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 2
Q292025Alkaline Hydrolysis and NGP
Given below are two statements :
Statement I: mathrmEt_2mathrmN-mathrmCH_2-mathrmCH_2-mathrmCl$\mathrm{Et}_2\mathrm{N}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{Cl}$ will undergo alkaline hydrolysis at a faster rate than mathrmEt_2mathrmCH-mathrmCH_2-mathrmCl$\mathrm{Et}_2\mathrm{CH}-\mathrm{CH}_2-\mathrm{Cl}$.
Statement II: In mathrmEt_2mathrmN-mathrmCH_2-mathrmCH_2-mathrmCl$\mathrm{Et}_2\mathrm{N}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{Cl}$, intramolecular substitution takes place first by involving lone pair of electrons on nitrogen.
In the light of the above statements, choose the most appropriate answer from the options given below:
A.textBoth Statement I and Statement II are incorrect$\text{Both Statement I and Statement II are incorrect}$
B.textStatement I is incorrect but statement II is correct$\text{Statement I is incorrect but statement II is correct}$
C.textBoth Statement I and Statement II are correct$\text{Both Statement I and Statement II are correct}$
D.textStatement I is correct but Statement II is incorrect$\text{Statement I is correct but Statement II is incorrect}$
Solution
### Core Logic
Statement I is correct because the nitrogen atom contains a lone pair situated at the beta$\beta$-position relative to the chlorine atom, promoting Neighboring Group Participation (NGP).
Statement II is correct because the lone pair on nitrogen attacks internally to kick out the chloride ion, forming a cyclic aziridinium ion intermediate. This quick intramolecular cyclization leads to an exceptionally rapid hydrolysis rate compared to standard aliphatic substitution.
### Pattern Recognition
Sees: Nitrogen with lone pair beta$\beta$ to a leaving group.
Shortcut: NGP (Neighboring Group Participation) accelerates substitution dramatically via intramolecular assistance.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Haloalkanes and Haloarenes
Q392025Ambident Nucleophiles Reactions
The products A and B in the following reactions, respectively are
mathrm A xleftarrow mathrm A g - mathrm N O _ 2 mathrm C H _ 3 - mathrm C H _ 2 - mathrm C H _ 2 - mathrm B r xrightarrow mathrm A g C N mathrm B$\mathrm {A} \xleftarrow {\mathrm {A g} - \mathrm {N O} _ {2}} \mathrm {C H} _ {3} - \mathrm {C H} _ {2} - \mathrm {C H} _ {2} - \mathrm {B r} \xrightarrow {\mathrm {A g C N}} \mathrm {B}$
### Core Logic
Both silver reagents exhibit significantly covalent bond characters:
- Reaction with mathrmAgNO_2$\mathrm{AgNO}_2$: 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:
mathrmA = mathrmCH_3-mathrmCH_2-mathrmCH_2-mathrmNO_2$\mathrm{A} = \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{NO}_2$
- Reaction with mathrmAgCN$\mathrm{AgCN}$: The covalent mathrmAg-mathrmC$\mathrm{Ag}-\mathrm{C}$ bond directs the nucleophilic attack to proceed through the lone pair on nitrogen, yielding an isocyanide compound:
mathrmB = mathrmCH_3-mathrmCH_2-mathrmCH_2-mathrmNC$\mathrm{B} = \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{NC}$
Hence, option (4) represents the correct combination.
### Pattern Recognition
Sees: Alkyl halide reacting with covalent silver salts of ambident anions.
Shortcut: Silver reagents (mathrmAgCN$\mathrm{AgCN}$ or mathrmAgNO_2$\mathrm{AgNO}_2$) drive bond formatting via the nitrogen center, producing isocyanides and nitroalkanes respectively.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Haloalkanes and Haloarenes
Q312025Substitution versus Elimination
Given below are two statements :
Statement (I): Alcohols are formed when alkyl chlorides are treated with aqueous potassium hydroxide by elimination reaction.
Statement (II) : In alcoholic potassium hydroxide, alkyl chlorides form alkenes by abstracting the hydrogen from the beta$\beta$-carbon.
