Nucleophilic substitution with AgCN$AgCN$ favors coordinate carbon bonding over nitrogen, yielding covalent isocyanides (R-NC$R-NC$).
Core Logic
Step 1: The starting material contains a double bond. Treating it with HCl$HCl$ leads to addition. According to Markovnikov's rule, the chloride ion attaches to the secondary position, creating chloride intermediate [A]$[A]$.
Step 2: Compound [A]$[A]$ reacts with AgCN$AgCN$. Since AgCN$AgCN$ is covalent, the lone pair on nitrogen acts as the attacking nucleophile, leading to substitution with an isocyanide group (-NC$-NC$) rather than a cyanide group (-CN$-CN$).
Step 1: Structural Synthesis
The intermediate [A]$[A]$ possesses a chlorine atom at the secondary carbon position. Substituting this chlorine with -NC$-NC$ provides the product corresponding to option (4).
The image outlines an addition reaction followed by substitution using silver cyanide.
Pattern Recognition
Distinguish between ionic vs covalent cyanide sources:
KCN / NaCN$KCN / NaCN \implies$ forms alkyl nitriles (R-CN$R-CN$)
AgCN$AgCN \implies$ forms alkyl isocyanides (R-NC$R-NC$)
Keywords:#product B formed in reaction sequence#JEE Main 2025 Evening Q34#Haloalkanes and Haloarenes substitution#AgCN nucleophilic substitution isocyanide#Hydrochlorination#Nucleophilic Substitution#Silver Cyanide Reaction
More Haloalkanes and Haloarenes Previous-Year Questions — Page 5
Rate of SNAr ∝ Number of electron-withdrawing groups (-I, -M) at ortho/para positions$$\text{Rate of } S_N\text{Ar} \propto \text{Number of electron-withdrawing groups (-I, -M) at ortho/para positions} $$
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$\text{NaOH at } 623\text{ K, } 300\text{ atm}$ (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)
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
Q42jee_main_2025_07_april_eveningPhysical 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$\text{Statement I is correct but statement II is incorrect}$
B.Statement I is incorrect but statement II is correct$\text{Statement I is incorrect but statement II is correct}$
C.Both statement I and statement II are incorrect$\text{Both statement I and statement II are incorrect}$
D.Both statement I and statement II are correct$\text{Both statement I and statement II are correct}$
Solution
Related Formula
μnet = √(μ₁² + μ₂² + 2μ₁μ₂ θ)$$\mu{\text{net}} = \sqrt{\mu_1^2 + \mu_2^2 + 2\mu_1\mu_2\cos\theta} $$Boiling point ∝ Dipole-dipole interactions + Van der Waals forces$$\text{Boiling point} \propto \text{Dipole-dipole interactions} + \text{Van der Waals forces}$$
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.
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$\mathrm{KCN} / \mathrm{NaCN} \rightarrow$ 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$-\mathrm{NC}$).
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^-$$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{Cl} \xrightarrow{\mathrm{HO}^-} \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{OH} + \mathrm{Cl}^-$$
Statement-II: The conversion proceeds well in the more polar medium.
CH₃-CH₂-CH₂-CH₂-Cl R₃N [CH₃-CH₂-CH₂-CH₂-NR₃]⁺Cl^-$$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{Cl} \xrightarrow{\mathrm{R}_3\mathrm{N}} [\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{NR}_3]^{+}\mathrm{Cl}^-$$
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
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.
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:
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
More Haloalkanes and Haloarenes Questions — jee_main_2025_28_jan_evening
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.
 "silver cyanide alkyl halide, isonitrile product substitution, chemoselective nucleophilic pathway")
 "ortho-isomer, para-isomer, dipole moment structural field")
 "ortho-isomer, para-isomer, dipole moment structural field")
 "ortho-isomer, para-isomer, dipole moment structural field")
 "silver cyanide alkyl halide, isonitrile product substitution, chemoselective nucleophilic pathway")
