CBSE Class 12 Chemistry Haloalkanes and Haloarenes NEET JEE Questions Set 02

Single Correct Answer

Question 1. The correct sequence of bond enthalpy of C–X bond is (NEET 2021)
(a) CH₃–Cl > CH₃–F > CH₃–Br > CH₃–I
(b) CH₃–F < CH₃–Cl < CH₃–Br < CH₃–I
(c) CH₃–F > CH₃–Cl > CH₃–Br > CH₃–I
(d) CH₃–F < CH₃–Cl > CH₃–Br > CH₃–I
Answer:
Ans. (c)
As the size of the halogens increases from F to I, bond dissociation enthalpy of C–X bond decreases, i.e., option (c) is correct.
CH₃–F > CH₃–Cl > CH₃–Br > CH₃–I

Teacher's Note:
1. Bond enthalpy falls as the C–X bond gets longer; that is why R–I is the most reactive alkyl halide.

 

Question 2. CH₃–Br + Nu⁻ → CH₃–Nu + Br⁻
The decreasing order of the rate of the above reaction with nucleophiles (Nu⁻) A to D is
[Nu⁻ = (A) PhO⁻, (B) AcO⁻, (C) HO⁻, (D) CH₃O⁻]
(a) D > C > A > B
(b) D > C > B > A
(c) A > B > C > D
(d) B > D > C > A
Answer:
Ans. (a)
Phenoxide ion is less resonance stabilized than AcO⁻ ion, therefore, nucleophilicity decreases in the order :
CH₃O⁻ (D) > HO⁻ (C) > PhO⁻ (A) > AcO⁻ (B)
Thus, option (a) is correct.

Teacher's Note:
1. The more the negative charge is delocalised by resonance, the weaker the nucleophile; CH₃O⁻ is stronger than HO⁻ because of the +I effect of CH₃.

 

Question 3. For the following :
(a) I⁻ (b) Cl⁻ (c) Br⁻
the increasing order of nucleophilicity would be
(a) Br⁻ < Cl⁻ < I⁻
(b) I⁻ < Br⁻ < Cl⁻
(c) Cl⁻ < Br⁻ < I⁻
(d) I⁻ < Cl⁻ < Br⁻
Answer:
Ans. (c)
Cl⁻ < Br⁻ < I⁻. In protic solvents, the larger, more polarisable iodide ion is the least solvated and hence the best nucleophile.

Teacher's Note:
1. In protic solvents nucleophilicity of halide ions runs opposite to basicity: I⁻ > Br⁻ > Cl⁻ > F⁻.

 

Question 4. The mechanism of S_N1 reaction is given as
R–X → R⁺X⁻ (ion pair) → R⁺ || X⁻ (solvent separated ion pair) →(Y⁻)→ R–Y + X⁻
A student writes general characteristics based on the given mechanism as :
(i) The reaction is favoured by weak nucleophiles.
(ii) R⁺ would be easily formed if the substituents are bulky.
(iii) The reaction is accompanied by racemization.
(iv) The reaction is favoured by non-polar solvents.
Which observations are correct? (JEE Main 2020)
(a) (ii) and (iv)
(b) (i) and (iii)
(c) (i), (ii) and (iii)
(d) (i) and (ii)
Answer:
Ans. (c)
The reaction is favoured by polar solvents. Therefore, option (iv) is not correct while the remaining three options are true. Thus, option (c) is correct.

Teacher's Note:
1. Polar protic solvents stabilise the carbocation and the leaving ion, which is exactly what an S_N1 reaction needs.

 

Question 5. Which of the following compounds is inactive towards S_N1 reaction? (JEE Main 2022)
(a) (CH₃)₃C–Cl
(b) CH₂=CH–CH₂Cl
(c) 1-Chlorobicyclo[2.2.1]heptane (Cl on the carbon shared by both rings)
(d) C₆H₅–CH(Cl)–CH₃
Answer:
Ans. (c)
A carbocation cannot be formed at the bridgehead and hence S_N1 reaction cannot occur at the bridgehead chloro compound. Thus, option (c) is correct.

