CBSE Class 12 Chemistry Haloalkanes and Haloarenes NEET JEE Questions Set 03

Single Correct Answer

Question 1. Which of the following will most readily give the dehydrohalogenation product? (JEE Main (online) 2018)
(a) Bromobenzene
(b) 1-Bromocyclohexene
(c) CH₂=CH–C(Br)=CH–Ph
(d) CH₂=CH–C(Br)=CH–CH₃
Answer:
Ans. (c)
Since triple bond is unstable in a small six-membered ring, therefore, bromide under option (b) cannot undergo dehydrohalogenation. Both bromides under option (c) and (d) have a β-hydrogens and hence can undergo dehydrohalogenation.
But bromide under option (c) upon elimination of HBr gives a triple bond which is conjugated with a benzene ring but bromide under option (d) gives a triple bond which is stabilized only by weak +I-effect of the CH₃ group.
(c) →(−HBr)→ CH₂=CH–C≡C–Ph (triple bond stabilized by resonance over benzene ring, more stable)
(d) →(−HBr)→ CH₂=CH–C≡C–CH₃ (triple bond stabilized by +I-effect of CH₃ group, less stable)
Thus, bromide under option (c) undergoes dehydrohalogenation most readily.

Teacher's Note:
1. When two substrates can both eliminate, compare the stability of the products: extra conjugation with a benzene ring makes the elimination easier.

Question 2. How many chiral compounds are possible on monochlorination of 2-methylbutane? (AIEEE 2012)
(a) 2
(b) 4
(c) 6
(d) 8
Answer:
Ans. (a)
2-Methylbutane on monochlorination gives four isomers of which I and III are chiral compounds.
I : ClCH₂–C*H(CH₃)–CH₂CH₃; II : CH₃–C(CH₃)(Cl)–CH₂CH₃; III : (CH₃)₂CH–C*H(Cl)–CH₃; IV : (CH₃)₂CH–CH₂–CH₂Cl
Hence, the number of chiral compounds is 2.

Teacher's Note:
1. The question asks for chiral compounds (constitutional isomers), not stereoisomers; do not count each enantiomer separately.

Question 3. 3-Methylpent-2-ene on reaction with HBr in presence of peroxide forms an addition product. The number of possible stereoisomers for the product is (JEE Main 2017)
(a) two
(b) four
(c) six
(d) zero
Answer:
Ans. (b)
Anti-Markovnikov's addition of HBr to 3-methylpent-2-ene gives product (I), CH₃–C*H(Br)–C*H(CH₃)–CH₂CH₃ (2-bromo-3-methylpentane).
Since compound (I) has two dissimilar chiral centres, therefore, it has 2² = 4 stereoisomers.

Teacher's Note:
1. With peroxide, Br goes to the less substituted carbon of the C=C; then count chiral centres and use 2ⁿ when they are dissimilar.

Question 4. Which of the following compounds produces an optically inactive compound on hydrogenation? (Each compound is a single enantiomer.) (JEE Main 2020)
(a) CH₃CH₂CH₂–CH(CH₃)–CH=C(CH₃)₂
(b) (CH₃)₂CH–CH(CH₃)–C(CH₃)=CH₂
(c) CH₃CH=CH–CH(CH₃)–CH₂CH₃
(d) (CH₃)₂CH–CH(CH₃)–CH=CH–CH₃
Answer:
Ans. (b)
(a) →(H₂/Pt)→ CH₃CH₂CH₂–CH(CH₃)–CH₂CH(CH₃)₂ : optically active
(b) →(H₂/Pt)→ (CH₃)₂CH–CH(CH₃)–CH(CH₃)₂ : optically inactive since the central carbon is attached to two identical groups
(c) →(H₂/Pt)→ CH₃CH₂CH₂–CH(CH₃)–CH₂CH₃ : optically active
(d) →(H₂/Pt)→ (CH₃)₂CH–CH(CH₃)–CH₂CH₂CH₃ : optically active
Thus, option (b) is correct.

Teacher's Note:
1. After hydrogenation, check whether two of the four groups on the old chiral carbon have become identical.

Question 5. In a mixture, two enantiomers are found to be present in 85% and 15% respectively. The enantiomeric excess (e.e.) is (West Bengal JEE 2015)
(a) 85%
(b) 15%
(c) 70%
(d) 60%
Answer:
Ans. (c)
Since the two enantiomers are present in 85% and 15% respectively, therefore, 15% out of 85% will form a racemic mixture with 15% of the other isomer. Therefore, enantiomeric excess (e.e.) = 85 − 15 = 70%.

