Match List-I with List-II. List-I (Conversion) List-II (Reagents, Conditions used) [cite: 248, 249] (A) Chlorobenzene ightarrow Phenol (I) Warm, H2O (B) p-Nitrochlorobenzene ightarrow p-Nitrophenol (II) (a) NaOH, 368 K; (b) H3O^+ (C) 2,4-Dinitrochlorobenzene ightarrow 2,4-Dinitrophenol (III) (a) NaOH, 443 K; (b) H3O^+ (D) 2,4,6-Trinitrochlorobenzene ightarrow 2,4,6-Trinitrophenol (IV) (a) NaOH, 623 K, 300 atm; (b) H3O^+ Choose the correct answer from the options given below:

Aryl halides are generally unreactive towards nucleophilic substitution due to resonance stabilization of the C-Cl bond. However, the presence of strong electron-withdrawing groups (-NO2) at ortho and para positions dramatically increases reactivity by stabilizing the intermediate carbanion: - (A) Chlorobenzene: Needs extreme conditions: NaOH at 623 K, 300 atm (Dow's Process) ightarrow (IV) - (B) p-Nitrochlorobenzene: One para -NO2 group softens required temperature to 443 K ightarrow (III) - (C) 2,4-Dinitrochlorobenzene: Two electron-withdrawing groups lower needed temperature further to 368 K ightarrow (II) - (D) 2,4,6-Trinitrochlorobenzene: Highly activated picryl chloride hydrolyzes smoothly with just warm water ightarrow (I) Matching sequences cleanly yields: (A)-(IV), (B)-(III), (C)-(II), (D)-(I). The more -NO2 groups present on the ring, the less aggressive the reagent/temperature setup required. Count -NO2 groups: 0 ightarrow 623K, 1 ightarrow 443K, 2 ightarrow 368K, 3 ightarrow warm water. Class 12 Chemistry: Haloalkanes and Haloarenes

Match List-I with List-II. List-I (Conversion) List-II (Reagents, Conditions used) [cite: 248, 249] (A) Chlorobenzene ightarrow Phenol (I) Warm, H2O (B) p-Nitrochlorobenzene ightarrow p-Nitrophenol (I

Aryl halides are generally unreactive towards nucleophilic substitution due to resonance stabilization of the C-Cl bond. However, the presence of strong electron-withdrawing groups (-NO2) at ortho and para positions dramatically increases reactivity by stabilizing the intermediate carbanion: - (A) Chlorobenzene: Needs extreme conditions: NaOH at 623 K, 300 atm (Dow's Process) ightarrow (IV) - (B) p-Nitrochlorobenzene: One para -NO2 group softens required temperature to 443 K ightarrow (III) - (C

Match List-I with List-II. List-I (Conversion) List-II (Reagents, Conditions used) [cite: 248, 249] (A) Chlorobenzene ightarrow Phenol (I) Warm, H2O (B) p-Nitrochlorobenzene ightarrow p-Nitrophenol (II) (a) NaOH, 368 K; (b) H3O^+ (C) 2,4-Dinitrochlorobenzene ightarrow 2,4-Dinitrophenol (III) (a) NaOH, 443 K; (b) H3O^+ (D) 2,4,6-Trinitrochlorobenzene ightarrow 2,4,6-Trinitrophenol (IV) (a) NaOH, 623 K, 300 atm; (b) H3O^+ Choose the correct answer from the options given below:

