CBSE Class 12 Chemistry Haloalkanes and Haloarenes MCQs Set 12

Answer: (d) X = o- and p-chlorotoluene, Y = trichloromethylbenzene

Explanation:
1. C₆H₅CH₃ + Cl₂/FeCl₃ (nuclear chlorination) → p-chlorotoluene (X, major product) + o-chlorotoluene (X, minor product)
2. C₆H₅CH₃ + Cl₂/hν (side chain chlorination) → C₆H₅CCl₃, trichloromethylbenzene (Y)
3. Thus, option (d) is correct.

Teacher's Note:
1. Lewis acid (FeCl₃) → ring substitution; light/heat → side-chain (free radical) substitution.

Answer: (b) from aniline by diazotisation followed by heating the diazonium salt with HBF₄

Explanation:
1. C₆H₅NH₂ →(NaNO₂/HCl, 273–278 K) C₆H₅N₂⁺Cl⁻ →(HBF₄) C₆H₅N₂⁺BF₄⁻ →(Δ) C₆H₅F + BF₃ + N₂
2. This is the Balz–Schiemann reaction.

Teacher's Note:
1. Direct fluorination of benzene is too violent to control, so aryl fluorides are always made via the diazonium fluoroborate.

Answer: (b) II < I < III

Explanation:
1. 1-Chlorobutane (III) with four carbon atoms has the highest boiling point. Since branching decreases the b.p., therefore, isopropyl chloride (II) has lower b.p. than 1-chloropropane (I). Thus, the order of increasing b.p. is II < I < III.

Teacher's Note:
1. Boiling point rises with molecular mass and falls with branching (smaller surface area, weaker van der Waals forces).

Answer: (d) C₂H₅–I

Explanation:
1. C₂H₅–I has the highest b.p. due to higher molecular mass.

Teacher's Note:
1. For the same alkyl group, b.p. increases in the order R–F < R–Cl < R–Br < R–I because van der Waals forces increase with size and mass of the halogen.

Answer: (b) CH₃Cl, CH₃F, CH₃Br

Explanation:
1. Due to shorter C–F bond distance than C–Cl, dipole moment of CH₃F is lower than that of CH₃Cl. However, due to lower electronegativity, C–Br bond is the least polar and hence dipole moment of CH₃Br is the lowest. In other words, option (b) is correct.

Teacher's Note:
1. Dipole moment = charge × distance, so the very short C–F bond outweighs fluorine’s higher electronegativity: CH₃Cl (1.86 D) > CH₃F (1.85 D) > CH₃Br (1.83 D).

Answer: (d) CH₃Cl

Explanation:
1. CCl₄ being symmetrical has zero dipole moment. The dipole moment of other halides increases as the number of halogen atoms decreases. Thus, CH₃Cl has the highest dipole moment.

Teacher's Note:
1. Adding more C–Cl bonds makes their bond dipoles partly cancel; in tetrahedral CCl₄ they cancel completely.

Answer: (c) 1,4-Dichlorobenzene

Explanation:
1. 1,4-Dichlorobenzene being symmetrical fits closely in the crystal lattice and hence has the highest melting point.

Teacher's Note:
1. Melting point depends on packing in the crystal, so the symmetrical para isomer melts highest even though the b.p. of the three isomers are close.

Answer: (b) nucleophilic substitution reaction

Explanation:
1. The nucleophile CN⁻ replaces Br⁻ from the alkyl halide, so it is a nucleophilic substitution reaction.

Teacher's Note:
1. A nucleophile (electron-rich CN⁻) attacks the electron-deficient carbon of C–X and the halide leaves.

Answer: (a) Ethyl bromide + alcoholic KCN

Explanation:
1. CH₃CH₂Br + KCN (alc.) → CH₃CH₂CN (propanenitrile) + KBr
2. KCN is ionic, so CN⁻ attacks through carbon and gives the nitrile; AgCN would give isocyanide.

Teacher's Note:
1. Count carbons: the CN carbon is included in the name, so ethyl bromide (2 C) gives propanenitrile (3 C).

Answer: (b) In X, CH₃CH₂ is directly linked to N

Explanation:
1. AgCN is covalent (Ag–C≡N:), so attack takes place through the lone pair on N. Since the product X is C₂H₅–N⇄C (ethyl isocyanide), therefore, CH₃CH₂ is directly attached to N.

Teacher's Note:
1. Isocyanides on hydrolysis give a 1° amine and formic acid, and on reduction give a 2° amine (here ethylmethylamine), which rules out (c) and (d).