Teacher's Note:
1. Enantiomeric excess = % of major enantiomer − % of minor enantiomer.

Question 6. A solution of (−)-1-chloro-1-phenylethane in toluene racemises slowly in presence of a small amount of SbCl₅ due to the formation of (JEE Main 2013)
(a) free radical
(b) carbanion
(c) carbene
(d) carbocation
Answer:
Ans. (d)
In presence of SbCl₅, (−)-1-chloro-1-phenylethane forms a carbocation as shown :
CH₃–CHCl–C₆H₅ + SbCl₅ → [CH₃–C⁺H–C₆H₅] SbCl₆⁻
Since the carbocation is a planar species, therefore, it can be attacked by SbCl₆⁻ either from the top or the bottom face with equal ease. As a result, a 50 : 50 mixture of two enantiomers of 1-chloro-1-phenylethane are formed, i.e., (−)-1-chloro-1-phenylethane undergoes racemization due to the formation of a carbocation intermediate.

Teacher's Note:
1. A Lewis acid pulls off Cl⁻; a planar benzylic carbocation can be attacked from either face, so racemization follows.

Question 7. (CH₃)₂CH–CH₂–CH₃ →(Cl₂, hν)→ N [isomeric products] C₅H₁₁Cl →(fractional distillation)→ M [isomeric products]
Give the number of N and M?
(a) 6, 6
(b) 6, 4
(c) 4, 4
(d) 3, 3
Answer:
Ans. (b)
Monochlorination of 2-methylbutane gives : ClCH₂–C*H(CH₃)CH₂CH₃ [(i) and (ii), (+) and (−) isomers]; (CH₃)₂CCl–CH₂CH₃ (iii); (CH₃)₂CH–C*HCl–CH₃ [(iv) and (v), (+) and (−) isomers]; (CH₃)₂CH–CH₂CH₂Cl (vi)
A total of 6 isomers will be formed. Since enantiomers have the same b.p., they cannot be separated by fractional distillation, i.e., (i) and (ii) will distil together and similarly (iv) and (v) will distil together. Thus, out of 6 isomers formed, four can be separated by fractional distillation. Hence N = 6, M = 4.

Teacher's Note:
1. Enantiomers have identical physical properties (b.p., solubility), so fractional distillation cannot separate them.

Question 8. The absolute configuration of the following compound is

Teacher's Note:
1. In a Fischer projection, if H is on a horizontal bond, read the direction of 1 → 2 → 3 and then reverse it.

Question 9. Which of the following biphenyls is optically active? (NEET Phase-I 2016)
(a) 2,2′-Diiodobiphenyl
(b) 2,6-Dimethylbiphenyl
(c) 3-Iodo-3′-nitrobiphenyl
(d) 2,2′-Dibromo-6,6′-diiodobiphenyl
Answer:
Ans. (d)
o-Substituted diphenyls having bulky groups on both the o-positions of each benzene ring are optically active since the two benzene rings due to steric hindrance lie in perpendicular planes and hence their mirror images are non-superimposable.
In (d), each ring carries two different bulky ortho groups (Br and I); the mirror images are non-superimposable (optically active).

Teacher's Note:
1. Biphenyl atropisomerism needs restricted rotation (bulky ortho groups) and no plane of symmetry (different groups on each ring).

Question 10. Which of the following compounds will show retention in configuration on nucleophilic substitution by OH⁻ ion? (JEE Main 2020)
(a) CH₃–CBr(H)–C₆H₁₃
(b) CH₃–CH(Br)–C₆H₅
(c) CH₃–CH(Br)–CH₃
(d) CH₃–CH(C₆H₅)–CH₂Br
Answer:
Ans. (d)
In compounds (a) and (b), nucleophilic attack by OH⁻ ion occurs on the chiral centre, hence inversion of configuration occurs if the reaction takes place by S_N2 mechanism. Compound (c) is not chiral.
Compound (d) contains a chiral centre but the reaction occurs on a carbon which is not chiral, therefore, retention of configuration occurs during its reaction with OH⁻ ion.

Teacher's Note:
1. Configuration at a stereocentre changes only if a bond to that stereocentre is broken.

