JEE Chemistry Haloalkanes and Haloarenes MCQs Set 03

Answer: (a) R–X + NaI (acetone) →

Explanation:
1. The Finkelstein reaction converts an alkyl chloride or bromide to an alkyl iodide using NaI in dry acetone.
2. NaCl or NaBr is insoluble in acetone and precipitates, which drives the reaction forward.

Teacher's Note:
1. Finkelstein gives R–I; the Swarts reaction (AgF, Hg₂F₂) gives R–F.
2. Acetone is the key solvent.

Answer: (a) Diphenylamine (C₆H₅–NH–C₆H₅)

Explanation:
1. The carbylamine test (CHCl₃ + alcoholic KOH, heat) is given only by primary amines, aliphatic or aromatic, which form foul-smelling isocyanides.
2. Ethylamine, cyclohexylamine and aniline are primary amines.
3. Diphenylamine is a secondary amine and does not respond.

Teacher's Note:
1. Primary amine: RNH₂ + CHCl₃ + 3KOH → RNC + 3KCl + 3H₂O.
2. Secondary and tertiary amines give no isocyanide.

Answer: (c) C₂H₅OH + NaCl →

Explanation:
1. Option 1 is the Lucas-type conversion of an alcohol to chloride.
2. Option 2 is addition of HBr to propene giving 2-bromopropane.
3. Option 4 is the Borodine-Hunsdiecker reaction giving CH₃Br.
4. Option 3 does nothing, because OH^- is a poor leaving group and NaCl is not an acid.

Teacher's Note:
1. An alcohol needs HX, PX₃, PX₅ or SOCl₂ to give a halide.
2. A neutral salt like NaCl does not work.

Answer: (b) CHCl₃

Explanation:
1. Bleaching powder supplies chlorine, which chlorinates the methyl group of acetone to CCl₃COCH₃.
2. This is hydrolysed by calcium hydroxide to CHCl₃ and calcium acetate.

Teacher's Note:
1. This is the laboratory preparation of chloroform.
2. Ethanol can be used in place of acetone.

Answer: (a) CH₃CH₂I

Explanation:
1. For the same alkyl group, boiling point rises R–F < R–Cl < R–Br < R–I because of larger size and polarisability.
2. Among the iodides, ethyl iodide has the larger molecular mass and surface area, so it boils higher (72 °C) than methyl iodide (42 °C).

Teacher's Note:
1. More electrons and more surface area mean stronger van der Waals forces.
2. Compare both the halogen and the alkyl chain.

Answer: (d) b > a > c > d

Explanation:
1. S_N1 rate depends on the stability of the carbocation formed.
2. (b) gives the aromatic tropylium cation (6 π electrons), which is exceptionally stable.
3. (a) gives the delocalisation-stabilised benzyl cation, and (c) gives a tertiary cation.
4. (d) would give a primary cation and rearrangement is slow, so it is the least reactive. The order is b > a > c > d.

Teacher's Note:
1. Aromatic carbocation > benzyl > tertiary > secondary > primary.
2. Neopentyl halides are very unreactive in both S_N1 and S_N2.

Answer: (c) Treating CHCl₃ with aqueous AgNO₃

Explanation:
1. On storage in light, CHCl₃ is oxidised to poisonous phosgene, COCl₂, and HCl.
2. Aqueous AgNO₃ then gives a white precipitate of AgCl, whereas pure chloroform gives none.

Teacher's Note:
1. Pure chloroform is covalent and gives no AgCl.
2. It is stored in full dark bottles with a little ethanol.

Answer: (b) a > b > c > d

Explanation:
1. Bromoacetone (a) is an α-halo ketone: the adjacent C=O stabilises the S_N2 transition state, so it is most reactive.
2. Among the rest, S_N2 rate falls with steric hindrance: primary n-butyl (b) > branched primary isobutyl (c) > secondary (d).

Teacher's Note:
1. Steric crowding at the carbon slows S_N2.
2. α-Carbonyl halides are especially reactive.

Answer: (a) D.D.T.

Explanation:
1. DDT, 1,1,1-trichloro-2,2-bis(4-chlorophenyl)ethane, is an insecticide.
2. Chloritone is a hypnotic and chloroform is an anaesthetic, not insecticides.