Teacher's Note:
1. A bridgehead carbocation cannot become planar, so it is not formed; the other three give stabilised 3°, allylic and benzylic cations.

 

Question 6. The organic chloro compound, which shows complete stereochemical inversion during a S_N2 reaction, is (AIEEE 2008)
(a) CH₃Cl
(b) (C₂H₅)₂CHCl
(c) (CH₃)₃CCl
(d) (CH₃)₂CHCl
Answer:
Ans. (a)
S_N2 reactions are accomplished by complete stereochemical inversion. Since 1° alkyl halides are most prone to S_N2 reactions, therefore, CH₃Cl undergoes complete stereochemical inversion.

Teacher's Note:
1. The least hindered halide (CH₃X > 1° > 2°) reacts most cleanly by the S_N2 path with back-side attack.

 

Question 7. The major product of the following reaction is

→ PhS⁻Na⁺, DMF → (JEE Main 2022)
(a)
(b)
(c)
(d)
Answer:
Ans. (a)
Alkyl halides undergo nucleophilic substitution more easily than aryl halides. Therefore, substitution occurs more easily in the side chain rather than in the ring.
Further, since S_N2 reactions occur with inversion of configuration (inversion at C–I, PhS⁻ enters from the side opposite to I), therefore, option (a) is correct.

Teacher's Note:
1. A polar aprotic solvent (DMF) and a strong nucleophile on a 2° benzylic halide point to S_N2: replace I with inversion and leave the aryl C–F untouched.

 

Question 8. In the following reaction sequence, the correct structure(s) of X is (are)
X →(i) PBr₃, Et₂O; (ii) NaI, Me₂CO; (iii) NaN₃, HCONMe₂→

(enantiomerically pure) (JEE Advanced 2018)
(a)
(b)
(c)
(d)
Answer:
Ans. (b)
In S_N2 reactions, inversion of configuration occurs at each chiral centre as shown :
X (Me wedge, OH wedge) →(PBr₃/Et₂O; S_N2, inversion of configuration at OH)→ Me wedge, Br hashed
→(NaI, Me₂CO; Finkelstein reaction, S_N2, inversion of configuration at Br)→ Me wedge, I wedge
→(NaN₃, HCONMe₂; S_N2, inversion of configuration at I)→ Me wedge, N₃ hashed (enantiomerically pure)
Three inversions amount to one overall inversion, so X has OH on the same face as Me (both wedges) at the same positions: option (b). Structure (d) is the mirror image of (b).

Teacher's Note:
1. Count the S_N2 steps: an odd number of inversions means the final group is on the face opposite to the original OH.

 

Question 9. In a S_N2 substitution reaction of the type
R–Br + Cl⁻ →(DMF)→ R–Cl + Br⁻
which one of the following has the highest relative rate? (AIPMT 2008)
(a) (CH₃)₃C–CH₂Br
(b) CH₃CH₂Br
(c) CH₃–CH₂–CH₂Br
(d) (CH₃)₂CH–CH₂Br
Answer:
Ans. (b)
Rates of S_N2 reactions are influenced by steric factors. Since CH₃CH₂Br has the least steric hindrance, therefore, it reacts at the fastest rate.

Teacher's Note:
1. Branching at the β-carbon also slows S_N2 sharply; neopentyl bromide is almost unreactive even though it is 1°.

 

Question 10. Consider the reaction,
CH₃CH₂CH₂Br + NaCN → CH₃CH₂CH₂CN + NaBr
This reaction will be the fastest in (NEET Phase-II 2016)
(a) ethanol
(b) methanol
(c) N,N′-dimethylformamide (DMF)
(d) water
Answer:
Ans. (c)
CH₃CH₂CH₂Br being a 1° alkyl halide reacts with NaCN by S_N2 mechanism which is favoured by polar aprotic solvent, i.e., DMF. Thus, option (c) is correct.

Teacher's Note:
1. Protic solvents (water, alcohols) hydrogen-bond to CN⁻ and slow it down; a polar aprotic solvent leaves the anion ‘naked’ and reactive.