In the light of the above statements, choose the most appropriate answer from the options given below:
A. Both Statement I and Statement II are incorrect
B. Statement I is incorrect but Statement II is correct
C. Statement I is correct but Statement II is incorrect
D. Both Statement I and Statement II are correct.
Solution
### Related Formula
textR-Cl + KOH_text(aq) rightarrow textR-OH + KCl quad (S_Ntext Nucleophilic Substitution)$\text{R-Cl} + KOH_{\text{(aq)}} \rightarrow \text{R-OH} + KCl \quad (S_N\text{ Nucleophilic Substitution})$textR-CH_2text-CH_2text-Cl + KOH_text(alc) rightarrow textR-CH=textCH_2 + KCl + H_2O quad (E2text Elimination)$\text{R-CH}_2\text{-CH}_2\text{-Cl} + KOH_{\text{(alc)}} \rightarrow \text{R-CH}=\text{CH}_2 + KCl + H_2O \quad (E2\text{ Elimination})$
### Core Logic
- **Statement I is incorrect:** Treatment of alkyl chlorides with aqueous KOH$KOH$ yields alcohols via a **nucleophilic substitution (S_N$S_N$) reaction**, not an elimination reaction.
- **Statement II is correct:** Alcoholic KOH$KOH$ acts as a strong base (R-O^-$R-O^-$ ions present), which preferentially abstracts a proton from the beta$\beta$-carbon atom, leading to dehydrohalogenation to form an alkene via an elimination pathway.
### Pattern Recognition
Remember: Aqueous medium = substitution (nucleophilic attack dominates due to highly hydrated, less basic hydroxide ions). Alcoholic medium = elimination (alkoxide acts as a bulky strong base to capture beta$\beta$-hydrogens).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Haloalkanes and Haloarenes
### Related Formula
textRate of S_NtextAr propto textNumber 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 textC-Cl$\text{C-Cl}$ bond. However, the presence of strong electron-withdrawing groups (-textNO_2$-\text{NO}_2$) at ortho and para positions dramatically increases reactivity by stabilizing the intermediate carbanion:
- (A) Chlorobenzene: Needs extreme conditions: textNaOH at 623text K, 300text atm$\text{NaOH at } 623\text{ K, } 300\text{ atm}$ (Dow's Process)
ightarrow$
ightarrow$ (IV)
- (B) p-Nitrochlorobenzene: One para -textNO_2$-\text{NO}_2$ group softens required temperature to 443text K$443\text{ K}$
ightarrow$
ightarrow$ (III)
- (C) 2,4-Dinitrochlorobenzene: Two electron-withdrawing groups lower needed temperature further to 368text K$368\text{ K}$
ightarrow$
ightarrow$ (II)
- (D) 2,4,6-Trinitrochlorobenzene: Highly activated picryl chloride hydrolyzes smoothly with just warm water
ightarrow$
ightarrow$ (I)
### Step 1: Final Match Alignment
Matching sequences cleanly yields:
(A)-(IV), (B)-(III), (C)-(II), (D)-(I).
### Pattern Recognition
The more -textNO_2$-\text{NO}_2$ groups present on the ring, the less aggressive the reagent/temperature setup required. Count -textNO_2$-\text{NO}_2$ groups: 0
ightarrow 623textK$0
ightarrow 623\text{K}$, 1
ightarrow 443textK$1
ightarrow 443\text{K}$, 2
ightarrow 368textK$2
ightarrow 368\text{K}$, 3
ightarrow textwarm water$3
ightarrow \text{warm water}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Haloalkanes and Haloarenes
Q422025Physical 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.textStatement I is correct but statement II is incorrect$\text{Statement I is correct but statement II is incorrect}$
B.textStatement I is incorrect but statement II is correct$\text{Statement I is incorrect but statement II is correct}$
C.textBoth statement I and statement II are incorrect$\text{Both statement I and statement II are incorrect}$
D.textBoth statement I and statement II are correct$\text{Both statement I and statement II are correct}$
Solution
### Related Formula
mutextnet = sqrtmu_1^2 + mu_2^2 + 2mu_1mu_2costheta $\mu{\text{net}} = \sqrt{\mu_1^2 + \mu_2^2 + 2\mu_1\mu_2\cos\theta} $textBoiling point propto textDipole-dipole interactions + textVan der Waals forces$\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^circ$180^{\circ}$, cancelling out completely:
mutextpara = 0 $\mu{\text{para}} = 0 $
Since mu_textmeta > 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 (mu = 0$\mu = 0$). They can never be more polar than any asymmetric ortho or meta structural isomer.
### 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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