Haloalkanes and Haloarenes Class 12: Reactions, Mechanisms and Board Answers

What Haloalkanes and Haloarenes is about

Weightage: This chapter is Unit VI, Haloalkanes and Haloarenes, which carries 6 of the 70 theory marks.Source: CBSE Chemistry (043) syllabus 2026-27, course structure, Unit VI Haloalkanes and Haloarenes

In real board papers: asked in 8 of the 8 board papers we hold (2023 to 2026), on average 6.1 marks per paper. See the questions

Key concepts

Classification, nomenclature and the C–X bond

NCERT section 6.1 Classification; 6.2 Nomenclature; 6.3 Nature of C–X Bond

Classification:

• By number of halogens: mono-, di-, tri- and polyhalogen compounds.
• Compounds with an sp³ C–X bond: alkyl halides (1°, 2° or 3° depending on the carbon bearing X), allylic halides (X on the carbon next to C=C) and benzylic halides (X on the carbon next to an aromatic ring).
• Compounds with an sp² C–X bond: vinylic halides (X on a C=C carbon) and aryl halides (X on a ring carbon).
• Dihalides: geminal (both X on one carbon, alkylidene halides) and vicinal (X on adjacent carbons, alkylene dihalides).

Naming: in the common system, alkyl halides are named as 'alkyl halide' (isopropyl chloride); in IUPAC they are haloalkanes (2-chloropropane). Haloarenes use locants or o-, m-, p-: 1,4-dibromobenzene (p-dibromobenzene).

The C–X bond: halogens are more electronegative than carbon, so the bond is polar, with $C^{\delta +}$ and $X^{\delta -}$. Down the group from F to I, the bond gets longer and weaker (bond enthalpy falls from C–F to C–I). This is why C–I breaks most easily in substitution reactions.

Worked example

Classify each as primary, secondary or tertiary halide, or as allylic, benzylic, vinylic or aryl halide: (a) CH₃CH(Cl)CH₃ (b) (CH₃)₃CBr (c) CH₂=CHCH₂Cl (d) C₆H₅CH₂Br (e) CH₂=CHCl (f) C₆H₅Cl. (3 marks)

Try it first, then show the solutionSolution

1. (a) Cl on a carbon attached to two other carbons: secondary (2°) alkyl halide. (½)
2. (b) Br on a carbon attached to three carbons: tertiary (3°). (½)
3. (c) Cl on the carbon next to C=C: allylic. (½)
4. (d) Br on the carbon next to the benzene ring: benzylic. (½)
5. (e) Cl on a doubly bonded carbon: vinylic. (½)
6. (f) Cl on a ring carbon: aryl halide. (½)

Answer: (a) 2° (b) 3° (c) allylic (d) benzylic (e) vinylic (f) aryl

Preparation of haloalkanes

NCERT section 6.4 Methods of Preparation of Haloalkanes

From alcohols: the –OH is replaced by halogen.

• Concentrated HCl with anhydrous ZnCl₂ (Lucas-type conditions) for 1° and 2° alcohols; 3° alcohols react just on shaking with concentrated HCl at room temperature. Reactivity of alcohols: 3° > 2° > 1°.
• PCl₃ or PCl₅: $3ROH+PC{l}_3\to 3RCl+H_{3}P{O}_3$; $ROH+PC{l}_5\to RCl+POC{l}_3+HCl$. PBr₃ and PI₃ are made in situ from red phosphorus and the halogen.
• Thionyl chloride is preferred: $ROH+SOC{l}_2\to RCl+S{O}_2+HCl$, because both by-products are gases that escape, leaving a pure alkyl chloride.

From hydrocarbons:

• Free radical halogenation of alkanes gives a mixture of isomers, so it is not a good method.
• Addition of HX to alkenes: an unsymmetrical alkene follows Markovnikov's rule (H goes to the carbon that already has more H atoms): $C{H}_{3}CH=C{H}_2+HBr\to C{H}_{3}CHBrC{H}_3$. Adding Br₂ in CCl₄ gives a vicinal dibromide; the brown colour of bromine disappears, a test for unsaturation.

Halogen exchange:

• Finkelstein reaction: alkyl chlorides or bromides + NaI in dry acetone give alkyl iodides; NaCl or NaBr precipitates in acetone, pushing the reaction forward (Le Chatelier).
• Swarts reaction: alkyl chlorides or bromides heated with a metallic fluoride such as AgF, Hg₂F₂, CoF₂ or SbF₃ give alkyl fluorides.