 

Question 11. In S_N1 reaction on chiral centres, there is (RE-AIPMT 2015)
(a) inversion more than retention leading to partial racemization
(b) 100% retention
(c) 100% conversion
(d) 100% racemization
Answer:
Ans. (a)
The leaving group still shields one face of the carbocation when the nucleophile attacks. Since inversion product exceeds the retention product, therefore, only partial racemization is observed, i.e., option (a) is correct.

Teacher's Note:
1. Complete racemization is the ideal S_N1 picture; in practice the ion pair gives a small excess of the inverted product.

 

Question 12. The incorrect statement about chirality is : (NEET 2022)
(a) The product obtained by S_N2 reaction of haloalkane having chirality at the reactive site shows inversion of configuration.
(b) Enantiomers are superimposable mirror images of each other.
(c) A racemic mixture shows zero optical rotation.
(d) S_N1 reaction yields 1 : 1 mixture of both enantiomers.
Answer:
Ans. (b)
Enantiomers are non-superimposable mirror images of each other. Hence statement (b) is incorrect.

Teacher's Note:
1. In ‘incorrect statement’ questions, test each option separately; the definition of enantiomers is the usual trap.

 

Question 13. 2-Chloro-2-methylpentane on reaction with sodium methoxide in methanol yields
(I) C₂H₅CH₂C(CH₃)₂–OCH₃
(II) C₂H₅CH₂C(CH₃)=CH₂
(III) C₂H₅CH=C(CH₃)–CH₃ (JEE Main 2016)
(a) all of these
(b) I and III
(c) III only
(d) I and II
Answer:
Ans. (a)
2-Chloro-2-methylpentane being a 3° alkyl halide can undergo either S_N1 or E₁ reaction. Further, the intermediate carbocation (I) can undergo elimination in two different ways, to give 2-methyl-2-pentene (major) and 2-methyl-1-pentene (minor).
CH₃CH₂CH₂–C(CH₃)₂Cl →(Ionization, −Cl⁻)→ CH₃CH₂CH₂–C⁺(CH₃)₂, 3° carbocation (I)
(I) →(CH₃ONa, S_N1)→ CH₃CH₂CH₂–C(CH₃)₂–OCH₃, 2-methoxy-2-methylpentane (substitution product)
(I) →(E₁, −H⁺)→ CH₃CH₂CH=C(CH₃)–CH₃, 2-methyl-2-pentene (major product) + CH₃CH₂CH₂–C(CH₃)=CH₂, 2-methyl-1-pentene (minor product)
Thus, all the three products are formed.

Teacher's Note:
1. A 3° halide in a protic solvent gives a mixture: the ether by S_N1 and both possible alkenes by E1.

 

Question 14. An alkene A on reaction with O₃ and Zn–H₂O gives propanone and ethanal in equimolar ratio. Addition of HCl to alkene A gives B as the major product. The structure of product B is (NEET 2019)
(a) H₃C–CH(Cl)–CH(CH₃)₂
(b) ClCH₂–CH₂–CH(CH₃)₂
(c) H₃C–CH₂–CH(CH₂Cl)–CH₃
(d) H₃C–CH₂–C(CH₃)₂–Cl
Answer:
Ans. (d)
Products of ozonolysis: (CH₃)₂C=O (propanone) and O=CHCH₃ (ethanal) (1 : 1).
Remove O atoms and join the carbon atoms by a double bond, the structure of the alkene is
CH₃CH=C(CH₃)₂ →(HCl, Markovnikov's addition)→ H₃C–CH₂–C(CH₃)₂–Cl
Thus, option (d) is correct.

Teacher's Note:
1. To rebuild an alkene from ozonolysis products, join the two carbonyl carbons by a C=C and drop the oxygens.