Aryl halides are generally unreactive towards nucleophilic substitution due to resonance stabilization of the C-Cl bond. However, the presence of strong electron-withdrawing groups (-NO2) at ortho and para positions dramatically increases reactivity by stabilizing the intermediate carbanion: - (A) Chlorobenzene: Needs extreme conditions: NaOH at 623 K, 300 atm (Dow's Process) ightarrow (IV) - (B) p-Nitrochlorobenzene: One para -NO2 group softens required temperature to 443 K ightarrow (III) - (C) 2,4-Dinitrochlorobenzene: Two electron-withdrawing groups lower needed temperature further to 368 K ightarrow (II) - (D) 2,4,6-Trinitrochlorobenzene: Highly activated picryl chloride hydrolyzes smoothly with just warm water ightarrow (I) Matching sequences cleanly yields: (A)-(IV), (B)-(III), (C)-(II), (D)-(I). The more -NO2 groups present on the ring, the less aggressive the reagent/temperature setup required. Count -NO2 groups: 0 ightarrow 623K, 1 ightarrow 443K, 2 ightarrow 368K, 3 ightarrow warm water. Class 12 Chemistry: Haloalkanes and Haloarenes

Match List-I with List-II. List-I (Conversion) List-II (Reagents, Conditions used) [cite: 248, 249] (A) Chlorobenzene ightarrow Phenol (I) Warm, H2O (B) p-Nitrochlorobenzene ightarrow p-Nitrophenol (I

Aryl halides are generally unreactive towards nucleophilic substitution due to resonance stabilization of the C-Cl bond. However, the presence of strong electron-withdrawing groups (-NO2) at ortho and para positions dramatically increases reactivity by stabilizing the intermediate carbanion: - (A) Chlorobenzene: Needs extreme conditions: NaOH at 623 K, 300 atm (Dow's Process) ightarrow (IV) - (B) p-Nitrochlorobenzene: One para -NO2 group softens required temperature to 443 K ightarrow (III) - (C

Match List-I with List-II. List-I (Conversion) List-II (Reagents, Conditions used) [cite: 248, 249] (A) Chlorobenzene…

