Previous year questions: Haloalkanes and Haloarenes

The questions below come from official CBSE board papers and from official Sample and Additional Practice Question Papers. Each heading carries the year label printed on that source. Marks are the marks printed for that question or part in the paper. Attempt each one first, then check the answer.

Q1 (CBSE Board 2025, 1 mark)

Which of the following molecules is chiral in nature? (A) 1-chloropropane (B) 2-chloropropane (C) 1-chlorobutane (D) 2-chlorobutane

Answer: (D) 2-chlorobutane. The carbon carrying chlorine is attached to hydrogen, chlorine, a methyl group and an ethyl group, so all four groups differ and the carbon is a stereocentre. The other three compounds have at least two identical groups on every carbon and are achiral. Marking scheme: one mark for identifying the chiral molecule and one for naming the four different groups on the carbon that carries chlorine.

Q2 (CBSE Board 2025, 1 mark)

The treatment of ethyl bromide with alcoholic silver nitrite gives (A) ethyl nitrite (B) nitroethane (C) nitromethane (D) ethene

Answer: (B) nitroethane. The nitrite ion can link through oxygen or through nitrogen. Silver nitrite is largely covalent, so nitrogen is free to donate its lone pair and the product is the nitroalkane. The silver halide that precipitates drives the reaction forward. Write the ambident nature of the nitrite ion first, then name the product and explain why the covalent silver salt gives attack through nitrogen.

Q3 (CBSE Board 2025, 3 marks)

Give a reason for each. (a) In chlorobenzene the C-Cl bond is shorter than the same bond in methyl chloride. (b) A Grignard reagent has to be made in the absence of moisture. (c) Hydrolysis of an optically active halide by the SN1 route gives a product with no optical activity.

Answer: (a) The halogen lone pair is conjugated with the ring in chlorobenzene, so the bond gains partial double bond character, and the sp2 carbon holds the shared pair more tightly. Both effects pull the atoms closer. (b) The carbon magnesium bond is strongly polar with carbon negative, so the reagent acts as a powerful base and any proton source converts it into a hydrocarbon. (c) The slow step leaves a flat sp2 cation, and a nucleophile attacks either face with equal chance, so the two enantiomers appear in equal amounts and the product is optically inactive. Allocation: one mark for each part, and part (c) needs the planar carbocation and the two faces of attack to earn full marks.

Q4 (CBSE Board 2024, 2 marks)

Give reasons. (i) Grignard reagents must be prepared under anhydrous conditions. (ii) Alkyl halides give alcohols with aqueous KOH, while with alcoholic KOH they give alkenes.

Answer: (i) Water protonates the carbanion-like carbon of the Grignard reagent, turning it into the corresponding hydrocarbon and removing the reagent before it can react. (ii) Aqueous KOH supplies hydroxide as a nucleophile in a strongly ionising solvent, so substitution gives the alcohol. Alcoholic KOH is a strong base in a poorly ionising solvent, so it removes a beta hydrogen and elimination gives the alkene. Two marks need two separate reasons, so write the aqueous case and the alcoholic case as separate paragraphs with the mechanism name in each.

Q5 (CBSE Board 2024, 2 marks)

(a) Out of chlorobenzene and 2,4,6-trinitrochlorobenzene, which is more reactive towards nucleophilic substitution, and why? (b) Which isomer of C4H9Cl has the lowest boiling point?

Answer: (a) 2,4,6-Trinitrochlorobenzene is far more reactive. The three nitro groups withdraw electron density strongly through the inductive and resonance effects, which stabilises the carbanion intermediate formed when the nucleophile adds. The chlorine then leaves more easily. (b) 2-Chloro-2-methylpropane, the tertiary isomer, has the lowest boiling point because it is the most branched molecule and has the least surface contact for van der Waals attractions. Part (a) needs the electron withdrawing nitro groups and the stabilised intermediate; part (b) needs branching and reduced surface contact.

