Haloalkanes and Haloarenes
Welcome to HSLC Guru! This page provides a complete English-medium guide to Class 12 Chemistry Chapter 10 — Haloalkanes and Haloarenes as per the ASSEB (Assam State School Education Board) syllabus. You will find a clear summary, detailed Question & Answer practice (1-mark, 2–3 mark, and 5–7 mark), MCQs, fill-in-the-blanks, true/false, and a glossary that builds confidence for HS Final examinations and competitive tests like NEET and JEE.
Summary
Classification and Nomenclature: Haloalkanes (alkyl halides) and haloarenes (aryl halides) are formed when hydrogen atoms of aliphatic or aromatic hydrocarbons are replaced by halogen atoms (F, Cl, Br, I). They are classified by the number of halogens (mono-, di-, poly-halogen compounds) and by the carbon bearing the halogen: primary (1°), secondary (2°), and tertiary (3°). Special structural classes include allylic halides (X attached to sp³ carbon next to a C=C), benzylic halides (X on sp³ carbon adjacent to a benzene ring), vinylic halides (X on sp² carbon of a C=C), and aryl halides (X directly on an aromatic ring). IUPAC names use the prefix halo- (chloro, bromo, iodo, fluoro) with the lowest locant; common names use alkyl halide format (e.g., isopropyl chloride for 2-chloropropane).
Methods of Preparation: Haloalkanes are commonly prepared from alcohols using HX (with ZnCl₂ for 1° alcohols — Lucas reagent), PCl₃, PCl₅, or SOCl₂ (Darzens method, gives pure product as SO₂ and HCl escape). From hydrocarbons, free-radical halogenation of alkanes gives a mixture, while addition of HX to alkenes follows Markovnikov’s rule (H adds to the carbon bearing more H atoms). In presence of peroxide, HBr addition follows the anti-Markovnikov (Kharasch / peroxide) effect via a free-radical mechanism. Allylic halogenation with NBS gives allyl/benzyl bromides. Halide exchange (Finkelstein and Swarts) converts one halide to another. Haloarenes are prepared by electrophilic aromatic halogenation of arenes with X₂ in presence of a Lewis acid (FeX₃ or AlX₃), or from arenediazonium salts via the Sandmeyer and Gattermann reactions.
Physical and Chemical Properties: Haloalkanes are colourless, sweet-smelling liquids/solids; their boiling points rise with size and atomic number of halogen and decrease with branching; they are insoluble in water but dissolve in organic solvents; densities of bromides and iodides are greater than water. The C–X bond is polar (δ+ on C, δ− on X), so the carbon is open to attack by nucleophiles and the molecule undergoes elimination, organometallic and reduction reactions. Nucleophilic substitution proceeds by two limiting pathways: SN2 (one-step, bimolecular, second-order, backside attack giving Walden inversion of configuration; favoured by 1° substrates and strong nucleophiles in polar aprotic solvents) and SN1 (two-step, unimolecular, first-order, via carbocation, racemisation; favoured by 3° substrates and polar protic solvents). Elimination follows E2 (concerted, anti-periplanar, second-order) or E1 (carbocation pathway), giving alkenes; Saytzeff’s rule states that the most substituted alkene predominates. Alcoholic KOH favours elimination, while aqueous KOH gives substitution.
Reactions with Metals and Haloarene Behaviour: Haloalkanes form organometallic compounds: with Mg in dry ether they yield Grignard reagents (RMgX); with Na in dry ether two molecules of haloalkane couple to give symmetrical alkanes (Wurtz reaction). The Wurtz–Fittig reaction couples an aryl halide with an alkyl halide using Na/dry ether to give an alkylarene; the Fittig reaction couples two aryl halides. Haloarenes show limited reactivity towards nucleophilic substitution due to resonance stabilisation of the C–X bond, partial double-bond character, sp² hybridised C, and instability of phenyl cation; they undergo nucleophilic substitution only under harsh conditions (e.g., chlorobenzene + NaOH at 623 K and 300 atm gives phenol — Dow’s process) or when activated by –NO₂ groups at ortho/para positions. They readily undergo electrophilic aromatic substitution (EAS) — halogenation, nitration, sulphonation and Friedel–Crafts reactions — predominantly at the ortho and para positions because halogens are deactivating but ortho/para directing. Uses and environmental impact: haloalkanes are valued as solvents (CHCl₃, CCl₄), refrigerants, propellants, anaesthetics (CHCl₃) and intermediates; however, polyhalogen compounds and chlorofluorocarbons (CFCs, freons) deplete the stratospheric ozone layer, while DDT pollutes the food chain — leading to the Montreal Protocol restrictions.