Teacher's Note:
1. DDT is made from chloral and chlorobenzene with conc. H₂SO₄.
2. It is now banned in many countries because it persists in the environment.

Answer: (c) The C–I bonds in iodoform are weak and the C–Cl bonds in chloroform are strong

Explanation:
1. Both haloforms are covalent.
2. The C–I bond (about 240 kJ/mol) is much weaker than C–Cl (about 330 kJ/mol), so iodoform releases I^- readily, giving yellow AgI.

Teacher's Note:
1. Bond strength order: C–F > C–Cl > C–Br > C–I.
2. Option 2 is true but does not explain the difference.

Answer: (b) R–H

Explanation:
1. Grignard reagents are strong bases and react with any compound containing acidic hydrogen (O–H, N–H, C≡C–H).
2. RMgX + R′NH₂ → RH + R′NHMgX.

Teacher's Note:
1. Grignard reagents are destroyed by water, alcohols and amines.
2. They give alkanes with active hydrogens.

Answer: (a) Phenyl isocyanide

Explanation:
1. Aniline is a primary aromatic amine, so it gives the carbylamine reaction.
2. C₆H₅NH₂ + CHCl₃ + 3KOH → C₆H₅NC + 3KCl + 3H₂O.

Teacher's Note:
1. Phenyl isocyanide has a very unpleasant smell.
2. The product is isocyanide, not cyanide.

Answer: (c) C₆H₅CH₂Cl

Explanation:
1. The fastest S_N1 reaction gives the most stable carbocation.
2. Benzyl chloride gives the delocalisation-stabilised benzyl cation.
3. The isopropyl cation (secondary) is less stable, ethyl is primary, and chlorobenzene does not ionise at all.

Teacher's Note:
1. Delocalisation stabilisation beats simple alkyl stabilisation.
2. Aryl halides do not undergo S_N1.

Answer: (c) CHI₃

Explanation:
1. Ethene + HBr gives x = C₂H₅Br.
2. Hydrolysis gives y = C₂H₅OH.
3. Ethanol with I₂/NaOH undergoes the iodoform reaction to give z = CHI₃.

Teacher's Note:
1. Ethanol is the only primary alcohol that gives iodoform.
2. Oxidation to CH₃CHO happens in situ.

Answer: (a) 1-Chloro-3-(trichloromethyl)benzene (meta product)

Explanation:
1. The CCl₃ group is strongly electron withdrawing (–I effect) and has no lone pair or π-bond to donate.
2. It is therefore a meta-directing, deactivating group, and electrophilic chlorination occurs at the meta position.

Teacher's Note:
1. Groups with δ+ carbon attached to the ring are meta directors.
2. Compare CH₃, which is ortho/para directing.

Answer: (d) CH₃CH₂COOCH₃

Explanation:
1. 1-Phenylethanol has the CH₃CH(OH)– group and acetophenone has the CH₃CO– group, so both respond.
2. Iodoacetaldehyde is converted by the base-iodine reagent into CI₃CHO, which cleaves to iodoform.
3. Methyl propanoate has neither group and does not respond.

Teacher's Note:
1. Esters of propanoic acid lack the methyl carbonyl unit.
2. Check each structure for CH₃CO– or CH₃CH(OH)–.

Answer: (b) C₆H₅CH₃

Explanation:
1. Magnesium forms the Grignard reagent C₆H₅CH₂MgBr.
2. Water (H₃O⁺) protonates the carbanion carbon, giving toluene, C₆H₅CH₃.

Teacher's Note:
1. Grignard plus acid or water gives the alkane.
2. Alcohols form only with a carbonyl compound or O₂.

Answer: (d) 4-Chloroacetophenone

Explanation:
1. This is Friedel-Crafts acylation.
2. Chlorine is an ortho/para director, and the bulky acylium ion attacks mainly the para position, giving 4-chloroacetophenone.

Teacher's Note:
1. Halogens deactivate the ring but direct ortho/para.
2. The para isomer dominates because of steric hindrance.