 

Question 15. The major product of the following reaction is

→(NaOEt, Δ)→ ? (JEE Main 2019)
(a)
(b) CH₃–C(CO₂CH₂CH₃)=CHCH₃
(c)
(d) CH₃CH₂–C(CO₂CH₂CH₃)=CH₂
Answer:
Ans. (b)
Elimination occurs in such a way that a more stable alkene is formed in accordance with Saytzeff rule.
C₂H₅O⁻ removes H from the CH₂ of the ethyl group while Cl⁻ leaves (−EtOH, −Cl⁻):
CH₃–CH₂–C(CH₃)(Cl)–COOCH₂CH₃ → CH₃–CH=C(CH₃)–COOCH₂CH₃, i.e., CH₃–C(COOCH₂CH₃)=CHCH₃
Thus, option (b) is correct.

Teacher's Note:
1. A 3° halide with a strong base (EtO⁻) on heating undergoes E2, not substitution; the more substituted (Saytzeff) alkene is the major product.

 

Question 16. The major product obtained in the following reaction is
(+)-C₆H₅–CH(Br)–CH₂–C₆H₅ →(t-BuOK, Δ)→ ? (JEE Main 2017)
(a) (+)-C₆H₅CH(Ot-Bu)CH₂C₆H₅
(b) (−)-C₆H₅CH(Ot-Bu)CH₂C₆H₅
(c) (±)-C₆H₅CH(Ot-Bu)CH₂C₆H₅
(d) C₆H₅CH=CHC₆H₅
Answer:
Ans. (d)
2° Alkyl halides exclusively undergo bimolecular elimination with a bulky strong base to form alkenes.
(+)-C₆H₅CH(Br)CH₂C₆H₅ →(tert-BuOK, Δ, −HBr)→ C₆H₅CH=CHC₆H₅ + tert-BuOH + Br⁻
Thus, option (d) is correct.

Teacher's Note:
1. t-BuO⁻ is a strong but very bulky base; it removes a β-H instead of attacking carbon, so substitution products are not expected.

 

Question 17. Correct order for reaction of the following alkyl halides with alcoholic KOH is
I : (CH₃)₂CH–CH(Br)–CH₃; II : CH₃–CH(Cl)–CH₂–CH₃; III : CH₃–CH(Br)–CH₂–CH₃; IV : CH₃CH₂Br (JEE Advanced 2013)
(a) I > II > III > IV
(b) I > III > II > IV
(c) IV > II > III > I
(d) I > IV > II > III
Answer:
Ans. (b)
The dehydrohalogenation of the given four alkyl halides (I-IV) give the alkenes (V-VIII): I → (CH₃)₂C=CHCH₃ (V); II → CH₃CH=CHCH₃ (VI); III → CH₃CH=CHCH₃ (VII); IV → CH₂=CH₂ (VIII).
Since alkene (V) is the most stable, therefore, alkyl halide (I) is most reactive.
Since alkyl halides (II and III) give the same alkene, but alkyl bromides are more reactive than alkyl chlorides, therefore, alkyl halide (III) is more reactive than alkyl halide (II). Of course, (alkene VIII) is the least stable and hence alkyl halide (IV) is the least reactive.
Thus, the overall reactivity decreases in the order : I > III > II > IV, i.e., option (b) is correct.

Teacher's Note:
1. In E2 the transition state has partial C=C character, so the halide that gives the more substituted alkene and has the better leaving group reacts faster.

 

Question 18. 1-Methylcyclohexan-1-ol →(20% H₃PO₄, 358 K)→ 'A' (major product)
1-Chloro-1-methylcyclohexane →((CH₃)₃CO⁻K⁺)→ 'B' (major product)
The products 'A' and 'B' formed in the above reaction are : (JEE Main 2021)
(a) A = 1-methylcyclohexene ; B = methylenecyclohexane
(b) A = methylenecyclohexane ; B = 1-methylcyclohexene
(c) A = methylenecyclohexane ; B = methylenecyclohexane
(d) A = 1-methylcyclohexene ; B = 1-methylcyclohexene
Answer:
Ans. (a)
Dehydration with H₃PO₄ at 358 K gives 1-methylcyclohexene (A, major product) + methylenecyclohexane (minor product). Due to Saytzeff rule, a more stable alkene 'A' is the major product.
With (CH₃)₃CO⁻K⁺ (−HCl), methylenecyclohexane 'B' is the major product and 1-methylcyclohexene is the minor product.
Due to steric hindrance by the bulky base, H from the less hindered CH₃ group is lost giving 'B' as the major product.
Thus, option (a) is correct.