List-I (Conversion)	List-II (Reagents, Conditions used) [cite: 248, 249]
(A) Chlorobenzene $ightarrow$ $ ightarrow$ Phenol	(I) $textWarm, textH_2textO$ $\text{Warm, } \text{H}_2\text{O}$
(B) p-Nitrochlorobenzene $ightarrow$ $ ightarrow$ p-Nitrophenol	(II) (a) $textNaOH, 368text K$ $\text{NaOH, } 368\text{ K}$; (b) $textH3textO^+$ $\text{H}3\text{O}^+$	(C) 2,4-Dinitrochlorobenzene $ightarrow$ $ ightarrow$ 2,4-Dinitrophenol	(III) (a) $textNaOH, 443text K$ $\text{NaOH, } 443\text{ K}$; (b) $textH_3textO^+$ $\text{H}_3\text{O}^+$	(D) 2,4,6-Trinitrochlorobenzene $ightarrow$ $ ightarrow$ 2,4,6-Trinitrophenol	(IV) (a) $textNaOH, 623text K, 300text atm$ ; (b) $textH_3textO^+$ $\text{H}_3\text{O}^+$ Choose the correct answer from the options given below: A. $text(A)-(II), (B)-(III), (C)-(I), (D)-(IV)$ B. $text(A)-(III), (B)-(IV), (C)-(II), (D)-(I)$ C. $text(A)-(IV), (B)-(III), (C)-(II), (D)-(I)$ D. $text(A)-(IV), (B)-(III), (C)-(I), (D)-(II)$ Solution & Explanation ### Related Formula $textRate of S_NtextAr propto textNumber of electron-withdrawing groups (-I, -M) at ortho/para positions$ ### Core Logic Aryl halides are generally unreactive towards nucleophilic substitution due to resonance stabilization of the $textC-Cl$ $\text{C-Cl}$ bond. However, the presence of strong electron-withdrawing groups ( $-textNO_2$ $-\text{NO}_2$) at ortho and para positions dramatically increases reactivity by stabilizing the intermediate carbanion: - (A) Chlorobenzene: Needs extreme conditions: $textNaOH at 623text K, 300text atm$ (Dow's Process) $ightarrow$ $ ightarrow$ (IV) - (B) p-Nitrochlorobenzene: One para $-textNO_2$ $-\text{NO}_2$ group softens required temperature to $443text K$ $443\text{ K}$ $ightarrow$ $ ightarrow$ (III) - (C) 2,4-Dinitrochlorobenzene: Two electron-withdrawing groups lower needed temperature further to $368text K$ $368\text{ K}$ $ightarrow$ $ ightarrow$ (II) - (D) 2,4,6-Trinitrochlorobenzene: Highly activated picryl chloride hydrolyzes smoothly with just warm water $ightarrow$ $ ightarrow$ (I) ### Step 1: Final Match Alignment Matching sequences cleanly yields: (A)-(IV), (B)-(III), (C)-(II), (D)-(I). ### Pattern Recognition The more $-textNO_2$ $-\text{NO}_2$ groups present on the ring, the less aggressive the reagent/temperature setup required. Count $-textNO_2$ $-\text{NO}_2$ groups: $0 ightarrow 623textK$ $0 ightarrow 623\text{K}$, $1 ightarrow 443textK$ $1 ightarrow 443\text{K}$, $2 ightarrow 368textK$ $2 ightarrow 368\text{K}$, $3 ightarrow textwarm water$ $3 ightarrow \text{warm water}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes Reference Study Guides 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 $S_N2$ $S_N2$ mechanism. Statement II: A secondary alkyl halide when treated with a large excess of ethanol follows $S_N1$ $S_N1$ 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 $S_N2 propto [R-X][Nu^-]$ $S_N2 \propto [R-X][Nu^-]$. Therefore, $S_N2$ $S_N2$ 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 $S_N2$ $S_N2$ 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 $S_N1$ $S_N1$ 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 ( $S_N2$ $S_N2$). Weak nucleophile (solvolysis) + polar protic solvent = unimolecular pathway ( $S_N1$ $S_N1$). ### 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 $C_4H_8Cl_2$ $C_4H_8Cl_2$, 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: $CH_3-CH(Cl)-CH_2-CH_3$ $CH_3-CH(Cl)-CH_2-CH_3$ (exists as 2 enantiomers: $d$ $d$ and $l$ $l$) Chlorination can occur at 4 different carbons: 1. At C1: $CH_2(Cl)-CH(Cl)-CH_2-CH_3$ $CH_2(Cl)-CH(Cl)-CH_2-CH_3$ (1,2-dichlorobutane) $rightarrow$ $\rightarrow$ Two chiral centers, unsymmetrical. Forms 4 optically active isomers (2 pairs of enantiomers). 2. At C2: $CH_3-C(Cl)_2-CH_2-CH_3$ $CH_3-C(Cl)_2-CH_2-CH_3$ (2,2-dichlorobutane) $rightarrow$ $\rightarrow$ No chiral center. Achiral (0 optically active). 3. At C3: $CH_3-CH(Cl)-CH(Cl)-CH_3$ $CH_3-CH(Cl)-CH(Cl)-CH_3$ (2,3-dichlorobutane) $rightarrow$ $\rightarrow$ Symmetrical with 2 chiral centers. Forms 3 stereoisomers: 1 meso (achiral) and 2 optically active (1 enantiomeric pair). 4. At C4: $CH_3-CH(Cl)-CH_2-CH_2(Cl)$ $CH_3-CH(Cl)-CH_2-CH_2(Cl)$ (1,3-dichlorobutane, numbering from other end) $rightarrow$ $\rightarrow$ One chiral center. Forms 2 optically active isomers (1 enantiomeric pair). 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 $sp^2$ $sp^2$ 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 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 Options 3 & 4: The halogen is attached to an $sp^3$ $sp^3$ hybridized carbon adjacent to a C=C double bond. These are allylic halides. 