CBSE Class 12 Chemistry Haloalkanes and Haloarenes MCQs Set 10

Answer: (c)

Explanation:
1. In a vinylic halide the halogen is bonded to an sp² carbon of a C=C double bond.
2. In (c) Br is attached directly to a doubly bonded ring carbon, so it is a vinylic halide; (a), (b) and (d) are allylic halides.

Teacher's Note:
1. Vinylic = halogen on the C=C carbon itself; allylic = halogen on the carbon next to the C=C. Count the position before deciding.

Answer: (a) Allyl

Explanation:
1. Br is attached to an sp³ carbon which is next to the C=C double bond, so it is an allylic position.

Teacher's Note:
1. The carbon carrying Br here is also tertiary, so watch out for the distractor “secondary” – it is not secondary.

Answer: (b) 1,2-dichloroethane

Explanation:
1. In vic-dihalides the two halogen atoms are on adjacent carbon atoms: ClCH₂–CH₂Cl.

Teacher's Note:
1. vic (vicinal) = halogens on neighbouring carbons; gem (geminal) = both halogens on the same carbon.

Answer: (b) gem-dihalide

Explanation:
1. Ethylidene chloride is CH₃CHCl₂; both Cl atoms are on the same carbon, so it is a gem-dihalide.

Teacher's Note:
1. “-idene” names (ethylidene, isopropylidene) indicate both halogens on one carbon; “ethylene dichloride” is the vic isomer.

Answer: (a) 2-Bromopentane

Explanation:
1. In 2-bromopentane, CH₃CH(Br)CH₂CH₂CH₃, Br is attached to an sp³ carbon of an alkyl group.
2. In vinyl chloride and chlorobenzene the halogen is on an sp² carbon; 2-chloroacetophenone is a halo-ketone, not an alkyl halide.

Teacher's Note:
1. First identify the hybridisation of the carbon holding the halogen: sp³ means haloalkane, sp² means vinylic or aryl halide.

Answer: (c) 1-Bromo-2-methylbutane

Explanation:
1. The longest chain containing the C–Br carbon has four carbons: BrCH₂–CH(CH₃)–CH₂–CH₃.
2. Numbering from the Br end and writing substituents alphabetically gives 1-bromo-2-methylbutane.

Teacher's Note:
1. An ethyl branch on C-2 usually means the chain was chosen wrongly – include the ethyl group in the main chain.
2. Prefixes are written in alphabetical order (bromo before methyl), so option (d) is wrong even though its numbers are right.

Answer: (b) 1-fluoro-4-methyl-2-nitrobenzene

Explanation:
1. The substituents get the lowest set of locants {1, 2, 4}; with equal sets, the group first in alphabetical order (fluoro) gets C-1.
2. So F is at C-1, NO₂ at C-2 and CH₃ at C-4: 1-fluoro-4-methyl-2-nitrobenzene.

Teacher's Note:
1. When several numberings give the same lowest locant set, give number 1 to the substituent that comes first alphabetically.

Answer: (b) SOCl₂

Explanation:
1. R–OH + SOCl₂ → R–Cl + SO₂↑ + HCl↑
2. The by-products are gases which escape, leaving almost pure alkyl chloride.

Teacher's Note:
1. The deciding point is purity: with SOCl₂ both by-products are gases, so no separation is needed.

Answer: (a) Both the compounds form same product on treatment with alcoholic KOH.

Explanation:
1. ClCH₂CH₂Cl and CH₃CHCl₂ both lose HCl with alcoholic KOH to give vinyl chloride, CH₂=CHCl (and finally ethyne).
2. With aq. NaOH, ethylene chloride gives ethane-1,2-diol while ethylidene chloride gives ethanal; ethylene chloride is a vic-dihalide and neither compound has a chiral carbon.

Teacher's Note:
1. Hydrolysis of a gem-dihalide gives a carbonyl compound (the gem-diol loses water), while a vic-dihalide gives a diol.

