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Haloalkanes and Haloarenes appeared 40 times across 3 years — 4.7% of Chemistry. This question is from Nucleophilic Substitution Reactions.

Year	2026	2025	2024	Total
Questions	11	16	13	40

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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.

A. (1)
The image outlines an addition reaction followed by substitution using silver cyanide.
B. (2)
The image outlines an addition reaction followed by substitution using silver cyanide.
C. (3)
The image outlines an addition reaction followed by substitution using silver cyanide.
D. (4)
The image outlines an addition reaction followed by substitution using silver cyanide.

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

Reference Study Guides

Haloalkanes And Haloarenes

More Haloalkanes and Haloarenes Previous-Year Questions

Q54 jee_main_2026_21_jan_morning Reactions of Haloalkanes

A hydrocarbon ‘P’ (

C₄H₈

) on reaction with HCl gives an optically active compound ‘Q’ (

C₄H₉Cl

) which on reaction with one mole of ammonia gives compound ‘R’ (

C₄H₁₁N

). ‘R’ on diazotization followed by hydrolysis gives ‘S’. Identify P, Q, R and S.

A. $P = CH₃ - CH₂ - CH = CH₂, Q = CH₃ - CH₂ - CH₂ - CH₂Cl, R = CH₃ - CH₂ - CH₂ - NH₂, S = CH₃ - CH₂ - CH(OH) - CH₃$
B. $P = Cyclobutane, Q = 1-Chlorobutane, R = Butan-1-amine, S = Cyclobutanol$
C. $P = CH₃ - CH = CH - CH₃, Q = CH₃ - CH₂ - CH(Cl) - CH₃, R = CH₃ - CH₂ - CH(NH₂) - CH₃, S = CH₃ - CH₂ - CH(OH) - CH₃$
D. $P = CH₃ - CH = CH - CH₃, Q = CH₃ - CH₂ - CH₂ - CH₂ - Cl, R = CH₃ - CH₂ - CH₂ - CH₂ - NH₂, S = CH₃ - CH₂ - CH₂ - CH₂ - OH$

Solution

Core Logic

Since P ( $C₄H₈$ $C_4H_8$) reacts with HCl to give an optically active compound Q ( $C₄H₉Cl$ $C_4H_9Cl$), P must be But-2-ene. Addition of HCl yields 2-chlorobutane, which possesses a chiral center.

CH₃-CH=CH-CH₃ HCl CH₃-CH₂-C^*H(Cl)-CH₃ (P is But-2-ene, Q is 2-chlorobutane)

Reactions of Haloalkanes diagram for Q54 - JEE Main 2026 Morning

Compound Q reacts with $NH₃$ $\mathrm{NH}_3$ to undergo nucleophilic substitution forming a primary amine R:

CH₃-CH₂-CH(Cl)-CH₃ NH₃ CH₃-CH₂-CH(NH₂)-CH₃ (R is Butan-2-amine)

Reactions of Haloalkanes diagram for Q54 - JEE Main 2026 Morning

Diazotization of primary aliphatic amines followed by hydrolysis yields an alcohol via a carbocation intermediate (S):

CH₃-CH₂-CH(NH₂)-CH₃ NaNO₂ / HCl / H₂O CH₃-CH₂-CH(OH)-CH₃ (S is Butan-2-ol)

Reactions of Haloalkanes diagram for Q54 - JEE Main 2026 Morning

Pattern Recognition

An optically active alkyl halide formed from a $C₄H₈$ $C_4H_8$ alkene with HX is a classic pointer to 2-halobutane originating from But-2-ene or But-1-ene. Diazotization of primary aliphatic amines ( $R-NH₂$ $R-NH_2$) with $NaNO₂/HCl$ $NaNO_2/HCl$ yields alcohols ( $R-OH$ $R-OH$) with possible rearrangements, though here a secondary carbocation is already stable.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes Class 12 Chemistry: Amines

Q59 jee_main_2026_21_jan_evening Nucleophilic Substitution ($S_N1$) and Carbocation Stability

The correct order of reactivity of the following benzyl halides towards reaction with

KCN

is:

Benzyl halide structures for Q59 - JEE Main 2026 Evening — Four substituted benzyl halide structures labelled a, b, c, d.

A. $(1) a > b > c > d$ $(1) \ a > b > c > d$
B. $(2) b > a > d > c$ $(2) \ b > a > d > c$
C. $(3) b > a > c > d$ $(3) \ b > a > c > d$
D. $(4) a > b > d > d$ $(4) \ a > b > d > d$

Solution

Core Logic

The reaction proceeds via an $SN1$ $S_N1$ mechanism for activated benzyl halides or nucleophilic substitution rate depends on carbocation stability / electronic effects of substituents (-OH, $-NH₂$ $-\text{NH}_2$, $-NO₂$ $-\text{NO}_2$, etc.).

Amino and hydroxy substituents strongly activate through +M effect.
Nitro groups strongly deactivate through -M effect.

