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Haloalkanes and Haloarenes appeared 38 times across 3 years — 4.4% of Chemistry. This question is from Preparation and Reactions of Styrene derivatives.

Year	2026	2025	2024	Total
Questions	11	15	12	38

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Choose the correct set of reagents for the following conversion: $Ethyl benzene 4-bromostyrene$ {{Q_IMG1}}

Conversion scheme of ethylbenzene to 4-bromostyrene for Q29 — The figure illustrates the multi-step conversion starting from ethylbenzene to yield a brominated styrene derivative.

Conversion scheme of ethylbenzene to 4-bromostyrene for Q29 — The figure illustrates the multi-step conversion starting from ethylbenzene to yield a brominated styrene derivative.

A. $Br₂/Fe; Cl₂, Δ; alc. KOH$
B. $Cl₂/Fe; Br₂/anhy. AlCl₃; aq. KOH$
C. $Br₂/anhy. AlCl₃; Cl₂, Δ; aq. KOH$
D. $Cl₂/anhy. AlCl₃; Br₂/Fe; alc. KOH$

Solution & Explanation

Core Logic

To synthesize 4-bromostyrene starting from ethyl benzene, we must carry out ring functionalization prior to developing the side-chain double bond:

Ring Bromination: Treatment of ethylbenzene with $Br₂$ $\text{Br}_2$ in the presence of $Fe$ $\text{Fe}$ (or $FeBr₃$ $\text{FeBr}_3$) acts via electrophilic aromatic substitution. The ethyl group is an ortho/para director, yielding 1-bromo-4-ethylbenzene as the major product owing to steric mitigation.
Side-Chain Halogenation: Free radical substitution with $Cl₂$ $\text{Cl}_2$ under thermal conditions ( $Δ$ $\Delta$) or UV light specifically chlorinates the benzylic position because the benzylic radical is exceptionally stable via resonance.
Elimination: Heating with alcoholic $KOH$ $\text{KOH}$ drives an $E2$ $E2$ elimination of the benzylic chloride, cleanly synthesizing the terminal alkene linkage of the styrene system.
The figure illustrates the multi-step conversion starting from ethylbenzene to yield a brominated styrene derivative.

Pattern Recognition

If you perform side-chain halogenation/alkene generation first, the ring substitution later would lack para-selectivity control and risk reacting across the alkene path. Hence, ring substitution MUST precede double bond creation.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes Class 11 Chemistry: Hydrocarbons

More Haloalkanes and Haloarenes Previous-Year Questions — Page 5

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

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).

. Statement (II): Boiling point of

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 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

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:

Physical Properties of Isomers vector 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).

Physical Properties of Isomers vector 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).

Physical Properties of Isomers vector 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).

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 :

Nucleophilic Substitution Reactions diagram for Q43 - JEE Main 2025 Evening — 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:

Nucleophilic Substitution Reactions solution diagram for Q43 - JEE Main 2025 Evening — 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:

Dihaloalkane starting reactant structure for Q38 — 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.

Elimination mechanism steps for conjugated diene synthesis — 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

Q34 jee_main_2025_28_jan_evening Nucleophilic Substitution Reactions

The product B formed in the following reaction sequence is :

Reaction sequence diagram for Q34 - JEE Main 2025 — 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=CH₂ + HCl arrow R-CHCl-CH₃

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).

Detailed mechanism step for reaction sequence of Q34 — 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 ( $R-NC$ $R-NC$)

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes

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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%)