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): $CH_2=CH-CH_2-Cl$ $CH_2=CH-CH_2-Cl$ is an example of allyl halide Reason (R): Allyl halides are the compounds in which the halogen atom is attached to $sp^2$ $sp^2$ hybridised carbon atom. In the light of the two above statements, choose the most appropriate answer from the options given below: A. $text(A) is true but (R) is false$ $\text{(A) is true but (R) is false}$ B. $textBoth (A) and (R) are true but (R) is not the correct explanation of (A)$ C. $text(A) is false but (R) is true$ $\text{(A) is false but (R) is true}$ D. $textBoth (A) and (R) are true and (R) is the correct explanation of (A)$ Solution ### Core Logic Assertion (A): $CH_2=CH-CH_2-Cl$ $CH_2=CH-CH_2-Cl$ 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 $sp^2$ $sp^2$ hybridized carbon atom. This statement is False. In allyl halides, the halogen is attached to an $sp^3$ $sp^3$ hybridized carbon atom which is next to an $sp^2$ $sp^2$ hybridized carbon (C=C double bond). ### Step 1: Conclusion Therefore, (A) is true but (R) is false. ### Pattern Recognition Allylic = $sp^3$ $sp^3$ C adjacent to C=C. Vinylic = $sp^2$ $sp^2$ 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. 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 $HNO_3$ $HNO_3$ (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 $NaOH$ $NaOH$, the presence of three strong electron-withdrawing $-NO_2$ $-NO_2$ groups activates the aromatic ring toward Nucleophilic Aromatic Substitution ( $S_NAr$ $S_NAr$). The $-Br$ $-Br$ is easily replaced by $-OH$ $-OH$ to form 2,4,6-trinitrophenol (picric acid). 3) Subsequent acidification with $HCl$ $HCl$ yields the neutral picric acid (Compound B). The image shows a reaction scheme starting from bromobenzene undergoing nitration followed by substitution. ### Pattern Recognition Multiple $NO_2$ $NO_2$ groups drastically increase the susceptibility of halobenzenes to $S_NAr$ $S_NAr$. Bromine is replaced completely by $OH^-$ $OH^-$ under alkaline conditions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes Class 12 Chemistry: Alcohols, Phenols and Ethers ← Previous Page 5 of 6 Next → More Haloalkanes and Haloarenes Questions — jee_main_2025_07_april_evening Practice all Haloalkanes and Haloarenes previous-year questions → The major product of the following reaction is: Given below are two statements : Statement I: will undergo Which among the following halides will generate the most stable The product B formed in the following reaction sequence is The products A and B in the following reactions, respectively Given below are two statements: Statement (I): is more polar The structure of the major product formed in the following Given below are two statements : Statement-I: The conversion proceeds Consider the following sequence of reactions: Consider the above sequence A hydrocarbon ‘P’ ( ) on reaction with HCl gives Given below are two statements : one is labelled as Given below are two statements: Statement I: High concentration of Identify A and B in the following reaction sequence. Identify structure of 2,3-dibromo-1-phenylpentane. The product (C) in the below mentioned reaction is: Example of vinylic halide is Given below are two statements : one is labelled as Given below are two statement one is labeled as Assertion The correct statement regarding nucleophilic substitution reaction in a chiral Number of optical isomers possible for 2-chlorobutane The total number of hydrogen atoms in product A and Total number of optically active isomers shown by , obtained SYSTEM Syllabus Analysis & Trend Mapping Dashboard Chapter Trends Formula Recognition Questions Feed Weightage Rankbit System Rankbit Dashboard Chapter Trends Formula Recognition Questions Feed Weightage 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%) Rankbit The ultimate ecosystem for IIT JEE and COMEDK. 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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

### 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

Option 2: The halogen is attached to an aromatic ring directly. This is an aryl halide.

Options 3 & 4: The halogen is attached to an

sp^3

hybridized carbon adjacent to a C=C double bond. These are allylic halides.

### 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. $text(A) is true but (R) is false$ $\text{(A) is true but (R) is false}$
B. $textBoth (A) and (R) are true but (R) is not the correct explanation of (A)$
C. $text(A) is false but (R) is true$ $\text{(A) is false but (R) is true}$
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).

### 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

Solution & Explanation

Reference Study Guides

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