Question & Answer
Very Short Answer Type (1 Mark)
Q1. What is a haloalkane?
Answer: A haloalkane is a compound formed when one or more hydrogen atoms of an alkane are replaced by halogen atoms; general formula CnH2n+1X.
Q2. Give an example of a vinylic halide.
Answer: Chloroethene (vinyl chloride), CH₂=CH–Cl, in which the halogen is bonded directly to an sp² carbon of a C=C double bond.
Q3. Write the IUPAC name of (CH₃)₃C–Br.
Answer: 2-bromo-2-methylpropane (a tertiary alkyl halide; common name tert-butyl bromide).
Q4. Name the reagent used in Darzens method to convert alcohol to alkyl chloride.
Answer: Thionyl chloride (SOCl₂) in presence of pyridine — the by-products SO₂ and HCl are gaseous, giving a pure alkyl halide.
Q5. What is Walden inversion?
Answer: The inversion of configuration about the chiral carbon during an SN2 reaction, in which the nucleophile attacks opposite to the leaving group like an umbrella turning inside out.
Q6. State Saytzeff’s rule.
Answer: In a dehydrohalogenation reaction the alkene that is most substituted (more alkyl groups on the C=C) is the major product because it is the most stable.
Q7. Define Grignard reagent.
Answer: An alkyl/aryl magnesium halide R–Mg–X formed by the action of haloalkane on magnesium turnings in dry ether; an extremely useful organometallic reagent in synthesis.
Q8. What is meant by anti-Markovnikov addition?
Answer: Addition of HBr to an unsymmetrical alkene in presence of peroxide, where Br adds to the carbon bearing more hydrogen atoms — opposite to Markovnikov’s rule (Kharasch effect).
Q9. Why is chlorobenzene less reactive towards nucleophilic substitution?
Answer: Because of resonance stabilisation, partial double-bond character of the C–Cl bond, sp² hybridisation of the ring carbon and the instability of the phenyl cation that would form.
Q10. Write the full form of CFC.
Answer: Chlorofluorocarbon — a class of polyhalogen compounds (e.g., CCl₂F₂, freon-12) responsible for stratospheric ozone depletion.
Q11. Why is iodoform CHI₃ used as an antiseptic?
Answer: Iodoform liberates free iodine on contact with skin, which is the actual antiseptic agent.
Q12. Name a polar aprotic solvent used in SN2 reactions.
Answer: Dimethyl sulphoxide (DMSO) or N,N-dimethylformamide (DMF) or acetone.
Short Answer Type (2–3 Marks)
Q11. Distinguish between primary, secondary and tertiary alkyl halides with examples.
Answer: A primary (1°) alkyl halide has the halogen attached to a carbon connected to only one other carbon, e.g., CH₃CH₂Cl (chloroethane). A secondary (2°) alkyl halide has the halogen on a carbon attached to two other carbons, e.g., (CH₃)₂CHCl (2-chloropropane). A tertiary (3°) alkyl halide has the halogen on a carbon bonded to three other carbons, e.g., (CH₃)₃CCl (tert-butyl chloride). Stability and reactivity of carbocation intermediate: 3° > 2° > 1°, so tertiary halides prefer SN1/E1 while primary halides undergo SN2/E2.
Q12. Explain Markovnikov’s rule with an example.
Answer: Markovnikov’s rule states that when an unsymmetrical reagent like HX adds to an unsymmetrical alkene, the negative part (X) attaches to the more substituted carbon (the carbon with fewer hydrogens), because the more stable carbocation is formed there. Example: CH₃–CH=CH₂ + HBr → CH₃–CHBr–CH₃ (2-bromopropane is the major product, not 1-bromopropane), because the secondary carbocation CH₃CH⁺CH₃ is more stable than the primary CH₃CH₂CH₂⁺.
Q13. Differentiate SN1 and SN2 reactions in three points.
Answer: (i) Mechanism: SN1 is two-step (slow ionisation forming carbocation, then fast nucleophile attack); SN2 is single-step concerted with backside attack. (ii) Kinetics: SN1 is first order, rate = k[RX]; SN2 is second order, rate = k[RX][Nu]. (iii) Stereochemistry & substrate: SN1 gives racemisation and is favoured by 3° substrates in polar protic solvents; SN2 proceeds with Walden inversion and is favoured by 1° substrates in polar aprotic solvents.
Q14. What is the Wurtz reaction? Give one example.
