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.Statement I is correct but statement II is incorrect$\text{Statement I is correct but statement II is incorrect}$
B.Statement I is incorrect but statement II is correct$\text{Statement I is incorrect but statement II is correct}$
C.Both statement I and statement II are incorrect$\text{Both statement I and statement II are incorrect}$
D.Both statement I and statement II are correct$\text{Both statement I and statement II are correct}$
Solution & Explanation
Related Formula
μnet = √(μ₁² + μ₂² + 2μ₁μ₂ θ)$$\mu{\text{net}} = \sqrt{\mu_1^2 + \mu_2^2 + 2\mu_1\mu_2\cos\theta} $$Boiling point ∝ Dipole-dipole interactions + Van 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°$180^{\circ}$, cancelling out completely:
μpara = 0$$\mu{\text{para}} = 0 $$
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:
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 (μ = 0$\mu = 0$). They can never be more polar than any asymmetric ortho or meta structural isomer.
Keywords:#dipole moment of dihalobenzene#JEE Main 2025 Evening Q42#boiling point trends isomers#para isomer symmetry cancellation#ortho-isomer#para-isomer#dipole moment structural field
More Haloalkanes and Haloarenes Previous-Year Questions — Page 3
D.(4) C₆H₅CH₃ [Br₂, Fe]Dark Ortho and para Bromotoluenes$(4)\text{ } C_6H_5CH_3 \xrightarrow[Br_2, Fe]{\text{Dark}} \text{Ortho and para Bromotoluenes}$
Solution
Core Logic
Analyzing each option:
(1) Alkyl substitution reaction with Br₂, hν$Br_2, h\nu$ (free radical substitution) targets the most stable free radical. The allylic or benzylic position is preferred. In the given structure Halogenation Reactions, the most stable radical is 3°$3^{\circ}$ allylic radical, leading to the major product Halogenation Reactions. The option portrays substitution at the terminal carbon, which is incorrect.
(2) Reaction involves diazotization followed by Sandmeyer reaction with Cu₂Br₂/HBr$Cu_2Br_2/HBr$Halogenation Reactions, converting aromatic amine to aryl bromide correctly.
(3) Free radical halogenation on toluene side-chain Halogenation Reactions selectively forms benzyl bromide. Correct.
(4) Electrophilic aromatic substitution of toluene with Br₂/Fe$Br_2/Fe$ in the dark correctly Halogenation Reactions produces ortho and para isomers. Correct.
Step 1: Final Conclusion
Reaction (1) is incorrectly represented as it gives a 1°$1^{\circ}$ radical product rather than the more stable 3°$3^{\circ}$ substituted major product.
Pattern Recognition
Free radical halogenation favors 3° > 2° > 1°$3^{\circ} > 2^{\circ} > 1^{\circ}$ substitution due to intermediate stability. Allylic and benzylic are even more favored. Always check if the halogen landed on the most substituted available carbon.
Chapter Mix
Class 12 Chemistry: Haloalkanes and Haloarenes
Qjee_main_2025_02_april_eveningChemical Reactions and Named Rules
Named Organic Transformations$$\text{Named Organic Transformations}$$
Core Logic
Let us systematically match each reaction in List-I to its standardized organic reaction name in List-II:
Reaction (A): Coupling of two aryl halides with sodium metal in dry ether to form biaryl is the classic Fittig reactionarrow$\rightarrow$(III).
Reaction (B): Conversion of benzene diazonium chloride to aryl halide using copper powder (Cu$\mathrm{Cu}$) in halogen acids like HCl$\mathrm{HCl}$ is the Gatterman reactionarrow$\rightarrow$(IV).
Reaction (C): Substitution of halogen in an alkyl halide with sodium iodide (NaI$\mathrm{NaI}$) in dry acetone solvent is the classic halogen exchange method called the Finkelstein reactionarrow$\rightarrow$(II).
Reaction (D): Replacement of the hydroxyl group in tertiary butyl alcohol with chlorine using conc. HCl$\mathrm{HCl}$ in the presence of anhydrous ZnCl₂$\mathrm{ZnCl_2}$ catalyst is the Lucas reactionarrow$\rightarrow$(I).
Step 1: Selection
Combining the selections, the correct sequence is:
(A)-(III), (B)-(IV), (C)-(II), (D)-(I)
This maps directly to option (2).