Answer: (b) C₂H₅ONO, C₂H₅NO₂

Explanation:
1. NaNO₂ is ionic; the nitrite ion attacks through oxygen to give ethyl nitrite, C₂H₅–O–N=O.
2. AgNO₂ is covalent; attack occurs through the lone pair on N to give nitroethane, C₂H₅NO₂.

Teacher's Note:
1. Nitrite is an ambident nucleophile: ionic KNO₂/NaNO₂ → alkyl nitrite (R–ONO); covalent AgNO₂ → nitroalkane (R–NO₂).

Answer: (b) It gives nitroethane on heating with aqueous solution of AgNO₂

Explanation:
1. C₂H₅Br + AgNO₂ (aq) → C₂H₅NO₂ (nitroethane) + AgBr

Teacher's Note:
1. Check the other options too: Na gives butane (Wurtz), alcoholic KOH gives ethene, and KSH gives ethanethiol.

Answer: (c) B > C > A

Explanation:
1. The reactivity in S_N1 reactions depends upon the stability of the intermediate carbocations. Since the stability of the carbocations derived from (A), (B) and (C) follows the order: allylic carbocation (from B) > 2° carbocation (from C) > 1° carbocation (from A),
2. therefore, S_N1 reactivity of these three bromides follows the order: B > C > A, i.e., option (c) is correct.

Teacher's Note:
1. For S_N1, always rank by carbocation stability: allylic/benzylic and 3° at the top, 1° at the bottom.

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

Explanation:
1. The order of reactivity in S_N2 reactions is: 1° > 2° > 3°.
2. Further, amongst 1° alkyl halides, unsubstituted alkyl halides are more reactive than substituted alkyl halides. Thus, the overall decreasing reactivity in S_N2 reactions is:
3. CH₃CH₂CH₂CH₂Cl (1°) (I) > (CH₃)₂CHCH₂Cl (1°) (III) > CH₃CH₂CHClCH₃ (2°) (II) > (CH₃)₃C–Cl (3°) (IV)

Teacher's Note:
1. In S_N2, even branching at the β-carbon (as in isobutyl chloride) slows the backside attack.

Answer: (c) PhCH₂Br > PhCHBrMe > PhCBrMe₂ > PhCBrMePh (S_N1)

Explanation:
1. More highly substituted the alkyl bromide, more easy is the S_N1 reaction. The correct order is: PhCBrMePh > PhCBrMe₂ > PhCHBrMe > PhCH₂Br.
2. Thus, option (c) is correct.

Teacher's Note:
1. Each extra methyl or phenyl group on a benzylic carbon further stabilises the carbocation, so S_N1 reactivity increases.

Answer: (a) 3-methyl-1-pentene

Explanation:
1. CH₃CH₂–C*H(CH₃)–CH=CH₂ (3-methyl-1-pentene)
2. It has a chiral carbon and hence, it is optically active. All other alkenes do not have a chiral carbon and hence are optically inactive.

Teacher's Note:
1. Look for a carbon with four different groups: here C-3 carries H, CH₃, C₂H₅ and CH=CH₂.

Answer: (b) rotation of polarized light

Explanation:
1. Enantiomers differ only in rotation of polarized light; they rotate it to the same extent but in opposite directions.

Teacher's Note:
1. Enantiomers have identical physical properties (m.p., b.p., solubility) and differ only in the direction of optical rotation and in reactions with other chiral species.

Answer: (a) 2 identical mesomers

Explanation:
1. Anti addition of Br₂ to trans-but-2-ene gives 2,3-dibromobutane in which the two Br atoms are placed so that the molecule has a plane of symmetry.
2. In Fischer projection both structures obtained appear as CH₃–CH(Br)–CH(Br)–CH₃ with both Br on the same side; they are superimposable, i.e., two identical mesomers (meso-2,3-dibromobutane).

Teacher's Note:
1. Remember: trans-alkene + anti addition → meso; cis-alkene + anti addition → racemic mixture.

Answer: (a) trans-1,3-Dibromocyclohexane

Explanation:
1. Due to electron-withdrawing effect of Br, addition of proton occurs at C₂ (the carbocation is formed at C-1, away from Br).
2. Further due to 1,3-diaxial interaction Br⁻ attacks from backside giving trans-1,3-dibromocyclohexane.