Q6 (CBSE Board 2023, 2 marks)

Why is the boiling point of o-dichlorobenzene higher than that of p-dichlorobenzene, but the melting point of the para isomer higher than that of the ortho isomer?

Answer: In the ortho isomer the two carbon chlorine bond moments do not cancel, so the molecule has a net dipole moment and its molecules attract each other through dipole dipole forces, which raises the boiling point. The para isomer is symmetrical, its bond moments cancel and its boiling point is lower. Its symmetry also lets it pack neatly into the crystal lattice, so its melting point is higher. A full answer names the dipole effect for the boiling point and the lattice packing for the melting point, and ties both to the symmetry of the isomer.

Q7 (CBSE SQP 2022-23, 2 marks)

Give a reason for each. (a) Electrophilic substitution of a haloarene gives the para product as the major product. (b) The product of an SN1 reaction is a racemic mixture.

Answer: (a) The halogen is bulky, so the electrophile is sterically blocked at the ortho position, while the para position stays open. Resonance stabilises the arenium ion at both ortho and para positions, so the para product dominates. (b) The planar carbocation formed in the slow step is attacked from both faces with equal chance, so the two enantiomers appear in equal amounts. Each reason carries one mark, so state the steric blocking at ortho and the resonance stabilisation shared by ortho and para.

Q8 (CBSE SQP 2024-25, 3 marks)

Various isomeric haloalkanes with the formula C4H9Cl undergo hydrolysis. Compound A is the most reactive by the SN1 mechanism. Identify A with a reason and write the mechanism.

Answer: A is 2-chloro-2-methylpropane, the tertiary chloride. Its ionisation gives a tertiary carbocation, stabilised by three alkyl groups through inductive donation and hyperconjugation, so the cation forms fastest and the SN1 rate is highest. Step 1: the carbon chlorine bond breaks slowly and reversibly to give the tert-butyl cation and chloride, and this is the rate determining step. Step 2: water attacks the planar cation from either face, and loss of a proton gives 2-methylpropan-2-ol. Both faces are attacked, so the alcohol is racemic if the halide was optically active. The mechanism marks usually split into the slow ionisation, the planar carbocation, and the attack from both faces giving equal enantiomers.

Q9 (CBSE SQP 2025-26, 3 marks)

Explain the following. I. Toluene with chlorine in sunlight gives benzyl chloride, while with chlorine in the dark it gives ortho and para chlorotoluene. II. The Finkelstein reaction is carried out in the presence of dry acetone.

Answer: I. Sunlight splits chlorine into atoms, which start a free radical chain at the methyl side chain, so benzyl chloride forms. In the dark a Lewis acid such as iron(III) chloride polarises the chlorine instead, and electrophilic substitution takes place on the ring at ortho and para positions. II. Dry acetone dissolves sodium iodide but not the sodium chloride or bromide produced. The halide salt precipitates, which removes a product and shifts the equilibrium towards the alkyl iodide. Acetone is also polar aprotic, which keeps SN2 fast. Name both the radical route in sunlight and the Lewis acid route in the dark, and add the precipitation of sodium chloride as the second mark in part II.

Q10 (CBSE Additional PQ 2022-23, 2 marks)

2-Bromooctane with aqueous NaOH gives 2-octanol. (a) Identify the mechanism and justify it. (b) What happens to the rate if the concentration of NaOH is halved, and if the concentration of 2-bromooctane is halved? The practice paper prints alcoholic NaOH here; with that reagent the main product would be an alkene, so the substitution product belongs with the aqueous reagent.

Answer: (a) The mechanism is SN2. Hydroxide is a strong nucleophile, the secondary carbon is open enough for backside attack, and the product has inverted configuration at that carbon. (b) Halving either reactant halves the rate, because the rate law is first order in the halide and first order in hydroxide, k[2-bromooctane][OH-]. Half a mark each for the mechanism, the inversion point and the two rate effects, so answer both rate questions explicitly with the rate law written out.