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)}
The reaction shows a dihalogenated aromatic derivative reacting with silver cyanide to generate a major product.
B.\text{Structure Option (2)}
The reaction shows a dihalogenated aromatic derivative reacting with silver cyanide to generate a major product.
C.\text{Structure Option (3)}
The reaction shows a dihalogenated aromatic derivative reacting with silver cyanide to generate a major product.
D.\text{Structure Option (4)}
The reaction shows a dihalogenated aromatic derivative reacting with silver cyanide to generate a major product.
Solution & Explanation
### Core Logic
The substrate contains two distinct carbon-halogen bonds: an aryl-bromide bond (mathrmAr-Br$\mathrm{Ar-Br}$) on the ring and an aliphatic alkyl-chloride bond (mathrmCH_2-Cl$\mathrm{CH_2-Cl}$) on the side chain.
1. **Aryl Halide Site (mathrmC_sp^2mathrm-Br$\mathrm{C}_{sp^2}\mathrm{-Br}$):** The bromine atom attached directly to the aromatic ring does not undergo standard nucleophilic substitution (S_N2$S_N2$ or S_N1$S_N1$) under normal conditions due to resonance stabilization, which gives the bond partial double-bond character.
2. **Alkyl Halide Site (mathrmC_sp^3mathrm-Cl$\mathrm{C}_{sp^3}\mathrm{-Cl}$):** The side-chain aliphatic carbon bond undergoes smooth, unhindered nucleophilic substitution.
When reacting with silver cyanide (mathrmAgCN$\mathrm{AgCN}$):
mathrmAgCN$\mathrm{AgCN}$ 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 (-mathrmNC$-\mathrm{NC}$)** 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:
* mathrmKCN / mathrmNaCN
ightarrow$\mathrm{KCN} / \mathrm{NaCN}
ightarrow$ ionic reagents
ightarrow$
ightarrow$ attacks via carbon to form a **Nitrile (-mathrmCN$-\mathrm{CN}$)**.
* mathrmAgCN
ightarrow$\mathrm{AgCN}
ightarrow$ covalent reagent
ightarrow$
ightarrow$ attacks via nitrogen to form an **Isonitrile (-mathrmNC$-\mathrm{NC}$)**.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Haloalkanes and Haloarenes
Keywords:#AgCN nucleophilic substitution#isonitrile vs nitrile formation#JEE Main 2025 Evening Q43#chemoselective alkyl substitution#silver cyanide alkyl halide#isonitrile product substitution#chemoselective nucleophilic pathway
More Haloalkanes and Haloarenes Previous-Year Questions — Page 3
Q442025Nucleophilic Substitution Reactions
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
### 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
Q2025Elimination 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-CH_2-CHX-R' xrightarrowtextexcess KOH/EtOH, Delta textConjugated Diene$R-CHX-CH_2-CHX-R' \xrightarrow{\text{excess } KOH/EtOH, \Delta} \text{Conjugated Diene}$
### 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.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Haloalkanes and Haloarenes
Q342025Nucleophilic Substitution Reactions
The product B formed in the following reaction sequence is :
The image outlines an addition reaction followed by substitution using silver cyanide.
A. (1)
B. (2)
C. (3)
D. (4)
Solution
### Related Formula
Markovnikov addition of HCl$HCl$ across an alkene:
R-CH=CH_2 + HCl rightarrow R-CHCl-CH_3$R-CH=CH_2 + HCl \rightarrow R-CHCl-CH_3$
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 implies$KCN / NaCN \implies$ forms alkyl nitriles (R-CN$R-CN$)
- AgCN implies$AgCN \implies$ forms alkyl isocyanides (R-NC$R-NC$)
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Haloalkanes and Haloarenes
Q342025Nucleophilic Aromatic Substitution
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. {{IMG_OPT1}}
B. {{IMG_OPT2}}
C. {{IMG_OPT3}}
D. {{IMG_OPT4}}
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
More Haloalkanes and Haloarenes Questions — jee_main_2025_24_jan_evening
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