Teacher's Note:
1. The −I effect of Br destabilises a positive charge on the carbon next to it, so the carbocation forms at the carbon farther away.

Answer: (c) (±)-Butan-2-ol

Explanation:
1. Racemic mixture, i.e., (±)-butan-2-ol is optically inactive due to equal and opposite rotation caused by (+) and (−)-enantiomers.

Teacher's Note:
1. A racemic mixture is optically inactive because of external compensation, even though each molecule in it is chiral.

Answer: (c) 4

Explanation:
1. CH₃CH=CHCH₂C*HBrCH₃ is an alkene having different substituents at each carbon atom of the double bond. Therefore, it exists in two geometrical isomers, i.e., cis- and trans.
2. Since the given alkene also has a chiral centre, therefore, each geometrical isomer has a pair of enantiomers. Thus, in all, there are 2 × 2 = 4 stereoisomers.

Teacher's Note:
1. Count stereo-elements: each C=C that can be cis/trans and each chiral carbon doubles the number of stereoisomers (2ⁿ, when no meso form is possible).

Answer: (d) ClCH₂–CH=CH₂

Explanation:
1. Allyl halides are more reactive than aryl and vinyl halides. Hence option (d) is correct.

Teacher's Note:
1. In vinyl and aryl halides the C–Cl bond has partial double bond character, whereas allyl chloride ionises to a resonance-stabilised allyl cation.

Answer: (c) II > I > III

Explanation:
1. I is a 2° chloride, II is a 1° chloride and III is an aryl chloride.
2. 1° alkyl halides are more reactive than 2° halides and both of them are more reactive than aryl halides. Thus, correct option is II > I > III, i.e., option (c) is correct.

Teacher's Note:
1. Aryl halides practically do not undergo S_N2: backside attack is blocked by the ring and the C–Cl bond has partial double bond character.

Answer: (d)

Explanation:
1. Aralkyl halides are more reactive than aryl halides, therefore, only the halogen in the side chain is displaced.
2. m-I–C₆H₄–CH₂Cl + NaCN →(DMF) m-I–C₆H₄–CH₂CN

Teacher's Note:
1. The ring iodine stays put: aryl halides need drastic conditions or electron-withdrawing groups for nucleophilic substitution.

Answer: (c) iv > iii > ii > i

Explanation:
1. Nitro groups at o- and p-positions withdraw electrons and stabilise the carbanion intermediate, so reactivity increases with the number of such NO₂ groups: iv (three) > iii (two) > ii (one).
2. NMe₂ is electron-releasing and deactivates the ring towards nucleophilic attack, so (i) is the least reactive.

Teacher's Note:
1. Electron-withdrawing groups at ortho/para positions activate aryl halides for nucleophilic substitution; electron-releasing groups do the opposite.

Answer: (c) C₆H₅CH₃

Explanation:
1. C₆H₅CH₂Br + Mg →(dry ether) C₆H₅CH₂MgBr →(H₃O⁺) C₆H₅CH₃ + Mg(OH)Br
2. The Grignard reagent is protonated by water to give toluene.

Teacher's Note:
1. Grignard reagents react with any compound having an acidic H (water, alcohols, acids) to give the hydrocarbon R–H.

Answer: (b) liberation of I₂

Explanation:
1. Iodoform’s antiseptic properties are due to the liberation of free iodine when it comes in contact with the skin, and not due to iodoform itself.

Teacher's Note:
1. Because of its objectionable smell, iodoform has now largely been replaced by other iodine-based antiseptics.

Answer: (c) (p-ClC₆H₄)₂CH–CCl₃

Explanation:
1. CCl₃CHO + 2 C₆H₅Cl →(conc. H₂SO₄) (p-ClC₆H₄)₂CH–CCl₃ + H₂O
2. The product is DDT, p,p′-dichlorodiphenyltrichloroethane.

Teacher's Note:
1. The CHO carbon becomes the CH joining the two p-chlorophenyl rings; the CCl₃ group is carried over unchanged.

Answer: (c) Non-biodegradable pollutant

Explanation:
1. DDT is a powerful insecticide. Due to its non-biodegradability, chemical stability and fat solubility, it gets accumulated in tissues of animals and causes toxicity.

Teacher's Note:
1. DDT is not broken down in the environment and builds up along food chains (biomagnification), which is why its use was banned in many countries.

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

Explanation:
1. Correct explanation: (+)- and (−)-2-butanol are chiral and rotate plane polarised light. The equal and opposite rotation caused by (+) and (−) enantiomers cancel each other and hence (±)-2-butanol is optically inactive.
2. Both statements are true, but R does not explain A.

