Given below are two statements : Statement-I: The conversion proceeds well in the less polar medium. Statement-II: The…

Q: Given below are two statements : Statement-I: The conversion proceeds well in the less polar medium. Statement-II: The conversion proceeds well in the more polar medium.

Analyzing the solvent effects on reaction kinetics: - In Statement-I, the reaction involves an anionic nucleophile (OH-), creating a highly localized charge density on the reactant side. The resulting transition state disperses this negative charge over a larger volume, lowering its charge density. Highly polar solvents strongly solvate the reactant ion, increasing the activation energy barrier. Consequently, less polar solvents accelerate this process. - In Statement-II, the reaction begins wit

Year	2026	2025	2024	Total
Questions	11	16	13	40

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Core Logic

Analyzing the solvent effects on reaction kinetics:

In Statement-I, the reaction involves an anionic nucleophile ( $OH⁻$ $OH^{-}$), creating a highly localized charge density on the reactant side. The resulting transition state disperses this negative charge over a larger volume, lowering its charge density. Highly polar solvents strongly solvate the reactant ion, increasing the activation energy barrier. Consequently, less polar solvents accelerate this process.
SN2 pathway charge density solvent dynamics part 1
In Statement-II, the reaction begins with neutral precursors ( $R₃N$ $R_3N$ and alkyl chloride). The resulting transition state develops partial charges ( $δ+$ $\delta+$ and $δ-$ $\delta-$) as the new bond forms, increasing its charge density relative to the reactants. Polar solvents stabilize this charged transition state, lowering the activation energy barrier. Thus, highly polar media accelerate this substitution pathway.
SN2 pathway charge density solvent dynamics part 1

Pattern Recognition

If the transition state concentrates charge relative to the reactants, polar solvents accelerate the reaction. If the transition state disperses charge, less polar solvents are favored.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes

More Haloalkanes and Haloarenes Previous-Year Questions — Page 8

Q74 jee_main_2024_30_jan_morning Classification

Given below are two statement one is labeled as Assertion (A) and the other is labeled as Reason (R). Assertion (A):

CH₂=CH-CH₂-Cl

is an example of allyl halide Reason (R): Allyl halides are the compounds in which the halogen atom is attached to

sp²

hybridised carbon atom. In the light of the two above statements, choose the most appropriate answer from the options given below:

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

Solution

Core Logic

Assertion (A): $CH₂=CH-CH₂-Cl$ $CH_2=CH-CH_2-Cl$ is an allyl halide. This statement is True. The halogen is attached to the carbon adjacent to the double bond (allylic position).

Reason (R): Allyl halides are compounds in which the halogen atom is attached to an $sp²$ $sp^2$ hybridized carbon atom. This statement is False. In allyl halides, the halogen is attached to an $sp³$ $sp^3$ hybridized carbon atom which is next to an $sp²$ $sp^2$ hybridized carbon (C=C double bond).

Step 1: Conclusion

Therefore, (A) is true but (R) is false.

Pattern Recognition

Allylic = $sp³$ $sp^3$ C adjacent to C=C. Vinylic = $sp²$ $sp^2$ C of the C=C itself.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes

Q jee_main_2024_31_jan_evening Nucleophilic Aromatic Substitution

Identify A and B in the following reaction sequence.

Nucleophilic Aromatic Substitution diagram for Q63 - JEE Main 2024 Evening — The image shows a reaction scheme starting from bromobenzene undergoing nitration followed by substitution.

Solution

Core Logic

When bromobenzene reacts with concentrated $HNO₃$ $HNO_3$ (nitration), the bromine atom is ortho/para directing. However, under drastic conditions with excess concentrated nitrating mixture, 1-bromo-2,4,6-trinitrobenzene is formed (Compound A).
When 1-bromo-2,4,6-trinitrobenzene (Compound A) is treated with $NaOH$ $NaOH$, the presence of three strong electron-withdrawing $-NO₂$ $-NO_2$ groups activates the aromatic ring toward Nucleophilic Aromatic Substitution ( $SNAr$ $S_NAr$). The $-Br$ $-Br$ is easily replaced by $-OH$ $-OH$ to form 2,4,6-trinitrophenol (picric acid).
Subsequent acidification with $HCl$ $HCl$ yields the neutral picric acid (Compound B).

