Given below are two statements :
Statement-I: The conversion proceeds well in the less polar medium.
mathrmCH_3-mathrmCH_2-mathrmCH_2-mathrmCH_2-mathrmCl xrightarrowmathrmHO^- mathrmCH_3-mathrmCH_2-mathrmCH_2-mathrmCH_2-mathrmOH + mathrmCl^-$$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{Cl} \xrightarrow{\mathrm{HO}^-} \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{OH} + \mathrm{Cl}^-$$
Statement-II: The conversion proceeds well in the more polar medium.
mathrmCH_3-mathrmCH_2-mathrmCH_2-mathrmCH_2-mathrmCl xrightarrowmathrmR_3mathrmN [mathrmCH_3-mathrmCH_2-mathrmCH_2-mathrmCH_2-mathrmNR_3]^+mathrmCl^-$$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{Cl} \xrightarrow{\mathrm{R}_3\mathrm{N}} [\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{NR}_3]^{+}\mathrm{Cl}^-$$
A.Both statement I and statement II are true
B.Both statement I and statement II are false.
C.Statement I is false but statement II is true
D.Statement I is true but statement II is false
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 (R_3N$R_3N$ and alkyl chloride). The resulting transition state develops partial charges (delta+$\delta+$ and delta-$\delta-$) 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.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Haloalkanes and Haloarenes
Keywords:#conversion proceeds well in the less polar medium#JEE Main 2025 Morning Q44#Organic Chemistry JEE Main 2025#Nucleophilic Substitution Reactions
More Haloalkanes and Haloarenes Previous-Year Questions — Page 6
Q67jee_main_2024_31_jan_eveningIUPAC Nomenclature of Haloalkanes
Identify structure of 2,3-dibromo-1-phenylpentane.
A.text(1) Structure A$\text{(1) Structure A}$
B.text(2) Structure B$\text{(2) Structure B}$
C.text(3) Structure C$\text{(3) Structure C}$
D.text(4) Structure D$\text{(4) Structure D}$
Solution
### Core Logic
Decode the IUPAC name: 2,3-dibromo-1-phenylpentane.
1) Parent chain is pentane (5 carbon chain: C1-C2-C3-C4-C5).
2) 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
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Haloalkanes and Haloarenes
Q68jee_main_2024_31_jan_morningElimination and Addition Reactions
The product (C) in the below mentioned reaction is:
CH_3-CH_2-CH_2-Br xrightarrow[Delta]KOH_(alc) A xrightarrow[Delta]HBr B xrightarrow[Delta]KOH_(aq) C$$CH_3-CH_2-CH_2-Br \xrightarrow[\Delta]{KOH_{(alc)}} A \xrightarrow[\Delta]{HBr} B \xrightarrow[\Delta]{KOH_{(aq)}} C$$
Q86jee_main_2024_31_jan_morningElimination and Substitution
CH_3CH_2Br + NaOH rightarrow textProduct A$$CH_3CH_2Br + NaOH \rightarrow \text{Product A}$$CH_3CH_2Br + NaOH / H_2O rightarrow textProduct B$$CH_3CH_2Br + NaOH / H_2O \rightarrow \text{Product B}$$
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:
CH_3CH_2Br + NaOH (textalc) rightarrow CH_2=CH_2 text (Ethene)$$CH_3CH_2Br + NaOH (\text{alc}) \rightarrow CH_2=CH_2 \text{ (Ethene)}$$
Hydrogen atoms in ethene (C_2H_4$C_2H_4$) = 4.
Reaction 2:
If the reagent is aqueous NaOH (NaOH / H_2O$NaOH / H_2O$), nucleophilic substitution (S_N2$S_N2$) occurs:
CH_3CH_2Br + NaOH (textaq) rightarrow CH_3CH_2OH text (Ethanol)$$CH_3CH_2Br + NaOH (\text{aq}) \rightarrow CH_3CH_2OH \text{ (Ethanol)}$$
Hydrogen atoms in ethanol (C_2H_6O$C_2H_6O$) = 6.
Total hydrogen atoms = 4 + 6 = 10.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Haloalkanes and Haloarenes
More Haloalkanes and Haloarenes Questions — jee_main_2025_24_jan_morning
We Map Every Repeating Question in Competitive Exams.
Say goodbye to generic mock test fatigue. RankBit uses smart analysis to group past exam questions into their foundational Repeating Question Types. Find chapter weightage, track repeating questions, and score higher with targeted practice.
Select Your Target Exam
Choose an exam track below to find formulas per chapter and patterns.
Syncing Exam Intelligence
Mapping formulas and patterns across all tracks…
PATH A — FULL LENGTH PRACTICE
Full Mock Test Hub
Simulate real NTA exam conditions with fully tracked mocks. Time yourself against past papers.
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
