Given below are two statements : Statement I: Et2N-CH2-CH2-Cl will undergo alkaline hydrolysis at a faster rate than Et2CH-CH2-Cl. Statement II: In Et2N-CH2-CH2-Cl, intramolecular substitution takes place first by involving lone pair of electrons on nitrogen. In the light of the above statements, choose the most appropriate answer from the options given below:

Statement I is correct because the nitrogen atom contains a lone pair situated at the beta-position relative to the chlorine atom, promoting Neighboring Group Participation (NGP). Statement II is correct because the lone pair on nitrogen attacks internally to kick out the chloride ion, forming a cyclic aziridinium ion intermediate. This quick intramolecular cyclization leads to an exceptionally rapid hydrolysis rate compared to standard aliphatic substitution. Sees: Nitrogen with lone pair beta to a leaving group. Shortcut: NGP (Neighboring Group Participation) accelerates substitution dramatically via intramolecular assistance. Class 12 Chemistry: Haloalkanes and Haloarenes

Given below are two statements : Statement I: Et2N-CH2-CH2-Cl will undergo alkaline hydrolysis at a faster rate than Et2CH-CH2-Cl. Statement II: In Et2N-CH2-CH2-Cl, intramolecular substitution takes p

Statement I is correct because the nitrogen atom contains a lone pair situated at the beta-position relative to the chlorine atom, promoting Neighboring Group Participation (NGP). Statement II is correct because the lone pair on nitrogen attacks internally to kick out the chloride ion, forming a cyclic aziridinium ion intermediate. This quick intramolecular cyclization leads to an exceptionally rapid hydrolysis rate compared to standard aliphatic substitution. Sees: Nitrogen with lone pair beta

Given below are two statements : Statement I: Et2N-CH2-CH2-Cl will undergo alkaline hydrolysis at a faster rate than Et2CH-CH2-Cl. Statement II: In Et2N-CH2-CH2-Cl, intramolecular substitution takes place first by involving lone pair of electrons on nitrogen. In the light of the above statements, choose the most appropriate answer from the options given below:

Statement I is correct because the nitrogen atom contains a lone pair situated at the beta-position relative to the chlorine atom, promoting Neighboring Group Participation (NGP). Statement II is correct because the lone pair on nitrogen attacks internally to kick out the chloride ion, forming a cyclic aziridinium ion intermediate. This quick intramolecular cyclization leads to an exceptionally rapid hydrolysis rate compared to standard aliphatic substitution. Sees: Nitrogen with lone pair beta to a leaving group. Shortcut: NGP (Neighboring Group Participation) accelerates substitution dramatically via intramolecular assistance. Class 12 Chemistry: Haloalkanes and Haloarenes

Given below are two statements : Statement I: Et2N-CH2-CH2-Cl will undergo alkaline hydrolysis at a faster rate than Et2CH-CH2-Cl. Statement II: In Et2N-CH2-CH2-Cl, intramolecular substitution takes p

Statement I is correct because the nitrogen atom contains a lone pair situated at the beta-position relative to the chlorine atom, promoting Neighboring Group Participation (NGP). Statement II is correct because the lone pair on nitrogen attacks internally to kick out the chloride ion, forming a cyclic aziridinium ion intermediate. This quick intramolecular cyclization leads to an exceptionally rapid hydrolysis rate compared to standard aliphatic substitution. Sees: Nitrogen with lone pair beta

Given below are two statements : Statement I: Et2N-CH2-CH2-Cl will undergo alkaline hydrolysis at a faster rate than…

Year	2026	2025	2024	Total
Questions	11	15	12	38

Q69 jee_main_2026_22_january_morning Properties of Halides

As compared with chlorocyclohexane, which of the following statements correctly apply to chlorobenzene? A. The magnitude of negative charge is more on chlorine atoms B. The C – Cl bond has partial double bond character C. C – Cl bond is less polar D. C – Cl bond is longer due to repulsion between delocalised electrons of the aromatic ring and lone pairs of electrons of chlorine. E. The C–Cl bond is formed using

sp²

hybridised orbital of carbon. Choose the correct answer from the options given below:

A. $A, C and E only$ $\text{A, C and E only}$
B. $B, C and D only$ $\text{B, C and D only}$
C. $A, D and E only$ $\text{A, D and E only}$
D. $B, C and E only$ $\text{B, C and E only}$

Solution

Core Logic

Let's analyze each statement regarding chlorobenzene vs. chlorocyclohexane:

A. Magnitude of negative charge: In chlorobenzene, the lone pair of Cl undergoes +M resonance with the ring. This delocalizes electron density into the ring, reducing the net negative charge on chlorine compared to the purely -I withdrawing Cl in chlorocyclohexane. (False) B. C-Cl double bond character: The +M effect (resonance) imparts partial double bond character to the C-Cl bond in chlorobenzene. (True) C. Less polar bond: Because resonance acts opposite to the inductive (-I) effect, the net dipole moment of chlorobenzene (1.5 - 1.6 D) is lower than that of chlorocyclohexane (approx 2.1 D). So the bond is less polar. (True) D. C-Cl bond is longer: Because of the partial double bond character, the C-Cl bond in chlorobenzene is actually shorter (169 pm) than in chlorocyclohexane (177 pm). (False) E. Hybridization: The carbon bonded to Cl in chlorobenzene is $sp²$ $sp^2$ hybridized, while in chlorocyclohexane it is $sp³$ $sp^3$ hybridized. (True)

Step 1: Conclusion

The correct statements are B, C, and E only.

Pattern Recognition

Resonance decreases bond length and decreases polarity in aryl halides compared to alkyl halides.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes

Q70 jee_main_2026_22_january_evening Dibromo Alkylation and Qualitative Tests

The dibromo compound [P] (molecular formula:

C₉H₁₀Br₂

) when heated with excess sodamide followed by treatment with dilute HCl gives [Q]. On warming [Q] with mercuric sulphate and dilute sulphuric acid yield [R] which gives positive Iodoform test but negative Tollen's test. The compound [P] is:

A.
B.
C.
D.

Solution

Related Formula

Gem/Vicinal Dibromide [P] excess NaNH₂ Terminal Alkyne [Q] [dil. H₂SO₄]HgSO₄ Methyl Ketone [R]

Core Logic

Step 1: Compound [R] gives positive Iodoform test and negative Tollen's test $\implies$ [R] is a methyl ketone (acetophenone derivative, $Ph-CO-CH₃$ $\text{Ph-CO-CH}_3$).

Step 2: Hydration of alkyne [Q] gives methyl ketone [R] $\implies$ [Q] is phenylacetylene ( $Ph-C$ $\text{Ph-C}\equiv\text{CH}$).

Step 3: Double elimination of [P] using excess $NaNH₂$ $\text{NaNH}_2$ gives phenylacetylene [Q] $\implies$ [P] is geminal dibromide 1,1-dibromo-1-phenylethane ( $Ph-C(Br)₂CH₃$ $\text{Ph-C(Br)}_2\text{CH}_3$).

Reaction flow of dibromo compound to methyl ketone for Q70 - JEE Main 2026 Evening

Pattern Recognition

Sees: Positive iodoform + negative Tollen's $\implies$ methyl ketone. Shortcut: Oxymercuration of terminal alkyne yields methyl ketone, identifying gem-dibromide at benzylic position.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

Q74 jee_main_2026_23_january_morning Structural Isomerism and Optical Activity

Consider all the structural isomers with molecular formula

C₅H₁₁Br

are separately treated with KOH (aq) to give respective substitution products, without any rearrangement. The number of products which can exhibit optical isomerism from these is ____.

Numerical Answer. Answer: 3 to 3

Solution

Core Logic

First, identify all structural isomers of $C₅H₁₁Br$ $C_5H_{11}Br$. Aqueous KOH will simply replace the $-Br$ $-Br$ with an $-OH$ $-OH$ group via nucleophilic substitution without rearrangement. Then, check which of the resulting $C₅H₁₁OH$ $C_5H_{11}OH$ products possess a chiral center.

Structural Isomerism and Optical Activity diagram for Q74 - JEE Main 2026 Morning

Step 1: Structural Isomers and Products

Pentyl skeleton isomers:

$n$ $n$-pentyl: 1-bromopentane $arrow$ $\rightarrow$ 1-pentanol (Achiral), 2-bromopentane $arrow$ $\rightarrow$ 2-pentanol (Chiral), 3-bromopentane $arrow$ $\rightarrow$ 3-pentanol (Achiral).
Isopentyl skeleton: 1-bromo-3-methylbutane $arrow$ $\rightarrow$ 3-methylbutan-1-ol (Achiral), 2-bromo-3-methylbutane $arrow$ $\rightarrow$ 3-methylbutan-2-ol (Chiral), 2-bromo-2-methylbutane $arrow$ $\rightarrow$ 2-methylbutan-2-ol (Achiral), 1-bromo-2-methylbutane $arrow$ $\rightarrow$ 2-methylbutan-1-ol (Chiral).
Neopentyl skeleton: 1-bromo-2,2-dimethylpropane $arrow$ $\rightarrow$ 2,2-dimethylpropan-1-ol (Achiral).

Step 2: Counting Optically Active Products

The products exhibiting optical isomerism are:

2-pentanol
3-methylbutan-2-ol
2-methylbutan-1-ol

Total structurally distinct products exhibiting optical isomerism is 3.

Pattern Recognition

To quickly spot a chiral center in a standard alkane chain, look for a carbon bonded to $-OH$ $-OH$, $-H$ $-H$, a methyl group, and a longer alkyl chain. Any symmetry kills chirality (like 3-pentanol).