Answer: The Wurtz reaction is the coupling of two molecules of an alkyl halide with metallic sodium in dry ether to form a higher symmetrical alkane: 2 R–X + 2 Na → R–R + 2 NaX. Example: 2 CH₃CH₂Br + 2 Na → CH₃CH₂–CH₂CH₃ (n-butane) + 2 NaBr. The reaction works best for symmetrical alkanes; using two different halides produces a mixture.
Q14a. Why does an aryl halide give an electrophilic substitution at ortho/para positions despite being deactivating?
Answer: Halogens have two opposing effects on the benzene ring: a strong electron-withdrawing inductive effect (–I) that deactivates the ring, and a weaker electron-donating mesomeric effect (+M) that operates through the lone pair on the halogen. The +M effect releases electron density at the ortho and para positions giving rise to extra resonance structures with negative charge there, so the intermediate arenium ion (sigma complex) for ortho/para attack is more stable than for meta attack. Thus halogens are deactivators (overall –I dominates) but ortho/para directors (+M selects positions).
Q14b. Give one similarity and one difference between the Wurtz and Wurtz–Fittig reactions.
Answer: Similarity: Both use sodium metal in dry ether to couple alkyl/aryl halides. Difference: The Wurtz reaction couples two alkyl halides to give a symmetrical alkane (R–R); the Wurtz–Fittig reaction couples an aryl halide with an alkyl halide to give an alkylarene (Ar–R), e.g., bromobenzene + ethyl bromide + Na → ethylbenzene.
Q15. Why are haloalkanes insoluble in water but soluble in organic solvents?
Answer: Although the C–X bond is polar, haloalkane molecules cannot form hydrogen bonds with water; the energy required to break the strong hydrogen-bond network of water is greater than the energy released by weak dipole–dipole interactions with haloalkane, so they are essentially insoluble. They mix freely with organic solvents because new solute–solvent interactions (London forces and weak dipole forces) are similar in magnitude to the solvent–solvent forces that are broken.
Q16. Briefly describe the role of CFCs in ozone layer depletion.
Answer: Chlorofluorocarbons (CFCs) such as CCl₂F₂ rise to the stratosphere unchanged because they are very stable. There, UV radiation cleaves a C–Cl bond producing reactive chlorine atoms (Cl•). Each Cl• destroys ozone in a chain reaction: Cl• + O₃ → ClO• + O₂; ClO• + O → Cl• + O₂. One Cl atom can destroy many thousand ozone molecules, thinning the protective ozone layer and increasing harmful UV radiation reaching the earth’s surface.
Q17a. What is the Sandmeyer reaction? Write a chemical equation.
Answer: When a benzenediazonium salt (ArN₂⁺Cl⁻) is treated with cuprous chloride (CuCl) or cuprous bromide (CuBr) dissolved in the corresponding halogen acid (HCl or HBr), the diazo group is replaced by the halogen to give an aryl halide. This is the Sandmeyer reaction. Example: C₆H₅N₂⁺Cl⁻ + CuCl/HCl → C₆H₅Cl + N₂ + CuCl. With CuCN/KCN, the cyanide replaces N₂ to give an aryl cyanide. The Gattermann variant uses Cu/HX in place of CuX.
Q17b. What is the Finkelstein reaction? Why is dry acetone used as the solvent?
Answer: The Finkelstein reaction converts an alkyl chloride or bromide into an alkyl iodide by reaction with sodium iodide in dry acetone: R–Cl + NaI → R–I + NaCl. NaI is soluble in dry acetone whereas NaCl and NaBr are insoluble; the formed NaCl/NaBr precipitates and the equilibrium shifts towards the right by Le Chatelier’s principle, giving high yields of alkyl iodide. The reaction is a classic SN2 process.
Q17c. What is the Swarts reaction? Give an example.
Answer: The Swarts reaction prepares alkyl fluorides by heating an alkyl chloride or bromide with a metallic fluoride such as AgF, Hg₂F₂, CoF₂ or SbF₃: CH₃–Br + AgF → CH₃–F + AgBr. It is the standard method to introduce fluorine in CFC and freon manufacture.
Long Answer Type (5–7 Marks)
Q17. Describe the methods of preparation of haloalkanes from alcohols, alkanes (free-radical) and alkenes (Markovnikov and anti-Markovnikov).
