Given below are two statements: Statement (I): is more polar than . Statement (II): Boiling point of 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:

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, cancelling out completely: Since mumeta > 0, the para-isomer is *less* polar than the meta-isomer. This directly falsifies Statement II. The geometric configurations map out as follows: Hence, Statement I is correct, but Statement II is incorrect. 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). They can never be more polar than any asymmetric ortho or meta structural isomer. Class 12 Chemistry: Haloalkanes and Haloarenes

Given below are two statements: Statement (I): is more polar than . Statement (II): Boiling point of is lower than the ortho-isomer, but it is more polar than the meta-isomer. In the light of the abov

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, cancelli

Given below are two statements: Statement (I): is more polar than . Statement (II): Boiling point of 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:

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, cancelling out completely: Since mumeta > 0, the para-isomer is *less* polar than the meta-isomer. This directly falsifies Statement II. The geometric configurations map out as follows: Hence, Statement I is correct, but Statement II is incorrect. 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). They can never be more polar than any asymmetric ortho or meta structural isomer. Class 12 Chemistry: Haloalkanes and Haloarenes

Given below are two statements: Statement (I): is more polar than . Statement (II): Boiling point of is lower than the ortho-isomer, but it is more polar than the meta-isomer. In the light of the abov

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, cancelli

Given below are two statements: Statement (I): is more polar than . Statement (II): Boiling point of is lower than the…

Year	2026	2025	2024	Total
Questions	11	16	13	40

Q jee_main_2025_08_april_evening Preparation and Reactions of Styrene derivatives

Choose the correct set of reagents for the following conversion:

Ethyl benzene 4-bromostyrene

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

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

Q30 jee_main_2025_29_jan_evening Nucleophilic Substitution Mechanisms

Which among the following halides will generate the most stable carbocation in Nucleophilic substitution reaction?

A. Allylic halide option (1)
B. Secondary halide option (2)
C. Secondary benzylic halide option (3)
D. Triphenylmethyl halide option (4)

Solution

Core Logic

The mechanism of $SN1$ $S_N1$ substitution proceeds via carbocation intermediate formation. Option (4) gives a triphenylmethyl carbocation ( $Ph₃C⁺$ $Ph_{3}C^{+}$), which is exceptionally stable due to extensive delocalization of positive charge across three phenyl rings (resonance stabilization via 9 canonical structures).

Nucleophilic Substitution Mechanisms diagram for Q30 - JEE Main 2025 Evening

Step 1: Stability Comparison

Stability sequence:

Ph₃C⁺ > benzylic > allylic > alkyl carbocations

Pattern Recognition

Look for maximum phenyl groups attached directly to the carbon bearing the leaving group to maximize resonance contribution.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes

Q29 jee_main_2025_28_jan_morning Alkaline Hydrolysis and NGP

Given below are two statements : Statement I:

Et₂N-CH₂-CH₂-Cl

will undergo alkaline hydrolysis at a faster rate than

Et₂CH-CH₂-Cl

. Statement II: In

Et₂N-CH₂-CH₂-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. $Both Statement I and Statement II are incorrect$
B. $Statement I is incorrect but statement II is correct$
C. $Both Statement I and Statement II are correct$
D. $Statement I is correct but Statement II is incorrect$

Solution

Core Logic

Statement I is correct because the nitrogen atom contains a lone pair situated at the $β$ $\beta$-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 $β$ $\beta$ to a leaving group. Shortcut: NGP (Neighboring Group Participation) accelerates substitution dramatically via intramolecular assistance.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes

Q39 jee_main_2025_28_jan_morning Ambident Nucleophiles Reactions

The products A and B in the following reactions, respectively are

A A g - N O _ 2 C H _ 3 - C H _ 2 - C H _ 2 - B r A g C N B

A. $CH₃ - CH₂ - CH₂ - ONO, CH₃ - CH₂ - CH₂ - NC$
B. $CH₃-CH₂-CH₂-ONO, CH₃-CH₂-CH₂-CN$
C. $CH₃ - CH₂ - CH₂ - NO₂, CH₃ - CH₂ - CH₂ - CN$
D. $CH₃ - CH₂ - CH₂ - NO₂, CH₃ - CH₂ - CH₂ - NC$

Solution

Core Logic

Both silver reagents exhibit significantly covalent bond characters:

Reaction with $AgNO₂$ $\mathrm{AgNO}_2$: 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:

A = CH₃-CH₂-CH₂-NO₂

Reaction with $AgCN$ $\mathrm{AgCN}$: The covalent $Ag-C$ $\mathrm{Ag}-\mathrm{C}$ bond directs the nucleophilic attack to proceed through the lone pair on nitrogen, yielding an isocyanide compound:

B = CH₃-CH₂-CH₂-NC

Hence, option (4) represents the correct combination.

Pattern Recognition

Sees: Alkyl halide reacting with covalent silver salts of ambident anions. Shortcut: Silver reagents ( $AgCN$ $\mathrm{AgCN}$ or $AgNO₂$ $\mathrm{AgNO}_2$) drive bond formatting via the nitrogen center, producing isocyanides and nitroalkanes respectively.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes

Q jee_main_2025_04_april_evening Substitution 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

β

-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

R-Cl + KOH(aq) arrow R-OH + KCl (SN Nucleophilic Substitution)

R-CH₂-CH₂-Cl + KOH(alc) arrow R-CH=CH₂ + KCl + H₂O (E2 Elimination)

Core Logic

Statement I is incorrect: Treatment of alkyl chlorides with aqueous $KOH$ $KOH$ yields alcohols via a nucleophilic substitution ( $SN$ $S_N$) 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 $β$ $\beta$-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 $β$ $\beta$-hydrogens).

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes

Solution & Explanation

Related Formula

Core Logic

Step 1: Spatial Alignments

Pattern Recognition

Chapter Mix

Reference Study Guides

More Haloalkanes and Haloarenes Previous-Year Questions — Page 4

Solution

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Stability Comparison

Pattern Recognition

Chapter Mix

Solution

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

More Haloalkanes and Haloarenes Questions — jee_main_2025_07_april_evening

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back