Step 1: Final Conclusion

The correct order of reactivity is $b > a > d > c$ $b > a > d > c$, corresponding to option (2).

Pattern Recognition

Sees: benzyl halide reactivity with cyanide. Trap: Confusing polar/inductive effects with resonance effects of substituents.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes

Q65 jee_main_2026_21_jan_evening Optical Isomerism and Composition Percentage

Given below are four compounds: (a) n-propyl chloride (b) iso-propyl chloride (c) sec-butyl chloride (d) neo-pentyl chloride Percentage of carbon in the one which exhibits optical isomerism is:

A. $(1) 52$ $(1) \ 52$
B. $(2) 56$ $(2) \ 56$
C. $(3) 46$ $(3) \ 46$
D. $(4) 40$ $(4) \ 40$

Solution

Core Logic

Among the given compounds, sec-butyl chloride (2-chlorobutane) is optically active and contains a chiral center. Molecular formula of 2-chlorobutane $arrow C₄H₉Cl$ . Molar mass $= 4(12) + 9(1) + 35.5 = 48 + 9 + 35.5 = 92.5 g/mol$ .

Step 1: Calculating Percentage of Carbon

% of C = (48)/(92.5) \times 100 = 51.89% \approx 52%

Pattern Recognition

Sees: optical isomerism identification combined with elemental percentage composition calculation. Trap: Selecting the wrong halogen derivative for optical activity.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes

Q59 jee_main_2026_22_january_morning Nucleophilicity Trends

The correct order of reactivity of

CH₃Br

in methanol with the following nucleophiles is

F⁻

,

I⁻

,

C₂H₅O⁻

and

C₆H₅O⁻

A. $I⁻ > C₆H₅O⁻ > F⁻ > C₂H₅O⁻$
B. $I⁻ > C₂H₅O⁻ > C₆H₅O⁻ > F⁻$
C. $I⁻ > C₂H₅O⁻ > F⁻ > C₆H₅O⁻$
D. $I⁻ > F⁻ > C₆H₅O⁻ > C₂H₅O⁻$

Solution

Related Formula

Rate \propto Nucleophilicity

Core Logic

The reaction of $CH₃Br$ $CH_3Br$ (a primary halide) occurs via the $SN2$ $S_N2$ mechanism. The rate depends directly on the nucleophilicity of the attacking species. Methanol is a polar protic solvent. In polar protic solvents, larger halide ions are better nucleophiles because they are less solvated. Therefore, $I^-$ $I^-$ is a stronger nucleophile than $F^-$ $F^-$. For the alkoxide and phenoxide, $C₂H₅O^-$ $C_2H_5O^-$ is a stronger nucleophile than $C₆H₅O^-$ $C_6H_5O^-$ because the negative charge on phenoxide is delocalized over the benzene ring, reducing its electron-donating ability.

Step 1: Final Order

Combining these factors, the overall order of nucleophilicity in methanol is:

I⁻ > C₂H₅O⁻ > C₆H₅O⁻ > F⁻

Pattern Recognition

In polar protic solvents: Size dominates (down a group, nucleophilicity increases). For species of similar size, less resonance stabilization means stronger nucleophile.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes

Q67 jee_main_2026_22_january_morning SN1 Mechanism

The correct order of the rate of reaction of the following reactants with nucleophile by

SN1

mechanism is: (Given: Structure I and II are rigid)

Four rigid structures reacting via SN1 diagram for Q67 - JEE Main 2026 Morning — Image depicts four bicyclic and aromatic structures labeled I through IV.

A. $IV < III < II < I$ $IV < III < II < I$
B. $III < I < II < IV$ $III < I < II < IV$
C. $II < I < III < IV$ $II < I < III < IV$
D. $I < II < III < IV$ $I < II < III < IV$

Solution

Related Formula

Rate of SN1 \propto Stability of intermediate carbocation

Core Logic

Evaluate the stability of the carbocations formed upon dissociation of the bromide ion:

Structure (I) and (II) form carbocations at bridgehead positions of rigid bicyclic systems. According to Bredt's rule, these are highly unstable because they cannot adopt the necessary planar $sp²$ $sp^2$ geometry. Among them, (I) has an additional methyl group which provides slight +I stabilization compared to (II). So, (II) is even less stable than (I).

Structure (III) forms a tertiary carbocation, but it's not a rigid bridgehead preventing planarity, so it's significantly more stable than I and II.

Structure (IV) is a highly stable trityl-like carbocation where the positive charge is stabilized by extensive resonance with the adjacent phenyl ring(s).

Step 1: Final Ordering

Stability order: $(II) < (I) < (III) < (IV)$ $(II) < (I) < (III) < (IV)$. Thus, the $SN1$ $S_{N}1$ rate follows the same order.

Pattern Recognition

Bridgehead halides effectively do not undergo $SN1$ $S_N1$ (or $SN2$ $S_N2$) reactions due to Bredt's rule. Resonance stabilized carbocations always dominate.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes

More Haloalkanes and Haloarenes Questions — jee_main_2025_28_jan_evening

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