How to Score Full Marks in the Board Exam
A complete bank of board-style questions with model answers for Haloalkanes and Haloarenes. Name the mechanism explicitly (SN1 or SN2), write structures and reaction arrows clearly, and use the NCERT-canonical reasons (resonance, sp^2 carbon, unstable phenyl cation for haloarenes; chloroform to phosgene for storage). For "how will you bring about" questions, state the reagent and conditions over the arrow.
1-Mark Questions (Definitions & Direct)
Q1. What is the Finkelstein reaction? Answer: The conversion of an alkyl chloride/bromide to an alkyl iodide by treating it with sodium iodide in dry acetone: R-X + NaI → R-I + NaX.
Q2. What is the Swarts reaction? Answer: The preparation of an alkyl fluoride by heating an alkyl chloride/bromide with a metallic fluoride such as AgF, HgF, CoF or SbF.
Q3. Why is the C-X bond polar? Answer: Because the halogen is more electronegative than carbon, it pulls the bonding electrons toward itself, giving carbon a partial positive charge and the halogen a partial negative charge.
Q4. Name the product when an alkyl halide reacts with alcoholic KOH. Answer: An alkene (by beta-elimination / dehydrohalogenation).
1-Mark Questions (continued)
Q5. What is a racemic mixture? Answer: A 50:50 mixture of two enantiomers; it is optically inactive because their equal and opposite rotations cancel.
Q6. State Saytzeff's rule. Answer: In dehydrohalogenation, the major product is the more substituted (more stable) alkene, formed by removing the hydrogen from the carbon with fewer hydrogen atoms.
Q7. Which compound forms when chloroform is left in air and light? Answer: Phosgene (COCl), a poisonous gas.
Q8. What is the general formula of a Grignard reagent? Answer: RMgX (an alkyl/aryl magnesium halide), formed from R-X and Mg in dry ether.
2-Mark Reasoning Questions
Q9. Why is dry acetone used in the Finkelstein reaction? Answer: Sodium iodide is soluble in acetone, but sodium chloride and sodium bromide are not. As the reaction proceeds, NaCl (or NaBr) precipitates out, which shifts the equilibrium toward the formation of the alkyl iodide.
Q10. Why are haloarenes less reactive than haloalkanes toward nucleophilic substitution? Answer: (i) In a haloarene, resonance gives the C-X bond partial double-bond character, making it shorter and stronger. (ii) The C-X carbon is sp hybridised and holds the bonding electrons more tightly. (iii) The phenyl cation that would form in an SN1 pathway is very unstable. All three factors make substitution very difficult.
2-Mark Reasoning Questions (continued)
Q11. Why must Grignard reagents be prepared under anhydrous conditions? Answer: Grignard reagents are extremely reactive toward protons. They react instantly with even traces of water (or any -OH/-NH) to give an alkane, destroying the reagent: RMgX + HO → R-H + Mg(OH)X. Therefore all apparatus and solvents must be perfectly dry.
Q12. Why does chloroform need to be stored in dark-coloured bottles filled to the brim, with a little ethanol added? Answer: Chloroform is slowly oxidised by air and light to the poisonous gas phosgene (COCl). Dark bottles exclude light, filling to the brim excludes air, and the added ethanol reacts with any phosgene formed to reduce its harmful effects.
Q13. Why is the boiling point of an alkyl halide higher than that of the corresponding alkane? Answer: The polar C-X bond gives the alkyl halide dipole-dipole interactions in addition to van der Waals forces, and the halide has a greater molecular mass. Both factors raise the boiling point above that of the non-polar parent alkane.
2-3 Mark Reasoning Questions
Q14. Distinguish between SN1 and SN2 mechanisms (any three points). Answer: (i) SN1 is a two-step mechanism through a carbocation; SN2 is a single step. (ii) SN1 follows first-order kinetics, rate = k[RX]; SN2 follows second-order kinetics, rate = k[RX][Nu]. (iii) SN1 proceeds with racemisation; SN2 proceeds with inversion of configuration. (iv) SN1 is favoured by tertiary halides (stable carbocation); SN2 by primary halides (least steric hindrance).