Worked example

(a) Why is thionyl chloride preferred for preparing alkyl chlorides from alcohols? (b) Write the equations for the Finkelstein and Swarts reactions using bromoethane. (3 marks)

Try it first, then show the solutionSolution

1. (a) $C{H}_{3}C{H}_{2}OH+SOC{l}_2\to C{H}_{3}C{H}_{2}Cl+S{O}_2+HCl$: both SO₂ and HCl are gases and escape, so the alkyl chloride is obtained pure. (1)
2. (b) Finkelstein: $C{H}_{3}C{H}_{2}Br+NaI\stackrel{\text{dry acetone}}{\to }C{H}_{3}C{H}_{2}I+NaBr$ (NaBr precipitates). (1)
3. Swarts: $C{H}_{3}C{H}_{2}Br+AgF\to C{H}_{3}C{H}_{2}F+AgBr$. (1)

Answer: Gaseous by-products give a pure product; CH₃CH₂Br + NaI → CH₃CH₂I + NaBr (acetone); CH₃CH₂Br + AgF → CH₃CH₂F + AgBr.

Preparation of haloarenes and physical properties

NCERT section 6.5 Preparation of Haloarenes; 6.6 Physical Properties

Haloarenes cannot be made from phenols with HX, because the C–O bond in phenols has partial double bond character and is hard to break.

• Electrophilic substitution: benzene or toluene + Cl₂ or Br₂ with a Lewis acid (iron or FeCl₃) in the dark gives chloro- or bromoarenes; toluene gives a mixture of o- and p-isomers, which can be separated because their melting points differ a lot. Iodination needs an oxidising agent (HNO₃ or HIO₄) to remove the HI formed, because the reaction is reversible. Fluorination is too violent.
• Sandmeyer reaction: aniline is diazotised with NaNO₂ + HCl at 273–278 K to give benzene diazonium chloride. Its solution with cuprous chloride or cuprous bromide replaces –N₂⁺ by Cl or Br: $C_6{H}_5{N}_{2}Cl\stackrel{C{u}_{2}C{l}_2}{\to }{C}_6{H}_{5}Cl+N_2$. Iodobenzene is obtained simply by shaking the diazonium salt with KI.

Physical properties:

• Haloalkanes have higher boiling points than hydrocarbons of similar mass, because of stronger dipole–dipole and van der Waals forces.
• For the same alkyl group: RI > RBr > RCl > RF (larger size and mass).
• Among isomers, branching lowers the boiling point (smaller surface area): 1-bromobutane > 2-bromobutane > 2-bromo-2-methylpropane.
• Para-dihalobenzenes have higher melting points than ortho and meta isomers, because their symmetry lets them fit better in the crystal lattice.
• Haloalkanes are only slightly soluble in water: the energy needed to break hydrogen bonds between water molecules is not regained by new attractions with the haloalkane. They dissolve in organic solvents.
• Bromo-, iodo- and polychloro-derivatives are denser than water.

Worked example

Arrange in increasing order of boiling point and explain: 1-bromobutane, 2-bromo-2-methylpropane, 2-bromobutane. Also explain why p-dichlorobenzene has a higher melting point than o-dichlorobenzene. (3 marks)

Try it first, then show the solutionSolution

1. All three are C₄H₉Br, so their masses are equal; branching makes the molecule more compact, reducing surface area and van der Waals forces. (1)
2. Order: 2-bromo-2-methylpropane < 2-bromobutane < 1-bromobutane. (1)
3. p-Dichlorobenzene is symmetrical and packs closely in the crystal lattice, so more energy is needed to melt it. (1)

Answer: 2-bromo-2-methylpropane < 2-bromobutane < 1-bromobutane; the para isomer packs better in the crystal.

Nucleophilic substitution: SN2 and SN1 mechanisms

NCERT section 6.7.1 Reactions of Haloalkanes (I) Nucleophilic substitution reactions

A nucleophile (electron-rich) attacks the δ+ carbon of R–X and the halide leaves. Examples: OH⁻ gives alcohols, OR′⁻ gives ethers, CN⁻ gives nitriles, NH₃ gives amines.

Ambident nucleophiles can attack through two atoms. KCN (ionic) gives alkyl cyanides (attack through C), while AgCN (mainly covalent) gives isocyanides (attack through N). Similarly KNO₂ gives alkyl nitrites (through O) and AgNO₂ gives nitroalkanes (through N).

SN2 (substitution, nucleophilic, bimolecular):

• One step. The nucleophile attacks from the side opposite the halogen; in the transition state the carbon is bonded to both, partly; then the halide leaves.
• Rate = k[R–X][Nu⁻], second order.
• Configuration is inverted (like an umbrella turning inside out).
• Bulky groups hinder the approach of the nucleophile, so reactivity is CH₃X > 1° > 2° > 3°.

