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JEE Mains Chemistry · Haloalkanes and Haloarenes

Reactivity Order in Nucleophilic Substitution

Rank halides for SN1 by the stability of the carbocation they would form, and for SN2 by how open the carbon is to backside attack; vinylic, aryl and bridgehead halides cannot form a usable cation at all.

All Haloalkanes and Haloarenes notesJEE Mains Chemistry strategy

Why this matters

Sixteen PYQs, one numerical, three from 2026, the largest page in the chapter. Seven rank halides for SN1 by the stability of the cation they form; four ask which halides cannot ionise at all or give no precipitate with silver nitrate; five compare SN2 rates, where crowding, a benzylic carbon, a neighbouring group or an adjacent oxygen changes the answer.

Concept 1 of 3: SN1 reactivity: ranking halides by carbocation stability

The slow step of SN1 is the halide leaving, so whatever makes the carbocation more stable makes the reaction faster. Resonance stabilises most (benzylic, allylic, a lone pair on an adjacent oxygen); alkyl groups help through hyperconjugation and their +I effect; electron-withdrawing groups make the cation worse.

Definition

  • Alkyl cations: 3° > 2° > 1° > CH3+\mathrm{CH_3^+}, by hyperconjugation and the +I effect.
  • Resonance: benzylic and allylic cations are stabilised, and each extra phenyl helps more: (C6H5)3C+>(C6H5)2CH+>C6H5CH2+\mathrm{(C_6H_5)_3C^+ > (C_6H_5)_2CH^+ > C_6H_5CH_2^+}. A cation that is both 3° and allylic is better still.
  • Ring substituents on a benzyl halide: p−OCH3>p−CH3>H>p−Cl>p−NO2\mathrm{p{-}OCH_3 > p{-}CH_3 > H > p{-}Cl > p{-}NO_2}. A donor para to the CH2\mathrm{CH_2} feeds the cation; a nitro group drains it.
  • An oxygen next to the carbon, as in CH3O−CH2−Cl\mathrm{CH_3O{-}CH_2{-}Cl}, gives the cation CH3O+=CH2\mathrm{CH_3O^+{=}CH_2}, so this primary halide ionises readily.
  • For the same carbon skeleton, the leaving group decides: R–I > R–Br > R–Cl > R–F.

SN1 order of alkyl halides; stability of phenyl-substituted cations

SN1: 3∘>2∘>1∘>CH3X(C6H5)3C+>(C6H5)2CH+>C6H5CH2+\text{SN1: } 3^\circ > 2^\circ > 1^\circ > \mathrm{CH_3X} \qquad \mathrm{(C_6H_5)_3C^+ > (C_6H_5)_2CH^+ > C_6H_5CH_2^+}

Worked example

Arrange for SN1 hydrolysis in aqueous ethanol: bromomethane, 1-bromopropane, 2-bromopropane, 2-bromo-2-methylpropane.
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The same idea in a real exam question:

JEE Mains · 2026 · 2 Apr 2026 Shift 1 · Q39Moderate

Example 1 · Haloalkanes and Haloarenes · Reactivity Order in Nucleophilic Substitution

Given below are two statements : Statement (I) : Benzyl chloride reacts faster in SN1S_{N}1 mechanism than ethyl chloride. Statement (II) : Ethyl carbocation intermediate is less stabilized by hyperconjugation than benzyl carbocation by resonance. In the light of the above statements, choose the correct answer from the options given below :

A para-chloro group slows SN1

Chlorine donates a lone pair (+R) but withdraws more strongly through its −I effect, so 4-chlorobenzyl chloride ionises a little more slowly than benzyl chloride. It is still much faster than 4-nitrobenzyl chloride.

Primary halides can still go SN1

The rule 3° > 2° > 1° applies to plain alkyl halides. A primary halide whose cation is stabilised by resonance, such as benzyl chloride, allyl chloride or CH3OCH2Cl\mathrm{CH_3OCH_2Cl}, ionises readily.
Drill 6 more on sn1 reactivity: ranking halides by carbocation stability

Concept 2 of 3: Halides that cannot ionise: vinylic, aryl and bridgehead

SN1 needs a flat carbocation with an empty p orbital. A vinylic or aryl halide has a strong, partly double C–X bond and would give a very unstable cation. A halogen at the bridgehead of a small cage cannot leave either, because the cage cannot flatten. Silver nitrate makes the test visible: AgX precipitates only when the C–X bond ionises.

