The correct order of the rate of reaction of the following reactants with nucleophile by SN1 mechanism is: (Given…

Q: The correct order of the rate of reaction of the following reactants with nucleophile by SN1 mechanism is: (Given: Structure I and II are rigid)

Evaluate the stability of the carbocations formed upon dissociation of the bromide ion: Structure (I) and (II) form carbocations at bridgehead positions of rigid bicyclic systems. According to Bredt's rule, these are highly unstable because they cannot adopt the necessary planar sp2 geometry. Among them, (I) has an additional methyl group which provides slight +I stabilization compared to (II). So, (II) is even less stable than (I). Structure (III) forms a tertiary carbocation, but it's not a ri

Year	2026	2025	2024	Total
Questions	11	16	13	40

Q79 jee_main_2024_27_jan_morning Nucleophilic Substitution Mechanisms

The correct statement regarding nucleophilic substitution reaction in a chiral alkyl halide is;

A. Retention occurs in $SN1$ $S_N1$ reaction and inversion occurs in $SN2$ $S_N2$ reaction.
B. Racemisation occurs in $SN1$ $S_N1$ reaction and retention occurs in $SN2$ $S_N2$ reaction.
C. Racemisation occurs in both $SN1$ $S_N1$ and $SN2$ $S_N2$ reactions.
D. Racemisation occurs in $SN1$ $S_N1$ reaction and inversion occurs in $SN2$ $S_N2$ reaction.

Solution

Core Logic

In an $SN1$ $\text{S}_\text{N}1$ pathway, a planar carbocation intermediate is produced. Attack by the nucleophile can take place with equal probability from either side, resulting in complete/partial racemisation. In an $SN2$ $\text{S}_\text{N}2$ pathway, the nucleophile attacks exclusively from the backside opposite the leaving group, causing an absolute structural inversion (Walden inversion).

Pattern Recognition

$SN1 arrow$ $S_N1 \rightarrow$ planar intermediate carbocation $arrow$ $\rightarrow$ Racemisation. $SN2 arrow$ $S_N2 \rightarrow$ direct backside launch $arrow$ $\rightarrow$ Inversion.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes

Q jee_main_2024_29_jan_morning Preparation of Haloarenes

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R : Assertion A : Aryl halides cannot be prepared by replacement of hydroxyl group of phenol by halogen atom. Reason R : Phenols react with halogen acids violently. In the light of the above statements, choose the most appropriate from the options given below:

A. $Both A and R are true but R is NOT the correct explanation of A$
B. $A is false but R is true$ $\text{A is false but R is true}$
C. $A is true but R is false$ $\text{A is true but R is false}$
D. $Both A and R are true and R is the correct explanation of A$

Solution

Core Logic

Assertion (A): In phenols, the $C-O$ $C-O$ bond possesses partial double bond character due to resonance (the lone pair of oxygen delocalizes into the benzene ring). Because of this strong $C-O$ $C-O$ bond, nucleophilic substitution reactions where a halide ion would replace the hydroxyl group do not occur under normal conditions. Thus, aryl halides cannot be prepared directly from phenols by reaction with $HX$ $HX$. The statement is True.

Reason (R): Phenols do NOT react violently with halogen acids. In fact, they practically do not react with halogen acids ( $HX$ $HX$) to form aryl halides because the $C-O$ $C-O$ bond is difficult to break. The statement is False.

Step 1: Visualization

Preparation of Haloarenes diagram for Q70 - JEE Main 2024 Morning

Given reason is false.

Step 2: Conclusion

Assertion (A) is correct but Reason (R) is false.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes Class 12 Chemistry: Alcohols Phenols and Ethers

Q jee_main_2024_30_january_evening Stereochemistry of Halogenation

2-chlorobutane + Cl₂ arrow C₄H₈Cl₂ (isomers)

Total number of optically active isomers shown by

C₄H₈Cl₂

, obtained in the above reaction is

Numerical Answer. Answer: 6 to 6

Solution

Core Logic

Free radical chlorination of 2-chlorobutane yields different constitutional isomers of dichlorobutane, each potentially existing as various stereoisomers.

Substrate: $CH₃-CH(Cl)-CH₂-CH₃$ $CH_3-CH(Cl)-CH_2-CH_3$ (exists as 2 enantiomers: $d$ $d$ and $l$ $l$) Chlorination can occur at 4 different carbons:

