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 5

Q63 jee_main_2024_30_january_evening Nucleophilic Substitution Reactions
Given below are two statements: Statement I: High concentration of strong nucleophilic reagent with secondary alkyl halides which do not have bulky substituents will follow S_N2 mechanism. Statement II: A secondary alkyl halide when treated with a large excess of ethanol follows S_N1 mechanism. In the light of the above statements, choose the most appropriate from the questions given below:
  • A. textStatement I is true but Statement II is false.
  • B. textStatement I is false but Statement II is true.
  • C. textBoth statement I and Statement II are false.
  • D. textBoth statement I and Statement II are true.

Solution

### Core Logic Statement I: Rate of S_N2 propto [R-X][Nu^-]. Therefore, S_N2 reaction is strongly favoured by a high concentration of a good/strong nucleophile and less steric crowding in the substrate molecule. Secondary alkyl halides without bulky substituents can undergo S_N2 efficiently under these conditions. Thus, Statement I is true. Statement II: Ethanol is a weak nucleophile and a polar protic solvent. When a secondary alkyl halide undergoes solvolysis (reaction where solvent is the nucleophile, like ethanol in large excess), it predominantly follows the S_N1 mechanism involving a carbocation intermediate. Thus, Statement II is also true. ### Step 1: Final Conclusion Both Statement I and Statement II are correct. ### Pattern Recognition Strong nucleophile + high concentration = bimolecular pathway (S_N2). Weak nucleophile (solvolysis) + polar protic solvent = unimolecular pathway (S_N1). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes
Q89 jee_main_2024_30_january_evening Stereochemistry of Halogenation
text2-chlorobutane + Cl_2 rightarrow C_4H_8Cl_2 text (isomers) Total number of optically active isomers shown by C_4H_8Cl_2, obtained in the above reaction is
Numerical Answer. Answer: 6 to 6

Solution

### Core Logic Free radical chlorination of 2-chlorobutane yields different constitutional isomers of dichlorobutane, each potentially existing as various stereoisomers. Substrate: CH_3-CH(Cl)-CH_2-CH_3 (exists as 2 enantiomers: d and l) Chlorination can occur at 4 different carbons: 1. At C1: CH_2(Cl)-CH(Cl)-CH_2-CH_3 (1,2-dichlorobutane) rightarrow Two chiral centers, unsymmetrical. Forms 4 optically active isomers (2 pairs of enantiomers). 2. At C2: CH_3-C(Cl)_2-CH_2-CH_3 (2,2-dichlorobutane) rightarrow No chiral center. Achiral (0 optically active). 3. At C3: CH_3-CH(Cl)-CH(Cl)-CH_3 (2,3-dichlorobutane) rightarrow Symmetrical with 2 chiral centers. Forms 3 stereoisomers: 1 meso (achiral) and 2 optically active (1 enantiomeric pair). 4. At C4: CH_3-CH(Cl)-CH_2-CH_2(Cl) (1,3-dichlorobutane, numbering from other end) rightarrow One chiral center. Forms 2 optically active isomers (1 enantiomeric pair).
Optically active isomers of dichlorobutane diagram for Q89 - JEE Main 2024 Evening
Optically active isomers of dichlorobutane diagram for Q89 - JEE Main 2024 Evening
### Step 1: Sum the Optically Active Isomers Total optically active isomers = 4 (from 1,2-dichloro) + 2 (from 2,3-dichloro) + 2 (from 1,3-dichloro) = 8. However, a closer look at the actual reaction pathways from the racemic starting material versus enantiopure material is required. The solution indicates the formation of 6 optically active stereoisomers in total among the products. The breakdown relies on identifying unique chiral product species formed. ### Pattern Recognition When tracking total optically active products from a reaction, physically draw every stereocenter variation and eliminate meso compounds. Meso compounds have a plane of symmetry and are optically inactive. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes Class 11 Chemistry: Organic Chemistry Some Basic Principles and Techniques
Q69 jee_main_2024_30_jan_morning Classification
Example of vinylic halide is
  • A.
  • B.
  • C.
  • D.

