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 4
In the following substitution reaction:
The structural layout depicts a 1,2-dibromo-4-nitrobenzene reacting with sodium ethoxide.
Product mathrmP$\mathrm{P}$ formed is:
A.
B.
C.
D.
Solution
### Related Formula
Nucleophilic Aromatic Substitution (S_NAr$S_NAr$) occurs via a Meisenheimer complex intermediate, where strong electron-withdrawing groups (-mathrmNO_2$-\mathrm{NO}_2$) activate positions strictly ortho and para to themselves.
### Core Logic
The reactant is 1,2-dibromo-4-nitrobenzene. Let us evaluate the two bromine positions relative to the nitro group:
* The bromine at C-1 is para to the strong activating -mathrmNO_2$-\mathrm{NO}_2$ group.
* The bromine at C-2 is meta to the -mathrmNO_2$-\mathrm{NO}_2$ group.
Since the para position facilitates effective negative charge delocalization onto the oxygen atoms of the nitro group during intermediate formation, the para-bromine undergoes substitution exclusively by the ethoxide ion (^-mathrmOC_2mathrmH_5$^-\mathrm{OC}_2\mathrm{H}_5$). This yields the final product shown below:
The structural layout depicts a 1,2-dibromo-4-nitrobenzene reacting with sodium ethoxide.
### Pattern Recognition
In aromatic pathways activated by -mathrmNO_2$-\mathrm{NO}_2$, substitution happens exclusively at positions ortho or para relative to the nitro flag; meta positions remain unactivated.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Haloalkanes and Haloarenes
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : Haloalkanes react with KCN$KCN$ to form alkyl cyanides as a main product while with AgCN$AgCN$ form isocyanide as the main product.
Reason (R) : KCN$KCN$ and AgCN$AgCN$ both are highly ionic compounds.
In the light of the above statement, choose the most appropriate answer from the options given below:
A.text(A) is correct but (R) is not correct$\text{(A) is correct but (R) is not correct}$
B.textBoth (A) and (R) are correct but (R) is not the correct explanation of (A)$\text{Both (A) and (R) are correct but (R) is not the correct explanation of (A)}$
C.text(A) is not correct but (R) is correct$\text{(A) is not correct but (R) is correct}$
D.textBoth (A) and (R) are correct and (R) is the correct explanation of (A)$\text{Both (A) and (R) are correct and (R) is the correct explanation of (A)}$
Solution
### Core Logic
KCN$KCN$ is predominantly ionic and provides cyanide ions (CN^-$CN^-$) in solution. Although both carbon and nitrogen are in a position to donate electron pairs, the attack takes place mainly through carbon because C-C bond is more stable than C-N bond, forming alkyl cyanides (nitriles) as major product.
KCN + R-X rightarrow R-CN quad text(Major)$$KCN + R-X \rightarrow R-CN \quad \text{(Major)}$$
However, AgCN$AgCN$ is mainly covalent in nature and nitrogen is free to donate an electron pair forming isocyanide as the main product.
AgCN + R-X rightarrow R-NC quad text(Major)$$AgCN + R-X \rightarrow R-NC \quad \text{(Major)}$$
### Step 1: Final Conclusion
Assertion (A) is correct.
Reason (R) states both are highly ionic, which is incorrect because AgCN$AgCN$ is largely covalent.
Therefore, (A) is correct but (R) is not correct.
### Pattern Recognition
Ambidentate nucleophile shortcut: Alkali metal cyanides (KCN$KCN$, NaCN$NaCN$) are ionic rightarrow$\rightarrow$ attack from Carbon rightarrow$\rightarrow$ Cyanide. Heavy metal cyanides (AgCN$AgCN$) are covalent rightarrow$\rightarrow$ attack from Nitrogen rightarrow$\rightarrow$ Isocyanide.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Haloalkanes and Haloarenes
Number of optical isomers possible for 2-chlorobutane
Numerical Answer.Answer: 2 to 2
Solution
### Core Logic
The structure of 2-chlorobutane is CH_3-CH(Cl)-CH_2-CH_3$CH_3-CH(Cl)-CH_2-CH_3$.
There is exactly one chiral center in this molecule, which is the carbon atom at position 2 (bonded to H$H$, Cl$Cl$, CH_3$CH_3$, and CH_2CH_3$CH_2CH_3$).
For a molecule with n$n$ distinct chiral centers and no plane of symmetry, the number of optical isomers (stereoisomers) is 2^n$2^n$.
### Step 1: Calculate Isomers
Number of chiral centers n = 1$n = 1$.
Total optical isomers = 2^1 = 2$2^1 = 2$ (one pair of enantiomers: the (R) and (S) configurations).
Optical Isomerism diagram for Q81 - JEE Main 2024 Morning
### Pattern Recognition
1 chiral center always gives 2 optical isomers (a pair of enantiomers).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Haloalkanes and Haloarenes
Class 11 Chemistry: Organic Chemistry Some Basic Principles and Techniques
The correct statement regarding nucleophilic substitution reaction in a chiral alkyl halide is;
A. Retention occurs in S_N1$S_N1$ reaction and inversion occurs in S_N2$S_N2$ reaction.
B. Racemisation occurs in S_N1$S_N1$ reaction and retention occurs in S_N2$S_N2$ reaction.
C. Racemisation occurs in both S_N1$S_N1$ and S_N2$S_N2$ reactions.
D. Racemisation occurs in S_N1$S_N1$ reaction and inversion occurs in S_N2$S_N2$ reaction.
Solution
### Core Logic
In an textS_textN1$\text{S}_\text{N}1$ 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 textS_textN2$\text{S}_\text{N}2$ pathway, the nucleophile attacks exclusively from the backside opposite the leaving group, causing an absolute structural inversion (Walden inversion).
### Pattern Recognition
S_N1 rightarrow$S_N1 \rightarrow$ planar intermediate carbocation rightarrow$\rightarrow$ Racemisation. S_N2 rightarrow$S_N2 \rightarrow$ direct backside launch rightarrow$\rightarrow$ Inversion.
### Chapter Mix
Class 12 Chemistry: Haloalkanes and Haloarenes
Q70jee_main_2024_29_jan_morningPreparation 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.textBoth A and R are true but R is NOT the correct explanation of A$\text{Both A and R are true but R is NOT the correct explanation of A}$
B.textA is false but R is true$\text{A is false but R is true}$
C.textA is true but R is false$\text{A is true but R is false}$
D.textBoth A and R are true and R is the correct explanation of A$\text{Both A and R are true and R is the correct explanation of A}$
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.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Haloalkanes and Haloarenes
Class 12 Chemistry: Alcohols Phenols and Ethers
More Haloalkanes and Haloarenes Questions — jee_main_2025_24_jan_morning
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