Answer: (a) (i) < (ii) < (iii) < (iv)

Explanation:
1. Density increases with the number and the atomic mass of halogen atoms.
2. Benzene (no halogen) < chlorobenzene (one Cl) < dichlorobenzene (two Cl) < bromochlorobenzene (Cl + heavier Br).

Teacher's Note:
1. For density, more halogen atoms and heavier halogens (Br, I) both raise the value.

Answer: (c) (iii) < (i) < (ii)

Explanation:
1. For isomeric alkyl halides, boiling point decreases with branching because the surface area and hence van der Waals forces decrease.
2. The most branched (CH₃)₃CBr has the lowest and the straight-chain 1-bromobutane the highest boiling point.

Teacher's Note:
1. Same molecular mass? Then compare shape: the more spherical (branched) isomer always boils lower.

Answer: (b) Cl⁺

Explanation:
1. Cl₂ + AlCl₃ → Cl⁺ + [AlCl₄]⁻; the electrophile Cl⁺ attacks the electron-rich benzene ring.

Teacher's Note:
1. The Lewis acid only generates the electrophile; the ring is always attacked by the positive species.

Answer: (a) Cl₂/UV light

Explanation:
1. Alkanes undergo free radical chlorination with Cl₂ in the presence of UV light (or heat), giving a mixture of isomeric monochloro products.

Teacher's Note:
1. Free radical halogenation of alkanes needs light or heat; Cl₂/Fe in the dark is for ring substitution of arenes.

Answer: (b) alcoholic KOH followed by NaNH₂

Explanation:
1. Alcoholic KOH removes one HBr to give vinyl bromide, CH₂=CHBr.
2. The second HBr is removed from the less reactive vinylic halide only by the stronger base NaNH₂, giving ethyne.

Teacher's Note:
1. Two dehydrohalogenations are needed; the second one, from a vinylic halide, needs a very strong base such as NaNH₂.

Answer: (b) (C) > (B) > (A)

Explanation:
1. Reaction with HX goes through a carbocation; stability 3° > 2° > 1°.
2. So the tertiary alcohol (C) reacts fastest and the primary alcohol (A) slowest: (C) > (B) > (A).

Teacher's Note:
1. This is the basis of the Lucas test: 3° alcohols give turbidity at once, 2° in a few minutes and 1° only on heating.

Answer: (b) (i), (ii), (iii)

Explanation:
1. An asymmetric carbon has four different groups attached to it.
2. In (i), (ii) and (iii) all four groups are different (D and H count as different); in (iv) two H atoms are attached, so it is not asymmetric.

Teacher's Note:
1. Isotopes count as different groups, so a C bearing D, I, Cl and Br is still chiral.

Answer: (a)

Explanation:
1. In (a) the positions of CH₃ and C₂H₅ are interchanged with respect to (A) while H and Br stay in place.
2. Interchanging any two groups gives the mirror image, so (a) is the enantiomer of (A); (b), (c) and (d) are only rotated drawings of (A) itself.

Teacher's Note:
1. One swap of two groups gives the enantiomer; two swaps bring you back to the same molecule.

Answer: (b) enantiomers

Explanation:
1. Lactic acid has one chiral carbon; its two forms are non-superimposable mirror images, i.e., enantiomers.

Teacher's Note:
1. A molecule with only one chiral centre can have only one pair of stereoisomers, and they are always enantiomers.

Answer: (c) inversion of configuration

Explanation:
1. In S_N2 the nucleophile attacks from the side opposite to the leaving group, so the configuration is inverted (Walden inversion).

Teacher's Note:
1. S_N2 = backside attack = inversion; S_N1 = planar carbocation = racemisation.

Answer: (d) (CH₃)₃C–I

Explanation:
1. All give the same tert-butyl carbocation; the rate then depends on the ease of breaking the C–X bond.
2. The C–I bond is the weakest and I⁻ is the best leaving group, so (CH₃)₃C–I reacts fastest.

