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

List-I (Conversion)	List-II (Reagents, Conditions used) [cite: 248, 249]
(A) Chlorobenzene $arrow$ $\rightarrow$ Phenol	(I) $Warm, H₂O$ $\text{Warm, } \text{H}_2\text{O}$
(B) p-Nitrochlorobenzene $arrow$ $\rightarrow$ p-Nitrophenol	(II) (a) $NaOH, 368 K$ $\text{NaOH, } 368\text{ K}$; (b) $H3O^+$ $\text{H}3\text{O}^+$	(C) 2,4-Dinitrochlorobenzene $arrow$ $\rightarrow$ 2,4-Dinitrophenol	(III) (a) $NaOH, 443 K$ $\text{NaOH, } 443\text{ K}$; (b) $H₃O^+$ $\text{H}_3\text{O}^+$	(D) 2,4,6-Trinitrochlorobenzene $arrow$ $\rightarrow$ 2,4,6-Trinitrophenol	(IV) (a) $NaOH, 623 K, 300 atm$ ; (b) $H₃O^+$ $\text{H}_3\text{O}^+$ Choose the correct answer from the options given below: A. $(A)-(II), (B)-(III), (C)-(I), (D)-(IV)$ B. $(A)-(III), (B)-(IV), (C)-(II), (D)-(I)$ C. $(A)-(IV), (B)-(III), (C)-(II), (D)-(I)$ D. $(A)-(IV), (B)-(III), (C)-(I), (D)-(II)$ Solution Related Formula $Rate of SNAr \propto Number of electron-withdrawing groups (-I, -M) at ortho/para positions$ Core Logic Aryl halides are generally unreactive towards nucleophilic substitution due to resonance stabilization of the $C-Cl$ $\text{C-Cl}$ bond. However, the presence of strong electron-withdrawing groups ( $-NO₂$ $-\text{NO}_2$) at ortho and para positions dramatically increases reactivity by stabilizing the intermediate carbanion: - (A) Chlorobenzene: Needs extreme conditions: $NaOH at 623 K, 300 atm$ (Dow's Process) $arrow$ $\rightarrow$ (IV) - (B) p-Nitrochlorobenzene: One para $-NO₂$ $-\text{NO}_2$ group softens required temperature to $443 K$ $443\text{ K}$ $arrow$ $\rightarrow$ (III) - (C) 2,4-Dinitrochlorobenzene: Two electron-withdrawing groups lower needed temperature further to $368 K$ $368\text{ K}$ $arrow$ $\rightarrow$ (II) - (D) 2,4,6-Trinitrochlorobenzene: Highly activated picryl chloride hydrolyzes smoothly with just warm water $arrow$ $\rightarrow$ (I) Step 1: Final Match Alignment Matching sequences cleanly yields: (A)-(IV), (B)-(III), (C)-(II), (D)-(I). Pattern Recognition The more $-NO₂$ $-\text{NO}_2$ groups present on the ring, the less aggressive the reagent/temperature setup required. Count $-NO₂$ $-\text{NO}_2$ groups: $0 arrow 623K$ $0 \rightarrow 623\text{K}$, $1 arrow 443K$ $1 \rightarrow 443\text{K}$, $2 arrow 368K$ $2 \rightarrow 368\text{K}$, $3 arrow warm water$ $3 \rightarrow \text{warm water}$. Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes Q42 jee_main_2025_07_april_evening Physical Properties of Dihalobenzenes Given below are two statements: Statement (I): The structural layouts show ortho and para halobenzene isomers (1,2-dichlorobenzene vs 1,2-dibromobenzene, and para-dibromobenzene). is more polar than The structural layouts show ortho and para halobenzene isomers (1,2-dichlorobenzene vs 1,2-dibromobenzene, and para-dibromobenzene). . Statement (II): Boiling point of The structural layouts show ortho and para halobenzene isomers (1,2-dichlorobenzene vs 1,2-dibromobenzene, and para-dibromobenzene). is lower than the ortho-isomer, but it is more polar than the meta-isomer. In the light of the above statements, choose the most appropriate answer from the options given below: A. $Statement I is correct but statement II is incorrect$ B. $Statement I is incorrect but statement II is correct$ C. $Both statement I and statement II are incorrect$ D. $Both statement I and statement II are correct$ Solution Related Formula $μnet = \sqrt(μ₁² + μ₂² + 2μ₁μ₂ θ)$ $Boiling point \propto Dipole-dipole interactions + Van der Waals forces$ Core Logic Let's analyze the visual structures alongside their scientific orientations: Statement (I) compares 1,2-dichlorobenzene and 1,2-dibromobenzene. Chlorine has a higher electronegativity than bromine, creating a larger bond dipole. The vacant d-orbital interactions do not invert this baseline dipole trend. Thus, 1,2-dichlorobenzene is more polar, making Statement I correct. Statement (II) evaluates dihalobenzene isomers. For the