Pattern Recognition

Multiple $NO₂$ $NO_2$ groups drastically increase the susceptibility of halobenzenes to $SNAr$ $S_NAr$. Bromine is replaced completely by $OH^-$ $OH^-$ under alkaline conditions.

Chapter Mix

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

Q jee_main_2024_31_jan_evening IUPAC Nomenclature of Haloalkanes

Identify structure of 2,3-dibromo-1-phenylpentane.

Solution

Core Logic

Decode the IUPAC name: 2,3-dibromo-1-phenylpentane.

Parent chain is pentane (5 carbon chain: C1-C2-C3-C4-C5).
Substituents:
Phenyl group at position 1.
Bromo groups at positions 2 and 3.

Option (3) correctly displays a 5-carbon straight chain. The first carbon attaches to the benzene ring (phenyl group), and the second and third carbons each hold a bromine atom.

IUPAC Nomenclature of Haloalkanes diagram for Q67 - JEE Main 2024 Evening

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes

Q jee_main_2024_31_jan_morning Elimination and Addition Reactions

The product (C) in the below mentioned reaction is:

CH₃-CH₂-CH₂-Br [Δ]KOH(alc) A [Δ]HBr B [Δ]KOH(aq) C

A. $Propan-1-ol$ $\text{Propan-1-ol}$
B. $Propene$ $\text{Propene}$
C. $Propyne$ $\text{Propyne}$
D. $Propan-2-ol$ $\text{Propan-2-ol}$

Solution

Step 1: Elimination to form Propene

CH₃-CH₂-CH₂-Br KOH (alc), Δ CH₃-CH=CH₂ (Compound A: Propene)

Step 2: Electrophilic Addition of HBr

Addition of $HBr$ $HBr$ follows Markovnikov's rule:

CH₃-CH=CH₂ + HBr Δ CH₃-CH(Br)-CH₃ (Compound B: 2-Bromopropane)

Step 3: Nucleophilic Substitution

Reaction with aqueous KOH leads to $SN2/SN1$ $S_N2/S_N1$ substitution of $Br^-$ $Br^-$ with $OH^-$ $OH^-$:

CH₃-CH(Br)-CH₃ KOH (aq), Δ CH₃-CH(OH)-CH₃ (Compound C: Propan-2-ol)

Pattern Recognition

Alc. KOH gives elimination (alkene). Aq. KOH gives substitution (alcohol). $HBr$ $HBr$ on unsymmetrical alkene gives Markovnikov addition.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes

Q86 jee_main_2024_31_jan_morning Elimination and Substitution

CH₃CH₂Br + NaOH arrow Product A

CH₃CH₂Br + NaOH / H₂O arrow Product B

The total number of hydrogen atoms in product A and product B is

Numerical Answer. Answer: 10 to 10

Solution

Core Logic

Reaction 1: If the reagent is alcoholic NaOH (implied due to formation of a distinct product A to contrast B), elimination occurs:

CH₃CH₂Br + NaOH (alc) arrow CH₂=CH₂ (Ethene)

Hydrogen atoms in ethene ( $C₂H₄$ $C_2H_4$) = 4.

Reaction 2: If the reagent is aqueous NaOH ( $NaOH / H₂O$ $NaOH / H_2O$), nucleophilic substitution ( $SN2$ $S_N2$) occurs:

CH₃CH₂Br + NaOH (aq) arrow CH₃CH₂OH (Ethanol)

Hydrogen atoms in ethanol ( $C₂H₆O$ $C_2H_6O$) = 6.

Total hydrogen atoms = 4 + 6 = 10.

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 8

Solution

Core Logic

Step 1: Conclusion

Pattern Recognition

Chapter Mix

Solution

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Chapter Mix

Solution

Step 1: Elimination to form Propene

Step 2: Electrophilic Addition of HBr

Step 3: Nucleophilic Substitution

Pattern Recognition

Chapter Mix

Solution

Core Logic

Chapter Mix

More Haloalkanes and Haloarenes Questions — jee_main_2025_24_jan_morning

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More Haloalkanes and Haloarenes Previous-Year Questions — Page 8

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Core Logic

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Core Logic

Chapter Mix

Solution

Step 1: Elimination to form Propene

Step 2: Electrophilic Addition of HBr

Step 3: Nucleophilic Substitution

Pattern Recognition

Chapter Mix

Solution

Core Logic

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