Definition

  • With alcoholic AgNO3\mathrm{AgNO_3}: R−X+Ag+→R++AgX↓\mathrm{R{-}X + Ag^+ \to R^+ + AgX\downarrow}. A precipitate forms at once with 3°, benzylic and allylic halides, slowly with 1°, and not at all with vinylic, aryl or bridgehead halides.
  • Vinylic halides would give a cation on an sp carbon; aryl halides would give a phenyl cation whose empty orbital lies in the ring plane, away from the π system. Neither forms.
  • Bridgehead halides in rigid bicyclic cages (Bredt's rule): the smaller the cage, the slower. 1-Halobicyclo[2.2.1]heptanes are slower than 1-halobicyclo[2.2.2]octanes, and both are far slower than an open-chain tertiary halide.
  • The exception runs the other way: a halide whose cation is aromatic ionises very easily. 3-Bromocyclopropene gives the cyclopropenyl cation (2 π electrons) and 7-bromocycloheptatriene gives the tropylium cation (6 π electrons).
HalideCation it would giveSN1 and the AgNO₃ test
(CH3)3C−Cl\mathrm{(CH_3)_3C{-}Cl}3° cation, stabilised by hyperconjugationFast; AgCl precipitates at once
C6H5CH2Cl\mathrm{C_6H_5CH_2Cl}Benzyl cation, stabilised by resonanceFast; AgCl precipitates
CH2=CHCH2Cl\mathrm{CH_2{=}CHCH_2Cl}Allyl cation, stabilised by resonanceFast; AgCl precipitates
CH3CH2CH2CH2Cl\mathrm{CH_3CH_2CH_2CH_2Cl}1° cation, unstableVery slow; precipitate only on long warming
CH2=CHCl\mathrm{CH_2{=}CHCl}Vinyl cation, charge on an sp carbonNo SN1; no precipitate
C6H5Cl\mathrm{C_6H_5Cl}Phenyl cation, empty orbital in the ring planeNo SN1; no precipitate
1-Bromobicyclo[2.2.2]octaneBridgehead cation that cannot become planarExtremely slow; no practical SN1
3-BromocyclopropeneCyclopropenyl cation, aromatic with 2 π electronsIonises readily; AgBr precipitates
Being tertiary is not enough: the cation must also be able to become planar.
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The same idea in a real exam question:

JEE Mains · 2022 · 28 June 2022 · Q45Moderate

Example 2 · Haloalkanes and Haloarenes · Reactivity Order in Nucleophilic Substitution

Which one of the following compounds is inactive towards SN1S_{N}1 reaction?

A tertiary bridgehead halide does not ionise

A halogen at the bridgehead of a small cage is tertiary, yet it barely reacts by SN1: the cage holds the carbon pyramidal, so the cation cannot flatten.

Some cyclic halides ionise because the cation is aromatic

3-Bromocyclopropene and 7-bromocycloheptatriene look like awkward substrates, but their cations are aromatic. They precipitate silver bromide readily.
Drill 3 more on halides that cannot ionise: vinylic, aryl and bridgehead

Concept 3 of 3: SN2 reactivity: crowding, neighbouring groups and benzylic halides

SN2 needs room behind the carbon for the nucleophile to come in. Every alkyl group on that carbon, or on the carbon next to it, gets in the way. Two effects speed SN2 up: a π system next to the carbon, which overlaps the p orbital of the transition state, and a lone pair inside the molecule that can attack first.

Definition

  • CH3X\mathrm{CH_3X} > 1° > 2° > 3°; tertiary halides practically do not react by SN2.
  • Branching on the next carbon also slows SN2: neopentyl halides (CH3)3C−CH2X\mathrm{(CH_3)_3C{-}CH_2X} are primary but react very slowly.
  • Benzylic and allylic halides react faster than ethyl halides by SN2, because the ring or C=C conjugates with the p orbital of the five-coordinate transition state.
  • Neighbouring group participation: in Et2N−CH2CH2−Cl\mathrm{Et_2N{-}CH_2CH_2{-}Cl} the nitrogen lone pair displaces chloride first, inside the molecule, to give a three-membered aziridinium ion; hydroxide then opens it. This makes it hydrolyse faster than Et2CH−CH2CH2−Cl\mathrm{Et_2CH{-}CH_2CH_2{-}Cl}.
  • The leaving group matters too: R–I > R–Br > R–Cl > R–F.