Question 11. Which of the following reactions will not produce a racemic mixture? (JEE Main 2020)
(a) CH₃CH₂CH=CH₂ →(HBr)→
(b) 1,3-Dimethylcyclohexene →(HCl)→
(c) CH₃–CH(CH₃)–CH=CH₂ →(HCl)→
(d) CH₃–CO–CH₂CH₃ →(HCN)→
Answer:
Ans. (c)
(a) CH₃CH₂CH=CH₂ →(HBr, Mark. addition)→ CH₃CH₂C*H(Br)–CH₃ : racemic mixture
(b) 1,3-Dimethylcyclohexene →(HCl, Mark. addition)→ 1-chloro-1,3-dimethylcyclohexane (chiral carbons) : racemic mixture
(c) CH₃–CH(CH₃)–CH=CH₂ →(H⁺)→ (CH₃)₂CH–C⁺H–CH₃ →(1,2-hydride shift)→ (CH₃)₂C⁺–CH₂CH₃ →(Cl⁻)→ (CH₃)₂C(Cl)–CH₂CH₃ : no chiral centre
(d) CH₃–CO–CH₂CH₃ →(HCN)→ CH₃–C*(OH)(CN)–CH₂CH₃ : racemic mixture
The product of reaction (c) does not contain a chiral centre, therefore, it represents a single compound and not a racemic mixture. Thus, option (c) is correct.

Teacher's Note:
1. Watch for carbocation rearrangements: here a hydride shift places Cl on a carbon carrying two CH₃ groups, which removes the chirality.

Question 12. Among the following compounds, I-IV, which one forms a yellow precipitate on reacting sequentially with (i) NaOH, (ii) dil. HNO₃, (iii) AgNO₃? (JEE Main 2021)
(a) I : 4-chlorotoluene
(b) II : 3-chlorotoluene
(c) III : 2-bromotoluene
(d) IV : 2-chlorobenzyl iodide (o-Cl–C₆H₄–CH₂I)
Answer:
Ans. (d)
In compounds (I, II and III) halogen is directly bonded to benzene ring and hence it cannot be displaced due to partial double bond character of C–Cl or C–Br bond.
In compound (IV), iodine is present in the side chain and hence will be hydrolysed by NaOH to give NaI which upon acidification with HNO₃ and subsequent treatment with AgNO₃ will give yellow ppt. of AgI.

Teacher's Note:
1. Aryl halides do not give the AgNO₃ test; only halogen in the side chain is released by aqueous NaOH. AgCl is white, AgBr pale yellow, AgI yellow.

Question 13. 2-Bromobenzyl chloride (o-Br–C₆H₄–CH₂Cl) →(NH₃, EtOH)→ ?
The product of the above reaction is (West Bengal JEE 2015)
(a) o-H₂N–C₆H₄–CH₂NH₂
(b) o-H₂N–C₆H₄–CH₂Cl
(c) o-Br–C₆H₄–CH₂NH₂
(d) o-EtO–C₆H₄–CH₂NH₂
Answer:
Ans. (c)
Aralkyl halides (ArCH₂Cl) are more reactive than aryl halides (Ar–Br), therefore, the halogen in the side chain is displaced to give the corresponding substitution product.
o-Br–C₆H₄–CH₂Cl →(NH₃, EtOH)→ o-Br–C₆H₄–CH₂NH₂
Thus, option (c) is correct.

Teacher's Note:
1. Under ordinary conditions the C–Br bond on the ring is not touched; only the benzylic C–Cl reacts.

Question 14. Bottles containing C₆H₅I and C₆H₅CH₂I lost their original labels. They were labelled as A and B for testing. A and B were separately taken in test tubes and boiled with NaOH solution. The end solution in each tube was made acidic with dilute HNO₃ and some AgNO₃ solution was added. Substance B gave a yellow precipitate. Which one of the following statements is true for this experiment?
(a) Addition of HNO₃ was unnecessary
(b) A was C₆H₅I
(c) A was C₆H₅CH₂I
(d) B was C₆H₅I
Answer:
Ans. (b)
Tube A. C₆H₅I →(NaOH, boil)→ No reaction →((i) dil. HNO₃, (ii) AgNO₃)→ No yellow ppt. of AgI
Tube B. C₆H₅CH₂I →(NaOH, boil; −C₆H₅CH₂OH)→ NaI →((i) dil. HNO₃, (ii) AgNO₃)→ AgI (yellow ppt.)
Thus, statement (b), i.e., A was C₆H₅I is correct.

