mostly racemisation with a little net inversion.
2. Predict the product of (CH₃)₃CBr with (a) aqueous KOH, (b) alcoholic KOH, (c) NaCN in DMSO. Answer: (a) Tertiary alcohol (SN1). (b) 2-Methylpropene (elimination dominates). (c) Mostly elimination with some substitution — a tertiary halide in a polar aprotic solvent with a strong base/nucleophile favours elimination.
3. Why are benzylic halides reactive in both SN1 and SN2? Answer: The benzylic carbocation is resonance-stabilised (SN1), and the benzylic carbon is also open to back-side attack with a transition state stabilised by the ring (SN2).
4. Arrange by reactivity toward SN1: C₆H₅CH₂Cl, CH₃CH₂Cl, (CH₃)₂CHCl, (CH₃)₃CCl. Answer: (CH₃)₃CCl > C₆H₅CH₂Cl > (CH₃)₂CHCl > CH₃CH₂Cl — tertiary carbocation is most stable, benzylic next by resonance, then secondary, then primary.
5. DDT was banned despite being an effective insecticide. Why? Answer: DDT is not biodegradable; it accumulates in fatty tissues and magnifies up food chains (biomagnification), harming birds and other wildlife long after spraying.

Board-pattern Questions with Answers

1. Write the IUPAC name of (CH₃)₂CHCH₂Br and CH₃CH(Cl)CH₃. Answer: 1-Bromo-2-methylpropane and 2-chloropropane.
2. Which is more reactive toward SN2: CH₃Br or CH₃CH₂Br? Why? Answer: CH₃Br — the methyl carbon has no alkyl groups blocking back-side attack, so the nucleophile approaches most easily.
3. How do you distinguish chloroform from carbon tetrachloride chemically? (Hint: think of the test for the C–H bond.) Answer: Chloroform reacts with alcoholic KOH and aniline to give phenyl isocyanide (carbylamine reaction, foul smell); CCl₄ has no hydrogen on carbon and does not.
4. Convert chlorobenzene to (a) phenol, (b) toluene (name the reaction). Answer: (a) NaOH at high temperature and pressure (Dow’s process), then acidification. (b) Wurtz–Fittig reaction with CH₃Cl and sodium in dry ether.
5. Why does p-dichlorobenzene have a higher melting point than o- or m-dichlorobenzene? Answer: The para isomer is symmetric and packs better in the crystal lattice, so more energy is needed to melt it.