Pattern Recognition
Named reaction matching questions are very straightforward. Keep a clear distinction between the Sandmeyer reaction (which uses cuprous halide, e.g. Cu₂Cl₂$\mathrm{Cu_2Cl_2}$) and the Gatterman reaction (which uses copper powder, Cu$\mathrm{Cu}$).
Chapter Mix
Class 12 Chemistry: Haloalkanes and Haloarenes
Q48jee_main_2025_02_april_eveningElimination and Addition Reaction Sequences
Consider the following sequence of reactions:
CH₃-CH₂-CH₂-CH(Br)-CH₃ alcoholic KOH P (Major Product) Br₂ Q$$\mathrm{CH_3-CH_2-CH_2-CH(Br)-CH_3} \xrightarrow{\text{alcoholic KOH}} \mathrm{P} \text{ (Major Product)} \xrightarrow{\mathrm{Br_2}} \mathrm{Q}$$
Consider the above sequence of reactions. 151~g$151~\mathrm{g}$ of 2-bromopentane is made to react. Yield of major product P$\mathrm{P}$ is 80%$80\%$ whereas Q$\mathrm{Q}$ is 100%$100\%$.
Mass of product Q$\mathrm{Q}$ obtained is _______ g.
Given molar mass in g~mol⁻¹$\mathrm{g~mol^{-1}}$ H: 1, C: 12, O: 16, Br: 80
Step 1: 2-bromopentane undergoes dehydrohalogenation via an E2 mechanism using alcoholic KOH$\mathrm{KOH}$. According to Saytzeff's rule, the more substituted alkene is the major product. Thus, pent-2-ene is the major product P$\mathrm{P}$.
Chemical reaction showing elimination of 2-bromopentane to form pent-2-ene
Step 2: Pent-2-ene undergoes electrophilic bromination with liquid bromine (Br₂$\mathrm{Br_2}$) to give 2,3-dibromopentane (product Q$\mathrm{Q}$):
Chemical reaction showing elimination of 2-bromopentane to form pent-2-ene
Step 1: Calculate Initial Moles of Reactant
Calculate the molar mass of 2-bromopentane (C₅H₁₁Br$\mathrm{C_5H_{11}Br}$):
Moles of P formed = 1 × 0.80 = 0.8~mol$$\text{Moles of P formed} = 1 \times 0.80 = 0.8~\mathrm{mol}$$
Since the conversion of P arrow Q$\mathrm{P} \rightarrow \mathrm{Q}$ has a yield of 100%$100\%$, the mole count remains stoichiometric:
Moles of Q formed = 0.8 × 1.00 = 0.8~mol$$\text{Moles of Q formed} = 0.8 \times 1.00 = 0.8~\mathrm{mol}$$
Step 3: Calculate Mass of Q
Product Q$\mathrm{Q}$ is 2,3-dibromopentane (C₅H₁₀Br₂$\mathrm{C_5H_{10}Br_2}$).
Calculate its molar mass:
Molar mass of Q = 5(12) + 10(1) + 2(80) = 60 + 10 + 160 = 230~ g~mol⁻¹$$\text{Molar mass of Q} = 5(12) + 10(1) + 2(80) = 60 + 10 + 160 = 230~\mathrm{g~mol^{-1}}$$Mass of Q = 0.8 × 230 = 184~g$$\text{Mass of Q} = 0.8 \times 230 = 184~\mathrm{g}$$
Pattern Recognition
Saytzeff vs Hofmann: Alcoholic KOH$\mathrm{KOH}$ is a small, non-bulky base, which selectively targets the internal secondary proton to yield the thermodynamic trans-alkene (pent-2-ene) as the major product rather than the terminal 1-alkene.
Chapter Mix
Class 12 Chemistry: Haloalkanes and Haloarenes
Qjee_main_2025_02_april_morningReactions of Alkyl Halides and Alkyne Hydration
An optically active alkyl halide C₄H₉Br$\mathrm{C_4H_9Br}$ [A] reacts with hot KOH dissolved in ethanol and forms alkene [B] as major product which reacts with bromine to give dibromide [C]. The compound [C] is converted into a gas [D] upon reacting with alcoholic NaNH₂$\mathrm{NaNH_2}$. During hydration 18 gram of water is added to 1 mole of gas [D] on warming with mercuric sulphate and dilute acid at 333K$333\mathrm{K}$ to form compound [E]. The IUPAC name of compound [E] is :
A.(1) But-2-yne$(1)\ \text{But-2-yne}$
B.(2) Butan-2-ol$(2)\ \text{Butan-2-ol}$
C.(3) Butan-2-one$(3)\ \text{Butan-2-one}$
D.(4) Butan-1-al$(4)\ \text{Butan-1-al}$
Solution
Related Formula
Dehydrohalogenation via alcoholic KOH follows E2 elimination mechanism:
Hydration of alkynes using HgSO₄/H₂SO₄$\mathrm{HgSO_4/H_2SO_4}$ yields ketones via keto-enol tautomerism.