Teacher's Note:
1. Optical activity of a pure enantiomer comes from its chirality; the inactivity of a racemic mixture is due to external compensation.

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

Explanation:
1. Correct assertion: Addition of HCl to propene in presence of peroxides gives 2-chloropropane (Markovnikov addition).
2. So A is false and R is true.

Teacher's Note:
1. The H–Cl bond is too strong to be broken homolytically by peroxide radicals, so HCl always adds by Markovnikov’s rule.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Removal of an allylic or benzylic H gives a free radical whose odd electron is delocalised by resonance over the double bond or the ring.
2. Because these radicals are stable, such hydrogens are abstracted easily. R explains A.

Teacher's Note:
1. This is why NBS or Cl₂/hν attacks the allylic or benzylic position selectively.

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

Explanation:
1. Correct reason: In DMSO, the halide ions are not solvated. As a result, free nucleophiles have far greater nucleophilicity and hence S_N2 reactions take place at a faster rate.
2. So A is true but R is false.

Teacher's Note:
1. Polar aprotic solvents (DMSO, DMF, acetone) speed up S_N2 because they leave the anion “naked”; protic solvents cage it by H-bonding.

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

Explanation:
1. Correct explanation: Alkyl halides are insoluble in H₂O since neither they form H-bonds with H₂O nor they break the H-bonds already existing in water.
2. Both statements are true, but R does not explain A.

Teacher's Note:
1. Solubility in water needs H-bonding with water; R–X molecules cannot provide enough energy to break water’s H-bonds.

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

Explanation:
1. Correct assertion: p-Dichlorobenzene has lower solubility than o- and m-isomers.
2. So A is false and R is true.

Teacher's Note:
1. Stronger lattice forces mean higher melting point and lower solubility – the para isomer has both.

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

Explanation:
1. Correct assertion: Aryl halides undergo nucleophilic substitution reactions with difficulty.
2. So A is false and R is true.

Teacher's Note:
1. Partial double bond character (resonance), sp² carbon and instability of the phenyl cation all make aryl halides unreactive.

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

Explanation:
1. Correct explanation: Chlorobenzene is less reactive towards nucleophilic substitution reaction due to partial double bond character of C–X bond (by resonance) which is very difficult to break.
2. Both statements are true, but R does not explain A.

Teacher's Note:
1. An o/p-NO₂ group activates the ring, but it is the resonance in chlorobenzene itself that explains its low reactivity.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. An NO₂ group at o- or p-position withdraws electrons by −R and −I effects and delocalises the negative charge of the intermediate carbanion onto its oxygen atoms.
2. The stabilised carbanion forms more easily, so substitution is faster. R explains A.

Teacher's Note:
1. A meta-NO₂ group cannot take the negative charge by resonance, so it activates much less.

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

Explanation:
1. Correct assertion: Nucleophilicity increases with polarizability (i.e., size). Thus, the correct order is: I⁻ > Br⁻ > Cl⁻ > F⁻.
2. So A is false and R is true.

Teacher's Note:
1. In protic solvents small F⁻ is heavily solvated, while large, soft I⁻ is the best nucleophile.

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

Explanation:
1. Ether molecules coordinate to Mg through the lone pairs on oxygen and stabilise (dissolve) the Grignard reagent.
2. Correct reason: Grignard reagents are insoluble in benzene. So A is true but R is false.

Teacher's Note:
1. The ether must also be dry: even traces of water destroy RMgX, converting it into the alkane.

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

Explanation:
1. 3CCl₄ + 2SbF₃ →(SbCl₅) 3CCl₂F₂ + 2SbCl₃ (Swarts reaction)
2. Correct explanation (of R): CCl₂F₂(g) →(hν) Cl·(g) + ·CClF₂(g)
3. Cl·(g) + O₃(g) → ClO·(g) + O₂(g); ClO· →(slow) Cl· + O·; ClO·(g) + O·(g) → Cl·(g) + O₂(g)
4. As a result of these reactions, O₃ layer is depleted in the stratosphere. Since O₃ absorbs UV (220–290 nm) light and thus protects the earth and its inhabitants from the harmful radiations of the sun.
5. Both statements are true but R is not the explanation of A.

Teacher's Note:
1. One Cl atom is regenerated in each cycle, so a single Cl atom can destroy thousands of O₃ molecules.