Teacher's Note:
1. Acid-catalysed dehydration gives the Saytzeff alkene; a bulky base such as t-BuOK gives the Hofmann (less substituted) alkene.

 

Question 19. The decreasing order of reactivity towards dehydrohalogenation (E₁) reaction of the following compounds is :
(A) ClCH₂CH₂CH₂CH₃   (B) ClCH₂CH₂CH=CH₂   (C) CH₃CH(Cl)CH₂CH₃   (D) CH₃CH(Cl)CH=CH₂ (JEE Main 2020)
(a) B > D > A > C
(b) B > D > C > A
(c) D > B > C > A
(d) B > A > D > C
Answer:
Ans. (c)
The reactivity of alkyl halides towards E₁ dehydrohalogenation depends upon the stability of the intervening carbocations. Here, the carbocations involved in this reaction are :
(i) C⁺H₂CH₂CH₂CH₃ (from A); (ii) C⁺H₂CH₂CH=CH₂ (from B); (iii) CH₃C⁺HCH₂CH₃ (from C); (iv) CH₃C⁺HCH=CH₂ ↔ CH₃CH=CHC⁺H₂ (from D)
Since their stability decreases in the order : (iv) > (ii) > (iii) > (i), therefore, the reactivity of the corresponding alkyl halides decreases in the order : D > B > C > A, i.e., option (c) is correct.

Teacher's Note:
1. In E1 the slow step is ionization, so rank the halides by the stability of the carbocation they form; an allylic cation (from D) is the most stable.

 

Question 20. The major product of the following reaction is

→(NaOMe, MeOH)→ ? (JEE Main 2018)
(a)
(b)
(c)
(d)
Answer:
Ans. (b)
NaOMe/MeOH brings about E₂ (trans-elimination) of HBr. The carbon next to C–Br on one side carries two CH₃ groups and no H, so the anti H is removed from the CH₂ on the other side, giving 3,3-dimethylcyclohexene.
Thus, option (b) is correct.

Teacher's Note:
1. MeO⁻ is a strong base; with a 2° halide next to a bulky quaternary carbon, E2 wins over S_N2, and the H removed must be anti to Br.

 

One or More Correct Answers

Question 21. The correct statement(s) for the following addition reactions is (are)
(i) trans-But-2-ene →(Br₂/CHCl₃)→ M and N
(ii) cis-But-2-ene →(Br₂/CHCl₃)→ O and P (JEE Advanced 2016)
(a) O and P are identical molecules
(b) bromination proceeds through trans-addition in both the reactions
(c) (M and O) and (N and P) are two pairs of enantiomers
(d) (M and O) and (N and P) are two pairs of diastereomers
Answer:
Ans. (b, d)
Bromination proceeds through trans-addition. Therefore, trans-addition to trans-alkenes gives meso-compound (M and N) while trans-addition to cis-alkenes gives racemic mixture (O and P).
(i) trans-2-Butene → meso-2,3-dibromobutane (M = N, identical meso compound)
(ii) cis-2-Butene → (2R,3R)- and (2S,3S)-2,3-dibromobutane (O and P, enantiomers)
Here, (M and O) and (N and P) are two pairs of diastereomers. Thus, options (b and d) are correct.

Teacher's Note:
1. Remember: anti addition of Br₂ — trans alkene gives meso, cis alkene gives racemic (±) product.

 

Passage-Based Questions

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.
An S_N2 reaction at an asymmetric carbon of a compound always gives
(a) an enantiomer of the substrate
(b) a product with opposite optical rotation
(c) a mixture of diastereomers
(d) a single stereoisomer.
Answer:
Ans. (d)
In S_N2 reactions, inversion of configuration occurs. Since the reactant and the product are not enantiomers, therefore, the sign of optical rotation may or may not change and hence a single stereoisomer is obtained, i.e., option (d) is correct.