Answer: (a) From alcohols: Alcohols react with halogen acids (HX): R–OH + HX → R–X + H₂O. Reactivity: HI > HBr > HCl, and 3° > 2° > 1° alcohols. ZnCl₂ (Lucas reagent) catalyses 1° alcohols. With phosphorus halides: 3 R–OH + PX₃ → 3 R–X + H₃PO₃; R–OH + PCl₅ → R–Cl + POCl₃ + HCl. With thionyl chloride (Darzens method): R–OH + SOCl₂ → R–Cl + SO₂↑ + HCl↑ — gives the purest product. (b) From alkanes — free-radical halogenation: CH₄ + Cl₂ →(hv or heat) CH₃Cl + HCl. The reaction proceeds through initiation, propagation and termination steps and gives a mixture of mono-, di-, tri- and tetra-halides; useful only for symmetrical alkanes. (c) From alkenes: Addition of HX gives an alkyl halide. By Markovnikov’s rule: CH₃–CH=CH₂ + HBr → CH₃–CHBr–CH₃ (major), because the secondary carbocation is more stable. By the anti-Markovnikov / peroxide / Kharasch effect (only with HBr, in presence of peroxide): CH₃–CH=CH₂ + HBr →(R–O–O–R) CH₃–CH₂–CH₂Br, because the reaction now proceeds by a free-radical mechanism in which the more stable secondary carbon radical bears the H and Br ends up on the terminal carbon. With HCl and HI no peroxide effect is observed.
Q18. Compare and contrast SN1 and SN2 mechanisms with the reaction of 2-bromobutane with hydroxide ion. Discuss kinetics and stereochemistry.
Answer: SN2 mechanism (bimolecular nucleophilic substitution): The hydroxide ion attacks the carbon bearing the leaving group on the side opposite to Br⁻; the C–OH bond forms while the C–Br bond breaks in a single concerted step through a five-coordinated transition state. The configuration at the chiral carbon is inverted (Walden inversion). For (R)-2-bromobutane, the product is (S)-2-butanol — pure inverted enantiomer. The reaction is second order overall: rate = k[CH₃CHBrC₂H₅][OH⁻]. It is favoured by 1° substrates, strong nucleophiles, polar aprotic solvents (DMSO, acetone) and weak/non-bulky nucleophiles. SN1 mechanism (unimolecular): The C–Br bond breaks first slowly to form a planar carbocation (rate-determining); then the hydroxide attacks rapidly from either face to give the alcohol. The product is a racemic mixture (50% R + 50% S) because the carbocation is sp² hybridised and planar. Kinetics: first order, rate = k[CH₃CHBrC₂H₅]. Favoured by 3° substrates (stable carbocation), polar protic solvents (water, alcohol), and weaker nucleophiles. Thus 2-bromobutane (a secondary substrate) can react by either pathway depending on conditions: in dilute alkaline alcohol solution at moderate temperature, SN2 dominates with hydroxide as a strong nucleophile; in polar protic solvents at higher temperature, SN1 may compete, giving racemic 2-butanol.
Q19. Explain the chemical reactions of haloarenes — limited reactivity, electrophilic aromatic substitution and nucleophilic substitution under harsh conditions, with examples.
Answer: Limited reactivity: Haloarenes (e.g., chlorobenzene) are far less reactive than haloalkanes towards nucleophilic substitution due to (i) resonance: the lone pairs on Cl interact with the ring giving partial double-bond character to C–Cl; (ii) shorter and stronger C–Cl bond; (iii) sp² hybridised ring carbon (more electronegative than sp³) holds the C–Cl pair tightly; (iv) the phenyl cation formed in SN1 is highly unstable; (v) electron-rich ring repels nucleophiles. Nucleophilic substitution under harsh conditions: Chlorobenzene + NaOH → sodium phenoxide at 623 K, 300 atm pressure (Dow’s process) which on acidification gives phenol. Activation by –NO₂ at o/p positions reduces the harshness: 2,4-dinitrochlorobenzene reacts with NaOH at 368 K to give 2,4-dinitrophenol; 2,4,6-trinitrochlorobenzene reacts with warm water to give picric acid. Electrophilic aromatic substitution (EAS): Halogens are deactivators (electron-withdrawing –I) but ortho/para directors (lone pair donating +M effect). (i) Halogenation: C₆H₅Cl + Cl₂/FeCl₃ → o- and p-dichlorobenzene. (ii) Nitration: C₆H₅Cl + conc. HNO₃/H₂SO₄ → o- and p-nitrochlorobenzene. (iii) Sulphonation: C₆H₅Cl + conc. H₂SO₄ → p-chlorobenzenesulphonic acid (major). (iv) Friedel–Crafts alkylation: C₆H₅Cl + CH₃Cl/anhyd. AlCl₃ → o- and p-chlorotoluene. (v) Friedel–Crafts acylation: C₆H₅Cl + CH₃COCl/AlCl₃ → o- and p-chloroacetophenone. Reaction with metals — Wurtz–Fittig: C₆H₅Br + 2Na + CH₃Br → C₆H₅–CH₃ (toluene) + 2NaBr; Fittig: 2 C₆H₅Br + 2Na → C₆H₅–C₆H₅ (biphenyl) + 2NaBr. Reduction: C₆H₅Cl + Ni/H₂ → C₆H₆ + HCl.