SN1 (substitution, nucleophilic, unimolecular):

• Two steps, usually in polar protic solvents (water, alcohols, acetic acid). Step 1 (slow): C–X breaks to give a carbocation. Step 2 (fast): the nucleophile attacks the carbocation.
• Rate = k[R–X], first order.
• Reactivity follows carbocation stability: 3° > 2° > 1° > CH₃X. Allylic and benzylic halides are very reactive by SN1, because their carbocations are stabilised by resonance.

For both mechanisms, with the same alkyl group the reactivity order is R–I > R–Br > R–Cl > R–F (weaker bond breaks faster).

Worked example

Which compound in each pair reacts faster by SN2, and why? (a) CH₃CH₂Br or (CH₃)₃CBr (b) CH₃CH₂CH₂Br or CH₃CH₂CH₂I (2 marks)

Try it first, then show the solutionSolution

1. (a) CH₃CH₂Br: it is a primary halide; the three methyl groups of (CH₃)₃CBr crowd the back of the carbon and block the nucleophile (steric hindrance). (1)
2. (b) CH₃CH₂CH₂I: the C–I bond is weaker than C–Br, so iodide leaves more easily. (1)

Answer: (a) CH₃CH₂Br (less hindered) (b) CH₃CH₂CH₂I (weaker C–I bond)

Stereochemistry: optical activity, chirality, inversion and racemisation

NCERT section 6.7.1 Reactions of Haloalkanes (c) Stereochemical aspects of nucleophilic substitution reactions

Plane-polarised light vibrates in one plane; it is produced by passing ordinary light through a Nicol prism. Optically active compounds rotate it: dextrorotatory (d or +) to the right, laevorotatory (l or −) to the left. The rotation is measured with a polarimeter.

Chirality: a carbon with four different groups is an asymmetric (stereogenic) carbon. A molecule whose mirror image cannot be superimposed on it is chiral and optically active, e.g. butan-2-ol and 2-chlorobutane. Molecules that can be superimposed on their mirror images are achiral, e.g. propan-2-ol.

Enantiomers: non-superimposable mirror images. They have identical physical properties except the direction of rotation. A 50:50 mixture is a racemic mixture (dl or ±), with zero optical rotation. Converting an enantiomer into a racemic mixture is racemisation.

Outcomes of a substitution at the chiral carbon:

• Retention: the product keeps the same arrangement of groups as the reactant (no bond to the chiral carbon broken).
• Inversion: the new group ends up on the side opposite the old one; this is what SN2 gives. For example, (−)-2-bromooctane with OH⁻ gives (+)-octan-2-ol.
• Racemisation: equal amounts of both forms. SN1 gives this because the carbocation is planar (sp²), so the nucleophile can attack from either face with equal chance.

Worked example

Optically active 2-bromobutane is hydrolysed under SN1 conditions. Is the product, butan-2-ol, optically active? Explain. (2 marks)

Try it first, then show the solutionSolution

1. In SN1, Br⁻ leaves first to give a carbocation, which is planar (sp²). (1)
2. The OH⁻ (or water) can attack this flat ion from either face equally, giving equal amounts of both enantiomers: a racemic mixture, which is optically inactive. (1)

Answer: No, the product is racemic (optically inactive), because the planar carbocation is attacked equally from both sides.

Elimination and reactions with metals

NCERT section 6.7.1 Reactions of Haloalkanes (II) Elimination reactions; (III) Reaction with metals

Elimination (β-elimination, dehydrohalogenation): heating a haloalkane with alcoholic KOH removes H from a β-carbon and X from the α-carbon, giving an alkene. With aqueous KOH, substitution (alcohol) is favoured instead.

Saytzeff (Zaitsev) rule: when more than one alkene can form, the more highly substituted alkene (more alkyl groups on the double bond carbons) is the major product. For 2-bromobutane: but-2-ene is the major product and but-1-ene the minor one.
$C{H}_{3}C{H}_{2}CH(Br)C{H}_3+KOH\stackrel{\text{alc}}{\to }C{H}_{3}CH=CHC{H}_3+KBr+H_{2}O$

Whether substitution or elimination wins depends on the alkyl halide, the strength and size of the base or nucleophile and the conditions. A bulky, strong base and a 3° halide favour elimination.

Grignard reagents: alkyl halides react with magnesium in dry ether to give alkylmagnesium halides, RMgX. The C–Mg bond is polar covalent with carbon δ−, so the R group acts as a nucleophile (carbanion-like). Grignard reagents react with any source of protons (water, alcohols, amines) to give hydrocarbons, so moisture must be avoided: $RMgX+H_{2}O\to RH+Mg(OH)X$.