Core Logic
Let's trace the full sequence line-by-row:
[A] is an optically active halide with formula C₄H₉Br arrow CH₃-CH(Br)-CH₂-CH₃$\mathrm{C_4H_9Br} \rightarrow \mathrm{CH_3-CH(Br)-CH_2-CH_3}$ (2-Bromobutane).
Reaction of [A] with hot ethanolic KOH produces [B] as the major product: CH₃-CH=CH-CH₃$\mathrm{CH_3-CH=CH-CH_3}$ (But-2-ene).
Treatment of [B] with Br₂$\mathrm{Br_2}$ yields a vicinal dibromide [C]: CH₃-CH(Br)-CH(Br)-CH₃$\mathrm{CH_3-CH(Br)-CH(Br)-CH_3}$ (2,3-Dibromobutane).
Reaction of [C] with alcoholic NaNH₂$\mathrm{NaNH_2}$ converts it via double dehydrohalogenation into gas [D]: CH₃-C≡ C-CH₃$\mathrm{CH_3-C\equiv C-CH_3}$ (But-2-yne).
Hydration of 1 mole of [D] with H₂O$\mathrm{H_2O}$ in the presence of Hg²⁺/H^+$\mathrm{Hg^{2+}/H^+}$ forms an enol intermediate that rapidly tautomerizes to compound [E]: CH₃-CO-CH₂-CH₃$\mathrm{CH_3-CO-CH_2-CH_3}$ (Butan-2-one).
Step 1: Visualization
Reaction roadmap step verification for Q27
Pattern Recognition
Whenever you see a 4-carbon chain undergoing terminal/internal dehydrohalogenation followed by hydration of the resulting alkyne, look closely at the configuration: symmetric or unsymmetric alkyne hydration both systematically lead to Butan-2-one because a stable ketone cannot form on position 1 via standard Kucherov hydration of an internal chain.
Chapter Mix
Class 12 Physics: Haloalkanes and Haloarenes
Class 11 Chemistry: Hydrocarbons
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
The reactions which cannot be applied to prepare an alkene by elimination, are:
A. Four different reaction schemes illustrating elimination, substitution, oxidation, and dehydrogenation conditions.
B. CH₃ - CH₂ - |BrCH - CH₃ KOH (aq.)$\mathrm{CH_3 - CH_2 - \underset{\overset{|}{\mathrm{Br}}}{\mathrm{CH}} - CH_3 \xrightarrow{\mathrm{KOH\ (aq.)}}}$
C. Four different reaction schemes illustrating elimination, substitution, oxidation, and dehydrogenation conditions.
D. Four different reaction schemes illustrating elimination, substitution, oxidation, and dehydrogenation conditions.
E. Four different reaction schemes illustrating elimination, substitution, oxidation, and dehydrogenation conditions.
Choose the correct answer from the options given below:
A.B & E Only$\text{B \& E Only}$
B.B, C & D Only$\text{B, C \& D Only}$
C.A, C & D Only$\text{A, C \& D Only}$
D.B & D Only$\text{B \& D Only}$
Solution
Core Logic
{{SOLUTION_IMG}}
Option (B) and (D) reaction are not able to form
alkene as a product.
Pattern Recognition
Aqueous KOH$\text{KOH}$ on alkyl halides favors substitution (forming alcohols) over elimination. Strong oxidizing mixtures like sodium dichromate oxidize secondary alcohols to ketones instead of dehydrating them.
Chapter Mix
Class 12 Chemistry: Organic Compounds Containing Halogens
Class 12 Chemistry: Alcohols, Phenols and Ethers
More Haloalkanes and Haloarenes Questions — jee_main_2025_07_april_evening
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
 "ortho-isomer, para-isomer, dipole moment structural field")
 "ortho-isomer, para-isomer, dipole moment structural field")
 "ortho-isomer, para-isomer, dipole moment structural field")