Q15. Why does an optically active alkyl halide give a racemic mixture in an SN1 reaction but an inverted product in an SN2 reaction? Answer: In SN1, the reaction goes through a planar carbocation that can be attacked by the nucleophile from either face with nearly equal probability, giving both enantiomers — a racemic mixture. In SN2, the nucleophile attacks from the side opposite the leaving group (backside attack) in a single step, flipping the configuration — the product is the inverted enantiomer (Walden inversion).
Q16. Why is the halogen in chlorobenzene ortho/para-directing even though it is deactivating? Answer: The halogen withdraws electron density through its inductive (-I) effect, which deactivates the whole ring (reactions are slower than for benzene). But its lone pair donates into the ring by resonance (+R), and this donation increases the electron density specifically at the ortho and para positions, so the incoming electrophile goes mainly there. Inductive controls the rate; resonance controls the position.
3-Mark Conversion Questions
Q17. How will you convert 1-chlorobutane to 1-iodobutane? Answer: By the Finkelstein reaction — treat 1-chlorobutane with sodium iodide in dry acetone: CHCHCHCHCl + NaI → CHCHCHCHI + NaCl↓.
Q18. How will you convert chlorobenzene to phenol? Answer: Heat chlorobenzene with aqueous sodium hydroxide at high temperature (~623 K) and high pressure (~300 atm) to give sodium phenoxide, then acidify to get phenol.
Q19. How will you bring about the conversion of propene to 1-bromopropane? Answer: Add HBr in the presence of an organic peroxide (the peroxide / Kharasch effect), which gives anti-Markovnikov addition: CH-CH=CH + HBr/(peroxide) → CH-CH-CHBr.
3-Mark Conversion & Product Questions
Q20. Give the products when 2-bromopropane reacts with (a) aqueous KOH and (b) alcoholic KOH. Answer: (a) With aqueous KOH (substitution): propan-2-ol, (CH)CHOH. (b) With alcoholic KOH (elimination): propene, CH-CH=CH.
Q21. What is the major product of dehydrohalogenation of 2-bromobutane, and why? Answer: The major product is but-2-ene (CH-CH=CH-CH). By Saytzeff's rule, the more substituted, more stable alkene predominates over but-1-ene.
Q22. How is toluene prepared from chlorobenzene? Name the reaction. Answer: By the Wurtz-Fittig reaction — chlorobenzene is treated with methyl chloride and sodium in dry ether: CHCl + CHCl + 2 Na → CH-CH (toluene) + 2 NaCl.
3-Mark Reaction & Reasoning Questions
Q23. Write the reaction and name the product when bromoethane reacts with sodium in dry ether. Answer: The Wurtz reaction couples two ethyl groups: 2 CHCHBr + 2 Na → CHCH-CHCH (n-butane) + 2 NaBr.
Q24. Why does R-I react faster than R-Cl in a nucleophilic substitution reaction? Answer: The C-I bond is weaker (lower bond enthalpy) than the C-Cl bond, so it breaks more easily, and iodide is a better leaving group. Hence R-I reacts faster than R-Cl.
Q25. Why is chloroform used as a solvent but not as an anaesthetic any more? Answer: Chloroform is a good solvent for fats, alkaloids and iodine. It was earlier used as an anaesthetic, but its use was discontinued because it is harmful (it is metabolised in the liver to poisonous phosgene and can damage the heart and liver).
3-Mark Reasoning Questions
Q26. Arrange the following in increasing order of SN2 reactivity: CHBr, (CH)CHBr, (CH)CBr, CHCHBr. Answer: SN2 reactivity decreases with increasing steric hindrance, so increasing order is: (CH)CBr < (CH)CHBr < CHCHBr < CHBr.
Q27. Why is the dipole moment of CHCl higher than that of CHF? Answer: Although fluorine is more electronegative, the C-F bond is shorter than the C-Cl bond. Dipole moment is the product of charge and distance, so the longer C-Cl bond gives CHCl a larger overall dipole moment than CHF (this is the order found in NCERT data).
Q28. Why does 2,4,6-trinitrochlorobenzene undergo nucleophilic substitution much more easily than chlorobenzene? Answer: The electron-withdrawing -NO groups at the ortho and para positions stabilise the negatively charged intermediate (carbanion) formed when the nucleophile attacks, so the substitution proceeds much more readily than in chlorobenzene, which lacks such stabilising groups.