Teacher's Note:
1. HNO₃ is needed to destroy excess NaOH, otherwise brown Ag₂O would precipitate and spoil the test.

Question 15. Which of the following compounds undergo nucleophilic substitution reaction most easily? (AIPMT Mains 2011)
(a) p-Nitrochlorobenzene
(b) p-Chlorotoluene
(c) p-Chloroanisole
(d) Chlorobenzene
Answer:
Ans. (a)
Presence of electron-withdrawing groups (i.e., NO₂, CN, etc.) at o- and p-positions w.r.t. Cl activates it towards nucleophilic substitution reactions while presence of electron-donating groups (i.e., CH₃, OCH₃, etc.) deactivates it towards nucleophilic substitution reactions.
Thus, p-nitrochlorobenzene undergoes nucleophilic substitution reaction most easily. Thus, option (a) is correct.

Teacher's Note:
1. An o/p-NO₂ group stabilises the negatively charged intermediate (carbanion) of nucleophilic aromatic substitution.

Question 16. Which of the following compounds will undergo substitution of the halogen atom most readily with CH₃ONa in CH₃OH?
(a) o-Fluoronitrobenzene
(b) o-Chloronitrobenzene
(c) o-Bromonitrobenzene
(d) o-Iodonitrobenzene
Answer:
Ans. (a)
In bimolecular nucleophilic aromatic substitution reactions the order of reactivity of aryl halides is : ArF > ArCl > ArBr > ArI. Thus, option (a) is correct.
The slow step is addition of the nucleophile; the highly electronegative F makes the carbon most electron-deficient and best stabilises the carbanion intermediate.

Teacher's Note:
1. In S_NAr the order of halogens is the reverse of S_N1/S_N2: F reacts fastest because C–X bond breaking is not the slow step.

Question 17. Which of the following will not undergo S_N1 reaction with OH⁻ ion? (NEET 2020)
(a) (CH₃)₃CCl
(b) C₆H₅CH₂CH₂Cl
(c) C₆H₅CH₂Cl
(d) CH₂=CH–CH₂Cl
Answer:
Ans. (b)
C₆H₅CH₂CH₂–Cl →(S_N1, does not occur)→ C₆H₅CH₂–C⁺H₂ + Cl⁻
The carbocation is primary and is not stabilized by benzene ring (the ring is not directly attached to the positive carbon). Thus, option (b) is correct.

Teacher's Note:
1. Benzylic stabilisation needs the ring on the carbon bearing the positive charge; one extra CH₂ in between removes it.

Question 18. Identify A and predict the type of reaction.
3-Bromoanisole (m-BrC₆H₄OCH₃) →(NaNH₂)→ A (NEET 2017)
(a) 2-Methoxyaniline (o-anisidine) and elimination-addition reaction
(b) 2-Bromoanisole and cine-substitution reaction
(c) Anisole and cine-substitution reaction
(d) 3-Methoxyaniline (m-anisidine) and substitution reaction
Answer:
Ans. (d)
Reaction occurs by benzyne mechanism.
NaNH₂ removes the H between OCH₃ and Br (−NaBr, −NH₃) to give a benzyne. NH₂⁻ adds to the benzyne carbon meta to OCH₃, because the carbanion formed (next to OCH₃) is stabilized by −I-effect of OCH₃ group.
Protonation by NH₃ (−NH₂⁻) gives 3-methoxyaniline; NH₂ ends up at the same position as Br, i.e., a substitution product.

Teacher's Note:
1. In benzyne reactions, the incoming NH₂ goes to the carbon that puts the negative charge next to the electron-withdrawing (−I) OCH₃ group.

Question 19. Which of the following can be used as the halide component for Friedel-Crafts reaction? (NEET Phase-II 2016)
(a) Chlorobenzene
(b) Bromobenzene
(c) Chloroethene
(d) Isopropyl chloride
Answer:
Ans. (d)
Aryl halides (i.e., chlorobenzene and bromobenzene) and vinyl halides (i.e., chloroethene) being less reactive than alkyl halides do not react with benzene to give the corresponding substitution products. Therefore, only isopropyl chloride can be used as the halide component in F.C. reactions.

Teacher's Note:
1. Friedel–Crafts needs a carbocation (or polarised complex); sp² C–X bonds of aryl and vinyl halides do not ionize with AlCl₃.