Teacher's Note:
1. Inversion of configuration does not always mean the sign of rotation changes; the product is a different compound, not the enantiomer of the substrate.

 

Question 23. 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.
In the solvolysis of 3-methyl-3-bromohexane, which of the following statements is not correct?
(a) it involves carbocation intermediate
(b) the intermediate involves sp² carbon
(c) polar solvents accelerate the reaction
(d) the rate of the reaction depends upon 3-methyl-3-bromohexane concentration
(e) it involves inversion of configuration
Answer:
Ans. (e)
Solvolysis means reaction with the solvent. Here, solvent acts as the nucleophile. 3-Methyl-3-bromohexane is a 3° alkyl halide, therefore, it undergoes solvolysis by S_N1 mechanism to afford a racemic product in which both retention and inversion have occurred and not only inversion.
CH₃CH₂–C(CH₃)(Br)–CH₂CH₂CH₃ →(−Br⁻)→ CH₃CH₂–C⁺(CH₃)–CH₂CH₂CH₃ (C in carbocation is sp²-hybridized) →(H₂O, −H⁺)→ (±)-3-methylhexan-3-ol

Teacher's Note:
1. Solvolysis of a 3° halide is S_N1: first order in the halide, faster in polar solvents, and racemisation rather than pure inversion.

 

Matrix Match Questions

Teacher's Note:
1. Peroxide or light/NBS signals a free radical path; a Lewis acid (FeBr₃) signals electrophilic substitution on the ring.

 

Integer and Numerical Answer Questions

Question 25. The total number of isomers including stereoisomers that could be obtained by replacing two hydrogen atoms of propane by two chlorine atoms are
Answer:
Ans. 5
Five, i.e., CH₃CH₂CHCl₂, (+) and (−) CH₃C*HClCH₂Cl, CH₃CCl₂CH₃, ClCH₂CH₂CH₂Cl

Teacher's Note:
1. 1,2-Dichloropropane has a chiral C-2, so it counts as two stereoisomers.

 

Question 26. How many chiral stereoisomers are possible for 2-bromo-3-chlorobutane?
Answer:
Ans. 4
CH₃–C*H(Br)–C*H(Cl)–CH₃
Since the compound has two different chiral carbons, therefore, 2² = 4 optical isomers are possible.

Teacher's Note:
1. 2ⁿ works only when the chiral centres are different; with identical ends (e.g., 2,3-dibromobutane) a meso form reduces the count.

 

Question 27. The total number of stereoisomers that can exist for M is
M : camphor (1,7,7-trimethylbicyclo[2.2.1]heptan-2-one) (JEE Advanced 2015)
Answer:
Ans. 2
M has two dissimilar chiral carbon atoms (C-1 and C-4, the bridgeheads), therefore, total number of stereoisomers theoretically possible = 2² = 4.
But due to the presence of a small bridge, the structure becomes rigid. As a result, rotation is not possible and the two chiral carbon atoms move together and thus behave as one chiral carbon. Consequently, the molecule M exists in two stereoisomers.

Teacher's Note:
1. In small bridged rings the configuration of one bridgehead fixes the other, so the pair counts as a single stereogenic unit.

 

Question 28. In the following monobromination reaction, the number of possible chiral products is
Enantiomerically pure 2-bromopentane, CH₃CH₂CH₂–C*H(Br)–CH₃ (1.0 mole) →(Br₂ (1.0 mole), 300°C)→ (JEE Advanced 2016)
Answer:
Ans. 5
In total seven products are formed (bromination at each carbon):
I : CH₃CH₂CH₂–C*H(Br)–CH₂Br (chiral); II : CH₃CH₂CH₂–CBr₂–CH₃ (achiral)
III and IV : CH₃CH₂C*H(Br)–C*H(Br)–CH₃, two diastereomers (both chiral)
V and VI : CH₃C*H(Br)–CH₂–C*H(Br)–CH₃, two forms (V is meso, achiral; VI chiral)
VII : BrCH₂CH₂CH₂–C*H(Br)–CH₃ (chiral)
Out of these 7 products, two products, i.e., II and V are achiral but the rest five are chiral and hence optically active.