Q20. Describe E1 and E2 elimination mechanisms with reference to dehydrohalogenation of 2-bromo-2-methylbutane and state Saytzeff’s rule.
Answer: Dehydrohalogenation removes H–X from adjacent carbons (β-elimination) using alcoholic KOH to form an alkene. E2 (bimolecular elimination): A single concerted step in which the base removes the β-hydrogen, the C–H and C–X bonds break simultaneously, and the C=C double bond forms; the H and X must be anti-periplanar (180°). Rate = k[RX][B⁻]; second order. Favoured by 2°/3° substrates and strong bases (alkoxides). E1 (unimolecular): Two steps — slow ionisation to a carbocation (rate-determining), then fast loss of β-proton by the solvent or weak base to give the alkene. Rate = k[RX]; first order. Favoured by 3° substrates, polar protic solvents, weak bases and elevated temperature. For 2-bromo-2-methylbutane (CH₃)(C₂H₅)C(Br)(CH₃) two β-H’s are available on different carbons; the alkene mixture contains 2-methyl-2-butene (more substituted) and 2-methyl-1-butene (less substituted). Saytzeff’s rule: The major product of an elimination is the more substituted (more stable) alkene; here 2-methyl-2-butene (a trisubstituted alkene) predominates over the disubstituted 2-methyl-1-butene because hyperconjugation stabilises the more substituted double bond. Anti-Saytzeff (Hofmann) products dominate only with bulky bases such as t-BuO⁻.
Q20a. Discuss the physical properties of haloalkanes — boiling point, density, solubility, dipole moment — and how they vary with the size of alkyl group and the halogen.
Answer: (i) Physical state and colour: Lower halomethanes (CH₃Cl, CH₃F) are gases; medium-sized members (C₂–C₁₈) are liquids; higher members are solids. They are colourless when pure, but many turn yellow on prolonged exposure to light. They have a sweet odour. (ii) Boiling and melting points: The boiling points of haloalkanes increase with increase in the size of the alkyl group and the size of the halogen (because of stronger London dispersion forces and higher polarisability). Order: R–I > R–Br > R–Cl > R–F. For isomers, branching reduces boiling point because the molecules become more compact and have less surface contact. (iii) Density: Bromides and iodides are denser than water; chlorides and fluorides of low molecular mass are less dense than water but density rises with increasing number of halogen atoms (CH₃Cl is lighter, CHCl₃ and CCl₄ are heavier than water). (iv) Solubility: Insoluble in water (cannot form H-bonds with water and the energy required to break water’s H-bond network is more than the energy released by dipole–dipole interactions); freely soluble in non-polar organic solvents (benzene, ether, ethanol). (v) Dipole moment: The C–X bond is polar (δ+ on C, δ− on X). Order of dipole moment: CH₃Cl (1.86 D) > CH₃F (1.85 D) > CH₃Br (1.83 D) > CH₃I (1.62 D); fluorine has a small bond length whereas iodine has a longer but less polar bond.
Q21. Write a note on uses and environmental impact of haloalkanes (CHCl₃, CCl₄, DDT, CFCs).
Answer: Uses: (i) Chloroform (CHCl₃) — solvent for fats, alkaloids and iodine, formerly used as anaesthetic, intermediate in dye industry. (ii) Carbon tetrachloride (CCl₄) — solvent for oils, fats, varnishes, industrial cleaning agent, fire extinguisher (pyrene) for electrical fires. (iii) Iodoform (CHI₃) — antiseptic. (iv) DDT (1,1,1-trichloro-2,2-bis(p-chlorophenyl)ethane) — first major synthetic insecticide. (v) Freons (CFCs) — refrigerants in air-conditioners and refrigerators, propellants in aerosol sprays. (vi) Polytetrafluoroethylene (Teflon) — non-stick coatings. Environmental impact: CCl₄ is highly toxic to liver and on photolysis in upper atmosphere generates Cl• radicals that destroy ozone. CHCl₃ is converted to phosgene (COCl₂) on exposure to air and light — a poisonous gas. DDT is non-biodegradable and accumulates in the food chain (bioaccumulation), harming fish-eating birds; banned in many countries. CFCs travel unchanged to the stratosphere where UV light produces Cl• that catalytically destroys ozone (Cl• + O₃ → ClO• + O₂; ClO• + O → Cl• + O₂), thinning the ozone layer and increasing skin cancer, cataracts and crop damage from UV. The Montreal Protocol (1987) phased out CFCs in favour of HFCs and HCFCs that are less harmful to ozone.