Teacher's Note:
1. When the alkyl group is the same, reactivity order is R–I > R–Br > R–Cl > R–F, set by C–X bond strength.

Answer: (c) Cl⁻ < Br⁻ < I⁻

Explanation:
1. In protic solvents, larger ions are less solvated and more polarisable, so nucleophilicity increases down the group: Cl⁻ < Br⁻ < I⁻.

Teacher's Note:
1. Nucleophilicity of halide ions in protic solvents runs opposite to their basicity – iodide is the weakest base but the best nucleophile.

Answer: (b) Electrophilic substitution reaction

Explanation:
1. FeCl₃ generates the electrophile X⁺, which substitutes a ring hydrogen; the CH₃ group directs it to ortho and para positions.

Teacher's Note:
1. Lewis acid + halogen + arene in the dark = ring (electrophilic) substitution; light + heat would instead halogenate the side chain.

Answer: (d) Both A and R are incorrect.

Explanation:
1. Thionyl chloride is preferred, because its by-products SO₂ and HCl are gases and escape, leaving pure alkyl chloride.
2. With PCl₃/PCl₅ the by-products H₃PO₃/POCl₃ remain in the mixture, so the alkyl halide is not obtained pure. Hence both A and R are incorrect.

Teacher's Note:
1. Remember the purity argument belongs to SOCl₂: both of its by-products are gases.

Answer: (b) (iii) > (ii) > (i)

Explanation:
1. The reactivity order of alcohols towards halogen acids is 3° > 2° > 1° as the stability of carbocations is of the order 3° > 2° > 1°.

Teacher's Note:
1. (i) is primary, (ii) is secondary and (iii) is tertiary – classify each alcohol first, then apply the carbocation stability order.

Answer: (d) CH₃CH₂–C(CH₃)₂–OH

Explanation:
1. As tertiary carbocation is more stable, so tertiary alcohols will yield the corresponding alkyl chloride on reaction with concentrated HCl at room temperature. While primary and secondary alcohols require the presence of a catalyst ZnCl₂.

Teacher's Note:
1. This is the basis of the Lucas test: tertiary alcohols give turbidity with conc. HCl/ZnCl₂ immediately, secondary ones in a few minutes and primary ones only on heating.

Answer: (b) S_N2 reaction

Explanation:
1. S_N2 reaction proceeds via formation of transition state which is formed easily in primary alkyl halide due to less steric hindrance.

Teacher's Note:
1. Link the class of halide to the mechanism: 1° → S_N2 (least hindered), 3° → S_N1 (most stable carbocation).

Answer: (a) Chlorobenzene

Explanation:
1. When a primary aromatic amine, is dissolved or suspended in cold aqueous mineral acid and treated with sodium nitrite, a diazonium salt is formed. When this freshly prepared diazonium salt is mixed with cuprous chloride, diazonium group is replaced by Cl. Then chlorobenzene is formed which is Y in this reaction (Sandmeyer reaction).

Teacher's Note:
1. Only the –N₂⁺ group is replaced, so exactly one Cl enters at the position where NH₂ was.

Answer: (c) C₆H₅–CH₂–CH(Cl)–CH₃

Explanation:
1. In this reaction, addition of HCl takes place on doubly bonded carbons in accordance with Markovnikov's rule, that is, addition of negative addendum will take place on that carbon which has lesser number of hydrogen.
2. C₆H₅–CH₂–CH=CH₂ + HCl → C₆H₅–CH₂–CH(Cl)–CH₃

Teacher's Note:
1. HCl adds across the side-chain C=C; it does not substitute on the ring, so options with Cl on the ring or a remaining C=C are ruled out.

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

Explanation:
1. For the same alkyl group, boiling point rises with the size and mass of the halogen because van der Waals forces increase: RI > RBr > RCl > RF.
2. R is also true (greater polarity and mass than the parent hydrocarbon), but it compares alkyl halides with hydrocarbons, so it does not explain A.