Question 20. Identify the major product
CH₃–CH(Br)–CH₂–CH(Br)–CH₂–CH₃ →(Mg, dry ether)→ ? (JEE Main 2022)
(a) CH₃–CH(MgBr)–CH₂–CH(MgBr)–CH₂CH₃
(b) 1-Ethyl-2-methylcyclopropane
(c) 1,2-Diethyl-4,5-dimethylcyclohexane
(d) 1,4-Diethyl-2,5-dimethylcyclohexane
Answer:
Ans. (b)
Mg inserts into one C–Br bond; the carbanion-like C–MgBr carbon then displaces Br from the carbon in the 1,3-position (−MgBr₂) closing a three-membered ring.
The product is 1-ethyl-2-methylcyclopropane.

Teacher's Note:
1. 1,3-Dihalides with Mg (or Zn) give cyclopropanes by intramolecular Wurtz-type coupling.

One or More Correct Answers

Question 21. Compound(s) that on hydrogenation produce(s) optically inactive compound(s) is (are) (each compound is a single enantiomer) (JEE Advanced 2015)
(a) H₃C–CH=CH–CH(Br)–CH₃
(b) CH₂=CH–CH(Br)–CH₂CH₃
(c) CH₂=C(CH₃)–CH(Br)–CH₃
(d) CH₂=CH–CH(Br)–CH₂CH₃ (mirror image of b)
Answer:
Ans. (b, d)
Compounds (b) and (d) being enantiomers upon hydrogenation give the same reduction product (I), CH₃CH₂–CH(Br)–CH₂CH₃ (3-bromopentane).
Compound (I) being symmetrical does not contain a chiral carbon and hence is optically inactive. Thus, options (b) and (d) are correct.
Options (a) and (c) give optically active products: (a) → CH₃CH₂CH₂–CH(Br)–CH₃ and (c) → (CH₃)₂CH–CH(Br)–CH₃, each with a chiral carbon.

Teacher's Note:
1. Hydrogenation turns a vinyl group into an ethyl group; if the chiral carbon already carries an ethyl group, it stops being chiral.

Passage-Based Questions

Question 22. Comprehension 2: An alkyl halide with β-hydrogen atoms when reacted with a base or a nucleophile has two competing routes : Substitution (S_N1 and S_N2) and elimination. Which route will be taken up depends upon the nature of the alkyl halide, strength and size of the base/nucleophile and reaction conditions. Thus, bulkier nucleophile prefers to act as a base and abstracts a proton rather than approaching a tetravalent carbon atom (steric reasons) and vice-versa.
Isopropyl bromide on heating with a concentrated solution of alcoholic (ethanolic) KOH predominantly gives
(a) Propene
(b) Propan-2-ol
(c) Propan-1-ol
(d) Isopropyl ethyl ether
Answer:
Ans. (a)
C₂H₅O⁻ Na⁺ is a strong base, therefore, heating isopropyl bromide with C₂H₅O⁻Na⁺/C₂H₅OH brings about dehydrohalogenation to yield propene.

Teacher's Note:
1. Alcoholic KOH (ethoxide) plus heat favours elimination; aqueous KOH favours substitution.

Question 23. Comprehension 2: An alkyl halide with β-hydrogen atoms when reacted with a base or a nucleophile has two competing routes : Substitution (S_N1 and S_N2) and elimination. Which route will be taken up depends upon the nature of the alkyl halide, strength and size of the base/nucleophile and reaction conditions. Thus, bulkier nucleophile prefers to act as a base and abstracts a proton rather than approaching a tetravalent carbon atom (steric reasons) and vice-versa.
2-Bromopropane is separately heated with aq. CH₃CO₂Na or with CH₃CH₂ONa/CH₃CH₂OH, the major product obtained in each case respectively are
(a) propene, isopropyl ethyl ether
(b) isopropyl acetate, propene
(c) isopropyl acetate, isopropyl ethyl ether
(d) propene in both the cases
Answer:
Ans. (b)
CH₃COO⁻ is a weak base and hence brings about substitution while C₂H₅O⁻ is a strong base and hence brings about elimination.

Teacher's Note:
1. Weak base, good nucleophile (CH₃COO⁻) gives substitution; strong base (C₂H₅O⁻) with a 2° halide gives mainly elimination.

