Choose the correct set of reagents for the following conversion: QIMG1

Q: Choose the correct set of reagents for the following conversion: QIMG1

### Core Logic To synthesize 4-bromostyrene starting from ethyl benzene, we must carry out ring functionalization prior to developing the side-chain double bond: 1. **Ring Bromination**: Treatment of ethylbenzene with Br2 in the presence of Fe (or FeBr3) 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. 2. **Side-Chain Halogenation**: Free radical substitution with Cl2 under th

Year	2026	2025	2024	Total
Questions	11	15	12	38

Q jee_main_2024_30_january_evening Stereochemistry of Halogenation

2-chlorobutane + Cl₂ arrow C₄H₈Cl₂ (isomers)

Total number of optically active isomers shown by

C₄H₈Cl₂

, 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: $CH₃-CH(Cl)-CH₂-CH₃$ $CH_3-CH(Cl)-CH_2-CH_3$ (exists as 2 enantiomers: $d$ $d$ and $l$ $l$) Chlorination can occur at 4 different carbons:

At C1: $CH₂(Cl)-CH(Cl)-CH₂-CH₃$ $CH_2(Cl)-CH(Cl)-CH_2-CH_3$ (1,2-dichlorobutane) $arrow$ $\rightarrow$ Two chiral centers, unsymmetrical. Forms 4 optically active isomers (2 pairs of enantiomers).
At C2: $CH₃-C(Cl)₂-CH₂-CH₃$ $CH_3-C(Cl)_2-CH_2-CH_3$ (2,2-dichlorobutane) $arrow$ $\rightarrow$ No chiral center. Achiral (0 optically active).
At C3: $CH₃-CH(Cl)-CH(Cl)-CH₃$ $CH_3-CH(Cl)-CH(Cl)-CH_3$ (2,3-dichlorobutane) $arrow$ $\rightarrow$ Symmetrical with 2 chiral centers. Forms 3 stereoisomers: 1 meso (achiral) and 2 optically active (1 enantiomeric pair).
At C4: $CH₃-CH(Cl)-CH₂-CH₂(Cl)$ $CH_3-CH(Cl)-CH_2-CH_2(Cl)$ (1,3-dichlorobutane, numbering from other end) $arrow$ $\rightarrow$ 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.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes Class 11 Chemistry: Organic Chemistry Some Basic Principles and Techniques

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

SN2

mechanism. Statement II: A secondary alkyl halide when treated with a large excess of ethanol follows

SN1

mechanism. In the light of the above statements, choose the most appropriate from the questions given below:

A. $Statement I is true but Statement II is false.$
B. $Statement I is false but Statement II is true.$
C. $Both statement I and Statement II are false.$
D. $Both statement I and Statement II are true.$

Solution

Core Logic

Statement I: Rate of $SN2 \propto [R-X][Nu^-]$ $S_N2 \propto [R-X][Nu^-]$. Therefore, $SN2$ $S_N2$ 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 $SN2$ $S_N2$ 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 $SN1$ $S_N1$ 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 ( $SN2$ $S_N2$). Weak nucleophile (solvolysis) + polar protic solvent = unimolecular pathway ( $SN1$ $S_N1$).

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes

Q 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 $sp²$ $sp^2$ 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

sp³

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

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):

CH₂=CH-CH₂-Cl

is an example of allyl halide Reason (R): Allyl halides are the compounds in which the halogen atom is attached to

sp²

hybridised carbon atom. In the light of the two above statements, choose the most appropriate answer from the options given below:

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

Solution

Core Logic

Assertion (A): $CH₂=CH-CH₂-Cl$ $CH_2=CH-CH_2-Cl$ 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 $sp²$ $sp^2$ hybridized carbon atom. This statement is False. In allyl halides, the halogen is attached to an $sp³$ $sp^3$ hybridized carbon atom which is next to an $sp²$ $sp^2$ hybridized carbon (C=C double bond).

Step 1: Conclusion

Therefore, (A) is true but (R) is false.

Pattern Recognition

Allylic = $sp³$ $sp^3$ C adjacent to C=C. Vinylic = $sp²$ $sp^2$ C of the C=C itself.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes

Q 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.
B.
C.
D.

Solution

Core Logic

When bromobenzene reacts with concentrated $HNO₃$ $HNO_3$ (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).
When 1-bromo-2,4,6-trinitrobenzene (Compound A) is treated with $NaOH$ $NaOH$, the presence of three strong electron-withdrawing $-NO₂$ $-NO_2$ groups activates the aromatic ring toward Nucleophilic Aromatic Substitution ( $SNAr$ $S_NAr$). The $-Br$ $-Br$ is easily replaced by $-OH$ $-OH$ to form 2,4,6-trinitrophenol (picric acid).
Subsequent acidification with $HCl$ $HCl$ yields the neutral picric acid (Compound B).

Pattern Recognition

Multiple $NO₂$ $NO_2$ groups drastically increase the susceptibility of halobenzenes to $SNAr$ $S_NAr$. Bromine is replaced completely by $OH^-$ $OH^-$ under alkaline conditions.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes Class 12 Chemistry: Alcohols, Phenols and Ethers

Choose the correct set of reagents for the following conversion: $Ethyl benzene 4-bromostyrene$ {{Q_IMG1}}

Solution & Explanation

Core Logic

Pattern Recognition

Chapter Mix

More Haloalkanes and Haloarenes Previous-Year Questions — Page 7

Solution

Core Logic

Step 1: Sum the Optically Active Isomers

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Identifying the functional groups

Chapter Mix

Solution

Core Logic

Step 1: Conclusion

Pattern Recognition

Chapter Mix

Solution

Core Logic

Pattern Recognition

Chapter Mix

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

Choose the correct set of reagents for the following conversion: Ethyl benzene 4-bromostyrene$$\text{Ethyl benzene} \longrightarrow \text{4-bromostyrene}$$ {{Q_IMG1}}

Solution & Explanation

Core Logic

Pattern Recognition

Chapter Mix

More Haloalkanes and Haloarenes Previous-Year Questions — Page 7

Solution

Core Logic

Step 1: Sum the Optically Active Isomers

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Identifying the functional groups

Chapter Mix

Solution

Core Logic

Step 1: Conclusion

Pattern Recognition

Chapter Mix

Solution

Core Logic

Pattern Recognition

Chapter Mix

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

Choose the correct set of reagents for the following conversion: $Ethyl benzene 4-bromostyrene$ {{Q_IMG1}}