Chapter Mix

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

Q61 jee_main_2026_24_january_morning Nomenclature of Haloalkanes

Match the List-I with List-II

List-I (Chloro derivative)	List-II (Example)
A. Vinyl Chloride	I. $CH₂=CH-CH₂Cl$ $CH_2=CH-CH_2Cl$
B. Benzyl chloride	II. $CH₃-CH(Cl)CH₃$ $CH_3-CH(Cl)CH_3$
C. Alkyl chloride	III. $CH₂=CHCl$ $CH_2=CHCl$
D. Allyl chloride	IV. Structure IV shows a benzene ring attached to a CH2Cl group.

Choose the correct answer from the options given below :

A. $A-IV, B-I, C-III, D-II$ $\text{A-IV, B-I, C-III, D-II}$
B. $A-III, B-IV, C-I, D-II$ $\text{A-III, B-IV, C-I, D-II}$
C. $A-III, B-IV, C-II, D-I$ $\text{A-III, B-IV, C-II, D-I}$
D. $A-I, B-II, C-IV, D-III$ $\text{A-I, B-II, C-IV, D-III}$

Solution

Core Logic

(A) Vinyl Chloride is a haloalkene where chlorine is directly attached to the $sp²$ $sp^2$ carbon of a double bond: $CH₂=CHCl$ $CH_2=CHCl$. So, A matches with III. (B) Benzyl chloride is a compound where chlorine is attached to an $sp³$ $sp^3$ hybridized carbon directly linked to a benzene ring ( $C₆H₅CH₂Cl$ $C_6H_5CH_2Cl$). Structure IV represents this. So, B matches with IV. (C) Alkyl chloride represents a standard saturated aliphatic chain attached to a chlorine atom, such as isopropyl chloride: $CH₃-CH(Cl)CH₃$ $CH_3-CH(Cl)CH_3$. So, C matches with II. (D) Allyl chloride features a chlorine attached to an $sp³$ $sp^3$ carbon that is adjacent to a carbon-carbon double bond: $CH₂=CH-CH₂Cl$ $CH_2=CH-CH_2Cl$. So, D matches with I.

Step 1: Final Conclusion

The correctly matched pairs are A-III, B-IV, C-II, D-I.

Pattern Recognition

Vinyl = directly on double bond. Allyl = one carbon away from double bond. Benzyl = one carbon away from phenyl ring.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes

Q67 jee_main_2026_24_january_morning Stereochemistry and Bond Strength

Given below are two statements: Statement I :

C-Cl

bond is stronger in

CH₂=CH-Cl

than

CH₃-CH₂-Cl

Statement II : The given optically active molecule,

Optically active alkyl chloride molecule — A chiral tertiary alkyl chloride with Phenyl, Methyl, Ethyl and Chloro groups attached to the central carbon.

on hydrolysis gives a solution that can rotate the plane polarized light. In the light of the above statements, choose the correct answer from the options given below

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

Solution

Core Logic

Statement I: $CH₂=CH-Cl$ $CH_2=CH-Cl$ is vinyl chloride. The lone pair on chlorine participates in resonance with the adjacent double bond. This imparts a partial double bond character to the $C-Cl$ $\text{C-Cl}$ bond, making it shorter and stronger than the pure single $C-Cl$ $\text{C-Cl}$ bond in $CH₃-CH₂-Cl$ $CH_3-CH_2-Cl$ (ethyl chloride).

Resonance in vinyl chloride — A chiral tertiary alkyl chloride with Phenyl, Methyl, Ethyl and Chloro groups attached to the central carbon.

Thus, Statement I is true.

Statement II: The given molecule is a chiral, tertiary alkyl chloride. When it undergoes hydrolysis, it follows an $SN1$ $S_N1$ mechanism because the resulting tertiary carbocation (stabilized by resonance from the phenyl ring and hyperconjugation) is very stable. An $SN1$ $S_N1$ reaction proceeds through a planar carbocation intermediate. Attack by water occurs from both faces with near equal probability, resulting in a racemic mixture.

A racemic mixture is optically inactive; it cannot rotate plane-polarized light. Therefore, Statement II is false.

Step 1: Final Conclusion

Statement I is true but Statement II is false.

Pattern Recognition

Vinyl and aryl halides have partial double bond character due to resonance. $SN1$ $S_N1$ at a chiral center leads to racemization (optically inactive product mix).

Chapter Mix

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

Solution & Explanation

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More Haloalkanes and Haloarenes Previous-Year Questions — Page 2

Solution

Core Logic

Step 1: Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Structural Isomers and Products

Step 2: Counting Optically Active Products

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

More Haloalkanes and Haloarenes Questions — jee_main_2025_28_jan_morning

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