Wurtz reaction: alkyl halides + sodium in dry ether give an alkane with double the number of carbons:
$2RX+2Na\to RR+2NaX$

Worked example

(a) Predict the major alkene formed when 2-bromopentane is heated with alcoholic KOH, and name the rule. (b) Why must Grignard reagents be prepared in anhydrous conditions? Write an equation. (3 marks)

Try it first, then show the solutionSolution

1. (a) Possible products: pent-1-ene and pent-2-ene. The more substituted alkene, pent-2-ene ($C{H}_{3}CH=CHC{H}_{2}C{H}_3$), is the major product: Saytzeff rule. (1½)
2. (b) Grignard reagents react with even traces of water, taking a proton to form the alkane and destroying the reagent: $C{H}_{3}MgBr+H_{2}O\to C{H}_4+Mg(OH)Br$. (1½)

Answer: (a) Pent-2-ene (Saytzeff rule). (b) Water converts RMgX into RH; CH₃MgBr + H₂O → CH₄ + Mg(OH)Br.

Reactions of haloarenes

NCERT section 6.7.2 Reactions of Haloarenes

Low reactivity towards nucleophilic substitution, because:
1. Resonance: the lone pair of Cl is delocalised into the ring, giving the C–Cl bond partial double bond character, so it is harder to break.
2. Hybridisation: the carbon is sp², which holds the bonding electrons more tightly; the C–X bond is shorter and stronger.
3. The phenyl cation would not be stabilised by resonance, so SN1 is ruled out.
4. The electron-rich ring repels the incoming nucleophile, and backside attack (SN2) is blocked by the ring.

Chlorobenzene gives phenol only under harsh conditions: aqueous NaOH at 623 K and 300 atm, followed by acidification. Electron-withdrawing groups (–NO₂) at ortho and/or para positions make substitution much easier: 4-nitrochlorobenzene reacts at 443 K, 2,4-dinitrochlorobenzene at 368 K, and 2,4,6-trinitrochlorobenzene just with warm water. A meta –NO₂ has no such effect, because it cannot take the negative charge by resonance.

Electrophilic substitution: halogen is ortho, para directing (its lone pair increases electron density at o and p by resonance), but deactivating (its strong −I effect withdraws electrons), so reactions are slower than in benzene. Halogenation, nitration, sulphonation and Friedel–Crafts alkylation and acylation give o- and p-products, with para usually major.

Reactions with sodium:

• Wurtz–Fittig: aryl halide + alkyl halide + Na in dry ether gives an alkylarene: $C_6{H}_{5}Cl+2Na+C{H}_{3}Cl\to C_6{H}_{5}C{H}_3+2NaCl$.
• Fittig: two aryl halides + Na give a diaryl: $2C_6{H}_{5}Cl+2Na\to C_6{H}_5{C}_6{H}_5+2NaCl$.

Worked example

Give two reasons why chlorobenzene is much less reactive than chloromethane towards nucleophilic substitution. How does a nitro group at the para position change this? (3 marks)

Try it first, then show the solutionSolution

1. Resonance gives the C–Cl bond of chlorobenzene partial double bond character, making it stronger and harder to break. (1)
2. The carbon is sp² hybridised, so the C–Cl bond is shorter and stronger; also the phenyl cation is unstable and the electron-rich ring repels nucleophiles (any one). (1)
3. A p-NO₂ group withdraws electrons and stabilises the negatively charged intermediate by resonance, so substitution happens at a much lower temperature (443 K instead of 623 K). (1)

Answer: Partial double bond character and sp² carbon (or unstable phenyl cation); a p-NO₂ group withdraws electrons and makes substitution far easier.

Polyhalogen compounds

NCERT section 6.8 Polyhalogen Compounds

• Dichloromethane (CH₂Cl₂): a solvent in paint removers, a propellant in aerosols, a process solvent for drugs and a metal-cleaning solvent. It harms the human central nervous system; low levels cause slightly impaired hearing and vision, and higher levels cause dizziness, nausea and tingling in the fingers and toes.
• Trichloromethane (chloroform, CHCl₃): a solvent for fats, alkaloids, iodine; used to make the refrigerant R-22. It was once used as a general anaesthetic but has been replaced by safer ones. In air and light it is slowly oxidised to phosgene (carbonyl chloride), a very poisonous gas, so it is stored in closed dark coloured bottles filled to the top: $2CHC{l}_3+O_2\stackrel{\text{light}}{\to }2COC{l}_2+2HCl$.
• Triiodomethane (iodoform, CHI₃): was used as an antiseptic, but its effect is due to the iodine it releases, not to iodoform itself; it has been replaced because of its objectionable smell.
• Tetrachloromethane (CCl₄): used to make refrigerants and aerosol propellants, and as a solvent; it can cause liver cancer, and when released it rises to the stratosphere and depletes the ozone layer.
• Freons (chlorofluorocarbons, such as Freon 12, CCl₂F₂, made from CCl₄ by Swarts reaction): stable, unreactive, non-toxic and easily liquefied; used in aerosols, refrigeration and air conditioning. In the stratosphere they start radical chain reactions that upset the ozone balance.
• DDT (p,p′-dichlorodiphenyltrichloroethane): the first chlorinated organic insecticide, effective against mosquitoes (malaria) and lice (typhus). Insects became resistant, it is highly toxic to fish, and because it is chemically stable and fat-soluble it is not metabolised quickly and builds up in fatty tissues along food chains.