para-isomer, individual bond dipoles are oriented at $180°$ $180^{\circ}$, cancelling out completely: $μpara = 0$ $$\mu{\text{para}} = 0 $$ Since $μmeta > 0$ $\mu_{\text{meta}} > 0$, the para-isomer is less polar than the meta-isomer. This directly falsifies Statement II. Step 1: Spatial Alignments The geometric configurations map out as follows: The structural layouts show ortho and para halobenzene isomers (1,2-dichlorobenzene vs 1,2-dibromobenzene, and para-dibromobenzene). The structural layouts show ortho and para halobenzene isomers (1,2-dichlorobenzene vs 1,2-dibromobenzene, and para-dibromobenzene). The structural layouts show ortho and para halobenzene isomers (1,2-dichlorobenzene vs 1,2-dibromobenzene, and para-dibromobenzene). Hence, Statement I is correct, but Statement II is incorrect. Pattern Recognition Dipole tracking rule: Para-substituted benzenes with identical groups possess a structural center of inversion, guaranteeing a net dipole moment of exactly zero ( $μ = 0$ $\mu = 0$). They can never be more polar than any asymmetric ortho or meta structural isomer. Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes Q43 jee_main_2025_24_jan_evening Nucleophilic Substitution Reactions The structure of the major product formed in the following reaction is : The reaction shows a dihalogenated aromatic derivative reacting with silver cyanide to generate a major product. A. \text{Structure Option (1)} B. \text{Structure Option (2)} C. \text{Structure Option (3)} D. \text{Structure Option (4)} Solution Core Logic The substrate contains two distinct carbon-halogen bonds: an aryl-bromide bond ( $Ar-Br$ $\mathrm{Ar-Br}$) on the ring and an aliphatic alkyl-chloride bond ( $CH₂-Cl$ $\mathrm{CH_2-Cl}$) on the side chain. Aryl Halide Site ( $Csp²-Br$ $\mathrm{C}_{sp^2}\mathrm{-Br}$): The bromine atom attached directly to the aromatic ring does not undergo standard nucleophilic substitution ( $SN2$ $S_N2$ or $SN1$ $S_N1$) under normal conditions due to resonance stabilization, which gives the bond partial double-bond character. Alkyl Halide Site ( $Csp³-Cl$ $\mathrm{C}_{sp^3}\mathrm{-Cl}$): The side-chain aliphatic carbon bond undergoes smooth, unhindered nucleophilic substitution. When reacting with silver cyanide ( $AgCN$ $\mathrm{AgCN}$): $AgCN$ $\mathrm{AgCN}$ is predominantly covalent. The lone pair on the nitrogen atom acts as the primary nucleophilic center rather than the carbon atom. Consequently, substitution at the aliphatic site yields an isonitrile ( $-NC$ $-\mathrm{NC}$) derivative as the major product, leaving the aryl bromide group completely untouched. Step 1: Structural Resolution The reaction progresses cleanly at the side-chain carbon: The reaction shows a dihalogenated aromatic derivative reacting with silver cyanide to generate a major product. Pattern Recognition Remember the key selectivity rule for cyanide nucleophiles: $KCN / NaCN arrow$ ionic reagents $arrow$ $\rightarrow$ attacks via carbon to form a Nitrile ( $-CN$ $-\mathrm{CN}$). $AgCN arrow$ $\mathrm{AgCN} \rightarrow$ covalent reagent $arrow$ $\rightarrow$ attacks via nitrogen to form an Isonitrile ( $-NC$ $-\mathrm{NC}$). Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes Q44 jee_main_2025_24_jan_morning Nucleophilic Substitution Reactions Given below are two statements : Statement-I: The conversion proceeds well in the less polar medium. $CH₃-CH₂-CH₂-CH₂-Cl HO^- CH₃-CH₂-CH₂-CH₂-OH + Cl^-$ Statement-II: The conversion proceeds well in the more polar medium. $CH₃-CH₂-CH₂-CH₂-Cl R₃N [CH₃-CH₂-CH₂-CH₂-NR₃]⁺Cl^-$ 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 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 Q jee_main_2025_28_jan_evening Elimination Reactions (E2) The major product of the following reaction is: The image details a multi-halogenated hydrocarbon chain reacting with excess ethanolic potassium hydroxide under heating conditions. A. 6-Phenylhepta-2,4-diene B. 2-Phenylhepta-2,5-diene C. 6-Phenylhepta-3,5-diene