Q21a. Describe the principal nucleophilic substitution reactions of a typical haloalkane (R–X) with: (i) aqueous KOH, (ii) alcoholic KOH, (iii) NaCN, (iv) AgCN, (v) KNO₂, (vi) AgNO₂, (vii) NH₃, (viii) RONa, (ix) RCOOAg, (x) LiAlH₄.
Answer: (i) aq. KOH: substitution gives an alcohol — R–X + KOH(aq) → R–OH + KX. (ii) alc. KOH: elimination gives an alkene — R–CH₂–CHX–R’ + KOH(alc) → R–CH=CH–R’ + KX + H₂O. (iii) NaCN (alcoholic): R–X + NaCN → R–CN (alkyl cyanide / nitrile, SN2 via C-attack — CN⁻ is an ambident nucleophile and the carbon end attacks because Na–CN is mostly ionic). (iv) AgCN (ethanolic): R–X + AgCN → R–NC (alkyl isocyanide / carbylamine — the more polarisable nitrogen end attacks because Ag–CN is largely covalent). (v) KNO₂: R–X + KNO₂ → R–O–N=O (alkyl nitrite). (vi) AgNO₂: R–X + AgNO₂ → R–NO₂ (nitroalkane). (vii) NH₃ (alcoholic, sealed tube): R–X + NH₃ → R–NH₂ (primary amine; further alkylation gives 2°, 3° amines and quaternary ammonium salts — Hofmann’s ammonolysis). (viii) RONa (Williamson synthesis): R–X + R’ONa → R–O–R’ + NaX (ether). (ix) RCOOAg: R–X + R’COOAg → R’COOR + AgX (ester). (x) LiAlH₄: reduction — R–X + 2[H] → R–H + HX (alkane).
Q22. Discuss in detail the optical activity of haloalkanes with reference to chirality, enantiomers, racemic mixture and the relation between configuration and SN1 / SN2 reaction.
Answer: Chirality: A carbon atom carrying four different groups is a stereogenic / chiral centre and the molecule lacks a plane of symmetry. The molecule and its mirror image are non-superimposable and are called enantiomers. Enantiomers rotate the plane of plane-polarised light by equal magnitudes but in opposite directions — one is dextrorotatory (+, d) and the other is laevorotatory (−, l). An equimolar mixture of two enantiomers is called a racemic mixture (±) and is optically inactive due to external compensation. The configuration around a chiral centre is described by the R/S system (Cahn–Ingold–Prelog priority rules): when the lowest-priority group is pointed away, if the remaining three priorities decrease clockwise the centre is R; anticlockwise, S. Effect of SN2: The transition state has the nucleophile and leaving group on opposite sides; the carbon hybridisation passes through sp² and emerges sp³ with the three other groups inverted (umbrella flip). Hence SN2 always proceeds with complete inversion of configuration (Walden inversion). Reaction of (R)-2-bromobutane with hydroxide therefore gives pure (S)-2-butanol. Effect of SN1: The first step generates a planar sp² carbocation that is achiral; the nucleophile then attacks the two faces of the cation with equal probability, producing equal amounts of the R and S products — that is, a racemic mixture. In practice some retention/inversion bias is seen due to ion-pair effects but ideally racemisation is complete.
Multiple Choice Questions (MCQs)
Q1. The IUPAC name of (CH₃)₂CHCH₂Cl is:
(a) isobutyl chloride (b) 1-chloro-2-methylpropane (c) 2-chloro-2-methylpropane (d) 2-methyl-1-chloropropane
Answer: (b) 1-chloro-2-methylpropane.
Q2. Which of the following undergoes SN1 reaction the fastest?
(a) CH₃Cl (b) CH₃CH₂Cl (c) (CH₃)₂CHCl (d) (CH₃)₃CCl
Answer: (d) (CH₃)₃CCl — gives the most stable tertiary carbocation.
Q3. Walden inversion is observed in:
(a) SN1 (b) SN2 (c) E1 (d) E2
Answer: (b) SN2.
Q4. Markovnikov’s addition of HBr to propene gives:
(a) 1-bromopropane (b) 2-bromopropane (c) propene-1-bromide (d) 1,2-dibromopropane
Answer: (b) 2-bromopropane.
Q5. Grignard reagent is prepared by reaction of haloalkane with:
(a) Na in dry ether (b) Mg in dry ether (c) Zn–Hg/HCl (d) LiAlH₄
Answer: (b) Mg in dry ether.