Teacher's Note:
1. The original stereocentre is untouched unless the substitution takes place on it; new stereocentres give diastereomers, and look for a meso form.

 

Assertion–Reason Questions

Question 29. Statement-1: Nucleophilic substitution on an optically active alkyl halide gives a mixture of enantiomers.
Statement-2: The reaction occurs according to S_N2 mechanism.
(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 according to S_N1 mechanism.

Teacher's Note:
1. A mixture of enantiomers (racemisation) is the mark of S_N1; S_N2 gives a single inverted product.

 

Question 30. Statement-1: Optically active 2-iodobutane on treatment with NaI in acetone undergoes racemization.
Statement-2: Reaction involves repeated Walden inversions.
(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. Each attack of I⁻ by S_N2 inverts the configuration; repeated inversions finally give equal amounts of both enantiomers.

Teacher's Note:
1. When the nucleophile and the leaving group are the same ion, S_N2 can go on back and forth, and the optical activity is slowly lost.

 

Question 31. Statement-1: The compound A shown below is optically active.

Statement-2: The compound shown below is mirror image of above compound A.

(JEE Main 2022)
(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 compound given in statement-2 is not a mirror image of compound A.
C-4 of A carries four different groups (H, NO₂, a cis-propenyl and a trans-propenyl group), so A is chiral and optically active. Rotating the compound of statement-2 in the plane shows it has the same configuration at C-4 as A, i.e., it is A itself.

Teacher's Note:
1. To test for mirror images, compare the spatial order of the same four groups; just redrawing a molecule in a new orientation does not make its enantiomer.

 

Question 32. Assertion: Reaction of alcohols with SOCl₂ is catalysed by the presence of a tertiary amine (R₃N).
Reason: Tertiary amine promotes the reaction by reacting with the by product HCl. (AIIMS 2014)
(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)
ROH + SOCl₂ → RCl + SO₂ + HCl. The tertiary amine (e.g., pyridine) neutralises HCl as it is formed, so the reaction is driven forward; reason explains the assertion.

Teacher's Note:
1. Pyridine with SOCl₂ removes HCl; SO₂ escapes as a gas, which is why this is the cleanest way to make pure alkyl chlorides.

 

Question 33. Assertion: Alkyl iodide can be prepared by treating alkyl chloride/bromide with NaI in acetone.
Reason: NaCl/NaBr are soluble in acetone while NaI is not. (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. (c)
Correct reason. NaCl/NaBr are insoluble in acetone while NaI is soluble.

Teacher's Note:
1. Finkelstein works because NaCl/NaBr precipitate from acetone, which shifts the equilibrium forward (Le Chatelier).

 

Question 34. Assertion: p-Dichlorobenzene is less soluble in organic solvents than the corresponding o-isomer.
Reason: o-Dichlorobenzene is polar while p-dichlorobenzene is non-polar. (AIIMS 2011)
(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. (b)
Correct explanation. Being symmetrical, p-dichlorobenzene molecules fit closely in the crystal lattice and hence has stronger intermolecular forces of attraction than those of o-dichlorobenzene. As a result, large amount of energy is needed to dissolve p- than o-isomer.

Teacher's Note:
1. The para isomer has the higher m.p. and lower solubility because of better crystal packing, not because of polarity.

 

Question 35. Assertion: Treatment of chloroethane with a saturated solution of AgCN gives ethyl isocyanide as the major product.
Reason: Cyanide ion (:CN⁻) is an ambident nucleophile. (AIIMS 2014)
(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. (b)
Correct explanation. AgCN being a covalent compound acts as a nucleophile through lone pair of electrons present on N to give ethyl isocyanide.
Ag–C≡N: + CH₃CH₂–Cl → CH₃CH₂NC + AgCl (ethyl isocyanide)

Teacher's Note:
1. Both statements are true, but the isocyanide forms because Ag–C is covalent, leaving only N to attack.