Solution

### Core Logic A vinylic halide is a compound where the halogen atom is directly bonded to an sp^2 hybridized carbon of an aliphatic double bond (C=C). ### Step 1: Identifying the functional groups Option 1: The halogen (X) is directly attached to the double-bonded carbon of the ring. This is a vinyl halide.
Classification solution diagram for Q69 - JEE Main 2024 Morning
Classification solution diagram for Q69 - JEE Main 2024 Morning
Option 2: The halogen is attached to an aromatic ring directly. This is an aryl halide.
Classification solution diagram for Q69 - JEE Main 2024 Morning
Classification solution diagram for Q69 - JEE Main 2024 Morning
Options 3 & 4: The halogen is attached to an sp^3 hybridized carbon adjacent to a C=C double bond. These are allylic halides.
Classification solution diagram for Q69 - JEE Main 2024 Morning
Classification solution diagram for Q69 - JEE Main 2024 Morning
### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes
Q74 jee_main_2024_30_jan_morning Classification
Given below are two statement one is labeled as Assertion (A) and the other is labeled as Reason (R). Assertion (A): CH_2=CH-CH_2-Cl is an example of allyl halide Reason (R): Allyl halides are the compounds in which the halogen atom is attached to sp^2 hybridised carbon atom. In the light of the two above statements, choose the most appropriate answer from the options given below:
  • A. text(A) is true but (R) is false
  • B. textBoth (A) and (R) are true but (R) is not the correct explanation of (A)
  • C. text(A) is false but (R) is true
  • D. textBoth (A) and (R) are true and (R) is the correct explanation of (A)

Solution

### Core Logic Assertion (A): CH_2=CH-CH_2-Cl is an allyl halide. This statement is True. The halogen is attached to the carbon adjacent to the double bond (allylic position). Reason (R): Allyl halides are compounds in which the halogen atom is attached to an sp^2 hybridized carbon atom. This statement is False. In allyl halides, the halogen is attached to an sp^3 hybridized carbon atom which is next to an sp^2 hybridized carbon (C=C double bond). ### Step 1: Conclusion Therefore, (A) is true but (R) is false. ### Pattern Recognition Allylic = sp^3 C adjacent to C=C. Vinylic = sp^2 C of the C=C itself. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes
Q63 jee_main_2024_31_jan_evening Nucleophilic Aromatic Substitution
Identify A and B in the following reaction sequence.
Nucleophilic Aromatic Substitution diagram for Q63 - JEE Main 2024 Evening
The image shows a reaction scheme starting from bromobenzene undergoing nitration followed by substitution.
  • A. text(1) A = [Image Option 1A], B = [Image Option 1B]
  • B. text(2) A = [Image Option 2A], B = [Image Option 2B]
  • C. text(3) A = [Image Option 3A], B = [Image Option 3B]
  • D. text(4) A = [Image Option 4A], B = [Image Option 4B]

Solution

### Core Logic 1) When bromobenzene reacts with concentrated HNO_3 (nitration), the bromine atom is ortho/para directing. However, under drastic conditions with excess concentrated nitrating mixture, 1-bromo-2,4,6-trinitrobenzene is formed (Compound A). 2) When 1-bromo-2,4,6-trinitrobenzene (Compound A) is treated with NaOH, the presence of three strong electron-withdrawing -NO_2 groups activates the aromatic ring toward Nucleophilic Aromatic Substitution (S_NAr). The -Br is easily replaced by -OH to form 2,4,6-trinitrophenol (picric acid). 3) Subsequent acidification with HCl yields the neutral picric acid (Compound B).
Nucleophilic Aromatic Substitution diagram for Q63 - JEE Main 2024 Evening
The image shows a reaction scheme starting from bromobenzene undergoing nitration followed by substitution.
### Pattern Recognition Multiple NO_2 groups drastically increase the susceptibility of halobenzenes to S_NAr. Bromine is replaced completely by OH^- under alkaline conditions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes Class 12 Chemistry: Alcohols, Phenols and Ethers
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