CBSE Class 12 Chemistry Haloalkanes and Haloarenes NEET JEE Questions Set 01

Single Correct Answer

Question 1. The major product of the following reaction is
(CH₃)₃C–CH=CH₂ + HI → ? (JEE Main 2022)
(a) (CH₃)₃C–CH₂CH₂I
(b) (CH₃)₃C–CH(I)–CH₃
(c) (CH₃)₂C(I)–CH(CH₃)₂
(d) (CH₃)₃C–C(CH₃)₂I
Answer:
Ans. (c)
H⁺ adds to give a 2° carbocation (CH₃)₃C–C⁺H–CH₃ (less stable).
1,2-Methyl shift converts it into the 3° carbocation (CH₃)₂C⁺–CH(CH₃)₂ (more stable).
Nucleophilic addition of I⁻ then gives (CH₃)₂C(I)–CH(CH₃)₂.
Thus, option (c) is correct.

Teacher's Note:
1. Whenever a 2° carbocation sits next to a carbon carrying methyl groups, look for a 1,2-methyl (or hydride) shift to a 3° carbocation.

Question 2. Identify the reagent(s) A and condition(s) for the reaction.
Cyclohexene →(A)→ 3,6-dichlorocyclohexene (JEE Main 2021)
(a) A = HCl ; Anhydrous AlCl₃
(b) A = HCl, ZnCl₂
(c) A = Cl₂, dark, Anhydrous AlCl₃
(d) A = Cl₂ ; UV light
Answer:
Ans. (d)
Cl₂/hν brings about allylic substitution: both allylic positions of cyclohexene are chlorinated, giving 3,6-dichlorocyclohexene, while the double bond is retained.

Teacher's Note:
1. Cl₂ in the dark adds across C=C; Cl₂ with UV light (or at high temperature) substitutes at the allylic carbon.

Question 3. In the reaction with HCl, an alkene reacts in accordance with the Markovnikov's rule to give a product 1-chloro-1-methylcyclohexane. The possible alkene is :
I : methylenecyclohexane; II : 1-methylcyclohexene; III : 4-methylcyclohexene (RE-AIPMT 2015)
(a) I
(b) II
(c) I + II
(d) III
Answer:
Ans. (c)
Methylenecyclohexane (I): H⁺ adds to the =CH₂ carbon giving the 3° carbocation at C-1 of the ring; Cl⁻ then gives 1-chloro-1-methylcyclohexane.
1-Methylcyclohexene (II): H⁺ adds to C-2 giving the same 3° carbocation at C-1, and hence the same product.
4-Methylcyclohexene (III) gives 1-chloro-3- or 1-chloro-4-methylcyclohexane. Hence I + II.

Teacher's Note:
1. Work backwards: find the carbocation that gives the product, then list every alkene that can form that carbocation.

Question 4. CH₂=C(CH₃)–CH=CH₂ —HBr (1 equiv.)→ ?
The major product of the above reaction is (JEE Main 2017)
(a) BrCH₂–CH(CH₃)–CH₂–CH₃
(b) (CH₃)₂C=CH–CH₂Br
(c) (CH₃)₂C(Br)–CH=CH₂
(d) (CH₃)₂CH–CH₂–CH₂Br
Answer:
Ans. (b)
In case of 1,3-dienes, the major product is the 1,4-addition product, i.e., option (b) is correct.
CH₂=C(CH₃)–CH=CH₂ + H⁺ → CH₃–C⁺(CH₃)–CH=CH₂ ↔ (CH₃)₂C=CH–C⁺H₂
Br⁻ attacks the terminal carbon: (CH₃)₂C=CH–CH₂Br

Teacher's Note:
1. Protonate the diene so that the allylic cation is tertiary; the 1,4-adduct (more substituted C=C) is the thermodynamic product.

Question 5. Product 'A' in the chemical reaction given below is
1-Methyl-1-vinylcyclohexane →(HBr, CCl₄)→ A (major product) (JEE Main 2021)
(a) 1-(2-Bromoethyl)-1-methylcyclohexane
(b) 1-Bromo-1,2-dimethylcyclohexane
(c) 1-(1-Bromoethyl)-2-methylcyclohexane
(d) 1-Bromo-1-(propan-2-yl)cyclohexane [1-bromo-1-isopropylcyclohexane]
Answer:
Ans. (d)
H⁺ adds to the terminal CH₂ of the vinyl group giving a 2° carbocation on the side chain, C⁺H(CH₃) attached to the ring carbon carrying CH₃.
1,2-Methyl shift (CH₃ moves from the ring carbon to the side-chain carbon) gives a 3° carbocation in the ring, carrying an isopropyl group.
Br⁻ attacks this carbon: 1-bromo-1-isopropylcyclohexane. Thus, option (d) is correct.