SN2 order of alkyl halides; leaving-group order

SN2: CH3X>1∘>2∘>3∘R−I>R−Br>R−Cl>R−F\text{SN2: } \mathrm{CH_3X} > 1^\circ > 2^\circ > 3^\circ \qquad \mathrm{R{-}I > R{-}Br > R{-}Cl > R{-}F}

Worked example

Arrange for reaction with NaI in acetone: chloromethane, 1-chlorobutane, 2-chlorobutane, 2-chloro-2-methylpropane.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 9 April 2024 · Q39Moderate

Example 3 · Haloalkanes and Haloarenes · Reactivity Order in Nucleophilic Substitution

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R) Assertion (A): SN2S_{N}2 reaction of C6H5CH2BrC_{6}H_{5}CH_{2}Br occurs more readily than the SN2S_{N}2 reaction of CH3CH2BrCH_{3}CH_{2}Br. Reason (R): The partially bonded unhybridized p-orbital that develops in the trigonal bipyramidal transition state is stabilized by conjugation with the phenyl ring. In the light of the above statements, choose the most appropriate answer from the options given below:

Primary does not guarantee fast SN2

Neopentyl halides are primary, but the tert-butyl group next door blocks the back of the carbon. They react far more slowly than 1-halobutanes.

Benzylic halides are fast by both mechanisms

Benzyl halides ionise easily (SN1) and also react fast by SN2, because the ring stabilises both the cation and the SN2 transition state. Do not assume that fast SN1 means slow SN2.
Drill 4 more on sn2 reactivity: crowding, neighbouring groups and benzylic halides

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

  • SN1 reactivity: ranking halides by carbocation stability

    SN1 order of alkyl halides; stability of phenyl-substituted cations

    SN1: 3∘>2∘>1∘>CH3X(C6H5)3C+>(C6H5)2CH+>C6H5CH2+\text{SN1: } 3^\circ > 2^\circ > 1^\circ > \mathrm{CH_3X} \qquad \mathrm{(C_6H_5)_3C^+ > (C_6H_5)_2CH^+ > C_6H_5CH_2^+}
  • SN2 reactivity: crowding, neighbouring groups and benzylic halides

    SN2 order of alkyl halides; leaving-group order

    SN2: CH3X>1∘>2∘>3∘R−I>R−Br>R−Cl>R−F\text{SN2: } \mathrm{CH_3X} > 1^\circ > 2^\circ > 3^\circ \qquad \mathrm{R{-}I > R{-}Br > R{-}Cl > R{-}F}

Reference tables (1)

Halides that cannot ionise: vinylic, aryl and bridgehead8 rows
HalideCation it would giveSN1 and the AgNO₃ test
(CH3)3C−Cl\mathrm{(CH_3)_3C{-}Cl}3° cation, stabilised by hyperconjugationFast; AgCl precipitates at once
C6H5CH2Cl\mathrm{C_6H_5CH_2Cl}Benzyl cation, stabilised by resonanceFast; AgCl precipitates
CH2=CHCH2Cl\mathrm{CH_2{=}CHCH_2Cl}Allyl cation, stabilised by resonanceFast; AgCl precipitates
CH3CH2CH2CH2Cl\mathrm{CH_3CH_2CH_2CH_2Cl}1° cation, unstableVery slow; precipitate only on long warming
CH2=CHCl\mathrm{CH_2{=}CHCl}Vinyl cation, charge on an sp carbonNo SN1; no precipitate
C6H5Cl\mathrm{C_6H_5Cl}Phenyl cation, empty orbital in the ring planeNo SN1; no precipitate
1-Bromobicyclo[2.2.2]octaneBridgehead cation that cannot become planarExtremely slow; no practical SN1
3-BromocyclopropeneCyclopropenyl cation, aromatic with 2 π electronsIonises readily; AgBr precipitates
Being tertiary is not enough: the cation must also be able to become planar.

Watch out for (6)

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