Answer: (a) II > I > IV > III

Explanation:
1. Benzyl chloride (II) is most reactive, since it is a benzylic halide reacting by S_N1 or S_N2.
2. 2-Phenylethyl chloride (I) is an ordinary primary alkyl halide.
3. Aryl halides are far less reactive, and the p-nitro compound (IV) reacts faster than the p-methoxy compound (III), where the donor OCH₃ deactivates the ring.

Teacher's Note:
1. Alkyl halides react much faster than aryl halides.
2. Electron-withdrawing groups speed nucleophilic aromatic substitution.

Answer: (b) (Prop-1-en-1-yl)cyclohexane

Explanation:
1. Zinc removes the two bromines from adjacent carbons, forming a double bond between them.
2. C₆H₁₁–CHBr–CHBr–CH₃ becomes C₆H₁₁–CH=CH–CH₃, which is (prop-1-en-1-yl)cyclohexane.

Teacher's Note:
1. Zn dehalogenation of a vicinal dibromide gives the C=C between the two carbons that carried the halogens.
2. The carbon skeleton is unchanged.

Answer: (b) Salicylic acid

Explanation:
1. With CCl₄ the Reimer-Tiemann intermediate is an –CCl₃ group, which hydrolyses to –COOH.
2. Acidification gives salicylic acid (2-hydroxybenzoic acid).

Teacher's Note:
1. CHCl₃ gives –CHO and CCl₄ gives –COOH.
2. The substitution is mainly at the ortho position.

Answer: (c) Both of these

Explanation:
1. Chlorine is a deactivating but ortho/para directing group.
2. Bromination therefore gives a mixture of o- and p-bromochlorobenzene, with the para isomer as the major product.

Teacher's Note:
1. Both isomers are formed.
2. Halogens direct o/p but deactivate by the –I effect.

Answer: (b) 1-Chloropropan-2-ol (CH₃CH(OH)CH₂Cl)

Explanation:
1. X is propan-1-ol, and Al₂O₃ dehydrates it to propene, Y.
2. Propene with Cl₂/H₂O forms a chlorohydrin: the chloronium ion is opened by water at the more substituted carbon (Markovnikov).
3. Cl goes to the terminal CH₂ and OH to the middle carbon, giving CH₃CH(OH)CH₂Cl.

Teacher's Note:
1. Halohydrin formation: halogen on the less substituted carbon, OH on the more substituted one.
2. Water is the nucleophile in excess.

Answer: (d) Triphenylmethane

Explanation:
1. This is a Friedel-Crafts alkylation.
2. CHCl₃ with AlCl₃ supplies the electrophile, and three benzene rings replace the three chlorines: 3C₆H₆ + CHCl₃ → (C₆H₅)₃CH + 3HCl.

Teacher's Note:
1. Each C–Cl bond can alkylate one benzene ring.
2. The product is triphenylmethane.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Assertion is true: the CCl₃^- carbanion is stabilised by the large, polarisable chlorine atoms and by d-orbital participation, and CF₃^- suffers lone-pair repulsion from fluorine.
2. Reason is a true fact, but electronegativity alone would predict the opposite, so it does not explain the assertion.

Teacher's Note:
1. The acidity of haloforms: CHI₃ > CHBr₃ > CHCl₃ > CHF₃.
2. Do not always attribute trends to electronegativity.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Assertion is true: electrophilic addition follows Markovnikov's rule because the benzylic carbocation is the more stable intermediate.
2. Reason is also a true statement, but the reaction here is ionic and involves a carbocation, not a radical, so the reason does not explain the product.

Teacher's Note:
1. Without peroxide, HBr adds via carbocation intermediates.
2. In presence of peroxide, the radical pathway gives the anti-Markovnikov product.

Answer: (c) A is true, but R is false.

Explanation:
1. Assertion is true: Saytzeff's rule favours the more substituted alkene, but-2-ene.
2. Reason is false: the C₃ hydrogen is on a more substituted carbon, and extra alkyl groups make it less acidic. The real reason is the greater stability of the more substituted alkene.

Teacher's Note:
1. Zaitsev (Saytzeff) products are the more stable alkenes.
2. Alkyl groups donate electrons and reduce acidity.