D. 2-Phenylhepta-2,4-diene Solution Related Formula Base-induced dehydrohalogenation follows the Zaitsev rule to maximize thermodynamic stability via conjugated double bond networks: $R-CHX-CH₂-CHX-R' excess KOH/EtOH, Δ Conjugated Diene$ Core Logic The reactant is a dihalide containing a phenyl substitution. Treating with excess alcoholic $KOH$ $KOH$ and heat induces double dehydrohalogenation via successive $E2$ $E2$ elimination pathways. The eliminations occur to yield the most stable, highly conjugated product where the double bonds are conjugated with each other and, if possible, with the aromatic phenyl ring system. Step 1: Eliminating and Tracking Conjugation Eliminating the first and second equivalents of $HBr$ $HBr$ sets up a conjugated diene system along the heptadiene chain. Tracing carbon numbers correctly from the end closest to the phenyl ring reveals that the conjugated diene centers sit across carbons 2 and 4, producing 2-Phenylhepta-2,4-diene. The image details a multi-halogenated hydrocarbon chain reacting with excess ethanolic potassium hydroxide under heating conditions. Pattern Recognition When dealing with excess elimination agents on dihalides, look for options that form a continuous conjugated diene structure (alternating double-single-double bonds). This conjugation offers significant thermodynamic stability, especially when directly extended from a phenyl group. Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes ← Previous Page 5 of 8 Next → More Haloalkanes and Haloarenes Questions — jee_main_2025_28_jan_morning Practice all Haloalkanes and Haloarenes previous-year questions → The major product of the following reaction is: Given below are two statements : Statement I: will undergo Which among the following halides will generate the most stable Match List-I with List-II. <div style="overflow-x: auto; margin: 1rem 0;"><table The product B formed in the following reaction sequence is Given below are two statements: Statement (I): is more polar The structure of the major product formed in the following Given below are two statements : Statement-I: The conversion proceeds Consider the following sequence of reactions: Consider the above sequence A hydrocarbon ‘P’ ( ) on reaction with HCl gives Which of the following reaction is NOT correctly represented ? Two products X and Y are formed in the following The correct order of reactivity of the following benzyl halides The correct order of reactivity of in methanol with the Match the List-I with List-II <div style="overflow-x: auto; margin: 1rem Given below are two statements: Statement I: High concentration of Given below are four compounds: (a) n-propyl chloride (b) iso-propyl The correct order of the rate of reaction of the Given below are two statements: Statement I : bond is As compared with chlorocyclohexane, which of the following statements correctly The dibromo compound [P] (molecular formula: ) when heated with Given below are two statement one is labeled as Assertion Consider all the structural isomers with molecular formula are separately The correct statement regarding nucleophilic substitution reaction in a chiral The total number of hydrogen atoms in product A and SYSTEM Syllabus Analysis & Trend Mapping Dashboard Chapter Trends Formula Recognition Questions Feed Weightage Rankbit System Rankbit Dashboard Chapter Trends Formula Recognition Questions Feed Weightage JEE Physics: Waves (+15.5%) \| Electrostatics: Concentric Shells (-29.7%) \| Modern Physics: Photoelectric Clones (+34.2%) \| Mathematics: Definite Integrals (+18.1%) \| Chemistry: Coordination Splitting (-11.4%) \| JEE Physics: Waves (+15.5%) \| Electrostatics: Concentric Shells (-29.7%) \| Modern Physics: Photoelectric Clones (+34.2%) \| Mathematics: Definite Integrals (+18.1%) \| Chemistry: Coordination Splitting (-11.4%) Rankbit The ultimate ecosystem for IIT JEE and COMEDK. 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List-I (Conversion)

