The product B formed in the following reaction sequence is :

Year	2026	2025	2024	Total
Questions	11	16	13	40

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

Markovnikov addition of $HCl$ $HCl$ across an alkene:

R-CH=CH₂ + HCl arrow R-CHCl-CH₃

Nucleophilic substitution with $AgCN$ $AgCN$ favors coordinate carbon bonding over nitrogen, yielding covalent isocyanides ( $R-NC$ $R-NC$).

Core Logic

Step 1: The starting material contains a double bond. Treating it with $HCl$ $HCl$ leads to addition. According to Markovnikov's rule, the chloride ion attaches to the secondary position, creating chloride intermediate $[A]$ $[A]$.

Step 2: Compound $[A]$ $[A]$ reacts with $AgCN$ $AgCN$. Since $AgCN$ $AgCN$ is covalent, the lone pair on nitrogen acts as the attacking nucleophile, leading to substitution with an isocyanide group ( $-NC$ $-NC$) rather than a cyanide group ( $-CN$ $-CN$).

Step 1: Structural Synthesis

The intermediate $[A]$ $[A]$ possesses a chlorine atom at the secondary carbon position. Substituting this chlorine with $-NC$ $-NC$ provides the product corresponding to option (4).

Detailed mechanism step for reaction sequence of Q34 — The image outlines an addition reaction followed by substitution using silver cyanide.

Pattern Recognition

Distinguish between ionic vs covalent cyanide sources:

$KCN / NaCN$ $KCN / NaCN \implies$ forms alkyl nitriles ( $R-CN$ $R-CN$)
$AgCN$ $AgCN \implies$ forms alkyl isocyanides ( $R-NC$ $R-NC$)

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes

More Haloalkanes and Haloarenes Previous-Year Questions — Page 8

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.

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

Q jee_main_2024_31_jan_evening IUPAC Nomenclature of Haloalkanes

Identify structure of 2,3-dibromo-1-phenylpentane.

Solution

Core Logic

Decode the IUPAC name: 2,3-dibromo-1-phenylpentane.

Parent chain is pentane (5 carbon chain: C1-C2-C3-C4-C5).
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

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes

Q jee_main_2024_31_jan_morning Elimination and Addition Reactions

The product (C) in the below mentioned reaction is:

CH₃-CH₂-CH₂-Br [Δ]KOH(alc) A [Δ]HBr B [Δ]KOH(aq) C

A. $Propan-1-ol$ $\text{Propan-1-ol}$
B. $Propene$ $\text{Propene}$
C. $Propyne$ $\text{Propyne}$
D. $Propan-2-ol$ $\text{Propan-2-ol}$

Solution

Step 1: Elimination to form Propene

CH₃-CH₂-CH₂-Br KOH (alc), Δ CH₃-CH=CH₂ (Compound A: Propene)

Step 2: Electrophilic Addition of HBr

Addition of $HBr$ $HBr$ follows Markovnikov's rule:

CH₃-CH=CH₂ + HBr Δ CH₃-CH(Br)-CH₃ (Compound B: 2-Bromopropane)

Step 3: Nucleophilic Substitution

Reaction with aqueous KOH leads to $SN2/SN1$ $S_N2/S_N1$ substitution of $Br^-$ $Br^-$ with $OH^-$ $OH^-$:

CH₃-CH(Br)-CH₃ KOH (aq), Δ CH₃-CH(OH)-CH₃ (Compound C: Propan-2-ol)

Pattern Recognition

Alc. KOH gives elimination (alkene). Aq. KOH gives substitution (alcohol). $HBr$ $HBr$ on unsymmetrical alkene gives Markovnikov addition.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes

Q86 jee_main_2024_31_jan_morning Elimination and Substitution

CH₃CH₂Br + NaOH arrow Product A

CH₃CH₂Br + NaOH / H₂O arrow Product 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:

CH₃CH₂Br + NaOH (alc) arrow CH₂=CH₂ (Ethene)

Hydrogen atoms in ethene ( $C₂H₄$ $C_2H_4$) = 4.

Reaction 2: If the reagent is aqueous NaOH ( $NaOH / H₂O$ $NaOH / H_2O$), nucleophilic substitution ( $SN2$ $S_N2$) occurs:

CH₃CH₂Br + NaOH (aq) arrow CH₃CH₂OH (Ethanol)

Hydrogen atoms in ethanol ( $C₂H₆O$ $C_2H_6O$) = 6.

Total hydrogen atoms = 4 + 6 = 10.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes

The product B formed in the following reaction sequence is : The image outlines an addition reaction followed by substitution using silver cyanide.

Solution & Explanation

Related Formula

Core Logic

Step 1: Structural Synthesis

Pattern Recognition

Chapter Mix

Reference Study Guides

More Haloalkanes and Haloarenes Previous-Year Questions — Page 8

Solution

Core Logic

Step 1: Conclusion

Pattern Recognition

Chapter Mix

Solution

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Chapter Mix

Solution

Step 1: Elimination to form Propene

Step 2: Electrophilic Addition of HBr

Step 3: Nucleophilic Substitution

Pattern Recognition

Chapter Mix

Solution

Core Logic

Chapter Mix

More Haloalkanes and Haloarenes Questions — jee_main_2025_28_jan_evening

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The product B formed in the following reaction sequence is :
The image outlines an addition reaction followed by substitution using silver cyanide.