3-Mark Reasoning & Uses
Q29. Why are freons (CFCs) being phased out? Answer: Freons are very stable and reach the stratosphere, where ultraviolet light releases chlorine free radicals that catalytically destroy ozone. This depletion of the ozone layer is the reason their use is being phased out under the Montreal Protocol.
Q30. Give the uses of (a) DDT and (b) carbon tetrachloride. Answer: (a) DDT was the first chlorinated organic insecticide, used against malaria-carrying mosquitoes; however, it is non-biodegradable and bioaccumulates, so its use is now restricted. (b) Carbon tetrachloride is used as an industrial solvent (for fats, oils) and was earlier used in fire extinguishers; its use is now restricted because it depletes the ozone layer.
Q31. Why is the antiseptic property of iodoform attributed to liberated iodine? Answer: Iodoform (CHI) is unstable and slowly liberates free iodine when in contact with skin or tissues; it is this free iodine that acts as the antiseptic, not the iodoform molecule itself.
5-Mark / Long-Answer Questions
Q32. Describe the SN1 and SN2 mechanisms with the help of an example each, including their kinetics and stereochemistry. Answer: SN2 (e.g. CHBr + OH-): the nucleophile OH^- attacks the carbon from the side opposite the leaving group (backside attack) in a single step, passing through a transition state in which both OH and Br are partially bonded; Br^- leaves and the configuration is inverted. Rate = k[CHBr][OH^-] (second order). SN1 (e.g. (CH)CBr + OH-): in the slow first step the C-Br bond ionises to give a stable tertiary carbocation; in the fast second step OH^- attacks the planar carbocation from either face, giving a racemic mixture. Rate = k[(CH)CBr] (first order).
Q33. Explain the term optical isomerism. Define chirality, enantiomers and racemisation. Answer: Optical isomerism is the property by which certain molecules rotate the plane of plane-polarised light. A molecule is chiral if it is non-superimposable on its mirror image; the commonest cause is a carbon bonded to four different groups. Enantiomers are the pair of non-superimposable mirror-image stereoisomers; one is dextrorotatory (+) and the other laevorotatory (-). Racemisation is the conversion of one enantiomer (or a pure compound) into an equal mixture of both enantiomers (a racemic mixture), which is optically inactive.
5-Mark / Long-Answer Questions (continued)
Q34. Account for the following: (a) alkyl halides are immiscible with water; (b) tert-butyl bromide undergoes SN1 faster than n-butyl bromide; (c) the boiling point of an alkyl halide decreases with branching. Answer: (a) To dissolve, an alkyl halide would have to break the strong hydrogen bonds between water molecules, but it can form only weak attractions with water; since the energy gained is less than required, it is immiscible. (b) tert-Butyl bromide forms a stable tertiary carbocation in the rate-determining ionisation step, so SN1 is fast; n-butyl bromide would form an unstable primary carbocation, so its SN1 is much slower. (c) Branching gives a more spherical molecule with a smaller surface area, so the van der Waals forces are weaker and the boiling point is lower.
Q35. How are haloalkanes prepared from (a) alcohols, (b) alkenes, and (c) by halogen exchange? Give one reaction for each. Answer: (a) From alcohols: R-OH + SOCl → R-Cl + SO↑ + HCl↑ (thionyl chloride gives a pure product). (b) From alkenes: addition of HBr by Markovnikov's rule, CH-CH=CH + HBr → CH-CHBr-CH. (c) Halogen exchange: the Finkelstein reaction, R-Cl + NaI (dry acetone) → R-I + NaCl.
Q36. Explain why haloarenes do not undergo nucleophilic substitution easily, and give two reactions they do undergo. Answer: Haloarenes are unreactive toward nucleophilic substitution because of (i) resonance giving the C-X bond partial double-bond character (shorter, stronger), (ii) the sp carbon holding electrons tightly, and (iii) the instability of the phenyl cation. However, they do undergo electrophilic aromatic substitution (the halogen is deactivating but ortho/para-directing; e.g. nitration gives o- and p-nitrochlorobenzene) and reactions with metals such as the Fittig reaction (two aryl halides + Na give a biaryl, e.g. biphenyl).