Match List-I with List-II: beginarray|l|l| hline beginarrayc textbfList-I \\ textbf(Reaction) endarray & beginarrayc textbfList-II \\ textbf(Name of reaction) endarray \\ hline text(A) quad text2Ar-X + text2Na xrightarrowtextDry Ether textAr-Ar + text2NaX & text(I) quad textLucas reaction \\ hline text(B) quad mathrmArN_2^+X^- xrightarrowmathrmCu / HCl mathrmArCl + mathrmN_2 uparrow + mathrmCuX & text(II) quad textFinkelstein reaction \\ hline text(C) quad mathrmC_2H_5Br + mathrmNaI xrightarrowtextDry Acetone mathrmC_2H_5I + mathrmNaBr & text(III) quad textFittig reaction \\ hline text(D) quad mathrmCH_3C(OH)(CH_3)CH_3 xrightarrowmathrmHCl / ZnCl_2 mathrmCH_3C(Cl)(CH_3)CH_3 & text(IV) quad textGatterman reaction \\ hline endarray Choose the correct answer from the options given below:

Solution & Explanation

### Related Formula textNamed 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 reaction** rightarrow **(III)**. - **Reaction (B)**: Conversion of benzene diazonium chloride to aryl halide using copper powder (mathrmCu) in halogen acids like mathrmHCl is the **Gatterman reaction** rightarrow **(IV)**. - **Reaction (C)**: Substitution of halogen in an alkyl halide with sodium iodide (mathrmNaI) in dry acetone solvent is the classic halogen exchange method called the **Finkelstein reaction** rightarrow **(II)**. - **Reaction (D)**: Replacement of the hydroxyl group in tertiary butyl alcohol with chlorine using conc. mathrmHCl in the presence of anhydrous mathrmZnCl_2 catalyst is the **Lucas reaction** 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. mathrmCu_2Cl_2) and the Gatterman reaction (which uses copper powder, mathrmCu). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes

More Haloalkanes and Haloarenes Previous-Year Questions — Page 6

Q67 jee_main_2024_31_jan_evening IUPAC Nomenclature of Haloalkanes
Identify structure of 2,3-dibromo-1-phenylpentane.
  • A. text(1) Structure A
  • B. text(2) Structure B
  • C. text(3) Structure C
  • D. text(4) Structure D

Solution

### Core Logic Decode the IUPAC name: 2,3-dibromo-1-phenylpentane. 1) Parent chain is pentane (5 carbon chain: C1-C2-C3-C4-C5). 2) Substituents: - Phenyl group at position 1. - Bromo groups at positions 2 and 3. Option (3) correctly displays a 5-carbon straight chain. The first carbon attaches to the benzene ring (phenyl group), and the second and third carbons each hold a bromine atom.
IUPAC Nomenclature of Haloalkanes diagram for Q67 - JEE Main 2024 Evening
IUPAC Nomenclature of Haloalkanes diagram for Q67 - JEE Main 2024 Evening
### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes
Q68 jee_main_2024_31_jan_morning Elimination and Addition Reactions
The product (C) in the below mentioned reaction is: CH_3-CH_2-CH_2-Br xrightarrow[Delta]KOH_(alc) A xrightarrow[Delta]HBr B xrightarrow[Delta]KOH_(aq) C
  • A. textPropan-1-ol
  • B. textPropene
  • C. textPropyne
  • D. textPropan-2-ol

Solution

### Step 1: Elimination to form Propene CH_3-CH_2-CH_2-Br xrightarrowtextKOH (alc), Delta CH_3-CH=CH_2 quad text(Compound A: Propene) ### Step 2: Electrophilic Addition of HBr Addition of HBr follows Markovnikov's rule: CH_3-CH=CH_2 + HBr xrightarrowDelta CH_3-CH(Br)-CH_3 quad text(Compound B: 2-Bromopropane) ### Step 3: Nucleophilic Substitution Reaction with aqueous KOH leads to S_N2/S_N1 substitution of Br^- with OH^-: CH_3-CH(Br)-CH_3 xrightarrowtextKOH (aq), Delta CH_3-CH(OH)-CH_3 quad text(Compound C: Propan-2-ol) ### Pattern Recognition Alc. KOH gives elimination (alkene). Aq. KOH gives substitution (alcohol). HBr on unsymmetrical alkene gives Markovnikov addition. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes
Q86 jee_main_2024_31_jan_morning Elimination and Substitution
CH_3CH_2Br + NaOH rightarrow textProduct A CH_3CH_2Br + NaOH / H_2O rightarrow textProduct B The total number of hydrogen atoms in product A and product B is
Numerical Answer. Answer: 10 to 10

Solution

### Core Logic Reaction 1: If the reagent is alcoholic NaOH (implied due to formation of a distinct product A to contrast B), elimination occurs: CH_3CH_2Br + NaOH (textalc) rightarrow CH_2=CH_2 text (Ethene) Hydrogen atoms in ethene (C_2H_4) = 4. Reaction 2: If the reagent is aqueous NaOH (NaOH / H_2O), nucleophilic substitution (S_N2) occurs: CH_3CH_2Br + NaOH (textaq) rightarrow CH_3CH_2OH text (Ethanol) Hydrogen atoms in ethanol (C_2H_6O) = 6. Total hydrogen atoms = 4 + 6 = 10. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes
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