Worked example

(a) Why is chloroform stored in closed, dark coloured bottles filled completely? Write the equation. (b) Give one environmental problem each caused by freons and DDT. (3 marks)

Try it first, then show the solutionSolution

1. (a) In light and air chloroform is oxidised to phosgene, a poisonous gas; dark bottles keep out light and full bottles keep out air. (1)
2. $2CHC{l}_3+O_2\stackrel{\text{light}}{\to }2COC{l}_2+2HCl$ (1)
3. (b) Freons reach the stratosphere and start radical chain reactions that deplete ozone; DDT is stable and fat-soluble, so it accumulates in animal tissues and is toxic to fish. (1)

Answer: (a) To prevent oxidation to poisonous phosgene: 2CHCl₃ + O₂ → 2COCl₂ + 2HCl. (b) Freons deplete ozone; DDT accumulates in fatty tissues and harms fish.

Chemistry in real life: Haloalkanes and Haloarenes

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• SN1 and SN2 Mechanisms
• Stereochemistry of Nucleophilic Substitution
• Reactions of Haloarenes

Haloalkanes and Haloarenes formula sheet

Alcohol + thionyl chloride

$ROH+SOC{l}_2\to RCl+S{O}_2+HCl$

Preferred: both by-products are gases, so the product is pure.

Alcohol + phosphorus halides

$3ROH+PC{l}_3\to 3RCl+H_{3}P{O}_3$

With PCl₅: $ROH+PC{l}_5\to RCl+POC{l}_3+HCl$.

Markovnikov addition of HBr

$C{H}_{3}CH=C{H}_2+HBr\to C{H}_{3}CHBrC{H}_3$

H goes to the carbon that already has more hydrogens.

Finkelstein reaction

$C{H}_{3}C{H}_{2}Br+NaI\stackrel{\text{dry acetone}}{\to }C{H}_{3}C{H}_{2}I+NaBr$

NaBr/NaCl precipitate in acetone and drive the reaction forward.

Swarts reaction

$C{H}_{3}Br+AgF\to C{H}_{3}F+AgBr$

Also with Hg₂F₂, CoF₂ or SbF₃.

Sandmeyer reaction

$C_6{H}_5{N}_{2}Cl\stackrel{C{u}_{2}C{l}_2/HCl}{\to }{C}_6{H}_{5}Cl+N_2$

Diazonium salt from aniline + NaNO₂ + HCl at 273–278 K; with Cu₂Br₂ gives bromobenzene.

SN2 hydrolysis

$C{H}_{3}Cl+O{H}^{-}\to C{H}_{3}OH+C{l}^{-}$

One step, inversion; rate = k[CH₃Cl][OH⁻].

Ambident nucleophiles

$RX+KCN\to RCN+KX;RX+AgCN\to RNC+AgX$

Likewise KNO₂ gives R–O–N=O (alkyl nitrite) and AgNO₂ gives R–NO₂ (nitroalkane).

Dehydrohalogenation (Saytzeff)

$C{H}_{3}C{H}_{2}CH(Br)C{H}_3+KOH\stackrel{\text{alc}}{\to }C{H}_{3}CH=CHC{H}_3+KBr+H_{2}O$

More substituted alkene is the major product.

Grignard reagent

$C{H}_{3}C{H}_{2}Br+Mg\stackrel{\text{dry ether}}{\to }C{H}_{3}C{H}_{2}MgBr$

With water: $RMgX+H_{2}O\to RH+Mg(OH)X$.

Wurtz reaction

$2C{H}_{3}Br+2Na\stackrel{\text{dry ether}}{\to }C{H}_{3}C{H}_3+2NaBr$

Doubles the carbon chain.

Chlorobenzene to phenol

$C_6{H}_{5}Cl+2NaOH\stackrel{623K,300atm}{\to }{C}_6{H}_{5}ONa+NaCl+H_{2}O$

Then acidify (H⁺) to get phenol.

Electrophilic substitution of chlorobenzene

$C_6{H}_{5}Cl+HN{O}_3\stackrel{conc.H_{2}S{O}_4}{\to }Cl{C}_6{H}_{4}N{O}_2+H_{2}O$

o- and p-nitrochlorobenzene; para is major.