Teacher's Note:
1. A 2° carbocation next to a quaternary carbon almost always rearranges by a 1,2-methyl shift.

Question 6. The product 'A' of the following sequence of reaction is
C₆H₅CH₂CH₃ →(Br₂/Fe)→ X →(Cl₂, Δ)→ Y →(Alc. KOH)→ A (JEE Main 2022)
(a) C₆H₅–C(Cl)=CH₂
(b) C₆H₅–C(Br)=CH₂
(c) p-Br–C₆H₄–C≡C–Cl
(d) p-Br–C₆H₄–CH=CH₂
Answer:
Ans. (d)
Br₂/Fe (nuclear halogenation): ethylbenzene → p-BrC₆H₄CH₂CH₃ (major product)
Cl₂, Δ (benzylic halogenation): p-BrC₆H₄CH₂CH₃ → p-BrC₆H₄CH(Cl)CH₃
Alc. KOH (dehydrohalogenation, −HCl): p-BrC₆H₄CH(Cl)CH₃ → p-BrC₆H₄CH=CH₂
Thus, option (d) is correct.

Teacher's Note:
1. Lewis acid (Fe/FeBr₃) means ring substitution; heat or light without a Lewis acid means side-chain (benzylic) substitution.

Question 7. Determine the product formed when 2,2-dimethyl-3-bromobutane reacts with ethanol at high temperature (JEE Main 2021)
(a) 2,3-dimethylbut-2-ene
(b) 3,3-dimethylbut-1-ene
(c) 3-ethoxy-2,3-dimethylbutane
(d) 3-ethoxy-2,2-dimethylbutane
Answer:
Ans. (a)
(CH₃)₃C–CH(Br)–CH₃ ionizes (−Br⁻) to the 2° carbocation (CH₃)₃C–C⁺H–CH₃ (less stable).
1,2-Methyl shift gives the 3° carbocation (CH₃)₂C⁺–CH(CH₃)₂.
At high temperature, C₂H₅OH removes a proton (−H⁺) to give (CH₃)₂C=C(CH₃)₂, 2,3-dimethylbut-2-ene.
Thus, option (a) is correct.

Teacher's Note:
1. With a weak base such as ethanol, high temperature favours E1 over S_N1, and the most substituted alkene is formed after rearrangement.

Question 8. In the given reaction sequence, the major product 'C' is
C₈H₁₀ →(HNO₃/H₂SO₄)→ A →(Br₂/hν)→ B →(Alc. KOH)→ C (JEE Main 2022)
(a) C₆H₅–C(NO₂)=CH₂
(b) p-O₂N–C₆H₄–CH=CH₂
(c) p-O₂N–C₆H₄–C(NO₂)=CH₂
(d) o-O₂N–C₆H₄–CH=CH₂
Answer:
Ans. (b)
From options, it is evident that the given compound is ethylbenzene (C₈H₁₀).
Nitration: C₆H₅CH₂CH₃ → p-O₂NC₆H₄CH₂CH₃ (A, major product)
Br₂/hν (benzylic halogenation): A → p-O₂NC₆H₄CH(Br)CH₃ (B)
Alc. KOH (dehydrohalogenation): B → p-O₂NC₆H₄CH=CH₂ (C)
Thus, option (b) is correct.

Teacher's Note:
1. The ethyl group is o,p-directing and the bulky side chain makes the para isomer the major product.

Question 9. The major product (P) in the following reaction is
Ph–CH=CH–CH₂Br →(HBr)→ P (JEE Main 2021)
(a) Ph–CH(Br)–CH(Br)–CH₂Br
(b) Ph–CH₂–CH(CH₂Br)–CH₂Br
(c) Ph–CH(Br)–CH₂–CH₂Br
(d) Ph–CH(Br)–CH=CH–Br
Answer:
Ans. (c)
Ph–CH=CH–CH₂Br + H⁺ → Ph–C⁺H–CH₂–CH₂Br (stabilized by resonance over Ph ring)
Br⁻ then gives Ph–CH(Br)–CH₂–CH₂Br
Thus, option (c) is correct.