Q6. Chlorobenzene reacts with NaOH at 623 K and 300 atm to give:
(a) benzene (b) phenol (c) benzoic acid (d) sodium benzoate
Answer: (b) phenol (Dow’s process).
Q7. Saytzeff’s rule predicts the formation of the:
(a) least substituted alkene (b) most substituted alkene (c) cis isomer (d) trans isomer
Answer: (b) most substituted alkene.
Q8. The reagent that converts ethanol cleanly to chloroethane is:
(a) Cl₂/hv (b) HCl alone (c) SOCl₂/pyridine (d) NaCl
Answer: (c) SOCl₂/pyridine (Darzens).
Q9. Polyhalogen compound responsible for ozone depletion is:
(a) DDT (b) CHCl₃ (c) CFC (d) CHI₃
Answer: (c) CFC.
Q10. Wurtz–Fittig reaction couples an aryl halide with an alkyl halide using:
(a) Mg/dry ether (b) Na/dry ether (c) Zn dust (d) AlCl₃
Answer: (b) Na/dry ether.
Q11. The order of reactivity of HX with alcohols is:
(a) HCl > HBr > HI (b) HI > HBr > HCl (c) HI > HCl > HBr (d) HCl > HI > HBr
Answer: (b) HI > HBr > HCl.
Q12. Reaction of alkyl halide with AgCN gives:
(a) alkyl cyanide (b) alkyl isocyanide (c) alkyl nitrite (d) ester
Answer: (b) alkyl isocyanide (carbylamine).
Q13. Williamson’s synthesis prepares:
(a) ether (b) ester (c) amine (d) alkene
Answer: (a) ether.
Fill in the Blanks
1. A haloalkane in which X is on a sp² carbon of C=C is called a __________ halide.
Answer: vinylic.
2. Lucas reagent is anhyd. ZnCl₂ + concentrated __________.
Answer: HCl.
3. SN1 reaction is favoured by polar __________ solvents.
Answer: protic.
4. The peroxide effect is observed only with __________ (HX).
Answer: HBr.
5. The catalyst used for halogenation of benzene is __________.
Answer: FeX₃ (anhydrous Lewis acid).
6. Reaction of R–X with Mg in dry ether produces __________.
Answer: Grignard reagent (R–Mg–X).
7. The conversion of chlorobenzene to phenol with NaOH at high temperature is called __________ process.
Answer: Dow’s.
True or False
1. Haloalkanes are generally soluble in water. (False) — they are insoluble; they cannot form H-bonds with water.
2. SN2 reaction proceeds with inversion of configuration. (True)
3. Chlorobenzene undergoes nucleophilic substitution as easily as chloroethane. (False) — much less reactive due to resonance.
4. Wurtz reaction uses sodium metal in dry ether. (True)
5. CFCs are environment-friendly and do not affect the ozone layer. (False) — they are major ozone-depleting substances.
6. Bulky bases promote Hofmann (anti-Saytzeff) elimination. (True)
7. Vinyl chloride and chlorobenzene have a partial double-bond character in C–Cl due to resonance. (True)
Important Distinguishing Reactions
Q. How will you distinguish between an alkyl chloride, an alkyl bromide and an alkyl iodide?
Answer: Heat the alkyl halide with alcoholic AgNO₃ and observe the precipitate: alkyl chloride gives a white precipitate of AgCl (soluble in NH₄OH), alkyl bromide gives a pale yellow precipitate of AgBr (sparingly soluble in NH₄OH), and alkyl iodide gives a yellow precipitate of AgI (insoluble in NH₄OH).
Q. How will you distinguish chlorobenzene from chloroethane?
Answer: Warm each compound with aqueous AgNO₃: chloroethane immediately gives a white precipitate of AgCl on warming; chlorobenzene gives no precipitate even on prolonged heating because the C–Cl bond in chlorobenzene has partial double-bond character and does not ionise easily.
Q. Why is the boiling point of an alkyl halide higher than that of an isomeric alkane?
Answer: Because the C–X bond is polar, alkyl halides experience dipole–dipole attractions in addition to London forces; halogen atoms are also more polarisable than H, increasing dispersion forces. Both factors raise the boiling point relative to the parent alkane.
Q. Why is allyl chloride more reactive than vinyl chloride towards nucleophilic substitution?
Answer: Allyl chloride (CH₂=CH–CH₂Cl) ionises to give a resonance-stabilised allyl carbocation, so it readily undergoes SN1. Vinyl chloride (CH₂=CH–Cl) cannot ionise easily because the C–Cl bond is shorter, stronger, has partial double-bond character (resonance with the C=C), and any vinyl carbocation formed would be very unstable.