Question 24. Comprehension 2: An alkyl halide with β-hydrogen atoms when reacted with a base or a nucleophile has two competing routes : Substitution (S_N1 and S_N2) and elimination. Which route will be taken up depends upon the nature of the alkyl halide, strength and size of the base/nucleophile and reaction conditions. Thus, bulkier nucleophile prefers to act as a base and abstracts a proton rather than approaching a tetravalent carbon atom (steric reasons) and vice-versa.
2-Bromopentane is heated with potassium ethoxide in ethanol. The major product obtained is
(a) 2-Ethoxypentane
(b) Pentene-1
(c) cis-Pentene-2
(d) trans-Pentene-2
Answer:
Ans. (d)
Heating with alc. C₂H₅OK brings about dehydrobromination to give trans-pentene-2 as the major product (Saytzeff alkene, and the trans isomer is more stable than the cis).

Teacher's Note:
1. Between cis and trans Saytzeff alkenes, the trans isomer (less steric strain) is the major product.

Matrix Match Questions

Teacher's Note:
1. An ambident nucleophile has two different donor atoms: NO₂⁻ (O or N) and CN⁻ (C or N).

Teacher's Note:
1. Overall substitution can still go through an elimination–addition route, as with chlorobenzene and NaNH₂.

Integer and Numerical Answer Questions

Question 27. The total number of alkenes possible by dehydrobromination of 3-bromo-3-cyclopentylhexane using alcoholic KOH is (IIT Paper I, 2011)
Answer:
Ans. 5
3-Bromo-3-cyclopentylhexane, CH₃CH₂–CH₂–C(Br)(C₅H₉)–CH₂CH₃, has three different types of α-hydrogens (on the propyl CH₂, on the ethyl CH₂ and on the ring CH) and hence can give three structurally isomeric alkenes.
CH₃CH₂CH=C(C₅H₉)–CH₂CH₃ (E + Z isomers); CH₃CH₂CH₂–C(C₅H₉)=CHCH₃ (E + Z isomers); CH₃CH₂CH₂–C(=C₅H₈)–CH₂CH₃ (cyclopentylidene, only one isomer)
But two of these can exhibit geometrical isomerism. Therefore, the total number of alkenes possible are five.

Teacher's Note:
1. Count each structural alkene, then check each for cis/trans (E/Z) isomerism.

Question 28. Total number of monobromo derivatives formed by the alkanes with molecular formula C₅H₁₂ (excluding stereoisomers) is. (JEE Main 2022)
Answer:
Ans. 8
CH₃CH₂CH₂CH₂CH₂Br, CH₃CH₂CH₂–CHBrCH₃, CH₃CH₂–CHBr–CH₂CH₃,
BrCH₂–CH(CH₃)–CH₂CH₃, (CH₃)₂CBr–CH₂CH₃, (CH₃)₂CH–CHBr–CH₃,
(CH₃)₂CH CH₂CH₂Br, (CH₃)₃CCH₂Br
Thus, in all, there are eight isomers.

Teacher's Note:
1. Go isomer by isomer: n-pentane gives 3, isopentane 4 and neopentane 1.

Question 29. In presence of sunlight, benzene reacts with Cl₂ to give product X. The number of hydrogens in X is. (JEE Main 2022)
Answer:
Ans. 6
Benzene hexachloride (C₆H₆Cl₆) has six hydrogen atoms.

Teacher's Note:
1. In sunlight, Cl₂ adds to benzene (three molecules) instead of substituting, so no H is lost.

Question 30. In the following reaction sequence, the amount of D (in g) from one mole of benzene is ............ (The yield (%) corresponding to the product in each step is given in the parenthesis.)
C₆H₆ →(CH₂=CH₂, H₃PO₄)→ A (80%) →(Br₂, Anhyd. AlBr₃)→ B (65%) →(NBS, hν, Δ)→ C (70%) →(Aq. KOH, Δ)→ D (80%)
Answer:
Ans. 58.53
The sequence of reactions is :
Benzene (1 mole) →(CH₂=CH₂/H₃PO₄, F.C. alkylation)→ ethylbenzene (A) (80%) →(Br₂/Anhyd. AlCl₃; Br is o, p-directing)→ p-bromoethylbenzene (B) (65%)
→(NBS, hν, Δ; benzylic bromination)→ 1-(p-bromophenyl)bromoethane (C) (70%) →(Aq. KOH, Δ; hydrolysis)→ 1-(p-bromophenyl)ethanol (D) (80%)
Amount of D (in moles) = \( \frac{1\times80}{100}\times\frac{65}{100}\times\frac{70}{100}\times\frac{80}{100} \) = 0.2912
Mol. formula of D = C₈H₉OBr
Mol. mass of D = 8 × 12 + 9 × 1 + 1 × 16 + 1 × 80 = 201 amu
∴ Amount of D (in g) = 0.2912 × 201 = 58.53

Teacher's Note:
1. Multiply all the fractional yields first, then convert moles to grams with the molar mass of the final product.