Teacher's Note:
1. Electrophilic addition goes through the most stable carbocation; a benzylic cation wins over one next to the electron-withdrawing CH₂Br group.

Question 10. In the addition of HBr to propene in the absence of peroxides, the first step involves the addition of
(a) H⁺
(b) Br⁻
(c) H^•
(d) Br^•
Answer:
Ans. (a)
In absence of peroxides, +vely charged electrophile, i.e., H⁺ adds first.

Teacher's Note:
1. Ionic (Markovnikov) addition starts with the electrophile H⁺; the peroxide (free radical) path starts with Br^•.

Question 11. The intermediate during the addition of HCl to propene in presence of peroxide is
(a) CH₃–ĊH–CH₂Cl
(b) CH₃–C⁺H–CH₃
(c) CH₃CH₂ĊH₂
(d) CH₃CH₂C⁺H₂
Answer:
Ans. (b)
Anti-Markovnikov's addition is observed only with HBr. Therefore, with HCl, even in the presence of peroxides, reaction occurs through a more stable carbocation intermediate, i.e., option (b) is correct.

Teacher's Note:
1. The peroxide effect works only for HBr; H–Cl is too strong to be cleaved by the radical chain.

Question 12. The major product of the following chemical reaction is
(CH₃)₂CH–CH=CH₂ + HBr —(C₆H₅CO)₂O₂→ ? (NEET 2021)
(a) (CH₃)₂CBr–CH₂–CH₃
(b) (CH₃)₂CH–CH₂–CH₂–Br
(c) (CH₃)₂CH–CH₂–CH₂–O–COC₆H₅
(d) (CH₃)₂CH–CH(Br)–CH₃
Answer:
Ans. (b)
In presence of peroxides, anti-Markovnikov's addition to alkenes occurs, i.e., option (b) is correct.
(CH₃)₂CH–CH=CH₂ + HBr —(C₆H₅CO)₂O₂, anti-Mark. addition→ (CH₃)₂CH–CH₂–CH₂–Br

Teacher's Note:
1. Benzoyl peroxide only starts the radical chain; it does not end up in the product.

Question 13. What will be the product in the following reaction?
Methylenecyclohexane →(NBS)→ ?
(a) 1-Bromo-2-methylcyclohexane
(b) 1-Bromo-4-methylcyclohexane
(c) 1-(Bromomethyl)cyclohexene
(d) 1-Bromo-3-methylcyclohexane
Answer:
Ans. (c)
NBS abstracts an allylic H from the ring carbon next to C=CH₂ (−HBr), giving allylic free radical I, with the double bond outside the ring (exocyclic).
Radical I is in resonance with radical II, in which the double bond is within the ring (endocyclic) and the odd electron is on CH₂.
Since endocyclic (within ring) double bond (as in radical II) is more stable than exocyclic (outside ring) double bond (as in radical I), therefore, initially formed less stable free radical (I) gets converted into the more stable free radical (II) which then reacts with Br₂ to give the product, 1-(bromomethyl)cyclohexene.

Teacher's Note:
1. NBS keeps the Br₂ concentration low, so substitution at the allylic position occurs instead of addition to C=C.

Question 14. The product of the reaction given below is
4-tert-Butylcyclohexene →(1. NBS/hν; 2. H₂O/K₂CO₃)→ X (JEE Main 2016)
(a) 4-tert-Butylcyclohexene
(b) 5-tert-Butylcyclohex-2-en-1-ol
(c) 5-tert-Butylcyclohex-2-en-1-one
(d) 5-tert-Butylcyclohex-2-ene-1-carboxylic acid
Answer:
Ans. (b)
Allylic bromination occurs through free radical intermediate. Abstraction of H from C-3 gives free radical Ia and from C-6 gives free radical Ib (numbering: C=C is C1=C2, tert-butyl at C-4).
Free radical (Ib) is stabilized by two equivalent resonance structures but free radical (Ia) is stabilized by two non-equivalent resonance structures. In other words, free radical (Ib) is more stable and thus gives the allylic bromide (II) on treatment with Br₂ (obtained from NBS and HBr).
Subsequent hydrolysis of II with H₂O/K₂CO₃ gives the allylic alcohol (III), 5-tert-butylcyclohex-2-en-1-ol. Thus, option (b) is correct.

