A hydrocarbon ‘P’ (C_4H_8$C_{4}H_{8}$) on reaction with HCl gives an optically active compound ‘Q’ (C_4H_9Cl$C_{4}H_{9}Cl$) which on reaction with one mole of ammonia gives compound ‘R’ (C_4H_11N$C_{4}H_{11}N$). ‘R’ on diazotization followed by hydrolysis gives ‘S’. Identify P, Q, R and S.
### Core Logic
Since P (C_4H_8$C_4H_8$) reacts with HCl to give an optically active compound Q (C_4H_9Cl$C_4H_9Cl$), P must be But-2-ene. Addition of HCl yields 2-chlorobutane, which possesses a chiral center.
mathrmCH_3-mathrmCH=mathrmCH-mathrmCH_3 xrightarrowmathrmHCl mathrmCH_3-mathrmCH_2-mathrmC^*mathrmH(mathrmCl)-mathrmCH_3 quad text(P is But-2-ene, Q is 2-chlorobutane)$$\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3 \xrightarrow{\mathrm{HCl}} \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{C}^*\mathrm{H}(\mathrm{Cl})-\mathrm{CH}_3 \quad \text{(P is But-2-ene, Q is 2-chlorobutane)}$$Reactions of Haloalkanes diagram for Q54 - JEE Main 2026 Morning
Compound Q reacts with mathrmNH_3$\mathrm{NH}_3$ to undergo nucleophilic substitution forming a primary amine R:
mathrmCH_3-mathrmCH_2-mathrmCH(mathrmCl)-mathrmCH_3 xrightarrowmathrmNH_3 mathrmCH_3-mathrmCH_2-mathrmCH(mathrmNH_2)-mathrmCH_3 quad text(R is Butan-2-amine)$$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}(\mathrm{Cl})-\mathrm{CH}_3 \xrightarrow{\mathrm{NH}_3} \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}(\mathrm{NH}_2)-\mathrm{CH}_3 \quad \text{(R is Butan-2-amine)}$$Reactions of Haloalkanes diagram for Q54 - JEE Main 2026 Morning
Diazotization of primary aliphatic amines followed by hydrolysis yields an alcohol via a carbocation intermediate (S):
mathrmCH_3-mathrmCH_2-mathrmCH(mathrmNH_2)-mathrmCH_3 xrightarrowmathrmNaNO_2 / mathrmHCl / mathrmH_2mathrmO mathrmCH_3-mathrmCH_2-mathrmCH(mathrmOH)-mathrmCH_3 quad text(S is Butan-2-ol)$$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}(\mathrm{NH}_2)-\mathrm{CH}_3 \xrightarrow{\mathrm{NaNO}_2 / \mathrm{HCl} / \mathrm{H}_2\mathrm{O}} \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}(\mathrm{OH})-\mathrm{CH}_3 \quad \text{(S is Butan-2-ol)}$$Reactions of Haloalkanes diagram for Q54 - JEE Main 2026 Morning
### Pattern Recognition
An optically active alkyl halide formed from a C_4H_8$C_4H_8$ alkene with HX is a classic pointer to 2-halobutane originating from But-2-ene or But-1-ene. Diazotization of primary aliphatic amines (R-NH_2$R-NH_2$) with NaNO_2/HCl$NaNO_2/HCl$ yields alcohols (R-OH$R-OH$) with possible rearrangements, though here a secondary carbocation is already stable.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Haloalkanes and Haloarenes
Class 12 Chemistry: Amines
Keywords:#hydrocarbon P C4H8#JEE Main 2026 Morning Q54#Haloalkanes and Haloarenes JEE Main 2026#Reactions of Haloalkanes JEE Main 2026
More Haloalkanes and Haloarenes Previous-Year Questions — Page 3
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.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
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 (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
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
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
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
More Haloalkanes and Haloarenes Questions — jee_main_2026_21_jan_morning
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