Glossary
| Term | Meaning |
|---|---|
| Haloalkane | Alkane in which one or more H atoms are replaced by halogen. |
| Haloarene | Aromatic hydrocarbon with halogen attached directly to the ring. |
| Allylic halide | X on sp³ carbon adjacent to a C=C bond. |
| Benzylic halide | X on sp³ carbon attached to a benzene ring. |
| Vinylic halide | X on sp² carbon of a C=C double bond. |
| Aryl halide | X bonded directly to sp² carbon of an aromatic ring. |
| Markovnikov’s rule | X of HX adds to the more substituted alkene carbon (more stable carbocation). |
| Peroxide effect | Anti-Markovnikov addition of HBr in the presence of peroxides via free-radical mechanism. |
| Darzens method | Conversion of alcohol to alkyl chloride using SOCl₂/pyridine. |
| Lucas reagent | Anhydrous ZnCl₂ + concentrated HCl; used to test/convert alcohols. |
| Finkelstein reaction | R–Cl/Br + NaI in dry acetone → R–I + NaCl/Br. |
| Swarts reaction | R–Cl + AgF/Hg₂F₂ → R–F + AgCl. |
| SN1 | Unimolecular nucleophilic substitution; first order; carbocation intermediate; racemisation. |
| SN2 | Bimolecular nucleophilic substitution; second order; concerted; Walden inversion. |
| Walden inversion | Inversion of configuration at the chiral centre during SN2 reaction. |
| E1 / E2 | Unimolecular / bimolecular elimination giving alkenes. |
| Saytzeff’s rule | Most substituted (more stable) alkene is the major elimination product. |
| Grignard reagent | R–Mg–X; alkyl/aryl magnesium halide; very reactive organometallic compound. |
| Wurtz reaction | 2 R–X + 2 Na (dry ether) → R–R + 2 NaX. |
| Wurtz–Fittig reaction | Aryl halide + alkyl halide + Na/dry ether → alkyl arene. |
| Fittig reaction | Two aryl halides + Na/dry ether → biaryl. |
| Sandmeyer reaction | ArN₂⁺ + CuX → ArX (X = Cl, Br, CN). |
| Dow’s process | Industrial conversion of chlorobenzene to phenol with NaOH at 623 K, 300 atm. |
| CFC (Freon) | Chlorofluorocarbon; ozone-depleting refrigerant/propellant. |
| DDT | 1,1,1-Trichloro-2,2-bis(p-chlorophenyl)ethane; non-biodegradable insecticide. |
| Montreal Protocol | 1987 international agreement to phase out ozone-depleting substances. |
| Phosgene | COCl₂; toxic gas produced when CHCl₃ reacts with O₂ in air and light. |
| Williamson synthesis | R–X + R’–O⁻Na⁺ → R–O–R’ + NaX (preparation of ethers). |
| Hofmann (anti-Saytzeff) rule | With bulky bases the least substituted alkene is the major elimination product. |
| Carbocation | Trivalent positively-charged carbon intermediate; stability 3° > 2° > 1° > methyl. |
| Ambident nucleophile | A nucleophile (e.g., CN⁻, NO₂⁻) able to attack through either of two atoms. |
| Friedel–Crafts reaction | EAS alkylation/acylation of arenes using R–X or R–COCl with anhyd. AlCl₃. |
| Lucas test | Reagent (anhyd. ZnCl₂ + conc. HCl) used to distinguish 1°, 2° and 3° alcohols by rate of turbidity. |
| Bioaccumulation | Build-up of non-biodegradable pollutants (e.g., DDT) in body tissues along the food chain. |
| Dipole moment | Vector measure of bond polarity; CH₃Cl has the highest, CH₃I the lowest among methyl halides. |
| Racemic mixture | 1:1 mixture of two enantiomers, optically inactive due to external compensation. |
| Enantiomer | One of a pair of non-superimposable mirror-image stereoisomers. |
| Chirality | Property of a molecule whose mirror image cannot be superimposed on the original. |
| Anti-periplanar | Geometric requirement (180°) for the leaving group and β-H in an E2 elimination. |
| Hyperconjugation | Stabilising delocalisation of σ(C–H) electrons into adjacent empty/π orbitals; explains stability of 3° > 2° > 1° carbocations and Saytzeff alkene. |
| Ozone hole | Localised severe thinning of stratospheric O₃, caused mainly by Cl•/Br• radicals from CFCs and halons. |
| Teflon | Polytetrafluoroethylene, –(CF₂–CF₂)n–; non-stick coating polymer. |