List-II (Reagents, Conditions used) [cite: 248, 249]

(A) Chlorobenzene

arrow

Phenol

(I)

Warm, H₂O

(B) p-Nitrochlorobenzene

arrow

p-Nitrophenol

(II) (a)

NaOH, 368 K

; (b)

H3O^+

arrow

2,4-Dinitrophenol

(III) (a)

NaOH, 443 K

; (b)

H₃O^+

(D) 2,4,6-Trinitrochlorobenzene

arrow

2,4,6-Trinitrophenol

(IV) (a)

NaOH, 623 K, 300 atm

; (b)

H₃O^+

Choose the correct answer from the options given below:

A. $(A)-(II), (B)-(III), (C)-(I), (D)-(IV)$
B. $(A)-(III), (B)-(IV), (C)-(II), (D)-(I)$
C. $(A)-(IV), (B)-(III), (C)-(II), (D)-(I)$
D. $(A)-(IV), (B)-(III), (C)-(I), (D)-(II)$

Solution

Related Formula

Rate of SNAr \propto Number of electron-withdrawing groups (-I, -M) at ortho/para positions

Core Logic

Aryl halides are generally unreactive towards nucleophilic substitution due to resonance stabilization of the $C-Cl$ $\text{C-Cl}$ bond. However, the presence of strong electron-withdrawing groups ( $-NO₂$ $-\text{NO}_2$) at ortho and para positions dramatically increases reactivity by stabilizing the intermediate carbanion:

- (A) Chlorobenzene: Needs extreme conditions: $NaOH at 623 K, 300 atm$ (Dow's Process) $arrow$ $\rightarrow$ (IV) - (B) p-Nitrochlorobenzene: One para $-NO₂$ $-\text{NO}_2$ group softens required temperature to $443 K$ $443\text{ K}$ $arrow$ $\rightarrow$ (III) - (C) 2,4-Dinitrochlorobenzene: Two electron-withdrawing groups lower needed temperature further to $368 K$ $368\text{ K}$ $arrow$ $\rightarrow$ (II) - (D) 2,4,6-Trinitrochlorobenzene: Highly activated picryl chloride hydrolyzes smoothly with just warm water $arrow$ $\rightarrow$ (I)

Step 1: Final Match Alignment

Matching sequences cleanly yields: (A)-(IV), (B)-(III), (C)-(II), (D)-(I).

Pattern Recognition

The more $-NO₂$ $-\text{NO}_2$ groups present on the ring, the less aggressive the reagent/temperature setup required. Count $-NO₂$ $-\text{NO}_2$ groups: $0 arrow 623K$ $0 \rightarrow 623\text{K}$, $1 arrow 443K$ $1 \rightarrow 443\text{K}$, $2 arrow 368K$ $2 \rightarrow 368\text{K}$, $3 arrow warm water$ $3 \rightarrow \text{warm water}$.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes

Q42 jee_main_2025_07_april_evening Physical Properties of Dihalobenzenes

Given below are two statements: Statement (I):

Physical Properties of Dihalobenzenes diagram for Q42 - JEE Main 2025 Evening — The structural layouts show ortho and para halobenzene isomers (1,2-dichlorobenzene vs 1,2-dibromobenzene, and para-dibromobenzene).

is more polar than

. Statement (II): Boiling point of

is lower than the ortho-isomer, but it is more polar than the meta-isomer. In the light of the above statements, choose the most appropriate answer from the options given below:

A. $Statement I is correct but statement II is incorrect$
B. $Statement I is incorrect but statement II is correct$
C. $Both statement I and statement II are incorrect$
D. $Both statement I and statement II are correct$