Assertion–Reason Questions

Question 31. Assertion: S_N1 reaction is basically a solvolysis reaction.
Reason: Polar protic solvents help the substrate to ionize and by the way get involved in S_N1 reaction. (AIIMS 2010)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (a)
Reason is the correct explanation of the assertion. A polar protic solvent ionizes the substrate and then itself acts as the nucleophile on the carbocation, so S_N1 is basically a solvolysis.

Teacher's Note:
1. In solvolysis the solvent (water, alcohol) is both the ionizing medium and the nucleophile.

Question 32. Assertion: Benzyl bromide when kept in acetone-water, produces benzyl alcohol.
Reason: The reaction follows S_N2 mechanism. (AIIMS 2003)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (c)
Correct reason. Benzyl halides undergo hydrolysis by S_N1 mechanism.

Teacher's Note:
1. Benzyl halides ionize easily in polar solvents because the benzyl carbocation is resonance stabilised.

Question 33. Assertion: 2-Bromobutane on reaction with sodium ethoxide in ethanol gives 1-butene as a major product.
Reason: 1-Butene is more stable than 2-butene. (AIIMS 2004)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (d)
Correct assertion. 2-Bromobutane on reaction with sodium ethoxide in ethanol gives 2-butene as the major.
Correct reason. 2-Butene is more stable than 1-butene.

Teacher's Note:
1. Ethoxide is a small base, so the Saytzeff (more substituted, more stable) alkene is the major product.

Question 34. Assertion: tert-Butyl chloride on treatment with sodium metal in dry ether gives 2,2,3,3-tetramethylbutane.
Reason: tert-Alkyl halides readily undergo Wurtz reaction. (AIIMS 2004)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (d)
Correct assertion. tert-Butyl chloride with Na in dry ether gives mainly isobutylene and isobutane, not 2,2,3,3-tetramethylbutane.
Correct reason. tert-Alkyl halides do not undergo Wurtz reaction readily; the strongly basic intermediate brings about elimination instead.

Teacher's Note:
1. Wurtz reaction works well only with 1° halides; bulky 3° halides mainly give alkenes and alkanes by elimination/disproportionation.

Question 35. Assertion: In comparison to ethyl chloride, it is difficult to carry out nucleophilic substitution on vinyl chloride.
Reason: Vinyl group is electron-donating. (AIIMS 2013, 2015)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (c)
Correct reason. In vinyl chloride the lone pair of Cl is in resonance with the C=C bond, giving the C–Cl bond partial double bond character; the sp² carbon also holds the bond more strongly. Hence substitution is difficult.

Teacher's Note:
1. Vinyl and aryl halides resist substitution for the same two reasons: resonance (partial C=Cl double bond) and sp² hybridisation of carbon.

Question 36. Assertion: The presence of nitro group facilitates nucleophilic substitution reaction in aryl halides.
Reason: The intermediate carbanion is stabilized due to the presence of the nitro group. (AIIMS 2006)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (a)
Reason is the correct explanation of the assertion. A nitro group at the o- or p-position withdraws electrons and delocalises the negative charge of the carbanion (Meisenheimer) intermediate, which speeds up the substitution.

Teacher's Note:
1. Only o/p-NO₂ can take up the negative charge by resonance; a meta NO₂ helps much less.

Question 37. Assertion: Chloral reacts with phenyl chloride to form DDT.
Reason: It is an electrophilic substitution reaction. (AIIMS 2007)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (a)
CCl₃CHO + 2 C₆H₅Cl →(conc. H₂SO₄)→ (p-ClC₆H₄)₂CH–CCl₃ (DDT) + H₂O. The protonated chloral acts as the electrophile and attacks the para position of chlorobenzene twice, so the reaction is an electrophilic substitution; reason explains the assertion.

Teacher's Note:
1. Cl on chlorobenzene is deactivating but o,p-directing, which is why the two rings in DDT are joined at their para positions.