Teacher's Note:
1. Hydrolysis with H₂O/K₂CO₃ only replaces Br by OH; it cannot oxidise the alcohol to a ketone or add a carbon to make an acid.

Question 15. The synthesis of alkyl fluorides is best accomplished by (JEE Main 2015)
(a) Finkelstein reaction
(b) Swarts reaction
(c) free radical fluorination
(d) Sandmeyer's reaction
Answer:
Ans. (b)
Swarts reaction is used to prepare alkyl fluorides: R–Cl/Br + AgF (or Hg₂F₂, CoF₂, SbF₃) → R–F

Teacher's Note:
1. Finkelstein gives iodides (NaI/acetone); Swarts gives fluorides (metallic fluorides). Direct fluorination is too violent to control.

Question 16. The compound,
C₇H₈ →(3Cl₂/Δ)→ A →(Br₂/Fe)→ B →(Zn/HCl)→ C
The compound C is (NEET 2018)
(a) o-Bromotoluene
(b) m-Bromotoluene
(c) p-Bromotoluene
(d) 3-Bromo-2,4,6-trichlorotoluene
Answer:
Ans. (b)
m-Bromotoluene (b) is formed.
C₆H₅CH₃ (M.F. = C₇H₈) →(3Cl₂/Δ, side chain chlorination)→ C₆H₅CCl₃ (A)
C₆H₅CCl₃ →(Br₂/Fe, nuclear bromination)→ m-BrC₆H₄CCl₃ (B) [CCl₃ is m-directing]
m-BrC₆H₄CCl₃ →(Zn/HCl, reduction)→ m-BrC₆H₄CH₃ (C)

Teacher's Note:
1. CH₃ is o,p-directing but CCl₃ is strongly electron-withdrawing and meta-directing; the order of steps decides the position.

Question 17. Ph–NH₂ →(HNO₂, 0°C)→ A →(HF, BF₃)→ B →(Δ)→ C, C is (AIIMS 2010)
(a) Ph–N⁺≡N BF₄⁻
(b) 4-Fluorobiphenyl
(c) 4,4′-Difluorobiphenyl
(d) Ph–F
Answer:
Ans. (d)
Ph–NH₂ →(HNO₂, 0°C)→ Ph–N⁺≡N (A) →(HF, BF₃)→ Ph–N⁺≡N BF₄⁻ (B)
Ph–N⁺≡N BF₄⁻ →(Δ)→ Ph–F (C) + BF₃ + N₂

Teacher's Note:
1. This is the Balz–Schiemann reaction, the standard route to aryl fluorides.

Question 18. Arrange the given set of compounds in order of increasing boiling points
I. 1-Chloropropane
II. Isopropyl chloride
III. 1-Chlorobutane (AIIMS 2015)
(a) II < III < I
(b) I < II < III
(c) II < I < III
(d) III < I < II
Answer:
Ans. (c)
For n-alkyl halides, the b.p. increases as the size of the alkyl chain increases, i.e., the b.p. of 1-chlorobutane (III) is higher than that of 1-chloropropane (I).
Further, among isomeric alkyl halides, the b.p. decreases with branching, i.e., the b.p. of isopropyl chloride (II) is lower than that of 1-chloropropane (I).
Thus, option (c), i.e., II < I < III is correct.

Teacher's Note:
1. Two rules settle most b.p. questions: bigger chain, higher b.p.; more branching, lower b.p. (smaller surface area).

Question 19. Increasing order of the boiling points of the major products, A, B and C of the following reactions will be
(i) CH₂=CH–CH₂CH₃ + HBr —(C₆H₅CO)₂O₂→ A
(ii) (CH₃)₂C=CH₂ + HBr → B
(iii) CH₃CH=CHCH₃ + HBr → C (JEE Main 2020)
(a) B < C < A
(b) C < A < B
(c) A < B < C
(d) A < C < B
Answer:
Ans. (a)
(i) Anti-Markovnikov addition: A = CH₃CH₂CH₂CH₂Br, n-butyl bromide (1°)
(ii) Markovnikov addition: B = (CH₃)₃CBr, tert-butyl bromide (3°)
(iii) C = CH₃CH₂CH(Br)CH₃, sec-butyl bromide (2°)
Since surface area decreases in the order : 1° > 2° > 3°, therefore, their bps decrease in the same order, i.e., A > C > B or B < C < A, i.e., option (a) is correct.