Friedel–Crafts on chlorobenzene

$C_6{H}_{5}Cl+C{H}_{3}Cl\stackrel{anhyd.AlC{l}_3}{\to }Cl{C}_6{H}_{4}C{H}_3+HCl$

1-chloro-2-methylbenzene (minor) and 1-chloro-4-methylbenzene (major).

Wurtz–Fittig reaction

$C_6{H}_{5}Cl+2Na+C{H}_{3}Cl\stackrel{\text{dry ether}}{\to }{C}_6{H}_{5}C{H}_3+2NaCl$

Fittig: $2C_6{H}_{5}Cl+2Na\to C_6{H}_5{C}_6{H}_5+2NaCl$.

Oxidation of chloroform

$2CHC{l}_3+O_2\stackrel{\text{light}}{\to }2COC{l}_2+2HCl$

Phosgene is poisonous; store in dark, full bottles.

Haloalkanes and Haloarenes: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the board asks Haloalkanes and Haloarenes

The question patterns to expect, and the method that works for each.

Tricks to remember Haloalkanes and Haloarenes

Short memory aids. Say each one aloud once, then use it on a question the same day.

• SN2 hates crowds, SN1 loves stable cations

  SN2: methyl > 1° > 2° > 3° (crowding blocks the back door). SN1: 3° > 2° > 1° > methyl (stable carbocation). Benzylic and allylic are fast by SN1.

  Use it when: Every reactivity order question.

• K is Carbon, Ag is the other atom

  KCN → cyanide (R–C≡N), AgCN → isocyanide (R–N≡C). KNO₂ → nitrite (R–O–N=O), AgNO₂ → nitro (R–NO₂). Ionic K salts attack through C or O; covalent Ag salts through N.

  Use it when: Ambident nucleophile questions.

• Alcoholic KOH Alkenes, Aqueous KOH Alcohols

  AAA: Alcoholic gives Alkene; aqueous gives alcohol.

  Use it when: Choosing between elimination and substitution.

• Saytzeff: the richer get richer

  In elimination, the double bond goes where it gets the most alkyl groups (H is removed from the carbon with fewer H atoms).

  Use it when: Predicting the major alkene.

• Fink = I, Swarts = F

  Finkelstein puts in Iodine (NaI, acetone); Swarts puts in Fluorine (AgF, SbF₃).

  Use it when: Halogen exchange questions.

Haloalkanes and Haloarenes: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Haloalkanes and Haloarenes

The slips examiners see most often, and what to write instead.

• ✗Mistake: Using aqueous KOH for elimination

  Wrong working

  CH₃CH₂Br + KOH(aq) → CH₂=CH₂ + KBr + H₂O

  Right working

  CH₃CH₂Br + KOH(alc) → CH₂=CH₂ + KBr + H₂O; with aqueous KOH: CH₃CH₂Br + KOH(aq) → CH₃CH₂OH + KBr

  ✓Do this: Write 'alc.' or 'aq.' on the arrow every time.

  Usually costs: The equation mark.

• ✗Mistake: Stating SN2 order as 3° > 2° > 1°

  Students mix up SN1 and SN2 orders.

  ✓Do this: Link SN2 to steric hindrance (back-side attack) and SN1 to carbocation stability.

  Usually costs: 1 mark in an ordering question.

• ✗Mistake: Saying SN1 gives inversion

  ✓Do this: SN1 goes through a planar carbocation, giving racemisation (in practice with some inversion); SN2 gives complete inversion.

  Usually costs: 1 mark in a stereochemistry question.

• ✗Mistake: Missing conditions on named reactions

  Wrong working

  C₆H₅Cl + NaOH → C₆H₅OH + NaCl

  Right working

  C₆H₅Cl + 2NaOH → C₆H₅ONa + NaCl + H₂O at 623 K and 300 atm, then H⁺ gives phenol

  ✓Do this: Conditions (dry ether, dry acetone, 623 K/300 atm, 273–278 K) are part of the answer.

  Usually costs: ½ mark per missing condition.

• ✗Mistake: Explaining haloarene reactivity only by 'benzene is stable'

  ✓Do this: Name the specific reasons: partial double bond character by resonance, sp² carbon, unstable phenyl cation, repulsion by the π cloud.

  Usually costs: 1 mark in a 2-mark reason.

NCERT exercise guide: Haloalkanes and Haloarenes

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Haloalkanes and Haloarenes: previous year questions

• Asked in 8 of the 8 board papers we hold (2023 to 2026), on average 6.1 marks per paper; 28 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Haloalkanes and Haloarenes

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

Every Haloalkanes and Haloarenes question, by year

Practice questions on Haloalkanes and Haloarenes

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQBoard level

   The reaction of an alkyl chloride with NaI in dry acetone is called:
   (a) Swarts reaction (b) Finkelstein reaction (c) Wurtz reaction (d) Sandmeyer reaction

   Hint

   Which one puts iodine in?