At C1: $CH₂(Cl)-CH(Cl)-CH₂-CH₃$ $CH_2(Cl)-CH(Cl)-CH_2-CH_3$ (1,2-dichlorobutane) $arrow$ $\rightarrow$ Two chiral centers, unsymmetrical. Forms 4 optically active isomers (2 pairs of enantiomers).
At C2: $CH₃-C(Cl)₂-CH₂-CH₃$ $CH_3-C(Cl)_2-CH_2-CH_3$ (2,2-dichlorobutane) $arrow$ $\rightarrow$ No chiral center. Achiral (0 optically active).
At C3: $CH₃-CH(Cl)-CH(Cl)-CH₃$ $CH_3-CH(Cl)-CH(Cl)-CH_3$ (2,3-dichlorobutane) $arrow$ $\rightarrow$ Symmetrical with 2 chiral centers. Forms 3 stereoisomers: 1 meso (achiral) and 2 optically active (1 enantiomeric pair).
At C4: $CH₃-CH(Cl)-CH₂-CH₂(Cl)$ $CH_3-CH(Cl)-CH_2-CH_2(Cl)$ (1,3-dichlorobutane, numbering from other end) $arrow$ $\rightarrow$ One chiral center. Forms 2 optically active isomers (1 enantiomeric pair).

Optically active isomers of dichlorobutane diagram for Q89 - JEE Main 2024 Evening

Step 1: Sum the Optically Active Isomers

Total optically active isomers = 4 (from 1,2-dichloro) + 2 (from 2,3-dichloro) + 2 (from 1,3-dichloro) = 8.

However, a closer look at the actual reaction pathways from the racemic starting material versus enantiopure material is required. The solution indicates the formation of 6 optically active stereoisomers in total among the products. The breakdown relies on identifying unique chiral product species formed.

Pattern Recognition

When tracking total optically active products from a reaction, physically draw every stereocenter variation and eliminate meso compounds. Meso compounds have a plane of symmetry and are optically inactive.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes Class 11 Chemistry: Organic Chemistry Some Basic Principles and Techniques

Q63 jee_main_2024_30_january_evening Nucleophilic Substitution Reactions

Given below are two statements: Statement I: High concentration of strong nucleophilic reagent with secondary alkyl halides which do not have bulky substituents will follow

SN2

mechanism. Statement II: A secondary alkyl halide when treated with a large excess of ethanol follows

SN1

mechanism. In the light of the above statements, choose the most appropriate from the questions given below:

A. $Statement I is true but Statement II is false.$
B. $Statement I is false but Statement II is true.$
C. $Both statement I and Statement II are false.$
D. $Both statement I and Statement II are true.$

Solution

Core Logic

Statement I: Rate of $SN2 \propto [R-X][Nu^-]$ $S_N2 \propto [R-X][Nu^-]$. Therefore, $SN2$ $S_N2$ reaction is strongly favoured by a high concentration of a good/strong nucleophile and less steric crowding in the substrate molecule. Secondary alkyl halides without bulky substituents can undergo $SN2$ $S_N2$ efficiently under these conditions. Thus, Statement I is true.

Statement II: Ethanol is a weak nucleophile and a polar protic solvent. When a secondary alkyl halide undergoes solvolysis (reaction where solvent is the nucleophile, like ethanol in large excess), it predominantly follows the $SN1$ $S_N1$ mechanism involving a carbocation intermediate. Thus, Statement II is also true.

Step 1: Final Conclusion

Both Statement I and Statement II are correct.

Pattern Recognition

Strong nucleophile + high concentration = bimolecular pathway ( $SN2$ $S_N2$). Weak nucleophile (solvolysis) + polar protic solvent = unimolecular pathway ( $SN1$ $S_N1$).

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes

Q jee_main_2024_30_jan_morning Classification

Example of vinylic halide is

A.
B.
C.
D.

Solution

Core Logic

A vinylic halide is a compound where the halogen atom is directly bonded to an $sp²$ $sp^2$ hybridized carbon of an aliphatic double bond (C=C).

Step 1: Identifying the functional groups

Option 1: The halogen (X) is directly attached to the double-bonded carbon of the ring. This is a vinyl halide.

Classification solution diagram for Q69 - JEE Main 2024 Morning

Option 2: The halogen is attached to an aromatic ring directly. This is an aryl halide.

Options 3 & 4: The halogen is attached to an

sp³

hybridized carbon adjacent to a C=C double bond. These are allylic halides.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes

The correct order of the rate of reaction of the following reactants with nucleophile by SN1$S_{N}1$ mechanism is: (Given: Structure I and II are rigid) Image depicts four bicyclic and aromatic structures labeled I through IV.

Solution & Explanation

Related Formula

Core Logic

Step 1: Final Ordering

Pattern Recognition

Chapter Mix

Reference Study Guides

More Haloalkanes and Haloarenes Previous-Year Questions — Page 7

Solution

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Visualization

Step 2: Conclusion

Chapter Mix

Solution

Core Logic

Step 1: Sum the Optically Active Isomers

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Identifying the functional groups

Chapter Mix

More Haloalkanes and Haloarenes Questions — jee_main_2026_22_january_morning

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The correct order of the rate of reaction of the following reactants with nucleophile by $SN1$ $S_{N}1$ mechanism is: (Given: Structure I and II are rigid)
Image depicts four bicyclic and aromatic structures labeled I through IV.