Teacher's Note:
1. All three products are isomeric C₄H₉Br, so only branching decides the order.

One or More Correct Answers

Question 20. The IUPAC name(s) of the following compound is (are)
H₃C–C₆H₄–Cl (para) (JEE Advanced 2016)
(a) 1-chloro-4-methylbenzene
(b) 4-chlorotoluene
(c) 1-methyl-4-chlorobenzene
(d) 4-methylchlorobenzene
Answer:
Ans. (a, b)
(a) and (b) are correct names.
Substituents are cited in alphabetical order, with the first-cited (chloro) getting the lower locant: 1-chloro-4-methylbenzene. Toluene is an accepted IUPAC parent name, so 4-chlorotoluene is also correct.

Teacher's Note:
1. With two substituents on benzene, list them alphabetically and give locant 1 to the one that comes first in the alphabet.

Passage-Based Questions

Question 21. Comprehension 1: The polarity of the carbon-halogen bond is responsible for the nucleophilic substitution reactions of alkyl halides which mostly occur by S_N1 and S_N2 mechanisms. The rates of S_N2 reactions are governed by steric factors while that of S_N1 reactions are governed by the stability of intermediate carbocations. Chirality has a profound role in understanding the mechanism of S_N1 and S_N2 reactions. Whereas S_N2 reactions of chiral alkyl halides are accompanied by inversion of configuration, S_N1 reactions are characterised by racemisation.
Which of the following is an example of S_N2 reaction?
(a) CH₃Br + OH⁻ → CH₃OH + Br⁻
(b) (CH₃)₂CHBr + OH⁻ → (CH₃)₂CHOH + Br⁻
(c) CH₃CH₂OH →(−H₂O)→ CH₂=CH₂
(d) (CH₃)₃C–Br + OH⁻ → (CH₃)₃COH + Br⁻
Answer:
Ans. (a)
CH₃Br has the least steric hindrance, so OH⁻ attacks from the back side in a single step: a typical S_N2 reaction.
(d) occurs by S_N1 (3° carbocation), (b) by a mixture of both, and (c) is an elimination (dehydration), not a substitution.

Teacher's Note:
1. Methyl and 1° halides react by S_N2; 3° halides react by S_N1.

Question 22. Comprehension 1: The polarity of the carbon-halogen bond is responsible for the nucleophilic substitution reactions of alkyl halides which mostly occur by S_N1 and S_N2 mechanisms. The rates of S_N2 reactions are governed by steric factors while that of S_N1 reactions are governed by the stability of intermediate carbocations. Chirality has a profound role in understanding the mechanism of S_N1 and S_N2 reactions. Whereas S_N2 reactions of chiral alkyl halides are accompanied by inversion of configuration, S_N1 reactions are characterised by racemisation.
S_N1 reaction of optically active alkyl halides leads to
(a) retention of configuration
(b) racemisation
(c) inversion of configuration
(d) none of these.
Answer:
Ans. (b)
The planar carbocation formed in the slow step is attacked by the nucleophile from both faces with nearly equal ease, giving a (±) mixture, i.e., racemisation.

Teacher's Note:
1. Planar intermediate means attack from both faces, which means racemisation.

Matrix Match Questions

Teacher's Note:
1. Link each name to its key reagent: Sandmeyer – Cu(I) salt; Balz-Schiemann – HBF₄; Hunsdiecker – RCOOAg + Br₂; Finkelstein – NaI/KI in acetone.

Integer and Numerical Answer Questions

Question 24. How many monochloro derivatives are possible when 3-methylheptane is subjected to free radical chlorination?
Answer:
Ans. 8
¹CH₃–²CH₂–³CH(⁸CH₃)–⁴CH₂–⁵CH₂–⁶CH₂–⁷CH₃
All the carbon atoms are different and hence eight monochloro derivatives are possible.

Teacher's Note:
1. Count the different kinds of H (non-equivalent carbons); with no symmetry, every carbon gives a different product.

Question 25. Excess chlorine is passed through boiling toluene, how many chloro derivatives would you get?
Answer:
Ans. 3
Three, i.e., benzyl chloride, benzal dichloride and benzotrichloride.