   Answer

   (b) Finkelstein reaction

2. Q2 · 1 mark · Assertion-reasonBoard level

   Assertion (A): Chlorobenzene is less reactive than chloromethane towards nucleophilic substitution.
   Reason (R): The C–Cl bond in chlorobenzene has partial double bond character due to resonance.
   (a) Both A and R are true and R is the correct explanation of A.
   (b) Both A and R are true but R is not the correct explanation of A.
   (c) A is true but R is false.
   (d) A is false but R is true.

   Hint

   Is resonance one of the reasons?

   Answer

   (a)

3. Q3 · 2 marks · Very short answerBoard level

   What are enantiomers? Draw (in words) the two enantiomers of 2-chlorobutane. (2 marks)

   Hint

   Non-superimposable mirror images.

   Answer

   Enantiomers are stereoisomers that are non-superimposable mirror images of each other (1). In 2-chlorobutane, C2 carries H, Cl, CH₃ and C₂H₅; one enantiomer and its mirror image (groups arranged in mirror order around C2) cannot be superimposed (1).

4. Q4 · 3 marks · Short answerBoard level

   Write equations for the preparation of (a) 2-bromopropane from propene, (b) chloroethane from ethanol using thionyl chloride, and (c) iodobenzene from benzene diazonium chloride. (3 marks)

   Hint

   Markovnikov; SOCl₂; KI.

   Answer

   (a) CH₃CH=CH₂ + HBr → CH₃CHBrCH₃ (1). (b) C₂H₅OH + SOCl₂ → C₂H₅Cl + SO₂ + HCl (1). (c) C₆H₅N₂Cl + KI → C₆H₅I + KCl + N₂ (1).

5. Q5 · 2 marks · Very short answerHarder board level

   Which compound in each pair undergoes SN1 faster? (a) (CH₃)₃CCl or CH₃CH₂CH₂CH₂Cl (b) C₆H₅CH₂Cl or C₆H₅CH₂CH₂Cl (2 marks)

   Hint

   Compare carbocation stability.

   Answer

   (a) (CH₃)₃CCl: it forms a stable 3° carbocation (1). (b) C₆H₅CH₂Cl: it forms a resonance-stabilised benzylic carbocation (1).

6. Q6 · 3 marks · Short answerHarder board level

   How will you convert (a) benzene to toluene, (b) ethanol to ethyl fluoride, (c) bromoethane to butane? (3 marks)

   Hint

   Use Wurtz–Fittig, Swarts and Wurtz.

   Answer

   (a) Benzene + Cl₂/FeCl₃ → chlorobenzene; then Wurtz–Fittig with CH₃Cl and Na in dry ether → toluene (1). (b) Ethanol + PBr₃ (or HBr) → bromoethane; then Swarts with AgF → ethyl fluoride (1). (c) 2CH₃CH₂Br + 2Na (dry ether) → CH₃CH₂CH₂CH₃ + 2NaBr (Wurtz) (1).

7. Q7 · 3 marks · Short answerHarder board level

   Explain why (a) the boiling points of haloalkanes are higher than those of alkanes of similar molecular mass, (b) RI > RBr > RCl in boiling point, and (c) neopentyl-type branching lowers the boiling point. (3 marks)

   Hint

   Think about polarity, size and surface area.

   Answer

   (a) Polar C–X bonds add dipole–dipole attractions, and halogens increase van der Waals forces (1). (b) Size and mass increase from Cl to I, so van der Waals forces increase (1). (c) Branching makes molecules more spherical, reducing surface contact and van der Waals forces (1).

8. Q8 · 5 marks · Long answerHarder board level

   (a) Describe the mechanism of SN1 hydrolysis of 2-bromo-2-methylpropane. (2) (b) An optically active alkyl halide gives an optically inactive product in water. Explain. (1) (c) Arrange C₆H₅CH₂Br, CH₃CH₂Br and (CH₃)₂CHBr in decreasing order of SN1 reactivity. (1) (d) Why are allylic halides reactive by SN1? (1) (5 marks)

   Hint

   Slow ionisation, then fast attack.

   Answer

   (a) Step 1 (slow): (CH₃)₃C–Br → (CH₃)₃C⁺ + Br⁻; step 2 (fast): (CH₃)₃C⁺ + OH⁻ → (CH₃)₃COH; rate = k[RBr] (2). (b) The planar carbocation is attacked from both faces equally, giving a racemic mixture (1). (c) C₆H₅CH₂Br > (CH₃)₂CHBr > CH₃CH₂Br (1). (d) The allylic carbocation is stabilised by resonance (1).