Teacher's Note:
1. Check whether R is talking about the same comparison as A – here A compares halogens, R compares halides with hydrocarbons.

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

Explanation:
1. KCN is ionic; CN⁻ attacks through the carbon atom (C–C bond is stronger than C–N), so methyl cyanide (CH₃CN) is the main product, not the isocyanide.
2. CN⁻ can attack through C or N, so it is an ambident nucleophile. A is incorrect, R is correct.

Teacher's Note:
1. KCN gives cyanides; covalent AgCN (attack through N) gives isocyanides.

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

Explanation:
1. Tertiary alkyl halides do not give the Wurtz product; with sodium they undergo dehydrohalogenation and give isobutane and isobutylene instead.
2. R correctly describes the Wurtz reaction. So A is incorrect and R is correct.

Teacher's Note:
1. Wurtz reaction works well only for primary alkyl halides; tertiary halides prefer elimination.

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

Explanation:
1. Hydrolysis of 2-bromooctane by OH⁻ is an S_N2 reaction; backside attack gives inversion of configuration.
2. An S_N2 reaction occurs in one step through a transition state, not through a carbocation. So A is correct but R is incorrect.

Teacher's Note:
1. Inversion is the signature of S_N2; a carbocation route would give racemisation instead.

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

Explanation:
1. 3° alkyl halides react fastest by S_N1 because they form the most stable (3°) carbocation.
2. R is true (first-order kinetics) but it does not explain why 3° halides react fastest.

Teacher's Note:
1. The reason for S_N1 reactivity order is carbocation stability (+I and hyperconjugation), not the rate law.

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

Explanation:
1. The electron withdrawing NO₂ group at o/p position lowers the electron density on the ring and stabilises the carbanion intermediate by resonance, so the nucleophile attacks more easily. R explains A.

Teacher's Note:
1. A meta-nitro group has much less effect because the negative charge of the intermediate cannot be delocalised onto it.

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

Explanation:
1. In chlorobenzene the Cl atom is ortho/para directing, so nitration gives mainly o- and p-nitrochlorobenzene. A is incorrect.
2. NO₂ is indeed a meta-directing group, but here the group already present (Cl) decides the orientation. R is correct.

Teacher's Note:
1. The orientation is decided by the substituent already on the ring, not by the incoming group.

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

Explanation:
1. Due to resonance the C–Cl bond in chlorobenzene acquires partial double bond character and becomes stronger and shorter than in chloroethane, so it is difficult to break. R explains A.

Teacher's Note:
1. Other reasons (sp² carbon, unstable phenyl cation, repulsion to the nucleophile) add to this, but resonance is the key one asked in exams.

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

Explanation:
1. The +R effect of halogen increases electron density at ortho and para positions, so the electrophile enters there.
2. Halogen also deactivates the ring by its −I effect; R is true but it explains the slow rate, not the o/p orientation.

Teacher's Note:
1. Halogens are the classic “deactivating but ortho/para directing” groups: −I controls rate, +R controls orientation.

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

Explanation:
1. Iodination of arenes is reversible because the HI formed reduces the aryl iodide back. An oxidising agent (HNO₃, HIO₄) is used to oxidise HI to I₂ and remove it.
2. So A is correct, but R is incorrect: the oxidising agent oxidises HI, not I₂.

Teacher's Note:
1. Oxidising agents remove HI by converting it to I₂; going from I₂ to HI would be a reduction.

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

Explanation:
1. Both statements are true. Inversion occurs because the nucleophile attacks from the side opposite to the leaving group (backside attack).
2. The single-step nature of the reaction does not by itself explain the inversion, so R is not the correct explanation of A.

Teacher's Note:
1. For A-R questions, ask "does R tell me WHY A happens?" – here the reason for inversion is backside attack, not the number of steps.