Teacher's Note:
1. In boiling toluene (no Lewis acid, heat/light) only the CH₃ group is chlorinated, step by step.

Question 26. Total number of compounds among the following having zero dipole moment is/are CCl₄, CH₃Cl, CH₂Cl₂, CHCl₃, o-, m- and p-dichlorobenzenes, benzal dichloride, benzotrichloride.
Answer:
Ans. 2
Two. CCl₄ and p-dichlorobenzene being symmetrical have zero dipole moment.

Teacher's Note:
1. Zero dipole moment needs a symmetrical arrangement in which the bond dipoles cancel.

Assertion–Reason Questions

Question 27. Statement-1: 1-Butene on reaction with HBr in the presence of a peroxide produces 1-bromobutane.
Statement-2: It involves formation of a primary radical. (Bihar Board 2012)
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation of Statement-1.
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation of Statement-1.
(c) Statement-1 is True, Statement-2 is False.
(d) Statement-1 is False, Statement-2 is True.
Answer:
Ans. (c)
Correct statement-2. The reaction occurs through the formation of a more stable 2° free radical.

Teacher's Note:
1. In the peroxide effect, Br^• adds to the terminal carbon so that the more stable 2° radical is formed.

Question 28. Statement-1: NBS is a specific reagent for allylic bromination.
Statement-2: Allylic bromination occurs through free radical intermediates.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation of Statement-1.
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation of Statement-1.
(c) Statement-1 is True, Statement-2 is False.
(d) Statement-1 is False, Statement-2 is True.
Answer:
Ans. (b)
Correct explanation. NBS gives a constant low concentration of Br₂ thereby favouring substitution and preventing addition reaction.

Teacher's Note:
1. NBS is chosen not because the reaction is radical but because it keeps [Br₂] very low, so addition to C=C is avoided.

Question 29. Statement-1: n-Butyl chloride has higher boiling point than n-butyl bromide.
Statement-2: C—Cl bond is more polar than C—Br bond.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation of Statement-1.
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation of Statement-1.
(c) Statement-1 is True, Statement-2 is False.
(d) Statement-1 is False, Statement-2 is True.
Answer:
Ans. (d)
Correct statement-1. n-Butyl bromide has higher boiling point than n-butyl chloride.

Teacher's Note:
1. Boiling points of alkyl halides depend mainly on van der Waals forces, which grow with the size and mass of the halogen: R–I > R–Br > R–Cl > R–F.

Question 30. Statement-1: CH₃Br + AgCN → CH₃NC + AgBr.
Statement-2: CN⁻ is an ambident ion.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation of Statement-1.
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation of Statement-1.
(c) Statement-1 is True, Statement-2 is False.
(d) Statement-1 is False, Statement-2 is True.
Answer:
Ans. (b)
Correct explanation. AgCN is a covalent molecule and hence attack occurs through the lone pair of electrons present on the nitrogen atom.

Teacher's Note:
1. KCN (ionic) gives alkyl cyanides; AgCN (covalent) gives isocyanides, because only the N lone pair is free.

Question 31. Statement-1: Butan-2-ol is optically active.
Statement-2: Its mirror image is non-superimposable on it.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation of Statement-1.
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation of Statement-1.
(c) Statement-1 is True, Statement-2 is False.
(d) Statement-1 is False, Statement-2 is True.
Answer:
Ans. (a)
Statement-2 is the correct explanation of statement-1. C-2 of butan-2-ol carries four different groups (H, OH, CH₃, C₂H₅), so the molecule is chiral.

Teacher's Note:
1. Optical activity requires a chiral molecule, i.e., one that is non-superimposable on its mirror image.

Question 32. Statement-1: Addition of Br₂ to 1-butene gives two optical isomers.
Statement-2: The product contains one asymmetric carbon. (Bihar Board 2012)
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation of Statement-1.
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation of Statement-1.
(c) Statement-1 is True, Statement-2 is False.
(d) Statement-1 is False, Statement-2 is True.
Answer:
Ans. (a)
Statement-2 is the correct explanation of statement-1. CH₃CH₂CH=CH₂ + Br₂ → CH₃CH₂C*H(Br)CH₂Br, which has one asymmetric carbon and exists as two optical isomers (d and l).

Teacher's Note:
1. One asymmetric carbon gives exactly two optical isomers (a pair of enantiomers).