The products A and B in the following reactions, respectively are

Q: The products A and B in the following reactions, respectively are

Both silver reagents exhibit significantly covalent bond characters: - Reaction with AgNO2: The bond between silver and oxygen is covalent, making the lone pair on the nitrogen atom the primary nucleophilic site. Attack via nitrogen yields a nitroalkane product: - Reaction with AgCN: The covalent Ag-C bond directs the nucleophilic attack to proceed through the lone pair on nitrogen, yielding an isocyanide compound: Hence, option (4) represents the correct combination. Sees: Alkyl halide reacting

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

Solution & Explanation

Core Logic

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_2025_28_jan_morning

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