9. Q9 · 3 marks · Short answerHOTS / competency

   Suppose a student heats 2-bromo-2-methylbutane with alcoholic KOH and obtains two alkenes. (a) Draw (in words) both alkenes. (b) Which is major and why? (c) What would be the main product with aqueous KOH? (3 marks)

   Hint

   Remove H from C1 or C3.

   Answer

   (a) 2-methylbut-2-ene, (CH₃)₂C=CHCH₃, and 2-methylbut-1-ene, CH₂=C(CH₃)CH₂CH₃ (1). (b) 2-methylbut-2-ene is major: it is more substituted (trisubstituted), as predicted by the Saytzeff rule (1). (c) 2-methylbutan-2-ol, by substitution (1).

10. Q10 · 2 marks · Very short answerHOTS / competency

    p-Nitrochlorobenzene reacts with aqueous NaOH at 443 K, but m-nitrochlorobenzene needs much harsher conditions. Explain. (2 marks)

    Hint

    Where can the negative charge of the intermediate go?

    Answer

    When OH⁻ adds, a negative charge spreads to ortho and para positions; a para –NO₂ can take this charge on its oxygen by resonance, stabilising the intermediate (1). A meta –NO₂ cannot delocalise the charge by resonance, so there is no such stabilisation (1).

11. Q11 · 2 marks · Very short answerHOTS / competency

    An alkane C₈H₁₈ gives only one monochloro product when chlorinated in sunlight. (a) Identify the alkane and give its IUPAC name. (b) Name the monochloro product. (c) Why is free radical chlorination usually a poor way to make a single haloalkane? (2 marks)

    Hint

    One product means every H atom is equivalent: think of a highly symmetrical molecule made only of CH₃ groups around a central C–C bond.

    Answer

    (a) (CH₃)₃C–C(CH₃)₃, 2,2,3,3-tetramethylbutane: all 18 H atoms are on equivalent CH₃ groups (1). (b) 1-chloro-2,2,3,3-tetramethylbutane (½). (c) Most alkanes have non-equivalent H atoms, so a mixture of isomers forms that is hard to separate (½).

12. Q12 · 1 mark · MCQHarder board level

    Which of the following is used as an antiseptic, its action being due to the iodine it releases?
    (a) CHCl₃ (b) CHI₃ (c) CCl₄ (d) CH₂Cl₂

    Hint

    Tri-iodo.

    Answer

    (b) CHI₃ (iodoform)

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Why are haloalkanes polar?Show answer

   Halogens are more electronegative than carbon, so the C–X bond has δ+ on carbon and δ− on halogen.

2. 2.Name the reaction of an alkyl bromide with NaI in dry acetone.Show answer

   Finkelstein reaction.

3. 3.What does AgCN give with an alkyl halide?Show answer

   An isocyanide (R–NC).

4. 4.Give the rate law for SN2 and SN1.Show answer

   SN2: rate = k[RX][Nu⁻]; SN1: rate = k[RX].

5. 5.What is a racemic mixture?Show answer

   An equimolar mixture of two enantiomers; it is optically inactive.

6. 6.State the Saytzeff rule.Show answer

   In dehydrohalogenation, the more highly substituted alkene is the major product.

7. 7.What is a Grignard reagent?Show answer

   An alkylmagnesium halide, RMgX, made from an alkyl halide and Mg in dry ether.

8. 8.Under what conditions does chlorobenzene give phenol?Show answer

   With aqueous NaOH at 623 K and 300 atm, followed by acidification.

9. 9.Is chlorine an activating or deactivating group, and how does it direct?Show answer

   Deactivating (−I), but ortho, para directing (+R).

10. 10.Why is DDT a problem in the environment?Show answer

    It is stable and fat-soluble, so it accumulates in fatty tissues along the food chain; it is toxic to fish and insects became resistant.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you classify any halide as 1°, 2°, 3°, allylic, benzylic, vinylic or aryl?
• ☐Can you write SOCl₂, PCl₃ and PCl₅ reactions with an alcohol?
• ☐Can you write the Finkelstein, Swarts and Sandmeyer reactions with conditions?
• ☐Can you draw the SN2 transition state and the SN1 carbocation steps?
• ☐Can you explain why SN1 gives racemisation and SN2 gives inversion?
• ☐Can you predict the major product of dehydrohalogenation using the Saytzeff rule?
• ☐Can you give four reasons for the low reactivity of haloarenes in nucleophilic substitution?
• ☐Can you write Wurtz, Wurtz–Fittig and Fittig reactions and two uses and one hazard each of CHCl₃, CCl₄, freons and DDT?

Haloalkanes and Haloarenes: frequently asked questions